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IN MEMORIAM 
BERNARD MOSES 




J 



UNIVERSITY EDITION. 



A 



THEORETICAL AND PEACTICAL 



TREATISE ON 



ALGEBRA; 



IN WHICH THE EXCELLENCIES OF THE DEMONSTRATIVE METHODS OF THE 

TKE^CH. AKE COMBINED WITH THE MOKE PRACTICAL OPERATrov« 

OF THE ENaLISn; AND CONCISE SOLUTIONS POINTED 

OUT AND PARTICULARLY INCU^CA,rjD. 



DESIGNED FOR SCHOOLS, COLLEGES AND PRIVATE STUDENTS 



BY H.N. EOBINSON, A. M 

:«RLV PROKESSSOR OK nlkK^j^r^CS IN THE tX. S. N^VT ' ^rTH^R 
TKE.X:SE O. .HXXHMEXZC; ..XUR.. PH.OSOPHV ; .STRo'xOMrTxC. 



EIGHTH FDITION. 

CINCINNATI: 

JACOB ERNST, NO. 183, MAIN STREET: 

ALBANY, NEW YORK • 

ERASTU3 H. PEASE & CO., 82, STATE STREET 

1850. t 






Entered, according to Act of Congress, in the year 1847, 

BY HORATIO N. ROBINSON", 

In the Clerk's Office of the District Court of the State of Ohio. 

^^ — wJ — 



BEW»lftW> «0W8 






STEREOTYPED BY JAMES & CO., 
CINCINNATI. 



PREFACE 



Some apology may appear requisite for offering a new book to the public on 
the science of Algebra — especially as there are several works of acknowledged 
merit on that subject already before the public, claiming attention. 

But the intrinsic merits of a book are not alone sufficient to secure its adop 
tion, and render it generally useful. In addition to merit, it must be adapted 
to the general standard of scientific instruction given in our higher schools ; it 
must conform in a measure to the taste of the nation, and correspond with the 
general spirit of the age in which it is brought forth. 

The elaborate and diffusive style of the French, as applied to this science, 
can never be more than theoretically popular among the English ; and the se- 
vere, brief, and practical methods of the English are almost intolerable to the 
French. Yet both nations can boast of men highly pre-eminent in this science, 
and the high minded of both nations are ready and willing to acknowledge the 
merits of the other ; but the style and spirit of their respective productions are 
necessarily very different. 

In this country, our authors and teachers have generally adopted one or the 
other of these schooFs, and thus have brought among us difference of opin- 
ion, drawn from these different standards of measure for true excellence. 

Very many of the French methods of treating algebraic science are not to be 
disregarded or set a^ide. First principles, theories and demonstrations, are the 
essence of all true science, and the French are very elaborate in these. Yet no 
effort of individuals, and no influence of a few institutions of learning, can 
change the taste of the American people, and make them assimilate to the 
French, any more than they can make the entire people assume French viva- 
city, and adopt French manners. 

Several works, modified from the French, have had, and now have consid- 
erable popularity, but they do not naturally suit American pupils. They are 
not sufficiently practical to be unquestionably popular ; and excellent as they 
are, they fail to inspire that enthusiastic spirit, which works of a more practi- 
cal and English character are known to do. 

At the other extreme are several English books, almost wholly practical, with 
little more than arbitrary rules laid down. Such books may in time make 
good resolvers of problems, but they certainly fail in most instances to make 
scientific algebraists. 

The author of this work has had much experience as a teacher of algebra, 

ni 



885995 



iv PREFACE 

and has used the diflerent varieties of text books, with a view to test their com- 
parative excellencies, and decide if possible on the standard most proper to be 
adopted, and of course he designed this work to be such as his experience and 
judgment would approve. 

One of the designs of this book is to create in the minds of the pupils a love 
for the study, which must in some way be secured before success can be at- 
tained. Small works designed for children, or those purposely adapted to per- 
sons of low capacity, will not secure this end. Those who give tone to public 
opinion in schools, will look down upon, rather than up to, works of this kind, 
and then the day of their usefulness is past. On the other hand, works of a 
high theoretical character are apt to discourage the pupil before his acquire- 
ments enable him to appreciate them, and on this account alone such works 
are not the most proper for elementary class books. 

This work is designed, in the strictest sense, to be both theoretical and prac- 
tical, and therefore, if the author has accomplished his design, it will be found 
about midway between the French and English schools. 

In this treatise will be found condensed and brief modes of operation, not 
hitherto much known or generally practised, and several expedients are system- 
atised and taught, by which many otherwise tedious operations are avoided. 

Some applications of the celebrated problems of the couriers, and also of the 
lights, are introduced into this work, as an index to the pupil of the subsequent 
utility of algebraic science, which may allure him on to more thorough investi- 
gations, and more extensive study. 

Such problems would be more in place in text books on natural philosophy 
and astronomy than in an elementary algebra, but the almost entire absence of 
them in works of that kind, is our apology for inserting them here, if 
apology be necessary. 

Quite young pupils, and such as may not have an adequate knowledge of 
physics and the general outlines of astronomy, may omit these articles of ap- 
plication ; but in all cases the teacher alone can decide what to omit and what 
to teach. 

Within a few years many new text books on algebra have appeared in difier- 
cnt parts of the country, which is a sure index that something is desired — 
something expected, — not yet found. The happy medium between the theo- 
retical and practical mathematics, or, rather, the happy blending of the two, 
which all seem to desire, is most difficult to attadn ; hence, many have failed 
in their efforts to meet the wants of the public. 

Metaphysical theories, and speculative science, suit tlie meridians of France 
and Germany better than those of the United States. But it is almost impos- 
sible to comment on this subject without being misapprehended ; the author 
of this book is a great admirer of the pure theories of algebraical science, for 
it is impossible to be practically skillful without having high theoretical acquire- 
ments. It is the man of theory who brings forth practical results, but it is not 
theory alone — it is theory long and well applied. 



PREFACE y 

Who will contend that Watt. Fitch, or Fu!t«n, were ignorant or inattentive 
to every theory concerning the nature and power of steam, vet they are only 
known as practical men, and it is almost in vain to look for any benefactors of 
raankmd, or any promoters of real science from those known only as theo- 
nste or among those who are strenuous contenders for technicalities and forms. 
V. e are led to these remarks to counteract, in some measure, if possible, that 
false impression existing in some minds, that a high standard work on al-ebra 
must necessarily be very formal in manner and abstrusely theoretical in mat' 
ter ; but m our view these are blemishes ratlier than excellencies. 

The author of this work is a great advocate for brevity, when not purchased 
at the expense of perspicuity, and this may account for the book appearing 
very small, considering what it is claimed to contain. For instance, we have 
only two formulas in arithmetical progression, and some authors have 20. 
We contend f.ke two are sufficient, and when weU understood cover the whole 
theory pertaining to the subject, and in practice, whether for absolute u.se or 
lasting improvement of the mind, are far better than 20. The great number 
only serves to confuse and distract the mind ; the two essential ones, can be 
remembered and most clearly and philosophically comprehended. The same 
remarks apply to geometrical progression. 

In the general theory of equations of the higher degrees this work is not too 
diffuse ; at the same time it designs to be simple and clear, and as much is 
given as in the judgment of the author would be acceptable, in a work as ele- 
mentary and condensed as this ; and if every position is not rigidly demonstra- 
ted, nothmg is left in obscurity or doubt. 

We have made special effbrt to present the beautiful theorem of Sturm in 
Buch a manner as to bring it direct to the comprehension of the student, and if 
we have failed in this, we stand not alone. 

The subject itself, though not essentially difficult, is abstruse for a learner 
and m our effort to render it clear we have been more circuitous and elaborate 
than we had hoped to have been, or at first intended. 

We may apply the same remarks to our treatment of Horner's method of 
solving the higher equations. 

Brevity is a great excellence, but perspicuity is greater, and, as a general 
thing, the two go hand in hand; and these views have guided us in preparin.- 
the whole work; we have felt bound to be clear and show the rationale of 
every operatio^i, and the foundation of every principle, at whatever cost. 

The Indeterminate and Diophantine analysis are not essential in a rc-ular 
course of mathematics, and it has not been customary to teach them in many 
institutions; for these reasons we do not insert them in our text book The 
teacher or the student, however, will find them in a concise form in a key to 
this work. -^ 



CONTENTS. 



iNXnODVCTIOX 



SECTION I. 

Addition 12 

Subtraction 16 

Multiplication 19 

Division 26 

Negative Exponents 27 

Algebraic Fractioxs 34 

Greatest Common Divisor 37 

Least Common Multiple 42 

Addition of Fractions 44 

Subtraction of Fractions 46 

Multiplication of Fractions 47 

Division of Fractions 49 

SECTION II. 

Equation of one unknown Quantity 52 

Question producing Simple Equations 62 

Equations of Two unknown Quantities 70 

Equations of Three or more unknown Quantities 76 

Problems producing Simple Equations of Two or more unknown quanti- 
ties '. 87 

Interpretation of negative values in the Solution of Problems 94 

Demonstration of Theorems 97 

Problem of the Couriers 98 

Application of the Problem of the Couriers 102 

SECTION III. 

IXVOLTTTIOX' 106 

Some application of the Binomial Theorem 108 

Evolution ". 113 

Cube root of Compound quantities 121 

Cube root of Numerals 123 

Brief method of Approximation to the Cube root of Numbers 124 

Exponential Quantities and Surds 1 28 

Pure EauATiONs 133 

Binomial Surds 141 

Problem producing pure Equations 147 

Problem of the Lights . •. 151 

Application of the Problem 152 

vii 



viii CONTENTS. 

SECTION IV. Page. 

Quadratic Equations 1 57 

Particular mode of completing a Square (Art. 99) 160 

Special Artifices in resolving Quadratics (Art. 106) 169 

Quadratic Equations containing two or more unknown quantities ... 175 

Questions producing Quadratic Equations 183 

SECTION V. 

Arithmetical Progression 189 

Geometrical Progression 195 

Harmonical Proportion 199 

Problems in Progression and Harmonical Proportion 200 

Geometrical Proportion 205 

SECTION VI. 

Binomial Theorem — its Demonstration, &c 213 

" '• its General Application 219 

Infinite Series 221 

Summation of Series . . • 225 

Reversion of a Series 231 

Exponential Equations and Logarithms 233 

Application of Logarithms 244 

Compound Interest 246 

Annuities • • 217 

SECTION VIL 

General Theory of Equations . 251 

Binomial Equations 2G0 

Newton's Method of Divisors (Art. 167) 264 

Equal Roots 267 

Transformation of Equations (by substitution) 270 

Transformation by Division . . . ._ , 275 

Synthetic Division 278 

General Properties of Equations 286 

Sturm's Theorem 297 

Newton's Method of Approximation •. 305 

Homer's « " " 307 

Solution of Equations 311 

Application of Equations to the Extraction of Roots 322 

APPENDIX. 

Specific Gravity 325 

Maxima and Minima ' 827 



ELEMENTS OF ALGEBRA. 



INTRODUCTION. 



DEFINITIONS AND AXIOMS. 

AxGEBRA is a general kind of arithmetic, an universal analysis, 
or science of computation by symbols. 

Quantify or magnitude is a general term applied to everything 
which admits of increase, (liminullon, and measurement. 

The measurement of quantity is accomplished by means of an 
assumed luiit or standard of measure ; and the unit must be the 
same, in kind, as the quantity measured. In measuring length, we 
apply length, as an inch, a yard, or a mile, &c. ; measuring area, 
we apply area, as a square inch, foot, or acre ; in measuring 
money, a dollar, pound, <fec., may be taken for the unit. 

Numbers represent the repetition of things, and when no ap- 
plication is made, the number is said to be abstract. Thus 5, 13, 
200, <fcc., arc numbers, but $5, 13 yards, 200 acres, are quanti- 
ties. 

In algebraical expressions, some quantities may he known, 
others unknown ; the known quantities are represented by the 
first or leading letters of the alphabet, «, 6, c, d, (fee, and the 
unknown quantities by the final letters, z, y, x, i(, &c. 

THE SIGNS. 

(1) The perpendicular cross, thus -|-, called plus, denotes ad- 
dition, or a positive value, state, or condition. 

(2) The horizontal dash, thus — , called minus, denotes sub- 
traction, or a negative value, state, or condition. 

(3) The diamond cross, thus X, or a point between two quan- 
tities, denotes that they are to be multiplied together. 

(9J 



10 ;feR01)U<^fQ^i; •, , 

(4) A, horizonta) |i?i© t^^:;?!t jjojat'abdvtj^ and(r, below, thus -i-, 
denotes division. Also, two quantities,' one above another, as 
numerator and denominator, thus -^ indicates that a is divided 

6 

hyb, 

(5) Double horizontal lines, thus =, represent equality. Points 
between terms, thus a : b :: c : d^ represent proportion, and are 
read as a is to 6 so is c to d. 

(6) The following sign represents root J ; alone it signifies 
square root. With small figures attached, thus ^^ ^\/ ^^, <fec., 
indicates the third, fourth^ fifth, S&c, root. 

Roots may also be represented by fractions written over a 

2. 1 1 

quantity, as a* a^ a^, &c., which indicate the square root, 
the third root, 2ind fourth root of a.* 

(7) This symbol, a^b, signifies that a is greater than b. 
This " , ct<Cl), signifies that a is less than b. 

(8) A vinculum or bar , or parenthesis ( ) is used to con- 
nect several quantities together. Thus a-\-b-\-cXx Q'e [a-]rb-\-c)x, 
denotes that a plus b plus c is to be multiplied by x. The bar 

y may be placed vertically, tlms, Avhich is the same 
as {a — d-\-e)y, or the same as ay — dy-\-ey without the 

vinculum. 

(9) Simple quantities consist of a single term, as a, b, ab, 3rr, 
(fee. Compound quantities consist of two or more terms con- 
nected by their proper signs, as a-\-x, ^b-\-'2,y, 7ab — 3xy-\-c, &c. 
A binomial consists of two terms ; a trinomial of three ; and a 
polynomial of many, or any number of terms above two. 

(10) The numeral which stands before a quantity is called its 
coefficient ; thus 3x, 3 is the coefficient of x, and indicates that 
three a?'s are taken. Coefficients may be literal, simple, or com- 
pound, as well as numeral ; thus abx, {a-{-b) x ; (c— -i+S) x. 
Here ab, (ff+^) and (c — <?4-2) may be considered coefficients of a:. 

(11) A measure of any quantity is that by which it can be 
divided without a remainder. 2 is a measure of 4, or any even 

* The adoption and utility of this last mode of notation, which ought to be 
exclusively used, will be explained in a subsequent part of this work. 



a 
—d 



EXERCISES ON NOTATION. 1 1 

number. 5a is the measure of 20ar. 3x is the measure of 12a?, 
or I2ax. 

A multiple of any quantity is that which is some exact num- 
ber of times that quantity ; thus 12 is a multiple of 3, or of 4, or 
of 6, and SOab is a multiple of 3a6, of 5a6, &;c. 

AXIOMS. 

Axioms are self-evident truths, and of coarse s^e above demon- 
stration ; no explanation can render them more clear. The fol- 
lowing are those applicable to algebra, and are the principles on 
which the truth of all algebraical operations yina//?/ rests: 

Axiom 1. If the same quantity or equal quantities be added to 
equal quantities, their sums will be equal. 

2. If the same quantity or equal quantities be subtracted from 
equal quantities, the remainders will be equal. 

3. If equal quantities be multiplied into the same, or equal 
quantities, the products will be equal. 

4. If equal quantities be divided' by the same, or by equal 
quantities, the quotients will be equal. 

5. If the same quantity be both added to and subtracted from 
another, the value of the latter will not be altered. 

6. If a quantity be both multiplied and divided by another, 
the value of the former will not be altered. 

7. Quantities which are respectively equal to any other quan- 
tity are equal to each other. 

8. Like roots of equal quantities are equal. 

9. Like powers of the same or equal quantities are equal. 



EXERCISES ON NOTATION. 

When definite values are given to the letters employed, we 
can at once determine the value of their combination in any alge- 
braic expression. 



12 * ELEMENTS OF ALGEBRA. 

Let«=5 6=20 c=4 d=l 
Then a-{-6— c=5-{-20— 4 or a-|-6— c=21 

— f-rf=--+l or-+(Z=5 

a 5 a 

«6-far+^==5X 20-1-5X4-}- 1 = 121 
2a4-86-{-2c-j-5c?=10-f60-h8-f-5=83 



SECTION I. 

ADDITION. 

(Art. 1.) Before we can make use of literal or algebraical 
quantities to aid us in any mallieraatical investig-ation. we must 
not only learn the nature of the quantities expressed, but how to 
add, subtract, multiply, and divide them, and subsequently learn 
how to raise them to powers, and extract roots. 

The pupil has undoubtedly learned in arithmetic, that quanti- 
ties representing different things cannot be added together ; for 
instance, dollars and yards of cloth cannot be put into one sum; 
but dollars can be added to dollars, and yards to yards ; units can 
be added to units, tens to tens, &,c. So in algebra, a can be 
added to a, making 2a ; oa can be added to 5a, making 8a. As 
a may represent a dollar, then 3a would be 3 dollars, and 5a 
would be 5 dollars, and the sum would be 8 dollars. Again, a 
may represent any number of dollars as well as one dollar ; for 
example, suppose a to represent 6 dollars, then 3a would be 18 
dollars, and 5a would be 30 dollars, and the whole sum would be 
48 dollars. Also, 8a is 8 times 6 or 48 dollars ; hence any num 
ber of a's may be added to any other number of a's by iinitmg 
their coefficients ; but a cannot be added to h, or 4a to 36, or to 
any other dissimilar quantity, because it would be adding unlike 
things, but we can write a-{-6 and 3a-i-36, indicating the addi* 
tion by the sign, making a compound quantify. 



ADDITION. * 13 

Let the pupil observe that a broad generality, a wide latitude 
must be given to the term addition. In algebra, it rather means 
uniting, condensing, or reducing terms, and in some cases, the 
sum may appear like difference, owing to the difference of signs. 
Thus, 4a added to — a is 3a ; that is, the quantities united can 
make only 3a, because the minus sign indicates that one a must 
be taken out. Again, lb-\-^b — 4Z/, when united, can give only 
6^, which is in fact the sum of these quantities, as 46 has the 
minus sign, which demands that it should be taken out; hence to 
add similar quantities we have the following 

Rule. Add the affirmative coefficients into one sum and the 
negative ones into another, and take their difference with the 
sign of the greater, to which affix the common literal quan- 
tity. 

EXAMPLES FOR PRACTICE. 
5a 17a? -]-5ab 6a-\-5b — 7crf + 8.ry 

2a 2x —Qab —6a— 46 —■2cd + Sxy 

Sum 7a 19jc — ab -\-b — 9cd-\-llxy 

5a-\-b cdy-\-ax 4a) — 6 

3a+c 2cdy — 'dax 2a:+10 

'7a--26-f c ^cdy-\-^ax ' — 3aH- 7 

—3a— 36 — 4c ^^.cdy^ax 6a?— 12 

Sum 12a— 46— 2c 9a:— 1 

N. B. Like quantities, of whatever kind, whether of powers or 
roots, may be added together the same as more simple or rational 
quantities. 

Thus 3a2 and Sa^ are lla^ and 7634-36=^=1 06^ No matter 
what the terms may be, if they are only alike in kind. Let the 
reader observe that 2(a+6)-}-3(a+6) must be together 5(a-|-6), 
that is, 2 times any quantity w^hatever added to 3 times the 
same quantity must be 5 times that quantity. Therefore, 
^Jx-{-y-\~^Jx-]ry='^ Jx-\-y, for Jx-\-y, which represents the 
square root of x-\-y, may be considered a single quantity. 

(Art. 2.) To find the sum of various quantities we have the 
following 
B 



14 



ELEMENTS OF ALGEBRA. 



Rule. Collect together all those that are alike, by uniting 
their coefficients, and then write the different sums, one after 
another, with their proper signs. 



1. 

Sxy 
2ax 

6ax 


EXAMPLES. 

2. 

9x^y 
^7x'y 
-\-3axy 
— 4a^ 

3axy — 2a^y 

5. 

\4ax—2x^ 
bax-\-3xy 
Sy^ —4ax 
3x^ +26 


3. 

Gxy—I2x^ 
—4x^ +3xy 
^4a^^2xy 
— 3a^+4a;2 


Sum Sax — 2xy 

4. 

4ax—lS0-}-B^x 
ba^ '\-3ax+9x' 
7a?y— 4^a?+90 


4xy-^Sx" 

6. 

9+10^«a;--5y 

2x+7jxy-{-5y 

5y-\-3jax+4y 

10 '-^4Jax-\-4y 


7ax+Sx^-]-7xy 







7. Add 2xy — 2a^, 3a^-\-xy, a~-{-xy, 4a^ — 3xy, 2xy — 2a^. 

Ans, 4a^-\-3xy. 

8. Add 8aV — 3xy, 5ax — 5xy, 9xy — bax, 2aV+a:y, 
&ax — 3xy. *^ns. 10a^x'^-{-5ax — xy. 

9. Add2a;^— lOy, 3jxy-\-l0x, 2x^y-\-25y, I2xy—Jxy, 
—Sy-\-\7jxy. i 

£ns. 2x^y\-\2xy-\-\9x-\-2x'' -\-l9^xy-\-7y. 

10. Add 2bx*—\2, 3oi^—2hx, 5x"—3jx. 37a;+12, 
a^+3. Ans. 9a;2+3. 

11. Add l()b^^3bx', 26V— 6^, I0^2bx% 6V— 20, 
3bx^-\-b\ Ans. \Qb''-\-3b''x—2bx''—lO. 

12. Add 2a2— 3aa:^+a^, 2«cr^— 13a?i/+8, lOo^— a?2/— 4. 

Ans. \2a^ — ax^-^-a? — l4xy-\-4. 

18. Add 9&c3— ISac^, ISftc^+ac, 9«c2— 24^>c^ 9ac2— 2. 

Ans* ac — 2 



ADDITION. 15 

14. Add 3m2— 1, 6am — 2m2+ 4, 7 — 8«m+2m2, and 
6m^+2am-\-l. Jins. Om^+U. 

15. Add 12a— 13a6+16aa:, 8— 4m4-2i/, — Ga+Ta^^-j- 
12t/ — 24, and lab — 16aa;+4m. 

Ans, 6a— 6a&+14i/+7a52— 16. 

16. Add 12ax^^Say\ —3Sax^—3ay*-\-lay\ S-\-l2ay\ 
^Qay^-\-l2—34ax'^-\-6ay^—day\ Ans. —2ay^-\-20. 

Add a-{-b and 3a — ob together. 

Add 6a:— 56+a+8 to —5a— 437+46-3. 

Add a-1-26-- 3c — 10 to 36 — 4a+5c+10 and 5&— c. 

Add ^a-\-b — 10 to c — d—a and — 4c+2a — 36 — ^7. 

Add 3a2+262— c to 2a6— 3a24-6c— 6. 

(Art. 3.) When similar quantities have literal coefficients, we 
may add them by putting their coefficients in a vinculum, and 
writing the term on the outside as a factor. 

Thus the sum of ax and bx is (aH-6)a;. 



1. 

Add ax-\-by^ 
2ca;+3ai/2 
4dx-{-ly^ 


3. 

2. 

ay-{-cx 
^ay-\-2cx 
4y +6a; 


Sum (a+2c+4rf)a:+(6+3«-i-7)i/2 

3. 

Add Zx-{-2xy 
bx-\-cxy 
(a -^b)x-\'2cdxy 


(4a+4)y-}-(3c-l-6)ar 

4. 

ax-\-ly 

lax — 3?/ 

—237 +4y 



Sum (a-|-26+3)a74-(2cf?+c+2)xy (8a— 2)a;+8?/ 

5. Add 8aa:+2(a?+a)+36, 9aa;+6(a?+a)— 96, and Ua;4- 
66— 7aa:— 8(a?+a). Ans, lOax+llar. 

6. Add {a-\-b)Jx and {c-\-2a^b)Jx together. 

T. Add 2^a\x-{-^ij)-\-2\, lSa—\^a\x-{-^y), — 15a3(ar4- 
53/)— 8. ^ns, 18a-fl3. 



16 ELEMENTS OF ALGEBRA. 

8. Add l7a{x-{-3ay)-\-l2a'b'c% S^lSay^8c^b^c\ — 7a(a?+ 



SUBTRACTION. 

(Art. 4.) We do not approve of the use of the term subtraction^ 
as applied to algebra, for in many cases subtraction appears like 
addition, and addition like subtraction. We prefer to use the 
expression, y^nc/m^ the difference. 

What is the difference between 12 and 20 degrees of north 
latitude ? This is subtraction. But when we demand the differ- 
ence of latitude between 6 degrees north and 3 degrees south, the 
result appears like addition, for the difference is really 9 de- 
grees, the sum of 6 and 3. This example serves to explain the 
true nature of the sign minus. It is merely an opposition to 
the sign plus; it is counting in another direction; and if we 
call the degrees north of the equator plus, we must call those 
south of it mimes, taking the equator as the zero line. 

So it is on the thermometer scale ; the divisions above zero 
are called plus^ those below minus. Money due to us may be 
called plus ; money that we owe should then be called minus, — 
the one circumstance is directly opposite, in effect, to the other. 
Indeed, we can conceive of no quantity less than nothing, as 
we sometimes express ourselves. It is quantity in opposite cir- 
cumstances or counted in an opposite direction ; hence the differ- 
ence or space between a positive a?id a negative quantity is 
their apparent sum. 

As a further illustration of finding differences, let us take the 
following examples, which all can understand : 

From 16 16 16 16 16 16 

Take 12 8 2 —2 —4 

Differ. 4 8 14 16 18 20 

Here the reader should strictly observe that the smaller the 
number we take away, the greater the remainder, and when the 
subtrahend becomes minus, it must be added. 







SUBTRACTIOIN 






From 


I2a 


12a 12a 12a 


12a 


12a 


Take 


20a 


16a 12a 9a 


6a 


a 



n 



Diff. -__8a —4a 3a 6a 11a 

When a greater is taken from a less, we cannot have a posi- 
tive ox plus difference, it must be minus. 

From 20a 10a 5a —5a —-10a 
Take Ua Ha 11a 11a —b —b —5a 

Diff. 9a —a -—Qa —Ua +6 ^>— 5a —5a 

Here it will be perceived that the difference between zero and 
any quantity is the same quantity with the sign changed. 

(Art. 5.) Unlike quantities cannot be written in one sum, (Art. 
1,) but must be taken one after another with their proper signs ; 
therefore, the difference of unlike quantities can only be ex- 
pressed by signs. Thus the difference between a and b is a — 6, 
a positive quantity if a is greater than 6, otherwise it is negative. 
From a take h — c, (observe that they are unlike quantities). 

OPERATION. 
From a+0+0 

Take 0+6— c 



Remainder, or difference, a — b-{-c 

This formal manner of operation may be dispensed with ; the 
ciphers need not be written, and the signs of the subtrahend need 
only be changed. 

From the preceding observation, we draw the following 

GENERAL RULE FOR SUBTRACTION, OR ALGEBRAIC DIF- 
FERENCES. 

Change the signs of the subtrahend^ or conceive them to be 
changed; then proceed as in addition. 

EXAMPLES. 

1. 2. 3. 

From 4a-{-2x — 3c Sax-{-2y a-\-b 

Take a-\-4x — 6c xy — 2i/ a — b 

Remainder, 3a — 2x-{-3c Sax — xy-{-4y 26 



18 



ELEMENTS OF ALGEBRA. 



From 
Take 


4. 5. 

2:^2—307+1/2 7a+2-- 
^x^^ix-\-a — a+2+ 


-5c 
c 


6. 

hx+ky 

Ix—hy 


Rem. 


Sx'+x+y^—a 8a * — 


■6c 


y 


From 
Take 


7. 

Sx'—'Sxy+2y'-i' c 
x^—6xy-\-Sy^—2c 




8. 

ax-\-'bx-\-cx 
x-Tax-\-Qx 


Diff. 


7x'-\-3xy^y''+Sc 


{c-\)x 


From 
Take 


9. 

ax-\-by-]-cz 
— mx — 7iy — pz 


10. 

a^b+c 


Diff. 


(a-Jrm)x-\-{b+n)y-\r{c-Jrp)z 


2a+26+2c 



(Art. 6.) From a take b. The result is a — b. The minus 
sign here shows that the operation has been performed ; b was 
positive before the subtraction ; changing the sign performed 
the subtraction; so changing the sign of any other quantity 
would subtract it. 

11. From 3a take {ab-\-x — c — y), considering the terms in 
the vinculum as one term^ the difference must be 3a — {ab-\-x — 
c — ^y), but if we subtract this quantity not as a whole, but term 
by term, the remainder must be 3a — ab — xArC-\-y, 

That is, when the vinculum is taken away, all the signs 
within the vinculum must be changed. 

12. From SOxy take {40xy—2b^-\-3c—4d). 

Rem. 2b^--l0xy—3c-{-id, 



13. From Jx-{-y+Sax-^l2 take '—{4jx+y—2ax-^b). 

Rem. 5ax — Sjx-Jry — 12 — b. 

14. Find the difference between 6/ — 2?/— 5 and — Sy^ — 
5y4-l2. -^ns. 14y+3?/— 17 

15. From 3a— &— 2a^+7 take S--3b-\-a-\-4x. , 

Mns. 2a4-2^>— 6a?— 1. 



MULTIPLICATION. 19 

16. From Sp-{-q+r — 2s take q^8r-\-2s — 8. 

w?ns. 3jo+9r — 4s-f8. 

17. From lSa^—2ax+9a^ take Sa^— 7aa;-— x^. 

18. From 20xy — S^a+Sy take 4xy+5a^ — y 

Ans. 162?/— 10a^+4y. 

19. From the sum of Qx'^y — Wax^, and Sx^y-\-^ax^, take 
Ax^y — Aax^-\-a. Diff. K^x^y — iax^ — a. 

20. From the sum of ISa^^-f-ScJa?— 3 and 24 — %a^b-\-2cdx 
take the sum of \2a^b — ^cdx — 8 and lQ-\-cdx — Aa^b. 

Diff. 12cc?ic4-13— fl26. 

21. From the difference between Sab — \2cy and — 3a6-f- 
4,cy take the sum of bab-^lcy and ab-{-cy, 

Diff. 5a6— lOcy. 
From 2a-\-2b take — a — b. 
From «x4-<^'''? take ax — bx. 
From a-[-c+6 take ft-f-c — 6. 
From Sx-{-2y-\-2 take 5a;+33/-f6. 
From 6a+2a:-|-c take 5a-jrQx — 3c. 
From — 4a— 2:e — 2 take — 6a — 2x — 2. 
From l2X'-2xy-\-3 take 7+6i/+10a?. 



MULTIPLICATION. 

(Art. 7.) The nature of multiplication is the same in arithmetic 
and algebra. It is repeating one quantity as many times as there 
are units in another ; the two quantities may be called factors, 
and in abstract quantities, either may be called the multiplicand ; 
the other will of course be the multiplier. 

Thus 4 X 5. It is indifferent whether we consider 4 repeated 
5 times or 5 repeated 4 times ; that is, it is indifferent which we 
call the multiplier. Let a represent 4, and b represent 5, then 
the product is aXbi or with letters we may omit the sign and 
the product will be simply ab. 



20 ELEMENTS OF ALGEBRA. 

The product of any number of letters, as ab c c/, is abed. 

The product oi x y z is xyz. 

In the product it is no matter in what order the letters are 
placed ; xy and yx is the same product. 

The product of axXby is axby or aba^y. Now suppose 
a=Q and b=8, then ab=4S, and the product of axXby would 
be the same as the product of 6xXSy or 48:n/. From this we 
draw the following rule for multiplying simple quantities : 

Multiply the coefficients together and annex the letters^ one 
after another^ to the product. 

EXAMPLES. 

1. Multiply Sx by 7a. Prod. 2\ax. 

2. Multiply 4i/ by ^ab Prod. I2aby. 

3. Multiply 3b by 5c, and that product by 10a?. 

Prod. I50bcx. 

4. Multiply Gax by 12by by lad. Prod. bOiaaxydb. 

5. Multiply Sac by lib by xy. 

6. Multiply af by pq by 4. 

In the above examples no signs were expressed, and of course 
plus was understood ; and it is as clear as an axiom that plus 
multiplied by plus must produce plus, or a positive product. 

(Art. 8.) As algebraic quantities are liable to be' affected by 
negative signs, we must investigate the products arising from 
them. Let it be required to multiply — 4 by 3, that is, repeat 
the negative quantity 3 times ; the whole must be negative, as 
the sum of any number of negative quantities is negative. 
Hence minus multiplied by plus gives minus, — aXb gives 
— aJ) ; also a multiplied by — b must give — ab, as we may 
conceive the minus b repeated a times. 

(Art. 9.) Now let it be required to multiply — i by — 3, that 
is, minus 4 must be subtracted 3 times ; but to subtract minus 
4 is the same as to add 4, (Art. 5,) giving a positive or plus 
quantity ; and to subtract it 3 times, as the — 3 indicates, will 
give a product of +12. 

That is, minus multiplied by minus gives plus. 



MULTIPLICATION. 21 

This principle is so important that we give another mode of 
illustrating it : 

Required the product of a — h by a — c. 

Here a — h must be repeated a — c times. 

If we take a — 5, a times, we shall have too large a product, 
as the multiplier a is to be diminished by c. 

That is a — h 

Multiplied by a 

Gives aa — ah^ which is too great by a — h repeated 

c times, or by at — c&, which must be subtracted from the former 
product; but to subtract we change signs, (Art. 5,) therefore the 
true product must be aa — ab — ac-\-ch. 

That is, the product of minus b by minus c gives plus be, 
and, in general, minus multiplied by minus gives plus. 

But plus quantities multiplied by plus give plus, and minus by 
plus, or plus by minus, give minus ; therefore we may say, in 
short, 

77m/ quantities affected by like signs, when multiplied toge- 
ther, give plus, and when affected by unlike signs give minus. 

(Art. 10.) The product of a into b can only be expressed by 
ab or ba. The product oi abed, <fec., is abed; but if b c and d 
are each equal to a, the product would be aaaa. 

The product of aa into aaa is aaaaa ; but for the sake of 
brevity and convenience, in place of writing aaa, we write a^. 
The figure on the right of the letter shows how many times the 
letter is taken as a factor, and is called an exponent. The pro- 
duct of c^ into a* is a repeated 3 times as a factor, and 4 times 
as a factor, in all 7 times ; that is, write the letter and add the 
exponents. 

EXAMPLES. 

"What is the product of (^ by a^ ? 
What is the product of x'^ by x^ 1 
What is the product of y"^ by y^ by if ? 
What is the product of a" by a"* ? 
What is the product of b''x^ by bx. 
What is the product of ac by ac^ by aV ? 



Ans. 


««. 


Ans, 


x''. 


Ans. 


y^\ 


Ans. a 


" + "». 


Ans. 


6V. 


Ans. 


c^c' 



22 ELEMENTS OF ALGEBRA. 

If adding numeral exponents is a true operation, it must be 
equally true when the exponents are literal. 

N. B. When the exponent is not expressed, one is understood, 
for a is certainly the same as a\ or once taken. 

(Art. 11.) Every factor must appear or be contained in a pro- 
duct. Thus ax^ multiplied by bx^ must be abx^. Now if a~Q 
and 6=10 the product would be QQx^, 

Multiply 2d' by la^. Product 2\aK 

From this we draw the following rule for the multiplication 
of exponential quantities. 

Multiply the coefficients and add the exponents of the same 
tetter. All the letters must appear in the product. 

EXAMPLES. 

Multiply 4a^ by 3a. Prod. 12«'. 

Multiply 2x^ by —2x^. Prod. — -6a^. 

Multiply 2x by Ix^ by 2a^y. Prod. 63aVy. 

What is the product of 2«a?S 4taxy<, labx ? 

Prod. bQd'x^by^ 

What is the product of 2a'S 3a'"a?, and ax 1 

Prod. 6a"+'"^-'a;^ 

Multiply Qa^x by 4a?. Prod. 36aV. 

Multiply llaWc^ by lac. Prod. l\Qa^bh\ 

Multiply lla^^^c by lQd'bh\ Prod. 110a^°6^V°. 

Multiply l^lb^c^x by bd^bxy\ Prod. GOba'^bh^x^yK 

Multiply Ula^cx"^ by 6la^b. Prod. 4Q97a^bcx\ 

Multiply niabh^x by 2(^b^c. Prod. 23ia'^b'*c^x. 
Multiply Qd^x by 6a?. 
Multiply 9ax^ by — 7ax. 
Multiply 7ax by — 4. 

Multiply Sac by — 2ca? by — 4c, Prod. 24ac^x. 

(Art. 12.) When one compound quantity is to be multiplied 
or repeated as many times as there are units in another, it is 
evident that the multiplicand must be repeated by every term of 
the multiplier. 



MULTIPLICATION. * 23 

Thus the product of a-{-b-i-c by x-\-y-\-z. 
It is evident that a-\-b-j-c must be repeated x times, then y 
times, then z times ; and the operation may stand thus : 

a-{-b-{-c 

x-{-y-\-z 

Product by x ax-\-bx-{-cx 
Prodlict by y o,y^by-\-cy 

Product by z az-\-bz-\-cz 



Entire Product ax-j-bx-\-cx-{-ay-{-by-\-cy-]-az-\-bz-\-cz. 

From the foregoing articles we draw the following general 
rule for the multiplication of compound quantities. 

Multiply all the terms of the multiplicand by each term of 
the multiplier, observing that like signs, in both factors, give, 
plus, and unlike, minus. 

Write each term of the product distinctly by itself, with its 
proper sign, and afterwards condense or connect the terms as 
much as possible, as in addition. 

EXAx^IPLES. 
1. 2. 

Multiply 2nx — 3x 3x-\' 2y 

By 2x +4y 42? — 5y 

Partial product 4ux/^—'63r^ 123^*+ 8xy 

2d part. prod. Saxy — VZxy — I5xy — lOy^ 

Whole prod. 4ax-\-Saxy — 6x'^ — I2xy 12ar^ — 7xy — lOy^ 

8. Multiply 2x^-{-xy^2y^ 

By '3x —3y 



Partial product Qx^-\-Sx~y — 6xy^ 

2d partial product — Qx^y — Sxy^-\-Gy^ 



Whole product 6x^ — 2x-y — 9xy^-{-6y^ 

Multiply 3a^—2ab-^^ by 2a — ib. 

Prod. Qc^—\Qa^b-\-Qab^-{-4b\ 
Multiply x^- — xy-^-y"^ bv x-{-y Prod x^-]ry^ 



24 ELEMENTS OF ALGEBRA. 

6. Multiply w — 3«c4-c^ by a — c. 

Prod. «='— 4«2c4-4ac2— c^ 
'2'. Multiply «-|-Z) by a-^h. Prod. a^-\-2ab-\-h^, 

8. Multiply x-\-y by x-\-y. Prod. a^-\-2xy-\-y^, 

9. Multiply a--h by «— Z/. ProJ. a^—2ab^b\ 

10. Multiply x — y by a; — 2/* Prod, a^ — 2xy-\-y^. 

(Art. 13.) By inspecting all the problems, from the 7th to the 
lOtli, we shall perceive that they are all binomial quantities, and 
the multiplicand and multiplier the same. 

But when a number is to be multiplied into itself the product 
is called a square. Now by inspecting the products, we find 
that the square of any binomial quantity is equal to plus ; the 
squares of the two parts and twice the product of the two parts. 

N. B. The product of the two parts will be plus or minus, 
according to the sign between the terms of the binomial. 

Let us now examine the product of a-\-h into a — h. 
a -\-h 2m +2n 

a — b 2m — 2n 



a^-\-ab 4m^-\-4mn 

— ab — b^ — imn — 4n^ 



Product a^ — b^ Am^ — in:- 



Multiply 2a-\-^b by 2a— 3b. Prod. 4a^—9b^. 

Multiply 3i/+c by By — c. Prod, dy^-^c^ 

Thus, by inspection, ive find the product of the sum and 
difference of two quantities is equal to the difference of their 
squares. 

The propositions included in this article are proved also in 
geometry. 

(Art. 14.) We can sometimes make use of binomial quantities 
greatly to our advantage, as a few of the following examples will 
show : 

1. Multiply a-\-b-{-c, by a-\-b-\-c. 

Suppose a-{-b represented by s, then it will be s-]-c. 



MULTIPLICATION. 25 

The square of this is s^-{-2sc-{-c^i restoring the value of 5, and 
we have {a-\-bY-\-2{a-\-b)C'{'cK 

2. Square x-]ry — z. Let x-]ry=s. 

Then {s—2:y=s^—2s2-i-z^={x-\-yy--2{x-\-y)2-\'2^ 

3. Multiply x-\-y-\-z by X'\-y — z. Prod. (x-{-yY — z* 

4. Multiply 2ocf^-^3x-\-2 by x^S. 

Prod. 2ar^— 1 9a^-|-26a:— 1 6 

5. Multiply ax-\-by by ax-\-cy. 

Prod. a^xl^-{'{ab-{'ac)3cy-\-cby* 

6. Multiply ix-\-y by hx — y. Prod, ^x^ — y*, 

If. Multiply a^+2a^b-\-2ab^ -{-b^ 
By (^—2a?b-\-2ab^ — 6^ 



a^-\-2(^b-\-2c^¥-\-c^b^ 

+2«462 +4«363^-4fl264^2a65 

— a^b'.^2a^b'^2ab^—b* 



Prod. ««— 6« 

Multiply ar2— ^x+f 
By ix-\-2 



+ 2 x"— x~\-i 



Product, |j:3+y3^_73j^4 

9. What is the product of a^-|-6"* by ff'*-|-6" ? 

10. What is the product of a?^ — ?a? by x^ — ^a?? 

v2?js. X* — ^xr^-\- I 0^2 — 

11. What is the product of 4x^-^8x^+l6x-{-S2 by 3a:--6? 

^ns. 12a;*—I92. 

12. What is the product of (^-{-a^b+ab^+b^ by «— i> ? 

.^??s. a*— 6*. 
3 



I 



I 



aa ELEMENTS OF ALGEBRA. 

DIVISION. 

(Art. 15.) Division is the converse of multiplication, the pro- 
duct being called a dividend, and one of the factors a divisor. If 
a multiplied by b give the product g6, then ab divided by a must 
give b for a quotient, and if divided by 6, give a. In short, if 
one simple quantity is to be divided by another simple quantity, 
the quotient must be found by inspection^ as in division of num- 
bers. 

EXAMPLES. 

1. Divide I6ab by 4a. *Ans. 4b. 

2. Divide 2lacd by 7c. Ans. Sad, 

3. Divide a¥c by ac. Ans. b^. 

4. Divide Saxy by 26c. Ans. — ^. 

In this last example, and in many others, the absolute division 
cannot be effected. In some cases it can be partially effected, 
and the quotients must be fractional. 

5. Divide Sacx^ by acy. Ans. . 

96 



6. Divide 7262^; by Sabx. Am, 

"7. Divide 21aby by llabx. Ans. 



a 



27y 



(Art. 16.) It will be observed that the product of the divisor 
and quotient must make the dividend, and the signs must con- 
form to the principles laid down in multiplication. The follow- 
ing examples will illustrate : 

8. Divide —dy by Sy. Ans, -—3. 

9. Divide — 9y by —-Sy. Ans, +3. 

10. Divide -\-9y by --3y. Ans. —3. 

• The term quotient would be more exact and technical here ; but, in re* 
suits hereafter, we shall invariably use the term Ans., as more brief and ele- 
gant, and it is equally ^vell understood. 



DIVISION. 27 

(Art. 17.) The product of cr* into a^ is a', (Art. 10,) that is, 
in multiplication we add the exponents ; and as division is the 
converse of multiplication, to divide powers of the same letter, 
we must subtract the exponent of the divisor from that of the 
dividend. 

Divide 2a^ by a\ 

Divide — a' by a*. 

Divide 16a:' by 4x. 

Divide 15axy^ by — Say, 

Divide 63a"* by 7a\ 

Divide 1203?** by — 3ax. 

Divide 7d^b by 2la^b\ Ans 

Divide — Sa'a::* by — Ic^s^. 
Divide UTa^^V by TSa^fic*. 
Divide 96a6c by I2c^bc^d, 
Divide c^bd^ by a^bh\ 

Divide 27c^b^cd^ by 2labcd, 

7 

Divide UabHd by 6a%(^. Jim, 1^. 

(Art. 18.) The object of this article is to explain the nature of 
negative exponents. 

Divide a* successively by a, and we shall have the following 
quotients : 

a^ a\ a, 1, -, ^, -3, &c. 

Divide c^ again, rigidly adhering to the principle that to 
divide any power of a by a, the exponent becomes one less, and 
we have 



Ans, 


2a\ 


Jtns. 


— a. 


Ans. 


40^, 


Ans. - 


-5xf, 


An3, Qa*^". 


Mns. --4ic«-'. 


7a^b 


1 


2ia^b* 


Sab' 


Ans, 


5 


Ans. 


36» 
2c* 


Ans. 


8 

d'cd' 


Ans. 


1 
abd'' 


9 
Ans, ;-a^b^d. 



28 ELEMENTS OF ALGEBKA. 

(^, a^ aS a", a~S a"^, a-^ &c. 

Now these quotients must be equal, that is, a^ in one series 
equals a^ in the other, and 

a^=c^, a=a}, l=a^ ~=a~* — „=a~^ -^^cr^ 

Another illustration. We divide exponential quantities by 
subtracting the exponent of the divisor from the exponent of the 
dividend. Thus a^ divided by a^ gives a quotient of a^~^=c^. 
a^ divided by a'=(^~'^=a~^. We can also divide by taking the 
dividend for a numerator and the divisor for a denominator, thus 

-^=—, therefore — ,=a""^ (Axiom 7.) 

From this we learn, that exponential factors maybe changed 
from a numerator to a denominator, and the reverse, by chang- 
ing the signs of the eocponents. 

Thus -=aX-^ .^ = JL ^^rrfn-n 

Ihus, ^ ax ^^_, g^3 ^„ X 

Divide a'bc by a^bh~\ Ans. a~^b~^c*. 

Observe, that to divide is to subtract the exponents. 

Divide liab^cd by dd'bc^ Ans. 'L—=lbdcr'c-\ 

Sac 3 

(Art. 19.) A compound quantity divided by asimple quantity, is 
effected by dividing each term of the compound quantity by the 
simple divisor. 

EXAMPLES. 

1. Divide Sax — 15a? by 3a?. Ans. a — 5. 

2. Divide Sx^^-Ux^ by 4a^. Ans, 2a;+3. 

3. Divide Sbcd-[-l2bcx — 9b^c by 36c. Ans. d-\-4x—Sb. 

4. Divide lax-\-Say — Ibd by — lad. 



M- DIVISION. 39 

ft. Divide I5c^bc — l5aco[^-^5a(P by — 5ac. . in -*.; 

.^ns. — 3ab-\Sa^ . 

6. Divide lOor'— 15a?2— 25a; by 5x. Ans. 2x^-~^x—^. 

7. Divide —\Oab-\-mab^ by •— 6a6. 

o 

8. Divide ^Qa%^-\-^Q(^b—Q(ib by — 12a&. 

^n«. — 3a6 — 5a+5. 
O. Divide lOrx — cry-\-'2crx by cr. 

10. Divide \Ovy-{-Ucl by 2rf. 

11. Divide iSay — \Sacd-{-2^a by 6a. 

12. Divide ma? — amx-\-'m by 7n. 

(Art. 20.) We now come to the last and most important ope- 
ration in division, the division of one compound quantity by an- 
other compound quantity. 

The dividend may be considered a product of the divisor into 
the yet unknown factor, the quotient ; and the highest power of 
any letter in the product, or the now called dividend, must be 
conceived to have been formed by the highest power of the same 
letter in the divisor into the highest power of that letter in tlie 
quotient. Therefore, both the divisor and the dividend must 
be arranged according to the regular powers of some letter. 

After this, the truth of the following rule will become obvious 
by its great similarity to division in numbers. 

Rule. Divide the first term of the dividend by the first term 
of the divisor, and set the result in the quotient.'^ 

Multiply the whole divisor by the quotient thus found, and 
subtract the product from the dividend. 

The remainder will form a neiv dividend, with which pro- 
ceed as before, till the first term of the divisor is no longer 
contained in the first term of the remainder. 

The divisor and remainder, if there be a remainder, are then 

* Divide theirs/ term of the dividend and of the remainders by ihe first 
term of the divisor ; be not troubled about other terms. 



30 ELEMENTS OF ALGEBRA, 

to be written in the form of a fraction, as in division of num- 
bers. 

EXAMPLES. 
Divide a^^2ab+b'' by a+6. 

Here, a is the leading letter, standing first in both dividend 
and divisor ; hence no change of place is necessaiy. 

OPERATION. 

a-f-6)a2+2a6-l-6-(a4-6 
a2-{- ab 



ab+b^ 
ab-hb^ 



Agreeably to the rule, we consider that a will be contained in 
a^ a times ; then the product of a into the divisor is a^-^-ab, and 
the^7's/ term of the remainder is ab, in which a is contained b 
times. We then multiply the divisor by 6, and there being no 
remainder, a-j-6 is the whole quotient. 

Divide c^-\-3a^x-\-3aa:^-\-x^ by x-\-a. 

As the highest power of a stands in the first term of the divi- 
dend, and the powers of a decrease in regular gradation from 
term to term, therefore we must change the terms of the divisor 
to make a stand first. 



OPERATION. 

c-f^)a'+3a2a?-l-3ax2-|-a^(a2-f2flx-fx* 
a^-\- c?x 



l€^X^3a3? 

"laH-Y^ax" 



ax^-\-3? 
ax^-\-2^ 



DIVISION. 81 



«3 ^2g 



— -3a*c4-3ac2 



ac^ — c* 
ac^ — c^ 



a«— 4a+4)a^— 6fl*+ 12a— 8(a--2 
a' — 4fl^-|- 4a 



— 2a2-l- 8a— 8 
— 20^-1- 8a— -8 



4. Divide 6a;*— 96 by 6a;— 12. ^ns. x^'\-2a*-{-4x-{'S 

5. Divide a^ — b^ by a — b. Ans. a-{-b, 

6. Divide 25a;«— a;*--2a:'— 80?^ by 5r»— 4a:*. 

Ans. 5x3+4a?«+3x+2. 

(Art. 21.) We may cast out equal factors from the dividend 
and divisor, without changing the value of the quotients, for 
amxy divided by am gives xy for a quotient ; cast out either of 
the common factors a or m from both dividend and divisor, and 
we shall still have xy for a quotient. This, in many instances, 
will greatly facilitate the operation. Thus, in the 4th example, 
the factor 6 may be cast out, as it is contained in all the terms ; 
and in the 6th example the factor x^ may be cast out ; the quo- 
tients will of course be the same. 

■y. Divide c^-\-4ax-{-43^-\-y^ by a+2a'. 

Ans. a4-2a'H f-—-, 

a-\-2x 

8. Divide 6a*4-9«'— -15a by 3a'''— 3a. 

(Observe Art. 21.) Ans, 2a2-f 2a-f-5. 

O. Divide x«— / by x^-\-2x?y-\-2xy^-\-y^. 

Arts, a^'-^2xh/-\'2xif'--f. 



32 ELEMENTS OF ALGEBRA. 

10. Divide ax'—{a^+b)3(^'\-b^ by ax-^. 

Jins. 'J? — ax — h, 

11. Divide 1 by 1— «. Ans. l+a+fl24-a^ &c., &c. 

12. Divide ar3-}-|i^V?^+l by -+i. 

4 4 2 

^m. 22?^— ^-{-2. 

N. B. We may multiply botli dividend and divisor by the 
same number as well as divide them. 

IS. Divide 1 — S^+lOiy^ — lOy^-j-Sy— y^ ^^^y i — 2y-\-y^. 

Ans. 1— 33/4-32/2— ^^ 

14. Divide «^4-4M by a^-^'iab-^^UK 

Ans. a^-\-2ab-\-2b\ 

15. Divide a?" — x'^-{-x^—xr-\-2x — 1 by x^-\-x — 1. 

Ans. x^—x'^+x^'—x-i-l. 

16. Divide a^ — x^ by a — a:. 

Ans. (t^-\-a^x-]-a^x^-\-ax^-{-x* 

17. Divide b^-{-y^ by b-]-y, 

Ans. ¥'-^^y-\-bY—by^'{-y^ 

18. Divide «'4-5a^2/-}-5«3/^+»/'' by a-\-y. 

Ans. tt"+4a?/-}-2/*. 

If more examples are desired for practice, the examples in 
multiplication may be tak'en. The product or answer may be 
taken for a dividend, and either one of the factors for a divisor ; 
the other will be the quotient. 

Also, the examples in division may be changed to examples in 
multiplication ; and these changes may serve to impress on the 
mind of the pupil the close connection between these two opera- 
tions. 

(Art. 22.) In the following examples the dividends and divi- 
sors are given in the form of fractions, and the quotients are the 
terms after the sign of equaUty. Let the pupil actually divide, 
and observe the quotients attentively. 

1. =x-\-a. 



DIVISION. 33 

X — a 
X — a ' . 

X — a I'll 

Hence we may conclude that in general a*"' — a"* is divisible by 
X — a, m being any entire positive number. 

'^"' ^"» 

That is, ^- =.T'"-'H-aa7"*~^+ - - - u:"^-x-\-a"'~\ 

X — a 

The quotient commencing with a power of a', one less than 
m, and ending with a power of a, one less than m. 

These divisions show, that the difference of two equal powers 
of different quantities is always divisible by the difference of 
their roots. 

(Art. 23.) By trial, that is, actual division, we shall find that 

x^ — a^ 

— 1 — =x — «. 
x-\-a 

x'—a" 3 „ , - , 
— i — = XT — ax^ + a^x — (t. 
x-\-a 

6 

= x^ — ax'^-]ra^'x^ — €^x^-\-a'^x — a^. 



x-\-ci 

Slc. &c. &c. &c. 

From which we learn that the difference of any two equal 
powers of different quantities, is also divisible by the sum of their 
roots when the exponent of the power is an even number. 

(Art. 24.) By actual division we find that 



x-^a 
a^-{-a^ 



=x^ — ax-{-a^. 

=x'^ — ax^-{-a^xi^ — a^x-{-a'^. 



x-\rct 

And in general, we may conclude that the sum of any two 
equal powers of different quantities, is divisible by the sum of 
their roots when the exponent of the power is an odd number. 



34 ELEMENTS OF ALGEBRA. 

In Alt. 22, if we make a=l the formulas become 



a;— 1 
a:^— 1 



X"* — 1 



x'+x-hl. 
=x"'-'-\'X"'-^-\-x"'-^, &c. 



X —1 

If we make a?=l, what will the formulas become? 

Make the same substitutions in articles 23 and 24, and exa- 
mine tlie results. 

By inspecting articles 22, 23, and 24, we find that 

{x-{-a){x — a)=3^ — a\ {x!^-jrax-{-a^){x — a)=x^ — a^ &c., 

for the product of the divisor and quotient must always produce the 
dividend. These principles point out an expedient of condensing 
a multitude of terms by multiplying them by the roots of the terms 
involved. Thus, x'^d=iax^-]rci^x'^zt:c^x-\-a'^, can be condensed to 
two terms by multiplying them by xdza, the root of the first and 
last term, with the minus sign where the signs are plus in the 
multiplicand, and with the plus sign where the signs are alter- 
nately plus and minus. See examples in Art. 22 and 24. 



ALGEBRAIC FRACTIONS. 

(Art. 25.) We shall be very brief on the subject of algebraic 
fractions, because the names and rules of operations are the same 
as numeral fractions in common arithmetic ; and for illustration, 
shall, in some cases, place them side by side. 

Case i. To reduce a mixed quantify to an improper fraC" 
tiorif multiply the integer by the denominator of the fraction^ 
and to the product add the numerator^ or connect it with its 
proper sign, -\- or — ; then the denominator being set under 
this sum, will give the improper Jraction required. 



ALGEBRAIC FRACTIONS. 85 

EXAMPLES. 
X 

1. Reduce 2| and ci-\r^ to improper fractions. 

An,. V and ^±f . 
o 

These two operations, and the principle that governs them, 
are exactly alike. 

9. Reduce 5 J and «+t- to improper fractions. 

An,. ya„d^\ 

8. Reduce 4 — j and a to improper fractions. 

^ , ax — b 

^ns. V and . 

^ X 

4. Reduce 5 — and 2b^ to improper fractions. 

25—4+1 22 , 2bc—'dx-\-a 

Ans. • = — and - 

5 5 c 

_ ^ , ^ , ab-\-x 

o, Keduce 5a-}- — y — to an improper fraction. 

6. Reduce 12-1 r — to an improper fraction. 

7. Reduce 4-|-2a:H — to an improper fraction. 

8. Reduce 5x — to an improper fraction. 

9. Reduce 3a — 9 Xq~" *^ ^" improper fraction. 

Ans. 



«+3* 

Case 2. 7%e converse of Case 1. T'o reduce improper 
fractions to mixed quantities, divide the numerator by the de- 
nominator, as far as possible, and set the remainder, {if any,) 



36 ELEMENTS OF ALGEBRA. 

over the denominator for the fractional part ; the two joined 
together with their proper sign, will be the mixed quantity 
sought, 

EXAMPLES. 

1. Reduce y and — ^ — to mixed quantities. 

X 

Jins. 5| and a-\-— . 
b 

a^-\-bx 

2. Reduce y and to mixed quantities. 

bx 
Ans. 2f and a-j- — . 

3. Reduce — ^ to amixed quantity. 

Jim, 5a+^-^. 

y 

2a2-.-262 

4. Reduce — to a whole or mixed quantity. 

Ans. 2a-f-26. 

, „ , 2a;3__2i/» , ^ 

5. Reduce — to a whole number. 

a?— 1^ 

Ans. 2(a?2-}-a7y+y2) ^y (Art. 22.) 

6. Reduce to a mixed quantity. 

^ _. , 10«2— 4rt+6 . ^ 

7. Jteduce to a mixed quantity. 

Oft 

« „ , 13:r-|-5 . , 

8. Reduce to a mixed quantity. 

-V « J 3a^— -12«a?+v— 9a: . , 

». Reduce to a mixed quantity. 

(Art. 26.) It is very desirable to obtain algebraic quantities in 
tlieir most condensed form. Therefore, it is often necessary to 
reduce fractions to their lowest terms ; and this can be done as 
in arithmetic, by dividing both numerator and denominator by 

v ■ 



ALGEBRAIC FRACTIONS. gf 

their obvious cominon factors, or for their final reduction, by 
their greatest common measure. If the terms have no common 
measure, the fraction is already to its lowest terms. 

The principle on which these reductions rest is that of divi- 
sion, explained in (Art. 21). 

Case 3. To find the greatest common measure of the term^ 
of afraction, divide the greater term by the less, and the last 
divisor by the remainder, and so on till nothing remains ; then 
the divisor last used will be the common measure required. 

But note, that it is proper to arrange the quantities according 
to the powers of some letter, as is shown in division. 

N. B. During the operation we may cast out, or throw in a 

factor to either one of the terms without affecting the common 

measure, as such a factor would make no part of the common 

measure, and the value of quantities is not under consideration. 

ab-\-b^ 
Thus, the fraction ^ ^^ has a-]rb for its greatest common 

measure; and this quantity is not affected by casting out the 

factor b from the numerator, and seeking the common measure 

a+b 
of the fraction — — —. 
c^ — b^ 

(Art. 27.) To demonstrate the truth of the rule for finding the 
greatest common measure, let us suppose D to represent a divi- 
dend, and d a divisor, q the first quotient and r the first re 
mainder. 

In short, let us represent successive divisions as follows : 

d)D{q 
dq 

r)d(q' 
rq' 



r')r{q" 
r'q" 



Now, in division, the dividend is always equal to the product 
of the divisor and quotient, plus the remainder, if any. 
D 



38 ELEiMENTS OF ALGEBRA. 



Therefore, 


r=-r'q" 


and 


d=rq'-{-r' 


and 


n^dq-\-r. 



As r=r'q", the last divisor r' is a factor in r (there being no 
remainder) ; that is, r' measures r. 

Now as r' measures r, it measures any number of times r, 
or rq'-\-r', ox d ; therefore r' measures d. 

Again, as r' measures d and r, it measures any number of 
times d -\-r ; that is, it measures J^'-f-r or D. 

Hence r', the last divisor, is a common measure to both D 

aid d, or of the terms fraction -^. 

a 

We have now to show that r' is not only the common mea- 
sure of D and d, but the greatest common measure. 

In division, if we subtract the product of the divisor and quo- 
tient from the dividend, we shall have the remainder. 

That is, D — dq=r, and (/ — rq'=^r'. 

Now every common measure of D and d is also a measure of 
2> — dq=r ; and every common measure of r and of d and r, is 
also a measure of d — rq'=r' ; that is, a measure of r'. But the 
greatest measure of r' is itself. This, then, is the greatest com- 
mon measure of D and d. 

EXAMPLES, 
1. Find the greatest common measure of the two terms of 

the fraction -r-i — z and with it, reduce the fraction to its lowest 

terms. 

CONSIDERATION AND OPERATION. 

The denominator has a' as a factor to all its terms, which is 
not a factor in the numerator ; hence this can form no part of 
the common measure, or the common measure will still be there 
if this factor is taken away. 

We then seek the common measure of a'* — 1 and a'-(-l» 



ALGEBRAIC FRACTIONS. ^ 



—02—1 



Hence a^-{-l is the common measure, which, used as a divisor 
to both numerator and denominator, reduces the fraction to 

2. Find the greatest common measure, and reduce the frao 



tion 






Divide this rem. by y^ xy^ — y^ 

aj2 — ^^3 

^n*. Common measure x — y. 

Fraction reduced — , , , • 
x^-\-y?y 

3. Find the greatest common measure and reduce the in.Z' 
a^-\''Zd^x^y-\-x'^y^ 
bc^xy-^-^ax^y'^ 

a^-\-x^y)a^-{-2a^x^-{-x'^y\a^-\-x'^y 
a*'\- €^s?y 

a'^xhj-^-xSf- 

Ans. Greatest common measure c^-\-xHf. Reduced frac- 
tion --I— ^ 



40 ELEMENTS OF ALGEBRA. 

Find the greatest common measure of a^-{-3a^b-{-3ab^'\~b' 
and a^c-{-2abc-]rb^c. 

Reject the common factor c in one of the quantities, 

a'-\-2ab-\-b^y+3a'b-\-SaI^-\-b%a-{-b 
a^-{-2a'b-{- ab^ 



a2Z,-{-2a62+63 
a^b-\-2ab^-^b'' 



4. Find the greatest common divisor, and reduce the fraction 
3a«~-2a— 1 



10 its lowest terms. 



4a«--.2a2— 3a-fl 

Here we find that neither term is divisible by the other ; but 
if these quantities have a common divisor, such divisor will still 
exist if we multiply one of the terms by any number whatever, 
to render division possible. 



Therefore take 4a^- 
Multiply it by 3 


-2a2— 


- 3a-l- 1 






3a2— 2a— l)12ft3- 
12a3- 


-6a2- 


- 9«+ 3(4a 

- 4a 




Multiply by 


2a^- 
3 


- 5a-i- 3 




3a2— 2a--l)6«2- 


-15fl+ 9(2 
- 4a— 2 




Divide by —1 1 




-lla+11 






«_l)3a?— 2a- 
3a2— 3a 


-l(3a+l 








a— 
a— 


-1 
-1 



Ans, Greatest common measure a— 1. Reduced fraction 
3a4-l 
4a*+2a— r 



ALGEBRAIC FRACTIONS. 41 



5. Find the greatest common measure and reduce the frac- 



tion 75 — ,.3^ ax^A-x^ *° ^^^ lowest terms 



Ans. Common divisor a^ — s^. Reduced fraction 



a — X 



6. Find the greatest common measure, and reduce the frac- 



tion — to Its lowest terms 



Arts. Common divisor a^ — y^. Red. frac. — 



a'+«y+y 



7. Reduce „ — to its lowest terms. 



a^—^ax-^-a? 
Ans» 



a-\-x 



8. Keduce — — — ; ; to its lowest terms. 

6a24-iiaa?4-3a?* 

3a— a? 
3a+x* 

a^boi^-\-2abx^ bx 

d. Reduce — - — ; — — — to its lowest terms. Ans, — . 

aV+2A* a 

.^ ^ , 6a3^+9ax2--12aa?— Sa^ . , 

10. Keduce to its lowest terms. 

Qax — 8a 

2a2+3a? 

Ans, — 

2 

11. What is the value of ? Ans, a^ — b^. 

a^-\-b^ 

12. Find the greatest common divisor of 12a* — 24a^b-{-l2a^b\ 
and 8(^b^—2iaW+24ab*'^^b\ Ans. 4(a^^2ab-{-b\J 

(Art. 28.) We may often reduce a fraction by separating both 
numerator and denominator into obvious factors, without the 
formality of finding the greatest common divisor. The follow- 
ing are some examples of the kind : 
4 



42 ELEMENTS OP ALGEBRA. 

1. Reduce — rr-rz to its lowest terms. 

a^-\-2ab-\-b'^ 

a^-^aH^ _ aja^^W) _ a{a—b){a^b) _a^--^h 
a2+2a6-f 62 {a-{.b)[a-\-b) («4-6)(«4-6) '" «+6 * 

^ 52^3 ^ 

2. Reduce — 7 — rr to its lowest terms. Ans. , . , . 

x^ — b^ a^4-6 

x^ 1 a? 1 

3. Reduce — - — to its lowest terms. Ans. ■ . 

xy^-y y 

cx~\~cx^ c~^cx 

4. Reduce ; — — to its lowest terms. Ans. . 

acx-\-abx ac-\-ab 

, „ / 2ir*— 16.r— 6 . , 

5. Reduce — ;; ^ to its lowest terms. Ans. #. 

3^3 — 24x — 9 ^ 

(Art. 29.) To find the least cojnmon multiple of two or 
more quantities. 

The least common multiple of several quantities is the least 
quantity in which each of them is contained without a remainder. 

Thus, the least common multiple of the prime factors, a, b, c, 
X, is obviously their product abcx. Now observe that the same 
product is the least common multiple also, when either one of 
these letters appears in more than one of the terms. Take a, 
for example, and let it appear with b, c, or x, or with all n^ 
them, as a, ab, c, ax, or «, b, ac, ax, the product abcx is still 
divisible by each quantity. Therefore, when the same factor 
appears in any number of the terms, it is only necessary that it 
should appear once in the product; that is, once in the least 
common multiple. If it should be used more than once, the 
product so formed would not be the least common multiple. 

From this examination, the following rule for finding the least 
common multiple will be obvious : 

Rule. Write the given quantities, one after another, and 
draw a line beneath them. Then divide by any prime factor 
that will divide two or more of them without a remainder^ 
bringing down the quotients and the quantities not divisibht 
to a line below. Divide this second line as the first, forming 



ALGEBRAIC FRACTIONS. 43 

a third, ^c, until nothing but prime quantities are left. Then 
multiply all the divisors and the remainders that are not divis- 
ible, and their product will be the least common multiple. 
N B. This rule is also in common arithmetic. 

EXAMPLES. 
1. Required the least common multiple of Sac, 4a^, I2ab, 
16ac, and ex. 

2a)Sac 4a^ I2ab Sac ex 



2c)4c 


2a 


66 


4c 


ex 


2)2 


a 


36 


2 


X 



I a Sb i X 

Therefore 2aX2cX2XaX3bXx=2ia'cbx. 

Here the divisor 2c will not divide 2a, but the coefficient of 
c will divide the coefficient of a, and we let them divide, for it 
is the same as first dividing by 2, and afterwards by c. From 
the same consideration we permit 2c to divide ex, or let the let- 
ter c in the divisor strike out c before x. 

By the rule we should divide by 2 and by c separately ; but 
this is a practical abbreviation of the rule. 

2. Required the least common multiple of 27a, Ibb, 9ab, and 
3a\ Ans. U^a^b. 

3. Find the least common multiple of (a^ — a^), 4(a — a:), 
and {a-]rx). Ans. 4(a^ — a^). 

4. Find the least common multiple of aoi^, bx, acx, and 
(^ — ST. Ans. {a^ — x^)acba^, 

5. Find the least common multiple of a-\-b, a — b, a^-\-ab-\-b^, 
and a^ — ab-{-b'^. Ans. a® — bK 

The least common multiple is useful many times in reducing 
fractions to their least common denominator. 

Case 4. To reduce fractions to a common denominator. 

(Art. 30.) The rule for this operation, and the principle on 
which it is founded, is just the same as in common arithmetic, 
merely the multiplication of numerator and denominator by the 



44 ELEMENTS OF ALGEBRA. 

same quantity. The object of reducing fractions to a common 
denominator is to add them, or to take their difference, as diffe- 
rent denominations cannot be put into one sum. 

Rule. Multiply each numerator by all the denominators, 
except its own, for a new numerator, and all the denominators 
for a common denominator. 

Or, find the least common midtiple of the given denomina- 
tors for a common denominator; then multiply each denomi- 
nator by such a quantity as will give the common denomina- 
tor, and multiply each numerator by the same quantity by 
which its denominator was multiplied, 

EXAMPLES. 

, „ ^ 2a ^ 36 

1. Reduce — and — to a common denommator. 

X 2c 

4ac , 3bx 

Ans, - — and . 

2ca? 2ca? 

2a 3a-l-26 

2. Reduce -— and to a common denominator. 

b 2c 

4ac , Zab-^2ab^ 

Ans. — — and • 

26c 26c 

„ _ - 5a 36 , ^ , 

3. Reduce — and — , and 4a to a common denommator. 

3a? 2c 

lOac , 9bx , 24cdx 

Mns. — — and and . 

6cx Qcx Qcx 

a x-\-l y 

4. Reduce -r, , —7 — , to fractions having a common de 

6 c x-\-a 

nominator. ^ , , ,, , ,v, , s 

Mns. acx-f-crc (bx-]rb)[x-\-a) bey 

bcx-i-abc bcx-[-abc ' bcx^fabc 

(Art. 31.) Case 5. Additionor finding the sum of fractions. 

Rule. Reduce the fractions to a common denominator; 
and the sum of the numerators, written over the common dejjo 
minator, will be the sum of the fractions. 



ALGEBRAIC FRACTIONS. 45 



EXAMPLES. 



^ ^ , 3a? 2a: , a? 

1. Add — , — , and - together. 

6Sx-{-S0x-\'25x 128a: 
•^"'- 105 =-[05- 

2. Add « and ^. Ans. "-2±p±. 

c be 



3. Add -, — and 



2' 3 a-\-x » 

Sa-\-3x+2a!'-\-2a^x-\-eaF+6x' 



Ans, 



Q{a-\-x) 



4. Add r and — — ;• A71S. — r — —-. 

«--6 a-{-b a^-^b"- 

K KAA o , «+3 , 2a— 5 14a— 13 

5. Add 2a H -— and 4a-\ — . Ans. 6aH -—— . 

5 4 20 

6. Add a r- and 0-] . Ans. a-JrbA ; . 

be be 



X — 2 2a? — 3 

7. Add 5a? H — — - and 4a> 



5a? 

5a:2— 16a;+9 



15a: 



Ans. 9x-\- 



3x b b—-x 

8. What is the sum of 26-f — , 7 and — -— ? 

5 b — X b 

Sb^x-'Sbx'-{-5x' 

y — 2 2v — 3 

9. What is the sum of by-\-- and 4v ? 

^3 5y 

6i/2— 16vH-9 



40 ELEMENTS OF ALGEBRA. 

10. What is the sum of 5a, ~a and — - — ? 

3a^ 4a 

- ^ , 8a?H-3aa?+6a' 
^ns. 5«+ —^ 

(Art. 32.) Case 6. Subtraction or finding difference* 
Rule. Reduce the fractions to a common denominator, and 
subtract the numerator of that fraction which is to he sub- 
tracted from the numerator of the other, placing the difference 
over the common denominator. 

EXAMPLES. 

, ^ Ix , 2a?—- 1* ^ 21a?— •42+2 \'7x-\-2 

1. From — take—-—. Ans, = — - — . 

2 3 DO 

1 1 X-\~'ll X'-^'IJ 

S. From take — ; — . Eq. fractions -v— -,, -r-A. 

X — y x-\-y x^ — 2/* x^ — if 

2v 
Difference or Jins, -5--^. 
a?2-~y* 

3. From % take ^. Diff. §~. 

3 7 21 

^ ^ 3a? , 2ar ^ 13a? 

4. From -— take —. Ans. — . 

, ^ 2a— b ^ 3a—4b 

5. From -- — subtract — -r — . 

4c So 

6a6— 36*— 12ac-i-166c 



Ans. 



126c 



Ua— 10 , „ , 3a— 5. 

6. From 3cH subtract 2a-| — . 

lo 7 

y. From a?4- JT^- subtract y~^ . Ans. a:— -3^,. 

x^+xy x^—xy x'^* 

a — h , 26 — 4a ^ 5ac? — bbd — 46c-f 8ac 
9. From -^ take -jj-. Mm. ^^ 



ALGEBRAIC FRACTIONS. 47 

9, From 3x+-t take x . Jins. 2x-\ 7 

be be 

a-\-b a — b 

10. Find the diflerence between — -r and — ^,-7-. 

a — b a-f-b 

4ab 
Mns. 



11. From ^^±^ take (^=^ ^ns. 4. 

xy xy 

Case 7. Multiplication of fractions. 

(Art. 33.) The multiplication of algebraic fractions is just the 
same in principle and in fact, as in numeral fractions, hence the 
rule must be the same. 

It is perfectly obvious, that f multiplied by 2 must be 4, and 
multiplied by 3 must be | ; and the result would be equally ob- 
vious with any other simple fraction ; hence, to multiply a frac- 
tion by a whole number, we must multiply its numerator. 

It is manifest that doubling a denominator without changing its 
numerator halves a fraction, thus h ; double the 2, and we have 
-4^ the half of the first fraction. 

Also |, double the 5 gives y'^, the half of |. In the same 
manner, to divide a fraction by 3 we would multiply its denomi- 
nator by 3, &LC. In general^ to divide a fraction by any num- 
ber, we must multiply the denominator by that number. 

Now let us take the literal fraction t» and multiply it by c, the 
product must be ~j- . 

a c 

Again, let it be required to multiply - by -. Here the mul- 
tiplication is the same as before, except the multiplier c is divided 
by d ; therefore if we multiply by c we must divide by d. But 

the product of j- by c is -j-; this must be divided by d, and we 
ac 



shall have -r-; for the true product of r by r~. 
bd ^ b d 






48 ELEMENTS OF ALGEBRA. 

From the preceding investigation we draw the following rule 
to multiply fractions : 

Rule. Multiply the numerators together for a new nume- 
rator^ and the denominators together for a new denominator, 

N. B. When equal factors, whether numeral or liteial, 
appear in numerators and denominators, they may be canceled, 
or left out, which will save subsequent reductions. 

EXAMPLES. 

1. Multiply ~ by- and ^. Ans. — ^. 

^ -' b ■^ X c ex 

In this example, b in the denominator of one fraction cancels 
b in the numerator of another. 

2. Multiply (^^ by -^ Ans. " 

3. Multiply -^ by -. An.. ~^. 

, , . , a^ — x^ , 2« ^ (a — x)a 

4. Multiply — - — by --— . Ans. ^. 

Zy a~\~x y 

5. Multiply* —, —— and . Ans, a. 

X x-\-y X — y 

x^\ , a;— 1 ^ ^ 3(a?2— 1) 

6. Multiply 3«, and ——7 together. Ans. -7 — r-rf-, 

^ ^ 2a a+b ^ 2{a+b) 

N. B. Reduce mixed quantities to improper fractions. 

a2__^ a'— 6^ 

7. What is the contimied product of — r-T~» 7~~7, ^^^ 

tt+o ax-f-ar 



a — X X 

4y^ 15v — 30 

8. Multiply —-^-—r by — ^- . Ans, 6y. 

^ -^ 5y — 10 '' 2y ^ 

* Se^^oe into factors when separation is obvious. 



ALGEBRAIC FRACTIONS. 41* 

9. Multiply -^^P^ by ^^-^, ^ns. — --^-. 

10. Multiply , by .—-5. Jlns. 3(«4-a:). 

fjp 'V^ ft —I— f) 

11. Required the continued product of -= r;., — 5- and 

. Ans, («+a?^ 

a — X 

13. Multiply a-l-T by a— |. .5ns. ! -^ — ^ — -^. 



13. Multiply -^— by -^-^X 



Ac ^ b-\-c X — b' 

Jn,. (^^±*!). 
b-\-c 

-- ,T , • , 3^ — 5x , 7a ^ 3aa? — 5a 

14. Multiply — -— by ^ „ „ . ./^n*. -p-3 — -• 

^ -^ 14 ^ 2x? — 3a; 4a;^ — 6 

• « *-r , . , 4a»— 166' ^ 56 ^56 

15. Multiply -^-^ by Q^.^g^^^^g.^^, •^^^•2^+46- 

Case 8 Division of Fractions, 

(Art. 34.) To acquire a clear understanding of division in 
fractions, let us return to division in whole numbers. 

The first principle to which we wish to call the attention of 
the reader, is, that if we multiply or divide both dividend and 
divisor of any sum in division, by any number whatever, we do 
not affect or change the quotient. (Art. 21.) 

Thus, 2)6(3 4)12(3 8)24(3 &c. 

The second principle to which we would call observation is, 
til at if we multiply any fraction by its denominator, we have the 
numerator for a product. 

Thus, I multiplied by 3 gives 1, the numerator, and | by 6 

gives 2, arid y multiplied by b gives a, <fec. 



50 ELEMENTS OF ALGEBRA. 

(JL C 

Now let it be required to divide j- by -. 

The quotient will be the same if we multiply both dividend 
and divisor by the same quantity. Let us multiply both terms 

by d, the denominator of the divisor, and we have — to be di- 
vided by the whole number c. But to divide a fraction by a 
whole number we must multiply the denominator by that num- 
ber. (Art. 33.) Hence -j- is th« true quotient required. 

We can mechanically arrive at the same result by inverting 
the terms of the divisor, and then multiplying the upper terms 
together for a numerator, and the lower terms for a denominator; 
therefore to divide one fraction by another we have the following 

Rule. Invert the tenns of the divisor, and proceed as in 
multiplication. 

EXAMPLES. 

1. Divide by — — r- *^n8. - — =-- ^ . 

2. Divide — by -. 

Operation: divisor inverted -X — = — j- An9. 

a ah 

^ ^. ., 15«6 , 10«c* 

»• Divide by -5 -.. 

a — X (t — ar 

^ . 15«6 ia-\-x){a — x) . 36(«-l-a?) 

Operation, X^^-^-^^^ '■. Ans. — ~-—^. 

* a — X lOffc 2c 

5. Divide — ^"r:^— by ^. Am. :^-^h\ 

x^ — 1hx-\-h^ •' X — h 

Operation, \ -r^. -r^Y.—y^^x'Arb'. 

' {x — b) (x — b) x-\-b 

* Divide into factors, in all such cases, and cancel. 



ALGEBRAIC FRACTIONS. 51 

2ax-{'X^ , X 2a-\-x 

6. Divide —5 r- by . Ans. 



ct> — x^ ^ a—x' ' a^-\-ax-^x^ 

^ ^. .J 14a?— 3 , lOrr— 4 ^ 70a:— 15 

7. Divide — by — ^rz — . Ans. — , 

5 • 25 10a: --4 

^ _. . - 9a;2— 3a: . x^ . 9a:— 3 

8. Divide bv -ir. ^ns. 



5*5' a: 

^ ^. .- 6a:— 7 , a:— 1 ^ 18a:— 21 

9. Divide — j-— by . .--^n*. 



a:+l -^ 3 * • ar^— 1 

«A T^- J ^+^^ u 2aa:+2aa:2 ^ 7 

10. Divide —T-x- by . ^ns, -— 



6a'' 



11. Divide -5 — , — ^ by — ■ -—. Jns. a^+x" 

or — 2ax-\-ar a — x 

12. Divide ?^t::?^ by <. ^ns. '-Ml^ 

5 5 y 

13. Divide 7-7— by r-r— • •^'i*- — 

a-{-b '' a-{-b in 

14. Divide 12 bv ^ ^ ^ — a. .-^n*. ^-r r~, 

a: a'+aa:+x* 

15. Divide by -. Jlns. b-\ 

XX a 

%A T\' i ^ — * u ^(^^ a M^ — *) 



Qc^x ^ id' * ISc^a:^ 



17. 



X ci 

Divide a by the product of — ; — into . 

x+y x^y 



Jins, 



^—y' 



X 



18. Divide -^, — .—r- by the product of — - — - into — r-r. 
2{a-\-b) ^ ^ 2a a-\-b 

Ans. 3a. 



53 ELEMENTS OF ALGEBRA. 

SECTION II. 

CHAPTER L 

Preparatory to the solution of problems, and to extended in- 
vestigations of scientific truth, we commenced by explaining the 
reason and the manner of adding, subtracting, multiplying, and 
dividing algebraic quantities, both whole and fractional, that the 
raind of the pupil need not be called away to the art of perform- 
ing these operations, when all his attention may be required on 
the nature and philosophy of the problem itself. 

For this reason we did not commence with problems. 

Analytical investigations are mostly carried on by means 

OF EQUATIONS. 

(Art. 35.) An equation is an algebraical expression, meaning 
that certain quantities are equal to certain other quantities. Thus, 
3-1-4=7; a-\-b=c; a?-l-4=10, are equations, and express that 
3 added to 4 is equal to 7, and in the second equation that a 
added to b is equal to c, &c. The signs are only abbreviations 
for words. 

The quantities on each side of the sign of equality are called 
members. Those on the left of the sign form ihe first member, 
those on the right the second. 

In the solution of problems every equation is supposed to con- 
tain at least one unknown quantity y and the solution of an equa- 
tion is the art of changing and operating on the terms by means 
of addition, subtraction, multiplication, or division, or by all these 
combined, so that the unknown term may stand alone as one 
member of the equation, equal to known terms in the other 
member, by which it then becomes known. 

Equations are of the first, second, third, or fourth dcgi*ee, 
according as the unknown quantity which they contain is of the 
first, second, third, or fourth power. 

ax-\-h=^ax is an equation of the first degree or simple 
equation. 



EQUATIONS. 53 

ax^-\-bx—-3ab is an equation of the second degree or quadratic 
equation. 

ax^-{-bx^-]-cx=2a'^b is an equation of the third degree. 

ax*-{-bx!^-{-cx^-\-dx=2ab^ is an equation of the fourth degree. 

We shall at present confine ourselves to simple equations. 

(Art. 36.) The unknown quantity of an equation may be 
united to known quantities, in four different ways : by addition, 
by subtraction, by multiplication, and by division, and further 
by various combinations of these four ways as shown by the 
following equations, both numeral and literal : 







NUMERAL. 


LITERAL. 


1st. 


By addition. 


x-\-Q-^lO 


x-\-a=b 


2d. 


By subtraction, 


ar— 8 = 12 


X — c=d 


3d. 


By multiplication, 


20ar=80 


ax=e 


4th. 


By division, 


:|=,e 


5=^+«. 



5th. x^-Q — 8-1-4 = 10-1-2 — 3, x-\-a-^b-[-c=d-\-c, &c., 
are equations in which the unknown is connected with known 
quantities by both addition and subtraction. 

X X 

2a:4--=21, ax-\--r='C, are equations in which the unknown 

o 

is connected with known quantities by both multiplication and 
division. 

Equations often occur, in solving problems, in which all of 
these operations are combined. 

(Art. 37.) Let us now examine how the w/2A;no2^'n quantity can 
be separated from others, and be made to stand by itself. 
Take the 1st equation, or other similar ones. 

.T-[-6=10 x-\-a=b 

Take equal quantities 6= 6 a=a from botli 

members, and .t=10 — 6 x=b — a the 

remainders must be equal. (Ax. 2.) Now we find the term 
added to x, whatever it may be, appears on the other side with 
a contrary sign, and the unknown term x being equal lo known 
terms is now known. 
e2 



fii ELEMENTS OF ALGEBRA. 

Take the equations x — 8 = 12 x — c=d 

Add equals to both memb. 8= 8 c=c 

Sums are equal a?=12-}- 8 x==d-{-c (Ax. 1.) 

Here again the quantity united to x appears on the opposite 
side with a contrary sign. 

From tliis we may draw the following principle or rule of 
operation : 

jlny term may be transposed from one member of an equa- 
tlon to the other, by changing its sign. 

Now 20a;=80. ax=e. If we divide both members by the 
coefficient of the unknown term, the quotients will be equal. 

(Ax. 4.) Hence a:=|^=4. x=-. 
" a 

That is, the unknown quantity is disengaged from known 

quantities, in this case, by division. 

X X 

Again, take the equations 2=IQ ; -,—g-^^' 

Multiply both members by the divisor of the unknown term, 
and we have .T= 16X4. x=gd-{-ad. Equations which must 
be true by (Ax. 3.), and here it will be observed that x is libe- 
rated by multiplication. 

From these observations we deduce this general principle : 

That to separate the unknown quantity from additional 
terms we must use subtraction ; from subtracted terms we 
must use addition ; jrom multiplied terms we must use divi- 
aion ; from divisors toe must use multiplication. 

In all cases take the opposite operation. 

EXAMPLES. 
1. Given 3a: — 4=7a: — 16 to find the value of a\ Ans. x=^3, 

a. Given 3a?+9 — 1 — 5.t=0 to find the value of x. 

Ans. x—i. 

3. Given 43/+7=?/+21 — S-f?/ to find y. Ans. y—5k 

4. Given 5ax — c=b — 3ax to find the value of x. 

Ans. 0*=-^-. 
8a 



EQUATIONS. 56 

a. Given ax^-\-bx=9x'^-\-cx to find the value of x in terms 

c b 

of a, b, and c. Ans. x= -. 

a — 9 

N. B. In this last example we observe that every term of the 
equation contains at least one factor of x ; we therefore divide 
every term by a?, to suppress this factor. 

(Art. 38.) In many problems, the unknown quantity xs 
often combined with known quantities, not merely in a simple 
manner, but under various fractional and compound forms. — 
Hence, rules can only embody general principles, and skill and 
tact must be acquired by close attention and practical application : 
but from the foregoing principles we draw the following 

General Rule. Connect and unite as much as possible all 
the terms of a similar kind on both sides of the equation. Then, 
to clear of fractions, midtiply both sides by the denominators^ 
one after another, in succession. Or, rnultiply by their con- 
tinued product, or by their least common multiple, (when such 
a number is obvious,) and the equation will be free of fractions. 

Then, transpose the unknown terms to the first member of 
the equation, and the known terms to the other. Then unite 
the similar terms, and divide by the coefficient of the unknoum 
term,, and the equation is solved. 

EXAMPLES. 

1. Given x-\rhx-{-^ — 7=6 — 1, to find the value of x. 
Uniting the known terms, after transposition, agreeably to the 
rule of addition, we find x-{-hx=^^. Multiply every term by 
2, and we have ^x-\-x=\S. Therefore x=Q. 

2. Given 2x-\-^x-{-lx — 3a=4Z>+3a, to find x. 

N. B. We may clear of fractions, in the first place, before we 
condense and unite terms, if more convenient, and among literal 
quantities this is generally preferable. 

In the present case let us multiply every term of the equation 
by 12, the product of 3X4, and we shall have 

24:s-|-9a^-|-4a?— 36a=486-f36a. 
Transpose and unite, and 37a? = 485 -|- 72a. 



56 ELEMENTS OF ALGEBRA. 
Divide by 37, and x= . 

3. Given 2a7-f-|^-{-4^=39, to find the value of x. 

Here are no scattering terms to collect, and clearing of frac- 
tions is the first operation. 

By an examination of the denominators, 12 is obviously their 
least common multiple, therefore multiply by 12. Say 12 
halves are 6 whole ones, 12 thirds are 4, 12 fourths are 3, &c. 

Hence, ar-l-4x+3.T=39X 12 

Collect the terms, 1 3a?=39 X 12 

Divide by 13, and x= 3 X 12 = 36, Ans 

N. B. In other books we find the numerals actually multiplied 
by 12. Here it is only indicated, which is all that is necessary. 
For when we come to divide by the coefficient of x^ we shall 
find factors that will cancel, unless that coefficient is prime to all 
the other numbers used, which, in practice, is very rarely the 
case. 

4. Given lx-\-\x-\-kx=a, to find x. 

This example is essentially the same as the last. It is identi- 
cal if we suppose «=39. 

Solution, 6,r-i-4x+3x=12« 

Or, 13x=12a 

Divide and ^'=-7^ 

Now if a be any multiple of 13, the problem is easy and 
brief in numerals. 

, ^. «, , 3.r — 11 5.T — 5 , 97 — Ix ,. . , 

5. Given 21 -{ r-- — = — - — -{ — to hrfd the value of .r. 

16 8 2 

Here 16 is obviously tlie least common multiple of the deno- 
minators, and the rule would require us to multiply by it, and 
f^uch an operation would be correct ; but in this c^se it is more 
easy to multiply by the least denominator 2, and then condense 
like terms. Thus, 



EQUATIONS. 57 

Multiply by 2, and we have 

42+— g — =— [-97—737. 

Recollect that we can multiply a fraction by dividing its deno- 
minator. Also observe that we can mentally take away 42 from 
both sides of the equation, and the remainders will be equal. 

(Ax. 2.) 

Then, ^ — = — - — [-55— 7a?, 

Multiply by 8, and 

3x— 11 = 10a:— 10+440— 562'; 
Transposing and uniting terms, we have 
490^=441 ; 
By division, x=9. 

6. Given |.T+2i + ll=fa-+17, to find x. 
If we commence by clearing of fractions, we should make 
comparatively a long and tedious operation. Let us first reduce 
it by striking out equals from both sides of the equation. We 
can take 1 1 from both sides without any formality of transposing 
or changing signs ; say drop equals from both sides, (Ax. 2.) and 
reduce the fraction %x=lx. 

All this can be done as quick as thought, and we shall have 

Multiply by 4, then i^-^+^^^la^+G ; 

^^+10=a:+24, or }^=x+U ; 
.5 5 

Hence, 7j?=70, or .r=10 ^ns. 

•7. Given ^a— b+lx+S+^a;— 10 = 100— 6— 7 to find the 
value of X. 

Collecting and uniting the numeral quantities, we have 

|a?+|a:+ ja:=94 ; 
Multiply every term by 60, and we have 
20a:+15a?+12a;=94.60 
Collecting terms, 47a7=94.60 

Divide both sides by 47, and x= 2.60=120 Jins. 



58 ELEMENTS OF ALGEBRA. 

(Art. 39.) When equations contain compound fractions and 
simple ones, clear them of the simple fractions first, and unite, 
as far as possible, all the simple terms. 

EXAMPLES. 

». Given — r a >i q ~ — ~^ — ^^ """ ^"^ value of x. 

Multiply all the terms, by the smallest denominator, 3. That 
is, divide all the denominators by 3, and 

^\x 39 

Multiply by 3 again, and Qx-\-l-\- =6a?-f"12. 

ZX'T' 1 

, 2Lr— 39 
Drop 6a?+7, and — - — ,— — =o. 

2x4- 1 

(J!lear of fractions, 21x — 39 = 10x4-5. 

Drop lOx and add 39, and we have llx=44, or a:=4. 

« ^. 7x4-16 x4-8 ^ , . ^ , .V, 1 f 

9. Given •■=- to find the value of x. 

21 4x— U 3 

Observe that — —^ — may be expressed in two parts, thus, 
</ 1 

7x 1 6 7x X 

— -4-— r* Observe also, that -^ — =~. Hence these terms may 

2121 213 

be dropped, the remainders must be equal. Transpose the 

16 x4-8 

minus term, then -—=- . 

21 4x — 11 

Clear of fractions, and 64x— 11 X 16=21x4-21X8. 

Drop 21x and observe tliat 11 X 16 is the same as 22X8. 

Then 43x— 22X8=21 X 8. ' Let «=8, 

Then 43x— 22«=21«. 

Transpose — 22« and 43x=43«. 

Hence x=«. But «=8. Therefore x=8. 

N. B. We operate thus, to call attention to the relation of 
quantities, and to form a habit of quick comparison, which will, 
in many instances, save much labor and introduce the pupil into 
the true spirit of the science. 



EQUATIONS. 99 

10. Given — = — \-~ to find the value of x. 

36 ax — 4 4 

By a slight examination we perceive that — is equal to ^x. 

oo 

Hence these terms may be left out, as they balance each other. 

Also, 3 g — g. 

5 40^—12 

Therefore -=- j-. 

9 5a? — 4 

Clear of fractions, and 25a: — 20 =36x — 108. 
Transpose 25a? — 36a:=20 — 108. 

Unite and change signs, and lla:=88 or a:=8, J^ns. 

,, r.- 20.T , 36 , 5a:+20 4a' 86 ^ . . 

11. Given - — h-^4-;^ r7i=^r--\-77^ to find or. 

25 ' 25 "^907—16 5 ' 25 

By taking equals from both sides, we have 
5x4-20 



9a:— 16 



=2. By reduction a: ==4. 



,« n- 3^ ^—1 a 20a:-l-13 ^ ^ , 

12. Given =6x — ■ to find x. 

4 2 4 

Multiply by 4, to clear of fractions, and 

3a:— 2a'-}-2=24.r — 20a: — 13. Reduced a:=5. 

(Art. 40.) When a minus sign stands before a compound 
quantity, it indicates tliat the whole is to be subtracted ; but we 
subtract by changing signs, (Art. 5). The minus sign before 

X 1 

— — — in the last example, does not indicate that the x is minus, 

but that this term must be subtracted. When the term is multi- 
plied by 4, the numerator becomes 2a: — 2, and subtracting it 
we have — 2a'-l-2. 

Having thus far explained, we give the following unwrought 
equations, for practice : 

'^ 3a: X 

IS. Given -^=t~1"24 to find the value of x. Jlns. 19^. 
2 4 

14. Given \x-\-\x=^\^ to find the value of x, Ans, 24. 



60 ELEMENTS OF ALGEBRA. 

15. Given ~^-}-_=20 =^ to find ar. ^ns. 9. 

16. Given -- — |— 1_=:16 ~ to find x. Ans, 13. 

CO 4 

«^ r^- o ^+3 , ,^ 12:^4-26 , , 
IT. Given 2^ — f-15= -' to find x. 

5 5 

Sns. x=12. 

18. Given x — =— to find x, Jins. a?=13. 

6 4 

19. Given hx-{-ix-]rlx-\-\x^ll to find x. Ans. x=60. 

20. Given lx-\-ix-\-\x=\^Q to find ar. ./^m. a^=120 

21. Given ^a:+^:r4-y^2a.'=90 to find a?. J3ns. ic=120. 

22. Given 52/+53/-|-t2/=82 to find y. Ans, y=84. 

23. Given 5xi-\x-f-ix^3i to find r. Ans. x=6. 

24. Given nx-\-^-j-r+-^+h=^^^ to find a?. 

3 4 6 24 

Ans. x=24. 

25. Given V+l+^+^+^^l^G to find y. Ans. y=5Q. 

•? 2 4 7 14 "^ 

26. Given ^^^^^^^ 29=^^^—30 to find x. Ans. x=2. 

a? +2 X — 2 

Tliere is a peculiar circumstance attending this 26th example, 
and the 4th example of Art. 42, which will cause us to refer to 
them in a subsequent part of this work. 

N. B. In solving equations 19, 20, 21, 22, and 23, use no 
larger numbers than those given, indicating and not performing 
numeral multiplications. 

(Art. 41.) Every proportion may be converted into an equa- 
tion. Proportion is nothing more than an assumption that the 
same relation or the same ratio exists between two quantities, 
as exists betw^een two other quantities. 

That is, ./? is to j9 as C is to D. There is some relation be- 
tween A and B. Let r express that relation, then B-=^rA. But 



EQUATIONS. Qi 

the relation between C and D is the same (by hypothesis) as 
between »^^ and B. Hence D=rC. 

Then in place of ^ : ^ : : CiD 
we have S.\rA:'. C:rC 

Multiply the extreme terms, and we have rCA, 

Multiply the mean terms, and we have rAC. 

Obviously the same product, whatever quantities may be re- 
presented by either .^, or r, or C. 

Hence, to convert a proportion into an equation, we have the 
following 

Rule. Place the product of the extremes equal to the pro- 
duct of the means. 

(Art. 42.) The relation between two quantities is not changed 
by multiplying or dividing both of them by the same quantity. 
Thus, a: b ::2a: 2b, or more generally, a:b::na: nb, for the 
product of the extremes is obviously equal to the product of the 
means. 

That is, a is to b as any number of times a is to the same 
number of times b. 

We shall take up proportion again, but Articles 41 and 42 are 
sufficient for our present purpose. 

EXAMPLES. 
1. Given Sx — 1 : 2a;+l ::Sx: x to find x. 
(By Art. 41.) 'Sx'—x^Qaf-^Sx, 
Transpose and unite, and we have 0=3x^4-42:. 
Divide by x, and 3a:-i-4=0 or x=: — 1, ^n». 

3 Sx 
3. Given - : — - : : 6 : 5x — 4 to find x. 

The first two terms have the same relation as i : ^x^ or as 
2 : X, Hence 2 : a; :: 6 : 6x — 4. 

Product of extremes and means, lOa? — 8=6a7 or x=2. 

„ „. (x—l)(x-\-l) x-\-l „ , 

3. Given ^^ ^^ ^ : -^ : : 2 : 1 to find x. 

aa 3a 

»in8. r— 3. 



62 ELEMENTS OF ALGEBRA. 

4. Given — — : {x — 5) : : | : | to find x, 

Ans. x=5. 

5. Given x-\-2 : a:: b : c to find the value of x. 

Ans. x= 2. 

c 

6. Given 2x — 3 : x — 1 : : 2a: : x-\- 1 to find the value of x. 

Ans. a?=3. 

; "y. Given .t+6 : 38— ^' :: 9 : 2 to find x. Ans. a?=30. 

8. Given x-\-4c : x — 11 : : 100 : 40 to find x. Ans. x=2\. 



qUESTIONS PRODUCING SIMPLE EQUATIONS. 

(Art. 43.) We now suppose the pupil can readily reduce a 
simple equation containing but one unknown quantity, and he is, 
therefore, prepared to solve the following questions. The only 
difiiculty he can experience is the want of tact to reason briefly 
and powerfully with algebraic symbols ; but this tact can only 
be acquired by practice and strict attention to the solution of 
questions. We can only give the following general direction : 

Represent the unknown quantity by some symbol or letter^ 
and really consider it as definite and known, and go over the 
same operations as to verify the answer when known. 

EXAMPLES. 

1. What number is that whose third part added to its fourth 
part makes 21 ? Ans. 36. 

The number may be represented by x. 

Then |a;+|ic=21. Therefore a;=36. 

2. Two men having found a bag of money, disputed about 
the division of it. One said that the half, the third, and the 
fourth parts of it made $130, and if the other could tell how 
much money the bag contained, he might have it all. How 
much money did the bag contain ? Ans. $120. 

(See equation 20, Art. 40.) 



EQUATIONS. e» 

3. A man has a lease for 20 years, one-third of the time past 
is equal to one-half of the time to come. How much of the time 
has passed ? 

Let x= the time past. 

Then 20 — x= the time to come. 

X go X 
By the question -= — - — . Therefore a?=12 Ans. 

4. What number is that, from which 6 being subtracted, and 
the remainder multiplied by 11, the product will be 121 ? 

Let x=^ the number. 

Then (^—6)11=121, or x— 6=11 by division. 

Hence x=^\l. 

5. It is required to find two numbers, whose difference is 6, 
such that if | of the less be added to \ of the greater, the sum 
shall be equal to I of the greater diminished by \ of the less. 

Let x= the less. Then x-\-Q= the greater. 

Ti t • 1 , ^+6 x-\-Q , 
By the question ix-\ — -— =-— \x. 

O o 

Drop Ix from both sides and add \x to both sides, and we 

have — 5-- =2, or a?=2, the less number. 
5 

We may clear of fractions in full, and then transpose and unite 
terms, but the operation would be much longer. 

6. After paying | and ~ of my money, I had $66 left ; how 
much had T at first ? Ans. $120. 

"y. After paying away ^ of my money, and k of what remained, 
and losing \ of what was left, I found that I had still $24. How 
much had I at first ? Jins. 60. 

^ -n-iiBi^WWhat number is that from which if 5 be subtracted, | of 
the remainder will be 40 ? Ans. 65. 

-. 4t*\^'' ^ "^^'^ ^^^^ ^ horse and a chaise for $200; h of the price 
of the horse was equal to \ of the price of the chaise. What 
was the price of each ? Ans. Chaise $120. Horse $80. 



64 ELEMENTS OF ALGEBRA. 

10. Divide 48 into two such parts, that if the less be divided 
by 4, and the greater by 6, the sum of the quotients will be 9. 

Arts. 12 and 36. 

11. An estate is to be divided among 4 children, in the fol- 
lowing manner : 

The first is to have $200 more than $ of the whole. 
The second is to have $340 more than \ of the whole. 
The third is to have $300 more than \ of the whole. 
And the fourth is to have $400 more than \ of the whole. 
What is the value of the estate? Jlns. $4800 

12. Find two numbers in the proportion of 3 to 4, whose 
sum shall be to the sum of their squares as 7 to 50. 

Ans. 6 and 8. 

N. B. When proportional numbers are required, it is generally 
most convenient to represent them by one unknown term, with 
coefficients of the given relation. Thus, numbers in proportion 
of 3 to 4, may be expressed by ^x and 4a^, and the proportion 
of a to 6 may be expressed by ax and bx. 

13. The sum of $2000 was bequeathed to two persons, so 
that the share of A should be to that of jB as 7 to 9. What was 
the share of each ? Ans. A's share $875, J5's share $1125. 

14. A certain sum of money was put at simple interest, and 
m 8 months it amounted to $1488, and in 15 months it amounted 
to $1530. AVhat was the sum 1 Ans. $1440. 

Let x=^ the sum. The sum or principal subtracted from the 
amount will give interest: therefore 1488 — x represents the 
mterest for 8 months, and 1530 — x is the interest for 15 months. 

Now whatever be the rate per cdnt. double time will give 
double interest, &c. Hence 8 : 15 :: 1488— x : 1530--a:. 

N. B. To acquire true delicacy in algebraical operations,*it is-< ♦ 
often expedient not to use large numerals, but let them be repre- 
sented by letters. In the present example let a=1488. The;i, ^ 
a-|-42=1530, and the proportion becomes 8: 15:: a — x:a-{- 

42— r. 






EQUATIONS. 65 

Multiply extremes, &c., 8a+8-42 — Sx=l6a — 15ic. "V 

Drop 8a and — Sx, We then have 8*42 =7a — 7x. 
Dividing by 7 and transposmg x=a — 48=1440, .4?ns. 

15. A merchant allows $1000 per annum for the expenses 
of his family, and annually increases that part of his capital 
which is not so expended by a third of it ; at the end of three 
years his original stock was doubled. What had he at first ? 

^ns. $14,800. 
Let x= the original stock, and «=1000. 
To increase any quantity by its 5 part is equivalent to multi- 

^X 'i.Ci 

plying it by |. Hence — - — is his 2d year's stock. 
3 

(See Universal Key to Algebra, page 17.) 

16. A man has a lease for 99 years, and being asked how 
much of it was already expired, answered that | of the time past 
was equal to f of the time to come. Required the time past and 
the time to come. 

Assume a— 99. Jins. Time past, 54 years. 

1.7. In the composition of a quantity of gunpowder 

The nitre was 10 lbs. more than f of the whole, 

The sulphur 4 2 lbs. less than ^ of the whole, 

The charcoal 2 lbs. less than i of the nitre. 

What was the amount of gunpowder ? w^ns. 69 lbs. 

18. Divide $183 between two men, so that ^ of what the first 
receives shall be equal to j\ of what the second receives. What 
will be the share of each ? ^ns, 1st, $63 ; 2d, $120. 

19. Divide the number 68 into two such parts that the difi*er- 
ence between the greater and 84 shall be equal to 3 times the 
difierence between the less and 40. .^ns. Greater, 42 ; Less, 26. 

20. Four places are situated in the order of the letters »/l, B, 
C, D The distance from ^ to I) is 34 miles. The distance 
from .^ to J? is to the distance from C to i) as 2 to 3. And | 
of the distance from j2 to B, added to half the distance from C 
to D, is three times the distance from B to C. What are the 
respectfte distances ? 

./2ns, From »^ to B=12 ; from B to C=4 ; from C to/;=18. 
6 



A 



,#*r-^ 



66 ELEMENTS OF ALGEBRA. 

^ ■=• ^^^41- 21. A man driving a flock of sheep to market, was met by a 
party of soldiers, who pkmdered him of \ of his flock and 6 
more. Afterwards he was met by another company, who took 
5 what he then had and 10 more: after that he had but 2 left. 
How many had he at first ? Ans. 45. 

y ra. Ajt'U. 22, A laborer engaged to serve for 60 days on these condi- 
tions : That for every day he worked he should have 75 cents 
and his board, and for every day he was idle he should forfeit 25 
cents for damage and board. At the end of the time a settlement 
was made and he received $25. How many days did he work, 
and how many days was he idle ? 

The common way of solving such questions is to let x= the 
days he worked ; then 60 — x represents the days he was idle. 
Then sum up the account and put it equal to $25. 

Another method is to consider that if he worked the whole 60 
days, at 75 cents per day, he must receive $45. But for every 
day he was idle, he not only lost his wages, 75 cents, but 25 
cents in addition. That is, he lost $1 every day he was idle. 
Now let x= the days he was idle. Then x= the dollars 
y he lost. And 45 — x=25 or ic=20 the days he was idle. 
>* " /'*-^^*^3. A boy engaged to carry 100 glass vessels to a certain 
place, and to receive 3 cents for every one he delivered, and to 
forfeit 9 cents for every one he broke. On settlement, he re- 
ceived 2 dollars and 40 cents. How many did he break ? 

Ans. 5. 

94. A person engaged to work a days on these conditions : 

For each day he worked he was to receive b cents, for each day 

he was idle he was to forfeit c cents. At the end of a days he 

received d cents. How many days was he idle ? 

Ans. , , days. 

25. It is required to divide the number 204 into two such 
parts, that | of the less being taken from the greater, the remain- 
der will be equal to f of the greater subtracted from 4 times the 
less. Ans. The numbers are 154 and 50.* 



EQUATIONS. 67 

(Art. 44.) We introduce this, and a few following problems, 
to teach one important expedient, not to say principle, which is, 
not always to commence a problem by putting the unknown 
quantity equal to a single letter. We may take 22", 3a?, or nx 
to represent the unknown quantity, as well as x, and we may 
resort to this ex^eAieniwhen fractional parts of the quantity are 
called in question, and take such a number of a?'s as may be 
divided without fractions. 

In the present example we do not put x= to the less part, as 
we must have | of the less part. It will be more convenient to 
put bx= the less part. Then f of it will be 2a7. Put a=204. 

SJ6. A man bought a horse and chaise for 341 {a) dollars. 
Now if J of the price of the horse be subtracted from twice the 
price of the chaise, the remainder will be the same as if | of the 
price of the chaise be subtracted from 3 times the price of the 
horse. Required the price of each. 

Ans. Horse $152. Chaise $189. 

N. B. Let 8j:= the price of the horse. 
Or let lx= the price of the chaise. 
Solve this question by both of these notations. 

27. From two casks of equal size are drawn quantities, which 
are in the proportion of 6 to 7 ; and it appears that if 16 gallons 
less had been drawn from that which is now the emptier, bnly one 
half as much would have been drawn from it as from the other. 
How many gallons were drawn from each ? Ans. 24 and 28. 

N. B. Let 6a? and 7a; equal the quantities drawn out. 

28. Divide $315 among four persons, A, B, C, and i), giving 
B as much and ^ more than A ; CI more than A and B toge- 
ther ; and D \ more than A, B and C. What is the share of 
each ? Ans. A $24. B $36. C $80, and D $175. 

If we take x to represent ./2's share, we shall have a very 
complex and troublesome problem.* But it will be more simple 
by making 6a?=.^'s share. 

* Taking x for ^'s share, and reducing their sum, gives Equation 24, 
Art. 40. 






68 ELEMENTS OF ALGEBRA. 

Thus, let 6x—£^s share. 

Then 9x=B''s share. 

And l5x-\-5x=C^s share. 

Also 35iC-l— — =2)'s share. 

4 



Sum 70^+^=315 

4 

280a;-{-35^=315X4 

315^=315X4 

x=4: Hence 6a;=24, w5's sh. 

29. A gamester at play staked ^ of his money, which he lost, 
but afterwards won 4 shillings ; he then lost ^ of what he had, 
and afterwards won 3 shillings ; after this he lost i- of what he 
had, and finding that he had but 20 shillings remaining, he left 
off playing. How much had he at first ? ^ns. 30 shillings. 

30. A gentleman spends | of his yearly income for the sup- 
port of his family, and | of the remainder for improving his 
house and grounds, and lays by $70 a year. What is his in- 
come? Jins. 9X70 dollars, or more generally, 9 times the 
sum he saves. 

31. Divide the number 60 (a) into two such parts that their 
product may be equal to tliree times the square of the less num- 
ber ? ^ns. 15 and 45, or |a= the less part. 

32. After paying away | and \ of my money, I had 34 {a) 
dollars left. What had I at first ? 

^ns, 56 dollars. General answer T:rX28. 

17 

33. My horse and saddle are together worth 90 (a) dollars, 
ar^d my horse is worth 8 times my saddle. What is the value 
of each? ^ns. Saddle $10. IJlorse $80. 

34. My horse and saddle are together worth a dollars, and 

my horse is worth n times my saddle. What is the value of 

a *^" 

each ? *^ns. Saddle — r-r. Horse 



n-i-1 n-t-1 



.tnf 



EQUATIONS. 69 

35. The rent of an estate is 8 per cent, greater this year than 
last. This year it is 1890 dollars. What was it last year? 

^ns, $1750. 

36. The rent of an estate is n per cent, greater this year than 
last. This year it is a dollars. What was it last year ? 

•^ns. ,^^ , dollars. 
100 +w 

37. j9 and B have the same income. */! contracts an annual 
debt amounting to i of it ; B lives upon | of it ; at the end of 
two years B lends to ,d enough to pay off his debts, and has 32 
(a) dollars to spare. What is the income of each ? 

Jlns. $280 or 4(35«). 

38. What number is that of which s» I and ~ added together 

make 73 (a)l a oa n i a ^^^ 

^ ' ' Ans. 84. General Ans. — -^ 

73 

39. A person after spending 100 dollars more than i of his 
income, had remaining 35 dollars more than | of it. Required 
his income. Ans. $450. 

40. A person after spending (a) dollars more than \ of his 
income, had remaining (6) dollars more than k of it. Required 
his income. 

Ans. — Yi — ■ <3ollars. 

41. There are two numbers in proportion of 2 to 3, and if 4 
be added to each of them, the sums will be in proportion of 5 
to 7 ? . Ans. 16 and 24. 

42. It is required to find a number such, that if it be increased 
by 7, the square root of the sum shall be equal to the square 
root of the number itself, and 1 more. Ans. 9. 

43. A sets out from a certain place, and travels at the rate of 
7 miles in 5 hours ; and 8 hours afterward B sets out from the 
same place in pursuit, at the rate of 5 miles in 3 hours. How 
long and how far must B travel before he overtakes A? 

Ans. 42 hours, and 70 miles. 



70 ' ELEMENTS OF ALGEBRA. 

SIMPLE EQUATIONS. 
CHAPTER IL 

(Art. 45.) We have given a sufficient number of examples, 
and introduced the reader sufficiently, far into the science pre- 
vious to giving instructions for the solution of questions contain- 
ing two or more unknown quantities. 

There are many simple problems which one may meet with 
in algebra which cannot be solved by the use of a single un- 
knoirn quantity, and there are also some which may be solved 
by a single unknown letter, that may become much more simple 
by using two or more unknown quantities. 

When two unknown quantities are used, two independent 
equations must exist, in which the value of the unknown letters 
must be the same in each. When three unknown quantities are 
used, there must exist three independent equations, in which the 
value of any one of the unknown letters is the same in each. 

In shorty there must be as many independent equations as 
unknovm. quantities used in the question. 

For more definite illustration let us suppose the following 
question : 

A merchant sends me a bill of iQ dollars for 3 pair of shoes 
and 2 pair of boots ; afterwards he sends another bill of 23 
dollars for 4 pair of shoes and 3 pair of boots, charging at the 
same rate. TPliat was his price for a pair of shoes, and what 
for a pair of boots ? 

This can be resolved by one unknown quantity, but it is far 
more simple to use two. 

Let X— the price of a pair of shoes, 

And ?/= the price of a pair of boots. 

Then by the question 3a?4-2y=16 

And ' 4:c+3i/=23. 

These two equations are independent; that is, one cannot be 
converted into the other by multiplication or division, notwith- 
standing the value of x and of y are the same in both equations. 

Having intimated that this problem can be resolved with one 



EQUATIONS. 71 

unknown quantity, we will explain in what manner, before we 
proceed to a general solution of equations containing two un- 
known quantities. 

Let x~ the price of a pair of shoes. 

Then 3a?= the price of three pair of shoes. 

And 16— 3a?= the price of two pair of boots. 

Consequently — - — = the price of one pair of boots. 

Now 4 pair of shoes which cost 4a?, and 3 pair of boots which 

48 9^ 

cost — - — being added together, must equal 23 dollars. 

That is, 4:p+24— ^a;=23. 

Or, 1 — ix=Q. Therefore a?=2 dollars, the price 

of a pair of shoes. Substitute the value of x in the expression 

— - — and we find 5 dollars for the price of a pair of boots. 

Now let us resume the equations, 

3:r+23/=16 {Jl) 
4a?-l-3y=23 [B) 

FIRST METHOD OF ELIMINATION. 

(Art. 4G.) Transpose the terms containing y to the right hand 
sides of the equations, and divide by the coefficients of a?, and 

16 2v 

From equation [A) we have x^= — -— ^ (C) 

o 

23 3^ 

And from [B) we have a?= — — ^ {D) 

Put the two expressions for x equal to each other. (Ax. 7.) 



And 



16— 2i/_23— 3i/ 
3 4 • 



An equation w^hich readily gives ^=^5, which, taken as the 
value of 2/, in either equation (C) or [D) will give x=2. 

This method of elimination, just explained, is called the 
method by comparison. 



72 ELEMENTS OF ALGEBRA. 

SECOND METHOD OF ELIMINATION. 

(Art. 47.) To explain another method of solution, let us again 
resume the equations : 

Sx+2y=lQ (^) 
4a?4-3y=23 (B) 

The value of x from equation (,/?) is x=i{l6 — 2y). 

Substitute this value for x in equation (jB), and we have 
4X 5(16— 2i/)+3i/=23, an equation containing only y. 

Reducing it, we find 2/=5 the same as before. 

This method of elimination is called the method by substitu- 
tion, and consists in finding the value of one unknown quantity 
from one equation to put that value in the other which will cause 
one unknown quantity to disappear. 

THIRD METHOD OF ELIMINATION. 

(Art. 48.) Resume again 3x-i-2y=l6 {d) 
4a:+3y=23 [b) 

When the coefficients of either x or y are the same in both 
equations, and the signs alike, that term will disappear by sub- 
traction. 

When the signs are unlike, and the coefficients equal, the term 
will disappear by addition. 

To make the coefficients of x equal, multiply each equation 
by the coefficient of x in the other. 

To make the coefficients of y equal, multiply each equation 
by the coefficient ofy in the other. 

Multiply equation {Jl) by 4 and 12x-\-Sy=Q4: 
Multiply equation {B) by 3 and l2x-\-9y=Q9 

Difference 2/ = 5 as before. 

To continue this investigation, let us take the equations 

2x-\-Sy=23 {A) 

5;r— 23/=10 {B) 

Multiply equation {A) by 2, and equation [B) by 3, and we 
have 4a;-}-63/=46 

15x~6y=30 



EQUATIONS. 73 

Equations in which the coefficients of y are equal, and the 
signs unUke. In this case add, and the i/'s will destroy each 
other, giving 19a?=76 

Or 07=4. 

This method of elimination is called the method by addition 
and subtraction. 

FOURTH METHOD OF ELIMINATION. 

(Art. 48.) Take the equations 2a;-(-3i/=23. {A) 
And 5x— 2t/==10. (i?) 

Multiply one of the equations, for example (^'■?), by some inde- 
terminate quantity, say m. 

Then 2ma7-{-3m?/=23w 

Subtract (i^) 5x'— 2?/=10 

Remainder, (C) (2m — 5)x-i-(3m+2)7/=-23m— 10 

As m is an indeterminate quantity, we can assume it of any 
\alue to suit our pleasure, and whatever the assumption may be, 
the equation is still true. 

Let us assume it of such a value as shall make the coefficient 
of 3/, (3m+2)=0. 

The whole term will then be times y, which is 0, and equa- 
tion (O) becomes 

(27?i— 5)a.— 23m— -10 * 
23m— 10 

But 3m-r2=0. Therefore 7W= — |. 
Which substitute for m in equation (jD), and we have 
_ — 23Xf— 10 _ — 23X 2— 30_--76^ 
^- __2Xf^ Z::2"><2— 15~— 19~ ' 

This is a French method, introduced by Bezout, but it is too 
indirect and metaphysical to be mjLich practised, or in fact much 
known. 

Of the other three methods, sometimes one is preferable and 
sometimes another, according to the relation of the coefficients 
and the positions in which they stand. 



74 ELEMENTS OF ALGEBRA. 

No one should be prejudiced against either method, and in 
practice we use either one, or modifications of them, as the case 
may require. The forms may be disregarded when the princi- 
ples are kept in view. 

(Art. 49.) To present these different forms in the most general 
manner, let us take the following general equations, as all par- 
ticular equations can be reduced to these forms. 

ax-\-hy =0 [A) 
a'x-\-b'y=c' [B) 

Observe that a and «', may represent very different quantities, 
so b and b' may be different, also c and c' may be different. In 
special problems, however, a may be equal to a', or be some 
multiple of it ; and the same remark may apply to the other 
letters. In such cases the solution of the equations are much 
easier than by the definite forms. Hence, in solving definite 
problems great attention should be paid to the relative values of 
the coefficients. 

First method. 

Transpose the terms containing y and divide by the coeffi- 
cients of a?, and 

x= ^ , also X'-= ^ (C) 

a a' ^ ' 

r«, r c — by c — b'y ... ^ . 

Therefore -= ^ (Axiom 7.) 

Clearing of fractions, give a'c — a'by=ac' — ab'y. 
Transpose, and {ab' — a'b)y—ac' — a'c. 

^ .... ac' — a'c 

By division V=-t; ri* 

^ ^ ab' — a'b 

When y is determined, its value put in either equation marked 
(C) will give X. 

Second method. 
From equation (.^) x= — ■ — ^. 



EQUATIONS. 70 

Which value of x substitute in equation (B) and 
a'c — a'by , ., 

Clearing of fractions and transposing a'c^ we have 

ab'y — a'by=ac' — a'c 

^ ac' — a'c 

Or V=-r, 77 

^ ab'—a'b 

The same value of y as before found. 

Third method. 
Multiply equation [A) by a\ and equation [B) by a. 
And a'ax-\-a'by=a'c. 

Also a'ax'\-ab'y=^ac' 



Difference {ab' — a'b)y=ac' — a'c 

Or V=—r. 77 same value as by the 

•^ ab' — a'6 ^ 

other methods. 

Fourth method. 
Multiply equation {A) by an indefinite number m, 
And amx-\-bmy=mc 

Subtract {B) a'x-\- b'y=c' 



And {am — a')x-]r{bm — b')y=mc — c'. 

Now the value of ?n may be so assumed as to render the 
coefficient of x=0, or am — «'=0. 

Then {bm — b')y=mc — c' 

_ mc — c' ,_, 



But am — a'=0, or m=5— . 

a 



76 ELEMENTS OF ALGEBRA. 

Put this resultant value of m, in equation (C), and 



a 



-ac' 



^, a'b — ab' 

bX b' 

a 



by multiplying both numerator and denominator by a. 

(Art. 50.) The principles just explained for elimination be- 
tween two quantities may be extended to any number, where the 
number of independent equations given are equal to the number 
of unknown quantities. For instance, suppose we have the 
three independent equations : 

ax-{-by-\-cz=d {Jl) 

a'x-^b'i/-{-c'z=d' (B) 

a"x-\-b"y-\-c"z=-d" (C) 

We can eliminate either x, or y, or z, (whichever may be 
most convenient in any definite problem) between equations {A) 
and (J5,) and we shall have a new equation containing only two 
unknown quantities. We can then eliminate the same letter be- 
tween equations [B) and (C,) or [A) and (C,) and have another 
equation containing the same two unknown quantities. 

Then we shall have two independent equations, containing two 
unknown quantities, which can be resolved by either of the four 
methods already explained. 

(Art. 51.) Another theoretical method, from the French, we 
present to the reader, more for curiosity than for any thing else. 

Multiply the first equation (^,) by an indefinite assumed num- 
ber 771. Multiply the second equation [B) by another indefinite 
number n, and add their products together. Their sum will be 

[am-\-a'n)x-\-{bm-\-b'n)y'\-{cm-\-c'n)z==dm-\-nd' 
Subtract eq. (C) a"x ■^b"y c"z=d". 

And {am-i-a'n--'a")x-{-{bm-\-b'n — b")y-\-{cm-\-c'n-^c")z 

=dm-\-nd' — d". 



EQUATIONS. 77 

As m and n are independent and arbitrary numbers, they can 
be so assumed that 

am-{-a'n — «"=0 and bm-\-b'n — 6"=0. 
Tlien am-\-a'n=a" (1) and bm-\-h'n=h" (2) 

dm-\-nd'—d" 

And 2:= i (3) 

cm-|-c'w — c ^ ' 

From equations (1) and (2) we cen find the values of m and 
y?, which values may be substituted in equation (3,) and then z 
will be fully determined. 

EXAMPLES FOR PRACTICE. 
I. Given < ,^ , ,^^ r to find tlie values of x and y. 

We can resolve this problem by either one of the four methods 
just explained. But we would not restrict the pupil to the very 
letter of the rule, for that in mnny cases might lead to operations 
unnecessarily lengtliy. 

If we take the third method of elimination, we should multi- 
ply the first equation h^ 12, the second by 8 ; but as the coefii- 
cients*of x contain the common factor 4, we can multiply by 3 
and 2, in place of 12 and^. . That is, multiply by i\\c fourth 
part of 12 and 8. 

In practice even this form need not be observed, we may de- 
cide on our multipliers by inspection only. 

Three times the 1st gives 24a:-[-l57/=204 

Twice the 2d gives 24a:+14?/— 200 

Difference gives 2/ =4 

Substituting this value of y in first equation, and 

8^+20 = 68 or a?=6. 

In solving this, we have used modifications of the 3d and 2d 
formal methods. 
g2 



78 ELEMENTS OF ALGEBRA. 

For exercise, let us use the 4th method. 

Smx-\- 5my = 68m 
Take 12a? + Ty =100 



(8r/i— 1 2)a?-}- (5m— 7)1/ = 68m— 1 00 

Assume 8m — 12=0. 

m, 68m — 100 

Then V——^ ;=r-' But m=|. 

rp, . 68X1—100 204—200 ^ ^ 

Therefore y^^—-.^-- =____= 4 ^n.. 

« r.. C5a?4-2?/ = 19 7 ^ , 

2. Given < « ^ r to nnd .t and y. 

llx — 6^=9 3 

If we multiply the first of these equations by 3, the coefiicients 
of y will be equal, and the equations become 

1507+ 61/ = 57, 
And Hx — 6^=9. 

To ehminate 3/, we add these equations (the signs of the terms 
cjDntaining y being unlike), and there results 

22a?=66, '• . 
.r=3. 

This value of x put in the 1st equation gives 

15-j-2i/=19, 
And ?/=2. 

3. Given -— _[_6?/=21 and ^^^+5x=23 to find a?andy. 

Clear of fractions and reduce. We then have 
a:+24^=76 
And 15X--1- y =63. 

In this case there are no abbreviations of the rules, as the 
coefficients of the unknown terms are prime to each other. 
Continuing the operation, we find x=4, 2/ =3. 



EQUATIONS. 79 

2t 3?/ 
4, Given x-\-y=\7 and -^=-j^ to find x and y. '^ 

Owing to the peculiarity of form in the 2d equation, it is most 
expedient to resolve this by the 2d method. 

From the 2d, x='^-. Then -^■\-y=.\'7. 
8 o 

Clearing of fractions, 9«/+8i/=17X8. 

Or, 173/=17X8, or ?/=8. 

Hence, a?=9. 



5. 



Cix+8v = 194> ^ , , 
Given < , , ^•^ ,«, f to find the values of x and y. 
C|t/+8a?=131 3 ^ 



Here we observe that both x and y are divided by 8, x in one 
equation, and y in the other ; also, x and y are both multiplied 
by 8. 

(Art. 51.) All such circumstances enable us to resort to many 
pleasant expedients which go far to teach the tnie spirit of al- 
gebra. 

X~\~1J 

Add these two equations, and — — ^-|-8(x-f2/)=325. 

Assume x-\-y=^s. 

Or let s represent the sum of x-\-y, then |«4-8«=325. 
Clear of fractions, and 5-1-645=325X8. 
Unite and divide by 65 and s=5X8. 

Or a:-l-y=5a. {^) By returning to the value of 5, and put- 
ting a=8. 

Multiply the 1st equation by 8, and 

a:-f642/=194a 

Subtract {A) x -{- y=5a 

Rem. 63y—l89a 

Divide by 63 and y=da=24. Whence a:r=2a=16. 
Let the pupil take any one of the formal rules for the solutioa 
of the 'preceding equations, and mark the difference. 

6. Given ^x-{-3y=2l and 5]/-j-3a?=29 to find x and y. 

Ans. ar=9. y=6. 



80 ELEMENTS OF ALGEBRA. 

^ 7. Given 4x-]ry==S4 and 4y-{-x=l(y, to find xandy, 

Jlns. a,'=8. 2/ =2. 
» 8. Given ix-\-ly~\4: and lx-\-ly— II, to find .r and y. 

Ans, a?=24. 2/=6. 

9. Given x-\-hj—^ and ^,r+?/=7 to find o^ and y. 

Ans. a?— 6. 2/=4. 

10. Given 4a?-j-7i/=99 and Uj-[-lx=^\ to find .t and y. 

Ans. x~l. y=l4, 

11. Given \^+Aa^^s=^~^27, 

A71S. a-^eo. y=40. 

12. Given --{--=6 and — [—=10, to find x and v. 

X y ^ y 

Multiply the first equation by c, the second by a, and we shall 
have 

ac , be 
X y 

ac , ad ,^ 
— 1-— =10^/. 
X y 



By subtraction [he — ad)~=Qc — 10a 
•/ 

,„j « be — ad 

I herafore —y. 

fie— lOrt ^ 



^^ ^. 147 147 „^ ,^. , 17 , 56 41 , ^' , , 

13. Given =28 M) and h— = -;r i^) to find 

X y ^ ^ X y 3 ^ ' 

the values of x and ?/. 

21 21 

Divide equation [A) by 7, and —4. 

X y 

114 

Divide this result by 21, and = — (C) 

X y 21 ^ ^ 



EQUATIONS. ;81 

Multiply (C) by 17, gives "-y=^ {O) 

73 219 

Subtract (/)) from [B) and we have — ^^~7pr' 

13 1 

Divide by 73, and -=—-=- or v=7. 
'' ^ 21 7 "^ 

Putting this value in equation (O) and reducing we find a:=3. 

14. Given —\ — = 1 and — | — =~H — to find the values 

X y y X y X 2 

^^ ^ ^^^^^ y Am. x=4. and 2/=2. 

,^ ..• Ca?-|-150: V— 50::3:2 ) ^ . , , 

15. Given ^ .^ •^, ,^^ _ ^ ^ to find a? and ;?/. 

ix — 50 :i/+lOO :: 5 : 9 ) 

.^n.5. a;=300. 7/=350. 



...„„„ ».H-.,+,=?t2|±fl 



to find rr and y. 



, , ., 151— I6.r , 9:rv— 110 

And ^^='—A i — 1— ^ — T- 

4y — 1 3?/ — 4 

Ans. rr=9. 2/=2. 

Note. — For solutions of examples 15 and 16, see Universal 
Key to the Science of Algebra, page 26. 

lY. Given < ."^'^ ''^ f to find the values of x and y. 

C — x-\-7y=23^ ^ 

Ana, a"=2. 2/=5. 

I§. Given \ \^'^'fJ'^ I to find x and y. 

Ans, a?=6. i/=15. 

19. Given x-\-y=S and o?^— 3/^=16 to find x and y. 

Ans. a^=5. 2/=3. 

20. Given A{x-\-y)=^^{x — y) and a;^ — ^7/^=36 to find a: and y. 

Ans. x—H' 2/=2ii. 
6 



82 



ELEMENTS OF ALGEBRA. 



21. Given x:y ::4:3 and x^ — y^=S7, to find x and y. 

Ans. a?=4. 2/=3. 

22. Given x-\-y—a and oi? — y^=ab to find x and y. 

^ a-\-h a — h 

Ans. x=——. y=-^. 



X~\~ 1 X 

23. Given = ^ and — r-r=l to find x and ?/. 

24. Given i(x-f 2)-{-8y.-31 and |(?/-f 5)+ 10a;= 192 to 
find the values of x and y. Ans. x=\9. y=B. 

25. Given 3x-\-7y=79 and 2y-\-ix=l9 to find the values 
of X and ?/. »^n5. a?=10. 2/~'^' 

26. Given |(.T4-3/) + '25=a" and ^{x-{-y) — 5=y to find 
the values of x and y. Ana. a?=85. i/=35. 

27. Given a?— 4=v+l and 5a?— •^=— — ?^+37 to find the 

•^ 3 4 6 ~ 

values of x and ?/. Ans. a?=8. 2/=3. 

2§. Given 4--^^=v— l^l and ^=^"+2 to find the 
6 -^ ^ 5 5 



values of x and y. 



»^m. a?=10. 2/=20. 



29. Given — — _+y— -=44— ^— and -— ^■ 



9a'--7 3V-I-9 4.r-i-5?/ ^ , , ^ n ^ 

— - — =-^ — ^ to find a? and y. Jins. a*=9. t/=4. 



30. Given 



23— .T 2 



. V— 3 „^ 73—3?/ 

1/4--:^^ =30 - 

^^a?— 18 3 



to find X and y. 



Jlns. a:=21. y=20. 



EQUATIONS. qSg 

CHAPTER III. 

Solution of Equations involving three or more U7iknown 
quantities. 

(Art. 52.) No additional principles are requisite to those given 
in articles 49 and 50. 

EXAMPLES. 

r x^ yj^ z= Q^ 

1. Given ] a?+22/+3z=16 [ to find x, y, and z. 
{ a;+3i/+4z=2I J 

By the 1st method, transpose the terms containing y and z in 
each equation, and 

x= 9 — y — z, 
x=\Q — 2y-^Sz, 
x=2\—3y—4z. 

Then putting the 1st and 2d values equal, and the 2d and 3d 
values equal, gives 

9 — y — z=\G—2y — 3z, 
16— 2i/— 3z=21— 31/ — iy. 

Transposing and condensing terms, and 
y=7-2z, 

Also, y—^ — ^y 

Hence, 5— z=7— 2z, or z=2 

Having z^2, we have 2/=5 — z=3, and having the values of 
both z and 3/, by the first equation we find x=^4. 

{ 2x-]-4y—3z=22 " 
4x-{-2y-\-5z=lS 
6a:+7?/— 2=63 J 

Multiplying the first equation by 2, 4x-\-Sy — 6z=44 
And subtracting the second, 4x — 2y-{- 5z=18 



2. Given \ 4x-{-2y-\-5z=lS [to find values of Xy y and z. 



The result is, (.^) l0y-^Uz==26 



84 ELEMENTS OF ALGEBRA. 

Then multiply the first equation by 3, 6x-\-l2y — 9^=66 
And subtract the third, 6x-\~ ly — z=63 

The result is, [B) 51/— 8z= 3 

Multiply the new equation [B) by 2, lOy — 16z= 6 

And subtract this from equation [A) lOy — llz=26 

^ The result is, 52-=20 

Therefore z= 4 

Substituting the value of 2: in equation (7?) and we find 3/=7. 
Substituting these values in the first equation, and we find a:=3. 

r 3.r+9j/+8z=41 1 
3. Given \ bx-\-iy — 22r=20 \ to find x, y and z. 
{ \lx-\-ly—Qz=^l J 

To illustrate by a practical example we shall resolve this ■by 
the principles explained in (Art. 51.) 

Zmx-\-9my-\-^mz=^i\m 
bnx -\-4:ny — 2nz —20n 

Sum {3m-]-5n)x-[-{dm-{-4n)y-{-{Sm—2n)z=4lm-\-20n 
Take 11a; +ly —Qz=37 

Rem. (3w-}-5n— 1 1) 0? — {7—dm—4n)y-{-{Sm—2n-\-6)z~ 
41?n+20n— 37. 



Assume ^m+5/i=:ll 


(1) 


And 9m-{-4n=7 


(2) 


41m+20?i— 37 

Then ^"=-5 1^ — r~~^ 

8w — 2n-\- 6 


(3) 


From equations (1) and (2) we find m= — f^ and n~\\ 
These values substituted in equation (3) we have 


—4lXj\-\-20X\'r 


—37 



— 8Xy%— 2Xtf+ 6 



Multiply both numerator and denominator by 11, and we shall 
^123 +520— 407 _ —10 _ 
have ^--Tl24-I~52"-f~66~--'i0~ 



■ " EQUATIONS. S5 

Putting this value of z in the 1st and 2d equations, we shall 
have only two equations involving x and i/, from which the 
values of these letters may be determined. 

These equations can be resolved with much more facility by 
multiplying the 2d equation by 4, then adding it to the 1st to 
destroy the terms containing z. 

Afterwards multiplying the 2d equation by 3, and subtracting 
the 3d equation, and there will arise two equations containing x 
and ?/, which may be resolved by one of the methods already 
explained. 

(Art. 53.) When three, four, or more unknown quantities with 
as many equations are given, and their coefficients are all prime 
to each other, the operation is necessarily long. But when sev- 
eral of the coefficients are multiples, or measures of each other, 
or unity, several expedients may be resorted to for the purpose 
of facilitating calculation. 

No specific rules can be given for mere expedients. Exam- 
ples alone can illustrate, but even examples will be fruitless to 
one who neglects general principles and definite theories. Some 
few expedients will be illustrated by the following 

EXAMPLES. 

r x-\-yi-z=si ^ 

1. Given \ x-\-y — z=25 \ to find x, y, and z. 



\ x-{-y — 2r=25 > to 
I x—y^z= 9 J 



Subtract the 2d from the 1st, and 2:2: =6. 
Subtract the 3d from the 2d, and 23/= 16. 
Add the 1st and 3d, and 2x=40. 

r x-\-y^-z=2Q \ 
2. Given -j x — y = 4 j- to find a?, y and z, 
\ X'-z = 6 J 

Add all three, and 3x=3Q or a:=12. 
X — y — z= 6 1 




Sy — X — ^z=12 \ to find a:, y 



and z. 



80 



ELEMENTS OF ALGEBRA. 



Assume x-j-y-\-z=s. Add this equation to each of the given 
equations, and we have 

2x= 6+s, (^) 

4i/=12-l-5, (B) 

8z=24+5. (C) 
Multiply {A) by 4, and (J5) by 2, and vjre have 

8a^=24+45, 

8i/=24-j-25, 

8z=24-l- 5, 

B-y addition, 8^=3 X 24+75. Or 5=72. 
Put this value of s in equation (./?) and vire have 
2:r=6-{-72. Or a;=3-|-36=39, &c. 

4. Given a?+i3/=100, i/-|-iz=100, 2-l-^a:=100 to find 
Xf i/f and z. 

Put a=100. ^ns. a;=64, y=^72, and z=84. 

u-\-v-\-x-\-z=li 
u-\-v-{-y-]-z=^l2 
u-{-x-{-y-\-z=lS 
v-\-x-\-y-{^z=l4: 
Here are Jive letters and five equations. Each letter has the 
same coefiicient, one understood. Each equation has 4 letters, 
z is wanting in the 1st equation, y in the 2d, &c. 
Now assume u-{-v-\-x-\-y-i-z=s. 
Then 5—^=10 ' {^) 
s—y=ii 
s — a?=12 
s — 1;=13 
s — M=14 



5. Given 



to find the value of each. 



Add, and 5s— 5=60 Or 5=15. 
Put this value of 5 in equation (.^), and 2;=5, <fec. 

6. Given x-{-y—ctt a?+z=Z>, y-{-z=c. 

Add the 1st and 2d, and from the sum subtract the 3d. 

»Sn8. x=^i{a-{-b — c), t/=^(a-}-c — b), z=i(b-{-c — aV 



V. Given 



EQUATIONS. 87 

2x=u-\-y-\-z 

3i/=u-\--x-{-z 

Az~u-\-x-\-y 

ii=^x — 14 

Ans. ?/=26, a;=40, 2/=30, 5r=24. 



. to find the value of ii, a:, y, 
and z. 



§. Given 



hx-\-iy-\-kz=Q2 



r a;=24. 
2r=120. 



9. Given -j y-\-a=2x-]-'Zz 
[ z-{-a=3x-\-3y 



Ans. 






2a;4- 2/— 2z=40 ra;=20. 

4y — x+3z=35 y=10. 

lO. Given ^ 3u-\- /=13 ^rw.. z= 5. 

y-\- u-\- ^=15 M= 4. 

3:r— 2/+3f— w=49 [ /= 1. 

(Art. 54.) Problems producing simple equations involving two 
or more unknown quantities. 

1. Find three numbers such, that the product of the 1st and 
2d, shall be 600; the product of the 1st and 3d, 300 ; and the 
product of the 2d and 3d, 200. 

Ans. The numbers are, 30, 20, and 10. 

2. Find three numbers, such that the^r^^ with 2 the sum of 
the second and third shall be 120 ; the second vrith \ the differ- 
ence of the third znd Jirst shall be 70 ; and the sum of the three 
numbers shall be 190. Ans. 50 ; 65 ; 75. 

3. A certain sum of money was to be divided among three 
persons, A, B, and C, so that A^s share exceeded i of the shares 
of B and C by $120 ; also the share of B exceeded | of the 
shares oi A and Cby $120; and the share of C, likewise, ex- 
ceeded I of the shares of A and B by $120. What was each 
person's share? 

Ans. A's share, $600; 5's, 480 ; and C's 360. 



88 ELEMENTS OF ALGEBRA. 

4. A and B, working at a job, can earn $40 in 6 days ; A and 
C together can earn $54 in 9 days; and B and C$80 in 15 
days. What can each person alone earn in a day? 

Let A earn ct% By^ and Cz dollars per day, then, 
By the question, 6a:-}- 6t/=40 
9a:-|- 92—54 
15y-fl5z=80' 

Dividing the equations by the coefficients of the unknown 
quantities, we have, i _a2 

a;4-z=6 
y+z=5|. 
See Problem 6. (Art. 53.) 

A man has 4 sons. The sum of the ages of the first, second 
and third is 18 years ; the sum of the ages of the first, second 
and fourth is 16 years; the sum of the ages of the first, third 
and fourth is 14 years ; the sum of the ages of tlie second, third 
and fourth is 12 years. What are their ages? See Problem 5. 
(Art. 53.) Jim, Their ages are, 8, 6, 4, 2. 

5. A^ B and C sat down to play, each one with a certain num- 
ber of shillings ; A loses to jS and C as many shillings as each 
of them has. Next B loses to A and C as many as each of 
them now has. Lastly, C loses to A and B as many as each of 
them now has. After all, each one of them has 16 shillings. 
How much did each one gain or lose ? 

Let x^= the number of shillings A had at first 
y= J?'s shillings, and 
z=. C's shillings. 
Then, by resolving the problem, we shall find re =26, i/=14, 
and 2:=8. Therefore, A lost 10 shilUngs, B gained 2, and C8. 
N. B. When the equations are found, divide the 1st by 4, the 
2d by 2, and then compare them with Ex. 3. (Art. 53.) 

6. A gentleman left a sum of money to be divided among four 
servants, so that the share of the first was \ the sum of the 
shares of the other three ; the share of the second, | of the sum 
of the other three ; and the share of the third, \ the sum of the 



EQUATIONS. 80 

Other three ; and it was found that the share of the last was 14 
dollars less than that of the first. What was the amount of 
money divided, and the shares of each respectively ? 

Ans, The sum was $120 ; the shares 40, 30, 24 and 26. 

Observe Prob. 7. (Art. 53,) in connection with this problem. 

7. A jockey has two horses, and two saddles -which are worth 

15 and 10 dollars, respectively. Now if the better saddle be put 
on the belter horse, the value of the better horse and saddle 
would be worth | of the other horse and saddle. But if the 
better saddle be put on the poorer horse, and the poorer saddle 
on the belter horse, the value of the better horse and saddle is 
worth once and j^j the value of the other. Required the worth 
of each horse 1 Ans. 65 and 50 dollars. 

8. A merchant finds that if he mixes sherry and brandy in 
quantities which are in proportion of 2 to 1, he can sell the mix- 
ture at 78 shillings jitr dozen; but if the proportion be 7 to 2 he 
can sell it at 79 shillings -per dozen. Required the price per 
dozen of the sherry and of the brandy ? 

Am, Sherry, 8l5. Brandy, 72s. 

In the solution of this question, put «=78. Then a+l=79. 

9. Two persons, A and j5, can perform a piece of work in 

16 days. They work together for four days, when A being 
called off, B is left to finish it, which he does in 36 days. In- 
what time would each do it separately ? 

Ans. A in 24 days ; ^ in 48 days. 

10. What fraction is that, whose numerator being doubled, 
and denominator increased by 7, the value becomes | ; but the 
denominator being doubled, and the numerator increased by 2, 
the value becomes |? Ans. |. 

11. Two men wishing to purchase a house together, valued 
at 240 {a) dollars ; says A to B, if you will lend me | of your 
money I can purchase the house alone ; but says B to A, if you 
lend me J of yours, I can purchase the house. How much 
money had each of them ? Ans. A had $160. B $120. 

8 



90 ELEMENTS OF ALGEBRA. 

12. It is required to divide the number 24 into two such parts, 
that the quotient of the greater part divided by the less, may be 
to the quotient of the less part divided by the greater, as 4 to 1. 

^ns. 16 and 8. 

13. A certain company at a tavern, when they came to settle 
their reckoning, found that had there been 4 more in company, 
they might have paid a shilling a-piece less than they did; but 
that if there had been 3 fewer in company, they must have paid 
a sliilhng a-piece more than they did. What then was the num- 
ber of persons in company, and what did each pay ? 

t^ns. 24 persons, each paid 7s. 

14. There is a certain number consisting of two places, a unit 
and a ten, which is four times the sum of its digits, and if 27 be 
added to it, the digits will be inverted. What is the number? 

Jns. 36. 

Note. Undoubtedly the reader has learned in arithmetic that 
numerals have a specific and a local value, and every remove 
from the unit multiplies by 10. Hence, if x represents a digit 
in the place of tens, and y in the place of units, the number must 
be expressed by lOx-{-y. A number consisting of three places, 
with X, y and z to represent the digits, must be expressed by 
\QQx-{-\Qy-\-z. 

15. A number is expressed by three figures ; the sum of these 
figures is 1 1 ; the figure in the place of units is double that in 
the place of hundreds, and when 297 is added to this number, 
the sum obtained is expressed by the figures of this number re- 
versed. What is the number 1 Ans. 326. 

16. To divide the number 90 into three parts, so that twice 
the first part increased by 40, three times the second part in- 
creased by 20, and four times the third part increased by 10, may 
be all equal to one another. 

Ans. First part 35, second 30, and third 25. 

l?". A person who possessed $100,000 (a,) placed the greater 
part of it out at 5 per cent, interest, and the other part at 4 per 



EQUATIONS. »1 

cent. The interest which he received for the whole amounted 
to 4640 [b) dollars. Required the two parts. 

£ns. 64,000 and 36,000 dollars. 

General Answer. (100& — 4a) for the greater part, and (5a — 
1006) for the less. 

18. A person put out a certain sum of money at interest at a 
certain rate. Another person put out $10,000 inore^ at a rate 1 
per cent, higher, and received an income of $800 more. A third 
person put out $15,000 more than the first, at a rate 2 per cent, 
higher, and received an income greater by $1,500. Required 
the several sums, and their respective rates of interest. 

Ans. Rates 4, 5 and 6 per cent. Capitals $30,000, $40,000 
and $45,000. 

10. A widow possessed 13,000 dollars, which she divided 
into two parts and placed them at interest, in such a manner, that 
the incomes from them were equal. If she had put out the first 
portion at the same rate as the second, she would have drawn 
for this part 360 dollars interest, and if she had placed the se- 
cond out at the same rate as the first, she would have drawn for it 
490 dollars interest. What were the two rates of interest? 

Ans, 7 and 6 per cent.* 

20. There are three persons, A^ B and C, whose ages are as 
follows : if -B's age be subtracted from ./?'s, the difference will 
be C's age ; if five times ^'s age and twice Cs age, be added 
together, and from their sum .^'s age be subtracted, the remain- 
der will be 147. The sum of all their ages is 96. What are 
their ages ? Ans. A's 48, B'a 33, C's 15. 

31. Find what each of three persons. A, B and C, is worth, 
from knowing, 1st, that what A is worth added to 3 times what 
B and C are worth, make 4700 dollars ; 2d, that what B is worth 
added to four times what A and C are worth make 5800 dollars; 
3d, that what C is worth added to five limes what A and B are 
worth make 6300 dollars. 

Ans. A is worth 500, B 600, C 800 dollars. 

See brief solution to these two problems, 18 and 19, in Key to Algebra. 



92 ELEMENTS OF ALGEBRA. 

Put s= the sum that Jl, B and C are wortli, to make an aux- 
iliary equation. 

22. Find what each of three persons, A^ B, C, is worth, 
knowing, 1st, that what A is worth added to / times whai^ and 
C are worth is equal to p; 2d, that what B is worth added to m 
times what A and C are worth is equal to q; 3d, that what C is 
worth added to n times what A and B are worth is equal to r. 
Let x^A's capital, y=B's, and z = C's. 
Then x-\-ly-^lz=p, is the first equation. 

Assume x-{- y-\- z=s. Multiply this equation by / and 
subtract the former, and (/ — l)x=ls-—p. 






By a similar operation, y= 

m — 1 

And, 



(A) 



By addition, 



ns — r 
■ w— 1 



Is — p 7ns — q ns — r 



This equation may take the following form : 

Now as the terms in parenthesis are fully determined, of 
known value, we may represent the first by cr, the second by h, 
and this last form becomes 

s=a5 — b 
By transposition, (fee. (a — 1)5=6 

b 



Therefore 



This known value of 5 put in each of the equations marked 
(A), and the values of x, y and z will be theoretically deter- 
mined. 



EQUATIONS. 93 

23. Three brothers made a purchase of $2000 (a;) the first 
wanted in addition to his own money i the money of the second, 
the second wanted in addition to his own j of the money of the 
third, and the third required in addition to his own | of the 
money of the first. How much money had each ? 

^ns. 1st had, $1280; 2d, $1440 ; and the 3d, $1680. 
Gen. Ans. 1st had ^|a; 2d, |f«; and the 3d f ^a. 
See Prob. 6. (Art. 53.) 

24. Some hours after a courier had been sent from A to B, 
which are 147 miles distant, a second was sent, who wished to 
overtake him just as he entered B, and to accomplish this he 
must perform the journey in 28 hours less time than the first did. 
Now the time that the first travels 17 miles added to the time the 
second travels 56 miles is 13 1 hours. How many miles does 
each go per hour 1 Ans. 1st 3, the 2d, 7 miles per hour. 

25. There are two numbers, such that ^ the greater added to 
I the lesser, is 13 ; and if h the lesser is taken from I the greater, 
the remainder is nothing. Required the numbers. 

Ans. 18 and 12. 

26. Find three numbers of such magnitude, that the 1st with 
the 5 sum of the other two, the second with ^ of the other two, 
and the third with | of the other two, may be the same, and 
amount to 51 in each case. Ans. 15, 33, and 39. 

27. .^ said to B and C, " Give me, each of you, 4 of your 
sheep, and I shall have 4 more than you will have left." B said 
to A and C, " If each of you will give me 4 of your sheep, I 
shall have twice as many as you will have left." C then said 
to A and B, " Each of you give me 4 of your sheep, and I shall 
have three times as many as you will have left." How many 
had each ? Ans. A Q, B %, and C 10. 

28. What fraction is that, to the numerator of which if 1 be 
added, the fraction will be | : but if 1 be added to the denomina- 
tor, the fraction will be | ? Ans. r^j. 

29. What fraction is that, to the numerator of which if 2 be 



94 ELEMENTS OF ALGEBRA. 

added, tlie fraction will be f ; but if 2 be added to the denomina- 
tor, the fraction will be j ? Ans. ~. 

30. What fraction is that whose numerator being doubled, and 
its denominator increased by 7, the value becomes | ; but the de- 
nominator being doubled, and the numerator increased by 2, the 
value becomes | ? Jins. J. 

31. If ./? give B $5 of his money, B will have twice as much 
money as A has left; and if B give A $5, A will have thrice as 
much as B has left. How much had each ? 

Alls, A $13, and J5 $11. 

32. A corn factor mixes wheat flour, which cost him 10 shil- 
lings per bushel, with bjirley flour, which cost 4 shillings per 
bushel, in such proportion as to gain 43| per cent, by selling the 
mixture at 1 1 shillings per bushel. Required the proportion. 

Ans. The proportion is 14 bushels of wheat flour to 9 of 
barley. 

33. There is a number consisting of two digits, which num- 
ber divided by 5 gives a certain quotient and a remainder of one, 
and the same number divided by 8 gives another quotient and a 
remainder of one. Now the quotient obtained by dividing by 5 
is double of the value of the digit in the ten's place, and the quo- 
tient obtained by dividing by 8 is equal to 5 times the unit digit. 
What is the number ? Ans. 41. 

Interpretation of negative values resulting from the solution 
of equations. 
(Art. 55.) The resolution of proper equations drawn from 
problems not only reveal the numeral result, but improper enun- 
ciation by the change of signs. Or the signs being true algebraic 
language, they will point out errors in relation to terms in com- 
mon language, as the following examples will illustrate : 

1. The sum of two numbers is 120, and their difference is 
160 ; what are the numbers ? 

Let X be the greater and y the less. Then . 
.T+?/ = 120 (1) 
x-^y = \m (2) 
The solution gives x=140, and y= — 20. 



EQUATIONS, 95 

Here it appears that one of the numbers is greater than the 
sum given in the enunciation, yet the sum of a; and y, in the al- 
gebraic sense, make 120. 

There is no swch abstract number as —20, and when minus 
appears it is only relative or opposite in direction or condition 
to plus, and the problem is susceptible of interpretation in an al- 
gebraic sense, but not in a definite arithmetical sense. 

Indeed we might have determined this at once by a considera- 
tion of the problem, for the difference of the two numbers is 
given, greater than their sum. But we can form a problem, an 
algebraic (not an abstract) problem that will exactly correspond 
with these conditions, thus : 

The joint property of two men amounts to 120 dollars, and 
one of them is worth 160 dollars more than the other. What 
amount of property does each possess ? 

The answer must be -1-140 and — 20 dollars ; but there is no 
such thing as minus $20 in the abstract ; it must be interpreted 
debt, an opposite term to positive money in hand. 

2. Two men, A and B, commenced trade at the same time ; 
A had 3 times as much money as B, and continuing in trade, A 
gains 400 dollars, and ^150 dollars; now A has twice as much 
money as B. How much did each have at first? 

Without any special consideration of the question, it implies 
that both had money, and asks how much. But on resolving the 
question with x to represent A's money, and y jB's, we find 

a;=_-300 _.5,.-i ,:, 

And y= — 100 dollars. 

That is, they had no money, and the minus sign in this case 
indicates debt ; and the solution not only reveals the numerical 
values, but the true conditions of the problem, and points out the 
necessary corrections of language to correspond to an arithmeti- 
cal sense, thus : 

A is three times as much in debt as B ; but A gains 400 dol- 
lars, and B 150 ; now A has twice as much money as B. How 
much were each in debt ? 

As the enunciation of this problem corresponds with the real 



96 ELEMENTS OF ALGEBRA. 

circumstance of the case, we can resolve the problem without a 
minus sign in the result. Thus : 

Let x= B's debt, then Sx= A's, debt 
150 — x= jB's money, 400 — 3a?= A''s money 
Per question, 400 — 3a7--=300--2x. Or a'=100. 

3. What number is that whose fourth part exceeds its third 
part by 12? ^725.-144. 

But there is no such abstract number as — 144, and we cannot 
interpret this as debt. It points out error or impossibility^ and 
by returning to the question we perceive that a fourth part of any 
number whatever cannot exceed its third part; it must be, its third 
part exceeds its fourth part by 12, and this enunciation gives the 
positive number, 144. Thus do equations rectify subordinate 
errors, and point out special conditions. 

4. A man when he was married was 30 years old, and his 
wife 15. How many years must elapse before his age will be 
three times the age of his wife? 

Ans, The question is incorrectly enunciated ; ll years before 
the marriage, not after, their ages bore the specified relation. 

5. A man worked 7 days, and had his son with him 3 days ; 
and received for wages 22 shillings. He afterwards worked 5 
days, and had his son with him one day, and received for wages 
18 shillings. What were his daily wages, and the daily wages 
of his son? 

Ans, The father received 4 shillings per day, and paid 2 shil- 
lings for his son's board. 

6. A man worked for a person ten days, having his wife with 
him 8 days, and his son 6 days, and he received $10.30 as com- 
pensation for all three; at another time he wrought 12 days, his 
wife 10 days, and son 4 days, and he received $13.20; at an- 
other time he wrought 15 days, his wife 10 days, and his son 12 
days, at the same rates as before, and he received $13.85. What 
were the daily wages of each 1 

Ans. The husband 75 cts., wife 50 cts. The son 20 cts. ex- 
pense per day. 



EQUATIONS. 97 

7. A man wrought 10 days for his neighbor, his wife 4 days, 
and son 3 days, and received $11.50 ; at another time he served 
9 days, his wife 8 days, and his son 6 days, at the same rates as 
before, and received $12.00 ; a third time he served 7 days, his 
wife 6 days, and his son 4 days, at the same rates as before, and 
he received $9.00. What were the daily wages of each ? 

Ans. Husband's wages,$l. 00; wifeO; son 50 cth. 

8. What fraction is that which becomes | when one is added 
to its- numerator, and becomes |- when 1 is added to its denomi-p' 
nator ? 

Ans. In an arithmetical sense, there is no such fraction. The 
algebraic expression, IZH, will give the required results. 

(Art. 58.) By the aid of algebraical equations, we are enabled 
not only to resolve problems and point out defects or errors in 
their enunciation, as in the last article, but we are also enabled 
to demonstrate theorems, and elucidate many philosophical truths. 
The following are examples : 

Theorem 1. It is required to demonstrate, that the half sum 
plus half the difference of two quantities give the greater of the 
^.wo, and the half sum minus the half difference give the less. 

Let x= the greater number, y= the less, 

»= their sum, d~ their difference. 

Then x-\-y= s {Jl) 

And X — 2/= ^ ■ (^) 

By addition, 2x= s-{- d -^ 

Or x=l8-\-hd 

Subtract [B) from {A) and divide by 2, and we have 
y=ls — hd 

These last two equations, which are manifestly true, demon- 
strate the theorem. 

Theorem 2. Four times the product of any two numbers, is 
equal to the square of their sum, diminished by the square of 
their difference. 

9 



98 ELEMENTS OF ALGEBRA 

Let x= the greater number, and y= the less, as in the last 
theorem. 2x=s -\-d 

And 2i/=s — d 



By multiplication 4xy=s^ — d- a demonstration of th( 
theorem. 

Many other theorems are demonstrable by algebra, but we de 
fer them for the present, as some of them involve quadratic equa- 
tions, which have not yet been investigated ; and we close the 
subject of simple equations by the following quite general prob- 
lem in relation to space, time and motion. 

To present it at first, in the most simple and practical manner, 
let us suppose 

Two couriers, A and B, 100 miles asunder on the same road 
set out to meet each other, A going 6 miles per hour and B 4. 
How many hours must elapse before they meet, and how far 
will each travel? 

Let x= A''s distance, y— B's, and t= the time. 
Then x-\-y=lOO (1) 

As the miles per hour multiplied by the hours must give the 
distance each traveled, therefore, 

x=6t and y=4t (2) 

Substitute these values in equation (1) and 
(6+4y= 100 

Therefore, ^=6^' ^^^ 

A J a, 100X6 .^ 100X4 ,^, 

And ^=6/=^:p4- 2/=4/=-^:j-^ (4) 

From equation (3,) we learn that the time elapsed before the 
couriers met was the whole distance divided by their joint mo- 
tion per hour, a result in perfect accordance with reason. From 
equations (4,) we ])erceive that the distance each must travel is 
the whole distance asunder multiplied by their respective mo- 
tions and divided by the sum of their hourly motions. 

Now let us suppose the couriers start as before, but travel in 
the same direction, the one in pursuit of the other. B having 



EQUATIONS. 99 

100 miles the start, traveling four miles per hour, pursued by 

A, traveling 6 miles per hour. How many hours must elapse 
before they come together, and what distance must each travel? 

Take the same notation as before. 

Then x — 3/ =100 (1.) As »^ must travel 100 miles more than 

B. But equations (2,) that is, x=Gt and y=4:t, are true under 
all circumstances. 

Then (6—4)/= 100 

And ^=J^=50 

6 — 4 

The result in this case is as obvious as an axiom. ^ has 100 
miles to gain, and he gains 2 miles per hour, it will therefore re- 
quire 50 hours. 

But it is the precise form that we wish to observe. It is the 
fact that the given distance divided by the difference of their mo- 
tions gives the time, and their respective distances must be this 
time multiplied by their respective rates of motion. 

Now the smaller the difference between their motions, the 
longer the time before one overtakes the other ; when the differ- 
ence is very small, the time will be very great ; when the differ- 
ence is nothing, the time will be infinitely great ; and this is in 
perfect accordance with reason ; for when they travel equally 
fast one cannot gain on the other, and they can never come to- 
gether. 

If the foremost courier travels faster than the other, they must 
all the while become more and more asunder ; and if they have 
ever been together it was preceding their departure from the 
points designated, and in an opposite direction from the one they 
are traveling, and would be pointed out by a negative result. 

(Art. 59.) Let us now make the problem general. 

Two couriers, A and B, d miles asunder on the same road, 
set out to meet each other; A going a miles per hour, B going 
b miles per hour. How many hours must elapse before they 
meet, and how far will each travel? 



100 ELEMENTS OF ALGEBRA. 

Taking the same notation as in the particular case, 
Let x= A's distance, 1/= jB's, and /= the time. 



Then x-\-y=d (1) 


x=at 


y^ht 


(2) 


Therefore {a^h)t=^ 


Or 


^ «+6 


(3) 


And x=al=^ — r^ 

nA-h 


y^ht= 


hd 
nA-h 


(4) 



If a=b, then x=ld and y=kd. A result perfectly obvious, 
the rates being equal. Each courier must pass over one half 
the distance before meeting. 

If «=0 a?= . — and yz=-—=d. That is, one 

will be at rest, and the other will pass over the whole distance. 

(Art. 60.) Now let us consider the other case, in which one 
courier pursues the other, starting at the same time from dif- 
ferent points. 

Let the line CD represent the space the couriers are asunder 
when the pursuit commences, and the point JE where they come 
together. C D E 

i i j ■ 

The direction from C towards B we call plus, the other direc- 
tion will therefore be minus. ♦ 
Now as in the 2d example, (Art. 58.) 

Put .r = C^=e'?'s distance 
y=DE=B^s distance 
Then x-—y=^Cn=d (1) 

As before, let t= the time. Then 
x=at y=bt (2) 

Therefore at — bt=d 

And t=J^ (3) 

For the distances we have 

"" (4) and y=il, (6) 



a— 6 ^ ^ *^ a — b 



EQUATIONS. 101 

By an examination of these equations, it will be percejve4 l-hat 
X and y will be equal when a is equal io']^' je\>U .still '^.ti^ls as 
a difference between them. This, is irj cons^qinpnpe^Qf ;c find ,y 
in that case being so very great th'^t 4'is, lost';iii'c&i:rfl)^i;i^c<i.; .6f> 
all values are great or small only in comparison with others, or 
with our scale of measure. 

To make this clear, let us suppose two numbers differ by one^ 
and if the numbers are small, the difference may be regarded as 
considerable ; if large, rnore inconsiderable ; if still larger, still 
more inconsiderable, &c. If the numbers or quantities be infin- 
itely great, tlie comparative small quantity may be rejected. 
Thus : 

5 and 6 differ by 1, and their relation is as 1 to 1.2. 

Also, 50 and 51 differ by 1, and their relation is as 1 to 1.02. 
500 to 501 are as 1 to 1.002, <fcc. The relation becoming nearer 
and nearer equality as the numbers become larger, and when the 
numbers become infinitely great the difference is comparatively 
nothing. 

When «=6 a — b—Q and ^=-7r a symbol of infinity. 

If we suppose b greater than «, a — b will become negative, and 
as X and y refer to the same point, that point must then be in the 
backward direction from that we suppose the couriers are mo- 
ving, and will show how far they have traveled since that event. 

If in the equations (3), (4) and (5), J=0, and at the same 

7 , 1 „ , , , , , 

tune a=b, then we shall have t=j'> x=- and y=~ : which 

shows that ^ is a symbol of indetermination, it being equal to 

several quantities at the same time. If d^O the two couriers 
were together at commencement ; and if they travel in the same 
direction, and equally fast, they will be together all the while, 
and the distances represented by x and y will be equal, and of 

all possible values. Hence ~ may be taken of any value what- 

x2 



102 ELEMENTS OF ALGEBRA. 

ever^^and may be made to take a particular valuer to correspond 
to aftg'!otfier cifoMainhie or condition.* 

^*i, 'i :* 5 yi 'K^' '5 i"! « ,> 5 amplication. 

(Art. 61.) We have hitherto considered CD a right line ; but 
the equations would be equally true, if we consider CD to be 
curved, and indeed we can conceive the line CD to wind about 
a perfect circle just forming its circumference, and the point £ 
upon the circle, CE being a little more than one circumference. 

This being understood. Equation (3,) (Art. 60,) gives us a so- 
lution to the following problems. 

1. The hour and minute hn?ids of a clock are together at 12 
6* clock. When are they next together? 

* The 26tli equation (Art. 40), if resolved in the briefest manner, will show 


the influence of the factor ^. In the equation referred to, add 30 to both 

members and divide the numerator of the second member by its denominator, 

and we have }-l=6. Drop l,and divide both members by 5, we then 

a;-|-l 
have — - — =1, or a;-l-l=x-|-2. Hence 1=2, a manifest absurdity. 
a;-l-3 

But all our operations, yea, and all our reasonings have been correct, but 

we did not pay sufficient attention to dividing the numerator by the denomi- 

6 (a;— 2) 

nator, which was —, r^-. Taking G for the quotient, which it would be in 

{X — Z) 

every case except when x — 2=0, leads to the absurdity; which absurdity, 

in turn, shows that x — 2=0, or x=2. 

As another illustration of the influence of this symbol, take the identical 

equation 100=100, or any other similar one. 

This is the same as 96-}-4=96-f-4 

Transposing, 4—4=96—96 

Resolving into factors, 1(4 — 4)=24(4 — 4) 

Dividing by the common factor, and 1=24 ; 

1 

But, to restore equality, — in this case must equal — , or 24. 


Hence we perceive that - is indeterminate, in the abstract, but may be ten* 

dered definite in particular cases. 



EQUATIONS. 103 

By the equation, t= this problem and all others like it 

are already resolved. All we have to do is to determine the 
values of d^ a, and b. 

There are 12 spaces (hours) round the dial plate of a clock ; 
hence d may represent 12. a and b are the relative motions of 
the hands, a moves 12 spaces or entirely round the dial plate 
while b moves one space. Hence a=12, 6 = 1, and a — 6=11. 

Consequently, ^=i-2 = ih. 5m. 27j\s. 

Again, We may demand what time the hour and minute 
hands of a dock are together between 3 and 4. 

From 12 o'clock to past 3 o'clock there are 3 revolutions to 

3X 12 
pass over in place of one, and the solution is therefore t= 

and so on for any other hour. 

2. WJiat time betweeti two and three o^ clock will the hour 
and minute hands of a dock maliC right angles with each other? 

Here the space that the one courier must gain on the other is 
two revolutions and a quarter, or 2ld. 

Hence /=i-f X2| = i-f X J = fl=2h. 27y\m. 

3. WJiat time between 5 and 6 will the two hands of a clock 
make a right line ? 

Here one courier must gain 5| revolutions, or d in the equa- 
tion must be multiplied by 51 = y. 

Hence, t=\\X'i=^\i. 

That is, the hands make a right line at 6 o'clock, a result man- 
ifestly true. 

This simple equation enables us to determine the exact time 
when the two hands of a clock shall be in any given position. 

We may apply this equation to a large circle, as well as to a 
small one ; it may apply to the apparent circular course of the 
heavens, as well as to a dial plate of a clock ; and the application 
is equally simple. 

The circle of the heavens, like all other circles, is divided into 



104 ELEMENTS OF ALGEBRA. 

360 degrees ; and the sun and moon apparently follow each other 
like two couriers round the circle. 

In one day the moon moves on an average 13°. 1764, (divisions 
of the circle,) and the sun apparently 0°. 98565, or not quite one 
division of the circle. The moon's motion being most rapid, 
corresponds to a in the equation, and the sun's apparent motion 
to b. Then «— 6 = 13°.1764— -0.98565=12°.19075 ; and the 
time required for one courier to gain on the other the required 

space, in this case a revolution of 360 degrees, or /= r= 

360 «— ^ 

To~i"o"n^-' which gives 29.5305887 days, or 29 days, 12 hours. 

44 minutes, 3 seconds, which is the mean time from one change 
of the moon to anotlier, called a synodic revolution. 

These relative apparent motions of the sun and moon round 
ihe circular arc of the heavens, are very frequently compared to 
the motions of the hour and minute hands of a clock round the 
dial plate; and from the preceding application of the same equa- 
tion we see how truly. 

We may not only apply this equation to the mean motions of 
the sun and moon, but it is equally applicable to the mean mo- 
tions of any two planets as seen from the sun. To appearance, 
the two planets would be nothing more than two couriers mo- 
ving in a circle, the one in pursuit of the other, and the time be- 
tween two intervals of coming together, (or coming in conjunc- 
tion, as it is commonly expressed,) Avill be invariably represen- 
ted by the equation 

_ d 
a — b 

To apply this to the motion of two planets, we propose the 
following problem : 

The planet Venus, as seen from the siaiy describes an arc of 
1° SQ' per day, and the earth, as seen from the same point, de- 
scribes an arc of 59'. ^t what interval of time will these two 
bodies come in a line with the sun on the same side ? 

Here rt=l°36' ^>=59' J=360° 

Therefore, a — i=37'; and as the denominator is 



EQUATIONS. 105 

minutes, the numerator must be reduced to minutes also ; hence 
the equation becomes » 

360X60 ,„„„j 
t= — -- — =583.8 days, nearly. 

We have not been very minute, as the motions of the planets 
are not perfectly uniform, and the actual interval between succes- 
sive conjunctions is slightly variable. Hence we were not par- 
ticular to take the values of a and b to the utmost fraction. A 
more rigid result would have been 583.92 days. Half of this 
time is the interval that Venus remains a morning and an even- 
ing star. 

(Art. 63.) This equation, as simple as it may appear, is one 
practical illustration of the true spirit and utility of analysis by 
algebra. 

The principles and relations of lime and motion are fixed and 

invariable, and the equation, /== j always represents 

that relation. 

If t can be determined by observation, as it may be in respect 
to the earth and the superior planets, the mean daily motions of 
the planets can be determined ; as f/=360°, a=59' 08" the mean 
motion of the earth, and suppose b the motion of Mars, for ex • 
ample, to be unknown. 

When unknown, represent it by x. 

Then t= or at — tx=d. 



Therefore x= — : — 



106 ELEMENTS OF ALGEBRA. 

SECTION III. 

INVOLUTION. 
CHAPTER L 

(Art. 64.) Equations, and the resolution of problems producing 
equations, do not always result in the first powers of the un- 
known terms, but different powers are frequently involved, and 
therefore it is necessary to investigate methods of resolving equa- 
tions containing higher powers than the first ; and preparatory 
to this we must learn involution and evolution of algebraic 
quantities. 

(Art. 65.) Involution is the method of raising any quantity to 
a given power. Evolution is the reverse of involution, and is 
the method of determining what quantity raised to a proposed 
power will produce a given quantity. 

As in arithmetic, involution is performed by multiplication, and 
evolution by the extraction of roots. 

The first power is the root or quantity itself ; 

The second power, commonly called the square, is the quan- 
tity multiplied by itself; 

The third power is llie product of the second power by the 
quantity ; 

The fourth power is the third power multiplied into the quan- 
tity, &c. 

The second power of a is a X« or a^ 

The third power is a^Xa or a^ 

The fourth power is a^Xa or a^ 

The second power of a^ is rt'*X«'* or c^ 

The third power of ii^ is «^X«'* or a^^ 

The 7Uh power of a'' has the exponent 4 repeated n times, 
or (&*, Therefore, to raise a simple literal quantity to any 
power, multiply its exponent by the index of the required 
power. 

Raise x to the 5th power. The exponent is 1 understood, 
and this 1 multiplied by 5 gives ar" for the 5th power. 





107 


£ns. 


x^^. 


Ans, 


y». 


Am. 


^. 


Ans. 


rpfnn 


Ans. 


(^X\ 


Arts. aWx\ 


Arts, 


c^y". 



INVOLUTION. 

Raise x^ to the 4th power 
Raise y' to the third power. 
Raise x"^ to the 6lh power. 
Raise a;" to the mth power. 
Raise as^ to the 3d power. 
Raise ab^x"^ to the 2d power. 
Raise cy to the 5th power. 

(Art. 66.) By the definition of powers the second power is 
any quantity multiplied by itself; hence the second power of 
ax is a^x^f the second power of the coefficient «, as well as the 
other quantity x ; but a may be a numeral, as 6a?, and its second 
power is 36x^ Hence, to raise any simple quantity to any 
power, raise the numeral coefficient, as in arithmetic, to the 
required power, and annex the powers of the given literal 
quantities. 

EXAMPLES. 

1. Required the 3d power of ^ax^. 

2. Required the 4th power of ^y^. 

3. Required the 3d power of — 2x. 

4. Required the 4th power of — ^x. 

Observe, that by the rules laid down for multiplication, the 
even powers of minus quantities must he plus, and the odd powers 
minus. 

^ . . , 1 . 2r/6« . 4a''b'^ 

5. Required the 2d power oi — — . Mns. 



Ans. 


Ila'x^ 


Ans. 


Wf- 


Ans. 


—Sx'. 


Ans. 


S\x\ 



5c •iSc' 

2rt 64a^ 

6. Required the 6th power of — — . An». ^. . 

oX i 4foX 

a^b 729a^^b^ 

7, Required the 6th power of -3—. Ans. g — . 

■^x X 

(Art. 67.) The powers of compound quantities are raised by 
the mere application of the rule for compound multiplication. 
(Art. 12.) 



108 



ELEMENTS OF ALGEBRA. 



/ 



Let a-\-b be raised to the 2d, 3d, 4th, &c. powers 
a +6 
a+b 





a -{-ab 
ab-\-b^ 


2d power or square 


, a^-\-2ab +b^ 
a-\-b 




a^-\-2a'b-\-ab^ 

a^b-\-2ab^-\-b^ 


3d power or cube, 


a^^^a^b^^ab^-\-¥ 
a-^b 




a'+2a'b^'^a^¥^-ab^ 

c^b-{-'Sa^b^^-^a¥-\-b'' 


The 4th power, 


«'^+4a='6-}-8«262-l-4«^'^-{-6^ 
«+6 




a^-\-Aa'b-\-Qa^b^-\-4d'b^+ab^ 

a'b-\-ia'b^-\-%a^b'-\-Aab'-^b^ 


The 5th power. 


«^-|-5a^6+ \QaW-\-\Oa^b''-^^ab''-\-b^ 



<tc., &c. 

By inspecting the result of each product, we may arrive ai 
general principles, according to which any power of a binomial 
may be expressed, widiout the labor of actual mulliplicatioa. 
This theorem for abbreviating powers, and its general application 
to both powers and roots, first shown by Sir Isaac Newton, has 
given it the name of Newton's binomial, or the binomial theorem. 



OBSERVATIONS. 



Observe the 5th power : «, being the first, is called the leading 
term; and b, the second, is called the following term. The sum. 
of the exponents of the two letters in each and all of the terms 
amount to the index of the power. In the 5th power, the sum 



INVOLUTION. 109 

of the exponents of a and 6 is 5 ; in the 4th power it is 4 ; in 
the 1 0th power it would be 10, &c. In the 2d power there are 
three terras ; in the 3d power there are 4 terms ; in the 4th 
power there are 5 terms ; always one more term than the index 
of the power denotes. 

The 2d letter does not appear in the first term ; the 1st letter 
does not appear in the last term. 

The highest power of the leading term is the index of the 
given power, and the powers of that letter decrease by one from 
term to term. The second letter appears in the 2d term, and its 
exponent increases by one from term to term as the exponent of 
the other letter decreases. 

The 8th power of (a-f-i) is indicated thus: [a-\-h)^. When 
expanded, its literal part, (according to the preceding observa- 
tions) must commence with c^ and the sum of the exponents of 
every term amount to 8, and they will stand thus : a^, d'b, a^b"^, 
a'b\ a'b\ (^b\ a^b\ ab\ b\ 

The coefficients are not so obvious. However, we observe 
that the coefficients of the first and last terms must be unity. 
The coefficients of the terms next to the first and last are equal, 
and the same as the index of the power. The coefficients in- 
crease to the middle of the series, and then decrease in the 
same manner, and it is manifest that there must be some law of 
connection between the exponents and the coefficients. 

By inspecting the 5th power of a+6, we find that the 2d co- 
efficient is 5, and the 3d is 10. 

^=10. 
2 

The third coefficient is the 2d, multiplied by the exponent of 
the leading letter, and divided by the exponent of the second 
letter increased by unity. 

In the same manner, the fourth coefficient is the third multi- 
plied by the exponent of the leading letter, and divided by the 
exponent of the second letter increased by unity ^ and so on from 
coefficient to coefficient. 
K 



fer^ .-.As-.^- L. 



110 ELEMENTS OF ALGEBRA. 

lOX 3 
The 4th coefficient is — - — =10 

3 

The 5th is i^= 5 

4 

5X1 
The last is — ^= 1 understood. 



Now let us expand (^a-\-bf 

For the 1st term write a^ 

For the 2d term write Sa'b 

For the 3d, ^^=28 2Sa'b^ 

For the 4th, ^^ 56a^6' 

o 

For the 5th, ^^ 70a^6^ 

4 

Now as the exponents of a and 6 are equal, we have arrived 
at the middle of the power, and of course to the highest coeffi- 
cient. The coefficients now decrease in the reverse order which 
they increased. 

Hence the expanded power is 

a«-f8«^6+28a«62+56a^6=^4-70a^6^-f56a'^6^+28a26e+8a6"+6^ 
Let the reader observe, that the exponent of 6, increased by 
unity is always equal to the number of terms from the beginning, 
or from the left of the power. Thus, h"^ is in the 3d term, &c. 
Therefore in finding the coefficients we may divide by the num- 
ber of terms already written, in place of the exponents of the 
second term increased by unity. 

If the binomial {a-\-h) becomes (a+1,) that is, when h be- 
comes unity, the 8th power becomes, 

a8_|.8a''-}-28a«-l-56a5+70a*+56a''+28a2+8a+l. 

Any power of 1 is 1, and 1 as a factor never appears. 
If a becomes 1, then the expanded power becomes, 

l+86+2862+5663+706^+5665+286«4-86^+6« 



INVOLUTION. HI 

The manner of arriving at these results is to represent the unit 
by a letter, and expand the simple literal terms^ and afterwards 
substitute their values in the result. 

(Art. 68.) If we expand [a — b) in place of {a-[-b,) the expo- 
nents and coefficients will be precisely the same, but the princi- 
ples of multiplication of quantities affected by different signs 
will give the minus sign to the second and to every alternate term. 

Thus the 6th power of {a — b) is 

a^—Ga^b + 1 5a'b''—20a'b''+ 1 5a''b^—Gab^+b\ 

(Art. 69.) This method of readily expanding the powers of a 
binomial quantity is one application of the " binomial theorem^^* 
and it was thus by induction and by observations on the result 
of particular cases that the theorem was established. Its rigid 
demonstration is somewhat difficult, but its application is simple 
and useful. 

Its most general form may arise from expanding (a-j-i)". 

When n=3, we can readily expand it; 

AVhen n=4, we can expand it ; 

When n= any whole positive number, we can expand it. 

Now let us operate with n just as we would with a known 
number, and we shall have 

{a-\-b)"=a^-\-na''~'b+n^^a''-^b\ &lc. 

We know not where the series would terminate until we 
know the value of n. We are convinced of the truth of the 
result when n represents any positive whole number; but let 7i 
be negative or fractional, and we are not so sure of the result. 
To extend it to such cases requires deeper investigation and 
rigid demonstration,which it would not be proper to go into at 
this time. We shall therefore content ourselves by some of its 
more simple applications. 

EXAMPLES. 

1. Required the third power of iix-\-2y. 

We cannot well expand this by the binomial theorem, because 
the terms are not simple literal quantities. But we can assume 
3a?=« and 2y==b. Then 

'6x-\-2y=a^b and (a-\-bf=a^-\.'Sa^b-^^a¥-\-b'^ 



112 



ELEMENTS OF ALGEBRA. 



Now to return to the values of a and 6, we have, 

3«2^=3 X 9a;2x 1y =54xy 
- 3a62=3 X 3ir X 43/2=36:ry . 

Hence (32;-l-2i/)='=27ar^+54a?2y-}-36;ri/2-|-'8^~ 

2. Required the 4th power of 2a^ — 3. 

Let x=2a^ 2/= 3* Then expand {x — yY^ and return the 
values of x and y, and we shall find the result, 

16a«— 96a6+216a*— 216a2+81. 

3. Required the cube of {a-\-b-\-c-\-d). 

As we can operate in this summary manner only on binomial 
quantities, we represent a-\-b by x, or assume x=a-\-by and 
yz=c-{-d. 

Then (a?4-»/)'=a;=^+3x2y-{-3;n/2-|-2/3. 

Returning the values of x and y, we have 

{a^bf-\-^[a-]rb)\c+d)-\-^a-\-h){c-\-df-\-{c-\-df. 

Now we can expand by the binomial, these quantities con- 
tained in parenthesis. 

4. Required the 4th power of 2a-\-^x. 

Ans. 16a^-i-96a3ic4-216«2a;2+216aa'3+8Ij:*. 

5. Expand (ic^+Si/^)^ 

Ans, a;^°-i-15j^i/2^90a?y+270a:y-l-405a;y+243y'». 

6. Expand {2a^-\-axf Jlns. 8a«+12a^a:4-6aV+a'ar* 



K* Expand (a; — 1)^ 
Ans. 



8. Expand (3iP— 5)^ 

9. Expand {a-\-2)\ 



10. Expand (1— da)^ 

.11. Expand {a-{-b-\-c)\ 

*2. Expand {a—2bf. 

18. Expand (1— 2a;)^ 



Am. 27x^— 135x2+2250:2— 125 
Arts. a*4-8a'-i-24a2-f32a+16. 
, „ , 3rt2 a% a* 



I 



EVOLUTION. 113 

EVOLUTION. 

CHAPTER n. 

(Art. 70). Evolution is the converse of involution, or the ex 
traction of roots, and the main principle is to observe how powers 
are formed, to be able to trace the operations back. Thus, to 
square «, we double its exponent, (Art. 65), and make it a^. 
Square this and we have «■*. Cube a^ and we have a^. Take 
the 4th power of x and we have x'*. The nth power of x^ is 
x^\ 

Now, if multiplying exponents raises simple literal quantities 
to powers, dividing exponents must extract roots. Thus, the 

2 

square root of a* is a^. The cube root of a^ must be a^ . The 
cube root of « must have its exponent, (1 understood,) divided by 

3, which will make a^* 

Therefore roots are properly expressed by fractional expo- 
nents. 

The square root of a is a*, and the exponents, 5, l, ^, &e. in- 
dicate the third, fourth, and fifth roots. The 6th root of a^ is 

5. 
x^; hence we perceive that the numerators of the exponent in- 
dicate the power of the quantity, and the denominator the root 
of that power. 

(Art. 71.) The square of ax is a^x^. We square both 
factors, and so, for any other powers, we raise all the factors to 
the required power. Conversely, then, we extract roots by 
taking the required roots of all the factors. Thus the cube root 
of Sx^ is 2x. 

The cube root of the factor 8 is 2, and of the factor x^ is x. 
The cube root of IQa^ cannot be expressed in a rational quantity, 
but it can be separated into factors, 8a''X2, and the cube root of 
the first factor can be taken, and the index of the root put over 
the other factor thus, 2aX2^, or 2ay2. In such cases, the 
radical sign is usually preferred to the fractional index, as making 
a more distinct separation between the factors. 
10 




114 ELEMENTS OF ALGEBRA. 

The square root of 64a^ is obviously 8a^, and from this and 
the preceding examples we draw the following 

Rule. For the extraction of the roots of monomials. Ex- 
tract the root of the numeral coefficients and divide the exponent 
of each letter by the index of the root, 

EXAMPLES. 

1. What is the square root of 49a^a;'' ? • Ans. laoc^, 

2. What is the square root of ^bc^W ? Ans. 5c^b. 

3. What is the square root of 20axl Ans. 2^bax. 

In 20, the square factor 4 can be taken out ; the other factor 
is 5. The square root of 4 is 2, which is all the root we can 
take ; the root of the other factors can only be indicated as in the 
answer. 

4. What is the square root of 12a^ ? Ans. 2a J3. 

5. What is the square root of 144«Va^i/^? Ans. I2ac^xy. 

6. What is the square root of 36^'* ? Ans. zbGx^. 

(Art. 72.) The square root of algebraic quantities may be 
taken with the double sign, as indicating either plus or minus^ 
for either quantity will give the same square, and we may not 
know which of them produced the power. For example, the 
square root of 16 may be either +4 or — 4, for either of them, 
when multiplied by itself, will produce 16. 

The cube root of ajo/ws quantity is always plus, and the cube 
root of a minus quantity is always minus. For -\-2a cubed 
gives +8a'', and — 2a cubed gives — 8a^, and a may represent 
any quantity whatever. 

EXAMPLES. 

1. What is the cube root of 125a^ ? Ans. 5a. 

2. What is the cube root of — 64ic« ? Ans. — 4a^. 

3. What is the cube root of -—21 6ay? Ans. —Qax^. 

4. What is the cube root of 729aV2 ? Ans. 9aV. 

5. What is the cube root of 32a^ ? Ans. 2alj4aF. 



EVOLUTION. 115 

6. What is the 4th root of 256a^x«? *^ns. zhiax!', 

7. What is the 4th root of 16ff ? ^ns. ±20*. 



8. What is the 4th root of 64xYl Ans. ±:^8xy, 
N. B. The 4th root is the square root of the square root. 

9. What is the 4th root of 20axl Ans. db(20)^a^a;^. 

10. What is the square root of 75 ? Ans. ±5^3^ 

75=25X3. 

11. Required the square root of \ . Ans. i-^-* 

32a^a?^ Aaa? 

12. Required the square root of -— . Ans. z^—~-. 

\%ax 3 

N. B. Reduce the fraction as much as possible, and then ex- 
tract the root. 

13. Required the square root of ,^^ . Ans. rh— r. 

128a 4 

14. Required the nth root of -^-— . Ans. ~t— 

3 -1 -1 



15. Required the wth root of j-. Ans. «« 6** c" 

1 



Observing that —=b~^c~^ 



CHAPTER ni. 
To extract roots of compound quantities. 

(Art. 72.) We shall commence this investigation by confining 
our attention to square root, and the only principle to guide us is 
the law of formation of squares. The square of a-\-b is a^-j- 
2ab-\-b'^. Now on the supposition that we do not know that the 
root of these terms is a-\-b, we are to find it or extract it out of 
the square 

a=-4-2a5+6^ 

We know that a\ the first teim, must have been formed by the 
multiplication of a into itself, and the next term is 2aXb. That 



116 ELEMENTS OF ALGEBRA. 

is twice the root of the first term into the second term of the root. 
Hence if we divide the second term of the square by twice the 
root of the first term, we shall obtain b, the second term of the 
root, and as b must be multiplied into itself to form a square, we 
add b to 2a, and have 2«4~^» which we call a divisor. 

OPERATION. 

a'-\-2ab-\-b%a+b. 



2a-]-b)2ab-\-b^ 
2ab-]-b^ 

We take a for the first term of the root, and subtract its square 
(«^) from the whole square. We then double a and divide 2ab 
by 2a and we find b, which we place in both the divisor and quo- 
tient. Then we multiply 2«+6 by b and we have 2ab-\-b^f to 
subtract from the two remaining terras of the square, and in this 
case nothing remains. 

Again, let us take a-\-b-\-c, and square it. We shall find its 
square to be 

a^-\-2ab-i-b''+2ac-\-2bc-\'d', 
a'-{-2ab-i-b^-^2ac-{-2bc-{-c\a+b-{-c 



2a-\-b 2ab-\-b^ 
2ab-\-b^ 



2a+2b-{-c 2ac4-26c+c2 

2ac-\-2bc-\-d' 



By operating as before, we find the first two terms of the root 
to be a-\-b, and a remainder of 2ac-\-2bc-{-c^. Double the root 
already found, and we have 2a-\-2b for a partial divisor. Divide 
the first term of the remainder 2ac by 2a, and we have c for the 
third term of the root, which must be added to 2a-\-2b to com- 
plete the divisor. Multiply the divisor by the last term of the 
root and set the product under the three terms last brought down, 
and we have no remainder. 



EVOLUTION. 117 

Again, let us take a+^-f-c to square; but before we square it 
let the single letter s=a-\-b. 

Then we shall have s-\-c to square, which produces 
«^+2sc+c^. To take the square root of this we repeat the first 
operation, and thus the root of any quantity can be brought into 
a binomial, and the rule for a binomial root will answer for a root 
containing any number of terms by considering the root already 
found, however great, as one term. 

Hence the following rule to extract the square root of a com- 
pound quantity. 

Arrange the terms according to the powers of some letter, 
beginning with the highest, and set the square root of the first 
term in the quotient. 

Subtract the square of the root thus found from, the first 
term and bring down the next two terms for a dividend. 

Divide the first term of the dividend by double of the root 
already found, and set the result both in the root and in the 
divisor. 

Multiply the divisor, thus completed, by the term of the root 
last found, and subtract the product fromthe dividend, and so on, 

EXAMPLES. 
1. What is the square root of 

a*-}-4a26+462— 4a2— 86-l-4(a2-f-26— 2 



2a2-f26 )ia^b-\-4:b'^ 
4a^b-\-4b^ 



2a2-{-46__2 — 4a2— 8&+4 

— .4a2__86+4 



2. What is the square root of 1 — 4b-\-4b^-]-2y — iby-\-y^l 

Ans. 1 — 2b-\-y, 

3. What is the square root of ^x^—ix^-^-l^x"^ — 6a;+9 ? 

Ans. 2x2—3?+ 3. 

4. What is the square root of 4a^^— 16x''+24a;2— 16a:+4 ? 

Ans. 2a;2— 4a?+2. 



1 18 ELEMENTS OF ALGEBRA. 

5. What is the square root of l6x^-\-24x^-{-89x^-{-60x-^l001 

6. What is the square root of 4a:^ — 16a;'4- 8^^*4-160;+ 4? 

Jins. 2x^ — 4a;— 2. 

7. What is the square root of 

x^-{-2xy-\-y''-\-Qxz-\-Qyz~{-Qz''1 Ans. x-\-y-^Zz, 

8. What is the square root of d^ — ab-\-\¥ ? 



Ans. a — lb. 



9. What is the square root of 







Ans. 


a b b a 

or J. 

b a a 



10. What is the square root of a?-^ — 2x^y^-{-y^ 1 

k i. 11 

Arts, x^ — y^ or y^ — x'\ 

(Art. 73.) Every square root will be equally a root if we change 
the sign of all the terms. In the first example, for instance, the 
root may be taken — c^ — 2&+2, as well as a^-1-26 — 2, for either 
one of these quantities, by squaring, will produce the given 
square. Also, observe that every square consisting of three 
terms only, has a binomial root. 

(Art. 74.) Algebraic squares may be taken for formulas, cor- 
responding to numeral squares, and their roots may be extracted 
in the same way, and by the same rule. 

For example, a-\-b squared is a^+2a6-}-6^ and to apply this 
to numerals, suppose «=40 and 6=7. 

Then the square of 40 is u^—lQQ{i 

2ab= 560 
b^= 49 



Therefore, (47)^=2209 

Now the necessary divisions of this square number, 2209, are 
not visible, and the chief difficulty in discovering the root is to 
make these separations. 



EVOLUTION. 119 

The first observation to malie is that the square of 10 is 100, 
of 100 is 10000, and so on. Hence, the square root of any 
square number less than 100 consists of one figure, and of any 
square number over 100 and less than 10000 of two figures, and 
so on. Every two places in a power demanding one place in 
its root. 

Hence, to find the number of places or figures in a root, we 
must separate the power into periods of two figures, beginning 
at the uniVs place. For example, let us require the square root 
of 22*09. Here are two periods indicating two places in the 
root, corresponding to tens and units. The greatest square in 22 
is 16, its root is 4, or 4 tens =40. Hence a=40. 

22 09(40+7^=47 
a^= 16 00 



2a-i-6=80-l-7=87 )609 
609 



Then 2a=80, which we use as a divisor for 609, and find it 
is contained 7 times. The 7 is taken as the value of b, and 
2a+&, the complete divisor, is 87, which multiplied by 7 gives 
the two last terms of the binomial square. 2ab-\-b^=fiQ(i-\-AQ 
=609, and the entire root 40+7=47 is found. 

Arithmetically, a may be taken as 4 in place of 40, and 1600 
as 16, the place occupied by the 16 makes it 16 hundred, and 
the ciphers are superfluous. Also, 2a may be considered 8 in 
place of 80, and 8 in 60 (not in 609) is contained 7 times, <fec. 

If the square consists of more than two periods, treat it as two, 
and obtain the two superior figures of the root, and when obtain- 
ed bring down another period to the remainder, and consider the 
root already obtained as one quantity, or one figure. 

For another example, let the square root of 399424 be ex- 
tracted. 



39-94-24 II 632 
36 



123 
1262 



394 
369 



25 24 
25 24 



120 ELEMENTS OF ALGEBRA. 

Ill this example, if we disregard the local value of the figures, 
we have «=6, 2a=12, and 12 in 39, 3 times, which gives b=S. 
Afterwards we suppose «=63, and 2a=126, 126 in 252, 2 times, 
or the second value of 6=2. In the same manner, we would 
repeat the formula of a binomial square as many times as we 
have periods. 

EXERCISES FOR PRACTICE. 

1. What is the square root of 8836? ^ns. 94. 

2. What is the square root of 106929? ^ns. 327. 

3. What is the square root of 4782969? £ns, 2187. 

4. What is the square root of 43046721? ^ns, 6561. 

5. What is the square root of 387420489? ^ns. 19683. 

When tliere are whole numbers and decimals, point off periods 
both ways from the decimal point, and make the decimal places 
even, by annexing ciphers when necessary, extending the deci- 
mal as far as desired. When there are decimals only, commence 
pointing off from the decimal point. 

EXAMPLES. 

1. What is the square root of 10-4976? ^ns, 3-24. 

2. What is the square root of 3271-4207? ^ns. 57-19+. 

3. What is the square root of 4795-25731 ? ^ns. 69-247+. 

4. What is the square root of -0036 ? Jins, -06. 

5. What is the square root of -00032754 ? .^ns. -01809+ 

6. What is the square root of -00103041? £ns. -0321. 
(Art. 75.) As the square of any quantity is the quantity mul- 

tiplied by itself, and the product of t by t- (Art. 64.) is -j^ ; 

hence to take the square root of a fraction we must extract the 
square root of both numerator and denominator, 

A fraction may be equal to a square, and the terms, as given, 
not square numbers ; such may be reduced to square numbers. 



f 



EVOLUTION. l^J^ 

EXAMPLES. 

What is the square root of //j ? 

Observe J,fj=zl^. Hence the square root is |. 

1. What is the square root of y^g ? ^ns. J. 

2i What is the square root of \\^1 Ans. |. 

3. What is the square root of l\\\ 1 Ans, |. 

4. What is the square root of f||j? Ans, |. 

When tlie given fractions cannot be reduced to square terms, 
reduce the value to a decimal, and extract the root, as in the last 
article. 

CHAPTER IV. 

To extract the cube root of compound quantities, 

(Art. 76.) We may extract the cube root in a similar manner 
as the square root, by dissecting or retracing the combination of 
terras in the formation of a binomial cube. 

The cube of a-{-h is a''-^^a^b-\-^ab^-\-b^ (Art. 67). Now 
to extract the root, it is evident we must take the root of the 
first term (a'*), and the next term is 3a^6. Three times the square 
of the first letter or term of the root multiplied by the 2d term 
of the root. 

Therefore to find this second term of the root we must divide 
the second term of the power (3a^6) by three times the square 
of the root already found («). 

da^)Sa'b{h 
Sa^b 



When we can decide the value of 6, we may obtain the com- 
plete divisor for the remainder after the cube of the first term is 
subtracted, thus : 

The remainder is 3a^b-\-dab^-\-b^ 

Take out the factor 6, and Sa^-\-2ab-\-b^ is the complete 

divisor for the remainder. But this divisor contains b, the very 

term we wish to find by means of the divisor ; hence it must be 

found before the divisor can be completed. In distinct algebraic 

11 



122 ELEMENTS OF ALGEBRA. 

quantities there can be no difficulty, as the terras stand separate, 
and we find b by dividing simply 3a^6 by 3a^ ; but in numbers 
the terras are mingled together, and h can only be found by triaL 

Again, the terras 3a2-}-3a6-|-62 explain the common arithme- 
tical rule, as 3a^ stands in the place of hundreds, it corresponds 
with the words : " Multiply the square of the quotient by 300," 
"and the quotient by 30," (3a,) &c. 

By inspecting the various powers of a-\-b, (Art. 67,) we draw 
the following general rule for the extraction of roots : 

Arrange the terms according to the poivers of some letter; 
take the required root of the first term and place it in the quo- 
tient : subtract its corresponding power from the first term, and 
bring down the second term for a dividend. 

Divide this term by tivice the root ^ke^-^j found for the square 
root, three times the square of it for the cube root, four times 
the third power for the fourth root, &c., and the quotient will 
be the next term of the root. Involve the whole of the root, 
thus found, to its proper power, which subtract from the given 
quantity, and divide the first term of the re7nainder by the same 
divisor as before: proceed in this manner till the whole root is 
determined. 

EXAMPLES. 

1. What is the cube root of x^-i-6x^--i09ir^-{-9Qx—64 ? 
a-G4-6a?s— 40a:^4-96:r— 64 (r^-l-2a:— 4 



Divisor 3x*) 6x^=lst remainder. 

x^-^6x^-{-l2x^-]-Sx^ ={x^-]-2xy 
Divisor 3a:''* ) — 12x''=2d remainder. 

x^-{-6x^ — 40a?='+96.T— 64 

2. What is the cube root of 270^+1 08a2-]-144«-|- 64 ? 

, Jins. 3a+4. 

8. What is the cube root of a'— 6alr-}-12aa?'-— Sa:^ ? 



EVOLUTION. 123 

4. What is the cube root of x^'—Sx^-\-5a^—3x--^l 1 

Ans. s? — X — 1.* 

5. What is the cube root of a^— 6a'6+2a62— SS^ ? 

Ans, a — 2b, 

3 1 

6. What is the cube Toot of 3?-{-3x-\ \ — -3 ? 

X X 



Ans. x-\ — . 

X 



7. Extract the fourth root of 

«4^8a3+24a2-|-32a4-] 6(a-l-2 



ic^) 80^, &c. 



a^-{-8a3+24a2+32a+16. 



(Art. 77.) To apply this general rule to the extraction of the 
cube root of numbers, we must first observe that the cube of 10 
is 1000, of 100 is. 1000000, &c.; ten times the root producing 
1000 times the power, or one cipher in the root producing 3 in 
the power ; hence any cube within 3 places of figures can have 
only one in its root, any cube within 6 places can have only two 
places in its root, &c. Therefore we must divide off the given 
power into periods consisting of three places, commencing at the 
unit. If the power contains decimals, commence at the unit 
place, and count three places each way, and the number of pe- 
riods will indicate the number of figures in the root. 

EXAMPLES. 

1. Required the cube root of 12812904. 
12-812-904(234 
a=2 a^= 8 



Divisor 3^2=12 )48 



12167 = (23)3 
3(23)2= 1587) 6459'~(4 



12 812 904=(234)» 



Jins, 


53. 


Ans. 


83. 


Arts. 


111. 


Ans. 


127. 


Ans. 


255. 


Ans. 


354. 


Ans. 


465. 



124 ELEMENTS OF ALGEBRA. 

Here 12 is contained in 48, 4 times; but it must be remem- 
bered that 12 is only a trial or partial divisor; when completed 
it will exceed 12, and of course the next figure of the root can- 
not exceed 3. 

The first figure in the root was 2. Then we assumed a— 2. 
Afterwards we found the next figure must be 3. Then we as- 
sumed a=23. To have found a succeeding figure, had there 
been a remainder, we should have assumed a=234, &;c., and 
from it obtained a new partial divisor. 

2. What is the cube root of 148877? 

3. What is the cube root of 571787? 

4. What is the cube root of 1367631 ? 

5. What is the cube root of 2048383 ? 

6. What is the cube root of 16581375 ? 
•y. What is the cube root of 44361864? 
8. What is the cube root of 100544625 ? 

(Art. 78.) The methods of direct extraction of the cube root 
of such numbers as have surd roots, are all too tedious to be 
much used, and several eminent mathematicians have given more 
brief and practical methods of approximation. * 

One of the most useful methods may be investigated as follows : 

Suppose a and a+c two cube roots, c being very small in 
relation to a. a^ and c^-\-^a^c-\-'dac'^-]r (^ are the cubes of the 
supposed roots. 

Now if we double the first cube («'), and add it to the second, 
we shall have 3<e-lr^a\+Za<^+,?. 

If we double the second cube and add it to the first, we shall 

^^""^ 3a»+6«2c+ 6ac^+2c». 

As c is a very small fraction compared to a, the terms con- 
taining c^ and c^ are very small in relation to the others, and 
the relation of these two sums will not be materially changed by 
rejecting those terms containing c^ and cS and the sums will 

*^^" ^® 3a'+3a^c 

And 30^'+ Sa^c 



EVOLUTION. 125 

The ratio of these terms is the same as the ratio of a-{-c to 

a+2c. 

c 



Or the ratio is 1- 



a-\-c 



(* 

But the ratio of the roots a to a-j-c, is 1 -| — . 

a 

Observing again that c is supposed to be very small in rela- 

c c 

tion to cr, the fractional parts of the ratios — ; — and - are both 
^ a-{-c ■ a 

small and very near in value to each other. Hence we have 

found an operation on two cubes which are near each other in 

magnitude, that will give a proportion very near in proportion to 

their roots ; and by knowing the root of one of the cubes, by this 

ratio we can find the otlier. 

For example, let it be required to find the cube root of 28, true 
to 4 or 5 places of decimals. As we wish to find the cube root 
of 28, we may assume that 28 is a cube. 27 is a cube near in 
value to 28, and the root of 27 we know to be 3. 

Hence a, in our investigation, corresponds to 3 in this exam- 
ple, and c is unknown; but the cube of a-\-c is 28, and a' 
is 27. 

Then 27 28 

2 2 



54 56 

Add 28 27 



Sums 82 : 83 : : 3 : a-\-c very nearly. 

Or (a-fo) = yj> =3-03658-1-, which is the cube root of 28, 
true to 5 places of decimals. 

By the laws of proportion, which we hope more fully to in- 
vestigate in a subsequent part of this work, the above propor- 
tion, 82 : 83 : ; a : a-]-c, may take this change : 
82 : I :: a : c 

Hence, c=/2. c being a correction to the known root, 
which, being applied, will give the unknown or sought root. 

From what precedes, we may draw the following rule for find- 
ing approximate cube roots : 
l2 



126 ELEMENTS OF ALGEBRA. 

Rule. Take the nearest rational cube to the given number , 
and call it an assumed cube ; or, assume a root to the given 
number and cube it. Double the assumed cube and add the 
given number to it ; also^ double the given number and add 
the assumed cube to it. Then, by proportion, as the first sum 
is to the second, so is the known root to the required root. Or 
take the difference of these sums, then say, as double of the 
assumed cube, added to the number, is to this difference, so is 
the assumed root to a correction. 

This correction, added to or subtracted from the assumed root, 
as the case may require, will give the cube root very nearly. 

By repeating the operation with the root last found as an as- 
sumed root, we may obtain results to any degree of exactness ; 
one operation, however, is generally sufficient. 

EXAMPLES. 
!• What is the approximate cube root of 120 1 

Ans. 4-93242+. 

2. What is the approximate cube root of 8-5 ? 

Jlns. 2-0408-1-. 

3. What is the approximate cube root of 63 ? 

Ans. 3-97905+. 

4. What is the approximate cube root of 515 ? 

Ans. 8-01559+, 

5. What is the approximate cube root of 16? 

The cube root of 8 is 2, and of 27 is 3 ; therefore the cube 
root of 16 is between 2 and 3. Suppose it 2-5. The cube of 
this root is 15*625, which shows that the cube root of 16 is a 
little more than 2*5, and by the rule 



31-25 


82 






16 


15-625 


: : 2-5 : to the reqi 




47-25 : 


47-625 


lired root. 


47-25 ; 


•375 


:: 2-5 : -01984 








Assumed root 


2-50000 






Correction 


•01984 




Approximate root 


2-51984. 



EVOLUTION. 127 

We give the last as an example to be followed in most cases 
where the root is about midway between two integer numbers. 

This rule may be used with advantage to extract the root of 
perfect cubes, when the powers are very large. 

EXAMPLE. 

The number 22*069-810-125 is a cube; required its root. 

Dividing this cube into periods, we find that the root must 
contain 4 figures, and the superior period is 22, and the cube root 
of 22 is near 3, and of course the whole root near3000; but less 
than 3000. Suppose it 2800, and cube this number. The 
cube is 21952000000, which being less than the given number, 
shows that our assumed root is not large enough. 

To apply the rule, it will be sufficient to take six superior 
figures of the given and assumed cubes. Then by the rule. 



219 520 
2 


220698 
2 


: 2800 
; : 2800 

5 

Assumed root. 
Correction, 




439040 
220698 


441396 
219520 




659738 


: 660916 : 
659738 




659738 


: 1178 : 
2800 






942400 
23i6 






669738)3298400( 
3298690 








2800 
5 




True root, 


2805 



The result of the last proportion is not exactly 5, as will be 
seen by inspecting the work ; the slight imperfection arises from 
the rule being approximate, not perfect. 

When we have cubes, however, we can always decide the unit 
figure by inspection, and, in the present example, the unit figure 






128 ELEMENTS OF ALGEBRA. 

in the cube being 5, the unit figure in the root must be 5, as no 
other figure when cubed will give 5 in the place of units. 

[For several other abbreviations and expedients in extracting 
cube root in numerals, see Robinson's Arithmetic] 

(Art. 79.) To obtain the 4th root, we may extract the square root 
of the square root. To obtain the 6th root, we may take the 
square root first, and then the cube root of that quantity. 

To extract odd roots of high powers in numeral quantities is 
very tedious and of no practical utility ; we therefore give no ex- 
amples. 

(Art. 80.) Roots of quantities may be merely expressed by 

radical signs. For example, the cube root of 16 may be ex- 

1 

pressed thus : ^16, or 16'^. If a cube factor is under the sign, 

that factor may be taken out by putting its root as a multiplier 

without the sign. In this example 16 has the cube factor 8, and 

3^16==^ ^8X2 =2 ^^2^ That is, twice the cube root of 2 is equal 

to the cube root of 16. Hence if the root of 2 is known, the 

root of 16 is equally known. The cube root of 40 is 2^40= 

y8X5-2 yj. 

In the same manner we may express the square root of any 
numbers. Thus, the square root of 18 is ^18 = ;y9X2=3^2. 
The square root of 24 is 2^6. 

Observe that we pick out the square or cube factors, as the 
case may be, and extract the root of such factors, placing the 
root without the sign. Of course the sign must remain over 
that factor whose root cannot be extracted. 

We give the following examples for practice : 

1. Reduce the square root of 75 to lower terms, or reduce 
J75. Ans. 5 J 3. 

2. Reduce ^IJSa^ to lower terms. .^ns. 7aj2'. 

3. Reduce Jl'2x^y to lower terms. ^dns, 2xj3y. 

4. Reduce ^JbAx'^ to lower terms. Ans. 3a? 1^23:. 

5. Reduce 4 ^T08 to lower terms. Ans. 12 ^jT, 

6. Reduce Jl^^a^x^ to lower terms. Ans, xjx — c^. 



EVOLUTION. 129 

7. Reduce ^J^2u? to lower terms. Jins. 2a V^- 



8. Reduce ^28«V to lower terms. ^ns. 2axjla, 

9. Reduce J^^ to lower terms. 

Where terms under the radical are fractional, it is expedient 
to reduce the denominator to a power corresponding to the radi- 
cal sign ; then by extracting the root there will be no fraction 
under the radical. 

The above example may be treated thus : 



We divided 4f into the factors /j and y ; the first factor is a 
square; the other factor, y, we multiply both numerator and 
denominator by 3, to make the denominator a square. 

In like manner reduce the following : 

10. Reduce ''^'^ to more simple terms. Ans. VJIQ* 

11. Reduce ^^Y ^^ "^^^^ simple terms. Ans. j V^* 

12. Reduce ^t^ ^^ "^^^® simple terms. Ans, i^^jQ* »5 



13. Reduce Jd'^-{-o^f)^ to more simple terms. 

Ans. a Vl+^. 

14. Reduce ^f to more simple terms. Ans. ^jQa. 

(Art. 81.) Radical quantities may be put into one sum, or the 
difference of two may be determined, provided the parts essen- 
tially radical are the same. 

Thus the sum of ^8 and ^72 is 8^2 and their 
difference is 4^2 

For 78=2^2 

And 772=672 

Sum 872 

Difference 4^2 

When radical quantities are not and cannot be reduced to the 
9 



139 ELEMENTS OF ALGEBRA. 

same quantity under the sign, their sum and difference can only 
be taken by the signs plus and minus. 

EXAMPLES. 



1. Find the sura and difference of Ji6a^x and J4a^x, 

Ans. Sum, Qajx ; difference, 2a Jx, 

J8. Find the sum and difference of ^128 and ^72. 

Ans. Sum, 14^2 ; difference, 2^2? 

S. Find the sum and difference of ^^JlS^ and ^^40. 

Ans. Sum, 5 ^^5 ; difference, ^^5. 

4. Find the sum and difference of ^^108 and 9'^^4. 

Ans, Sum, 12 ^^4; difference, 6 ^^4. 

5. Find the sum and difference of ^| and ^|. 

Ans. Sam, f ^2 ; difference, J^2. 

6. Find the sum and difference of ^^56 and ^<yi89. 

j9?is. Sum, 5 ^^7 ; difference, ^J7. 



■y. Find the sum and difference of S^arb and S^lGa'^b. 

Ans. Sum, {V2a"-^^a) jl \ difference, (12«-— 3a)76T 

(Art. 82.) We multiply letters together by writing them one 
after another, as abxy. If they are numeral quantities, their 
product appears as a number; if two or more of them are nu- 
meral, the product of these quantities will appear as a number. 

This fundamental principle of multiplication may be applied to 
the multiplication of surds. Let it be required to multiply 
5^2 by 3^7. Here suppose «=5, Z>=3, x=^2, y=»j7. 
Then the product of 5J2~ by d^Y is abxy or 1572X7= 
15^14. Hence, for the multiplication of quantities affected by 
the same radical sign, we draw the following 

Rule. Multiply the rational parts together for the rational 
part of the prodttct, and the radical parts together for the 
radical part of the product. 



EVOLUTION. 131 

EXAMPLES. 

1. Required the product of 5^5 and 3^8. 

Product reduced. Ans. 30^10 

2. Required the product of 4^l2"and 3^27 JJns. 2i Jq 

3. Required the product of 3 72" and 2^8] Ans. 24. 

4. Required the product of 2''^T4 and 3 ^47 

Ans. 12 yY. 

5. Required the product of 2^5"and 2^To. Ajis. 20^2. 

(Art. 83.) When two quantities are affected by different radi 
cal signs, tlieir product can only be indicated, unless we first re 
duce them to the same root. 

The product of ^Jct into ^^b can only be indicated thus, 
JaX^^b, unless we reduce them to the same root by means 
of their fractional indices, thus : 

a''=a^ b'=b^ , 

Here it is obvious that a and b may appear under the same 
root, ^ or ^ ^, if we take a, 3d power, in place of a ; and 6, 2d 
power, in place of b. 

Therefore the product of a^ into b^ is {a^b^)'^. 

Hence the following more general rule to multiply radical 
quantities together : 

Reduce the, surds to the same root, if necessary ; then mul- 
tiply the rational quantities together, and the surds together ; 
then annex the one product to the other for the whole product : 
which may be reduced to more simple terms if necessary. 

EXAMPLES. 

1 s 

1. Required the product of {a-^-by and {a-\-b)^ . 

Ans. (flr-fi)^^ 

2. Required the product of ^7 and ^^1. Ant, (7)^ 



132 ELEMENTS OF ALGEBRA. 

3. Required the product of 2^3 and 3 ^^4. 

^ns. 6V432I 

4. Required the product of ^Jl5 and ^10. 

^m. V225000. 

(Art. 84.) Division, being the converse of multiplication, one 
operation will point out the other, and without further comment, 
we may give the following rule for the division of radicals : 

Rule. Reduce the surds to the same root, when necessary ; 
and divide the rational part of the dividend by the rational 
part of the divisor, and the surd part of the dividend by the 
surd part of the divisor, and annex the quotients Jor the whole 
quotient ; which may be reduced if necessary, 

EXAMPLES. 

1. Divide 4^50 by 2^57 Ans, 2jTo, 

2. Divide 6 VTOO by 3 V"5". Ans. 2 ^20^ 

3. Divide ^T by ^jY. 

Let a=7. Then the example is, to divide the a^ by a^, or 

3 2 1 

a^ by a^ ; quotient a°. As we always subtract the exponents 

of like quantities to perforni division (Art. 17) ; therefore 7^ must 
be the required quotient. 

4. Divide 6^54 by 372'. J9ns. QJs'. 

5. Divide {aWd^y by d^. Ans, {abf. 

6. Divide [IQa'-^l'la^xf by 2«. Ans, {4.a—2x)K 
(Art. 85.) In the course of algebraical investigations, we might 

fall on the square root of a minus quantity, as J — a, J — I, 
^ — b, <fec., and it is important that the pupil should readily 
understand that such quantities have no real existence ; for no 
quantity, either plus or minus, multiplied by itself will give 
minus a, minus b, or minus any quantity whatever ; hence there 
is no value to J — a, &c., and such symbols are said to be irrU" 
tional or imaginary. 



EVOLUTION. ' 133 

(Art. 86.) The square and cube root of any quantity, as a, 
being expressed by J a and ^Ja, and as by involving the root 
we obtain the power, hence the square of Ja, is a; and the 
cube of XI a, is a. Hence, removing the sign involves to the 
corresponding power. 

EXAMPLES. 

1. What is the square of ^JSax ? ^ns. Sax. 

2. What is the cube of ^^Qy^ ? JJns. 6y\ 



3. What is the square of Ja^ — ar' ? »^ns. a^— ar^. 

4. What is the square of \/-— ? ^ns. -^. 

^ 6-{-x 6+x 

5. What is the cube of ^Jl-\-al Jins. l+a. 

(Art. 87.) When we have two or more quantities, and the 
radical sign not extending over the whole, involution will not 
remove the radical, but will change it from one term to another. 
Thus, Jx-\-a the square will be x-\-2aJx-\-c^ ; the radical 
sign is still present, but not in the first term. 



PURE EQUATIONS. 
CHAPTER IV. 

(Art. 88.) Pure equations in general are those wherein a com- 
plete power of the unknown quantity is concerned, and no special 
artifice is requisite to render the power complete. 

The unknown quantity may appear in one or in several terms ; 
when it appears in several, its exponents will be regular, descend- 
ing from a higher to a lower value, or the reverse. 

In such cases we must reduce the equation by evolution. 

Like rool(S of equal quantities are equal. (Ax. 8.) 

EXAMPLES. V 

1. Given 3a?^ — 9=66 to find the value of x. 

Solution, 3x'^=-75 a?^=25. Hence, x=±:5. _,.it. 
M 



134 ELEMENTS OF ALGEBRA. 

We place the double sign before 5, as we cannot determine 
whether 25 was produced by the square of +5 or of — 5. 

In practical problems, the nature of the case will commonly" 
determine, but in every abstract problem we must take the 
double sign. 

S5. Given ^ — 6a;-|-9=a^ to find the value of x. 
By evolution, x — 3=zba. Hence, a?=3±a, Ans. 

3. Given x^ — ^-^-^^ — l^=c^W to find x. 

Take cube root and x — \—ah\ and x^=-ahAr\, Ans, 

4. Given a?'* — 2ii?^-l-l = 16 to find the value of x. 

Jlns, x=zhjb, or ±^—-3. 

5. Given x^ — 4a;+4=64 to find the value of x. 

Jins. ic=10, or — 6. 

6. Given x^y^'\-2xy-\-l=4x^y^ and x=2y to find the values 
of a? and y. Ans. iP==t:^2, y=dz^J2, 

T. Given ar^-|-?/^=13 and x^ — 2/^=5 to find the values of x 
and 2/. Ans. x=d=^3 y=dc:2. 

(Art. 89.) The unknown quantity of an equation is as likely 
to appear under a radical sign as to be involved to a power. In 
such cases we free the unknown quantity from radicals by involu- 
tion (Art. 86.), having previously transposed all the terms not 
under the radical to one side of the equation, the radical being 
on the other. 



9. Given >Jx-\- l=a — 1 to find the value of x. 
By involution, x-\-l==a^ — 2a-\-l, Hence, x=(^ — 2a. 



lO. Given Jl2-]-x:=3'\-Jx to find the value of x. 
Square both sides, and we have 

l2+x=9-Jrejx+x. 
Drop 9-{-x and 3=6 Jx; or Jx=^. a;=|, Arts 



II. Given Jx — 16=8— ^a; to find x. Jins. a:==25. 



PURE equations/ Ijfe 

12. Given — -,=- = zJl to find x, 

'J X X 

Multiply by Jx, observing that x divided by x gives 1 ; and 
we have x — aa:=l, 

Or (l— a)a?=l, 

Therefore x=- . 

1 — a 

«a n- V^+28 V^+38 , . ,,, , - 

13. Given — ,= =— f= to find the value of x» 

V:f+4 Va;-f-6 

By clearing of fractions, we have 

a:+34/H-168=a;4-42V^+152. 
Reducing, 16=8^a? 

By division, 2= ^x, or 4=a;. 

To call out attention and cultivate tact, we give another solu- 
tion. Divide each numerator by its denominator, and we have 

1+_J1.^_=1,. 32 



Va;+4 ' ^0^+6 
Drop unity from both sides, and divide by 8 ; we then have 

3 4 

Va?+4~Vx+6 

Clearing of fractions, 3^cr4-18=4^a?-[-l6 

Dropping equals, 2= Jx. Hence, a:=4. 

2<z 

14. Given Jx-\- Ja-{-x—-p=== to ^nd x. Jim. x=^\a 
sj a-\-x 

1a^' 



15. Given x-^ J a^^x^^-j====^ to find x. 

>^ a^-\-xr 

Ans. x=^aj^ 

16. Given x-^a=Jd^-\-X'j¥+x^ to find x. 

*^ns. a?= — 

4a 



136 ELEMENTS OF ALGEBRA. 

17. Given ^^= =_^^__ to find x.* Ans. a;=6. 

V6:cH-2 4:JQx-\-Q 

4+a; 



18. Given V64+a?2— 8x=— _:!= to find x. Ans, a'=3. 



19. Given ^/5+a;-i-^a;= — to find x. Ans. a:=4. 

ao. Given Jx-\rjx — Jx — Jx=~{ —v—j- ) to find x. 

An,. »=-. 

81, Given -5^=1+^^-1^ to find x. 
^bx-\-a 3 

Because a^ — 6^=(a+6)(a — 6) 

We infer that 5a>— 9=(^5a?+3) (75a>-3) 

5a?— 9 
Therefore, . . o =V^^ — ^* '^^® given equation then 



^^5^-3 
Now assume J^x — 3=i/. {A) 



becomes J^x — 3 _ , 



Then 2/=l+5y. Consequently, ?/=2. Returning to equa- 
tion {A)y we have J^x — 3=2 

Jox=5, Therefore, x=5y Ans, 

23. Given ^= — =-^^^ to find x, Ans, x— — . 

Jax-^b Sjax-{-5b « 

(Compare 22 with examples 13, and 17.) 
28. Given (l+a;7^+12)^ = l+a? to find x, Ans. a?=2. 



24. Given V^^ + V^ ^g, to find x. Ans. x^^ 

Jix-\-i—j4.x 9 

* See 2d solution to Equation 13. 



PURE EQUATIONS. ' 137 

25. Given a^ — 2ax-{-x^=b, to find x. Ans. x=^ — Jb. 

26. Given ? ^ =^ to find a;. 

l_^l_^a;2 14-^1— a?2 ^ 

4 

27. Given -r — - — r-r=5» to find a;, Ans. x=6, 

x^ — 2x-\-\ 

28. Given -^^ZlL^l + ^Zl^zi to find a?. Arts. a?=3. 

29. Given J^-h Ja?— 9= to find x. Arts, a?=25. 

30. Given h(.^-Z:±=l^^ to find ar. ./^n*. a^=4. 



31. Given -^^ ^ = to find the value of a?. 

^a? — Jx — a ^ ^ 

2n+l * 

Multiply the first member, numerator and denominator, by 
(V^+V^ — ^)» thenboth members by a, and extract square root. 



32. Given — ! — -^ ! — =Z>, to find a*. 

Assume a-\-x=y. Then the equation becomes 



« , T^ ±« 



y±Jti:^=b, Hence, y 



2/ ^26—^* 

And x=±a 



'l±726--^2 
J2E^^ 



(Art. 90.) To resolve the following examples, requires a de- 
gree of tact not to be learned from rules. Quickness of percep- 
tion is requisite, as well as sound reasoning. Quickness to per- 
ceive the form of binomial squares, and binomial cubes, and a 
12 



138 ELEMENTS OF ALGEBRA. 

readiness to resolve quantities into simple or coinpound factors, 
as the case may require. 

1. Given a;^-l-2a?=9-|-— to find the value of a?. 

Multiply by x, and x^-\-23?=9x-^\S, 
Separate into factors, thus : (a;-l-2)a?^=(a;-f-2)9. 
Divide by the common factor, jr4-2, and ar^=9, or a;=rb3. 



3. 



'Given .< 21 —oa C ^^ ^"^ *^^® values of x and y. 



Add the two equations together, and we have 

a;2+2a73/+2/2=36. 
Extract square root, and a?+i/=d=6. {A) 

From the first equation we have (a;+3/)a?=12. (J5) 
Divide equation [B) by (.^), and a?=±2. 

This example required perception to recognise the binomial 
square, and also to separate into factors. 

8. Given a^-{-y'^= and xy=^ to find the values 

^ X — y X — y 

of X and y. 

From the first equation subtract twice the second, and 

Therefore, (a? — ■y)'=l, and x — 2/=l. 
Continuing the operation, we shall find a?=3, and 2/=2. 

4. Given x^y-{-xy'^=\QQ and a;^+^'^= 189, to find the values 
of X and y. Ans. x=b or 4 ; i/=4 or 5. 

To resolve this problem, requires the formation of a cube, or 
to resolve quantities into factors. 

5. Given x^-\-y^=={x-\-y)xyy and a:+2/=4, to find the 
values of x and y. Ans, a?=2 ; 2/=2. 

6. Given x-]ry : a? :: 7 : 5, and xy-\-y^=^\2Q, to find the 
values of x and y. Ans. x=±15, 2/=±6. 



, PURE EQUATIONS. 139 

T. Given x — y : 1/ :: 4 : 5, and a;^ -4-4^^=181, to find the 
values of x and y. Arts. a;=±9, 2/=±5. 

8. Given Jx-{-Jy : Jx — Jy : : 4 : 1, and a>— -y=16, to 
find the values of x and y, Ans, a?=25, y=9. 

sc^ X I 

9. Given -^r-f-T^+T^^* *o find x. Ans. x=n\. 

9 3 4 

10. Given x-\-y : o.-^ :: 3 : O ^^ ^^^ ^ ^^^ 

And a?3__2^3=,56 5 n 

Ans. a:=4, i/=2. 

11. Given x?y'\-xy^=^0 ] 

1 1 5 I to find X and y. 

Observe that xy(x-{-y)=a^y-{'xy*. 

Clear the 2d equation of fractions, and y+a? or x-\-y=i—^. 

Now assume a:+y=5, and xy=-'p. Then the original equa- 
tions become 5/? =30 

And 6s=5p 

Equations which readily give s and /), and from them we de- 
termine X and y. 

N. B. When two unknown quantities, as x and y, produce 
equations in the form of 

x-\-y^8 (1) 

And ic?/=jo (2) 

such equation can be resolved in the following manner : 

Square (1), and x^\-1xy-\-y'^=^8^ 
Subtract 4 times (2) ^xy =4p 

DifT. is x^ — 2xy-{-y^=s'^ — 4p 



By evolution x — y=Js^ — 4p (3) 

Add equation (1) and (3), and we have, 

2x=s+j7^^^ (4) 

Sub. (3) from (1), and 2y=s^jjs^ — ip (5) 



140 ELEMENTS OF ALGEBRA. 

To verify equations (4) and (5), add them and divide by 2, 
and we have x-\-y=s. Multiply (4) by (5), and divide by 4, 
and we have xy=p. 

(Art. 91.) No person can become very skilful in algebraic 
operations as long as he feels averse to substitution; for judi- 
cious substitution stands in the same relation to common algebra, 
as algebra stands to arithmetic. The last example is an illustra- 
tion of this remark. To acquire the habit of substituting, may 
require some extra attention at first, but the power and advantage 
gained will a thousand fold repay for all additional exertion. 

As a general principle, whenever x and y, or any other two 
letters combine in the form of x-\-y and xy, or factors of these 
terms, put x-\-y=s, the sum of the two letters, and xy=^p^ their 
product. In some of the following examples this substitution 
will be expedient. 

12. Given xArJ^-\-y^ 19^ ^^ ^^^ ^ 

And ar^+3^-1-2/ =133 3 

Put X -Yy =s, and J^y^P 
Then s +;? = 19 (^) 

And s2__p2^i33 (^) 

Divide [B) by (.^), and we have s—p—7, &c. 

*^ns. x=9 or 4, 2/=4 or 9. 

13. Given x*-{-23c^y^+y'^=l29Q—4xy{s^-}-xy-\-y^) and 
X — ^2/=4, to find X and y. 

Put sP-{-y^=s, and xy=p 
Then the first equation becomes 5^=1296 — ip{s-\-p) 
Multiply and transpose, and s^-{'4sp-\-4p^=l296 

vSquare root 5+2/)— ±36 

But s-\-2p=x'^+2xy-\-y''=±:36 

Therefore a?+i/=±6, or ±^—36. 

Rejecting imaginary quantities, we find x=5 or ~1, and 
t/=l or — 5. 

14. Given ~=G, and a?-fv+a?v=ll» to find the values 

x—y 

of X and y. Ans. ic=5 or 1, y=^\ or 5. 



PURE EQUATIONS. 14% 

15. Given x^-\-y^=2xy{x-\-y), and xy=l6, to find the 
values of x and y. ^ns. a?=2^5+2, 2/=2^5 — 2. 

16. Given x^-\-y^^a, and afy-{-xy^=a, to find the relative 
values of X and y- •^^*' a?=y. 

IT. Given cc+y : a? :: 5 : 3, and xy—6, to find a? and y. 

.^ns. a?=±3, 2/=±2. 

18. Given 3:+?/ : a? :: 7 : 5, and xy-\-y^ =^126^ to find a: 
and y. *dns, a?=it:15, ?/=±6. 

19. Given x^-\-y^=a, and xy=h, to find the values of 
X and 2/. "^^^^ x—^h»Jci-\r2b-\-iJa — 26. 

(Art. A)* Equations in the form of x'^ — 2ax^-]ra^=i>, require 
for their complete solution, the square root of an expression in 
the form of azkzjb ; for by extracting the square root of tlie 
equation, we have 

x^-^a=dizjb 

/ ~ 

Hence ^=a/ «=t V^ 

The right hand member of this equation is an expression well 
known among mathematicians as 

A BINOMIAL SURD. 

Expressions in this form may or may not be complete powers ; 
and it is very advantageous to extract the root of such as are 
complete, for the roots will be smaller, and more simple quantities, 
in the form of a'db^6', or of Ja'zt:»Jh'. 

Let us now investigate a method of extracting these roots ; and, 
for the sake of simplicity, let us square 34-^7. 

By the rule of squaring a binomial, we have 9 -{-6^74-7, 
Or, 16+677"; 

Conversely, then, the square root of I6-I-677, is 3+^77 

* That the same Articles may number the same in both the School and Col- 
lege Edition, we shall designate all additional Articles, in this volume, by 
A, B, &c. 



142 ELEMENTS OF ALGEBRA. 

But when a root consists of two parts, its square consists of the 
sum of the squares of the two parts, and twice the product of 
the two parts. 

Now we readily perceive that 16 is the sum of the squares of 
the two parts expressing the root; and 6^7 , the part containing 
the radical, is twice the product of the two parts. 

To find what this root must be, let x represent one part of the 
root, and y the other : 

Then a^-\-y^=16, (1) 

And 2xy=6j7, (2) 

Add equations (1) and (2), and extract square root, and we 
''"« .r+y=^l6+6Jl (3) 

Subtract equation (2) from (1), and extract square root, and 
we have x — y=,^ IQ — 6^7. (4) 

Multiply (3) and (4), and we have 

:r2— 7/2=^256— 252=^T=2 ; (5) 
Add (1) and (5), and we have 

2;i32=18, or a?=3 ; 

Sub. (5) from (1), and 2y^=l4, or y==j7^ 

Whence x-{-y, or the square root of 16-|-6^7 is S-{-^7. 
We shall now be more general. 

Take two roots, one in the form of a^Jb, 

and one in the form of Ja±Jb ; 

Square botn, and we shall have a^dz2aJb-\-bj 

And adt2^ab-{'b. 

In numerals, and, in short, in all cases, the sum of the squares 
of the two parts of the root, as (a^~\-b), in the first square, and 
{a-\-b)j in the second, contain no radical sign ; and the sum of 
these rational parts may be represented by c and c', and the 
squares represented in the form of 



PURE EQUATIONS. }0 

* . **■ 

c zt.2ajb, 

or of c'±z2jab. 

Hence, generally, if we represent the parts of the roots by x 
and y, we shall have a^-\-y^= the sum of the rational parts, and 
2xy= the term containing the radical. 

The signs to x and y must correspond to the sign between 
the terms in the power. If that sign is minus, one of the signs 
of the root will be minus ; it is indifferent which one. 

EXAMPLES. _ 

1. What is the square root of 11+672? Jns, 3+^2. 

2. What is the square root of 7+4^^? Ma. 2-1-73^ 

3. What is the square root of 7 — 2^15? 

Ms. V^— >/2 or 72— 75I 

4. What is the square root of 94-{-4275"? Ms. 7+3^5^ 

5. What is the square root of 28+10^3? Ms. S-f-^SL 

6. What is the square root of 



np + 2m^ — 2 m Jnp-\-m^l 
In this example put a=^7ip-\-7n^, and x and y to represent 
the two parts of the root. 

Then x'^y'-^m^-i-a, 

and 2xy = — 2m J a. 

Ms. ±:{jnp+m^ — m). 



7. What is the square root of bc-\-2bJbc — b^l 

Ms. dsz{b-['Jbc—b^). 

8. What is the sum of ^16+307^-}-^16— -30^1^? 

Ms. 10. 

9. Whatis the sum of ,^11+6^2 and /^7— 2^10? 

Ans. 3-1- 75^ 

10. What IS thesum of ^31 + 127^and ^—1+4^115? 

Ms. 8. 



144 ELEMENTS OF ALGEBRA. 

In a similar manner we may extract the cube root of a bino- 
mial surd, when the expression is a cube ; but the general solu- 
tion involves the solution of a cubic equation, and, of course, 
must be omitted at this place; and, being of little practical utility, 
we may omit it altogether. 

(Art. 92.) Fractional exponents are at first very troublesome 
to young algebraists ; but such exponents can always be ban- 
ished from piire equations by substitution. For the exponents 
of all such equations must be multiples of each other; otherwise 
they would not be pure, but complex equations. 

To make the proper substitution, put the lowest exponent of 
any letter, as x, equal to a simple letter, say P ; and the lowest 
exponentof any other letter, as y, equal to another simple letter, 
say Q. And let this be a general rule. 

EXAMPLES. 

£ 1 11 

1. Given x^-\-y^^=^Q, and x^ •\-y^=-20, to find the values 

of X and y. 

By the above direction, put x^=F, and y^'=Q* 
Squaring these auxiliaries, or assumed equations. 

And a?^=P2, and y^=-QK 
Now the original equations become 
P + Q=6 (1) 
i>24.^=20 (2) 
By squaring equation (1), P2_}-2P§-|-Q2=36. 
Subtracting equation (2), wehave 2P$=16. 
Subtracting this last from equation (2), and we have 

By extracting square root P — Q=±2 
But by equation (1), P-\-Q— 6 

Therefore, 7^=4 or 2, and ^=2 or 4, 

2 i 

Hence, a?3=4 or 2, and y* =2 or 4. 
Square root x' =2 or (2)* ' 

3 

Cubing gives a? =8 or (2)^ 



2/=32 or 1024, 



PURE EQUATIONS. l4d 

2. Given xy'^-\-y =21, and xY+y^=^^^^ to find the values 
of X and y. 

By comparing exponents in the two equations, we perceive 
that if we put xy^=Fj and y=Q, the equations become 

Solved as the preceding, gives P=18, ^=3. 
From which we obtain x—2, or j^j, t/=3 or 18. 

3. Given x^^-{-x^y^== 208 | 

„% r to find the values of x and y 

And i/2+a:^2/"' = 1053 j 

2 2 

Assume x^=P, and y^=Q. 

By squaring and cubing these assumed auxiliary equations, 
we have 4 2 

Seek the common measure (if there be one) between 208 and 
1053. 

From the above substitution, the given equations become 
P3-|-P2^= 208 = 13.10 (1) 

Separate the left hand members into factors, and 

P^(P-}-$) = 13.16 (3) 

Q2($4-P) = 13.81 (4) 

Divide equation (4) by (3), and we have 

^' 81 ^ ^9 

•^= — . iiiXtractrng square root 0=2 

Or ^'=— :-• Substitute this value of Q, in equation (1), 

And /^H— -=13.16 
4 

Or 4P34-9PQ3=13.64 
That is, 13/^=13.64 
Hence, P3^64 or P=4 

13 



146 ELEMENTS OF ALGEBRA. 

But a;"=/^=64. Therefore, a:=db8 

As Q-=-^ and P=4, Q=9 and ?/==t:27. 

3333 "> 

4. Given x"-]-x*y*-{-y" = lOOd —a I 

3 3 Mo find X and y. 

And x^-{-xY^+y^ =582193=6 j 

Put a7^=P and y^ = Q 

Then x^--=P^ and 3/^=^2 

And x3=/^ and 3/=^=^ 
Our equations then become 

P2 4. pQ JL.c^z=: 1 Equations having no fractional exponents, 
and are of the same form as in Problem 



P*-VF-Q''-\-Q'=b \ 



12. (Art. 91.) 

Arts. x—Si or 16, ^=16 or 81. 



5. Given x-fa;*i/^ = 12 

, , r to find the values of x and y. 

And y\-x^'y^= 4 J 

Arts, a:=9, t/==l. 

6. Given x-\-x^y^=a \ 

J J r to find the values of x and y. 

And y-\-x^y^=b J 

a-\-b ^ a-\-b 

7. Given a:2_j_;^4^4=fl{ J 

^ ^ r to find the vahies of x and y. 

And »/'^4-.''?'2/*=^ i 

8. Given Jx-\-Jy : ^:c — Jy :: 4 : 1, and x — y==l6, to 
l^nd the vahies of x and y. £n.s. a'=25, y=9. 

1 1 

9. Given x''-\-y''= 5 j. ^^ ^^^^j ^j^^ ^.^^^^^^ ^^ ^ ^^^^^1 ^^ 



And 0? +?/ =13 

•^ns. x=9 or 4, y=i or 9. 



PURE EQUATIONS. I47 

10. Given x-\-y : x — y :: 3 : 1 ^ to find the values of 
And a;^— 7/3=56 \ ^ and y. 

Ans, x=i, 2/=2. 
11. 



Given 
And 


X -\-y 

1 1 

x^+y~^ 


=35 1 

■ to find the values of x 
=5 

Ans. x= 


and 

27, 


y- 


^s. 






CHAPTER V. 










Problems 


producing Pure 


Equations, 









(Art. 93.) We again caution the pupil, to be very careful not to 
involve factors, but keep them separate as long as possible, for 
greater simplicity and brevity. The solution of one or two of 
the following problems will illustrate. 

1. It is required to divide the number of 14 into two such parts, 
that the quotient of the greater divided by the less, may be to the 
quotient of the less divided by the greater, as 16 : 9. 

*^ns. The parts are 8 and 6. 

Let x= the greater part. Then 14 — x=' the less. 

X 1 4 X 

Per question, — : '- : : 16:9. 

14 — X X 

Multiply extremes and means, and "- — = — ^ ^ 

14 — X X 

Clearing of fractions, we have 9x^=16(14 — xy 

By evolution, 3:r=4(14 — a:) =4. 14 — ix 

By transposition, 7.r=4.14 

By division, .t=4.2=8, the greater part. 

14 X 

Had we actually multiplied by 16, in place of indi- 

X 

eating it, the exact value and form of the factors would have been 
lost to view, and the solution mighthaverun in\o nnadfected qua- 
dratic equation. 

The same remark may be applied to many other problems, 
and many are put under the head of quadratics that may be re- 
duced by pure equations. 



148 ELEMENTS OF ALGEBRA. 

2. Find two numbers, whose difference, multiplied by the 
difference of their squares, is 32, and whose sum, multiplied by 
the sum of their squares, gives 272. 

If we put x= the greater, and y= the less, we shall have 
{x-y){x'^f)^ 32 (1) 
And (^+?/)(a;2+2/2)=272 (2) 
Multiply these factors together, as indicated, and add the equa- 
tions together, and divide by 2, and we shall have 
a.3-l-y3_152 (3) 
If we take (1) from (2), after the factors are multiphed, we 
shall have 2xy'^-\-2x^y—2'i0, or xy{x-\-y) = \'2Q (4) 

Three times equation (4) added to equation (3) will give a 
cube, &c. A better solution is as follows : 

Let x-\-y= the greater number, and x — y= the less. 
Then 2x= their sum, and 2^/= their difference. 
Also, 4a?2/= equal the difference of their squares, and 
2x^4-2y= the sum of their squares. 

By the conditions, 2yX4xy= 32 

And 2a:(2a;2+2?/2)=272 

By reduction, xy^= 4 
And x'-i-xy^'^es 

By subtraction, x^ =64 or x=4 

Hence, 2/ = l, and the numbers are 5 and 3. 

We give these two methods of solution to show how much 
depends on skill in taking first assumptions. 

3. From two towns, 396 miles asunder, two persons, A and B, 
set out at the same time, and met each other, after traveling as 
many days as are equal to the difference of miles they traveled 
per day, when it appeared that .^ had traveled 216 miles. How 
many miles did each travel per day? Let x=^^s rate, and y= 
B's rate. 

Then x — y= the days they traveled before meeting. 

By question, {x — y)x=2lGj and {x — y)y=lSO. 

216 180 6 5 

Consequently, = or -=-. 

X y ^ y 



PURE EQUATIONS. 149 

Therefore, y=^oc, which substitute in the first equation, and 
we have {x—lx)x=2\G, or --=216=6X6X6. 
By evolution, a:=36; therefore y=SO. 

4. Two travelers, j9 and B, set out to meet each other, A 
leaving the town C, at the same time that B left/). They traveled 
the direct road between C and £) ; and on meeting, it appeared 
that A had traveled 18 miles more than B, and that .-^ could have 
gone -6's distance in 1 5| days, but B would have been 28 days 
in going ./^'s distance. Required the distance between C 
and n. 

Let x= the number of miles A traveled. 
Then x — 18= the number B traveled. 

=J1 s daily progress. 
15^- 

X 

— =-B's daily progress. 
28 

X 18 X 

Therefore, x : x — 18 : : — — -- : -^ 
15^ 28 

, J x^ 4(x— 18)2 

And — =-5: '-. 

28 63 

Divide the denominators by 7, and extract square root, and we 
have 

Therefore, a?==72 ; and the distance between the two towns 
is 126 miles. 

5. The difference of two numbers is 4, and their sum multi- 
plied by the difference of their second powers, gives 1600. 
What are the numbers ? Ajis. 12 and 8. 

6. What two numbers are those whose difference is to the 
less as 4 to .3, and their product, multiplied by the less, is 
equal to 504? J?2S. 14 and 6. 



150 ELEMENTS OF ALGEBRA. 

7. A man purchased a field, whose length was to its breadlli 
as 8 to 5. The number of dollars paid per acre was equal 
to the number of rods in the length of the field; and the 
number of dollars given for the whole was equal to 13 times 
the number of rods round the field. Required the length and 
breadth of the field. 

,^ns. Length 104 rods, breadth 65 rods. 
Put 8x=the length of the field. 

8. There is a stack of hay, whose length is to its breadth as 
5 to 4, and whose height is to its breadth as 7 to 8. It is worth 
as many cents per cubic foot as it is feet in breadth ; and the 
whole is worth at that rate 224 times as many cents as there are 
square feet on the bottom. Required the dimensions of the stack. 

Put 5x = the length. 

^ns. Length 20 feet, breadth 16 feet; height 14 feet. 

9. There is a number, to whicli if you add 7, and extract the 
square root of the sum, and to which if you add 16 and extract 
the square root of the sum, the sum of the two roots will be 9. 
What is the number ? *^ns. 9. 

Put x^ — 7= the number. 

10. »/5 and B carried 100 eggs between them to market, and 
each received the same sum. If *^ had curried as many as 
B, he would have received 18 pence for them; and if B had 
taken as many as JJ, he would have received 8 pence. How 
many had each ? ^ns. A 40, and B 60. 

11. The sum of two numbers is 6, and the sum of their cubes 
is 72. What are the numbers ? Ans. 4 and 2. 

12. One number is c^ times as much as another, and the pro- 
duct of the two is IP-. What are the numbers ? 

h 
Ans. — and ab. 
a 

13. The sum of two numbers is 100, the difl'erence of 
their square roots is 2. What are the numbers? 

Ans. 36 and 64. 



PURE EQUATIONS. 151 

Put a;= the square root of the greater number, 
And y= the square root of the less number ; or 
Put x-{-y= the square root of the greater, &;c. 

14. It is required to divide the number 18 into two such parts, 
that the squares of those parts may be to each other as 25 to 16. 

Let x= the greater part. Then 18 — x= the less. 
By the condition proposed, x^: (18 — x)^::25: 16. 
Therefore, 1 6a;2=25(l 8— a:")2 
By evolution, 4x=± 5(18 — x) 
If we take tlie plus sign, as we must do by tlie strict enuncia- 
tion of the problem, we find x — \0. Then 18 — x=S. 
And (10)2; (8)2::25: 16 
If we take the minus sign, we shall find a?=90. 

'J'hen 18— a:=18— 90=— 72. 
And (90)^: ( — 72)^:: 25: 16; a true proportion, correspond- 
ing to the enunciation ; but 18 in tliis case is not tlie number 
divided, it is the difference between two numbers whose squares 
are in proportion of 25 to 16. 

15. It is required to divide the number a into two such 
parts that the squares of those parts may be in proportion of b 
to c. 

Let x= one part, then a — a;= the other. 
By the condition, aP' : [a — x)^ ::b:c 
Therefore, cx^=zh{a — xY 

By evolution, Jcx=dcjb{a — x) 

Faking the plus sign, x=-—^,—f~ and a — x~ ,, . , . 

Taking the minus sign, Xz=^—~' — j~ and a — x= -77 — ~-. 

slb—sj^ V^ — V^ 

Prob. 14, is a particular case of this general problem, in which 
«=18,6=25, and c=16; and substituting these values in the re- 
sult, we find a?=10, and .r=90, as before. 

If we take 6=c, the two divisions will be equal, each equal 
io 5G, when the plus sign is used ; hut when the minus sign is 



152 ELEMENTS OF ALGEBRA. 

J ajh ajb , . ^ . ^ . 

used, x=-jr~- — 7'=--^> a symbol of infinity, as the denomi- 

nator is contained in the numerator an infinite number of times. 

(Art. 58.) The other part, a — x— ,, *^ , = y—, also a 

Jb—Jc 

symbol of infinity ; and the two parts, 

ctjb a^c _ a{Jb — ,Jc) _ 

Jb—Jc Jb^Jc {Jb^Jc)'""' 

It may appear absurd, that the two parts, both infinite and 
having a ratio of equality, (which they must have, if b—c) can 
still have a difference of a. But this apparent absurdity will 
vanish, when we consider that the two parts being infinite in 
comparison to our standards of measure, can have a difference 
of any finite quantity which may be great, compared with 
small standards of measure, but becomes nothing in comparison 
with infinite quantities. See (Art. 60.) 

Application of the foregoing Problem. 

(Art. 94.) It is a well established principle in physics that 
light and gravity emanating from any body, diminish in inten- 
nty as the square of the distance increases. 

Two bodies at a distance from each other, and attracting at a 
given point, their intensities of attraction will be to each other 
as the masses of the bodies directly and the squares of their 
distances inversely. Two lights, at a distance from each other, 
illuminating at a given point, will illuminate in proportion to the 
magnitudes of t!ie lights directly, and the squares of their dis- 
tances inversely. 

These principles being admitted — 

16. Whereabouts on the line between the earth and the moon 
will these two bodies attract equally, admitting the mass of the 
earth to be 75 times that of the moon, and their distance asunder 
30 diameters of the earth ? 

Represent the mass of the moon by c, 

and the mass of the earth by by 
.their distance asunder by o. 



PURE EQUATIONS. 153 

The distance of tlie required point from the earth's centre, 
represent by x. Then the remaining distance will be [a — x). 

Now by the principle above cited, y? : [a — xj- ::b:c. 

This proportion is the same as appears in the preceding gei> 
eral problem ; except that we have here actually made the ap- 
plication, and must give the definite values to «, b and c. 

As before, x= ,, , — j- and a — x= — 77-^ — 7- 
Jb-{-Jc Jb-\-Jc 

a=30, /;=75, c=l. 

x= , ^ -i — =26.9, nearlv. Hence, a — x=3.1, nearly. 

7754-1 

If we take the second values for tlie two distances, from the 
general result, namely, x=.—f-r ^ , and a — x— ,, , » and 
give tlie numeral values, we shall have 

x= , i ^. =33.9, «— ^=— 3.9, nearly. 
775—1 

These values siiow that in a line beyond the moon, at a 
distance of 3.9 the diameters of the earth, a body would be 
attracted as much by the earth as by the moon, and the value 
of (a — x) being minus, shows that the distance is now counted 
the other way from the moon, not as in the first case towards 
the earth ; and the real distance, 30, corresponding to a in the 
general problem, is now a difference. 

We may make very many inquiries concerning the intensity 
of attraction on this line, on the same general principle. 

For example, we may inquire, ivhereabouts, on the line be- 
tween the earth and moon will the attraction of the earth 6e 16 
times the attraction of the moon? 

Let x= the distance from the earth. 
Then a — x— the distance from the moon. 

The attraction of the earth will be represented by — • 



154 ELEMENTS OF ALGEBRA. 

(J 

The attraction of the moon at the same point will be 

By the question, 



6_ 16c 

x^ {a — xf 



By evolution, -^^— =±— -^^ — 

•^ X a — X 

Clearing of fractions, ajb — Jbx=4tjcx. 
Usinor tlie plus si^n, x=—, , , ^ , -=20.5, nearly. 

Using the minus sign, a;= =55.7, nearly, or 

25.7 diameters of the earth beyond the moon. 

Observe that the 4 wliic-h stands as a factor to Jc is the 
square root of 16, the number of times the intensity of the 
earth's attraction was to exceed that of the moon. 

If we propose any other number in the place of 16, its 
square root will appear as a factor to Jc ; we may therefore 
inquire at what distance the intensity of tlie earth's attraction 
will be 11 times that of the moon, and the answer will be from 
the earth in a line through the moon, 

_^_ and -^^L—. 
Jb-\-Jnc Jb — Jnc 

The same application that we have made of this general prob- 
lem to the two bodies, the earth and the moon, may be made 
to any two bodies in the solar system ; and the same application 
we have made to attraction may be made to liglit, whenever 
we can decide the relative intensity of any two lights at any 
assumed unify of distance. 

(Art. 95.) This problem may be varied in its application to 
meet cases where the distances are given, and the compara- 
tive intensities of light or attraction are required. 

For example, the planet Mars and the moon both transmit 
the sun's light to the earth by reflection, and we now inquire 



PURE EQUATIONS. 155 

the relative intensities of their lights at given distances, and 
in given positions. 

If the surface of Mars and that of the moon were equal, 
they would receive the same light from the sun at equal distance 
from that luminary; but at different distances equal surfaces 
would receive light reciprocally proportional to the squares of 
their distances. 

The surfaces of globular bodies are in proportion to the squares 
of their diameters. Now let M represent the diameter of Mars 
and m the diameter of the moon. Also, let R represent the dis- 
tance of Mars from the sun, and r the distance of the moon from 
the sun. 

Then the quantity of light received by Mars may be expressed 

by —p^ ; and the relative quantity received by the moon 

by --. But these lights, when reflected to the earth, must be 

diminished by the squares of the distances of tliese two bodies 
from the earth. Now if we put J) to represent the distance of 
Mars from the earth, and d the distance of the moon, we shall 

have ^^j-- for the relative illumination by Mars when the whole 

enlightened face of that planet is towards the earth, and -y^ for 

the light of the full moon. 

When the whole illuminated side of Mars is turned towards 
the earth, which is the case under consideration, (if we take the 
whole diameter of the body,) it is then in opposition to the sun, 
and gives us light, we know not how much, as we have no 
standard of measure for it ; but we can make a comparative mea- 
sure of one by the otlier, and therefore the light of Mars in this 
position may be taken as unity, and in comparison with this let 
us call the light of the full moon x. 



Then 



B'D' ' rht' 



Therefore ^=(£)(72)(^)- 



156 ELEMENTS OF ALGEBRA. 

As the value of a fraction depends only on the relation of the 

numerator to the denominator, to find the numeral value of a?, it 

will be sujfficient to seek the relation of m to M, of M to r, and 

of n to d. 

tn 43 
il/=4000 miles nearly, and m=2150 ; hence, -^==-^ 

R 144 
i?=144000000, and r=95000000 ; or — =-7r- 

r 95 

Z>=144000000— 95000000=49000000 or ^=1?5^ 

d 24 



£/=240000 

Therefore, :r=Q^-y(^^) ( -1^ ) =27611 



)(^)' 



That is, in round numbers, the light of the full moon is twenty- 
seven thousand six hundred times the light of Mars, when that 
planet is brightest, in its opposition to the sun. 

We will add one more example by the way of farther illustra- 
tion. 

What comparative amount of solar light is reflected to the 
earth by Jupiter and Saturn, when those planets are in opposi- 
tion to the sun ; — the relative diameter of Jupiter being to that 
of Saturn as 111 to 83, and the relative distances of the Earth, 
Jupiter and Saturn, from the sun, being as 10, 52 and 95, re- 
spectively ? 

Ans. Taking the light reflected by Saturn for unity, that by 
Jupiter will be expressed by 24. y'/^ nearly. 

The philosophical student will readily perceive a more ex- 
tended application of these principles to computing the relative 
light reflected to us by the different planets ; but we have gone to 
the utmost limit of propriety, in an elementary work like this. 

From Art. 94th to the end of this chapter can hardly be said 
to be algebra ; it is natural philosophy, in which the science of 
algebra is used ; however, we would offer no apology for thus 
giving a glimpse of the utility, the cui bono^ and the application 
of algebraic science. 



QUADRATIC EQUATIONS. t5T 

SECTION IV. 

QUADRATIC EQUATIONS. 
CHAPTER I. 

(Art. 96.) Quadratic equations are either simple or compound. 
A simple quadratic is that which involves the square of the un- 
known quantity only, as ax^=b ; which is one form of pure equa- 
tions, such as have been exhibited in the preceding chapter. 

Compound quadratics, or, as most authors designate them, ad- 
jected quadratics, contain both the square and the first power 
of the unknown quantity, and of course cannot be resolved as 
simple equations. 

All compound quadratic equations, when properly reduced, 
may fall under one of the four following forms : 

(1) x2+2ax=6 

(2) x^-—2ax=b 

(3) x^-^2ax=—b 

(4) x--\-2ax=—b 

If we take x-\-a and square the sum, we shall have 

x'^-\-2ax-\-a^ 
If we take x — a and square, we shall have 

x^ — 2ax-\-a^ 
If we reject the 3rd terms of these squares, we have 

ar-\-2ax, and a?^ — "lax 
The same expressions that we find in the first members of the 
four preceding theoretical equations. 

It is therefore obvious that by adding d^ to both sides of the 
preceding equations, the first members become complete squares. 
But in numeral quantities how shall we find the quantity corres- 
ponding to a^ ? We may obtain a^ by the formal process of 
taking half the coefficient of the first power of a?, or the half of 
2a or — 2a, which is a or — a, the square of either being a^. 

Hence, when any equation appears in the form of a?^db2aa?= 
±5 ,we may render the first member a complete square, and effect 
a solution by the following 
O 



158 ELEMENTS OF ALGEBRA. 

Rule. Md the square of half the coefficient of the lowest 
power of the unknown quantity to the first inember to complete 
its square; add the same to the second member to preserve the 
equality. 

Then extract the square root of both members, and we shall 
have equations in the form of 

xd^a=zhJbT^ 
Transposing the known quantity a and the sohition is accom- 
plished. 

In this manner we find the values of x in the four preceding 
equations, as follows : 



1) x= — a±z»Jb-\-d' 



(2) x= adtiJb-\-a^ 

(3) x=^ adtja^'^ 

(4) x='-a±Jd'—b 

When b is greater than a^ equations (3) and (4) require the 
square root of a negative quantity, and there being no roots to 
negative quantities, the values of x in such cases are said to be 
imaginary. 

The double sign is given to the root, as both plus and minus 
will give the same power, and this gives rise to two values 
of the unknown quantity ; either of which substituted in the 
original equation will verify it. 

After we reduce an equation to one of the preceding forms, 
the solution is only substituting particular values for a and b ; 
but in many cases it is more easy to resolve the equation as an 
original one, than to refer and substitute from the formula. 

(Art. 97.) We may meet with many quadratic equations that 
would be very inconvenient to reduce to the form of x^~\-^ax=b; 
for when reduced to that form 2a and b may both be 
troublesome fractions. 

Such equations may be left in tlie form of 
ax^-i-bx=c 
An equation in which the known quantities, a, b, and c, are all 
whole 7iumberSf and at least a and b prime to each other. 



QUADRATIC EQUATIONS. 159 

We now desire to find some method of making the first mem- 
ber of this equation a square, without making fractions. We 
therefore cannot divide by «, because b will not be divided by a, 
the two letters being prime to each other by hypothesis. But 
the first term of a binomial square is always a square. There- 
fore, if we desire the first member of our equation to be convert- 
ed into a binomial square, we must render the first term a 
square, and we can accomplish that by multiplying every term 
by a. 

The equation then becomes 

Put y=ax. Then y^-\-by=ca 
Complete the square by the preceding rule, and we have 

We are sure the first member is a square ; but one of the terms 
is fractional, a condition we wished to avoid ; but the denomina- 
tor of the fraction is 4, a square, and a square multiplied by a 
square produces a square. 

Therefore, multiply by 4, and we have the equation 
4y'-\-4by-\-b^=:4ca-\-b^ 
An equation in which the first member is a binomial square and 
not fractional. 

If we return the values of y and y^ this last equation becomes 
4a^x'-]-4abx-^b^=iac+b^ 

Compare this with the primitive equation 
ax^-\-bx—c. 

We multiplied this equation first by a, then by 4, and in ad- 
dition to this we find b^ on both sides of the rectified equation, b 
being the coefficient of the first power of the unknown quantity. 
From this it is obvious that to convert the expression ax^-\-bx 
into a binomial square, we may use the following 

Rule 2. Multiply by four times the coefficient of x\ and 
add the square of the coefficient of x. 

To preserve equality, both sides of an equation must be mul- 



160 ELEMENTS OF ALGEBRA. 

tiplied by the same factors, and the same additions to both sides. 
We operate on the first member of an adfected equation to 
make it a square, we operate on the second member to preserve 
equality. 

(Art. 98.) For the following method of avoiding fractions in 
completing the square, the author is indebted to Professor T. J. 
Matthews, of Ohio. 

Resume the general equation a:i(^-\-hx==c 

Assume a?=- Then aot^=-~ and bx= — 
a a a 

7/2 7)7/ 

The general equation becomes — j- — =c 

Or ii^-\-bu—ac 

Now when b is even, we can complete the square by the first 
rule without making a fraction. In such cases this transforma- 
tion is very advantageous. 

When b is not even, multiply the general equation by 2, and 
the coefficient of x becomes even, and we have 
2ax2-{-2bx=2c (1) 

Assume x=—- Then 2ax^=-- and 2oa:=— — 
2a 2a 2a 

With these terms, equation (1) becomes 
u^ , 2bu „ 

Or u^-{- 2bu=4ac 
Complete the square by the first rule, and we have 
tc'-{-2bu-{-b^=4ac-^b^ 

An equation essentially the same as that obtained by completing 
the square by the rule under (Art. 97.) ; for we perceive the sec- 
ond member is the same as would result from that rule ; hence 
this method has no superior advantage except when b is even, in 
the first instance. 

(Art. 99.) The foregoing rules are all that are usually given 
for the resolution of quadratic equations ; but there are sotne 



QUADRATIC EQUATIONS. 161 

intricate cases in practice that we may meet with, where 
neither of the preceding rules appear practical or convenient. 
To m'aster these with skill and dexterity, we must return to a 
more general and comprehensive knowledge of binomial squares. 
x^-\-2ax-\-a^ is a simple and complete binomial square. Let 
us strictly examine it, and we shall perceive, 
1st. That it consists of three terms; 

2d. Two of its terms, theirs/ and the third, are squares; 
3d. The middle term is twice the product of the square 
roots of the first and last term. 

Now let us suppose the third term, a^ to be lost, and we have 
only x^-^-'Hax. We know these two terms cannot make a square, 
as a binomial square must consist of three terms.* 

We know also that the last term must be a square. 
Let it be represented by t^. 
Then, by hypothesis, x^-\-1ax-^t^ is a complete binomial 
square. 

It being so, 2xt=:i2ax, by the third observation above. 
Therefore, t=a and t^=a^ 
Thus a^ is brought back. 
!• Again, 4a^-{-4ab are the first and second terms of a bi- 
nomial square ; what is the 3rd term ? 
Let t^ represent the third term. 
Then 4a^-^4ab-\-t^ is a binomial square. 
Hence, 4at-—4ab or t=b and t^=b^ 
That is, t^ represented the 3d term, and b^ is the identical 3d 
term, and 4a'^-^4ab-{-b^ is the actual binomial square whose root 
is 2a-\-b. 

2. 363/^4-361/ are the first and 2d terms of a binomial square, 
what is the 3d term ? JJns. 9. 

3. -| 1-9 are the 2d and 3d terms of a square, what is the 

X I 

first ? ^ns. 



X 



2. 



* In binomial surds two terms may make a square, and this may condemn 
the technicality here assumed ; but it is nothing against the spirit of this ar- 
ticle. 

14 



162 ELEMENTS OF AJ.GEBKA. 

4. — 49 are the 1st and 2d terms of a binomial square, 

49 
what is the 3d ? ^ns. — . 

5. 9^/2 — Gy are the 1st and 2d terms of a binomial square, 
what is the 3d? ,^ns. 1. 

6. ax^-\-bx are the 1st and 2d terms of a binomial square, 

b^ 
what is the 3d ? tllns. -—. 

4a 

7. 81a;2 — ^ are the 1st and 3d terms of a binomial square, 
what is the 2d or middle term ? *^ns. ±18. 

8. y^ — Sx^y are the 1st and 2d terms of a binomial square, 
what is the 3d ? ^ns. 16a?. 

\2x 
O. 7^"1~36 are the 2d and 3d terms of a binomial square, 

a^ 
what is the 1st ? ^ns. -—. 

361 

V^ 
10. ^—-\-SQ are the 1st and 3d terms of a binomial square, 
361 

what is the middle term ? J3ns. rh-r^. 

^ 11. If x-\~—-; are the 2d and 3d terms of a binomial square, 
16 

what is the 1st term ? JJns. 4x^. 

12. The 1st term of a binomial square is-— the 2d term is 

4V^ 
±12, what is the 3d term ? ^ns. -^. 

XT 

(Art. 100.) Adfected quadratic equations, after being reduced 
to the form of a^-\-2ax==^bi can be resolved without any formality 
of completing the square, by the following substitution : 



QUADRATIC EQUATIONS. 163 

Assume x=y — a 
Then x^^'if'—lay^c^ 

And 1ax=- -\-2ay — 2a^ 



By addition, x^-\-2ax—y'^ — a^~b 



Hence, y=zt^b-\-a 



And x= — a±jJb-\-a^y the same result as 
may be found in equation (1), (Art. 96.) 

Rule for Substitution. .Assume the value of the unknown 
quantity equal to another unknown, annexed to half the coeffi- 
cient of the inferior power with a contrary sign. 

(Art. 101.) For further illustration of the nature of quadratic 
equations, we shall work and discuss the following equation : 
Given x^-\-ix=-QOj to find x. 
Completing the square, (Rule 1st.) a;^+4cr+4=64. 
Extracting square root, a;-{-2=it8. 

Hence, it=6 or x= — 10. 
That is, either plus 6, or minus 10, substituted for x in the 
given equation, will verify it. 

For 6^+4 X 6=60. Also, (— 10)^ — 4X 10=60 
If a? =6 then x — 6=0 
If a?=— 10 then a?+10=0 
Multiply these equations together, and we have 
X — 6 
X +10 
a?2— . Qx 

10a?— 60 



Product, a^_}-4a7~60=0 

Transpose, and a:^+4a;=60, the original equation. 

Thus we perceive, that a quadratic equation may be considered 
as the product of two simple equations, and these values of x in 
the simple equations are said to be roots of the quadratic, and this 
view of the subject gives the rationale of the unknown quantity 
having two values. 



164 ELEMENTS OF ALGEBRA. 

In equations where but one value can be found, we infer that 
the other value is the same, and the two roots equal, or one of 
them a cipher. 

EXAMPLES FOR PRACTICE. 

1. Given x^ — ^x — 7=33, to find x. Ans. x=lO or — 4. 

2. Given X" — 20a?=— 96, to find x. Mns, 12 or 8. 

3. Given x^-\-Qx-\-l=Q2, to find x. 

4. Given 2/^+12^=589, to find y, 

5. Given ?/2— 6?/-|-10=65, to find y. 

6. Given a:2+12a:+2=110, to find x. 
■y. Given x^ — 14ic=51, to find x. 

8. Given a?2+6a;+6=9, to find x. 

9. Given x'^-\-%x—\2,\o find x. 

10. Given a72_[_i2a;=10,to finda?. 

The ^reader will observe that the preceding examples are in, 
or can be immediately reduced to the form of x^±i2ax=^b, and of 
course their solution is comparatively easy. The following are 
mostly in the form of ax'^-\-hx=c. 

11. Given 5a:;+ 4a? =204, to find a?. 



Ans. 


7 or —13. 


Ans. 


19 or —31. 


Ans. 


11 or —5. 


Ans. 


6 or —18. 


Ans. 


17 or —3. 


Ans, 


— 3±2V3. 


Ans., 


— 4±2V7. 


Ans. 


— 6±746. 



u u 

According to (Art. 98,) put x=-. Then 6x^=-- and 

5 o 

4x= — , and the equation becomes — -|--— =204. 
5 5 5 

Clearing of fractions, w^+4w=1020. 

Completing the square and extracting the root, we have, 

w+2=±32, or w=30 or 

K 

But x=-. Therefore, x= 6 or 
5 

12. Given 5a?'+4a?=273, to find x. 

13. Given 7a;'— 20iP=32, to find x, 

14. Given 25a;2— 20a^=— 3, to find x, 

15. Given 21^?'— 292x=— 500, to find x. 



—34 






34 
~"5"' 


Ans. 




Ans, 


7or- 


-n- 


Ans, 


4 or 


7 


Ans, 


ior 


-^• 


Ans, 


llif 


or 2. 



QUADRATIC EQUATIONS. 105 

16. Given 2x'^ — 5a?=I17, to find x. 
Here, as 5 or b of the general equation is not even, we must 
multiply the whole equation by 2, to apply the above principle ; 
or we may take Rule 2. (Art. 97.) 

Multiply by 8, and add 5^ or 25 to both members. 

Then 16a;2— 40ir+25=961 

Square root, 4x — 5=zb31. Hence, a:=9 or — 6|. 

(Art. 102.) It should be observed that all quadratic equations 
can be reduced to the form of or^dz:2ax=^b, or, as most authors 
give it, x^±px=q; but when the terms would become fractional 
by such reduction, we prefer the form ax^dtJ)x=±Ci for the sake 
of practical convenience, as mentioned in (Art. 97.) 

(Art. 103.) It is not essential that the unknown quantity 
should be involved literally to its first and second powers ; it is 
only essential that one index should be double that of the other. 
In such cases the equations can be resolved as quadratics. For 
example, x^ — 4x^=621 is an impure equation of the sixth 
degree, yet with a view to its solution, it may be called a quad- 
ratic. For we can assume y=x^; then 1/^=a;^ and the equa- 
tion becomes y"^ — 4i/=621, a quadratic in relation to y, giving 
t/=27, or —23. 

Therefore, ar^=27 or —23 



And x=3 ory— 23. 

There are other values of a:; but it would be improper to seek 
for them now; such inquiries belong to the higher order of 
equations. 

3 

For another example, take x^ — x^ =56, to find the values 
of x. 

Here we perceive one exponent of a? is (/oM&/e that of the other; 
it is therefore essentially a quadratic. 

Such cases can be made clear by assuming the lowest power 
of the unknown quantity equal to any simple letter. In the 

3 

present case assume y=x- ; then y^=x^f and the equation is 



166 ELEMENTS OF ALGEBRA. 

By Rule 2, 4?/2—4i/-{- 1=225 
By evolution, 2y — l=rt:15 
Hence, y—S or — 7 

3 3 

And by returning to the assumption y=x'^ we find a?2=8, 
or a;'^=2. Hence, a?=4 ; or, by taking the minus value of y, 
r=V49^ 

(Art. 104.) When a compound quantity appears under differ- 
ent powers or fractional exponents, one exponent being double 
ihat of the other, we may put the quantity equal to a single letter, 
ind make its quadratic form apparent and simple. For example, 
suppose (lie values of x were required in the equation 



Assume ^2.r^-}-3a:-l-9=2/ 
Then by involution, 2.T2-f3a;-l-9=?/- {Ji) 
And the equation becomes y'^ — 5?/=6 [B) 

Which equation gives y=6 or — 1. These values of y, sub- 
stituted for?/ in the equation [A), give 2j?^-)-3a;-t-9=36 

Or 2rc2^3.T+9=l 
From the first of these we find a: =3 or — 4^ 
From the last, we find x—l[ — 3±^ — 55,) imaginary quan- 
tities. 

EXAMPLES. 

1. Given {x-\-\2f -\-{x-\-\2Y =^(a to find the values of x. 

Ans. x—4 or 69. 

2. Given (a:4-«)'^-|-2/;(a:4-«)* =3^2, to find the values of x. 

Ans. x=h^ — a or 816' — a* 

* It is proper to remark, that in many instances it would be difficult to verify 
the equation by taking the second values of x, as by squaring, the minus quan- 
tity becomes plus, and in returning the values, there is no method but trial to 
decide whether we shall take a plus or a minus root. Hence, these second 
answers are sometimes called roots of solution. In many instances hereafter, 
we shall give the rational and positive root only. 



QUADRATIC EQUATIONS. 167 



Ans. x=f or |. 



3. Given 9a?4-4-|-2^9a.'4-4=15, to find the values of x. 

An 

4. Given (lO+:i?)^— -(10+:r)^-=2, to find x. 

Ans. x=Q, 

5. Given (a^— 5)=^--3(a^— 5J2=40, to find x. 

Ans. x=d. 

6. Given 2{l-{-X'-x^)—{[-{-x--x')^ + '^==0, to find x. 

Ans. x==^-{-'^j4\. 

7. Given rr+16— 3(a^4-16f = 10, to find x. Ans. x=^d. 

8. Given Sx^"— 23?"=8, to find x. Ans. a:="^2. 

9. Given x'^+x'^=756, to find a;. ./^ns. a^=243. 

8 16 

10. Given -, ito^I+t;; -^ to find a^. Ans. x=S or 1. 

11. Given 4a;'+^'^=39, to find x. Ans. 0^=^729. 

12. Given .'c2—2a;-f6(a;'-^— 2^+5)^ = 11, to find x. 

Ans. x=l. 

x^ \2x 

13. Given -— 777= — 32, to find the value of x. 

"ol 19 ^ . ^ ^ 

Ans. a?=152 or 76. 

If much difficulty is found in resolving this 13th example, the 
pupil can observe the 9th example, (Art. 99). 

14. Given 81a:2+17+^=99, to find the values of x. 

XT 

Ans. a:=l, or — 1, or — \. 

Observe that the 1st and 3d terms of the first number are 
squares, see (Art. 99.) 

15. Given 81.^2-1-174-4=^+— +15» to find a'. 

x^ cr X 

Ans. x=2 or — If. 

4 955 

16. Given 25x^+6+5-^=-^, to find the values of x. 

Ans. x=2, or — 2, or — --. 
15 



168 ELEMENTS OF ALGEBRA. 

4x^ Sx 
17. Given — -4-— =6|, to find the values of x. 

^ns. x=7 or — 11^-. 

(Art. 105.) Equations of the third, fourth, and higher degrees, 
can be resolved as quadratics, provided vve can find a compound 
quantity in the given equation in.volved to its Jirst and second 
power, vt^ith known coefficients. 

To determine in any particular case, whether such a com- 
pound quantity is involved in the equation, we must transpose 
all the terms to the first member, and if the highest power of 
the unknown quantity is not even, multiply every term of the 
equation by the unknown letter to make it even, and then extract 
the square root, to two or three terms, as the case may require ; 
and if we find a remainder to be any multiple or any aliquot 
part of the terms of the root, a reduction to the quadratic form 
is effected ; otherwise it is impossible, and the equation cannot 
be resolved as a quadratic. 

For example, reduce the following equation to the quadratic 
form, if it be possible. 

1. Given a?^~-8«a;='+8aV4-32a3a;—9a^=0, to find the values 
of X by quadratics. 

OPERATION. 

a?4^8aa;3+8a2.c2-l-32a=^a;— .9a''=0 {pc'^-Aax) 
x" 



2x^^4ax -^Sax^+Sa^'x^ 
—Sax'+lGa'x^ 



—Sa^x^-\-S2a^x—'9a* 
This remainder can be put into this form : 
— 8a2(a;2^4aa?)— 9a^ 
Now we observe the original equation can be written thus : 
(^x^—.4axy^Sa%x'—'iax)—9a!^=0 
By putting x^ — 4ax=y we have 

?/^ — 8tt^i/=9a'* a quadratic. 
Completing the square^^ — 8a^3/-|-16a''=25a'' 
By evolution y — 4a^=zt5a^ 
Hence i/=9a^ — «^ 



QUADRATIC EQUATIONS. ' 169 

Or x^ — 4ffa:=9a^ or — a^ 

Completing the square x^ — 4ax-l-4a^= 1 3a^ or 3a^ 
By evolution x — 2«=±«^13 or «^3 
Hence x may have the four following values {2a-\-atJV6]y 
(2a—aj'vS), {2a-{-a^S), {2a—ajS). Either of which being 
substituted in the original equation will verify it. 

2. Reduce x^-\-2ax'^-\-5a'^x-{- 40^=0 to a quadratic. 

As the highest power of x is not even, we must multiply by x 
to make it even. Then 

x'+2ax^-\-5a'x^-]-4a'x=0 
By extracting two terms of the square root, and observing the 
remainder, the part that will not come into the root, we find that 
{x''-\-axy-h4a\x2-^ax)=0 
Divide by {x^-\-ax) and x^-{-ax-\-4a^=0 a quadratic. 

3. Given x^-{-2x'^—7a^—Sx-{-l2=0, to find the values of x. 
This equation may be put in the following form : 

w3ws. a:=l or 2, or — 2 or — 3. 

4. Given sc^ — 8a;^-}-19 x — 12=0, to find the values of x. 

Ans, x=\ or 3 or 4. 

5. Given a?''— 10r^-l-35ar^— 50a;4-24=0, to find the values 
of X. Am. x=\, 2, 3 or 4. 

6. Given x'^—-2x^-\-x=l^2, to find the values of x. 

Ans. x=4 or — 3. 

■y. Given y^ — 2cif-\-{c^ — 2)'if-\-2cy=^c^, to find the values 

(Art. 106.) The object of this article is to point out a few lit- 
tle artifices in resolving quadratics, which apply in particular 
cases only, but which at times may save much labor. It is there- 
fore proper that they should be presented, though some minds 
prefer uniformity to facility. 
15 



170 ELEMENTS OF ALGEBRA. 

For example, take equation (B) (Art. 104.) 

1. y^—.5y=Q Put 2a=5 
Then /— 2ai/=2a-hl. 

Add a^ to both sides to complete the square, (Rule 1.) 
And ?/'— 2«?/-i-a'=a2-f-2a+l 
By evolution, y — a=dt:{a-\-l.) 

Hence, y=2a-\-l=z6 or — 1 

2. Given ?/^ — 7y=S, to find y. ^ns. y=S or — 1. 

3. Given a;^-|-ll:i'=26, to find the values of x. 

Assume 2a=ll ; then 4«-j-4==26. 
Now put these values in place of the numerals, and complete 
the square, and x'^-{-2ax-^a^—a^ -{-4a-\-i. 

By evolution, a:+a=±(a+2) Hence, x=2 or — 13. 

4. Given x^ — 17a?=60, to find the values of x. 

Assume 2«=17; then 6^4-9=60 
And x^—'2ax-{.a''=d'-\-6a+9. 
By evolution, x — a=±(a+3.) Hence, ir=20 or — 3. 

5. Given a;^-|-19:c=92, to find the values of x. 

Assume 2a=19 ; then 8a+ 16=92 
Putting these values and completing the square we have 
a^-\-2ax-]-a^=a^+Sa-{-lG 
x-\-a=diz{a-\-4:) or a?=4 or — 23. 

Observe that in the preceding equations we invariably put the 
coefficient of the first power of the unknown quantity equal 2a. 
Then if v^^e find the absolute term in the second member of the 
equation equal to 2«-{- 1 

or 4«-}- 4 

or 6a-\' 9 

or 8rt+16 

Or, in general, m2a-\rm^. That is, any multiple of 2« plus 
the square of the same multiplier equal to the second member, 
then the equation can be resolved in this manner ; for in fact one 



QUADRATIC EQUATIONS. 171 

of the roots of the equation is this multiplier of 2«, and the other 
root is db(2a+7w), m being the multiplier, and it may represent 
any number, integral or fractional ; but there is no utility in ope- 
rating by this method unless m is an integer, and not very large. 
To present a case where m is fractional, we give the following 
equation ; x^ — 9a:=Y, to find the value of x. 

Put 2a=9; then 5X2fl+~Y' ^"^ ^^® equation becomes 
x^ — 2ax=a-\-l. 

Therefore, x — a=±(a+5). Hence, x= — 2 or 2a+i=9|. 

(Art. 107.) When the roots of the equation are irrational or 
surd, of course this method of operation will not apply ; but we 
can readily determine whether the roots will be surd or not. For 
example, take the equation x''-{-ldx=iO. 

Put 2«=13; then 4a-[-4=30 And 6a+9=48 
From this, we observe that one of the roots of the equation 
lies between 2 and 3. 

(Art. 108.) When the roots of an equation are irrational or 
surd, no artifice will avail us, and we must conform to set rules ; 
but when the roots are small integers, we can frequently find 
some method to avoid high numeral quantities ; but special artifi- 
ces can only be taught by examples, not by precept. The follow- 
ing are given as examples : 

1. Given a?2-f-9984a?=l 60000, to find the values of x. 

Observe that 9984=10000—16 

Put 2a=10000; then 32a=160000 

These substitutions transform the equation to 

x^-\-(2a-^iG)X=32a 

Completing the square by (Rule 1) and 
a^-|-(2a— 16)a,'-l-(rt— 8)2=a2_|_i6^^g4 

By evolution, x-{-{a — 8)=±(a+8) , * 

Hence, x=lQ or — 2a= — 10000. ^ * 

2. Given a;^-j-45a7=900O, to find the values of x. 

If we put 2«=45, the multiplier and its square, requisite to 



172 ELEMENTS OF ALGEBRA. 

produce 9000, is so large that it is not obvious, and of course 
there will be no advantage in adopting this method ; at the same 
time, we wish to avoid the high numerals we must encounter by 
any set rule of solution. 

We observe that 45 X 200=9000. Put a=45 
Then x^+ax=200a 
Complete the square by (Rule 2,) and 



By evolution, 2x-{-a=^a{a-\-800)=j45XS4b 

Multiply one of the factors, under the radical, by 5, and divide 
the other by 5, and the equivalent factors will be 225 X 169, both 
squares. Taking their root, resuming the value of a, and the 
equation becomes 

2a?+3.15=13.15 
Drop 3.15 from both sides 
And 2a?=10.15 or x=75, Jlns, 

3. Given \Qx^ — 225a;=225, to find the values of x. 
This equation is found in many of the popular works on 
algebra, and in several of them the common method of resolving 
it may be seen. 

Observe that 225=15X15. 
Put a=15; then a-\-\ — lQ^ and the equation becomes 
{a-\-\)x''—a'x=a^ 
Completing the square by (Rule 2j, and 
4(a+])V— 4(a+l)aV+a''=a^+4a3-|-4a2 
By evolution, 2{a-^\)x — a^=a^-\-2a 
Transpose a^ and divide by 2, and we have 
{a-\-\)x=a^-\-a—a{a-{-l) 
Divide by («H-1) and a?=«=15, Ans, 

We give one more example of the utility of representing nume- 
rals, or numeral factors, by letters, in reducing the following 
equation : 



QUADRATIC EQUATIONS. 173 

M n- 18 , 81— a^ a;'— 65 

4. Given _+_^_=.__ to find ^. 

By examining the numerals, we find 9, and several multiples 
of 9. Therefore, let a— 9, and using a in the place of 9 the 
equation becomes 

x^ ax Sa 

Clearing of fractions, we have 

1 6a^+8a''x-^Sx^^x'—G5x^ 

Transposing all to one side, and arranging the terms according 
to the powers of x, we have 

x^i-Sx^—Gox^—Sa^x—lGc^^^O {x^+4x) 

x' 



8x'i-lQx^ 



—Slx^—Sa^x—lGa^ 
Or -^a\x''-\-Sx-\-l6) 

Therefore, by (Art. 105,) the equation becomes 

{x^-{-4xy—a%x^-{-Sx-\-lQ)^0 
Or {x+iyx'^a^x+if 
By division, x^=a'^ 

And a?=±«=±9, *^ns. 

The preceding examples may be of service in reducing some 
of the following 

EXAMPLES. 

1. Given a:^-f-ll^=80, to find x. Ans. a:=5, or — 16. 

^ ^. . 3:r— 3 ^ , 3a?— 6 ^ , 

2. Given 5a? ^=2a?+ — -— , to find x. 

X — 3 2 

Ans. a?=4, or — 1. 

x ir-l~l 13 

3. Given —r-^-\ =--7r-, to find x, Ans. x=2. 

a-f- 1x6 

p2 



174 ELEMENTS OF ALGEBRA 

72tX 
4. Given 7x-{-— — r-=50, to find x. Ans. a:=2, or 



in — 5-— 50, to find X. Ans. a:=2, or . 

1 \J—oX 4 



5. Given (~-\-yj-^(—\-y\=^Q, to find the values of ?/. 

Ans. y=S or 2, or — Sdr^S. 

4 2 

6. Given x-i-jr7x^=i4, to find the vaUies of x. 

Ans. a:=db8 or d=(— ll)i 

•y. Given ^/^-f-ll+^i/^-jli 1+2=44, to find the values of ?/. 

c/!??2S. ^=±5 or ±^38? 

8. Given 14+2^—- — ;z=3?4- ' , to find the values of a\ 

X^~-7 3 

Ans. a;=28 or 9. 

9. Given 32^2—9^—4=80, to find the values of x. 

Ans. x=7 or — 4. 

10. Given --^'--= — -^L_ to find x. Ans. x=4. 

4-1-7^^ Vx 

11. Given -^— _ _-J+x— 2==24— 3^-, to find the values 

it'— O 



of X. Ans. x=Q or i 

IQ 14 2x 22 

12. Given -^=_- to find the values of x. 

X x^ 9 

Ans. x—3 or yi-. 

13. Given — ^ — ^ jTH" ='^ — •^' ^o find the values of x. 

X ——OX~j~.7 

Ans. x=l or —28. 

14. Given mi^ — 2mxjn=nx^ — 7nn, to find x. 

n Jmn 
Ans. x=~^ r-. 

17^3 

15. Given x'^-\-—--—S4x-{-'i6, to find the values of x. 
(See Exms. Art. 99.) Ans. x=2, or —2, or — 8, or — ^. 



QUADRATIC EQUATIONS. I75 

16. 

N.B. Put ( -4— )==v^ Jins. x=Q or —9. 



_i 

Given < -—. — ( ■—, — V f = . ^ , to find the values of a:. 
l\-\-x\\-\-x/ 3 12 



ly. Given 3/* — i2/^==y^+82/+12, to find the vahies of y. 
(See Art. 99.) Ans. y=3 or —2. 

8 16 

18. Given 7- 7rs=l-|-7ir :^> to find the values of x. 

(2x — if {2x — 4)'* 

Ms* x—d or 1. 

19. Given ( ' ^ j =x — 2, to find the values of x, 

Ans. x—^ or 3. 



CHAPTER II. 

Quadratic Equations^ containing two or more unknown 
quantities. 

(Art. 108.) We have thus far, in quadratics, considered equa- 
tions involving only one unknown quantity; but we are now 
fully prepared to carry our investigations farther. 

Two equations, essentially quadratic, involving two unknown 
quantities, depend for their solution on a resulting equation of the 
fourth degree. 

This principle may be shown in the following manner : 

Two equations, essentially quadratic, and in the most general 
form, involving two unknown quantities, may be represented 
thus : 

Oif-\-axy-\-by'^-\-cx-\-dy-{-e =0, 

x^-\-a'xy -\-h'y'^-\-c'x-\-d'y-]re'=0. 

We do not represent the first terms with a coefficient, as any 
coefiicient may be reduced to unity by division ; and a, b, c, &;c., 
and a', b\ &c. may represent the result of such division ; and, 
of course, may be of any value, whole or fractional, positive or 
negative. 



176 ELEMENTS OF ALGEBRA. 

Arranging the terms, in the above equations, in reference to x, 
we have 

c^^[ay\-c)x-]rhy^-^dyJ^e =0, (1) 

a?-\-{a'y-\-c')x-\-b"y'^-\-d'y-\-e'=Q, (2) 
By subtracting (2) from (1), we have 

[(a — ci']y-\-c — c'~\x-{-{b — b')y'^-\-{d — d')y-\-e — e'=0; 

Therefore .^M^^kHfT:^. 

This expression for a?, substituted in either equation (1) or (2), 
will give a Ji72al equation, involving only one unknown quan- 
tity, y. 

But to effect this substitution would lead to a very complicated 
result; and as our object is only to show the degree to which 
the resulting equation will rise, we may observe that the ex- 
pression for the value of x is in the form of — r— — -. This 

ry-{-s 

put in either of the equations (1) or (2), its square, or the ex- 
pression for sc^, will be of the fourth degree ; and no term can 
contain ?/ of a higher degree than the fourth. 

Therefore, in general, the resolution of tivo equations of the 
second degree, involving two unknown quantities, depends upon 
that of an equation of the fourth degree involving one unknown 
quantity. 

(Art. 109.) Two or more equations, involving two or more 
unknown quantities, can be resolved by quadratics, when they 
fall under one of the following cases : 

1st. One of the equations only may be quadratic; the othei 
must be simple, or capable of being reduced to a simple form. 

2d. The equations must be similar in form, or the unknown 
quantities similarly involved or combined in a similar manner, 
as they combine in regular powers ; or, 

3d. The equations must be homogeneous ; that is, the expo- 
nents of the unknown quantities must make the same sum in 
every term. 

In the first and second cases, we eliminate one quantity in 



QUADRATIC EQUATIONS. 177 

one equation and substitute its value in the other, or perform an 
equivalent operation, by rules already explained. 

In the third case, we throw in a. factor to one unknown quan- 
tity, to make it equal to the other, or assume it to be so ; but 
these principles can only be explained by 

EXAMPLES. 

aa^-\-bxy -j- cy'^ =c, 
a'x^-\-b'xy -\-c'y^ =/. 

These are homogeneous equations, for the exponents of the 
unknown quantities make the same sum 2, in every term. Id 
such cases, assume x—vy ; then the equations become 

av^y'^-\-bvy^-\-cy^=e; or ?/= — ^-r—. — ; — 
•^ -^ ^' '^ av^-{-bv-{-c 

f 

a'v^y^-\-b'vy^-\-c'y^=f or y'^ — "^ 



Hence, 



a'v^-]-b'v-{-c 

f 



av--{-bv-{-c a'v^-{-b'v-{-c' 

An equation involving the 1st and 2d powers of v, and, of 
course, a quadratic. 

The solution of this equation will give v. Having v, we 
have y^ and y, and from vy we obtain x. 

For a particular example, we give the following : , 

1. Given S 4^^'— 2xi/-12 •) ^^ ^^^ ^^^ ^^^^^^ ^^ ^ ^^^^ 

l2y'-\-3xy^ SS 
Put x—vy ; then the equations become 
4y2^2 — 2i'2/^=12, or y^=- 



And 2f+3vy':^ 8, or 3/ -gl^g^ 

Whence, —-5 =—-■-—■. Dividing by 2, then clearing ol 

2v^ — V 2-\-Sv 

fractions, we have 

64-9^=8^2—41; or Sy'^— 13i;=6. 

13 



178 ELEMENTS OF ALGEBRA. 

This last equation gives v=^2 or -— |. 

Q Q 

Omitting the negative value y^==— ——==-== I, 

Therefore, ?/=zbl, and x^vy—zt2. 

2. Given 2a? — 3?/=l, and 2ar^H- a:?/— 5^/^=2 0, to find the 
values of x and y. 

These equations correspond to the first observation, one of 
them only being quadratic, the other simple ; and the solution 
is effected by finding the value of x in the first equation. Sub- 

stituting that value -^^ — in the 2d, and reducing, we have 

2y^-]-7y=d9i which gives y==3. Hence, a? =5. 

3. Given x'^-\-y^ — x — y—78, and xy-\-x-\-y—39, to find 
the values of x and y. 

In these equations x and y are similarly involved, not equally 
involved ; nor are the equations homogeneous. In cases of this 
kind, as we have before remarked, a solution by a quadratic can 
be effected, but no general or definite rule of operation can be 
laid down; the hitherto acquired skill of the learner, and his 
power of comparison to discern the similarity, will do more than 
any formal rules. 

To resolve this example, we multiply the 2d equation by 2, 
and add the product to the first ; we then have 

x'-\-2xy+y^-{-x-\-y=l5e, or {x-\-yy-\-{x-\-y)=l56. 
Put x-\-y=s ; then s^-{-s=l56, a quadratic, which gives s, 
or x-{-y==l2. This value of x-^-y, taken in the second equa- 
tion, gives xy=21. From this sum and product of x and y, 
we find a?=9 or 3, and 2/=3 or 9. 

Again, after we multiply the second equation by 2, if we sub- 
tract it from the first, we shall have 

x^-—2xy-\-y^-^3x^3y=Q 
or {x—yf^3{x-\-y)=^Q 
or (a:— ?/)2=--3Xl2=36 
or X — ^2/ ==±6 
But x-\-y=l2 

Hence, 2a7=18 or 6, and a?=9 or 3, as before. 



QUADRATIC EQUATIONS. 179 

(Art. 110.) There are some equations to which the foregoing 
observations do not immediately apply, or not until after reduc- 
tions and changes take place. The following is one of them. 

' .2 , „_13 1 



4. Given J "" ' •" y l jq fj^^^j ^j^g values of x and y. 



y 

Here neither of the equations is simple, nor are both letters 
similarly involved, nor are the equations homogeneous ; yet we 
can find a solution by a quadratic, because the two equations 
have a common compound factor, which taken away, will bring 
the equations far within the limits or condition laid down ; and 
this remark will apply to all problems that can be resolved by 
quadratics not seemingly within the limits of the three con- 
ditions. 

12 

From the first of these equations, we have y- 



From the second, y- 



x^-\-x 
18 



x'+l 



12 18 

Hence, -YT——~wiir' Divide the denominators by (a;+ 1] 

and the numerators by 6, and we have 

2 3 



a quadratic equation. 



X x^ — x-\-i 

Clearing of fractions, and 1x' — 2a?+2=3jp 
or 2.T^ — 5a? ^ — 2. 
(Rule 2.) 16a;2__^^_|_25=25— 16=9. 

We write A to represent the second term. It is immaterial 
what its numeral value may be, as it always disappears by taking 
the root. 

By evolution, 4a: — 5=db3 
Hence, x=2 or 5. 

T> , ^2 12 ^ 12 12X4 ,^ ^ 

The following is of a similar character : 



180 ELEMENTS OF ALGEBRA. 

1 C S^^ — y^ — {^~\~y)'^ 8 ^ to find the values of x 
^^^^ C {x-'yflx+y)=d2l audi/. 

8 
Divide the first equation by {x-{-y) and x — y — 1— {^). 

x-jry 

32 

Divide the second by {x-\-y) and {x — yf=-j^ (^)« 

Put x-{-y=Sj and transpose minus 1, in equation (^), and 
x—y=-4-l. By squaring, (a?— 1/)2= -f -|-i. 

o So 

32 

Equation (B) gives (x — yy= — 

s 

Therefore, '^+'-^+l='l. 

S^ s ' s 

Clearing of fractions, and transposing 325, we have 
61— 165+ 5^=0 
By evolution, 8 — s=0 or s=8. That is, a;+2/=8, which 
value, put in equation (j^j, gives x — ^2/ =2. 
Whence, x=5 and y=S. 

MISCELLANEOUS EXAMPLES. 

1. Given x=2y^ and ^[x — y)=5, to find the values of x 
and y. Ans. x=\S or 122. 

2/= 3 or — 2^. 

2. Given 2:r-|-2/=22, and a;z/+2i/^=120, to find the values 
of X and ?/. .^ns. x=S, y=Q. 

3. Given :z?4-3/ ' x — y : : 13 : 5, and a?-(-?/^=25, to find 
the values of x and y. Jins. x=9j and y=4. 

M p- Q x^-\-y^='\^xy ^ to find the values of x and ?/. 

ix -\-y =12 5 -'^^*- a:=8 or 4, 2/=4 or 8. 

5. Given 7.'^-\-2xy-\-y^=l2Q^2x—-2y, and x?/— ?/'=8, to 
find the values of x and 3/. 

C a?=6or9, or~-9±V5. 
.^ns. ^ ^^^ or 1, or —Sdr^S 



QUADRATIC EQUATIONS. 181 

-xy 



^31^ -{-xy =567 
6. Given S _i_o 2_ao c to ^^^ the values of x and y 



■ /lY,^ ^ — ±4^2 or ±14. 
•^'^*- ^ ±3^2 or ±10. 






Given ,3^ 3 _ 3^ 



find the values of x 
I 2xy-\-3a^=dy^ — 3 3 and y. 

Ans. aj=±2, y— ±3. 
6 ,5 

C3a^H-a^t/ = 687 , , , 

8. Given < . 210 __i ca c to find the values of x and y. 

Ans, a:=±4, 2/=±5. 

In the first four examples, one of the equations is simple ; in 
the 5th and 6th, x and y are similarly involved ; and the 6th, 
7th and 8th are homogeneous. 

(Art. 111.) When we have fractional exponents, we can re- 
move them, as explained in (Art. 92.) ; but in some cases it may 
not be important to do so. 

EXAMPLES. 

3 2 11 

1. Given x^-\-y^=3x and x^-\-y^=Xi to find the values 
of X and y, 

' Put x^=^P\ then x=P'' and x^-==P^ 

And2/^=Q; then y=^ and 1/^=^2 

Now the primitive equations become 

P3^Q2=3P2^ and P-\-q=P' 
From the 1st, Q^={z—P)P^ 
From the 2d, Q =(P-^l)P 
By squaring, Q'={P—iyP^ 

Put the two values of Q^ equal to each other, rejecting or 
dividing by the common factor P^, and we have 
Q 



182 ELEMENTS OF ALGEBRA. 

(P— 1)2=3— P 
or P2— 2P-fl=3— P 
or P^—Pz=1 
Hence, P =2 or 1, and x—\ or 1, and 2/=8. 

2. Given x'^+?/^-f2;r'+2?/^=23, and a:V'=^6, to find 
the values of x and y. ^ns. x=21 or 8, ?/=8 or 27. 

11 2 2^ 2 2 

3. Given a:'' +^^=8, and a;-^ +7/34-373^3 =259, to find the 
values of x and y. 

_ C a;=125 or 27, 
^ 1/= 27 or 125. 

£221 21 

4. Given a73+?/5_|_3j3_|_^5_26, and x^y^==^S, to find the 
values of a; and y. Ans. x=S, i/=32. 

1 
a? 4x* 33 

5. Given — 1 — r=-7- and x — v=5» to find the values of x 

y yh "^ 

and y, Ans. a:=9, 2/=4. 

6. Gijven \ 1 1 ~ r ^^ find the values of x and y. 

\y — 2a?y= 4 J ^ns. ;r=2|, ;y=16. 

•y. Given a?(/+l)+3/(a?^+l)+2a:V=55, and a?/ + 

va;^=30, to find the values of x and ?/. ^ C a:==4 or 9, 

^ y=9 or 4. 

2 3_ Jl 1 

§. Given x^y^==2y^ and 8a?^ — y^ = \i, to find the values 
of x and t/. /9 S ^=2744 or 8, 

^i/=9604 or 4. 

(Art. 112.) No additional principles, to those already given, 
are requisite for the solution of problems containing three or 
more unknown quantities in quadratics. As in simple equations, 
we must have as many independent equations as unknown 
quantities. 

As auxiliary to the solution of certain problems, particularly in 
geometrical progression, we give the following problem : 



QUADRATIC EQUATIONS. 183 

Given x+y=s, and xy=p, to find the values of x^+y^, 
x^+2/^5 x'*-l-y, and x^+y^ expressed in terms of s and p. 

Squaring the first, x^-\-^xy-\-y'^=s^ 
Subtract twice the second, 2xy =2/? 

1st result, a?2-)-?/2 =s^ — 2p {A) 

Again, 
{x-{-y)[3(F'-\-y'^)—x^-{-x^y-\-xy^-{-y^=s^ — 2ps 
Subtract xy[x-\-y) = ps 

2d result, x''-]-y^=s^-^'^ps {B) 

Square (^), and x'+2xhf-\-y'^=s'*—is^p-{-4p'' 

Subtract 2a?y = 2/ 

3d result, x'-i-y^=s^ — 45^^+2/ (C) 

Multiply (^) by {B) and 

x^-{-x^y^-\-x'^y^-\-y^=8^ — 5s^p-\-Qsp^ 
Subtract x'^y^{x-\-y) = sp^ 

4th result, x^-^-y^ =s^ — 5s^p-{-6sp^ (D) 

CHAPTER III. 

Questions producing Quadratic Equations. 

(Art. 113.) The method of proceeding to reduce the question 
into equations, is the same as in simple equations ; and, in fact, 
many problems which result in a quadratic may be brought out 
by simple equations, by foresight and skill in notation. Others 
again are so essentially quadratic, that no expedient can change 
their form. 

EXAMPLES. 

1. A person bought a number of sheep for $240. If there 
had been 8 more, they would have cost him $1 a-piece less. 
What was the cost of a sheep, and how many did he purchase ? 

• 240 

Let x=^ the number of sheep ; then — ;-=cost of one. 

X 



184 ELEMENTS OF ALGEBRA. 

If he had x-\-S sheep, — r— =cost of one. 

T, , .240 240 , , 

By the question, = — r-^-rl 

X iC-pO 

Clearing of fractions, 240a;+1920=240a;+a;2-j-8a; 
Or a:2+8a?=1920 

Resolving gives a:=40 or — 48 ; but a minus number will not 
apply to sheep ; the other value only will apply to the problem 
as enunciated. 

This question can be brought into a simple equation thus : Let 
X — 4= the number of sheep, then 8 more would be expressed 
by ic-1-4, and the equation would be 
240 240 



X — 4 x-\-4 



1. Put a=240. 



Then -^=-^+1 
X — 4 x-i-4 

Clearing of fractions, ax-\-4a=ax — 4a-}- a;^ — 16. 

Transposing, a;2=8a-l-16=8(a-i-2)=16X 121 
Extracting square root, a:=4X 11=44. Hence, x — 4=40, the 
number of sheep. Divide 240 by 40, and we have $6 for the 
price of one sheep. 

(Art. 114.) In resolving problems, if the second member is 
negative after completing the square, it indicates some impossi- 
bility in the conditions from which the equation is derived, or an 
error in forming the equation, and in such cases the values of the 
unknown quantity are both imaginary. 

2. For example, let it be required to divide 20 into two such 
parts that their product shall be 140. 

Let a?= one part, then 20 — x= the other 
By conditions, 20a: — a?^=140 

Or, a:2__20a:=— 140 
Completing the square, a? — 20a:-{-100= — 40 

By evolution, x — 10=±2^-— le 

Or, a; =10 ±2^^^^^ 



QUADRATIC EQUATIONS. 185 

This result shows an impossibility ; there are no such parts 
of 20 as here expressed. It is impossible to divide 20 into two 
such parts that their product shall be over 100, the product of 
10 by 10, and so on with any other number. Tke product of 
two parts is the greatest possible, when the parts are equal. 

3. Find two numbers, such that the sum of their squares being 
subtracted from three times their product, 11 will remain; and 
the difference of their squares being subtracted from twice their 
product, the remainder will be 14. 

Let a?= the greater number, and y=ihe less. 
By the conditions, 'Sxy — x^ — y^=^^ 
And 2xy^x^+y^=U 

These are homogeneous equations ; therefore, put x=vy ; 
Then 3vy''-^vy—y^=ll {A) 

And 2vy^'—vy+y^=U {B) 

Conceive (.^) divided by {B) and the fraction reduced, we have 

Clearing of fractions and reducing, we find 
^ *' 3i;2__20v=— 25. 

5 

A solution gives one value of v, - 

Put this value in equation (.^), and we have 

Multiply by 9. and 4t5y''—26y^^9y^=l 1 X 9, 
Or, Ilt/2=11X9, 

y^= 9 or y=S' Hence, a?=5. 

^ 4. A company dining at a house of entertainment, had to pay 
$3.50 ; but before the bill was presented two of them went away ; 
in consequence of which, those who remained had to pay each 
20 cents more than if all had been present. How many persons 
dined? . ' Ans. 7. 

16 



186 ELEMENTS OF ALGEBRA. 

«S. There is a certain number, which being subtracted from 
22, and the remainder multiplied by the number, the product will 
be 117. What is the number? ^ns. 13 or 9. 

6. In a certain number of hours a man traveled 36 miles, but 
if he had traveled one mile more per hour, he would have taken 
3 hours less than he did to perform his journey. How many 
miles did he travel per hour? Ans. 3 miles. 

•y. A person dies, leaving children and a fortune of $46,800, 
which, by the will, is to be divided equally among them ; but it 
happens that immediately after the death of the father, two of the 
children also die ; and if, in consequence of this, each remaining 
child receive $1950 more than he or she was entitled to by the 
will, how many children were there ? Ans. 8 children. 

8. A gentleman bought a number of pieces of cloth for 675 
dollars, which he sold again at 48 dollars by the piece, and gain- 
ed by the bargain as much as one piece cost him. What was 
the number of pieces 1 Ans. 15. 

This problem produces one of the equations in (Art. 107.) 

9. A merchant sends for a piece of goods and pays a certain 
sum for it, besides 4 per cent, for carriage ; he sells it for $390, 
and thus gains as much per cent, on the cost and carriage as the 
12th part of the purchase money amounted to. For how much 
did he buy it? Ans. $300. 

10. Divide the number 60 into two such parts that their pro- 
duce shall be 704. Ans. 44 and 16. 

11. A merchant sold a piece of cloth for $39, and gained 
as much per cent, as it cost him. What did he pay for it ? 

Ans. $30. 

12. A and B distributed 1200 dollars each, among a certain 
number of persons. A relieved 40 persons more than B, and 
B gave to each individual 5 dollars more than A. How many 
were relieved by A and B P Ans. 120 by A, and 80 by B. 

This problem can be brought into a pure equation, in like man- 
ner as (Problem 1 .) 



QUADRATIC EQUATIONS. 187 

13. A vintner sold 7 dozen of sherry and 12 dozen of claret 
for ^50, and finds that he has sold 3 dozen more of sherry for 
j610 than he has of claret for ^6. Required the price of each? 

J2ns. Sherry, £2 per dozen; claret, ^3. 

14. .^ set out from C towards Dj and traveled 7 miles a 
day. After he had gone 32 miles, B set out from D towards 
C, and went every day J^- of the whole journey ; and after 
he had traveled as many days as he went miles in a day, he met 
^, Required the distance from C to 2). 

^ns. 76 or 152 miles ; both numbers will answer the con- 
dition. 

15. A farmer received $24 for a certain quantity of wheat, 
and an equal sum at a price 25 cents less by the bushel for a 
quantity of barley, which exceeded the quantity of wheat by 16 
bushels. How many bushels were there of each ? 

Jlns. 32 bushels of wheat, and 48 of barley. 

16. ^ and B hired a pasture, into which Jl put 4 horses, and 
B as many as cost him 18 shillings a week ; afterwards B put 
in two additional horses, and found that he must pay 20 shillings 
a week. At what rate was the pasture hired? 

^ns. B had six horses in the pasture at first, and the price 
of the whole pasture was 30 shillings per week. 

IV. A mercer bought a piece of silk for ^16 45., and the num- 
ber of shillings he paid per yard, was to the number of yards as 
4 to 9. How many yards did he buy, and what was the price 
per yard. £ns. 27 yards, at 12 shillings per yard. 

18. If a certain number be divided by the product of its 
two digits, the quotient will be 2, and if 27 be added to the num- 
ber, the digits will be inverted. What is the number? 

^ns. 36, 

19. It is required to find three numbers, whose sum is 33, 
such that the difference of the first and second shall exceed the 
difference of the second and third by 6, and the sum of whose 
squares is 441. Jlns. 4, 13, and 16. 



188 ELEMENTS OP ALGEBRA. 

20. Find those two numeral quantities whose sum, product, 
and sum of their squares, are all equal to each other. 

Ans, No such numeral quantities exist. In a strictly algebraic 
sense, the quantities are 

|±§V^»and|=FW— 3- 

21. What two numbers are those whose product is 24, and 
whose sum added to the sum of their squares is 62? 

Ans. 4 and 6. 

22. It is required to find two numbers, such that if their pro- 
duct be added to their sum it shall make 47, and if their sum be 
taken from the sum of their squares, the remainder shall be 62? 

Ans. 7 and 5. 

23. The sum of two numbers is 27, and the sum of their 
cubes 5103. What are their numbers? Ans. 12 and 15. 

24. The sum of two numbers is 9, and the sum of their fourth 
powers 2417. What are the numbers? Ans. 7 and 2. 

25. The product of two numbers multiplied by the sum of 
their squares, is 1248, and the difference of their squares is 20 
What are the numbers? Ans. 6 and 4 

Let x-\-y=i\ie greater, and x — y=\he less. 

26. Two men are employed to do a piece of work, which 
they can finish in 12 days. In how many days could each do 
the work alone, provided it would take one 10 days longer than 
the other? Ans. 20 and 30 days. 

27. The joint stock of two partners, A and B, was $1000. 
A's money was in trade 9 months, and jB's 6 months ; when 
they shared stock and gain, A received $1,140 and B $640. 
What was each man's stock? 

Ans. A's stock was $600 ; ^'s $400. 

28. A speculator from market, going out to buy cattle, met 
with four droves. In the second were 4 more than 4 times the 
square root of one half the number in the first. The third con- 
tained three times as many as the first and second. The fourth 
was one half the number in the third and 10 more, and the whole 



^.- f 



ARITHMETICAL PROGRESSION. 189 

number in the four droves was 1121. How many were in each 
drove? ^7is. 1st, 162 ; 2d, 40 ; 3d, 606 ; 4th, 313. 

29. Divide the number 20 into two such parts, that the pro- 
duct of their squares shall be 9216. ^ns. 12 and 8. 

30. Divide the number a into two such parts that the produ/*t 
of their squares shall be b. 

dns. Greater part |-}--^a2— 476V. 
Less part -^-(a^^Ajly, 

31. Find two numbers, such that their product shall be equa'i 
to the difference of their squares, and the sum of their squares 
shall be equal to the difference of their cubes. 

Jins. rfc^V'sTand |(5±V5)". 



SECTION V. 

ARITHMETICAL PROGRESSION. 
CHAPTER I. 

A series of numbers or quantities, increasing or decreasing by 
the same difference, from term to term, is called arithmetical pro- 
gression. 

Thus, 2, 4, 6, 8, 10, 12, <fcc., is an increasing or ascending 
arithmetical series, having a common difference of 2 ; and 20, 
17, 14, 11, 8, &c., is a decreasing series, having a common dif- 
ference of 3. 

(Art. 115.) We can more readily investigate the properties of 
an arithmetical series from literal than from numeral terms. Thus 
let a represent the first term of a series, and d the common dif- 
ference. Then 



190 ELEMENTS OF ALGEBRA. 

a,(a+t/),(«+2t?),(«-|-3t/),(a4-4f/), &c., represent an ascend- 
ing series ; and 

a, [a — d),(a — 2d), {a — 3d), {a — id), &.C., represent a descend- 
ing series. 

Observe that the coefficient of d, in any term is equal to the 
number of the preceding terms. 

The first term exists without the common difference. All 
other terms consist of the first term and the common difference 
multiplied by one less than the number of terms from the Jirst. 

Wherever the series is supposed to terminate, is the last term, 
and if such term be designated by L, and the number of terms 
by n, the last term must be «+(« — l)d, or a — (n — 1)^/, accord- 
ing as the series may be ascending or descending, which we draw 
from inspection. 

Hence, Z=a=b(n — l)d {Jl) 

(Art. 116.) It is manifest that the sum of the terms will be the 
same, in whatever order they are written. 

Take, for instance, the series 3, 5, 7, 9, 11, 

And the same inverted, 11, 9, 7, 5, 3. 

The sums of the terms will be 14, 14, 14, 14, 14. 



Take the series «, a-{- d, a-\-2d, a-\-3d, a-\-id. 

Inverted, a-\-4:d, a-\-M, a-^2d, a-f- d, a 

Sums will be 2a+4d, 2a-{-4d, 2a-\-4:d, 2a-{-4d, 2a-\-4d, 
Here we discover the important property, that, in an arithmeti- 
cal progression, the sum of the extremes is equal to the sum of 
any other two terms equally distant from the extremes. Also, 
that twice the sum of any series is equal to the extremes, or 
first and last term repeated as many times as the series contains 
terms. 

Hence, if S represents the sum of a series, and n the num- 
ber of terms, a the first term, and L the last term, we shall 
have 2S=n{a-\-L) 

Or S=l{a+L) {B) 

The two equations [A) and {B) contain five quantities, a, d. 



ARITHMETICAL PROGRESSION. 191 

Z, n, and S; any three of ihem being given, the other two can 
be determined. 

Two independent equations are sufficient to determine two un- 
known quantities, (Art. 45,) and it is immaterial which two are 
unknown if the other three are given. 

By examining the two equations 

L=a-\-{n—l)d [A) 

We perceive that the value of any letter, L for example, can be 
drawn from equation (B) as well as from (A). 

It can also be drawn from either of the equations after n or a 
is eliminated from them. Hence, the value of L may take foKr 
different forms. The same may be said of the other letters, 
and there being five quantities or letters and four different 
forms to each, the subject of arithmetical progression may in- 
clude twenty different equations. But we prefer to make no 
display with these equations, believing they would add dark- 
ness rather than light, as they are all essentially included in the 
two equations, (A) and (B), and these can be remembered literal- 
ly and philosophically, and the entire subject more surely under- 
stood. 

These two equations are sufficient for problems relating to 
arithmetical series, and we may use them without modification 
by putting in the given values just as they stand, and afterwards 
reducing them as numeral equations. 

EXAMPLES. 

1. The sum of an arithmetical series is 1455, the first term 
5, and the number of terms 30. What is the common difiference? 

Jlns. 3. 

Here *S'=:1455, «=5, n=SO. L and d are sought. 
Equation (5) 1455=(5+Z)15. Reduced X=92 
Equation {A) 92=5 +29£/. Reduced £?=3, Jim, 



192 ELEMENTS OF ALGEBRA. 

2. The sum of an arithmetical series is 567, the first term 7, 
and tlie common difference 2. What is the number of terms? 

^ns. 21. 
Here s=5G7, a=7, d=2. L and n are sought. 
Equation [A) L=7-\-2n—2=5-{-2n 

Equation {B) 6Q7={7-\-5-]-2n)^=Qn-\-n^ 

Or n2+6/i+9=576 

n+3=24, or n=21,^ns. 

3. Find seven arithmetical means between 1 and 49. 

Observe that the series must consist of 9 terms. 
Hence, a=l, Z=49, 72=9. 

^ns. 7, 13, 19,25, 31, 37,43. 

4. The first term of an arithmetical series is 1, the sum of the 
terms 280, the number of terms 32. What is the common dif- 
ference, and the last term? *^ns, d=l, Z=162. 

5. Insert three arithmetical means between I and 5* 

Ans. The means are |, y\, \\, 

6. Find nine arithmetical means between 9 and 109. 

Jins, (?=10. 

7. What debt can be discharged in a year by paying 1 cent 
he first day, 3 cents the second, 5 cents the third, and so on, in- 
creasing the payment each day by 2 cents? 

Ans. 1332 dollars 25 cents. 

8. A footman travels the first day 20 miles, 23 the second, 26 
he third, and so on, increasing the distance each day 3 miles. 
How many days must he travel at this rate to go 438 miles? 

Arts, 12. 

9. What is the sum of n terms of the progression of 1,2, 3, 

4, 5, <fec. ? /I e '^/i I N 

10. The sum of the terms of an arithmetical series is 950, 
the common difference is 3, and the number of terms 25. What 
is the first term ? Am. 2. 



^ .m>^i 



ARITHMETICAL PROGRESSION. 193 

11. A man bought a certain number of acres of land, paying 
for the first, $5 ; for the second, $| ; and so on. When he came 
to settle he had to pay $3775. How many acres did he pur- 
chase, and what did it average per acre ? 

Ans. 150 acres at $35^ per acre. 

Problems in Arithmetical Progression to which the precediji^g 
formulas, (A) and (P), do not immediately apply. 

(Art. 117.) When three quantities are in arithmetical progres- 
sion, it is evident that the middle one must be the exact mean 
of the three, otherwise it would not be arithmetical progression ; 
therefore the sum of the extremes must be double of the mean. 

Take, for example, any three consecutive terms of a series, as 
«-l-2J, a-\-^d, a+4cZ; 
and we perceive by inspection that the sum of the extremes is 
double the mean. 

When there are four terms, the sum of the extremes is equal 
to the sum of the means, by (Art. 116.) 

To facilitate the solution of problems, when three terms are 
in question, let them be represented by [x — y), x, {x-\-y), y being 
the common difference. 

When four numbers are in question, let them be represented 
by {x — 37/), {x — y), {x+y), [x-\^3yy, 2y being the common dif- 
ference. 

So in general for any other number, assume such terms that 
the common difference will disappear by addition. 

1. There are five numbers in arithmetical progression, the 
sum of these numbers is 65, and the sum of their squares is 
1005. What are the numbers ? 

Let a:= the middle term, and y the common difference. Then 
X — 2y, x — y, x, x-\-y, a?+2i/, will represent the numbers, 
and their sum will be 5a:=65, or x=13. Also, the sum of 
their squares will be 

5a;2-fl0i/2=1005 or .^2-1-2^=201. 
But a'2=169; therefore, 2?/2=32, 2/'=16 or 2/=4. 

Hence, the numbers are 13 — 8=5, 9, 13, 17 and 21. 
17 



■^--^mr^^' 



194 ELEMENTS OF ALGEBRA. 

9. There are three numbers in arithmetical progression, their 
sum is 18, and tlie sum of their squares 158. What are those 
numbers? Ans. 1, 6 and 11. 

3. It is required to find four numbers in arithmetical progres- 
sion, the common difference of which shall be 4, and their con- 
tinued product 176985. Ans. 15, 19, 23 and 27. 

4. There are four numbers in arithmetical progression, the 
sum of the extremes is 8, and the product of the means 15. 
What are the numbers? Ans. 1, 3, 5, 7. 

5. A person travels from a certain place, goes 1 mile the first 
day, 2 the second, 3 the third, and so on ; and in six days after, 
another sets out from the same place to overtake him, and travels 
uniformly 15 miles a day. How many days must elapse after 
the second starts before they come together? 

Ans. 3 days and 14 days. 
Reconcile these two answers. 

6. A man borrowed $60 ; what sum shall he pay daily to can- 
cel the debt, principal and interest, in 60 days; interest at 10 per 
cent, for 12 months, of 30 days each? 

Ans. $1 and f of a cent. 

•y. There are four numbers in arithmetical progression, the 

sura of the squares of the extremes is 50, the sum of the squares 

of means is 34 ; what are the numbers? Ans. 1, 3, 5, 7. 

8. The sum of four numbers in arithmetical progression is 
24, tlieir continued product is 945. What are the numbers ? 

Ans. 3,5,7,9. 

9. A certain number consists of three digits, which are in 
arithmetical progression, and the number divided by the sum of 
its digits is equal to 26; but if 198 be added to the number its 
digits will be inverted. What is the number ? Ans. 234. 



GEOMETRICAL PROGRESSION. 195 

CHAPTER II. 

GEOMETRICAL PROGRESSION. 

(Art. 118.) When numbers or quantities differ from each 
other by a constant multiplier in regular succession, they consti- 
tute a geometrical series, and if the multiplier be greater than 
unity, the series is ascending ; if it be less than unity, the series 
is descending. 

Thus, 2:6: 18 : 54 : 162 : 486, is an ascending series, the 
multiplier, called the ratio, being three ; and 81 : 27 : 9 : 3 : 1 : 
i : i, &c., is a descending series, the multiplier or ratio being |. 

Hence, a : ar : ar^ : ar^ : ar"^ : ar^ : ar^ : &c., may represent 
any geometrical series, and if r be greater than 1, the series is 
ascending, if less than 1, it is descending. 

(Art. 119.) Observe that the ^rs^ power of r stands in the 
2d term, the 2d power in the 3d term, the third power in the 
4th term, and thus universally the power of the ratio in any 
term is one less than the number of the term. 

The first term is a factor in every term. Hence the 10th 
term of this general series is ar^. The 17th term would be ar^^. 
The nth term would be ar"~^ 

Therefore, if n represent the number of terms in any series, 
and Zthe last term, then L=ar'^^ (1) 

(Art. 120.) If we represent the sum of any geometrical series 
by s, we have 

s^=^a-\-ar-\-ar^-\-ar^ -\- &:c. . . ar'^^^+ar""'. 
Multiply this equation by r, and we have 

rs=ar-\-ar^-\-ar^-{- &c. af''~^-\-ar'^. 
Subtract the upper from the lower, and observe that 
Lr=ar^\ then (r — l)s=Zr — a. 

Therefore, s= ^^"""^, • (2) 

r — 1 ^ ' 

As these two equations are fundamental, and cover the whole 

subject of geometrical progression, let them be brought together 

for critical inspection. 



196 ELEMENTS Or ALGEBRA. 

Z=ar»- (1), S^^ (3). 

These two equations furnish the rules given for the operations 
in common arithmetic. 

Here we perceive five quantities, a, ?•, n, L and S^ and any 
three of them being given in any problem, the other two can be 
determined from the equations. 

To these equations we may apply the same remarks as were 
made to the two equations in arithmetical progression (Art. 116.) 

Equation (2), put in words, gives the following rule for the 
sum of a geometrical series ; 

Rule. Multiply the last term by the ratio ^ and from the 
product subtract the first term, and divide the remainder by the 
ratio less one, 

EXAMPLES FOR THE APPLICATION OF EQUATIONS 
(1) AND (2). 

V 1. Required the sum of 9 terms of the series, 1, 2, 4, 8, 16, 
&c. JJns. 511. 

2. Required the 8th term of the progression, 2, 6, 18, 54, 
&c. Jins. 4374. 

3. What is the sum of ten terms of the series 1, f , J, &c. ? 

/ffjo " 1 7 4 7 5 

<l» Required two geometrical means between 24 and 192. 
N. B. When the two means are found, the series will consist 
of four terms ; the first term 24 and the last term 192. 
By equation (1) Z=ar"-^ 
Here a— 24, Z=192, n=4, and the equation becomes 
192=24r=' or r=2. 
Hence, 48 and 96 are the means required. 

5. Required 7 geometrical means between 3 and 768. 

^ns, 6, 12, 24, 48, 96, 192. 

6. The first term of a geometrical series is 5, the last term 
1215, and the number of terms 5, What is the ratio 1 tdns. 3. 



GEOMETRICAL PROGRESSION. 197 

•y. A man purchased a house, giving $1 for the fust door, $2 
for the second, $4 for the third, and so on, there being 10 doors. 
What did the house cost him ? Jins, $1023. 

(Art. 121.) By Equation (2), and the Rule subsequently given, 
we perceive that the sum of a series depends on the first and last 
terms and the ratio, and not on the number of terms ; and 
whether the terms be many or few, there is no variation in the 
rule. Hence, we may require the sum of any descending series, 
as 1, 5, 1, \i &c., to infinity, provided ive determine the last 
term. Now we perceive the magnitude of the terms decrease 
as the series advances ; the hundredth term would be extremely 
small, the thousandth term very much less, and the infinite term 
nothing ; not too small to be noted, as some tell us, but absolutely 
nothing. 

Hence, in any decreasing series, when the number of terms 
is conceived to be infinite, the last term, X, becomes 

0, and Equation (2) becomes ,9= -• 

By change of signs s=- — ;-• 

This gives the following rule for the sum of a decreasing infi- 
nite series : 

Rule. Divide the first term by the difference between unity 
and the ratio, 

EXAMPLES. 

1. Find the value of 1, |, ~, &c., to infinity. 

a=l, r=|. ^ns. 4. 

2. Find the exact value of the decimal .3333, (fee, to infinity. 

Jins, I. 

This may be expressed thus : to + to o» ^^' Hence, 

0' ' 10 



3. Find the value of .323232, &;c., to infinity. 

«=T(fo' «^=T7Voo ; therefore r=^^-^. Ans. 



32 

99' 



4. Find the value of .777, &c., to infinity. Jins. |. 

r2 




198 ELEMENTS OF ALGEBRA. 

5. Find the value of | : 1 : | : ^3, &c. to infinity. 

6. Find the value of 5 : f : |, Sic. to infinity. ^ns. 7h 

7. Find the value of the series ^, 3'^, &c., to infinity ? ^ns. |. 

8. What is the value of the decimal .71333, Sic, to infinity? 

9. What is the value of the decimal .212121, <&;c., to infinity? 

(Art. 122.) If we observe the general series, (Art. Il8.)a: ar: 
ar^ : ar^ : ar^, Sic, we sliall find, by taking three consecutive 
terms anywhere along in the series, that ihe product of the ex- 
tremes will equal the square of the mean. Hence, to find a 
geometrical mean between two numbers, we must multiply them 
together, and take the square root. If ive take four consecutive 
terms, tlte product of the extremes ivill be equal to the product 
of the means. 

(Art. 123.) This last property belongs equally to geometrical 
proportion, as well as to a geometrical series, and the learner must 
be careful not to confound proportion with a series. 

a : ar : : b : br, is a geometrical proportion, 72ot a continued 
series. The ratio is the same in the two couplets, but the mag- 
nitudes a and b, to which the ratio is applied, may be very dif- 
ferent. 

We may suppose a : ar two consecutive terms of one series, 
and b : br any two consecutive terms of another series having 
the same ratio as the first series, and being brought togetiier they 
form a geometrical proportion. Hence, the equality of the ra- 
tio constitutes proportion. 

To facilitate the solution of some difficult problems in geomet- 
rical progression, it is desirable, if possible, to express several 
terms by two letters only, and have them stand sym,metrically. 

Three terms maybe expressed by x : Jxy : y, or by x^ : xy : 
y^ as the product of the extremes are here evidently equal to the 
square of the mean. 

To express four terms with x and y symmetrically, we at first 



GEOMETRICAL PROGRESSION. 199 

write P : X : : y : Q. The first three being in geometrical progres- 

^2 2 

sion, gives P3/=x''^ or /*= — . In the same manner, we find Q=~ 

And taking these values of P and Q we have— : a?:: 3/: — to rep- 
resent four numbers symmetrically with two letters. 

Taking three numbers as above, and placing them between P 
and Q^ thus, P : x^ : xy : y^ : Q, we have five numbers ; and 

by reducing P and Q into functions of x and 1/, we have—: x^ : 



y 



ay : y^ : — , for five terms symmetrically expressed. 

x^ a^ ^2 ,.3 

Six numbers thus, — ,: — : x : : y : ~ : '-z 

y^ y ^ X XT 

Sometimes we may more advantageously express unknown 
numbers in geometrical proportion by iP, xy^ xy^^ &c. ; x being 
the first term, and y the ratio. 

HARMONICAL PROPORTION. 

(Art. 124.) When three magnitudes, a, b, c, have the relation 
of a : c : : a — b : b — c ; that is, the first is to the third as the dif- 
ference between the first and second is to the difference between 
the second and third, the quantities a, b, c, are said to be in har- 
monical proportion, 

(Art. 125.) Four magnitudes are in harmonical proportion 
when the first is to the fourth as the difference between the first 
and second is to the difference between the third and fourtli. 
Thus, a, b, c, d, are in harmonical proportion when aid:: 
a — b : c — d, or when a : d: : b — a : d — c. 

An harmonical mean between two numbers is equal to twice 
their product divided by their sum. For a : x : b representing 
three numbers in harmonical proportion, we have by the definition, 
(Art.124.) a: b : : a — x : x — b. 

Therefore, ax — ab=ab — bx or a:=— r-7. 

a-^b 



200 ELEMENTS OF ALGEBRA. 

1. Find the liarmonical mean between 6 and 12. \8ns. 8. 

2. Find the third harmonical number to 234 and 144. 

£ns, 104. 

3. Find the fourth harmonical proportion to the numbers 24 : 
16: 4. dns, 3. 

4. There are four numbers in harmonical proportion, the first 
is 16, the third 3, the fourth 2. The second is lost ; find it. 

Ans. 8. 

PROBLEMS IN GEOMETRICAL PROGRESSION, AND 
HARMONICAL PROPORTION. 

1. The sum of three numbers in geometrical progression is 
26, and the sum of their squares 364. What are the numbers ? 
Let the numbers be represented by x : Jxy : y. 
Then x-\-Jxy'-\-y= 26=a (1) 

And a?24- xy-{'y^=ZQi=b (2) 

Transpose, Jxy in Equation (1) and square, we have 
x^-\-2xy-\-y'^=c^ — 2aJxy-\-xy (3) 

Or a-2+ xy-i-y^=a'~2ajxy (4) 

The left hand members of Equations (2) and (4) are equal ; 

hence, , — 

a^ — 2aJxy—o. 

^2 J 

Therefore, Jxy=^-- — =6 (5) 

This gives the second term of the progression, and now from 
equations (1) and (5) we find ie=2, 2/=18, and the numbers 
are 2, 6, 18. 

52. The sum of four numbers in geometrical progression is 15, 
(a), and the sum of their squares 85 (6). What are the 
numbers ? 

Let tlie numbers be represented as in (Art. 123.) 



Then |+^-h2/+|-=«(l) 
And J+^+2/^+g=M2) 



GEOMETRICAL PROGRESSION. 201 

Assume x-\-y^=s 
xy=p 
Then by (Art. 112.) 

And xr^-\-y^=s^ — 3sp 
Transposing {x-\-y) in Equation (I), and {x^-{-y^) in Equation 
(2), we have 

'^^t=:a^s (3) and '^^+t=b-8'+2p (4) 

y ' a? ^ ' y^ ^ x^ ^ ' 

Square (3) and transpose 2xy or 2jo and 

|+J=(a-s)=-2;> (5) 

The left hand members of equations (4) and (5) are equal, there- 
fore, {a—sY—2p^h'—s''-{-2p 

Or a^—2as-\-2s^—^p=h (6) 

Clear equation (3) of fractions, and x^-\-y^—ap — ps. 

That is, 6'^ — 3sp=ap — ps or p=-— - — (7) 

Ct~\~ -is 

Put this value of p in equation (6) and reduce, we have. 

Or as''-\-bs=^{a'^b) 

Taking the given values of a and b we have, 
155-2+85i'=70X15 
Or 3*2 -{-17s =2 10, an equation which gives s=6. 
Put the values of a and s in equation (7), and/?=8. 
That is, a?-|-i/=6, and xy=%, from which we find x=2, and 
i/=4 ; therefore, the required numbers are 

1, 2, 4 and 8, Ans. 

.3. The arithmetical mean of two numbers exceeds the geo- 



202 ELEMENTS OF ALGEBRA. 

metrical mean by 13, and the geometrical mean exceeds the har- 
monical mean by 12. What are the numbers? 
Let X and y represent the numbers. 
Then l{x-\-y)=^ the arithmetical mean, ijxy= the geome- 
trical mean, (Art. 112.) and — r^= the harmonical mean. 

Let «=12; 

Then, by the question, i{x^y) = Jxy-{-a-\-\ (1) 

And Jxy—^^ -\-a (2) 

^ "^ x-\-y ^ ' 

By onr customary substitution, these equations become 
is=^^-{-a-i-l (3) 

And Jv=-^\a . (4) 

Take the value of s from equation (3) and put it into equation 
(4), dividing the numerator and the denominator by 2, and we 
have , ;j , ,_, 

Clearing of fractions, we shall have 

Drop equals, and Jp=:^[a-\-l)a (6) 

Put tliis value of Jp in equation (3) and we have 

Or s=2(a4-l)2 (7) 

For the sake of brevity, put («+l)=^; squaring equation 
(6) and restoring the values of s and p in equations (6) and ^7), 
and we have xy=a^b^ {Jl) 

x-{-y=2b^ (B) 
Square (JB) and 

a^-\-2xy^7f=ib^ 
Subtract 4 
times (^) 



I 4.xy ^ia'b^ 



And x''-^2xy+y^=4b%b'--a^)=4b%b-\-a){b^a) (C) 



GEOMETRICAL PROGRESSION. 203 

As «=12 and bz=l3, b-{-a=25, and b — a=l. 
Therefore, (C) becomes (a?— ?/f= 46^X25 X 1. 
By evolution, x — y=2bX5 
Equation {B) x-\-y=2¥ 
By addition 2x=2b^-^\.Qb 

Or x= 62-{- 56=(Z>+5)6=18X13=234 

By subtraction, 2y=2b^ — 106 

y^ Z>2— 5Z>=(6--5)6=^ 8X13=104. 
A more brief solution is the following : 
Let X — y and x-{-y represent the numbers. 
Then x= the arithmetical mean, Ja^ — y'^=^ the geomelri- 

X "—IJ 

cal mean, (Art. 112), and —= the harmonical mean. By 

the question, 

a-— 13==^"^^=^(1), and ^^^y -[-n^.Jx'—xf (2) 



The right hand members of equations (1) and (2) being the 
same, therefore, —-\-\2-=^x — 13. 

X 

By reduction, y^=2DX. 
Put this value of y^ in equation (1), and by squaring 

x"—2Qx-\-{l3f=x^—2bx, or a;=(13)2=169. 
Hence, ,y=65, and the numbers are 104 and 234. 

-1. Divide the number 210 into three parts, so that the last 
shall exceed the first by 90, and the parts be in geometrical pro- 
gression. Ans. 30, 60, and 120. 

5. The sum of four numbers in geometrical progression is 
30 ; and the last term divided by the sum of the mean terms is 
1|. What are the numbers ? Ans, 2, 4, 8, and 16. 

6. The sum of the first and third of four numbers in geo- 
metrical progression is 148, and the sum of the second and 
fourth is 888. What are the numbers ? 

Ans, 4, 24, 144, and 864. 



204 ELEMENTS OF ALGEBRA. 

7. It is required to find three numbers in geometrical progres- 
sion, such that their sum shall be 14, and the sum of their squares 
84. Ans. 2, 4, and 8. 

8. There are four numbers in geometrical progression, the 
second of which is less than the fourth by 24 ; and the sum of 
the extremes is to the sum of the means, as 7 to 3. What are 
the numbers ? Ans. 1, 3, 9 and 27. 

9. The sum of four numbers in geometrical progression is 
equal to the common ratio +1, and the first term is ^-^. AVhat 
are the numbers ? Ans. ^^, y\, y q, f-^. 

10. The sum of three numbers in harmonical proportion is 
26, and the product of the first and third is 72. What are the 
numbers ? Ans. 12, 8, and 6. 

11. The continued product of three numbers in geometrical 
progression is 216, and the sum of the squares of the extremes 
is 328. What are the numbers ? Ans. 2, 6, 18. 

12. The sum of three numbers in geometrical progression is 
13, and the sum of the extremes being multiplied by the mean, 
the product is 30. What are the numbers ? 

Ajis. 1, 3, and 9. 

13. There are three numbers in harmonical proportion, the 
sura of the first and third is 18, and the product of the three is 
576. What are the numbers ? Aiis. 6, 8, 12. 

14. There are three numbers in geometrical progression, the 
difference of whose difference is 6, and their sum 42. What 
are the numbers ? A?7S. 6, 12, 24 

15. There are three numbers in harmonical proportion, the 
difference of M'hose difference is 2, and three times the product 
of the first and third is 216. What are the numbers ? 

Ans. 6, 8, and 12. 

IC. Divide 120 dollars between four persons, in such a 
way, that their shares may be in arithmetical progression ; and 
if the second and third each receive 12 dollars less, and the 



GEOMETRICAL PROPORTION. 205 

fourth 24 dollars more, the shares would then be in geometri- 
cal progression. Required each share. 

Arts. Their shares were 3, 21, 39, and 57, respectively. 

IV. There are three numbers in geometrical progression, 
whose sum is 31, and the sum of the first and last is 26. What 
are the numbers ? *^n8. 1, 5, and 25. 

18. The sura of six numbers in geometrical progression is 
189, and the sura of the second and fifth is 54. What are the 
numbers ? Ans. 3, 6, 12, 24, 48 and 96. 

19. The sura of six numbers in geometrical progression is 
189, and the sum of the two means is 36. What are the num- 
bers ? dns. 3, 6, 12, 24, 4^ and 96. 

CHAPTER III. s i- ^ ; ^ : : ^\-f si : v> 

PROPORTION. 

(Art. 176.) We have given the definition of geometrical pro- 
portion in (Art. 41.) and demonstrated the most essential prop- 
erty, the equality of the products between extremes and means. 
We now propose to extend our investigations a little farther. 

Proportion can only exist between magnitudes of the same 
kind, and the number of times and parts of a tirae, that one 
measures another, is called the ratio. Ratio is always a num- 
ber, and not a quantity, 

(Theorem 1.) If two magnitudes have the same ratio as 
two other Sf the first two as numerator and denominator may 
form one member of an equation ; and the other two magnitudes 
as numerator and denominator will form the other member. 

Let A and B represent the first two magnitudes and r their ratio. 

Also C and D the other two magnitudes, and r their ratio. 

Then, — =r and — =r Therefore, (Ax. 7) __=-— 

(Theorem 2.) Magnitudes which are proportional to the 
same proportionals, are proportional to each other. 

Suppose a: b : : P : Q ) Then we are to prove that 



and c: d : : F : Q > a: b: : c: d 

and x: y :: F : Q ) and a: b:: x: y, &c. 



m(^ ELEMENTS OF ALGEBRA. 

b Q 
From the first proportion, — =— 

From the second, — = — 

c F 

Hence, (Ax. 7) — = — or a : b : : c : d 
^ a c 

In the same manner we prove a : b : : x : y 
And c : d : : X : y 

(Theorem 3.) If four magnitudes constitute a proportion, 
the Jirst will be to the sum of the first and second, as the 
third is to the sum of the third and fourth. 

By hypothesis, a : b : : c : d ; then we are to prove that 
a : a-{-b : : c : c+cZ. 

By the given proportion,— =-. Add unity to both mrmbers, 

and reducing them to the form of a fraction we have = '-. 

* a c 

Throwing this equation into its equivalent proportional form, we 

have, I , , , 

a : a-f-b : : c : c-\-d. 

N. B. In place of adding unity, subtract it, and we shall find 

that 7 1 

a : a — b : : c : c — a. 

or a : b — a : : c : d — c. 

(Theorem 4.) If four m^agnitudes be proportional, the sum 
of the first and second is to their difference, as the sum of 
the third and fourth is to their difference. 

Admitting that « : 6 : : c : c?, we are to prove that 

a-\-b : a — b : : c-\-d : c — d 
From the same hypothesis, (Theorem 3.) gives 
a : a-\-b : : c : c-\-d 
And a : a — b : : c : c — d 

Changing the means, (which will not afl!(ect the product of the 
extremes and means, and of course will not destroy proportion- 
ality,) and we have, 



GEOMETRICAL PROPORTION. %(yi 

a : c : : a-{-b : c-\~d 

a : c : : a — b : c — d 
Now by (Theorem 2.) a-rb : c-[-d : : a — b : c — d 
Changing the means, a-{-b : a — b : : c-{-d : c — d 

(Theorem 5.) Jf four magnitudes he proportional, like 
powers or roots of the same, will he proportional. 

Admitting a : b : : c : d, we are to show that 
11 II 

n 71 71 n 

«" : 6'' : : c" : d'\ and a : b : : c : d 

(I c 
By the hypothesis, y^-j- Raising both members of thw 

equation to the nth power, and ' ''■' 

h^~d'' 
Changing this to the proportion a" : 6" : : c^ : d'^ 

a c 
Resuming again the equation 7=-^> and taking the nth root 



a c 
of each member, we have — - =— ,-• Converting this equation 





6" 


rf" 




into its 


equivalent proportion, we have 






1 1 
a : h : 


1 

n 

: c : 


1 

r 



vfif,:> 



Now by the first part of this theorem, we have . 

mm m _)n 

« " : 6 '* : : c " : cZ " , 7n representing any 
power whatever, and n representing any root. 

(Theorem 6.) If four magnitudes be proportional, also 
four others, their compound, or product of term by term, will 
form a proportion. 



208 ELEMENTS OF ALGEBRA, 



Admitting that 


a: b: 


:c .d 


And 


x: y: 


: m : n 


We are to show that, 


ax : by : 


: mc : ni 


From the first proportion, 


a c 
b~~d' 




From the second, 


x_m 





y n 

Multiply these equations, member by member, and 

ax_mc 
by nd 
^ Or ax : by : : mc : nd. 

The same would be true in any number of proportions. 
(Theorem 7.) Taking the same hypothesis as in (Theorem 
6.) we propose to show, that a proportion may be formed by di- 
viding one proportion by the other, term by term. 
By hypothesis, a : b : : c : d 
And X : y : : m : 71 
Multiply extremes and means, ad=bc (I) 
And nx=my (2j 

Divide (1) by (2), and -X -=- X - 
^ ' ^ ' X n m y 

Convert these four terms, which make two equal products, into 
a proportion, and we shall have 

abed 
X ' y ' ' m ' n 

By comparing this with the given proportions, we find it com- 
posed of the quotients of the several terms of the first propor- 
tion divided by the corresponding term of the second. 

(Theorem 8.) If four magnitudes be proportional, we may 
multiply the first couplet or the second couplet, the antecedents 
or the consequents, or divide them by the same factor, and the 
results will be proportional in every case. 



GEOMETRICAL PROPORTION. 209 

Suppose a : b :: c : d 

Multiply ex. and means, and .... ad=bc (1) 

Multiply this eq. by m, and .... mad=mbc 

Now, in this last equation, ma may be considered as a single 
term or factor, or md may be so considered. So, in the second 
member, we may take mb as one factor, or mc. Hence we may 
convert this equation into a proportion in four different ways. 

Thus, as ma : mb : : c : d 

or as a : b :: mc : md 

or as ma : b : : mc : d 

or as a : mb : : c : md 

If we resume the original equation (1), and divide it by any 
number, m, in place of multiplying it, we can have, by the same 
course of reasoning, 

a b _ 

— : — :: c : d 
m m 

, c d 

a : o : : - - : — 

m m 

— : b : : — : d 
m, m 

b d 

a : — :: c : — 

m m 

The following examples are intended to illustrate the practi- 
cal utility of the foregoing theorems : 

EXAMPLES. 

1. Find two numbers, the greater of which shall be to the 
less, as their sum to 42 ; and as their difference is to 6. 

Let x=\[\e greater, r/=the less. 
18 



310 ELEMENTS OF ALGEBRA. 



Then, per question, 


lx:y: 


:x-\-y:42 
ix-^y: 6 


0) 
(2) 


(Theorem 2.) 


x+y : 


42 : : x—y : 


6 


Changing the means 


x-\-y : 


X — y : : 42 : 


6 


(Theorem 4.) 


2x 


2y ::48: 


36 


(Art. 42.) 


x : 


y:: 4: 


3 


(Theorem 2.) 


4 


• 3 : : ar— 2/ : 


6 


And 


4 


3 ::x-{-y: 


42 



From these last proportions, x — y= 8 

And x-\-y=5Q. Hence, a?=32, y=24. 

2. Divide the number 14 into two such parts, that the 
quotient of the greater divided by the less shall be to the quo- 
tient of the less divided by the greater, as 16 to 9. 

Let x= the greater part, and 14 — a;=the less. 

X 14— i3? 

By the conditions, — : : : 16 : 9 

14 — X X 

Multiplying terms, x^ : (14 — xf : : 16 : 9 

Extracting root, x : (14 — x) : : 4:3 (Theor. 5.) 

Adding terms, x : 14 : ; 4:7 

Dividing terms, x : 2 : : 4 : 1 

Therefore, x=S. 

3. There are three numbers in geometrical progression whose 
sum is 52, and the sum of the extremes is to the mean as 10 to 
3. What are the numbers ? j^ns. 4, 12, and 36. 

Let X, xy, xy^ represent the numbers. 

Then, by the conditions, x-\-xy-\-xy^=62 (1) 

And xy^-\-x : xy : : 10 : 3 

(Art. 42.) y^+l : y : : 10 : 3 

Double 2d and 4th, y^-\-l : 2y : : 10 : 6 

Adding and sub. terms, y^-^2y-\-l : y^ — 2y-]-l : : 16 : 4 

Extracting square root, y+1 ' y — 1 : : 4 : 2 

Adding and sub. terms, y : 1 : : 3 : 1 Hence, y—S. 

52 52 52 

From equation (1), ^=j^_=__- =_=4. 



GEOMETRICAL PROPORTION. 211 

4. The product of two numbers is 35, the diflerence of their 
cubes, is to the cube of their difference as 109 to 4. What are 
the numbers ? Ans, 7 and 6. 

Let X and y represent the numbers. 
Then, by the conditions, a?y=35, and x^ — y^ : [x — yY ' : 109:4 
Divide by [x—y) (Art. 42.) and x^-\-xy-\-y^ : {x—yf : : 109:4 
Expanding, and a^-\-xy-\-y^ : a^ — 2xy-\-y^ : : 109:4 

(Theorem 3.) ^xy : {x^yf : : 105:4 

But Sxy, we know from the first equation, is equal to 105. 
Therefore, {x — yy=4, or x — ^=2, 

We can obtain a very good solution of this problem by putting 
x-^y== the greater, and x — ^?/= the less of the two numbers. 

5. What two numbers are those, whose difference is to their 
sum as 2 to 9, and whose sum is to their product as 18 to 77? 

^ns. 1 1 and 7. 

O. Two numbers have such a relation to each other, that if 4 
be added to each, they will be in proportion as 3 to 4 ; and if 4 
be subtracted from each, they will be to each other as 1 to 4. 
What are the numbers ? - ^ns. 5 and 8. 

v. Divide the number 16 into two such parts that their pro- 
duct shall be to the sum of their squares as 15 to 34. 

Jlns. 10, and 6. 

8. In a mixture of rum and brandy, the difference between 
the quantities of each, is to the quantity of brandy, as 100 is to 
the number of gallons of rum; and the same difference is to the 
quantity of rum, as 4 to the number of gallons of brandy. 
How many gallons are there of each 1 

Jins. 25 of rum, and 5 of brandy 

9. There are two numbers whose product is 320 ; and the 
difference of their cubes, is to the cube of their difference, as 61 
to 1. What are the liumbers? ^ns. 20 and 16. 

10. Divide 60 into two such parts, that their product shall be 
to the sum of their squares as 2 to 5. ^ns. 40 and 20. 



212 ELEMENTS OF ALGEBRA. 

11. There are two numbers which are to each other as 3 to 
2. If 6 be added to the greater and subtracted from the less, 
the sum and the remainder will be to each other, as 3 to 1. 
What are the numbers ? Ans. 24 and 16. 

12. There are two numbers, which are to each other, as 16 
10 9, and 24 is a mean proportional between them. What are 
the numbers 1 Ans. 32 and 18. 

13. The sum of two numbers is to their difference as 4 to 1, 
and the sum of their squares is to the greater iis 102 to 5. 
What are the numbers ? Ans. 15 and 9. 

14. If the number 20 be divided into two parts, which are to 
each other in the duplicate ratio of 3 to 1, what number is a 
mean proportional between those parts ? 

Ans, 18 and 2 are the parts, and 6 is the mean proportion 
between them. 

15. There are two numbers in proportion of 3 to 2 ; and if 

6 be added to the greater, and subtracted from the less, the results 
will be as 3 to 1. What are the numbers? Jins. 24 and 16. 

16. There are three numbers in geometrical progression, the 
product of the first and second, is to the product of the second 
and third, as the first is to twice the second; and the sum of 
the first and third is 300. What are the numbers ? 

Ans. 60, 120, and 240. 

IT, The sum of the cubes of two numbers, is to the differ- 
ence of their cubes, as 559 to 127 ; and the square of the first, 
multiplied by the second, is equal to 294. What are the num- 
bers ? Arts. 7 and 6. 

18. There are two numbers, the cube of the first is to the 
square of the second, as 3 to 1 ; and the cube of the second is 
to the square of the first as 96 to 1. What are the numbers ? 

Ans. 12 and 24. 



BINOMIAL THEOREM. 213 



SECTION VI 



CHAPTER I. 



INVESTIGATION AND GENERAL APPLICATION OF 
THE BINOMIAL THEOREM. 

(Art. 127.) It may seem natural to continue right on to the 
higher order of equations, but in the resolution of some cases in 
cubics, we require the aid of the binomial theorem ; it is there- 
fore requisite to investigate that subject now. 

The just celebrity of this theorem, and its great utility in the 
higher branches of analysis, induce the author to give a general 
demonstration : and the pupil cannot be urged too strongly to 
give it special attention. 

In (Art. 67.) we have expanded a binomial to several powers 
by actual multiplication, and in that case, derived a law for form- 
ing exponents and coefficients when the power was a whole 
positive number ; but the great value and importance of the 
theorem arises from the fact that the general law drawn from that 
case is equally true, when the exponent is fractional or negative, 
and therefore it enables us to extract roots, as well as to ex- 
pand powers. 

(Art. 128.) Preparatory to our investigation, we must prove 
the truth of the following theorem : 

If there be two series arising from different modes of ex- 
panding the same, or equal quantities, with a varying quan- 
tity having regular powers in each series; then the coefficients 
of the same ptowers of the varying quantity in the two series 
are equal. 

For example, suppose 
Jl-{-Bx-\-Cx'-\-Dx\&Lc. =a-]-bx-\-cx^^-da?, &c. 
This equation is true by hypothesis, through all values of x. 
It is true then, when a;=0. Make this supposition, and A=a, 
Now let these equal values be taken away, and the remainder 
divided by x. Then again, suppose x=^0, and we shall find 
B=h, In the same manner we find C=c, D=d, -S=:e, &c. 



214 ELEMENTS OF ALGEBRA. 

(Art 129.) A binomial in the form of a-{-x may be put in 
the form of «xf 1-] — J; for we have only to perform 
the multiplication here indicated to obtain a-\-x. Hence, 

1 + - ) > it will be sufficient to mul- 
tiply every term of the expanded series by a™ for the expansion 
of (a-\-x)^, but as every power or root of 1 is I, the first term 

1 + -- j is 1, and this multiplied by a" 

must give a^ for the first term of the expansion of {a-{-x)^y what- 
ever m may be, positive or negative, whole or fractional. 

X 

As we may put x in place of — , we perceive that any bino- 
mial may bo^reduced to the form of (1+a^), which, for greater 
facility, we shall operate upon. 

(Art. 130.) Let it be required to expand (l+a;)"', when m is 
a positive whole number. By actual multiplication, it can be 
shown, as in (Art. 67.) that the first term will be 1, and the 
second term mx. For if m=2, then 

{l-JrxY={l-{-xY=i-\-2x, &c. 
If ?7i=3, (l+ar)'"=l+3:r, &c. 
And in general, {l-\-xy"=l-\rmx-\-Jlocr^y-\'Bx% &c. 

The exponent of x increasing by unity every term, and ./?, 
JB, C, &c., unknown coefficients, which have some law of de- 
pendence on the exponent m, which it is the object of this investi- 
gation to discover. 

(Art. 131.) Now if m is supposed to be a fraction, or if m=-, 
the expansion of (1+a;)"* will be a root in place of a power, and 

we must expand (l-J-x)*^. i 

For example, let us suppose r=2, then {l-\-ooY = {\-\-x) , 



BINOMIAL THEOREM. 215 

and to examine the form the series would take, let us actually 
undertake to extract the square root of (1-j-a?) by the common 
rule. 

OPERATION. 



l-{-x(i^hx—^x' 



2A 



1 1 



2+^x X 

x-\-\qi?' 

Thus we perceive that in case of square root, the first term of 
the series must be unity, and the coefficient of the second term 
is the index of the binomial, and the powers of x increase by 
unity from term to term. 

We should find the same laws to govern the/o7'm of the series, 
if we attempted to extract cube, or any other root ; but, to be 
general and scientific, we must return to the literal expression 

Now as any root of 1 is 1, the first term of this root must be 

1, and the second term will have some coefficient to x. Let 

that coefficient be represented by p ; and as the powers of x will 

increase by unity every term, we shall have 
I 

{l-{-xy=^l-\-px-]-j9x\ &c. 
Take the r power of both members, and we shall have 
l+x=-^{l-[-px+^x^ &C.Y 

As r is a whole number, we can expand this second mem- 
ber by multiplication ; that is, by (Art. 130.), the second mem- 
ber must take the following form 

l-\-x=\-i-rpx-{-^'x\ Sic. 

Drop 1, and divide by x, and we have 

By (Art. 128.) l=:rp or jo=- ; 



216 ELEMENTS OF ALGEBRA. 

That is, the coefficient represented by /; must be equal to the 
index of tlie binomial. 



Therefore {l-]rx)'' = !•}-- x-{-£x^-{- Bx^-\-, &c. ; the same 
general form as when the exponent was considered a whole 
number. 

(Art. 132.) If we take m=- we have to expand the ro6t of 
a power. The first term must be 1, and the second term will 
contain a', with some coefficient, and the coefficients of x will 
rise liigher and higher every term. 

n 

That is, {i+x)r =i-{-px-\-Mx^, &c. Take the r power of 
both members, and {l-{-x)^={l-\-px, <fec.)''. 

Expanding both members, as in (Art. 130), 

l-\-nx-{-ax'^, SLc. = l-\-rpx~{-^x^, &c. 
Now, by (Art. 128), 7i=rp or p=-. 



Therefore, (1 -fjc) ^= 1 -j- -x-\-^a^+Bx\ &c. ; the first two 

terms following the same law, relative to the exponents, as in the 
former cases. Now let us suppose w negative. Then 

(l+a;)'" will become (l+a?)-^= (Art. 18.) 

Or by (Art. 130.) ^ 



l-{-mx-\-^x^, &c. 
By actual division, l-{-ma7+.^a?^, &c.) 1 (1 — mx-jr^d'a^, &c. 

— mx — £xi^ 
— mx — m^x^ 

That is, (l-|-a?)~^=l — mx-\-A'x^, which shows that the same 
general law governs the coefficient of the second term, as in the 
former cases. 

Hence it appears that whether the exponent 7n of a binomial 



BINOMIAL THEOREM. 



217 



be positive or negative, whole or fractional, the same general 
form of expression must be preserved. 

That is, in all cases {l-\-x)'^=l-\-mx-\-^x^-\-BaP, &c. 

(Art. 133.) For clearness of conception, let the pupil bear in 
mind that the coefficients of an expanded binomial quantity 
depend not at all on the magnitudes of the quantities themselves, 
but on the exponent. Thus, {a-\-b) to the 5th, or to any other 
power, the coefficients will be the same, whether a and b are 
great or small quantities, or whatever be their relation to each 
other. 

(Art. 134.) The equation {l-\-xy''=l-\-mx-[-Ja^-{-Bx\&c,y 
is supposed to be true, therefore it must be true, if we square 
both members. But we have only a portion of one member. 
We have, however, as much as we please to assume, and suffi- 
cient to determine the leading terms of its square, which is all 
that we desire. Square both members, and (l-i-2^-i-a:^)'"= 

By expanding the second member, and arranging the terms 
according to the powers of x, we shall have 



(1 +2a:-f a;2)'«=l +2ma:-l-^^ 



^+2m^ 



3 . 2C 1 



(fee. 



(1) 



Now if we assume y=2x-\-x^, the first member of this equa- 
tion becomes {l-\-y)^. If we expand this binomial into a serieSf 
it must have the same coefficients as the expansion of [l-\-x)^, 
because the coefficients depend entirely on the exponent m 
(Art. 133.) 

Therefore, {l-{-y)^=l -\-my-\-Ay''-\-By^-\- Cy\ &c. 

Put the values of y, y^, y^, <fcc., in this equation, and arrange 
the terms according to the powers of x, and we have 



(l+2a?-H^'^ 



■■l-{-2mx-\-^^ 



a;2+ 



4^ 
8^ 



^'+121? 
16C 



x% (fee. (2) 



The left hand members of equations (1) and (2), are identical, 
19 



# 



v 



218 ELEMENTS OF ALGEBRA. 

and the coefficients of like powers in the second member must 
be equaL (Art. 128.) 

Therefore, 4£-\-m=m^-\-2M, or A= — - — =w»— x — 

And 8 B-\-4A='lmJi-^%B 

- Ji{r)\ — 2) 7??— 1 m— 2 
Therefore B= ^ — -^=m. — — (3) 

O lit A 

By putting the coefficients of the fourth powers of x equal, 
we have 

14C-l-1254-.^=2wi2?+^2 

To obtain the value of C from this equation, in the requisite 
form, is somewhat difficult. 

We must make free use of the preceding values of A and j5, 
which are alone sufficient ; but, to facilitate the operation, we 
shall make use of the following auxiliary. 

Assume P=m — 2, then P-l-l=m — 1, 

mP^-m ?n— 1 
and — =m,—--^Ji (a) 

Also, by inspecting equation (3), we perceive that 

^B=dP, and 2mJ5=I^l^ (Jj) 

o 

By putting the values of 125 and 2mB in the primitive equa- 
tion, and dividing every term by A, and in the second member 
taking the value of A from equation (a), and we shall have 

14C 2mP mP+m 

A -f*^-^^- 3 "^ 2 

Multiply by 6, and substitute the value of 

24P+ 6=24m — 42, because P=m — 2, and we have 

84 C 

+24m— 42=4mP-f3mP+3m. 
Ji 

Collecting terms, and dividing by 7, gives 
•2 * 



>!**^* 



BINOMIAL THEOREM. 219 

But dm — 6=3P, which substitute and transpose, and 

Ji 

These results, showing the vahies of A, J9, and C, in terms 
of the exponent of a binomial, clearly point out one invariable 
law of connection and dependence of coefficients and exponents 
one upon another ; thus arriving at the following general expan- 
sion of a binomial quantity through a rigid demonstration, includ- 
ing every case. 

(l+a:)"*=l+ma?+wi— - — x^-\-m—- — x\ &,c. (1) 

2 2 3 

In this equation we may write any quantity, whole or frac- 
tional, in place of a?, and the expansion will be the same. Now 

X 

in place of a; write -, and we have 
a 

, , X , m — 1 a^ , m — 1 m — 2 a;^ „ 
l+m-+m.-— .^+m.^-.— --3,&c. (2) 

Multiply both members of this equation by a™, and we have 
(«+a?)'"=a'»+?na™-'a:-|-m.— -— «"»-V, &c. (3) 

Either formula, (1), (2), or (3), may be used ; one may be 
more convenient than another, in particular cases of application. 

When m is a whole positive number the series will terminate ; 
all other cases will result in an infinite series. 

APPLICATION. 

1. Required the square root of {\-\-x). 

Apply formula (1), making m=| ; then the derelopmen* 
will be 

/, I \5 t I 1 a> , oX OtuX 

{l-\-x) =l-[-^rr— — 4- 



('+!)• 



2.4 ' 2.4.6 2.4.6.8 



The law of the series, in almost every case, will become appa- 
rent, after expanding three or four terms, provided we keep the 
factors separate. 



220 ELEMENTS OF ALGEBRA. 

The above will be the coefficients for any binomial square 
root (Art. 133.) ; hence the square root of 2 is actually expressed 
in the preceding series, if we make a;=l. 

Then (1-f 1)^=1+-^-— -—, &c. The square root of 3 is 
expressed by the same series, when we make a: =2, &c. 

2. Required the cube root of {a-{-b) or its equal, a( \-\ — J. 

Formula (2) gives 

5/1. *\^ ^/,_i_^ 262 2.56« 2.5.86^ . \ 
« (^+«) =« (^+3^-3:6^^+3:6:9^3-3:^X12^.' &c.j 

This expresses the cube root of any binomial quantity, or any 

quantity that we can put into a binomial form, by giving the 

proper values to a and b. For example, required the cube root 

1 
of 10, or its equal, 8+2. Here a=8, 6=2. Then a'* =2, 

and -=?. 
a 

1 2 2 5 

Therefore, K^-^yIT'^QA^'^^qJa^' ^"^'"^ ^^ ^^® ^"^® '^^^^ 

of 10, and so for any other number. 

3. Expand -r into a series, or its equal, {a — b)~^. 

Am. -+- 2+-3+-4, &c. 
a c? a^ a^ 

4. Expand (a^+ft*) into a series, or its equal, a( 1+— 2) • 

Ans, air- ;T-^+TTrT» &c. 

2a Sc^ 16a^ 

_i 

5. Expand (/(c^-j-x^) ^ into a series. 



INFINITE SERIES. 221 

6. Expand {a^ — x'^)^ into a series. 

Jlns, J~( a^ — -5 — -r-^ — -=—.y Sic. I 

-4 

■y. Expand {a-{-y) 

a* a^ ' a^ a' a^ 
8. Find the cube root of .. , .. . 



b' 2b' 1469 



CHAPTER II. 

OF INFINITE SERIES. 

(Art. 136.) An infinite series is a continued rank, or progres- 
sion of quantities in regular order, in respect both lo magnitudes 
and signs, and they usually arise from the division of one quantity 
by another. 

The roots of imperfect powers, as shown by the examples in 
the last article, produce one class of infinite series. Some of 
the examples under (Art. 121.) show the geometrical infinite 
series. 

Examples in common division may produce infinite series foi 
quotients ; or, in other words, we may say the division is con- 
tinuous. Thus, 10 divided by 3, and earned out in decimals, 
gives 3.3333, Sic, without end, and the sum of such a series is 
3^. (Art. 121.) 

(Art. 137.) Two series may appear very different, which arise 
from the same source ; tlius I, divided by l-\-a, gives, as we may 
see, by actual division, as follows: 

l-\-a)l (1 — a-\-a^ — «'^-j-«'', &c. without end. 

— a 
— a — tt^ 

t2 






222 ELEMENTS OF ALGEBRA. 



Also, ft-f 1)1 ( ,4— „ J, &c. without end. 



■+-J 



These two quotients appear very different, and in respect to 
single terms are so ; but in these divisions there is always a re- 
mainder, and either quotient is incomplete without the remam- 
der for a numerator and the divisor for a denominator, and when 
these are taken into consideration the two quotients will be 
equal. 

We may clearly illustrate this by the following example : — 
Divide 3 by 1+2, the quotient is manifestly 1 ; but suppose them 
literal quantities, and the division would appear thus : 
l_j_2)3 (3— 6-1-12, &c. 
3-f-6 

—6 
—6—12 



12 

12-1-24 

—24 

Again, divide the same, having the 2 stand first. 
2-1-1)3 (i-|-l-f,&c. 
3+1 



3 

2 




1- 


_3 
4 




3 

4 




3_L.3 



Now let us take either quotient, with the real value of its re- 
mainder, and we shall have the same result. 

Thus, 3+12=15; and — 6, and the remainder — 24 divided 






INFINITE SERIES. 223 

by 3, gives — 8, which makes — 14 ; hence, the whole quotient 
is 1. 

Again, |-|-|=V» and— I— i=i. 

Hence, \^ — 1=|=1, the proper quotient. 

If we more closely examine the terms of these quotients, we 
shall discover that one is diverging, the other converging, and 
by the same ratio 2, and in general this is all a series can show, 
the degree of convergency, 

(Art. 138.) We convert quantities into series by extracting 
the roots of imperfect powers, as by the binomial, and by actual 
division, thus : 

1. Convert — ; — into an infinite series. 
a-\-x 

Thus, a-\-x)a (1— -+-,— ^4", &c. 

a-\-x 



— X 

^ 

a 



a 

2. Convert into an infinite series. 

a — X 

Jins. I-l-^+^,+^,&c. 

Observe that these two examples are the same, except the 
signs of X : when that sign is plus the signs in the series will be 
alternately plus and minus ; when minus, all will be plus. 

\-\~x 

3. What series will arise from ? 

1 — X 

.^ns. \-{-2x-\-2x'-\-2x\ &c. 
Observe that in this case the series commences with 2x. The 

unit is a proper quotient, and the series arises alone from 

the remainder after the quotient 1 is obtained. 






. ** 

#>>» 



224 ELEMENTS OF ALGEBRA. 

4. What series will arise from -^-, — - ? 



a^^:^) «-l-K^-S+S-J &e. 



a 



.-+. 



o^ 



Observe, in this example, the term a:, in the numerator does 
not find a place in the operation ; it will be always in the r^ 

mainder ; therefore, ^ will give the same series. 

5. What series will arise from dividing 1 by 1 — a+a^ or 

from ,-— ? ./Ins, 1+a—- «=*— a''+aH«^— «^— a'°, A'C. 

1 — a-f-a^ ' 

In this example, observe that the signs are not alternately plus 
and minus, but two terms in succession plus, then two minus ; 
this arises from there being two terms in place of one after the 
minus sign in the divisor. 

6, What series will arise from ~ ? 

1 — r 

Ans, a-]rcir-\-ai^-\-ar^-\-ar*t &c. 

Observe that this is the regular geometrical series, as appears 
in (Art. 118.) 

1 . 



■y. What series will arise from 



1—1 

Ms. 1-f 1-f-l + l, &c. 

That is ^ is 1 repeated an infinite number of times, or infinity, 
a result corresponding to observations under (Art. 60.) 



■^'If 



SUMMATION OF SERIES. 225 

8. What series will arise from the fraction j- ? 

a — b 

h,hh .¥h , b'h . 
a a^ «"^ ' or 

If a=6, this series will be -, repeated an infinite number of 

(I 
times. 

This series can also be expanded by the binomial theorem, 

for ^=h(a — b) . This observation is applicable to seve- 

a — 6 ^ ' 

ral other examples. 

(Art. 139.) A fraction of a complex nature, or having com- 

l y^ 

pound terms, such as - — —g, may give rise to an infinite 

series, but there will be no obvious ratio between the terms. 
Some general relation, however, will exist between any one term 
and several preceding terms, which is called the scale of rela- 
tion, and such a series is called a recurring series. Thus 
the preceding fraction, by actual division, gives l + aJ-f-Sx^-f- 
13a;''-|-41a?^+121a;% &c., a recurring series, which, when carried 
to infinity, will be equal to the fraction from which it is derived. 

Expand , mto a series 

Am. l+3x-\-ix'-\-7x^+Ux*-\-lS3p, <fcc. 

CHAPTER III. 

SUMMATION OF SERIES. 

(Art. 140.) We have partially treated of this subject in geo- 
metrical progression, in (Art. 121) ; the investigation is now 
more general and comprehensive, and the object in some respects 
different. There we required the actual sum of a given number 
of terms, or the sum of a converging infinite series. Here the 
series may not be in the strictest sense geometrical, and we may 
not require the sum of the series, but what terms or fractional 
quantities will produce a series of a given convergency. 
15 



226 ELEMENTS OF ALGEBRA. 

The object then, is the converse of the last chapter; and for 
every geometrical series, our rule will be drawn from the sixth 

example in that chapter ; that is, , a being the first term 

of any series, and r the ratio. We find the ratio by dividing 
any term by its preceding term. 

Hence, to find what fraction may have produced any geomet- 
rical series, we have the following rule: 

Rule. Divide the first term of the series by the alge- 
braic difference between unity and the ratio. 

EXAMPLES. 

1. What fraction will produce the series 2, 4, 8, 16, &c? 

2 
Here a=2, and r=2 ; therefore, Ans. 

2. What fraction will produce the series 3 — 9-}- 27 — 81, &c.? 
Here a=3, and r= — 3 ; then — r=3, 

Hence, — =^r-nr '^^s. 

1 — r 1+3 

3. What fraction will produce the series y\, y|^, &;c. ? 

Here «=yo» and r=r^-^; therefore, 

y\ ^10 3 3 1^ 

r±?i:Xlo=lo=:i=9=3' -^^^^ 

4. What fraction will produce the series, 

l__?-}-^'-_^', &o. ? [See example 1, (Art. 134.)] 
a a^ a^ ^ r v /j 

cc \ a 

Here g=1, and r= ; then = — j — , *^ns. 

a i_i_^ a-\-x 

a 

5. What fraction will produce the series l+2ar-h2a7^+2a?^ 
&c. ? [See example 3, (Art. 138.) and the observation in con- 
nection.] 



SUMMATION OF SERIES. 227 

Here o=2a?, r=x;' then - — '—, will give all after the first 
1 — X ° 

, , 2x l-\-x 

term; therefore, l-j-:; =:; , ^ns. 

1 — X 1 — X 

^ „.u r • 11 1 ^ . d db . db^ dW 

6. What fraction will produce the series 5-H — ^ r-, 

&c. ? Jins. -f-7 . 

^ „r, n • Ml t , ' d , adx , a^dx^ „ 

7. What fraction will produce the series j-\ — — — | — ^^, &c. 



'b — ax' 

8. What fraction will produce the series 

„, 2b\ W 26\ , ' 2ab 

2b +-^ „— 4-, &c. ^ns. —TT* 

a (£- a^ ' a-\-b 

9. What fraction will produce the series 

\-\-a^a^—a^-^a^-\-a'—a^—a}\&LQ..1 

See example 5, (Art. 138.) 

Put \-\-a=b\ then d?-\-a^=a^b, and a^-\-a''=aV), and the 
series hecomes b — a^b-\-a^b — , &c. 

. ,. . . . . b l-\-a 1 
Hence, the fraction required is — -; — -o=-r-, — ^=-. ; — 5« 

10. What fraction will produce the series x-\-x^-{-x^i &c. ? 

1 ,7' 

If x=l, this expression becomes - — 7 — ^i' ^ symbol of in- 
finity. 

11. What fraction will produce the series 1 + |4"2"5> ^^' ^ 

1 5 



Jins. 



1—1 5-3 



Hence, the sum of this series, carried to infinity, is 2|. 



228 ELEMENTS OF ALGEBRA. 

In the same manner, we may resolve every question in (Art. 
121.) 

12. What is the sum of the series, or what fraction will pro- 
duce the series 1 — a-\-a^ — a*~{-a^ — a'-\-(^ — , &c. ? 

13. What is the value of the infinite series 

6,6.6 



10^(10/^(10)^ 



&;c. ^ns. 



RECURRING SERIES. 

(Art. 141.) We have explained recurring series in (Art. 139.) 

and it is evident that we cannot find their equivalent fractions 

by the operation which belongs to the geometrical order, as no 

common relation exists between the single terms. The fraction 

-j — ^, by actual division, gives the series l-\-Sx-{-4x^-\-7x'^ 

+ lla;^-j~18.r^ &c., without termination ; or, in other words, the 
division would continue to infinity. 

Now, having a few of these terms, it is desirable to find a 
method of deducing the fraction. 

There is no such thing as deducing the fraction, or in fact no 
fraction could exis4 corresponding to the given series, unless 
order or a law of dependence exists among the terms ; therefore 
some order must exist, but that order is not apparent. 
Let the given series be represented by 

.^4-^-{-c4-i}+^+/; &c. 

Two or three of these terms must be given, and then each suc- 
ceeding term may depend on two or three or more of its pre- 
ceding terms. 

In cases where the terms depend on (wo preceding terms we 
may have 

C=mBx-\-7iJ2x^ 

n=mCx-\-nBx^ 

E=mDx-[-n Cy? 

&c.=&c. 



(1) 



(2) 



RECURRING SERIES. 229 

In cases where the terms, or law of progression depend on 
three preceding terms we may have 

E=mDx-\-n Cx^-^rBx^ 
F=m Ex+nDx'^-^r Cx^ 
&c.=&;c. 

The reason of the regular powers of x coming in as factors, 
will be perfectly obvious, by inspecting any series. 

The values of m, n and r express the unknown relation, or law 
that governs the progression, and are called the scale of relation. 
We shall show how to obtain the values of these quantities in a 
subsequent article. 

(Art. 142.) Let us suppose the series of equations (1), to be 
extended indefinitely, or, as we may express it, to infinity, and 
add them together, representing the entire sum of *^-{-B-{' 
C+A &c., to infinity, by S ; then the first member of the 
resulting equation must be (S—M — B), and the other member 
is equally obvious, giving 

S—^^B=mx(S--^)-{-n3c'S 

AArB—^Jix , V 

Hence, o=- j (a) 

1 — mx — nar 

In the same manner, from equations (2), we may find 
A +^+ C-~{Ji^ B)mx—An x^ 
~~ 1 — mx — na?^ — rx^ 

(Art. 143.) The /orm of a series does not depend on the 
value of iP, and any series is true for all values of x. Equations 
(I) then, will be true, if we make x=\. 

Making this supposition, and taking the first two equations of 
the series (1), we have 

C=mB-\-nA } (c) 

And /)=mC-{-nB ] 

In these equations, A, B, C, D, are known, and m and n un- 
known ; but two unknown quantities can be determined from two 
equations ; hence m and n can be determined. 
U 



"S^. 



230 ELEMENTS OF ALGEBRA. 

When the scale of relation depends upon three terms, we 
take three of the equations (2), making x=lt and we determine 
w, n, and r, as in simple equations. 

EXAMPLES. 

1. What fraction will produce the series 

l-\-Sx-h'^x^-^7x^-{-Ux\ &c.? 
To determine the scale of relation, we have w(?=l, ^=3, 
C=4, and jD=7. Then from equations (c), we have 
n+3n2=4 
3nH-4m=7 
These equations give m=l, and w=l. 
Now to apply the general equation (a), we have td=l, B=3x. 

Then S=- -=- ^=- -^, ^ns. 

1 — mx — nc(r 1 — x — or 1 — x — ar 

2. What fractional quantity will produce the series 

l-^6x-{-l2xF-\-482^-+ 120a;S &c., to infinity ? 
Here .^=1, ^=6^7, 7/i=l, n=6. 

Hence, by applying equation («), we find — j for the 

expression required. 

3. What quantity will produce the series 

l_(-3a;-|-5:c2^-7a::2+9a?^ to infinity? l-{-x 

^ 4. What quantity will produce the infinite series 
l^4x+Q3c^+nx^+2Sx^-i-mx^, &c.? in which m=2, n=:— 1, 
r=3. 

[Apply equation (6).] ^ns. A___^__. 

5. What fraction will produce the series 

l+a?+5ar'+13x3+41a;^+121ar5, &c. ? 

1— a; 



^ns. 



l^2X'-'3a^' 



REVERSION OF A SERIES. 
O. What fraction will produce the series 

l+Sx+2x^-'X^Sx^—2x^+x% &c.? 

Jlns. 



231, 



1 — x-\-x^ 



REVERSION OF A SERIES. 

(Art. B.) To revert a series is to express the value of the un- 
known quantity in it, (which appears in the several terms under 
regular powers,) by means of another series involving the powers 
of some other quantity. 

Thus, let X and y represent two indeterminate quantities, and 
the value of y be expressed by a series involving the regular 
powers of x. That is 

y=ax-^bx'^-\-cx^-\'dx'^ &c. 

To revert this, is to obtain the simple value of re, by means 
of another series containing only the known quantities, «, 6, c, 
d, &c. and the powers of y. 
To accomplish this,asst<me 

x^Ay-\-By'^-\'Cy^, &c.* 

Substitute the assumed value of x for a?, x^^ x^, &c., in the pro- 
posed series, and transposing y, we shall have 



0= 



r«^ 


y+aB \f-\-aC 


y^+aD 


—1 


-\-bd'\ -\-2b^B 


-\-2bAC 


' +c^^ 


+bB'- 




-{-Sc^'B 






+dA'' 



y+, &c. 



* By examining previous authors, we have found none that explain the 
rationale of such assumptions : but these are points on which the learner 
requires the greatest profusion of light. We explain thus ; the term x must 
have some value, positive, negative, or zero, and the series Ai/-\-Bi/2-\-Ci/3, 
&c., can be positive and of any magnitude. It can be negative, and of any 
magnitude, by giving the coefficients A, B and C, negative values. It can be 
zero by making y=0. Therefore it can express the value of x, whatever that 
may be ; and because the powers of y are regular, the substitution of this 
value of X for the several powers of x, in the primitive series, will convert 
that series into regular powers of y, which was the object to be accomplished. 






232 



ELEMENTS OF ALGEBRA. 



*'As every term contains y, we can reduce the equation by 
dividing by y ; uud afterwards, in consideration that the equation 
must be true for all values of y : making 2/=0, we shall have 



a.^— 1=0 



or 



a 



Continuing the same operation and mode of reasoning as in (Art. 
128.), and we shall find, in succession, that 

a 



{^) 



B=— 



C= 



2)=- 



W—ac 



5b'^5ahc+aH 



&c. = &c. 



Substituting these values of Jlf Bj C, &c., in the assumed 
equation and we have the value of a? in terms of a, 6, c, &c., 
and the powers of y as required, or a complete reversion of the 
series. 

EXAMPLES. 

1. Revert the series y=x-{-3f-\-x^f &c. 
Here a=l 6=1 c=l, &c. 
Assume x=»^y-{-By^-\-Cy^-\-&c, • 

Result, x=y — y^-\-'if — 2/^+y' — <fec. 

«. Revert the series x=y — %- +^ — r <fec. 

Here a=l 5= — | c=5 d= — k &c. 

Assume yz=Jlx-{-B3i?-]rCx^-\-Dx'^-\- &c., and find 
the values of w?, B, C, &c., from the formulas [F). 

Result. y=x+-^+_^+-^_&e. 



LOGARITHMS. 233 

3. Revert the series y=x-{'dx'^-\-6x^-{-7x^-]r^s^, <fec. 

Result a;=2/--33/2+13?/'— 53^. 

In case the given series is of the form of x=ay-{-by^-\-cy^i 
&c., the powers of y varying by 2, the equations [F) will not 
apply, and we must assume y=jlx-\-Bx^-{-Cx\ &c., and sub- 
stitute as before, and we shall find 

^= 1 

a 

-4 



(«) 



3b^—ac 






In common cases, after the coefficients, as far as D, are deter- 
mined, the law of continuation will become apparent, especially 
if the factors are kept separate. 

CHAPTER IV. 
EXPONENTIAL EQUATIONS AND LOGARITHMS. 

(Art. 144.) We have thus far used exponents only as known 
quantities ; but an exponent, as well as any other quantity, may 
be variable and unknown, and we may have an equation in the 
form of a^=b. 

This is called an exponential equation, and the value of x can 
only be determined by successive approximations, or by making 
use of a table of logarithms already determined. 

(Art. 145.) Logarithms are exponents. A given constant 
number may be conceived to be raised to all possible powers, and 
thus produce all possible numbers ; the exponents of such pow- 
ers are logarithms, each corresponding to the number produced 

Thus, in the equation a'=6, x is the logarithm of the number 
b ; and to every variation of x, there will be a corresponding 
20 



234 ELEMENTS OF ALGEBRA. 

variation to b ; a is constant, and is called the base of the sys- 
tem, and differs only in different systems. 

The constant a cannot be 1, for every power of 1 is 1, and tlie 
variation of x in that case would give no variation to b ; hence, 
the base of a system cannot be unity ; in the common system it 
is 10. 

In the equation 10^=2, x is, in value, a small fraction, and 
is the logarithm of the abstract number 2. 

In the equation a^=b, if we suppose a?=l, the equation becomes 
a^=a; that is, the logarithm of the base of any system is unity. 

If we suppose x=0, the equation becomes a°=l ; hence, the 
logarithm of 1 is 0, in every system of logarithms. 

(Art. 146.) The logarithms of two or more numbers added 
together give the logarithm of the product of those numbers^ 
and conversely the difference of two logarithms gives the lo- 
garithm of the quotient of one number divided by the other. 

For we may have the equations a^=b, av=^b\ and (f=b". 

Multiply these equations together, and as we multiply powers 
by adding the exponents the product will be 
a^-{v^z=^l)b'b" 

Hence, by the definition of logarithms, x-\-y-{-z is the loga- 
rithm for the number represented by the product bb'b". Again. 

divide the first equation by the second, and we have a=^-!'=- ; 

and from these results we find that by means of a table of loga- 
rithms multiplication may be practically performed by addition, 
and division by subtraction, and in this consists the great utility 
of logarithms. 

(Art. 147.) In the equation a^=b, take a=10, and x succes- 
sively equal to 0, 1, 2, 3, 4, &c. 

Then 10°=1, 10^=10, 10^=100, 10=^=^1000, &c. 
Therefore, for the numbers 
1, 10, 100, 1000, 10000, 100000, &c., we have for corres- 
ponding logarithms 

0, 1, 2, 3, 4, 5, &c. 



LOGARITHMS. 235 

Here it may be observed that the numbers increase in geometri- 
cal progression, and tlieir logarithms in arithmetical progression. 

Hence the number which is the geometrical mean between two 
given numbers must have the arithmetical mean of their lo- 
garithms, for its logarithm. 

On this principle we may approximate to the logarithm of any 
proposed number. For example, we propose to find the lo- 
garithm of 2. 

This number is between 1 and 10, and the geometrical mean 
between these two numbers, (Art. 122.), is 3.16227766. The 
arithmetical mean between and 1 is 0.5; therefore, the number 
3.16227766 has 0.5 for its logarithm. 

Now the proposed number 2 is between 1 and 3.162, &c.,and 
the geometrical mean between these two numbers is 1.778279, 
and the arithmetical mean between 0. and 0.5 is 0.25 ; therefore, 
the logarithm of 1.778279 is 0.25. 

Now the proposed number 2 lies between 1.778279, and 
3.16227766, and the geometrical mean between them will fall 
near 2, a little over, and its logarithm will be 0.375. Continuing 
the approximations, we may at length find the logarithm of 2 to 
be 0.301030, and in the same manner we may approximate to 
the logarithm of any other number, but the operation would be 
very tedious. 

(Art. 148.) We may take a reverse operation, and propose a 
logarithm to find its corresponding number; thus, in the general 
equation a^=n; x may be assumed, and the corresponding value 
of n computed. 

Thus suppose 3?=^^; then (10)'*=n, or 10^=n^''. 



Hence, 72=^^1000=1.995262315. 

That is, the number 1.9952, &c., (nearly 2) has 0.3 for its 
logarithm. In the same way we may compute the numbers cor 
responding to the logarithms 0.4, 0.5, 0.6, 0.7, &c. 

(Art. 149.) We may take another method of operation to find 
the logarithm of a number. 



236 ELEMENTS OF ALGEBRA. 

Let the logarithm of 3 be required. 
The equation is 10^=3, the object is to find x. 

It is obvious that x must be a fraction, for 10^=10 ; there- 
fore, 

Let x=—. Then 10'"=3, or 10=3'='. Here, we perceive, 

X 

that X' must be a little more than 2. Make x'=2-{- — Then 

X"' 

1 1 

left *^ 10 

3^X3 =10; or 3 =— . 



Or 



m-- 



Here we find by trial that x" is between 10 and 11 ; take it 10, 
and a?'=2-l-~ ; hence, a?=:4 j=0.476, for a rough approxima- 
tion for the logarithm of 3. A more exact computation gives 
0,4771213; but all these operations are exceedingly tedious, and 
to avoid them, mathematicians have devised a more expeditious 
method by means of a converging series ; which we shall inves- 
tigate in a subsequent article. 

(Art. 150.) It is only necessary to calculate directly the lo- 
garithms of prime numbers, as the logarithms of all others may 
be derived from these. Thus, if we would find the logarithm 
of 4, we have only to double that of 2 ; for, taking the equation 
(10)^=2, and squaring both members we have, (10)^=^=4; or 
taking the same equation, and cubing both members, we have 
(10)^^=8 ; which shows that twice the logarithm of 2 is the 
logarithm of 4, and three times the logarithm of 2 is the loga- 
rithm of 8 ; and, in short, the sum of two or more logarithms 
corresponds to the logarithm of the product of their numbers, 
(Art. 142.), and the difference of two logarithms corresponds to 
the logarithm of the quotient, which is produced from dividing 
one number by the other. 

a 
Thus, the logarithm of ^ = log. a — log. b. The abbrevia- 
tion, (log. a), is the symbol for the logarithm of a. 



LOGARITHMS. 237 

(Art. 151.) As the logarithm of 1 is 0, 10 is 1, 100 is 2, &c., 
we may observe that the whole number in the logarithm is one 
less than the number of places in the number. 

The whole number in a logarithm is called its characteristic, 
and is not given in the tables, as it is easily supplied. For ex- 
ample, the integral part of the logarithm of the number 67430 
must be 4, as the number has 5 places. The same figures will 
have the same decimal part for the logarithm when a portion of 
them become decimal. 

Thus, 67430 logarithm 4.82885 

6743.0 3.82885 

674.30 2.82885 

67.430 1.82885 

6.7430 0.82885 

.67430 —1.82885 

For every division by 10 of the number, we must diminish 
the characteristic of the logarithm by unity. 

The decimal part of a logarithm is always positive ; the index 
or characteristic becomes negative, when the number becomes 
less than unity. ^ 

By reference to (Art. 18.), we find that — =10~', -7r-==T7j2= 

10~^ &c. That is : fractions may be considered as numbers 
with negative exponents, and logarithms are exponents ; there- 
fore the logarithm of — , or .1, is — 1 ; of — — , or .01, is — 2, 

&c. If any addition is made to .01 the logarithm must be more 
than — 2 ; but, for convenience, we still let the index remain 
— 2, and make the decimal part plus. Hence the index alone 
must be considered as minus. 

Negative numbers have no logarithms ; for, in fact, as we 
have before observed, there are no such numbers. 

(Ai;t. C.) We now design to show a practical method of com- 
puting logarithms. Ttie methods hitherto touched upon were 
only designed to be explanatory of the nature of logarithms ; but, 
to calculate a table, by either of those methods, would exhaust the 



238 ELEMENTS OF ALGEBRA. 

patience of the most indefatigable. To arrive at easy practical 
results requires the clearest theoretical knowledge, and we must 
therefore frequently call the attention of students to first prin- 
ciples. . 

The fundamental equation is a^=b, in which a is the con- 
stant base and x is the logarithm of the number b ; and b may 
be of any magnitude or in any form, if it be essentially jt?osf/ive. 

Now we may take «=l-l-c and 6=1-| — 

Then the fundamental equation becomes (l-j-c)^=(l+- j (1) 

Here :r=log. (lH-j^)==log.(^)=log.(jD-{-l)-log.;>(2) 

Raise both members of equation (1) to the nth power, then 
we shall have 



(l+c)-=(i+i)' 



Expand both members into a series by the Binomial Theorem, 
Then \-\-nxc-\-nx . — - — &-\-nx . — - — . — r— <r+ 



nx — 1 nx — 2 nx — 3 . , ,, , i 1 . '* — 1 1 

2 * 3 



na^.— „— .— :r— .-j-c'*+, &c. = l-{-M.--{-,i._^.--f 



n^\ n— 2 1 , n-—\ n—2 n--3 1 , , 

^•-2— 3-/-^^-^-3--^-/+' ^'' 

Drop unity, from both members, and divide by n, then we have 

/ , nx — 1 „ , nx — 1 nx — 2 „ , nx — 1 nx — 2 nx — 3 . 
<'+-2-^ +-3 3-' +-3- • -3- • —4-' 

\ _ In — 1 1 n — 1 n — 2 1 , n— -1 n— 2 n — 3 1 
+&^C'J ~ ^+-2~'p"*"~2~*~3~y'^"2"*~3~*"T~*^* 

+, &c. 

This equation is true for all values of n. It is, therefore, true 



LOGARITHMS. 239 

when n=0. Making this supposition and the above equation 
reduces to 

As c remains a constant quantity for all variations of x and /?, 
the series in the vinculum may be represented by the symbol M. 

Thea .i^=i_i,+l,_l,+l--, &e. (3) 



Take the value of x from equation (2), and substitute it in 
equation (3), 

I 1 4_ 1 __ ' 



Then [log.(;)+l)—log.7)]M=;p-—,+ -3— —,-!-, &c. (4) 



Again, we may have the fundamental equation (l-}-c)^= 
( 1 V in which y is the logarithm of ( 1 j, the same as x 

was of ( 1-|-- V 

V pJ 

Or 2/=log.^^^^=log.(;3— 1)— log, ;?. (5) 

Operating on the equation (l-fc)^=l , the same as we 

did on equation (1), we shall find 

pog.(p-l)-log. ;,]iV/=^_l._l_l-_, &e. (6) 
Subtract equation (6) from equation (4), and we obtain 
[log.(p+l)-log.(;,-l)]il/=2(iH-l5+^,+l-„ &c.) (7) 

Dividing by M, and considering that log.f ^^- — - J=log.(jD+lV 
log.(jo — 1), the equation can take this form 



240 ELEMENTS OF ALGEBRA. 

, /p+l\ 2 /I , 1 , 1 , 1 , 1 „ \ 

As p may be any positive number, greater than 1, make 
-—— r=lH — . Then p=2z-{-lj and equation (8) becomes 

2/1 ^ f 1 ^ \ 

By this last equation we perceive that the logarithm of {z-[-\) 
will become known when the log. of z is known, and some 
value assigned for the constant M. Baron Napier, the first dis- 
coverer of logarithms, gave M the arbitrary value of unity, for 
the sake of convenience. 

Then, as in every system of logarithms, the logarithm of 1 is 
0, make 2'=1, in equation (9), and we shall have 

'°S.2=2(i+3i^+^^,+. &c.)=.0.693147]8. 

This is called the Napierian logarithm of 2, because its magni- 
tude depends on Napier's base, or on the particular value of M 
being unity. 

Having now the Napierian logarithm of 2, equation (9) will 
give us that of 3. Double the log. of 2 will give the logarithm 
of 4. Then, with the log. of 4, equation (9) will give the loga- 
rithm of 5, and the log. of 5 added to the log. of 2, will give the 
logarithm of 10. 

Thus the Napierian logarithm of 10 has been found to great 
exactness, and is 2.302585093. 

The Napierian logarithms are not convenient for arithmetical 
computation, and Mr. Briggs converted them into the common 
logarithms, of which the base is made equal to 10. 

To convert logarithms from one system into another, we 
may proceed as follows : 

Let e represent the Napierian base, and a the base of the 
common system, and N any number. 



m -^ 



LOGARITHMS. 241 

Also, let X represent the logarithm of iVy corresponding to 
the base a, rind y the logarithm of iV, corresponding to the 
base c. 

Then «^=iV, 

and ey =A\ 

Now, by inspecting these equations, it is apparent that if the 
base a is greater than the base e, the log. x will be less than the 
iog.y. 

These equations give «^=e". 

Taking the logarithms of both members, observing that x and 
y are logarithms already, we have 

a:log. a=i/log. e. 

This equation is true, whether we consider the logarithms 
taken on the one base or on the other. Conceive them taken on 
the common base, then 

log. a=l, and x=y\og.e (10) 

X 

or \os.e=-, 

^ y 

In this equation x and y must be logarithms of the same num- 
ber, and therefore if we take a?=l, which is the logarithm of 
10, in the common system, y must be 2.302585093, as pre- 
viously determined. 

Hence log.e=^_L^=0.434394482 . . . . (11) 

This last decimal is called the modulus of the common sys- 
tem ; for by equation (10) we perceive that it is the constant 
multiplier to convert Napierian or hyperbolic logarithms into 
common logarithms. 

But equation (9) gives Napierian logarithms when ^/=1; 
therefore the same equation will give the common logarithms by 
causing M io disappear, and putting in this decimal as a factor. 

Equation (9) becomes the following formula for computing 
ommon logarithms : 
21 



# «* 



242 



0.86858896 



ELEMENTS OF ALGEBRA. 
log.(2:-l-l)— log.2:= 
1 



fctl+3(2^W^5(2il7+' ^'')^^^ 



To apply this formula, assume z=lO. Then 
log. z=l, and 22:4-1=21. 



21 

2P=441 

441 



0.86858896 

0.041361379-f-l = 0.041361379 

93792-f-3 == 31264 

212-^5 = 42 



.041392685=sum of series. 
Io2:.10= 1.0 



Iog.(2:+l)= L041392685=log.ll. 

If we make z=99, then {z-\-l) = iOO, and log.(^+l)=2, 
and 2zH-l=I99. 



199 
39601 



0.86858896 

436477-M =0.00436477 
11^3= 4 



0.0043648 i=sum of series. 

Hence 2.00000— log.99=0.00436481 
By transposition log.99=l. 995635 19 

Subtract Iog.ll=1.04139269=log.ll 

log. 9=0.95424234 
ilog.9=log. 3=0.477121 17=log.3. 

Thus we may compute logarithms with great accuracy and 
rapidity, using the formula for the prime numbers only. 

By equation (11) we perceive that the logarithm of the Na- 
pierian base is 0.434294482 ; and this logarithm corresponds to 
the number 2.7182818, which must be the base itself. 

We may also determine this base directly : 

In the fundamental equation (1), the base is represented by 
(l+c). In equation (A), c must be taken of such a value as 



LOGARITHMS. 243 

(A f,Z fA 

shall make the series c — — -j-^ — j+j &c., equal to 1. But to 
^ o 4 

determine what that value shall be, in the first place, put 

V=C h , &C. 

Now by reverting the series (Art. B.), we find that 

But, by hypothesis, the series involving c equals unity, that is, 
2/=l. Therefore 

By taking 12 terms of this series, we find (l+c)=2.7182818, 
the same as before. 

If we take equation (3), making M=i, we find the Napierian 
logarithm, or ^=1-1-^+1^^+, &c. 

This series will not converge rapidly unless p is a. large num- 
ber. But by equation (2) x—\og.{p-\-l) — log. p. 

Hence log.(;?+l)— log.;?=-— — ,-|-^^— ,+, &c. 

If in this equation we make j9=l, we shall find the Napierian 
log. of 2 equal to the series 

1 — -H-- — -+-, &c., to infinity. 

But we have already found the same logarithms equal to the 
series ^V q"l~q7qp"i"q7^~f"» ^^' ) » therefore these two series 

are equal to each other, and because the former did not rapidly 
converge, it became necessary to obtain the latter 



4 ' " 






244 ELEMENTS OF ALGEBRA. 

USE AND APPLICATION OF LOGARITHMS. 

(Art. 152.) The sciences of trigonometry, mensuration, and 
astronomy alone, can develop the entire practical utility of lo- 
garithms. The science of algebra can only point out their 
nature, and the first principles on which they are founded. To 
explain their utility, we must suppose a table of logarithms formed, 
corresponding to all possible numbers, and by them we may re- 
solve such equations as the following : 

1. Given 2''= 10 to find the value of x. 

If the two members of the equation are equal, the logarithms 
of the two members will be equal, therefore take the logarithm 
of each member ; but as a: is a logarithm already, we shall have 
X log. 2=log. 10. 

Or ^=l^li=_i_ =3.3319+ 
log. 2 .30103 ^ 



.-S-l / 



2. Given (729)^=3, to find the value of x. 

^ Raise both members to the x power, and 3^—729=9', 
Or 3=^=3^ Hence, x—G. 

3. Given a^-{-b^=c, and a^ — b^=d, to find the values of x 
and y. 

By addition, 2a^=cH- J. Put c-\rd=2m ; 

Then a'^—m. Take the logarithm of each mem- 

ber, and x log. 6!=log. ??i, or x=j^ . 

j By subtracting the second equation from the first and 

- . - , „ ^ , log. n 

makmff c — a=2rt, we shall find y=, . 

^ ^ log. a 

4# Given (216)^^=12, to find the value of a:. 



jins. xz- ^' 



5. Given ; — =e, to find the value of x» 

a 



log. 12 



log. m — ^log. a , . 1 . / 1 I \ 

Mns, x== — ^— i T^ — m bemg equal to (ae-l-c). 

log. o 



LOGARITHMS. 245 



6. Given 4^ =16, to find the value of x. Ans, a?=6. 

•y. Given 6 = — ^^ — — to find the value of x. 
71 

_ 18 ]og.24+log.l7— 3 log.71 

3 log. 6 

§. Required the result of 23.46 multiplied by 7.218, and the 
product divided by 11.17. 

OPERATION. 

23.46 log. 1.37033 

7.218 loff. 0.85842 



Sum . 
11.17 Subtr. 

Result,. ... 15.16 



2.22875 
log. L. 04805 



log. 1.18070. 



N. B. Roots can be extracted by logarithms in the following 
manner : 

Let the cube root of 12 be required, and the root represented 
by X. 

Then we shall liave the equation 

a>=12^ 
Take the log. of both members, and we then have 

log. x=- log. 12. 

log. 12 = 1.07918 
log. 37=0.35972 

Taking the number corresponding to this logarithm from a 
table of logarithms, we find x, or the cube root of 12, to be 
2.2887. In the same manner, by the aid of a table of logarithms, 
we may obtain the exact or approximate value of any proposed 
root of any' number whatever. 
v2 



246 ELEMENTS OF ALGEBRA. 

CHAPTER V. 

COMPOUND INTEREST. 

(Art. 153.) Logarithms are of great utility in resolving some 
questions in relation to compound interest and annuities ; but for 
a full understanding of the subject, the pupil must pass through 
the'foUowing investigation: 

Let p represent any principal, and r the interest of a unit of 
this principal for one year. Then 1 -[-r would be the amount 
of$l,oriei. Put.^=l+r. 

Now as two dollars will amount to twice as much as one dol- 
lar, three dollars to three times as much as one dollar, &;c. 
Therefore, \ : A :: A : ./5^=the amount in 2 years, 
And \ '. A. :: A^ : ./5^— the amount in 3 years, 
&c. &c. 

Therefore, A"" is the amount of one dollar or one unit of the 
principal in n years, and p times this sum will be the amount for 
p dollars. Let a represent this amount ; then we have this gen- 
eral equation, 

pA^'^^a. 

In questions where n, the number of years, is an unknown 
term, or very large, the aid of logarithms is very essential to a 
quick and easy solution. 

For example, what time is required for any sum of money 
to double itself at three per cent, compound interest? 

Here a=2;9,and ./?= (1.03), and the general equation becomes 
/»(1.03)"=2;) 
Or (1.03)"=2. Taking the logarithms 

„Iog.(1.03)=Iog.2, or „=^_-^^ll^^=^«-^=23.45 

years nearly. 

2. A bottle of wine that originally cost 20 cents was put awa} 
for two hundred years : what would it be worth at the end of 
that time, allowing 5 percent, compound interest? 



ANNUITIES. 247 

This question makes the general equation stand thus : 
(20 cts. being ^ of a dollar) l(1.05)200= a 
Therefore, ' (1.05)200=5« 

Taking the logarithms 200 log. (1.05)=log.5+log. a 
Hence log. a=200 log. (1.05)-~log. 5. Mns. $3458.10. 

3. A capital of $5000 stands at 4 per cent, compound interest; 
what will it amount to in 40 years 1 Ans. $24005.10. 

4. In what time will $5 amount to $9, at 5 per cent, com- 
pound interest ? Ans, 12.04 years. 

5. A capital of $1000 in 6 years, at compound interest, 
amounted to ^1800; what was the rate per cent? 

Ans. log. (l4-r)= ^^' ' or 10y\ nearly. *. 

6. A certain sum of money at compound interest, at 4 per 
cent, for four years, amounted to $350.95| ; what was the sum? 

Ans, $300. 

7. How long must $3600 remain, at 5 per cent, compound in- 
terest to amount to as much as $5000, at 4 per cent, for 12 years 1 

Jins. 16 years, nearly. 

ANNUITIES. 

(Art. 154.) An annuity is a sum of money payable peri- 
odically, for some specified time, or during the life of the re- 
ceiver. If the payments are not made, the annuity is said to be 
in arrear, and the receiver is entitled to interest on the several 
payments in arrear. 

The worth of an annuity in arrear, is the sum of the several 
payments, together with compound interest on every payment 
after it became due. 

On this definition we proceed to investigate a formula to be 
applied to calculations respecting annuities. 

Let p represent the annual principal or annuity to be 



248 ELEMENTS OF ALGEBRA. 

paid, and l-\-r=Ji, the amount of annuity of principal for one 
year, at the given rate r. 

Let ?2 represent the number of years, and put ^' to represent 
the entire amount of the annuity in arrear. 

It is evident, that on the last payment due, no interest could 
accrue, and therefore the sum will be p. The preceding pay- 
ment will have one year's interest ; it will therefore be p^ ; the 
payment preceding that will have two years' compound interest ; 
and, of course, will be represented hy pJP, (Art. 153.) Hence 
the whole amount of A' will be 

A'=p-\-p.i-\-pJP^pJi\ &c., to pA^-\ 

This is a geometrical series, and its sum (Art. 120.) is 

^^ jo^"--p_ ;?[(l+r)"— 1] 

This general equation contains four quantities, ,B\ p, r, and n ; 
any three of them being given in any question, the others can be 
found, except r. 

EXAMPLES. 

1. An annuity of $50 has remained unpaid for 6 years, at 
compound interest on the sums due, at 6 per cent., what sum is 
now due ? 

By the ffeneral equation, 

50[(l.06)«-l] 

*^ Tog 

Taking the log. of both members, we have 

log. Jl'= log. 50+log. [(1.06)«— l]--log. .06. 

The value of (1.06)®, as found by logarithms, is 1.41852, from 
which subtract 1, as indicated, and take the log. of the decimal 
number .41852, we then have 

log.^'=1.69897-}-(— 1.62172)— (—2.7781 51)=2.54218, 

From which we find, ./2'=$348.56 Aiu. 

2. In what time will an annuity of $20 amount to $1000, at 
4 per cent., compound interest? 



ANNUITIES. 249 

The equation applied, we have 

20C(1£4)»-1] 
.04 

Dividing by 20, and multiplying by 0.4, we have 

2=(1.04)»— 1 or (1.04)"=3. 

log. 3 .477121 ^„ 
Ans. n=-^ — 2^^=.^^^ =28 years, 
log. 1.04 .017033 ^ 

3. What will an annuity of $50 amount to, if suffered to 
remain unpaid for twenty years, at 3^ per cent, compound in- 
terest? ^ns. $1413.98. 

4. AVIiat is the present value of an annuity or rental of $50 
a year, to continue 20 years, discounting at the rate of 3^ per 
cent., compound interest ? 

N. B. By question 3d, we find that if the annuity be not paid 
until the end of 20 years, the amount then due would be $1413.98. 
If paid now, such a sum must be paid as, put out at compound 
interest for the given rale and time, will amount to $1413.98. 

Now if we had the amount of $1 at compound interest for 20 
years, at 31 per cent., that sum would be to $1 as $1413.98 is 
to the required sum, $713.50. 

(Art. 155.) To be more general, let us represent the present 
worth of an annuity by F. By (Art. 153.) the amount of one 
dollar for any given rate and time, is ^"; ^^ being l-\-r and n 
the number of years. By (Art. 154.) the value of any annuity 
p remaining unpaid for any given time, n years, at any rate of 

compound mterest r, is ^ or Ji'. 

Now by the preceding explanation we may have this propor- 
tion : 

A^ : 1 ::./?': P, or P=='-^ (1) 

Hence, to find the present worth of an annuity, we have this 
Rule. Divide the amount of the annuity supposed unpaid 

for the given number of years, by the amount of one dollar for 

the same number of years. 



250 ELEMENTS OF ALGEBRA. 

If in equation (1) we put the value of Jl', we shall have 
P^n=:i- i-. Divide both members by A^, and we have 

P 

P=CI^ (2) 

r 

This last equation will apply to the following problems : 

5. The annual rent of a freehold estate is p pounds or dollars, 
to continue forever. What is the present value of the estate, 
money being worth 5 per cent., compound interest ? 

Here, as n is infinite, the term, -^ becomes 0, and equation 

(2) becomes P=i-=-^=20p ; that is, the present value of the 
estate is worth 20 years' rent. 

6. The rent of an estate is $3000 a year ; what sum could 
purchase such an estate, money being worth 3 per cent., com- 
pound interest ? Jlns. $100000. 

7. What is the present value of an annuity of $350, assigned 
for 8 years, at 4 per cent. ? Ans. $2356.46. 

8. A debt due at this time, amounting to $1200, is to be dis- 
charged in seven annual and equal payments-; what is the 
amount of these payments, if interest be computed at 4 per cent. ? 

Ans. $200, nearly. 

9. The rent of a farm is $250 per year, with a perpetual lease. 
How much ready money will purchase said farm, money being 
worth 7 per cent, per annum 1 Ans. $3571^. 

10. An annuity of $50 was suffered to remain unpaid for 
20 years, and then amounted to $1413.98 ; what was the rate 
per cent., at compound interest ? 

N. B. This question is the converse of problem 3, and, of 
course, the answer must be 83 per cent. But the general equa- 
tion gives us 

U13.98=52C(l±r)^J=i]. 



GENERAL THEORY OF EQUATIONS. 261 

Or 28.2796=^^— t:l— i; 
r 

an equation from which it is practically impossible to obtain r, 
except by successive approximations. 



SECTION VII. 

CHAPTER I. 
GENERAL THEORY OF EQUATIONS. 

(Art. 156.) In (Art. 101.) we have shown that a quadratic 
equation, or an equation of the second degree, may be conceived 
to have arisen from the product of two equations of the first de- 
gree. Thus, if a?=a, in one equation, and x—b in another 
equation, we then have 

X — a=0, 
and X — b = 0; 



By multiplication, x^ — {a-\rb)x-]-ab=0. 

This product presents a quadratic equation, and its two roots 
are a and b. 

If one of the roots be negative, as x= — «, and x=b, the 
resulting quadratic is 

x^-\-{a—b)x—ab=0. 

If both roots be negative, then we shall have 
x''+{a-{-b)x-\-ab=0. 

Now let the pupil observe that ihe exponent of the highest 
power of the unknown quantity is 2 ; and there are two roots. 
The coefficient of the first -power of the unknown quantity is 
the algebraic sum of the two roots, with their signs changed; 
and ihe absolute term, independent of the unknown quantity, 
is the product of the roots {the sign conforming to the rules of 
multiplication). 



< 



252 ELEMENTS OF ALGEBRA. 

When the coellicients and absolute term of a quadratic are not 
large, and not fractional, we may determine its roots by inspec- 
tion, by a careful application of these principles. 

EXAMPLES. 

Given a;^— 20a;+96=0, to find x. 

The roots must be 12 and 8, for no other numbers will make 
— 20, signs changed, and product 96. 

Given y^ — 6y — 55=0 to find?/. Roots 11 and — 5. 

Given x^ — Qx — 40=0 to find x. Roots 10 and — 4. 

Given 3tr^-\-(Sx — 91=0 to find x. Roots 7 and — 13. 

Given y^ — 5y — 6=0 to find y. Roots 6 and — 1. 

Given y^-\-l'2y — 589=0 to find ^. 

Here it is not to be supposed that we can decide the values of 
the roots by inspection ; the absolute term is too large ; but, 
nevertheless, the equation has two roots. 

Let the roots be represented by P and Q. 
, From the preceding investigation 

^+^=— 12 (1) 

And P^=— 589 (2) 

By squaring eq. (1) P'^-^'lPq^q-^ 144 
4 times eq. . . • (2) 4PQ =—-2356 

By subtraction, 'p^—^pq^q-=. ~2500 

By evolution, P— Q=±50 

But P-i-Q=— 12 

Hence 7^=19 or — 31, and ^=—31 or +19, the true 
roots of the primitive equation ; and thus we have another 
method of resolving quadratics. 

(Art. 157.) In the same manner we can show that the product 
of three simple equations produce a cubic equation, or an equa- 
tion of the third degree. Conversely, then, an equation of the 
third degree has three roots. 

The three simple equations, x=a, x==b, x~c,* may be put 

* Of course, x cannot equal different quantities at one and the same time ; 
and these equations must not be thus understood. 



GENERAL THEORY OF EQUATIONS. 253 

in the form of x — a=0, x — 6=0, and x — c=0, and the pro- 
duct of these three give 

{x — a){x — 6)(a?— c)=0 ; 
and by actual multiplication, we have 

x^ — {a-{-b-\-c)x^-^{ab-\-ac-{-bc)x — abc=0. 

If one of the simple equations be negative, as x= — c, or 
X'\-c=0, the product or resulting cubic will be 

x^ — (a+6 — c)x^-\-{ab — ac — bc)x-\-abc=0. 

If two of them be negative, as x= — b and a;= — c, the 
resulting cubic will be 

x^-\-{b-{-c — a)x'^-\- {be — ab — ac)x — abc=0. 

If all the roots be negative, the resulting cubic will be 
x^-{-{a-{-b-\-c)x^-\-{ab-{-ac-\-bc)x-{-abc=0. 

Every cubic equation may be reduced to this form, and con- 
ceived to be formed by such a combination of the unknown term 
and its roots. 

By inspecting the above equations, we may observe 

1st. The first term is the third power of the unknown 
quantity. 

2cl. The second term is the second power of the unknown 
quantity^ with a coefficient equal to the algebraic sum of the 
roots, with the contrary sign. 

3d. The third term is the first power of the unknown 
quantity f with a coefficient equal to the sum of all the products 
which can be made, by taking the roots two by two. 

4th. The fourth term is the continued product of all the roots, 
with the contrary sign. 

It is easy, then, to form a cubic equation which shall have 
any three given numbers for its roots. 

Assuming x for the unknown quantity, what will the equation 
be which shall have 1, 2 and 3 for its roots ? 

^ns. a;3— (H-24-3)ar^-l-(2+3-4-6)a?~6=0 ; 
Or x^—6x^-{-nx=6. 

Find the equation which shall have 2, 3, and — 4 for its roots. 

^ns. a;^—ir2— 14^+24=0 
W 



254 ELEMENTS OF ALGEBRA. 

Find the equation which shall have — 3, — 4, and -}-7 for its 
roots. dns. a^^^Ox^ — 37a: — 84=0. 

Or a;2— 37a?— 84=0. 

These four general cases of cubic equations may all be repre- 
sented by the general form. 

Thus: a;=^+j9a?2+//a?+r=0, (1) 

(Art. 158.) When the algebraic sum of three roots is equal to 
zero, equation (1) takes the form of 

x^-\-qx-\-r=0 (2) 

Equation (1) is a regular cubic, and is not susceptible of a 
direct solution, by Cardan's rule, until it is transformed into 
another wanting the second term, thus making it take the form 
of equation (2). To make this transformation, conceive one of 
the roots, or x, in equation (1), represented by ii-\-v. 

Then x^=v?-{-^iv'v-]-Zuv^ -^v^ 

qx = qu -^-qv 

r = r 

By addition, and uniting the second member according to the 
powers of u, we shall have 

u^+{Sv-\-p)u''-\-{3v''-{-2pv-\-q)u-\-{v^+pv^-\-qv-\-r)=0, 

for the transformed equation. But the object was to make such 
a transformation that the resulting equation should be deprived 
of its second power; and to effect this, it is obvious that we 
must make the coefficient of u^ equal zero, or 3v-\-p=0. 
Therefore, v= — ijo. 

Hence, we perceive that if x, in the general equation (1), be 

put equal to u — ~, there will result an equation in the form of 

o 

i^-\-qu-\'r=0, or the form of equation (2). 

As x=u — ^, and if «, &, and c represent the roots of equa- 

o 

tion (1), or the values of a?, the roots of (2), or values of u will be 
a4-5jo, 6+5/), and c+^/). 



GENERAL THEORY OF EQUATIONS. 255 

•'^ 
EXAMPLES. 

1. Transform the equation a^ — 9a?^+26a; — 30=0, into another 

wanting the second term. 

By the preceding investigation, we must assume 

x=u-\-3. Here p= — 9 ; therefore, — J;J=3. 

2Qx= 26M+78 

—30 = —30 



Sum, =u^ — u — 6=0, the equation required. 

2. Transform the equation x^ — 6ar+ 10a? — 8=0, into another 
not containing the square of the unknown quantity. 

Put x=u-\-2. Result, v? — 2w— 4=0 

3. Transform x^ — 'dx^-\-Qx — 12=0, into another equation, 
wanting the second power of the unknown quantity. 

Put a?=w+l. Result, i^^+St/— 8=0. 

(Art. 159. We have shown, in the last article, that any regular 
cubic equation containing all the powers of the unknown quan- 
tity can be transformed into another equation deficient of the 
second power ; and hence all cubic equations can be reduced to 
the form of a?+Zpx^'iq. 

We represent the coefficient of x by 3/?, and the absolute term 
by 2q, in place of single letters, to avoid fractions, in the course 
of an investigation. 

Now, if we can find a direct solution to this general equation, 
it will be a solution of cubic equations generally. 

The value of x must be some quantity ; and let that quantity, 
whatever it is, be represented by two parts, v-\-y., or let 
x=v-\-y- Then the equation becomes 

By expanding and reducing, we have 

^^+2/^+3 iyy+p) {v -\-y)=2q. 

Now as we have made an arbitrary division of x into two parts, 
V and y, we can so divide it, that 

vy-{-p=0 



256 ELEMENTS OF ALGEBRA. 

This hypothesis gives 

v'-\-y'=^ 2q, {A) 

And vy=—p, [B) 

Here we have two equations, (^) and (B), containing two un- , 
known quantities, similarly involved, which admit of a solution 
by quadratics. (Art. 108.) Hence we obtain v and ?/, and their 
algebraic sum is x. 

From equation (B), 

This substituted in equation (^), gives 

Or, v^ — 2gv^—p^, a quadratic. 

Hence v=(^q±Jq^-i-p^y 

And y={q^j7^'y 

Or, as y^^ — — — • 



— > 



1 1 

Therefore x'={q+ J^^')' +{q— Jf+ff ■ ■ (C) 

Or, .=(,±V?+/) -(^Jr^^ • • (D) 

These formulas are familiarly known, among mathematicians, as 
Cardan's rule. 

(Art. 160.) When p is negative, in the general equsflion, and 
its cube greater than q^^ the expression Jq^ — p^ becomes imagi- 
nary ; but we must not conclude that the value of x is therefore 

imaginary ; for, admitting the expression tjq^ — p^ imaginary, it 
can be represented by a J — 1, and the value of a?, in equation 
(C), will be 



GENERAL THEORY OF EQUATIONS. 257 
'«'=(?+«v'^)*+(?-aV^)' 

Now by actually expanding the roots of these binomials by the 
binomial theorem, and adding their results together, the terms 
containing J — 1 will destroy each other, and their sum will be 
a real quantity ; and, of course, the value of x will become real. 
If in any particular case it becomes necessary to make the series 

converge, change the terms of the binomial, and make --s/— 1 

stand first, and 1 second. 

EXAMPLES. 

Given x^ — 6a; =5. 6, to find the value of x. 

Here, Sp= — 6, and 2q—5.6, or p= — 2, and q=2.S, 
Then 



a;=(2.8-f 77.84— 8) +(2.8—77.84—8) , by equation (C) 



Or ir=(2.8+.47— 1) +(2.8— .47— 1)' 



Or =(1+17—1) +(i--| /— 1) 

V(2.8) ^ '^ ^ ^ ■'^^ ^ 

Expand the binomials by the binomial theorem, (Art. 135.) 
and for the sake of brevity, represent ^7 — ^ ^7 ^» 

Then b'=—^\, and b'=J^X^\. 



(l+i7-l) = l+6. (l-i7-l) = l-6. 
22 



258 ELEMENTS OF ALGEBRA. 

^ ^ ^ ^' 3.6 ^3.6.9 3.6.9.12 ' 

Sum. =^-<|e*0-<3S?2*>' ^"^ 

=24-0.004535—0.000034=2.0045. 
Therefore, 

3-y^=2.0045 or a?=(2.0045)(2.8)'=2.8256, nearly. 

(Art. 161.) Every cubic equation of the form of 
X^±:px==dzq 
has three roots, and their algebraic sum is 0, because the equa- 
tion is wanting its second term. (Art. 157.) 

If the roots be represented by «, Z>, and c, we shall have 
a-\-h-\-c==0, and abc=±q. 

If any two of these roots are equal, as b=c, then «= — 25 (1), 
and ab^=z:hq (2). Putting the value of a taken from equation 
(1), into equation (2), and we have — 2b^=dzq. 

Hence, in case of there being two equal roots, such roots must 
each equal the cube root of one half the quantity represented 
byq, 

EXAMPLES. 

The equation x^ — 48a?=128 has two equal roots; what are 
the roots 1 

Here, — 26^=128, or ¥='^M', therefore, 6=— 4. 

Two of the roots are each equal to ■ — 4, and as the sum of the 
three roots must be 0, therefore — 4, — 4, -\-%, must be the 
three roots. 

If the equation x^ — 27x=:54 have two equal roots, what are 
the roots ? Ans, — 3, — 3, and +6. 

Either of these roots can be taken to verify the equation ; and 
if they do not verify it, the equation has not equal roots. 

(Art 162.) If a cubic equation in the form of 



GENERAL THEORY OF EQUATIONS. 259 

have two equal roots, each one of the equal roots will be 

equal to . 

i(p±Jf-3g). 

The other root will be twice this quantity subtracted from jO, 
because the sum of the three roots equal/?. (Art. 157.) 

This expression is obtained from the consideration that the 
three roots represented by a, b, and c, must form the following 
equations: (Art. 157.) 

a-{-b-j-c=p (1) 

ab-{-ac-\-bc=q (2) 

abc—r (3) 

On the assumption that two of these roots are equal, that is, 
a=b, equations (1) and (2) become 

2a-fc=jO (4) 

And a^-^2ac=q (5) 

Multiply equation (4) by 2 c, and we have 

4(^-{-2ac=2ap (7) 

Subtract (5) a^-^2ac= q (8) 

And we have 3a^ =2ajo — q. 

This equation is a quadratic, in relation to the root a, and a 
solution gives a=^{p±:Jp^ — 3^). 

(Art. 163.) A cubic equation in the form of x'^±px=dzq can 
be resolved as a quadratic, in all cases in which q can be resolved 
into two factors, m and n, of such a magnitude that m^-\-p=n. 

For the values of p and ^, in the general equation, put the 
assumed values, mn=q, and p=n — m^. 

Then we have x^-\-nx — m^x=mn. 

Transpose — m^a?, and then multiply both members by x, and 
x*-\-na^=m^x^'\-mnx. 

Add - - to both members, and extract square root ; 
4 

Tl Yl Tl 

Then a^+-=wa?-|--. Drop -, and divide by a?, and a?=w. 



260 ELEMENTS OF ALGEBRA. 

Therefore, if such factors of q can be found, the equation is 
already resolved, as x will be equal to the factor m. 

EXAMPLES. 

1. Given x^-\-Qx=SS, to find the values of x. 

Here, wn=88 =4X22, 42-f-6=22. Hence, a;=4. 

^^, a. Given a;^H-3a;=14, to find oiie value of x. 

2X7=14, 22+3=7. Hence, a:=2. 

3. Given a?^-}-6a?=45, to find one value of x. Ans. ic=3. 

4. Given a? — 13a;= — 12, to find x. Ans, a;=3. 

5. Given 2/^+48?/ =104, to find ^. Ans. y=2. 

In the above examples we have given only one answer, or one 
root ; but we have more than once observed, that every equation 
of the third degree must have three roots. Take, for example, 
the 4th equation. We have found, as above, one of its roots to 
be 3. Now we may conceive the equation to have been the 
product of three factors, one of which was {x — 3); therefore the 
equation must be divisible by x — 3 without a remainder, (other- 
wise 3 cannot be a root) ; and if we divide the equation by x — 3, 
the quotient must be the product of the other two factors. 

Thus, x—S)x^—ldx-\-12{x''+3x-^4 
x'^—Sx^ 



3:i'2_13a; 
3a^— 9x 



— 4a:-{-12 

— 4a:4-12 

By putting a^-\-3x — 4=0, and resolving the equation, we find 
07=1, or — 4, and the three roots are 1, 3, — 4. Their sum is 
0, as it should be, as the equation is deficient of its second term. 
In the general equation 

x^-{-px^-jrqx-\-r=0. 

If p and q are each equal to 0, at the same time, the equation 
becomes ^''+r=0, a binomial equation. 



GENERAL THEORY OF EQUATIONS. 201 

Every binomial equation has as many roots as there are 
units in the exponent of the unknown quantity. 

Thus a?3_{_8=o, and 3?=^— 8=0, or a?5+l=0, and a;='— 1=0, 
&c., are equations which apparently have but one root, but a 
full solution will develop three. 

Take, for example, ;r'=r8 

By evolution, x =2 



2x2—8 
2^2— .4x 



4x— 8 



Now by putting a;^+2a:+4=0, and resolving the equation, 
we find a?= — 1-|-V — 3, and a:= — 1 — J — 3; and the three 
roots of the equation a^ — 8 = 0, are 2, — 1 + ^^/ — 3, and 
— 1 — J — 3, two of them imaginary, but either one, cubed, will 
give 8. 

The three roots of the equation ir^-{-l=0, are 

1 , 1 ,__ ^ 1 1 ^ 

-1' 2-^2>/^' ^^^^ 2~2^-"^- 

CHAPTER II. 
GENERAL THEORY OF EQUATIONS— Continued. 

(Art. 164.) In the last Chapter we confined our investigations 
to equations of the second and third degrees ; and if they are 
well understood by the pupil, there will be little difiiculty, in 
future, as many of the general properties belong to equations of 
every degree. 

All the higher equations may be conceived to have been formed 
by the multiplication of the unknown quantity joined to each 
of the roots of the equation with a contrary sign, as shown in 
(Art. 157.). 



262 ELEMENTS OF ALGEBKA. 

Let a, b, c, d, e, <fec., be roots of an equation, and a? 
its unknown quantity, then the equation may be formed by the 
product of {x — a)[x — h){x — c), &c., which product we may 
represent by 

x'''-\-Ax^-'^Bx'^-^ Tx-\-U=0. 

Now it being admitted that equations can be thus formed by 
the multiplication of the unknown quantity joined to its roots, 
conversely, when any of its roots can be found, such root, with 
its contrary sign joined to the unknown term, will form a com- 
plete divisor for the equation ; and by the division the equation 
will be reduced one degree, and conversely. 

Tf any quantity, connected to the unhiown quantity by the 
sign plus or minus, divide an equation without a remainder, 
such a quantity may be regarded as one of the roots of the 
equation. 

The product of all the roots form the absolute term U. 

(Art. 165.) Eve)^ equation having unity for the coefficient 
of the first term, and all the other coefficients, whole numbers, 
can have only ivhole numbers for its commensurable^ roots. 

This being one of the most important principles in the theory 
of equations, its enunciation should be most clearly and distinctly 
understood. Such equations may have other roots than whole 
numbers; but its roots cannot be, among the definite and irre- 
ducible fractions, such as |, J, y , &c. Its other roots must be 

among the incommensurable quantities, such as ^2, {3)**, &c., 
i. e., surds, indeterminate decimals, or imaginary quantities. 

To prove the proposition, let us suppose j a commensurable 

but irreducible fraction, to be a root of the equation 

Jl, B, Slc, being whole numbers. 
Substituting this supposed value of x, we have 

* Commensurable numbers are all those that measure or can be measured by 
unity ; hence, all whole numbers and definite fractions are commensurable. — 
Surds, and imaginary quantities, arc incommensurable. 



GENERAL THEORY OF EQUATIONS. 263 

Transpose all the terms but tlie first, and multiply by b^-\ and 
we have 

Now, as a and b are prime to each other, b cannot divide «, 
or any number of times that a may be taken as a factor, for j- 



being irreducible, -X« is also irreducible, as the multiplier a 

will not be measured by the divisor b ; therefore y- cannot be 

expressed by whole numbers. Continuing the same mode of 

reasoning, — cannot express whole numbers, but every term in 

the other member of the equation expresses whole numbers. 

Hence, this supposition that the irreducible fraction r is a root 

of the equation, leads to this absurdity, that a series of whole 
numbers is equal to another quantity ^/m^ must contain a fraction. 

Therefore, we conclude that any equation corresponding to 
these conditions cannot have a definite commensurable fraction 
among its roots. 

(Art. 166.) Any equation having fractional coefficients, can be 
transformed to another in which the coefficients are all whole 
numbers, and that of the first term unity. 
For example, take the equation 

m n p 

Assume x=-^—, and put this value of x in the equation, 
mnp 

A„d -|^+^+-^+£=.0. 
m^ny^ m^ny^ mny p 

Multiply every term by mhi^p^, and we have 

y'^-\-anpy^-\-bm^np'^y-\-cm^n^p^=-Q. 



SJ64 . ELEMENTS OF ALGEBRA. 

When m, n, and p have common factors., we may put x equal 
to y divided by the least common multiple of these quantities, 
as in the following examples: 

U%lu UX C 

Transform the equation x^-\ ■-{- — -\ — =0, into another 

pm m p 

which shall have no fractional coefficients, and that of the first * 

term unity. 

To eflfect this, it is sufficient to put x=—. With this value 

pm 

of X the equation becomes 

p^m^ p^m^ pm^ p 

Multiplying every term by phn^, we obtain 

y^-r ay^ ■\- hphny -{- cp^m^=0 

for the transformed equation required. 

5^3 3^ i^/p J 

Transform the equation x'^-{ — 7r+-:i-+^r7"f'T7;=0> i"to an- 

6 4 24 12 

other, having no fractional coefficients. 

Besult, y +201/8+ 12.24i/2+7(24)22/4-2(24)»=0. 

(Art. 167.) Now as every commensurable root consists of 
whole numbers, and as the coefficients are all whole numbers, 
each term of itself consists of whole numbers, and the commen- 
surable roots are all found among the whole number divisors of 
the last term ; and if these divisors are few and obvious, those 
answering to the roots of the equation may be found by trial. If 
the factors are numerous, we must have some systematic method 
of examining them, such as is pointed out by the following rea- 
soning : 

Take the equation x'^-{-Ax^-{-Bx^-}-Cx-{-D=0. 

Let a represent one of its commensurable roots. Transpose 
all the terms but the last, and divide every term by a. 

_ =-^c^^^a^-^Ba-^C. 
a 

But, since a is a root of the equation, it divides D without a 



GENERAL THEORY OF EQUATIONS. 265 

remainder, the left hand member of this last equation is therefore 
a whole number, to which transpose C, also a whole number, 

and represent —-{-C by N. 

Then JV^—a'—^a'—Ba, 

Divide each term by a, and transpose Bj and we have 

N 
a 

The right hand member of this equation is an entire quantity, 

(not fractional), therefore the other member is also an entire 

quantity ; let it be represented by N', and the equation again 

divided by a. 

N' 
Then — = — a — £. 

a 

Transpose — ./3 ; reasoning the same as before, we can repre- 
sent the first member by iV", and we then have 

N" 
Divide by a, and — = — 1. This must be the final result, in 

case « is a root. 

From these operations we draw the following rule for deciding 
what divisors of the last term are roots of an equation. 

Rule. Divide the last term by the several divisors, [each 
designated by a,) and add to the quotient the coefficient of the 
term involving x. 

Divide this sum by the divisors (a), and add to the quotient 
the coefficient of the term involving x^. 

Divide this sum by the divisors (a), and add to the quotient 
the coefficient of the term involving a^. 

And thus continue until the first coefiicient, *^, is transposed, 
and the sum divided by a ; the last quotient will be minus one^ 
if a is in fact a root. 

23 



266 



ELEMENTS OF ALGEBRA. 



EXAMPLES. 
1. Required the commensurable roots (if any) of the equation 



w 



CO 



CO 



I 



" 


7 






_ 








o 

'a 


c^ 


7 












CO 


o 


CO 


CO 

7 












1 




"rj* 


c^ 


-^ 


CO 


1 


CO 

+ 


I—I 
1 


1 


J3 




1 






1 








1 












riH 


-^3 
















1 


c 


CO 


o 


lo 


tJ< 


•? 


c^ 


1— » 

1 


1 


03 




1— t 

1 




. 


1 


+ 


w 


s" 


2 


1 

T 


^ 


CO 


1 


+ 


1 


2 


0) 

o 


I— I 

1 


QO 
1 


00 

1 


1 


T 






.2 
o 


1 


1 


1 


1 


1 






^ 


CD 


N 


,_4 


c» 


CO T}< 


•? 


r^ 


o 


^c 


1 
1 


1 

O 

1 


1— 1 

1 


1— 1 

1 


1 + 


1 


1 


EC 

O 


n ; and, be 
roots. 


CO 

1 


7 












CL, 


.2 o 

j3 


N 


1 

00 


CO '- 


t 












1 


7 


1 1 


1 










o 2- 


r-^ CD 

1 1 


7 












> 


, roots of 
^ree, these 






•> 




^ 




CO 


^ 


S^ '^ 






-2 




B 






o 


u "^ 


o -< 




.2 




a 




o 


•5 


i -B 




3 t3 

5^ <J 








o 




a 




t3 




"O 




•^ 






OJ 




CO 




-•si^ 




rt *t3 



EQUAL KOOTS. 267 

2. Required the commensurable roots of the equation 
a:3— 6a^-flla?--6=0. Ans. 1,2,3. 

3. Required the commensurable roots of the equation 
a;4__6^2__i6^_j_2i=o. Ans. 3 and 1. 

Here the student might hesitate, as one regular term of the 
equation is wanting, or rather the coefficient of a;'^ is ; hence, 
the equation is a?*±0.r='— 6a?2--16a:+21=0. 
Go through the form of adding 0. 

4. Required the commensurable roots of the equation 
a'4_.6^+5:c2_|_23;_io=o. Ans, —1, +5. 

As the commensurable roots are only two, there must be two 
incommensurable roots ; and they can be found by dividing the 
given equation by 57+1, and that quotient by x — 5, and resolv- 
ing the last quotient as a quadratic. 

EQUAL ROOTS. 

(Art. 168.) In any equation, as 

x'-^Ax'-YB;£'-^Cx^-\-Bx-\-E^^,'' .... (1) 
the roots may be represented by a, Z>, c, rf, e, and either one, put 
in the place of a?, will verify the equation. 

Now, let y represent the difference between any two roots, as 
a — 6 ; then y=a — 6, and by transposition h-\-y=^a. But as a 
will verify the equation, it being a root, its equal, (6+2/)i sub- 
stituted for X, will verify it also. That is, 

(6-f3/)^+^(6+3/)^+J5(6+3/)^+q6+2/)^+i)(6+2/)+^=0. 

By expanding the powers, and arranging the terms according 
to the powers of y, we have 

W -\-^b^y -\-\^bY +io^*y -\-^by^-{-y^ 

Ab^-^4.Abhj-\- 6My-{- 4Aby^+Ay* 
Bb'-\-dBb'y-h 3Bbf+Bf 
Cb^Jf-2Cby-{-Cy^ 
Db^Dy 

We might have been more general, and have taken x™~\-Ax'f^—^, &c., for 
the equation ; but, in our opinion, we shall be better comprehended by taking 
an equation definite in degree ; the reasoning is readily understood as general. 



=0. 



:0. 



268 ELEMENTS OF ALGEBRA. 

Now, as & is a root of the equation, the first column of this 
transformation is identical with the proposed equation, on sub- 
stituting the root b for x. Hence, the first column is equal to 
zero ; therefore, let it be suppressed, and the remainder divided 
hyy. 

We then have 

4M^ -\-6M^y-]-4My^-{-^y^ 
SBb^-\-3Bby-{-By^ 
2Cb +Cy 
B ^ 

On the supposition that the two roots a and b are equals y 
becomes nothings and this, last equation becomes 

5b^-\-4M^-{-3Bb^-\-2Cb-\-B=0. 

As 6 is a root of the original equation, x may be written in 
place of b ; then this last equation is 

5x'-^4^x^-{-SBaf-\-2Cx-\-n=0 (2) 

This equation can be derived from the primitive equation by 
the following 

Rule. Multiply each coefficient by the exponent of x, and 
diminish the exponent by unity. 

Equation (2) being derived from equation (!)» by the above 
rule, may be called a derived polynomial. 

(Art. 169.) AVe again remind the reader that b will verify the 
primitive equation (1), it being a root, and it must also verify 
equation (2) ; hence, b at the same time must verify the two 
equations (1) and (2). 

But if b will verify equation (1), that equation is divisible by 
(^x — 6), (Art. 164.), and if it will verify equation (2), that equa- 
tion also, is divisible by {x — 6), and {x — b) must be a common 
measure of the two equations (1) and (2). That is, in case 
the primitive equation has two roots equal to b. 

(Art. 170.) To determine whether any equation contains equal 
roots, take its derived polynomial by the rule in (Art. 168.), and 
seek the greatest common divisor (Art. 27.), [which designate by 



GENERAL THEORY OF EQUATIONS. 269 

(Z)),] of the given equation and its first derived polynomial ; 
and if the divisor D is of the first degree, or of the form of x — A, 
then the equation has two roots each equal to h. 

If no common measure can be found, the equation contains no 
equal roots. If D is of the second degree, with reference to ar, 
put D=Oj and resolve the equation ; and if D is found to be in 
the form of (x — hy ; then the given equation has three roots 
equal to h. 

If I) be found of the form of {x — h)(x — A'), then the given 
equation has two roots equal to A, and two equal to h'. 

Let D be of any degree whatever ; put D=0, and, if possible, 
completely resolve the equation ; and every simple root of D is 
twice a root in the given equation ; every double root of D will 
be three times a root in the given equation, and so on. 

EXAMPLES. 

1. Does the equation .t"* — 2x^ — lx'^-{-Wx — 12 = contain 
any equal roots, and if so, find them ? 

Its derived polynomial is ^x^ — Qx^ — 14a?+20. 

The common divisor, by (Art. 27.), is found to be x — 2 '•> 
therefore, the equation has two roots, equal to 2. 

The equation can then be divided ^zz.'{ce by x — 2, or once by 
i^x — 2)^ or by oi? — 4a;+4. Performing the division, we find 
the quotient to be x^-^-'^oc — 3, and the original equation is now 
separated into the two factors, 

(a:2_4a?-l-4)(:i"2+2a:— 3) =0. 

The equation can now be verified by putting each of these 
factors equal to zero. From the first we have already a:=2, 
and 2, and from the second we may find x=\ or —3; hence, 
the entire solution of the equation gives 1, 2, 2, — 3 for the four 
roots. 

2. The equation x^-\-2x'^—Ux^—Sx^-\-2Qx-]-\Q=0 has two 
equal roots ; find them. Jlns. 2 and 2, 

3. Does the equation x^ — 2x'^-\-*Sx^ — lx^-\-Sx — 3=0 contain 
equal roots, and how many ? 

Ans, It contains three equal roots, each equal to 1. 
x2 



270 ELEMENTS OF ALGEBRA. 

4. Find the equal roots, if any, of the equation 

iT^+a^— 16a?+20=-0. ^ns 2 and 2. 

5. Find the equal roots of the equation 

Ans. Two roots equal to 1, and two roots equal — 2. 

6. Find the equal roots of the equation 

a;^__5a;2+lO:r— 8=0. 

Ans. It contains no equal roots. 

(Art. 171.) Equations which have no commensurable roots, or 
those factors of equations which remain after all the commensu- 
rable and equal roots are taken away by division, can be resol- 
ved only by some method of approximation, if they exceed the 
third or fourth degree. It is possible to give a direct solution in 
cases of cubics and in many cases of the fourth degree ; but, in 
practice, approximate methods are less tedious and more conve- 
nient. 

JVe may transform any equation into another whose roots 
shall be greater or less than the roots of the given equation by 
a given quantity. 

Suppose we have the equation 

3(^^Ax''-\-Bx'-{-Cx'-{-Dx-\-E=0, (1) 

and require another equation, whose roots shall be less than those 
of the above by a. 

Put x=a-\-y, and, of course, the equation involving y will 
have roots less than that involving a?, by a, because y=x — «, or 
y is less than x, by a. 

In place of rr, in the above equation, write its equal {a-\-y), 
and we have 

{a^-yY^Jl{a^-yY-\-B[a^yy^C{a^yY-^D{a^-y)-hE==Q. 

By expanding and arranging the terms according to the powers 
of y, we shall have, as in (Art. 168.), 



' 


EQUAL ROOTS. 






271 


[a-\-yf=a^ -\-5a'^ 


y+lOa^ 


2/2+ lOa^ 


i/^+5« 


^+2/"' 




^{a+yy=^a'-\-4£a' 


+6^a2 


-\-AAa 


+^ 






B{a-]-yY=BQ?-\-ZBa^ 


+3i?a 


B 




=0 


C{a-^yf=Ca^-\-lCa 


+c 


• • < • 


(2) 


D{a-\-y) =Da -^D 
E . . , ^E 











After a little observation, these transformations may be made 
very expeditiously, for the first perpendicular column may be 
written out by merely changing x to «, in the original equation, 
and then, each horizontal column run out by the law of the 
binomial theorem. 

Thus a^ becomes Sa'', and this, again, lOa^ &lc. 

Now, the first column of the right hand member of this equa- 
tion consists entirely of known quantities ; and the coefficients 
of the different powers of y are known ; hence we have an 
equation, involving the several powers of y, in form of equa- 
tion (1), 

Or, y^-{-Jl'y'^-{-By-{-C'y^-i-I)'y-\-E=0', the equation 
required; ^', J5', &;c., representing the known coefficients of 
the different powers of y* 

In commencing this subject, we took an equation definite in 
degree, for the purpose of giving the pupil mjDre definite ideas ; 
but it is now proper to show the form of transforming an equation 
of the most general character. 

For this purpose, let us take the equation 

x'^+^x'^-'+Bx'^-'' Gx-{-JI=0. 

Now let it be required to transform this equation into anothc- 
whose roots shall be less than the roots of this equation by a. 

Put x=a-\-y, as before ; then 



272 



JLLEMfiNTS OF ALGEBRA. 



+ 







^ 


• 




53 




^ • 








1 


W 




s 


1 




53 


g 




1 




X 


CQ . 


1 


N 


•— • 




S 








S 


s 










H 


- 


H 


- ; 



^ 



+ 



1 


CO . 


s 


^ 1. 


« 


« Q 


^ 


R^ O 




/- N ^^ ^ 


'^ 


« CO 


1 


s J^ 






+ 


+ + 


1 


1 1 


? 


s s 


Q 


53 « 


^ 


f=q O 



Cb 



^ ^ 



e? 



2 ^ 



^ B 

i - 

a> 



M 03 

a o 

^ o 
bo 

rt O 

is o 

IS «s 

cS -^ 



^IS 



CD O 

^ II 
O) li 



** 03 C 

2 S T 



Si 






s 

7 
g 






hie a 
•5 .^ 



§5 






>5 



ft 



<^ «S 






5i ^ 



^ -< 



5>i <^ 









^ .2 



•^S E-. 



EQUAL ROOTS. 273 

If we should desire to make the third term (counting from the 
highest power of y) of equation (2) to disappear, we must 

Put 10«2+4.^«+^=0; and this involves the 
solution of an equation of the second degree, to find the definite 
value of a. To make the fourth term disappear would require 
the solution of an equation of the third degree ; and so on. 

If a is really a root of the primitive equation, then x=a, ?y=0, 
and each perpendicular column of the transformed equation is 0. 

If we designate the first perpendicular column of the general 
transformed equation by X, and the coefficienls of the succeeding 
columns by 

X' X" X' 



1 1.2 1.2.3' 



&c., we shall have, 



X" X'" 

X+X'3,+-2/^+— 2/= r=0- 

The coefficients of the different powers of i/, as X', X", X'", &c. 
are called derived polynomials ^ because each term of X' can be 
derived from the corresponding term of X ; and each term of X" 
can be derived from the corresponding terms of X', by the law of 
the binomial theorem^ as observed in the first part of this article. 
But, to recapitulate : 

X is derived from the given equation by simply changing x 
to a. 

X' is derived from X by multiplying each of the terms of X' 
by the exponent of a, in that term, and diminishing that expo- 
nent by unity, and dividing by the exponent of y increased by 1. 

X" is derived from X', in the same manner as X' is derived 
from X ; and so on. 

X' is called the first derived polynomial ; X" the second, &c. 

•To show the utility of this theorem, we propose to transform 
the following equations : 

1. Transform the equation 

into another, which shall not contain the 3d power of the un- 
known quantity. 

18 • 



274 ELEMENTS OF ALGEBRA. 

12 
By (Art. 172.), put x=i/-\-— or ... . x=3-\-y 

Here a=3 and m=4. 
X =(3)^— 12(.3)3-}-17(3)2— 9(3)4-7 ... or. . .X =—110 
X' =4(3)3— 36(3)2+34(3)— 9 or. . .X' =—123 

^=6(3)2-36(3)+ 17 or. . .^'=— 37 

X'" X'" 

Trs-<'y-'-' ..or..._=..0 

Therefore the transformed equation must be 

y— 372/2— 1232/—1 10=0. 

2. Transform the equation 

a;3_6^2_^13^_12=0, 

into another wanting its second term. Put x=2-\-y. 

X =(2)=^— 6(2)2+13(2)— 12 or. . .X =—2 

X' =3(2)2— 12(2) + 13 . or. . .X' =+1 

Y=3(2)^-6 or...^= 

X'"' , . X'" 

2:3=' •••• ^^•••2:3= ' 

Therefore the transformed equation must be 
2/^+3/— 2=0. 

3. Transform the equation 

a;4__4^__8.r+32=0, 
into another whose roots shall be 2 less. 

Put x=2-{-y. Result, y+4?/»— 24i/=0. 

As this transformed equation has no term independent of y, 
y=0 will verify the equation ; and x=2 will verify the original 
equation, and, of course, is a root of that equation. 



• EQUAL ROOTS. "» ^ 275 

4. Transform the equation 

into another whose roots shall be greater by 3. 

Put x=—3-{-y. Result, y^-\-iy'^-{-9y'^ — i2y=^0, 

5. Transform the equation 

a^4__8a?3_|.^_^ 82a:— 60= 0, 
into another wanting its second term. 

Result, y— 23?/2+22i/+60=0. 

(Art. 173.) We may transform an equation by division, as 
well as by substitution, as the following investigation will show. 

Take the equation 

x'-\-^x^i-Bx^-}-Cx-{-n=0 (1) 

If we put x=a-{-y, in the above equation, it will be trans- 
formed (Art. D.) into 

As x=a-\-y, therefore y^=x — a ; and put this value of y in 
equation (2), we have 

(^'-«r-l-|^'(^-«)^+^(^-«r+X'(rr~«)+X=0. . .(3) 

Now it is manifest that equation (3) is identical with equation 
(1), for we formed equation (2) by transforming equation (l),and 
from (2) to (3) we only reversed the operation. 

Now we can divide equation (3), or in fact equation (1), by 
{x — a), and it is obvious that the first remainder will be X. 

Divide the quotient, thus obtained, by the same divisor, {x — a), 
and the second remainder must be X'. 

Divide the second quotient by {x — a), and the third remainder 

X'' 
must be -—. 
2 

X'" 
The next remainder must be — — , &c., &c., according to the 

2.U 

degree of the equation. 



276 ELEMENTS OF ALGEBRA. 

Now if we reserve these remainders, it is manifest tliat they 
may form the coefficients of the required transformed equation ; 
taking the last remainder for the Jirst coefficient ; and so on, in 
reverse order. 

For illustration, let us take the third example of the last article. 

a:=2+i/, or y=x — 2. 
a:_2)a;''— 4a;''— 8a?+32(ar'— 2a?2— 4a;— 16 - 



— 2a;3— 8a? 






— 2ar'+4a;2 






__4a;2___8a? x- 


-2)a;«- 


-2a?2— 4a:— 16(a;2— 


— 4a;2_|_8a, 


a;^- 


-2x^ 


— 16a?4-32 




__4a;— 16 


— 16a;+32 




— 4a;-f 8 


0=X 


24-X 


jr— 2)a;2_4(^_j_2 


a;— 2)a:+2(l 


x^—2x 




a-— 2 


2a;— 4 
2ar— 4 


2.3 


•=s 





Hence the transformed equation is 

y+4i/3zb0i/2— 24i/+0=0 ; 
or, ?/^+4y — 24i/=0, as before. 

For a further illustration of this method, we will again operate 
on the first example of the last article. 



EQUAL ROOTS. 277 

a;— 3)a;'*— 12a;'5-f 170^2— 9a?+7(a;3--9a;2— lOa:— 39 



— 9a?34-17a;2 
— 9x34-27a;2 



—lOa^— 9^ 
—10x2+300? 



— 39ic+ 7 
— 39a?+117 



— 110=X. 1st Remainder. 



a;— 3)r^--9a;=— 1 Oa?— 39(a;^— 6:r— 28 



—Gx^'—lOx 
— 6a;2+18a7 



— 28ic— 39 

— 28iC+84 



— 123=X'. 2d Remainder. 

a:— 3)a^ — GiP— 28(a;— 3 x — 3).'r— 3(1 

x^ — 3a? x—3 



„ , « 0=——. 4th Rem. 

— 3a?+ 9 2.3 



X" 
— 37=-—. 3d Remainder. 



Hence 2/''±0i/^—-37y2_i23?/— 110=0, must be the trans- 
formed equation. 

We shall have a 4th remainder, if we operate on an equation 
of the 4th degree ; a 5th remainder with an equation of the 5th 
degree ; and, in general, n number of remainders with an equa- 
tion of the nth degree. 



278 ELEMENTS OF ALGEBRA. 

But to make this method sufficiently practical, the operator 
must understand 

SYNTHETIC DIVISION. 

(Art. 174.) Multiplication and division are so intimately blended, 
that they must be explained in connection. For a particular 
purpose we wish to introduce a particular practical form of per- 
forming certain divisions ; and to arrive at this end, we commence 
with multiplication. 

Algebraic quantities, containing regular powers, may be 
multiplied together by using detached coefficients, and annexing 
the proper literal powers afterwards. 

EXAMPLES. 

1 Multiply a^+2aa;+a;^ by a-\-x. 
Take the coefficients. Thus 
1+2+1 
1 + 1 



1+2+1 
1+2+1 



Product, .... 1+34-3+1 
By annexing the powers, we have 

«3+3«2a;-|-3aa?2+ar'. 

2. Multiply x^-^-xy-^-y"^ by x^ — xy-{-y^. 

As the literal quantities are regular, we may take detached 
coefficients, thus : 

1 + 1 + 1 

1—1 + 1 ^ '% 



1 + 1 + 1 
_1_1-_1 

1+1+1 

Product,. . . . 1+0+1+0+1 



SYNTHETIC DIVISION. 279 

Here the second and fourth coefficients are ; therefore the 
terms themselves will vanish ; and, annexing the powers, we shall 
have for the full product 

3. Multiply 3.t2— 2a?— 1 by 4a;-f-2. 
3—2—1 

4+2 



12—8 — i 
6—4—2 



12—2—8—2 
Product, . . . 12a?=^— 2a:^— 8a:— 2. 

4. Multiply x^ — aa^^+aV — a^x-^a'^ by x-{-a. 
1—1 + 1—1-1-1 
1 + 1 



1—1 + 1—1+ 1 
+ 1-1 + 1-1 + 1 

1+0+0+0+0+1 



As all the coefficients are zero except the first and last, there- 
fore the product must be 

(Art. 175.) Now if we can multiply by means of detached 
coefficients, in like cases we can divide by means of them. 

Take the last example in multiplication, and reverse it, that is, 
divide x^-]-a^ by x-\-a. 

Here we must suppose all the inferior powers of x^ and c^ 
really exist in the dividend, but disappear in consequence of their 
coefficients being zero ; we therefore write all the coefficients of 
the regular powers thus : 



280 ELEMENTS OF ALGEBRA. 

Divisor. Dividend. Quotient. 

1 + 1)1+0+0+0+0+1(1—1-1-1—1+1 

1+1 

-1+0 

—1—1 



1+0 

1+1 



-1+0 

—1—1 



1+1 
1+1 



Annexiig the regular powers to the quotient, we have x'^ — ax^-\- 
c^y? — c^x-\- a^, for the full quotient. 

2. Divide a^- Sa^^+lOo^i^— 10a263+5a6^— 6^ by o^- 
2a6+62. 

l_-2+l)l— 5+10— 10+5— 1(1— 3+3— 1 
1—2+ 1 

__3+9— 10 
—3+6— 3 



3—7+5 
3—6+3 

—1+2-^ 
—1+2—1 

These coefficients are manifestly the coefficients of a cube ; 
therefore the powers are readily supplied, and are 

N. B. If we change the signs of the coefficients in the divisor, 
except the jSrst, and then addihe product of those changed terms, 
we shall arrive at the same result. 

Perform the last example over again, after changing the signs 
of the second and third terms of the divisor. Thus, 



SYNTHETIC DIVISION. 281 

14-2— 1)1— 5+10— 10+5—1(1— 34-3— 1 

1+2— 1 

Sum . . *— 3+ 9—10 
__3_ 6+ 3 



Sum .... * 3— 7+5 
3+ 6—3 



Sum ''— J +2—1 

__ 1—2+1 

Sum *~0+0 

3. Divide a?=^— Ga^+llx- 6 by x^2. 
Change the sign of the second term of the divisor. 
1+2)1—6+11—6(1—4+3 
1+2 
—4+11 
—4— 8 



3—6 
3+6 



Let the reader observe, that when the first figure of the divisor 
is 1, the first figure of the quotient will be the same as the first 
figure of the dividend ; and the succeeding figures of the quotient 
are the same as the first figures of the partial dividends. 

Now this last operation can be contracted. 

Write down the figures of the dividend with their proper signs, 
and the second figure of the divisor^ with its sign changed, on 
the right. Thus 

1—6+11—6(2.= Divisor 
2— 8+6 



(l_-4+3) 

The first figure, 1 , is brought down for the first figure oj 
the quotient. 

The divisor, 2, is put under — 6 ; their sum is - — 4, which 
multiplied by 2, and the product — 8 put under the next term 
24 



2S2 ELEMENTS OF ALGEBRA. 

the sum of +11 — 8 is 3, which multiplied by 2, gives 6, and the 
sum of the last addition is 0, which shows that there is no 
remainder. 

The numbers in the lower line show the quotient^ except the 
last ; that shows the remainder, if any. 

This last operation is called synthetic division. 

4. Divide x^-\-2x^ — 8a? — 24 by x — 3. 

COMMON METHOD. 

a?— 3)a?3-l-2x2— 8a;— 24(ir2^5^_|_7 
x^— 3x2 



Qx"-^ Qx 

^x'^—lfiX 



7a?— 24 
7a?— 21 

— 3 

SYNTHETIC METHOD. 

1-1-2— 8—24(3 
3+15-1-21 

(1+5+7)— 3 

Now we are prepared to work the examples in (Art. E.) in a 
more expeditious manner. 

Transform again, the equation x"^ — Ax^ — 8a?+32=0, to an- 
other, whose roots shall be less by 2. 

This equation has no term containing ar^, therefore the coeffi- 
cient of a? must be taken =0, if we use Synthetic Division. 

FIRST OPERATION. 

l—4±0— 8+32 (2 
2—4—8—32 



(1—2—4—16), 0=X. 

SECOND OPERATION. 

1—2—4—16(2 
2db0— 8 



(1-j-O— 4),-24=X'. 



SYNTHETIC DIVISION. 283 

THIRD OPERATION. 

IdzO —4 (2 

2+4 

X" 
(1+2) 0=±-. 

FOURTH OPERATION. 

1+2(2 
2 



X'" 

Hence our transformed equation is y'*-{-4y^ — 24i/:i=0, as 
before. 

To transform an equation of the fourth degree, we must have 
four operations in division; an equation of the nth degree n 
operations, as before observed. 

Bict these operations may be all blended in one. Thus 

1 __4 -i-0 — 8 32 (2 
2 —-4 — 8 —32 



.2 —.4—16 =X 

2 0—8 



—4 —24 =X' 
2 +4 

2 

~~~ X'" 

^=-2:3- 

We omit the first column, except in the first line, as there are 
no operations with it. 

The pupil should observe the structure of this operation. It 
is an equation of the 4th degree, and there are four sums in ad- 
dition, in the 2d column ; three in the next ; two in the next, 
&c., giving the whole a diagonal shape. 



284 ELEMENTS OF ALGEBRA. 

Transform the equation x"^ — 12a:^+17cr2 — 9x-\-7=0, into an- 
other whose root shall be 3 less. 

OPERATION. 

1 —12 +17 — 9 + 7 (3 
+ 3 —27 — 30 —117 

=X 



r 



— 9 


—10 


— 39 —110: 


+ 3 


—18 


— 84 


— 6 


—28 


— 123=X' 


+ 3 


— 9 




— 3 


—37 = 


_X" 
2 


3 







2.3 

Hence the transformed equation is 

y+Oi/'- 37y2_i23^__l 10=0. 

Transform the equation x^ — 12a? — 28=0, into another whose 
roots shall be 4 less. 




4 

T 


—12 

+ 16 

4 


—28 (4 
+ 16 

— 12=X 


4 


32 




T 


36= 


=X' 


4 






12 = 


_X" 
■ 2 * 





Hence the transformed equation must be» 2/^+12i/^+36y — 
12=0, on the supposition that we put i/=x — 4. 

Transform the equation x^ — 10a?^+3a? — 6946=0, into another 
whose roots shall be 20 less. 

Put ar=20+2/. 



SYNTHETIC DIVISION. 285 



—10 
20 


3 

200 

203 
600 

803 


—6946 (20 
4060 


10 


—2886 


20 




30 

20 




50 





The three remainders are the numbers just above tlie double 
lines, which give the following transformed equation : 
2/^-h50i/2+803i/— 2886=0. 

Transform this equation into another, whose roots shall be 3 
less. Put 2/==3-l-z. 



50 


803 


—2886 (3 


3 


159 


+2886 


53 


962 
168 





3 




56 


1130 




3 







Hence the second transformed equation is 

;r=^-f592:2-f-1130z=0. 
This equation may be verified by making z=^0 ; which gives 
y=^ and :c=20-l-3=23. 

Thus we have found the exact root of the original equation by 
successive transformations ; and on this principle we shall here- 
after give a general rule to approximate to incommensurable roots 
of equations of any degree ; but before the pupil can be prepared 
to comprehend and surmount every difficulty, he must pay more 
attention to general theory, as developed in the following 
Chapter. 



286 ELEMENTS OF ALGEBRA. 

CHAPTER III. 

GENERAL PROPERTIES OF EQUATIONS. 

(Art. 176.) dny equation, having only negative roots, ivill 
have all its signs positive. 

If we take — «, — b, — c, &;c., to represent the roots of an 
equation, the equation itself will be the product of the factors ; 

{x-\-a), {x-\-b), {x-\-c), &c., =0 : 

and it is obvious that all its signs must be positive. 

From this, we decide at once, that the equation x^-^-'Ha:^-}- 
6x-rQ=0 ; or any other numeral equation, having all its signs 
plus, can have no rational positive roots. 

(Art. 177.) Sitrdsj and imaginary roots, enter equations by 
pairs. 

'J'ake any equation, as 

x''-\-Ax'-\-Bx^-^ ar-l-7)=0, 

and suppose [a-\- Jb) to be one of its roots, then (a — Jb) must 
be another. 

In place of x, in the equation, write its equal, and we have 

{a-\'jl)y^A{a^fbY+B[a-\-J~bf-^C{a+Jb)-\-D=Q. 

By expanding the powers of the binomial, we shall find some 
terms rational and some surd. The terms in which the odd 
powers of Jb are contained will be surd ; the other terms 
rational ; and if we put R to represent the rational part of this 
equation, and SJb to represent the surd part, then we must have 

/?-}- Sjb=0. 

But these terms not having a common factor throughout, cannot 
equal 0, unless we have separately J?=0, and *S'=0 ; and if this 
be the case we may have 

Ji.-^SJb=0. 

This last equation, then, is one of the results of {a-]rjb), 
being a root of the equation. 



GENERAL PROPERTIES OF EQUATIONS. 287 

Now if we write {a — Jb) in place of x, in the original equa- 
tion, and expand the binomials, using the same notation as 
before, we shall find 

But we have previously shown that this equation must be true ; 
and any quantity, which, substituted for rr, reduces an equation 
to zero, is said to be a root of the equation; therefore {a — Jb) 
is a root. 

The same demonstration will apply to {-{-sjci), ( — /y/«), to 
-\-tJ — a, — J — «, and to imaginary roots in the form of 



(«-f6^— 1), i^a—h^—i). 

(Art. 178.) If we change the signs of the alternate terms of 
an equation^ it will change the signs of all its roots. 
At first, we will take an equation of an even degree. 
If « is a root to the equation 

x'-}-^x'-{-Bx'-i-Cx-\-I)=0 . (1) 

then will — a be a root to the equation 

x'—Ax^-{-Bx^—Cx-}-I)=0 (2) 

Write a for x, in equation (1), and we have 

a'-\-^a'-{-Ba^-{-Ca-{-I)=0 (3) 

Now write — a for x, in equation (2), and we have 

a'-\-Jia'+Bd'-{-Ca-\-I)=0 (4) 

Equations (3) and (4) are identical ; therefore if «, put for x in 
equation (1), gives a true result, — a put for x in equation (2), 
gives a result equally true. 

We will now take an equation of an odd degree. 

If the equation x^-[-Jx^-{-Bx-\- Cr=0, 

have a for a root, then will the equation ** 

x^---j2x^-{. Bx—C=0, 
have — a for a root. 

From the first a^-\-^a'^+Ba-\-C=0. 

From the second < — a^ — ^a^ — Ba — C=0. 



286 ELEMENTS OF ALGEBRA. 

This second equation is identical with the first, if we change 
all its signs, which does not essentially change an equation. 

The equation ai^+a?^— 19a:2+lla?+30=0, has ~1, 2, 3, 
and — 5, for its roots ; then from the preceding investigation 
we learn that the equation 

must have 1 — 2, — 3, and +5 for its roots. 

(Art. 179.) If we introduce one positive root into an equa- 
tion, it will produce at least one variation in the signs of its 
term; if two positive roots, at least two variations. 

The equation a^-]rX-\-l=0, having no variation of signs, can 
have no positive roots. (Art. 176.) Now if we introduce the 
root +2, or which is the same thing, multiply by the factor x — 2, 

a^-\- x+i 
X — 2 



x^-\- x^-}- X 
—2x2— 2a:— 2 

Then x^ ^ x""' — x—2=0 ; 

and here we find one variation of signs from -{-x^ to — a^, and 
one permanence of signs through the rest of the equation. 

If we take this last equation and introduce another positive 
root, say +5» or multiply it by x — .5, we shall then have 

1—1 —1 —2 
1—5 



1—1 —1 —2 

—5 +5 +5 +10 

.x.4_6a;3+4^-f3a;+10=0. 

Here are two variations of signs, one from -\-x'^ to — 6x^, and 
another from — 6^:^ to +4x^. 

And thus we might continue to show that every positive root, 
introduced into an equation, will produce at least one variation 
of signs. But we must not conclude that the converse of this 
proposition is true. 



GENERAL PROPERTIES OF EQUATIONS. 289 

Every positive root will give one variation of signs ; but 
every variation of signs does not necessarily show the existence 
of a positive root. 

For an equation may have 

for roots ; then the equation will be expressed by the product of 
the factors 

As one of these terms, (— 2ax), has the minus sign, it will 
produce some minus terms in the product ; and there must neces- 
sarily be variations of signs ; yet there is no positive root. At 
the same time, the whole factor in which the minus term is found, 
must be plus, whatever value be given to a', as it is evidently 
equal to {x — df-\rh'^, the sum of two squares. 

The equation 

has two variations of signs, and two permanences, but the roots 
are all imaginary, viz., 

(2+^ZT), (2—7—1"), (— l + ^ZT), and (— l—^HT). 

If it were not for imaginary roots, the number of variations 
among the signs of an equation would indicate the number of 
plus roots : and this number, taken from the degree of the equa- 
tion, would leave the number of negative roots ; or the number 
of permanences of signs would at once show the number of 
negative roots. 

To determine a priori the number of real roots contained in 
any equation, has long bafHed the investigations of mathemati- 
cians ; and the difficulty was not entirely overcome until 1829, 
when M. Sturm sent a complete solution to the French Academy. 
The investigation is known as Sturm's Theorem, and will be pre- 
sented ill the following Chapter. ^ 

LIMITS TO ROOTS. 

(Art. 180.) All positive roots of an equation are comprised 
between zero and infinity ; and all negative roots between zero 
• 25 



290 ELEMENTS OF ALGEBRA. 

and minus infinity ; but it is important to be able at once to 
assign much narrower limits. 

We have seen, (Art. 179.), that every equation, having a posi- 
tive root, must have at least one variation among its signs, and 
at least one minus sign. 

If the highest power is minus, change all the signs in the 
equation. 

Now we propose to show that the greatest positive root must 
he less than the greatest negative coefficient plus one. 

Take the equation 

It is evident, that as the first term must be positive for all de- 
grees, X must be greater and greater, as more of the other terms 
are minus : then x must be greatest of all when all the other 
terms are minus, and each equal to the greatest coefficient, (Z> 
being considered the coefficient of x^). 

Now as A, B, &c., are supposed equal, and all minus, we shall 
have 

x'-^Jl{x^-\-x''^-x-\-l)^{i. 

For the first trial take x=^Ji, and transpose the minus quantity, 
and we have 

Divide by .^^, and we have 

Now we perceive that the second member of the equation is 
greater than the first, and the expression is not, in fact, an equa- 
tion. x^=Jl proves X not to be large enough. 

For a second trial put £P=w^-{-l. 

Then (^-M)^=^[(^-{-l)^-f(c^-l-l7-+(^4-l)4-l] 

Dividing by (./^-f-l)', we have 

^+i^(^+if^(.'^4-if^(.^-i-i)' 
We retain the sign of equality for convenience, though the 



GENERAL PROPERTIES OF EQUATIONS. 291 

members are not equal. The second member consists of terms 
ill geometrical progression, and their sum, (Art. 120), is 

1 — —^- — - . Hence the first member is greater than the second, 

(^+1) 

which shows that (ji-{-l) substituted for x, is too great. But ^ 
was too small, therefore the real 'value of x, in the case under 
consideration, must be more than .^ and less than [ji-\-l). 

That is, the greatest positive root of an equation^ in the 
most extreme case, must be less than the greatest negative 
coefficient plus one. 

In common cases the limit is much less. 

From this, we at once decide that the greatest positive root of 
the equation x^ — ^x'^-\-lx^ — 8^:^ — 9x — 12=0, is less than 13. 

Now change the second, and every alternate sign, and we 
have the equation 

a^5+3x^+7x-'^+*8x2~9;r+12=0. 

The greatest positive root, in this equation, is less than 10 ; but, 
by (Art. 178.), the greatest positive root of this equation is the 
greatest negative root of. the preceding equation ; therefore 10 is 
the greatest limit of the negative roots of the first equation ; and 
all its roots must be comprised between +13 and — 10 ; but as 
this equation does not present an extreme case, tlie coeflicients 
after the first are not all minus, nor equal to each other ; there- 
fore the rea/ limits of its roots must be much witliin +13 and 
—10. In fact, the greatest positive root is between 3 and 4, and 
the greatest negative root less than 1. 

If it were desirable to find the limits of the least root, put 

0:=-, and transform the equation accordingly. Then find, as 

y 

just directed, the greatest limit of y, in its equation ; which will, 
of course, correspond to the least value of x in its equation. 

(Art. 181.) If we substitute any number less than the least 
root, for the unknown quantify, in any equation of an even 
degree, the result will be positive. And if the degree of the 
equation be odd, the result will be negative. 

Let a, b, c, &c., be roots of an equation, and x the unknown 



292 ELEMENTS OF ALGEBRA. 

quantity. Also, conceive a to be the least root, b the next 
greater, and so on. Then the equation will be represented by 
[x — a){x — b){x — c){x — d), &c., =0. 

Now in the place of x substitute any number h less than a, and 
the above factors will become 

(^h—a){h—b){h'--c){h-^d), &c. 

Each factor essentially negative, and the product of an even 
number of negative factors, is positive ; and the product of an 
odd number is negative ; therefore our proposition is proved. 

Scholium. — If we conceive h to increase continuously, until it 
becomes equal to a, the first factor will be zero ; and the product 
of them all, whether odd or even, will be zero, and the equa- 
tion will be zero, as it should be when A becomes a root. 

If h increases and becomes greater than a, without being 
equal to b,the result of substituting it for x will be negative, 
in an equation of an even degree, and positive in an equation 
of an odd degree. 

For in that case the first factor will be positive, and all the 
other factors negative ; and, of course, the signs of their product 
will be alternately minus and plus, according as an even or odd 
number of them are taken. 

If h is conceived to increase until it is equal to b, then the 
second factor is zero, and its substitution for x will verify the 
equation. If h becomes greater than b, and not equal to c, then 
the first two factors will be positive ; the rest negative ; and the 
result of substituting h for x will give a positive or negative 
result, according as the degree of the equation is even or odd. 

If we conceive h to become greater than the greatest root, 
then all the factors will be positive, and, of course, their product 
positive. 

For example, let us form an equation with the four roots ■ — 5, 
2, 6, 8, and then the equation will be 

(a;+ 5) (a^— 2) (a:— 6) (;r~8) =0, 
Or. . . . .r*— lla;=^— -4a?2+284a:--480=0. 
(The greater a negative number is, the less it is considered.) 



GENERAL PROPERTIES OF EQUATIONS. 293 

Now if we substitute — 6 for x^ in the equation, the result must 
be positive. Let — 6 increase to — 5, and the result will be 0. 
Let it still increase, and the result will be negative, until it has 
increased to +2, at which point the result will again be 0. 

If we substitute a number greater than 2, and less than 6, for 
X, in the equation, the result will again be positive. A number 
between 6 and 8, put for x, will render the equation negative ; 
and a number more than 8 will render the equation positive ; 
and if the number is still conceived to increase, there will be no 
more change of signs, beccmse we have passed all the roots. 

If in any equation we substitute numbers for the unknown 
quantity, which differ from each other by a less number than 
the difference between any two roots, and commence with a 
number less than the least root, and continue to a number greater 
than the greatest root, we shall have as many changes of signs 
in the results of the substitution as the equation has real roots. 

If one real root lies between two numbers substituted for the 
unknown quantity, in any equation, the results will necessarily 
show a change of signs. 

If one, or three, or any odd number of roots, lie between the 
two numbers substituted, the results will show a change of signs. 

If an even number of roots lie between the two numbers sub- 
stituted, the results will show no change of signs. 

In the last equation, if we substitute — 6 for x, the result 
will be plus. 

If we substitute -|-3, the result will also be plus, and give no 
indication of the two roots — 5 and +2, which lie between. 

(Art. 182.) If an equation contains imaginary roots, the 
factors pertaining to such roots will be either in the form of 
(a?2+a), or in the form of [(a? — af^b^'], both positive, whatever 
numbers may be substituted for x, either positive or negative ; 
hence, if no other than imaginary roots enter the equation, all 
substitutions for x will give positive results, and of course, no 
changes of sign. It is only when the substitutions for x pass 
real roots that we shall find a change of signs. 
z2 



294 ELEMENTS OF ALGEBRA. 

CHAPTER IV. 
GENERAL PROPERTIES OF EQUATIONS—Continued. 

(Art. 183.) If we take any equation which has all its roots 
real and unequal, and make an equation of its first derived poly- 
nomial, tlie least root of this derived equation will be greater than 
the least root of the primitive equation, and less than the next 
greater. 

If the primitive equation have equal roots, the same root will 
verify the derived equation.* 

We shall form our equations from known positive roots. 

Let tz, b, c, dj &,c.., represent roots ; and suppose a less than 
h, b less than c, &:c., and x the unknown quantity. An equation 
of the second degree is 

x^—[a-^b)x-\-ab={)'. 
Its first derived polynomial is 

2x—'[a-\-b). 

If we make an equation of this, that is, put it equal to 0, we 
shall have 

- a^b 

Now if b is greater than «, the value of x is more than a, 
and less than 6, and proves our proposition for all equations of 
the second degree. If we suppose «=Z>, then x==a^ in both 
equations. 

An equation of the third degree is 

x'-^{a-{-b-{-c)x'-{-[ab-\-ac-]-bc)x—abc=0> .... (I) 

Its first derived polynomial is 

^x'-^2[a-\-b-\-c)x-{-ab-]-ac-\-bc=Q (2) 

This equation, being of the second degree, has two roots, and 
only two. 

• To ensure perspicuity and avoid too abstruse generality, we operate on 
Cijuations definite in degree ; the result will be equally satisfactory to the 
learner, and occupy, comparatively, but little space. 



GENERAL PROPERTIES OF EQUATIONS. 295 

Now if we can find a quantity which, put for x, will verify 
equation (2), that quantity must be one of its roots. If we try 
two quantities, and find a change of signs in the results, we are 
sure a root lies between such quantities. (Art. 181.) Therefore 
we will try a, or write a in place of x» As h and c are each 
greater than a, we will suppose that 

c=a-{-h'. 
With these substitutions, equation (2) becomes 

3a^^2{da-\-h-\-h')a-{-Sa'-^2ah-\-2ah'-\-hh'^0; 

Reduced, gives -{-hh'=0. 

Therefore a cannot be a root ; if it were we should have 0=0. 
If we now make a substitution of b for a*, or rather {a-]-h) 
for Xy and reduce the equation, we shall find 

It is apparent that this quantity is essentially minus, as h' 
is greater than h. Hence, as substituting a for x, in the equa- 
tion, gives a small plus quantity, and b for x gives a small 
minus quantity, therefore one value of x, to verify equation (2), 
must lie between a and b. 

This proves the proposition for equations of the third degree : 
and in this manner we may prove it for any degree ; but the labor 
of substituting for a high equation would be very tedious. 

If we suppose «=&, and put c=a-^h\ and then substitute a 
in place of a?, we shall find equations (1) and (2) will be verified. 

Therefore in the case of equal roots, the equation and its 
first derived polynomial will have a common measure, as before 
shown in (Art. 168). 

If all the roots of an equation are equal, the equation itself 
may be expressed in the form of 

(x— «)™=0. 
Its first derived polynomial, put into an equation, will be 
m[x — «)'«-^=0. 

It is apparent that the primitive equation has m roots equal to 
a; and the derived equation, (m — 1), roots also equal to a. 



296 ELEMENTS OF ALGEBRA. 

Lastly, take a general equation, as 

x'^-^Ax-^-'-^-Bx'^-'^ i?:r-l-/S'=0. 

Its first derived polynomial, taken for an equation, will be 

mx-^-^-\-{m—\)£x'^~'- 72=0. 

We may suppose this general equation composed of the 
factors 

{x — d){x — h){x — c), &c., =0, 

and also suppose h greater, hut insensibly greater, than a ; c 
insensibly greater than 6, &c. Then the equation will be nearly 

and its derived polynomial, 

m{x — a)'«-'=0, 
cannot have a root less than (a), the least root of the primitive 
equation ; but its root cannot equal «, unless the primitive equa- 
tion have equal roots ; therefore it must be greater. 

By the same mode of reasoning we can show that the greatest 
root of an equation is greater than the greatest root of its derived 
equation ; hence the roots of the derived equation are interme- 
diate, in value, to the roots of the primitive equation, or contained 
within narrower limits. 

(Art. 184.) If we take any equation, not having equal roots, 
and consider its first derived polynomial also an equation, 
and then substitute any quantity less than the least root of 
either equation, for the unknown quantity, the result of such 
substitution ivill necessarily give opposite signs. 

Let a, b, c, &c., represent the roots of a primitive equation, 
and a', b', &c., roots of its first derived polynomial ; x the un- 
known quantity. Then the equation will be 

{x — a)(x — b){x — c), &LQ., to m factors =0 ; 
the derived equation will be 

{x — a'){x — b'){x — c'), &c., to (m — 1) factors =0. 

Now if we substitute h for x, and suppose h less than either 
root, then every factor, in both equations, will be negative. 

The product of an even number of negative factors is positive, 
and the product of an odd number is negative ; and if the factors, 



GENERAL THEORY OF EQUATIONS. 297 

in the primitive equation are even, those in the derived equation 
must be odd. 

Hence any quantity less than any root of either equation, will 
necessarily give to these functions opposite signs. 

(Art. 185.) Now if we conceive h to increase until it becomes 
equal to a, the least root, the factor (a^— -«) will be 0, and reduce 
the whole equation to 0. Let h still increase and become 
greater than a, and not equal to a\ (which is necessarily greater 
than a, (Art. 184.), and the factor {x — a) will become plus, while 
all the other factors, in both equations, will be minus, and, of 
course, leave the same number of minus factors in both functions, 
which must give them the same sign. Consequently, in passing 
the least root of the primitive equation a variation is changed into 
a permanence. 

Sturm's Theorem. 

(Art. 186.) If we take any equation not having equal roots, and its 
first derived polynomial, and operate with these functions as tliough 
their common measure was desired, reserving the several remain- 
ders with their signs changed, and make equations of these func- 
tions, namely, the priraiUve equation, its first derived polynomial 
and the several remainders with their signs changed, and then 
substitute any assumed quantity, h, for x, in the several functions, 
noting the variation of signs in the result ; afterwards substitute 
another quantity, A', for x, and agam note the variation of signs ; 
the difference in the number of variation of signs, resulting 
from the two substitutions, will give the number of real roots 
between the limits h and h'. 

If * — QQ and -fOfO are taken for h and h', we shall have the 
whole number of real roots ; which number, subtracted from the 
degree of the equation, will give the number of imaginary roots. 

DEMONSTRATION. 
Let X represent an equation, and X' its first derived polyno- 
mial. 

In operating as for common measure, denote the several quo- 

♦ Symbols of infinity. 



298 ELEMENTS OF ALGEBRA. 

tients by Q, Q', Q", &c., and the several remainders, with their 
signs changed, by R, R', R", &c. 

In these operations, be careful not to strike out or introduce 
minus factors, as they change the signs of the terms ; then a 
re-change of signs in the remainder would be erroneous. 

From the manner of deriving these functions, we must have 
the following equations : 

X' ^ X' 

R ^ R 
R_ _R;; 

R' ^ R" 

&c. &c. 
Clearing these equations of fractions, we have 

X =X' Q — R 

X'=RQ' >— R' 

R =R'Q" — R" I, ... (A) 

R'=R"Q"— R" 

As the equation X=0 must have no equal roots, the functions 
X and X' can have no common measure (Art. 168.), and we shall 
arrive at a final remainder, independent of the unknown quantity, 
and not zero. 

Proposition 1. No two consecutive functions, in equations (A), 
can become zero at the same time. 

For, if possible, let such a value of h be substituted for x, as 
to render both X' and R zero at the same time ; then the second 
equation of (A) will give R'=0. Tracing the equations, we must 
finally have the last remainder R^ =0 ; but this is inadmis- 
sible ; therefore the proposition is proved. 

Prop. 2. IVIien one of the functions becomes zero, by giving 



GENERAL PROPERTIES OF EQUATIONS. 299 

a particular value to x, the adjacent functions between which it 
is placed must have opposite signs. 

Suppose R' in the third equation, (A), to become (I, then the 
equation still existing, we must have Il--= — R". 

The truth of Sturm's Theorem rests on the facts demonstrated 
in Arts. 184, 185, and in the two foregoing propositions. 

If we put the functions X, X', R, R', &;c., each equal to , 
that is, make equations of them, and afterwards substitute any 
quantity for x, in these functions, less than any root, the first 
and second functions, X and X', will have opposite signs, (Art. 
184.) ; and the last function will have a sign independent of a?, 
and, of course, invariable for all changes in that quantity. The 
other functions may have either plus or minus, and the signs 
have a certain number of variations. 

Now all changes in the number of these variations must 
come through the variations of the signs in the primitive func- 
tion X. A change of sign in any other function will produce 
no change in the number of variations in the series. 

For, conceive the followitig equations to exist: 



(B) 



Now take x—h, yet h really less than any root of the equa- 
tions, (B), and we may have the following series of signs : 




X 


=-. 


— " 


X' 


= 


+ 


R 


= 


— 


R' 


= 


— 


R" 


= 


— 


R" 


= 


+ , 



(C) 



Or we mav have any other order of signs, restricted only to the 
fact that the signs of the two first functions must be opposite, and 
the last mvariable, or unaffected by all future substitutions. 
Here are three variations of signs. 



300 ELEMENTS OF ALGEBRA. 

Now conceive h to increase. No change of signs can take 
place in any of the equations, unless h becomes equal and passes 
a root of that equation ; and as there are no equal roots, no two 
of these functions can become at the same time (Prop. 1.) ; 
hence a change of sign of one function does not permit a change 
in another ; therefore by the increase of h, one of the functions, 
(C), will become 0, and a further increase of h will change its 
sign. 

In the series of signs as here represented, X' cannot be the 
first to change signs, for that would leave the adjacent functions, 
X and R, of the same sign, contrary to Prop. 2 ; nor can the 
function R' be the first to change sign, for the same reason. 

Hence X or R or R" must be first to change sign. 

If we suppose X to change sign, the other signs remaining the 
same, the number of variations of signs is reduced by unity. 

If R or R" change sign, the number of variations cannot be 
changed ; a permanence may be made or reduced, and all cases 
that can happen with three consecutive functions may be ex- 
pressed by the following combinations of signs ; 

, + ± - ) 
Or — ± + ^ 

either of which gives one variation and one permanence. 

Now as no increase or decrease in the number of variations of 
signs can be produced by any of the functions changing signs, 
except the first, and as 'that changes as many times as it has real 
roots, therefore the changes in the number of variations of signs 
show the number of real roots comprised between /* and h'. 

If h and h' are taken at once at the widest limits of possibility, 
from — infinity to + infinity, the number of variations of signs 
will indicate the number of real roots; — and this number, 
taken from the degree of the equation, will give the number of 
the imaginary roots. 

(Art. 187.) The foregoing is a full theoretical demonstration 
of the theorem ; but the subject itself being a little abstruse^ 
some minds may require the following practical elucidation. 



GENERAL PROPERTIES OF EQUATIONS. 301 

Form an equation with the four assumed roots, 1, 3, 4, 6. 
The equation will be 

(ic— l)(a?— 3)(a:— 4)(a:— 6)=0 ; 
orX = x^—l4c(^-\-Q7x'—l26x-\-72=0 Roots 1, 3, 4, 6. 
X'= 4x^ — 42ic2-l-I34a:— 126 =0- . i?oo^s 2, 3.3, 5 nearly. 

R =13a^— 91a;+153 Boots 2.8, 4.1 nearly. 

R'=:70a;— 252 • • • • Hoot 3.6 

R"= + 

Let the pupil observe these functions, and their roots, and see 
that they correspond with theory. The least root of X is less 
than the least root of X'. (Art. 183.) The roots of any func- 
tion are intermediate between the roots of the adjacent functions. 
This corresponds with (Prop. 2.) ; for if three consecutive func- 
tions have the same sign as — , — , — , or -}-, +» "|-» the middle 
one cannot change first and correspond to (Prop. 2.) ; but signs 
change only by the increasing quantity passing a root, and it must 
pass a root of one of the extreme functions first ; therefore the roots 
of X' must be intermediate in value between the roots of X and 
R ; and the roots of R intermediate in value between the roots 
of X' and R' ; and so on. But the roots of X' are within nar- 
rower limits than the roots of X (Art. 183.) ; therefore the roots 
of all the functions are within the limits of the roots of X. 

We will now trace all the changes of signs in passing all the 
roots of all the functions. 

We will first suppose x or A = ; which is less than any 
root ; then as we increase h above any root, we must change the 
sign of that function, and that sign only. 

We represent these changes thus : 
2A 



302 




ELEMENTS OF ALGEBRA. 










X 


X' 


R 


R' 


R'' 




'^When 


x=0 


+ 


— 


+ 


— 


+ . . 


• • 4 variations 


t( 


a;=l.l 


— * 


— 


+ 


— 


+ . « 


.3 




(( 


a:=2.1 


— 


+* 


+ 


— 


+ . . . 


.3 




(( 


a:=2.9 


— 


+ 


* 


— 


+ •• • 


.3 




n 


a;=3.1 


+* 


+ 


— 


— 


+ • • 


.2 




(( 


x=SA 


+ 


^ 


— 


— 


+ • . 


.2 




(( 


cr=3.7 


+ 


— 


— ■ 


+* 


+ . • 


.2 




(( 


a;=4.1 


* 


— 


-— 


+ 


+ . • 


.1 




(( 


x=iAl 


— 


— 


+* 


+ 


+ . . . 


. 1 




a 


a;=5.1 


— - 


+* 


+ 


+ 


+ •• 


.1 




a 


a;=6.1 


+* 


+ 


4- 


+ 


H-. . . 


.0 





We commenced with 4 variations of signs, and end with 
variations, after we have passed all the roots ; therefore the real 
roots, in the primitive equation, must be 4 — 0=4. 

By this it can be clearly seen, by inspection, that the changes 
of sign in all the functions, except the first, produce no change 
in the number of variations. 

In making use of this theorem we do not go through the inter- 
mediate steps, unless we wish to learn the locality of the roots as 
well as their number. We may discover their number by sub- 
stituting a number for x less than any root, and then one greater ; 
the diflference of the variations of signs will be equal to the num- 
ber of real roots. 

If we take — qo and +QC, the sign of any whole function 
will be the same as that of its first term. 



* In making this table, we did not really substitute the numbers assumed for 
X, as we previously determined the roots ; and as passing any root changes the 
sign in that function, we write a star against that sign which has just 
changed. 



APPLICATION OF STUKM'S THEOREM. 303 

CHAPTER V. 

APPLICATION OF STURM'S THEOREM. 

(Art. 188.) In preparing the functions, remember that we are 
at liberty to suppress positive numeral factors. 

EXAMPLES. 

1. How many real roots has the equation x^-\-9x=^Q 1 

Here X = a^+Qx—Q 

R=-- x+l 

R'=— 
Now for X substitute — co or — 100000, and we see at once 
^^^' - X X' R R' 

— + -T — 2 variations. 
Again, for x put +co or -f- 100000, and the resulthig signs 

must be , , ... 

-f- + — — 1 variation. 

Hence the above equation has but one real root; and, of course, 

two imaginary roots. 

To find a near locality of this root, suppose x=0, and the 

signs will be , « • .- 

° — -f- — ' — 2 variations. 

x=l + + — — 1 variation. 

Hence the real root is between and 1 

Now as we have found x, in the equation x^-\-9x — 6=0, to 

be less than 1, x^ may be disregarded, and 9x — 6=0, will give 

us the first approximate value of x; that is, x=.6, nearly. 

2. How many real roots has the equation x^ — 3^;^ — 4=0 ? 

X= a;4_3^___4 

X'=^4x^'-^Qx 
R =-f 25 
If x= — CO -f- •— +2 variations. 

x=-\-o^ -{- + +0 variation. 

Hence there must be 2 real roots, and 2 imaginary roots. 



304 ELEMENTS OF ALGEBRA. 

3. How many real roots has the equation x^ — 4x^ — 621=0 ? 
(See Art. 103.) 

X^x^--4x^'-GU 

X'=x^—2oi^ 

R=+625 
When x= — co + — +2 variations. 

When x=-\-co -{- + +0 variation. 

Hence there are two real roots and four imaginary roots. 

4. How many real roots has the equation x^ — 15a?+21=0? 

^ns. 3. 

5. How many real roots are contained in the equation 

a^—5x^-hSx^l=01 Am, 1. 

6. How many real roots are containtjd in the equation 







2a^^--13a?24- 10:ir— 19=0 ? 




Ans, 2. 


T. 


Find the number and situation of the roots of the equation 
a,3^11a^_102a?+181=0. 






X= a?3_|_ii^__io2x+181 
X'=3a:2_^22^— 102 
R= 122a:— 393 

R=+ 






Putting 


a?=— CO — -f — 

a;=-l-co + + -}- 




3 variations. 
variation. 



Hence all the roots are real. 

To obtain the locality of these roots there are several principles 
to guide us; there is (Art. 180), but the real limits are much 
narrower than that article would indicate, unless all the coeffi- 
cients after the first are minus, and equal to the greatest. 

A practised eye will decide nearly the value of a positive root 
by inspection; but by (Art. 183.) we learn that the root of R, 
or 122a? — 393=0, must give a value to x intermediate between 
the roots of the primitive equation. 

From this we should conclude at once that there must be a 
root between 3 and 4. 



# . 



NEWTON'S METHOD OF APPROXIMATION. 305 

Substituting 3 for x^ in the above functions, we have 
+ — — H- 2 variations. 
ir=4 •\- 4" -f- + variation. 
Hence there are two roots between 3 and 4. 

As the sum of the roots must be — 11, and the two positive 
roots are more than 6, there must be a root near — 17. 

As there are two roots between 3 and 4, we will transform 
the equation, (Art. 175.), into another, whose roots shall be 3 less; 
or put a:=3-|-t/. Then we shall have 

X= 2/3+202/2—93/4-1=0 
X'= 3?/2+40?/— 9 
R =1221/ —27 

R'=-}- 

The value of y, in this transformed equation, must be near 
the value of y in the equation 122i/=27, (Art. 183.) ; that is, y 
is between .2 and .3 

?/=.2 gives + — — + 2 variations. 
?/=.3 gives H- 4- + + variation. 
Hence there are two values of y between .2 and .3 ; and, of 
course, two values of x between 3.2 and 3.3. 

We may now transform this last equation into another whose 
roots shall be .2 less, and further approximate to the true values 
of a?, in the original equation. 

Having thus explained the foregoing principles, and, in our 
view, been sufficiently elaborate in theory, we shall now apply it 
to tlie solution of equations, commencing with 

NEWTON'S METHOD OF APPROXIMATION. 

(Art. 189.) We have seen, in (Art. 175.), that if we have any 
equation involving x^ and put a:=«+2/, and with this value 
transform the equation into another involving y, the equation 
will be 

XH-X't/+~>+|^y» r=0- 

If a is the real value of x then i/=0, and X=0. 
26 



306 ELEMENTS OF ALGEBRA. 

If a is a very near value to x, and consequently y very 
small, the terms containing y'^, y^, and all the higher powers of 
y, become very small, and may be neglected in finding the ap- 
proximate- value of y. 

Neglecting these terms, we have 

X+X'y=0, 

Or y=_J, (1) 

In the equation x=a-\-y^ if a is less than a:, y must be posi- 
tive ; and if y is positive in the last equation, X and X' must 
have opposite signs, corresponding to (Art. 184.). 

Following formula (1), we have an approximate value of y ; 
and, of course, of x. The value of x, thus corrected, again call 
«, and find a correction as before ; and thus approximate to any 
required degree of exactness. 

EXAMPLES. 

1. Given 3a;^+4a:^ — 5a: — 140=0, to find one of the approxi- 
mate values of x. 

By trial we find that x must be a little more than 2. 
Therefore, put 0^=2+1/. 

X= 3(2)5-1- 4(2)=^— 5(2)— 140 . • . or . . . X =— 22 
X'=15(2)*-|-12(2)2— 5 . .or. . . X'= 283. 

X 22 

Whence y= — ^,=^^^=0.07 nearly. 
X 283 

For a second operation, we have 

.r=2.07+y. 

X= 3(2.07)5-f 4(2.07)3— 5(2.07)--l40... By log. X=— 0.854 

X'=l5(2.07)^-}-12(2.07)2— 5 By log. X'= 321.82 

Hence the second value, or 2/=— ^—— =0.00265-}- 

3218.2 

And a:=2.07265-|- 

2. Given x^-\-2x^ — 23iC=70, to find an approximate value 
of X. Ans. 5.1345-i-. 



HORNER'S METHOD OF APPROXIMATION. 307 

3. Given x^ — 3x^4- 75a: =10000, to find an approximate 
value of X, Ans. 9.886-[-. 

4. Given 3a;* — 35a^ — \\x^ — 14a?4-30=0, to find an approxi- 
mate value of X. £ns. 11.998+. 

5. Given 5x^ — 3x^ — 2a;=1560, to find an approximate value 
of X, Ans. 7.00867-f. 

CHAPTER VI. 
HORNER'S METHOD OF APPROXIMATION. 

(Art. 190.) In the year 1819, Mr. W. G. Horner, of Bath, 
England, published to the world the most elegant and concise 
method of approximating to roots of any yet known. 

The parallel between Newton's and Horner's method, is 
this ; both methods commence by finding, by trial, a near value 
to a root. 

In using Horner's method, care must be taken that the number, 
found by trial, be less than the real root. Following Newton's 
method we need not be particular in this respect. 

In both methods we transform the original equation involving 
a?, into another involving y, by putting a;=r+t/, as in (Art. 175), 
r being a rough approximate value of x, found by trial. 

The transformed equation enables us to find an approximate 
value of y, (Art. 189.). 

Newton's method puts this approximate value of y to r, and 
uses their algebraic sum as r was used in the first place ; again 
and again transforming the same equation, after each successive 
correction of r. 

Horner's method transforms the transformed equation into 
another whose roots are less by the approximate value of y ; and 
again transforms that equation into another whose roots are less, 
and so on, as far as desired. 

By continuing similar notation through the several transforma- 
tions we may have 



808 ELEMENTS OF ALGEBRA. 

X =r-{-y 
y=s-\-z 
z = t-\-z' 

Z'=^U-\'Z" 

&c. &c. 

Hence a:=r-|-5-f-/, &;c. ; r, 5, /, &c., being successive figures 
of the root. Thus if a root be 325, r=300, s=20, and ^=5. 

On the principle of successive transformations is founded the 
following 

Rule for approximating to the true value of a real root 
of an equation. 

1st. Find by Sturm^s Theorem^ or otherwise, the value of 
the first one or two figures of the root, which designate by r. 

2d. Transform the equation (Art. 175), into another whose 
roots shall be less by r. 

3d. JVith the absolute term of this transformed equation for a 
dividends and the coefficient of y for a divisor, find the next 
figure of the root. 

4th. Transform the last equation into another, whose roots 
shall be less, by the value of the last figure determined; and 
so proceed until the whole root is determined, or sufficiently 
approximated to, if incommensurable. 

Note 1, In any transformed equation, X is a general symbol 
to represent the absolute term, and X' represents the coefficient 
of the first power of the unknown quantity. If X and X' be- 
come of the same sign, the last root figure is not the true one, 
and must be diminished. 

Note 2. To find negative roots, change the sign of every alter- 
nate term, (Art. 178.) : find the positive roots of that equation, 
and change their signs. 

(Art. 191.) We shall apply this principle, at first, to the solu- 
tion of equations of the second degree ; and for such equations 
as have large coefficients and incommensurable roots, it will fur- 
nish by far the best practical rule. 



HORNER'S METHOD OF APPROXIMATION. 399 

EXAMPLES. 
1. Find an approximate root of the equation 

x2^x — 60=0. 
"VVe readily perceive that x must be more than 7, and less 
than 8, therefore r=7. 

Now transform this equation into another whose roots shall be 
less by 7. 

Operate as in (Art. 175.), synthetic division 

1 1 .-_60 (7 

' 7 56 



8 
7 

15 



Trans., eq. y^ -}- Iby — 4=0 

Here we find that y cannot be far from y"!-, or between .2 and 
.3 ; therefore transform the last equation into another whose 
roots shall be .2 less ; thus, 



1 15 




-4 (.2 


0.2 




3.04 


15.2 


—0.96 


2 






15.4 






second transformed equation, 


therefore, is 


z2+I5.4z- 


-0.96=0 


.1.. ... • i_ 1 


-I" . 


.... ..... -96 



To obtain an approximate value of z, we have -— — or 0.06. 

In being thus formal, we spread the work over too large a 
space, and must inevitably become tedious. To avoid these diffi- 
culties, we must make a few practical modifications. 

1st. We will consider the absolute term as constituting' the 
second member of the equation ; and, in place of taking the 
algebraic sum of it, and the number placed under it, we will take 
their difference. 



310 



ELEMENTS OF ALGEBRA. 



2(1. We will not write out the transformed equations ; that is, 
not attach the letters to the coefficients ; we can then unite the 
whole in one operation. 

3d. Consider the root a quotient ; the absolute term a dividend, 
and, corresponding with these terms, we must have divisors. 

In the example under consideration, 8 is the first divisor ; 15 
is the^rs^ fria/ divisor ; 15.2 is the second divisor, and 15.4 is 
the second trial divisor; 15.46 is the third divisor, &c. 

Let us now generalize the operation. The equation may be 



represented by 



x^-{-ax^=n 



Transform this into another whose roots shall be less by r ; that 
equation into another whose roots shall be less by s, &c., &c. 



1 


SYNTHETIC 

a 3 
r a 

. . a-\~r 
r 


DIVISION. 

n, ( r+s 


1st divisor, . . 


71' 


1st trial divisor, 


. .«+2r 


n" 
&c. 


2d divisor, . . 


. .a-\-2r-\-s 




2d trial divisor, 


. .«-4-2r4-2s 
&c. 





In the above we have represented the difference between n 
and [ar-\-r^) by ?i', &c. As n', n", n"\ &c., with their corre- 
sponding trial divisors, will give 5, ^, w, &c., the following for- 
mula will represent the complete divisors for the solution of all 
equations in the form of 



HORNER'S METHOD OE APPROXIMATION 



311 



1st divisor, . 

add . 

2d divisor, . 

add . 

3d divisor, . 

add . 

4th divisor, . 
&c. 



zt:a-\- r 

r-f- s 

±a-\-2r-\- s 



'-1- t 



±a-f2rH-2s-f- t 

t+u 

dza+2ri-28-\-2t-\-u 



Equations which have expressed coefficients of the highest 
power, as 

the formulas will be : 
1st divisor, 
add 



2d divisor, 
add 

3d divisor, 
&c. 



rta-j- cr 

cr-{- cs 

±a+2cr-l- cs 

cs-\-ct 

=t:«+2cr+2c5+cf 
&c. 



To obtain trial divisors we would add cr only, in place of 
(cr-\-cs)i Sic. 

We will now resume our equation for a more concise solution. 



I. 


x^ 


-x=6Q 








n 


r stu 




1 


60 


( 7.262 




7 


56 




1st divisor, . 


. 8 


4 




add . 


7.2 


304 


2d divisor, . 


15.2 




96 


add . • 


26 




9276 


3d divisor, . 


. 15.46 


324 


add . 


'62 




31044 


4th divisor, . 


. 15.522 


1356 



312 ELEMENTS OF ALGEBRA. 

We can now divide as in simple division, and annex the quo- 
tient-figures to the root, thus: 

, _ 15522)1356 (08734 
124176 



11424 

10865 



559 
465 



94 
Hence ar=7.2620873+. 

2. Find x, from the equation x^ — 700a;=59829. 
On trial, we find x cannot exceed 800 ; therefore, r=700. 

n rsi 
fl-h r —700-1-700= 59829(777 

rH- s 700-1- 70=770 00)000 

fl-l-2r-i- s =770 770)59829=n' 

s-{-f=70-f7 77 5390 



a-|-2r-|-2s-l-^ =847 847)5929=n" 

5929 



Hence, a;=777. 

3. Find x from the equation x^ — 1283a?= 16848. 
By trial, we find that x must be more than 1000, and less 
than 2000; therefore, r=1000. 



a-1- r= —283 
r-\- s= 1200 


n rsiu 
16848(1296 
—283 


a-f2r-l- s= 917 
si-t 290 


917)2998 
1834 


a-{-2r-^2s-]-t= 1207 
^-1-7^=96 


1207)11644 
10863 


1303 


1303) 7818 
• 7818 



Hence, a:=1296. 



313 



HORNER'S METHOD OF APPROXIMATION. 
4. Given x^ — 5j?=8366, to find x. 

By trial, we find x must be more than 90, and less than 100. 
Therefore, a-\- r = 85 ) 8366 ( 94=a; ■ 
r-\-s . . 94 765 



a-l-2r-|-s=179 ) 716 
716 

5. Find x, from the equation o.'^ — 375a?+ 1904=0. 
Here the first figure of the root is 5. 





5 
—375 


—1904(5.1480052207 
—1850 


1st divisor. 


—370. 
5.1 


—5400 
—3649 


2d divisor. 


—364.9 
.14 


—175100 
—145903 


3d divisor, 


—364.76 

48 


—2919600 
—2917696 


4th divisor. 


—364.712 
8 


—190400 
— 1823519 


5th divisor. 


—364.7039 


—80491 
—72941 




—7550 
—7294 




—256 
—255 



—1 

6. Given x''-\-lx—UM^O, to find x. Ans. 31.2311099. 

7. Given a.^— 21a?=214591760730, to find x. Jins. 463251. 

It might be difficult for the pupil to decide the value of r, as 
applied to the last example, without a word of explanation : x 
must be more than the square root of the absolute term, that is, 
27 



314 ELEMENTS OF ALGEBRA. 

more than 400000 ; then try 500000, which will be found too 
great. 

(Art. 192.) When the coefficient of the highest power is not 
unity, we may (if we prefer it to using the last formulas for di- 
Tisors), transform the equation into another, (Art. 166.), which 
shall have unity for the coefficient of the first term, and all the 
other coefficients whole numbers. 

8. Given 7x^—'Sx=37o. 

Put x=- and we shall have y^ — 3i/=2625. 

One root of this equation is found to be 52.7567068-|-> one- 
seventh of which is 7.536672-|- ; the approximate root of the 
original equation. -^ 

9. Given 7^2— 83ic+187=0, to find one value of a:. 

^ns. 3.024492664 

10. Given x^ — y\a?=8, to find one value of x. 

Ans. 2.96807600231 

11. Given 4a?^+Ja7=j, to find one value of x. 

Ans. Ans. .14660+ 

12. Given \x^-\-%x=jj, to find one value of x. 

Ans. .6042334631 

13. Given 115 — 3a?^ — 7x=0, to find one value of x. 

Ans. 5.13368606 

(Art. 193.) We now apply the same principle of transforma- 
tion to the solution of equations of the third degree. 

EXAMPLES. 
1. Find one root of the equation 

a:3—a;.2_j_70^— 300=0. 
We find, by trial, that one root must be between 3 and 4. 



HORNER'S METHOD OF APPROXIMATION. 315 





'S 

(a 

s 




i 


4th Coefficient. 




1 


—1 
3 

2 
3 

3 


70 

**76 

I5J 

91 =X' 


—300 (3. 

228 


3 

3 

CO 

i 

ST. 



— 72=X» 


0- 


1 


8 
0.7 

8.7 

7 


91 
6.09-| 

* *97.09 
6.58J 


s 
— 72 (0.7 
67.963 


i 


— 4.037=X 


3 

so 




9.4 

7 

10.1 


103.67=X' 


, ■-' 










t 




1 


10.1 
.03 


103.67 
.3039-^ 

*103.9739 , 
.3048J 

104.2787=X' 


— 4.037 (0.03 

— 3.119217 




10.13 
3 


— 0.917783=X 




10.16 
3 

laTo 






1 


10.19 

.008 

10.198 

8 


104.2787 

.081584 


0.917783(0.008 
0.8.34882272 




*104.360284 
81648 


.082900728 




10.206 

8 


104.441932 





10.214 



316 ELEMENTS OF ALGEBRA. 

The terms here marked X' are trial divisors ; we have pre- 
fixed stars to the numbers that we may call complete divisors. 
We rest here with the equation 

(2"f-l-10.214(z'74-104.4419z"— 0.0829=0. 

The value of z" is so small that we may neglect all its powers, 

except the first, and obtain several figures by division, thus : 

104 ) 829 ( 797 

728 



101 
936 

"74 

r stu 
Hence, one approximate value of a? is .... 3.738797-1- • 
(Art. 193.) We may make the same remarks here as in (Art. 
191.), and, as in that article, generalize the operation. 

Let x^-\-Aaf-]rBx=N represent any equation of the third 
degree, and transform it into another whose roots shall be r less ; 
thus, 

1 



A 

r 

r-^A 

r 


B 

{r-^A)r 

%r-hA)r-{-B 

{2r-^A)r 
'Sr'4-2Ar-\-B 


= N ( r 
r'-{-Ar'-\-Br 

N' 


2r-i-A 
r 





3r-}-^ 
The transformed equation is 

2/'+(3r-l-^)y2_^(3r^-i-2^r4-^)i/=iV'. 
If we put (3r-l-^)=:^', {^i^-\-2Ar-^B)=B\ 

and JSr^r^—Ar'^—Br^^', 

we shall have . . . y^-\-A'y^-\-B'y—N', an equation similar 
to the primitive equation. 

If we transform this equation into another whose roots shall be 
less by «, we shall have 

z'-{-{Ss-{-A')z^'\-{Ss^+2A's-i'B')z=N'\ 
Or, . . z^-\-A"z^'{'B"z=N" ; an equation also similar to 



HORNER'S METHOD OF APPROXIMATION. 317 

the jfirst equation. And thus we may go on forming equation 
after equation, similar to the first, whose roots are less and less. 

The quantities N', N", &c., are the same as X in our previous 
notation, and the quantities i?', B", &c., are the same as the 
general symbol, X' ; but we have adopted this last method of 
notation to preserve similarity. 

Observe, that as [r-^-Ay-^-B is the first complete divisor, N 
is the number considered as a dividend and r the quotient; and 

therefore . . . r= , nearly. 

The next equation gives us 

N> N' 



nearly. 



And the next, t= 



{8-\-A')s-\-B' s(5+3r-l-^)-f^' 

N" N" very nearly. 



{t-\-A")t-{-B" /[f+3(r-}-s)-i-^]-fJ?" 

_ N'" ^ N^ 

^ {u-\-A"')u-\-B"~u[u-{-^{r-\-s-\-t)-\-jr\-\-B"' 
&c. = &c. <fcc. 

The denominators of these fractions are considered complete 
divisors, and the quantities, B\ B", B'", are considered trial 
divisors. The further we advance in the operation the nearer 
will the trial and true divisors agree. 

Before the operation is considered as commenced, we must 
find the first figure of the root (r) by trial. Then the operator 
can experience no serious difliculty, provided he has in his mind 
a clear and distinct method of forming the divisors ; and these 
may be found by the following 

Rule. 1st. Write the number represented by B, and directly 
under it, write the value 0/* r(r+A) ; the algebraic sum of these 
two numbers is the first complete divisor. 

2d. Directly under the first divisor write the value of r^, and 
the sum of the last three numbers is B\ or the first trial divisor. 

3d. Find by trial, as in simple division, how often B' is con- 
tained in N', calling the first figure s, [making some allowance 
for the augmentation of B'), and s will be a portion of the 
root under trial. 

2b2 . 



318 ELEMENTS OF ALGEBRA. 

4th. Take, tilt value, of the expression (3v-\rs-\-A)s and add 
it to the first trial divisor ; the sum is the second divisor (if 
we have really the true value of s). 

In general terms; 
Under any complete divisor, write the square of the last figure oi 
the root ; add together the three last columns, and their sum is 
the next trial divisor. With this trial divisor decide the next 
figure of the root. 

Take the algebraic sum, of three times the root previously- 
found, the present figure under trial, and the coefficient ./?, and 
multiply this sum by the figure under trial, and this product, 
added to the last trial divisor, gives the next complete divisor. 

EXAMPLES. 
1. Given a;34-2a:24-3a?= 13089030, to find one value of x. 
By trial, we soon find that x must be more than 200 and less 
than 300 ; therefore we have 

r=200, ^=2, B=3, 
By the rule, 

N rs 

B 3 40403 ) 13089030 ( 235 

r{r^A) 404O0 -j 80806 

1st divisbr 40403 [ 139763 ) 500843 =N' 

r* 40000 J 419289 

1st trial divisor . . B' =120803 163108 ) 815540=iV^' 
{3r-{-s-\-Jl)s . . . 18960 1 815540 

2d divisor 139763 [ 

s^ 900J 

2d trial divisor . . i?"=l 59623 

(3r+3s-|-/-l-^)< . . 3485 

3d divisor 163108 

Hence a;=235. 



HORNER'S METHOD OP APPROXIMATION. 3 19 

2. Given ar^+173a;= 14760638046, to find one value of x. 
Here ./2=0, jB=173, and we find, by trial, that x must be 
more than 2000, and less than 3000 ; therefore r=2000. 

B 173 

rir-^-A) 4000000 >| 

1st divisor 4000173 i 

r^ 4OOOOOOJ 

B' 12000173 

{3r+s-\-^)s 2560000 ^ 

2d divisor 14560173 i" . 



160000 J 



B" 17280173 

(3r4-3s4-^-i-c^)^ . . 362500-1 

3d divisor 17642673 | 

250oJ 

B"' 18007673 

(3/?+w4-^)w .... 22059 

4th divisor 18029732 

4000173 ) 14760638046 ( 2453=a; 
8000346 

14560173 ) 67602920 =iV' 
58240692 

17642673 ) 93622284 =N" 
88213365 

18029732) 54089196=iV^" 
54089196 

3. Given a^-\-2x^ — 23a? =70, to find an approximate value 
ofir. ' Ans. a;=5.134578-i-. 

4. Given a^ — 17x^+423?= 185, to find an approximate value 
of X. Ans, a;=1502407+. 



320 ELEMENTS OF ALGEBRA. 

(Art. 194.) When the coefficient of the highest power is not 
unity, we may transform the equation into another, (Art. 166.), 
in which the coefficient of the first term is unity, and all the 
other coefficients whole numbers ; but it is more direct and con- 
cise to modify the rule to suit the case. 

If the coefficient of the first power is c, the first divisor will 
be {cr+A)r-\-B, in place of (r+^)r+i?. 

In place of (3r-f-s-|-.^)s, to correct the first trial divisor, we 
must have (3cr+cs+«/?)s ; and, in general, in place of using 3 
times the root already found, we must use 3c times the root ; and, 
in place of the square of any figure, as i^, s% &c., we must use 
cr^, cs^ &c. 

EXAMPLES. 
5. Find one root of the equation, Sx^-\-2x^-\-4x=75. 
By trial, we find that x must be more than 2, and less than 3 ; 
therefore ^^^^ ^^^ ^^2, B=4. 

N r stu 

B 4. 75 ( 2.577 

(cr4-^)r 16. -j 40 

1st divisor 20. [ 35=A^' 

cr" 12. J 29375 

B' 48. 5625=i\r" 

{^cr-\-cS'\'A)8 . .10.751 5038579 

2d divisor 58.75 f .586421=iV'" 

cs" 75 J .517301099 

B" 70.25 69119901 

*{^cR-{-ct-^A)t . 1.7297 1 

3d divisor 71.9797 j Continue, by simple division, thus : 

c^ __}^^ 739 ) 6911 ( 935 

B" 73.7241 6651 

{^cR-{-cu-\-Ji)u . 176057 "^ 

4th divisor 73.900157 ' 221 

Hence, 3^=2.577935+. 

• ii is a symbol to represent the entire root, as far as determined. 



HORNER'S METHOD OF APPROXIMATION. 321 

6. Find one root of the equation, ^x^ — 6a;^-|-3x= — 85. 

Ans. x= — 2.16399—. 

v. Find one root of the equation, \2x^-\-x^ — 5a^=330. 

^ns. a:=3.036475-l-. 

8. Find one root of the equation, ^x^-\-Qx'^ — 7x=2200. 

Ans. :r=7.10735364-. 

9. Find one root of the equation, Zx^ — 30?^ — 2a?=1560. 

Ans. a:=7.0086719-|-. 

(Art. 195.) This principle of resolving cubic equations may 
be applied to the extraction of the cube root of numbers, and 
indeed gives one of the best practical rules yet known. 

For instance, we may require the cube root of 100. This 
gives rise to the equation 

3(^-\-Jlx^-\-Bx=\m', 
in which .^=0, and B=0, and the value of x is the root 
sought. 

As A and B are each equal to zero, the rule under (Art. 193.) 
may be thus modified. 

1st. Keeping the symbols as in (Art. 193.), and finding r by 
trial, r^ will be the first divisor, and 3r^ is B', or the first trial 
divisor. 

2d. By means of the dividend {so called), and the first trial 
divisor, we decide s the next figure of the root. 

3d. Then (3r-\-s)s ; that is, three times the portion of the 
root already found, with the figure under trial annexed, and 
the sum multiplied by the figure under trial, will give a sum, 
which, if written two places to the right, under the last trial 
divisor, and added, will give the next complete divisor. 

4th. After we have made use of any complete divisor, write 
the square of the last quotient figure under it ; the sum of the 
three preceding columns is the next trial divisor ; which use, 
and render complete, as above directed, and so continue as far as 
necessary.* 

• III case of approximate roots after three or four divisors are found, we 
may find two or three more figures of the root, with accuracy, by simple division. 
22 



322 ELEMENTS OF ALGEBRA. 

We may now resolve the equation 

a;3=:i00, r=4. 



1st divisor . • . 
B'=6r^ . . 

(3r-f5)s . . 

2d divisor . . . 
s^ . . . 


.16 

.48 
. 756^ 

• 5556 
. 36J 

.6348 
. 5536^ 


r stu 
100 ( 4.6415889+ 
64 

36 
33336 




2664000 
2561344 


3d divisor . . . 


. 640336 ' 
16. 


102656 
64602721 


B" .... 


645888 
13921] 


38053279 
32308321 


4th divisor . . . . 


64602721 ' 
1, 


5744958 
5169331 




64616643 
646166 


575627 
516933 




58694 
58154 



2. Extract the cube root of 673373097125. 



1st divisor . 


. 64 


iV 


rstu 


B' . . 


.192 


673373097125 


( 8705 • 


{3r+s)s 
2d divisor . 


. 1729] 

.20929 
49. 


512 
161373 






146503 


rNoTE. — To deter- 


B" . . 

{sR+ty 


. 22707 
. 15696] 

.2286396 


14870097 
13718376 ' 


mine 5, we hav« 
192)1613( 

Some allowance 
made for the in- 
^ crease of 192. 


3d divisor . 


1151721125 




36j 


1151721125 




B'" 


2302128 
131425 








230344225 





HORNER'S METHOD OF APPROXIMATION. 323 

S. Extract the cube root of 1352605460594688. 

Jlns. 110592. 

4. Extract the cube root of 5382674. ^ns. 175.25322796. 

5. Extract the cube root of 15926.972504. 

^ns, 25.16002549 

6. Extract the cube root of 91632508641. 

Ms, 4508.33859058. 



7, Extract the cube root of 483249. 



Ms. 78.4736142. 



(Art. 196.) The method of transforming an equation into an- 
other, whose roots shall be less by a given quantity, will resolve 
equations of any degree ; and for all equations of higher degrees 
than the third; we had better use the original operation, as in 
(Art. 192.), and attempt no other modification than conceiving the 
absolute term to constitute the second member of the equation ; 
and the difference of the numbers taken in the last column in 
place of their algebraic sum. 

The following operation will sufficiently explain : 

1. Find one value of x from the equation 
a:4_^3;z;2 4-75^=:10000. 



±0 
9 

9 
9 

18 
9 

27 
9 

36 



—3 

81 



75 

702 



= 10000 
6993 



r 

(9 



78 777 
162 2160 

240 2937 
243 

483 



3007 =iV' 



(Continued on the next page.) 



324 ELEMENTS OF ALGEBRA. 



CI. 


1 


36 

.8 

36.8 

8 

37.6 

8 

38.4 

8 

39.2 


483. 
29.44 


2937. 
409.952 . 


s 
=3007. ( 0.8 
2677.5616 


? 

ii 
m 


512.44 

30.08 


3346.952 
434.016 


329.4384=iV^" 


3 

1 


542.52 
30.72 

573.24 


3780,968 

Take the coefficients to their nearest 
unit. 


I 


39 
+ 
39 
+ 
39 


573 
3 

576 
3 

579 


3781 
46 

3827 
46 

3873 


t 
= 329.4384 ( 0.08 
306.16 




23.2784=iV"' 




1 


39 


579 


3873 
3 


u 
= 23.2784(0.00600-1- 
23.256 



3876 224 

3 



3879 

Hence, . a:=9.88600-|-. 

N. B. We went through the first and second transformations in full. Had 
we been exact, in the third, we should have added .08 to 39.2, and multiplied 
their sum, (39.28), by .08, giving 3.1424 ; we reserve 3. only to add to the 
next column. By a similar operation we obtain 46. to add to the next column. 

EX AMP I. E s. 

1 . Given x—x2—a^—x*-\-500=0, to find one value of a;. Ans. 4.46041 671 

2. Given x"— 5a;3-{-9a;=2.8, to find one value of x. Ans. .3297105.5072 

3. Given 20a;-j-lla;2-*[-9^^ — a:^=4, to find one value of a;. 

Am. .17968402502 

4. Required the 5th root of 5000 ; or, in other terms, find one root of the 
equation a;5=5000. Ans. 5.49280-|- 

/ 8x \2 
6. Given ^"^^^y^JTi J ^° ^"^ °"® ^^^^® °^ ^- ^^**- 2.120003355 



APPENDIX 



Those who have taken but a superficial view of the science of algebra, com- 
monly regard it only as a means of more easily resolving arithmetical prob- 
lems. They do not, at once, recognize that it is a powerful engine for philo- 
sophical investigations. We have shown this, in some degree, in our 
application of the problems of the couriers and the lights ; and the few pages 
now left us, we shall devote solely to the application of algebra to philosophical 
truths ; not for the purpose of elucidating philosophy, but for impressing upon 
the mind the power and utility of algebra. 

With this object in view, we propose to investigate the subject of 

SPECIFIC GRAVITY. 

Gravity is weight. Specific gravity is the specified weight of one body, 
compared with the specified weight of another body (of the same bulk), taken 
as a standard. 

Pure water, at the common temperature of 60° Fahrenheit, is the standard 
for solids and liquids ; common air is the standard for gases. 

Water will buoy up its own weight. If a body is lighter than water, it 
will float ; if heavier than water, it will sink in water. 

If a body weighs 16 pounds, in air, and when suspended in water weighs 
only 14 pounds, it is clear that its bulk of water weighs 2 pounds ; and the 
body is 8 times heavier than water ; therefore the specific gravity of this body 
is 8, water being 1. 

If the specific gravity of a body is n, it means that it is n times heavier 
than its bulk of water. Therefore — 

If we divide the weight of any body by its specific gravity, the quotient 
will be, the weight of its bulk of wafer. 

On this fact alone we may resolve all questions pertaining to specific gravity. 

EXAMPLES. 

1. Two bodies, whose weights were A and B, and specific gravities a and b, 
were put together in such proportions as to make the specific gravity of the 
compound mass c. What proportions of A and B were taken ? 

A quantity of water, equal in bulk to A, must weigh — 

r» 

A quantity « « « « B, « « — 

A quantity of water, equal in bulk to (A-\-B) , will weigh 

Therefore, j — _=-lt-; Or, bcA-\-acB==abA-\-abB ; 

a b c 

Or, b(c—a)A=a(b—<:)B. 
2C (325) 



226 APPENDIX. 

Hence the quantities of each must be reciprocal to these coefficients ; or if we 

take one, or unity of By we must take -ji J units of A. 

2. Hiero, king of Sicily, sent gold to his jeweler to make him a crown ; he 
afterwards suspected that the jeweler had retained a portion of the gold, and 
substituted the same weight of silrer, and he employed Archimedes to ascer- 
tain the fact, who, after due reflection, hit upon the expedient of specific gravity. 

He found, by accurately weighing the bodies both in and out of water, that 
the specific gravity of gold was 19, of silver 10.5, and of the crown 16.5. 
From these data he found what portion of the king's gold was purloined. Re- 
peat the process. ** 

The preceding problem is the abstract of this, in which A may represent the 
weight of the gold in the crown, B the weight of the silver, and {A-\-B) the 
weight of the crown; a=:19, 5=10^, c=162. 

Then if we take £=1, one pound, one ounce, or any unity^ of weight, of 

silver, the comparative weight of the gold Will be expressed by -A ). 

That is, for every ounce of silver in the crown, there were 43^ ounces of 
gold. If clearer to the pupil, he may resolve this problem as an original one, 
without substituting from the abstract problem. 

3. I wish to obtain the specific gravity of a piece of wood that weighs 10 
pounds ; and as it will float on water, I attach 21 pounds of copper to it, of a 
specific gravity of 9. The whole mass, 31 pounds, when weighed in water, 
weighs only 4 pounds ; hence 27 pounds of the 31 were buoyed up by the 
water ; or we may say, the same bulk of water weighed 27 pounds. Required 
the specific gravity of the wood. 

Let s represent the specific gravity of the wood. 

10 
Then — = the weight of the same bulk of water. 

21 
And ~^= the weight of water of the same bulk as the copper. 

10 7 30 

Hence, }--=27, Or, 5=:— ==.405, nearly. 

4. Granite rock has a specific gravity of 3. A piece that weighs 30 ounces, 
being weighed in a fluid, was found to weigh only 21.5 ounces. What was 
the specific gravity of that fluid ? 

The weight of the fluid, of the same bulk as the piece of granite, was evi- 
dently 8.5 ounces. Let s represent its specific gravity. 

8.6 30 

Then — = the weight of the same bulk of water ; also --=10= the 

weight of the same bulk of water. ^ 

8.5 
Hence — =10, Or, .s=.85 .4n5., which indicates impure alcohoL 
s 



APPENDIX. 



327 



5. The specific gravity of pure alcohol is .797 ; a quantity is offered of the 
specific gravity of .85, what proportion of water does it contain*? 

Let A= the pure alcohol, and W = the water. 

Then 4,+W=^-i^. 

Ans. The resolution of thip equation shows 1 portion of water to 2.255-f- 
portions of alcohol. 

6. There is a block of marble, in the walls of Balbeck, 63 feet long, 12 wide, 
and ] 2 high. What is the weight of it in tons, the specific gravity of marble 
being 2.7 and a cubic foot of water 62^ pounds, Ans. 683y\ tons. 

7. The specific gravity of dry oak is 0.925 ; what, then, is the weight of a 
dry oak log, 20 feet in length, 3 feet broad, and 2^ feet deep ? Ans. 867l|- lbs. 

We may now change the subject, to make a little examination into maxima 
and minima. For tliis purpose, let us examine Problem 2 (Art. 114). 

1. Divide 20 into two such parts tliat their product shall be 140. It may 
be impossible to fulfil this requisition, therefore we will change it as follows : 

Divide 20 into two such parts, that their product will be i\iQ greatest possible. 

Let x-\-y= one part, and x — ?/= the other part. 

Then 2a;=20, and a;=IO, and the product, x^ — y'^, is evidently the 
greatest possible when t/=0. Hence the two parts are equal, and the greatest 
product is 100, or the square of one half the given number. 

2. Given the base, m, and the perpendicular, n, of a plane triangle, to find 
the greatest possible rectangle that can be inscribed in the triangle.* 

Let ABC hQ the triangle, BC=m, a 

AF=n, AD=x, and AE=x', De be a 
very small distance, so that x' is but insen- 
sibly greater than x. 

As D, comparatively, is not far from the 
vertex, it is visible, that the rectangle 
a'h'dd' is greater than the rectangle abed. 

If we conceive the upper side of the 
rectangle to pass through U, in place of 
D, and we represent AD' by x, and Ae 
by x'f it is visible that the rectangle 
e'fg'h' is less than the rectangle efgh. °ff o'<y~f s,i^' kit c. 

If we subtract the rectangle abed from the rectangle a'b'c'd', we slkall have 
a positive remainder. 

If we subtract the rectangle efg h from the rectangle e'fg'h', we shall have 
a negative remainder. 

* We do not introduce this problem to show its solution ; it belongs to the calculus, 
and, in its place, is extremely simple ; we introduce it to show a principle of reasoning 
extensively used in the higher mathematics, and, perchance our illustration may aid 
a pupil in his progress in the calculus. 




388 ELEMENTS OF ALGEBRA. 

The rectangle abed cannot be the greatest possible, so long as we can have 
a positive remainder by subtracting it from the next consecutive rectangle 
immediately below. 

After we pass the point on the line AF where the greatest possible rectangle 
comes in, the next consecutive rectangle immediately below, will become less ; 
and by subtracting the upper from it, the difference will be negative. 

Hence, when abed becomes the greatest possible rectangle, the difference 
between it and its next consecutive rectangle can be neither plus nor minus, 
Dut must be zero. 

Therefore, it is manifest, that if we obtain two algebraical expressions for the 
two rectangles abed axvA a'b'e'd', and put their difference equal to 0, a resolu- 
tion of the equation will point out the position and magnitude of the maximum 
rectangle required. 

Put the line ab=i/, and a'b'=t/'. As AD=x and AE=x', DF=7i — x 
Siud EF=n — a/. The rectangle abcd=y(n — a;), and a'b'dd'=y'(n — x'). 

From the consideration just given, the maximum must give 
y>{n~x>)—y{n—x)={). 

By proportional triangles, we have x : y :: n : m or, y = — . 

By a like proportion, we have y>=—x>. 

Put these values of y and y in the above equation, and, dividing by — , 

we have x'(n — x')=x(n — x) ; 

Or, x^ — x'^ '==n{x — x'). 

By division, x -\- x' =n 

As x' is but insensibly greater than ar, 2a:'=« ; which shows ihat AD is one 
half J.F, and the greatest rectangle must have just half the altitude of the triangle. 

3. Required the greatest possible cylinder thai can he cut from a right cone. 

Conceive the triangle (of Prob. 2.) to show the vertical plane cut through 
the vertex of the cone, and ah-=y the diameter of the required cylinder. Then, 
the end of the cylinder is .7854?/2, and its solidity is .l^b^y^(n — x). The next 
consecutive cylinder is .7854_y'2(n — x'). Hence y'^{n — x')=y^(n-^x). 

By similar tnangles x : y : : n : m, Or, y 2= — ~ and y'2=z—-xi, 

Hfcpce, x'^(n~x')z=x'^(n~x), Or, a;3~x'3=n(a;2— x'^) ; 
Divide both members by (x — x'), and x^-{-xx'-\-x'^=n(x-\-x'). 
As x=x' infinitely near, 3a^=2nx, or, x=^n; which shows that the 
altitude of the maximum cylinder is ^ the altitude of the cone. 

In this way all problems pertaining to maxima and minima can be resolved ; but 
the notation and language of the calculus, in all its bearings, is preferable to this. We 
had but a single object in view — that of showing the power of algebra. 






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THE UNIVERSITY OF CALIFORNIA UBRARY