# Full text of "A theoretical and practical treatise on algebra : in which the excellencies of the demonstrative methods of the French are combined with the more practical operations of the English ; and concise solutions pointed out and particularly inculcated"

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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. ^^ 885995 THE UNIVERSITY OF CALIFORNIA UBRARY