)>■■■* 1 IN MEMORiAM FLORIAN CAJORl i '. Wffni^d^- ^ r<^C^C; ^^/rt( '^f^Ly "^^^ 6 '^(^ ■/ Digitized by the Internet Archive in 2008 with funding from IVIicrosoft Corporation http://www.archive.org/details/completealgebradOOfickrich abixiBon^ ^Ipxtn €ouxni^, THE COMPLETE A L G E B E A, DESIGNED FOR USB IN SCHOOLS, ACADEMIES, AND COLLEGES. BY JOSEPH FICKLIN, Ph.D., FB07B8S0B OT ICATHSUATICS IN THE UNIVBR8ITT OP THE STATE OF MISSOUHL IVISON, BLAKEMAN, TAYLOE & CO., NEW YORK AND CHICAGO. ROBINSON'S Shorter Course. FIRST BOOK IN ARITHME TIC. Primary. COMPLETE ARITHMETIC. In One volume* COMPLETE ALGEBRA. ARITHMETICAL PROBLEMS. Oral and Written. ALGEBRAIC PROBLEMS. KEYS to Complete Arithmetic and Problems, and to Complete Algebra end Problems^ in separate volumes^ for Teachers, AHtlnnetic, orai: and written, usually taught iiv THREE hooks, is now offered, complete and thorough, in ONE hooh, " the complete arithmetic' * This Complete Arithmetic is also published in two volumes. TART J. and PART 11, are each bound separately^ and in cloth. Copyright, 1874, by DANIEL W. FISH. Electrotyped by Smith & McDougal, 82 Beekman St., N. Y. PREFACE. THE author of this treatise on Algebra has undertaken the difficult task of preparing a work complete in one vol- ume, which shall be sufficiently thorough for classes in Colleges and Universities, and at the same time sufficiently elementary for classes in Common Schools and Academies. To accomplish this desirable end the work has been so arranged that certain chapters and parts of chapters may be omitted by classes pursu- ing an elementary course. The aim has been : 1. To treat each subject in harmony with the present modes of mathematical thinking ; 2. To make every statement with such brevity and precision that the student can- not fail to understand the meaning ; 3. To give a clear and rig- orous demonstration of every proposition ; 4. To present one difficulty at a time, and just at that stage of the student's progress when he is prepared to understand its treatment ; 5. To treat with special care those subjects which have been found by expe- rience to present peculiar difficulties ; G. To make the work thor- oughly practical as well as thoroughly theoretical ; 7. To present each subject in such a manner as to create a love for the study. In the arrangement of subjects the author has departed widely from the beaten track ; but he feels confident that the plan he has adopted will commend itself to the experienced and thought- ful teacher. To facilitate frequent reviews, '' Synopses for Review " have been placed at convenient intervals throughout the work. To avoid making the present work too voluminous. Con- tinued Fractions, Reciprocal Equations, Elimination by the Method of the Greatest Common Divisor, and Cardan's formula for cubic equations have been omitted. These subjects are treated in the Appendix to the author's "Book of Algebraic Problems." IV PEEFACE. In preparing the present treatise the author has first consulted his own experience as a teacher, and the book has been mainly written to meet the wants of his own classes ; but he does not hesitate to acknowledge that he has received great assistance from many sources. A part of the material used in the chapters on Positive and Negative Quantities, Greatest Common Divisor and Least Common Multiple, Fractions, Simple Equations, In- equalities, Theory of Exponents, Mathematical Induction, and the sections on Permutations, Combinations, and Logarithms, has been taken from Prof. Todhunter's excellent treatise on Alge- bra. The works of Bertiand, Young, Peacock, Euler, Bland, Goodwin, and Wrigley have been consulted with advantage. While the author has availed himself of such material in the books named as suited his purposes, it will be found that much of that so taken has long since become common property, having assumed a stereotyped form ; and that other portions have been very much modified. It will be found, also, that the present treatise contains a large amount of new and original matter, which has not been inserted because it was novel, but because it served to simplify and elucidate the subject. Special attention is called to the full and thorough manner in which the subject of Factoring is treated ; to the demonstration of the Lemma, upon which the Binomial Theorem depends ; to the classification and treatment of Kadical Quantities ; to the treat- ment of Quadratic Equations, Higher Equations, Simultaneous Equations, Ratio, Proportion, Progressions, Interpolation, Recur- ring Series, Reversion of Series ; and to the Theory of Equations. The chapter on " Logarithms and Exponential Equations " is almost entirely the work of Prof. James M. Greenwood, A. M., Superintendent of the Public Schools of Kansas City, Mo., and formerly Prof, of Math, in the North Missouri State Normal School; and the "Synops'^s for Review "have nearly all been prepared by Prof. George S. Bryant, A. M., of Christian College, Columbia, Mo. To these and other able and experienced teachers the author is also indebted for many valuable suggestions in rela- tion to other portions of the work. UXIVBRSITT OP THE StATB OP MISSOURI, ) THE AUTHOB. Columbia, January^ 1875. ) SUGGESTIONS TO TEACHERS. 1. If the problems in the book are not sufficiently numerous or sufficiently varied, make some of your own, or take some from the book of "Algebraic Problems," made to accompany this volume. 2. The Synopses for Review should be placed upon the black-board, and dwelt upon until the topics embraced in the review are thoroughly fixed in the mind of the student. To illus- trate the manner of conducting a review, suppose the synopsis on page 10 is under consideration. Let the student point to the word "Algebra," and define it ; then to "Algebraic Quantity," and define it ; then to the two kinds of Algebraic Quantity — " Known and Unknown" — and define them ; and so on. 3. The following chapters and parts of chapters may be omitted by classes pursuing an elementary course : That part of Chapter IV from Art. 125 to Art. 128 inclusive, and from Art. 133 to Art. 136 inclusive ; Chapter VIII ; Chapter XIV ; that part of Chapter XVI from Art. 441 to Art. 457 inclusive ; Chapter XVII ; Arts. 482 and 483 of Chapter XVIII ; all of Chapter XX after Geometrical Progressions ; Chapter XXI ; Chapter XXII ; Chapter XXIII. CONTENTS CHAPTER I. DEFINITIONS AND NOTATION. PAGE DEPnnTioKS 1 Axioms 7 PAGB Notation 7 Synopsis yoR Review. 10 CHAPTER 11. FUNDAMENTAL PROCESSES. Addition 11 Subtraction 15 Synopsis for Review , 17 Multiplication 18 Synopsis FOR Review 26 Division 26 Factoring 33 Synopsis for Review 40 CHAPTER HI. POSITIVE AND NEGATIVE QUANTITIES. Relation BETWEEN Positive AND Nega- I A Negative. Quantity not less than TivE Quantities 43 I Zero in the Arithmetical Sense. . 47 CHAPTER IV. GREATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE. Greatest Common Divisor 48 General Rule 50 Least Common Multiple. Synopsis for Review 60 CHAPTER V. FRACTIONS. Definitions and Fundamental Princi- | Combinations op Fractions 67 The Signs of Fractions 79 Synopsis for Review 79 PLES 61 Reduction of Fractions 64 CHAPTER VI. DEFINITIONS AND GENERAL PRINCIPLES RELATING TO EQUATIONS. Definitions and Pbinoifles 80 I Transformation of Equations CHAPTER VII. SI^IPLE EQUATIONS. Simple Equations with One Unknown Quantity 8i Simple Equations with Two Unknown Quantities 99 Simple Equations with any Number of Unknown Quantities 108 Synopsis for Review 120 Discussion of Problems 120 Zero and Infinity 126 Finite, Determinate, and Indetermi- nate Quantities 126 Synopsis for Review 132 CONTENTS. Vll CHAPTER VIII. VANISHING FBACTIONS — INDETERMINATE EQUATIONS AND PROBLEMS — INCOMPATIBLE EQUATIONS. Vanishing Fractions 133 | Incompatible Equations 139 Indeterminate Equations 134 An Impossible Problem 141 XjTDETKBMiNATBi Problems 137 ' Synopsis for Review 141 CHAPTER IX. INEQUALITIES. Theorems 142 I Equation and Inequality Combined. .145 Solution of an Inequalttt 145 I Synopsis for Review 140 CHAPTER X. INVOLUTION AND EVOLUTION. Involution .147 I Higher Roots of Quantities 163 Evolution 151 I Synopsis for Review 165 CHAPTER XI. THEORY OP EXPONENTS. Basis of Theory (276) 165 j Negative Exponents , PosiTivB Fractional Exponents .167 Synopsis for Review 171 CHAPTER XII. RADICAL QUANTITIES. Definitions 172 Reduction of Radical Quantities . . 173 Combinations of Radical Quantities. 183 Involution of Radical Quantities. . . .192 Evolution of Radical Quantities 194 Reduction of Fractions having Surd Denominatobs 197 Propositions Relating to Irrational Quantities 803 Simplification of Complex Radical Quantities 206 Imaginary Quantities 210 Radical Equations 215 Synopsis for Review 218 CHAPTER XIII. QUADRATIC EQUATIONS WITH ONE UNKNOWN QUANTITY. Definitions and Principles 221 Incomplete Quadratic Equations 223 Complete Quadratic Equations 226 Theory of Quadratic Equations 230 Discussion of the Equation x^ +px=q.2i'(i Problem op the Lights 245 Quadratic Expressions a48 Synopsis for Review ^....250 CHAPTER XIV. HIGHER EQUATIONS WITH ONE UNKNOWN QUANTITY, The Two Forms aa?'=c, cKC'°+6a?'=c.... 252 | Problems 254 CHAPTER XV. SIMULTANEOUS EQUATIONS. Pairs of Equations Involving Radical DEFiNmoNS 255 Pairs op Equations one of which is op THE First and the other of the Second Degree 255 Particular Systems 260 Quantities 269 Groups with more than Two Unknown Quantities 273 Synopsis for Review ^ 379 VUl CONTENTS. CHAPTER XVI. RATIO, PROPORTION, AND VARIATION. Ratio Pbopobtion . .280 Variation 293 Synopsis fob Review 298 CHAPTER XVII. MATHEMATICAL INDUCTION. Theobems 299 | Synopsis roR Review. .801 CHAPTER XVIII. PERMUTATIONS — COMBINATIONS— BINOMIAL FORMULA- HIGHER ROOTS. -EXTRACTION OP Permutations 302 Combinations 305 BiNOMIAI. FOBMlTIiA 307 The n'A Root of Quantities 315 The n'ARootof a Numbee 316 Synopsis fob Review 317 CHAPTER XIX. IDENTICAL EQUATIONS. Pbopbrties of Identical Equations. ..318 I Decomposition of Rational Fbaction8.819 Undetebmined Coefficients 319 1 Synopsis fob Review 321 CHAPTER XX. SERIES GekebaIi Definitions 822 Abithmetical Pboobession 323 Arithmetical Mean 329 Geoicetrical Progression 331 Geometrical Mean 336 Series BY THE Differential Method. 341 Intebpolation 844 DEVELOPMBirr of Expressions into Se- BiES 846 Recubring Series 850 Reversion OF Series ....854 The Binomial Formula for any Expo- nent 357 Synopsis for Review 361 CHAPTER XXI. LOGARITHMS AND EXPONENTIAL EQUATIONS. Logarithms The Two Principal Systems. Exponential Equations 873 Synopsis for Review 374 CHAPTER XXII. COMPOUND INTEREST AND ANNUITIES. Compound Interest 375 Amount fob nYeabs 375 Annuities 876 Synopsis fob Review 877 CHAPTER XXIII. THEORY OF EQUATIONS. Definitions 378 General Properties 878 Transformation op Equations 389 Theorem of Descartes 396 Derived Functions 400 Roots Common to Two Equations 408 Equal Roots 403 Limits of the Roots of an Equation. .404 Sturm's Theorem 409 HoRNER^s Method of Approximation.. 414 ALGEBRA. CHAPTER I. DEFmiTIOI^S AlsTD NOTATIOISr. DEFINITIONS. 1. Algebra is that branch of Mathematics in which quan« tities are represented by letters, or by a combination of letters and figures, and in which the relations of quantities to each other and the operations to be performed are indicated by Signs, The letters, figures, and signs are called Symbols, 2. Algebraic Language consists in the use of algebraic symbols. 3. An Algebraic Quantity or Expression is one expressed in algebraic language. There are two kinds of algebraic quantities — known and unknown. 4. Known Quantities are those whose values are given. They are represented by numbers or the leading letters of the alphabet. 5. Unknown Quantities are those whose values are not given. They are represented by the final letters of the alphabet. 6. The sign + is called the plus sign, and signifies that the quantity to which it is prefixed is to be added. Thus, a + b signifies that b is to be added to a, and is read a plus b. IS a represents 9, and b represents 3, then a -{- b is equal to 12. 7. The Sum is the result obtained by addition. 3 DEFINITIOJS'S. 8. The sign — is called the minus sign, and signifies that the quantity to which it is prefixed is to be siiUracted. Thus, a — h signifies that h is to be subtracted from «, and is read a minus b. If a is 9, and b is 3, then a — bw equal to 6. 9. Oriie Mernainder or I>ifference is the result ob- tained by subtraction. 10. The sign x is called the sign of multiplication, and signifies that the quantity which precedes it is to be multiplied by the one which follows it. Thus, a x b signifies that a is to be multiplied by b, and is read a multiplied by b, or a into b. The sign of multiplication is often omitted. Thus, ab is equivalent to a x i. Sometimes a point is used instead of the sign X . Thus, a'b \s, equivalent to a x b. The sign of multiplication must not be omitted when the numbers are expressed by figures. Thus, 45 is not equivalent to 4x5. 11. OClie JProduct is the result obtained by multiplica- tion. 12. The sign -^ is called the sign of division, and signifies that the quantity which precedes it is to be divided by the one which follows it Thus, a -r- b signifies that a is to be divided by b, and is read a divided by b. The expression t is equivalent to a-^b, 13. The Quotient is the result obtained by division. 14. The sign = is called the sign of equality, and signifies that the quantities between which it is placed are equal. Thus, a = J signifies that a is equal to b, and is read a equals b, or a is equal to b. 15. An Equation consists of two expressions connected by the sign of equality. Thus, x-{-y = a, a-^-b^^c — d, are equations. The First Member of an equation is the quantity on the left of the sign of equality, and the Second Member is the quan- DEFINITIONS. 3 tity on the right of the sign. Thus, in the equation, x-}-yz=a—b, X 4- y is the first member, and a — b the second member. 16. The sign > or < is called the sign of inequality, and signifies that the quantities between which it is placed are ttn- equal, the opening being turned toward the greater. Thus, ay b signifies that a is greater than b, and is read a is greater than b ; and 5 < a signifies that b is less than «, and is read b is less than a. 17. An Inequality consists of two expressions connected by the sign of inequality, and its members are named as those of an equation. 18. When an expression is inclosed by a Parenthesis { ), the operations which are indicated in that expression are to be regarded as performed, and the parenthesis is to be regarded as expressing the result. Thus, the expression (a + Z>) (c — d) indi- cates that the sum of a and b is to be multiplied by the difibrence between c and d. The vinculum , the brackets [ ], and the brace \ \ hav^ the same signification as the parenthesis Thus, a -\-b X c — d \^ equivalent to \a -\-b){c — d). The vinculum is sometimes placed in a vertical position. Thus, d is equivalent to (a + 5 — c) d. a — c 19. The Terms of an expression are the parts which are connected by the sign + or the sign — . Thus, a, b, c, and d are the terms of the expression a -{■ b — c -\- d. 20. A Polynomial is an expression containing two or more terms. 21. A Binoinial is a polynomial containing only two terms. Thus, abc -f a; is a binomial. 22. A Trinomial is a polynomial containing only three terms. Thus, ab •{- ac -- be is a trinomial 4 DEFINITIONS. 23. A 31onoinial is an expression which does not contain parts connected by the sign + or the sign — . Thus, abc is a monomial. 24. When one quantity is the product of two or more other quantities, each of the latter is called a Factor of the product. Thus, fl, h, and c are factors of the product ahc. 25. A Numerical Factor is one which is expressed by a figure, or figures. 26. ^ Literal Factor is one which is expressed by a letter, or letters. 27. When a product contains one factor which is numerical, and another which is literal, the former factor is called the Coefficient of the latter. Thus, in the product 'tabc, 7 is the coefficient of dbc. The term coefficient is sometimes applied to any factor. Thus, in the product '7 abc, 7« may be called the coefficient of he, and in the product ahc, a may be called the coefficient of be, or ab the coefficient of c. 28. A Foiver of a quantity is the product of factors each of which is equal to that quantity. Thus, « x « is the second power of a; « x « X « is the third power of a; a x a x a x a is the fourth power of a ; and so on. The first power of a is a. 29. An Fxponent is a number placed on the right of, and a little above a quantity, and indicates how many times the quantity is to be used as a factor. Thus, a^ is equivalent to ^ X a\ a^ is equivalent to a x a x a', «* is equivalent to T,xaxaxa\ and so on. If no exponent is expressed, 1 is understood. Thus, a is equivalent to a\ The product of n factors each equal to a is expressed by a", and n is called the exponent of a. The second power of a, that is, a^, is often called the square of a, and the third power of a, that is, a^, is often called the DEFINITIONS. cube of a. The expression a^ is read a to the fourth power, or t)riefly, a to the fourth ; and a" is read a to the n^\ 30. The Square Hoot of any given quantity is that quantity which has the given quantity for its square or second potoer ; the cuhe root is that which has the given quantity for its cule or third power ; ihQ fourth root is that which has the given quantity for ii^ fourth power ; the fifth root is that which has the given quantity for its fifth power ; and so on. 31. The Hadical Sign ^ indicates that some root of the quantity to which it is prefixed is to be found. 33. The Index of the root is the number placed above the radical sign. The square root of a is denoted thus, ^«, or simply thus, j^a ; the cube root of a is denoted thus, ^a ; the fourth root of a is denoted thus, ^a ; and so on. 33. The symbols employed in Algebra are classified as fol- lows: 1. Symbols of Quantity are letters and other characters used to represent quantities. 2. Symbols of Operation are the signs +, — , x, -i-, /y/, and the exponential sign. 3. Symbols of Helation are the signs =, >, <, and others to be explained hereafter. 4. Symbols of Aggregation are the signs ( ), [ ], I i, — , and |. 34. Similar or Like Quantities are those whose lit' eral parts are identical. Thus, ^ab, lOab, 12ab, and 'Zbah are similar. 35. I>i8similar or Unlike Quantities are those whose literal parts are different. Thus, 4:ab and lOa^ are dissimilar. Remark. — An exception must be made in those cases where letters are considered as coefficients. Thus, Oic^ and bj^ are similar if a and b are con- sidered as coefficients. 6 DEFINITIONS. 36. Each of the hteral factors of a term is called a Dinien- 8 ion of the term, and the Def/7*ee of a term is equal to the number of its dimensions. The degree of a term, therefore, is equal to the sum of the exponents of its literal factors. Thus, 6aWc is of the sixth degree. 37. A Homogeneous Polynomial is one whose terms are all of the same degree. Thus, 7a^ + Za?h + ^^ahc is homo- geneous. 38. Hie Numerical Value of an algebraic expression is the number obtained by substituting for each letter its numeri- cal value, and then performing the indicated operations. Thus, if a = 5 and J = 6, the numeiical value of the expression 3a — U is 3. Some terms of frequent use in Algebra are here defined. 39. A JProposition is the statement of a truth, or of something to be done. Propositions are of the following kinds: Axioms, theorems, lemmas, problems, postulates, corollaries, scholiums. 1. An Axiom is a self-evident truth. 2. A TJieorem is a truth requiring demonstration. 3. A Lemma is an auxiUary theorem used in the demonstra- tion of another theorem. 4. A Problem is a question proposed for solution. 0. A Postulate assumes the possibility of the solution of some problem. 6. A Corollary is an obvious consequence deduced from one or more propositions. 7. A SchoUtim is a remark upon one or more propositions. 40. An Hypothesis is a supposition, made either in the enunciation of a proposition, or in the course of a demonstra- tion. 41. A H^ormula is a theorem expressed in algebraic lan- guage. NOTATION^. 42. AXIOMS. 1. The whole is equal to the sum of all its parts. 2. If equal quantities be added to equal quantities, the sums will be equal. 3. If equal quantities be subtracted from equal quantities, the remainders will be equal. 4. If equal quantities be multiplied by the same or by equal quantities, the products will be equal. 5. If equal quantities be divided by the same or by equal quantities, the quotients will be equal. 6. Quantities that are equal to the same quantity are equal to each other. NOTATION. 43. Algebraic Notation consists in representing quan- tities, operations, and relations by means of symbols. EXAMPLES IX NOTATION. 44. Express, in algebraic language, the following eight state- ments : 1. The second power of a, increased by twice the product of b and c, diminished by the second power of c, and increased by d, is equal to m times x. Ans, a^ + 2bc — r? ■\- d = mx. 2. The quotient arising from dividing a by the sum of % and h, is equal to twice h diminished by c. Ans. 7 = 25 — - c. X -{- b 3. One-third of the remainder obtained by subtracting four from six times x, is equal to the quotient arising from dividing five by the sum of a and b. ^ 6a; — 4 5 •^ Ans. — - — — a-\-b 8 NOTATION. 4. Three-fourths of the sum of x and five, is equal to three- sevenths of l, diminished by seventeen. Am, I (a: + 5) = ^5 - 17. 5. One-ninth of the sum of three times x and J, added to one- third of the sum of twice x and four, is equal to the product of fl, J, and c, ^^^ 1 (2^; + 4) + ^ (So: + J) = abc, 6. The quotient arising from dividing the sum of a and h by the product of c and d, is greater than p times the sum of m, n, x, and y* -« ^ + ^ ^ / . . . \ ^ Ans. — -j— > p{m -\- n -\- X -^^ y). 7. The square root of the sum of a and h is equal to m times the cube root of the remainder obtained by subtracting y from x, Ans. ^/a ■\- h =. m'^x — y. 8. The square root of x, diminished by the square root of y, is equal to n times the sum of the fourth root of a and the fourth root of h. Ans. ^x — ^y — n {%/a + yh). 45. Express, in common language, the following six algebraic expressions : ^ a -{■ X X m b c~ a -\- b' Ans. The quotient arising from dividing the sum of a and x by b, increased by the quotient of x divided by c, is equal to the quotient of m divided by the sum of a and b. 2. 3a2 -^ [^ — c){d-{-e)=x — y. 3. 3a^ + b — c{d-{-e)=x — y. a -\- X a — y _ m 5H-d + c 3 a + b ^ Vdb-h^Vc . , m 0. —777 =z ox -\ . a -{-2b n NOTATION. ^ 46. NUMEKICAL VALUES. If « = 1, b = 3, cz=4:, d = 6, e = 2, and /= 0, find the numerical yalue of each of the following ten expressions : 1. a + 2b + 4c. 2. 3b ^6d — 2e. 3. ab + 2bo + Sed. 4. ac + 4:cd — 2 Je. 5. abc + 45^ + ec — ^. 6. ^2 + J2 ^ c2 +/2. r, cd 4:be cd T "^ 'S^" ~ 24* 8. c4 — 4c8 + 3c — 6. 9 ^ + ^^ * 2c — 3a 10. ^(27b) - V(2c) -f V(2c). 11. Find the value of (x -{- y){x^ y), when a; = 8 and y = 5. Ans. 39. 12. Find the value of x + y x x ^ y, when a; = 8 and t/ = 5. ^W5. 43. 13. Find the value of x ^{x^ — Sy) -{- y ^{x^ + Sy)y when X = 5 and y = 8. Ans. 26. 14. Find the value of (b — x) {^a + &) + ^\(a—b) {x^-y)\, when a = 16, 5 = 10, xz=z 5, and ^ = 1. ^^5. 76. ^7^5. 23. -47i5. 35. Ans. 63. Ans. 88. Ans. 92. Ans. 26. Ans. 15. Ans. 6. Ans. 5. Ans. 9. 15. Find the value of a; in the equation x = ^ when a = 10, 5 = 5, c = 4, c? = 2, and 6 = 3. Ans. X = 20. 10 DEFINITIONS AND NOTATION. 47. SYNOPSIS FOR REVIEW. Algebba. Algebraic Quantity. \ ^"<^^' ( Unknown. Sum. Remainder. Product. Quotient. Equation | Inequality j Firet Member. Second Member. First Member. Second Member. Names op Expressions. o EH < Eh o HH iz; eg Terrm. Polynomial Symbols Binomial. Trinomial. Monomicd . . . Numerical. Factor \ Literal. CoeflBcient. Known : a, b, c, d, etc. ' Of quantity • . . ■( Unknown : t, u, v, w, X, y, z. {"T i ? X, .J "T-y "71 V , Exponent. Of relation, . . . | =, >, <. . Of aggregation. | ( ), H j [ ] ^ — > I • Similar Quantities. Dissimilar Quantities. Degree of a Term. Homogeneous Polynomial. Numerical Value. Proposition Axiom. Theorem. Problem. Postulate. CoroUary. L Scholium. Hypothesis. Formula. CHAPTEE II. FUNDAMENTAL PROCESSES. ADDITION 48. Addition is the process of finding the simplest expres- sion for the sum of two or more quantities. 49. A JPositive Term is one which is preceded by the ^gn + . When a term has no sign prefixed, the sign + is un- derstood. 50. A Negative Term is one which is preceded by the sign — . ORDER OF TERMS. 51. The value of a polynomial whose terms are positive is the same in whatever order the terms may be written. Thus, a-\-h-\-c=a-\-c-\-h=h-\-c-{-a=c-\-a-\-b=c-\-h-\-a=l)-\-a-^c. 53. When a polynomial contains both positive and negative terms, we may write the former tenns first in any order, and the latter after them in any order. Thus, a-\-b—c—e=za-\-h—e—c =b-i-a — c — e=b-\-a — e — c. 53. In some cases we may vary the order of terms still further. Thus, if a = 10,b = 6, and c = o, then a -\- b — c = a — c -{- b z=:b — c •\- a. But, if a = 2, b = Q, and c = b, the expression a — c -{- h presents a difficulty, because we are apparently required to sub- tract 5 from 2. It will be convenient to agree that such an ex- pression as a — c -\-by when c is greater than a, shall be under- stood to be equivalent to a -\- b — c. At present we shall not use such an expression except when c is less than a -\- b. In like manner we shall consider —b-\-a as equivalent to a — b. We shall recur to this point in Chapter III. 12 FUNDAMENTAL PROCESSES. REDUCTION OF SIMILAR TERMS. 54, When two or more terms of a polynomial are similar, it may be reduced to a simpler form. 1. Let it be required to simplify the expression ^a% — Sa^c + 9ac2 — %a^b + "id'c — 6R This expression may be written thus : 4a2J — 2a2J + la^c — Sa^c + 9flc2 — 6^ (53). Now Aa^b — 2a^ = 2a% and ^a^c — dah = 4:0.^6. Hence the given expression may be reduced to 2a^b + 4a2c + Qac^ — 6&2. 2. Let it be required to simplify the expression 2a^b(^ — ^a^b(^ + Wbc^ — 8a3^>c2 4- \\a%(?. We write the positive terms in one column and the negative terms in another thus, %a^b(? - ^a%& 19a3^>c2 __ V2a^b(^ = Wbc\ 3. Let it be required to simplify the expression ^abc -f Sa^J + 2a^ — ha^b — Za^b. Arranging the terms thus, 4:abc + 3fl2J — 5^25 -f 'ZaJ^b — Sa^b and uniting, we obtain Aabc -f 5a^ — Sa^. But^bc + 6a^—Sa^=^abc+5a^—5a^—da^z=4abc—Sa^, RULE. L Reduce the positive similar terms to one term by addition, n. In like manner reduce the negative similar terms to one term. in. Then subtract the less result from the greater, and to the remainder prefix the sign of the greater. ADDITION. 13 EXAMPLES. Keduce each of the following expressions to its simplest form : 1. 10^4 -f 3a* + 6a* — a* — 5a*. Ans. UaK 2. ba^b + 3 ^/aFi — "tab + 17a5 + 2 V~d^ — Wb — SV~a¥c — 10«^ + 9«'^- ^ns. Sa^b — 3 V^c. 3. 3a — 2a — 7c + 3c + 2a + 4c — 3a. Ans. 0. 4. %h — 8ac2 + 15J3c + 8ac + 9ac2 — 24Z>8c. Ans. ac? + 8ac. 6. 6ac2 — bab^ + 7ac2 — ^a¥ — 13ac2 -f- Uab\ Ans. lOal^. 6. d^b — dab^-{- Sa^ + 5c — Sa^b + 8a^+ 2a^ +c +ab^—Sc. 55, To find the sum of two or more quantities. 1. Let it be required to find the sum of c — d -{- e and a — b. Adding c -\- e to a — J, we obtain a — b-\-c + e', we have, however, added d too much to a — b\ hence, in order to obtain the correct sum, d must be subtracted from a — b-{-c-\-e. We thus obtain a — b-\-c-\-e — d. Therefore, to find the sura of quantities, all the terms of which are unlike, write them in any order, prefixing to each term its proper sign. 2. Let it be required to find the sum of aS + 3a2 — 4aJ 2a2 — 3aJ + J2 — c, and a^ + 2a5 — bb'^ + 3c. Their sum, after reducing, is a^ + ^a? — bab — 46^ ^ 2c (54). BZTLE. I. Write similar terms, with their proper sigfis, in the same colum7i. II. Reduce each column (54), and to the results annex those terms which cannot be reduced, prefixing to each its proper sign. EXAMPLES. (1.) (2.) 7a; 4_ Sab + 2c IQa^^ -f be — 2abc — Sx — Sab — be — ^aW + %c + Qabc bx — "dab 4- 9c — lOa^^ — 12 Z> c + 3a^c Sum 9a; — 9a5 + 6c. 2a^l^ — 2bc + '^abc. 14 FUNDAMENTAL PROCESSES. 3. Find the sum of 4:a — ob -\- Sc — 2dj a -^ b — 4:C + bd, 3a — 7J + 6c + 46?, and a-\- ^b — c — Id. Ans. 9« — 75 + 4c. 4. Find the sum of a^ ^ 2^^ — 3x -^ 1, 2a:^ — 3x^ + 4a; — 2. dx^ -{-^ •\- 5, and ^ — dx^ — 5x + d. Aris. lOa:^ — 4a; + 13. 5. Find the sum of x^ — dxy -{- y^ -\- x -\- y — 1, 2x^ -{■ 4:xy— 'Sy2 — 2x — 2y + 3, 3a;2 _ 5^^ _ 4y2 ^^x + 4:y — 2, and 6x^ + lOxy -\- 5y^ -j- X -\- y. Ans. 122^^4- Gxy — y^ + Zx -f 4?/. 6. Find the sum of a;^ — 2a3? + a% a? + 3ffa;2^ and 27^ — ax^. A71S. ^ + aH, 7. Find the sum of 4aJ — ar^^ Sar^ — 2aJ, and 2aa; + 25a;. Ans. 2ab + 2a;2 + 2aa; + 25a;. 8. Find the snm oi a + b -\- c -\- dy a -\- b ■}- c — d, a -\-b — c + d^ a — b -{- c -\- df and — a-i-b + c-\-d. Ans. 3a + 35 + 3c + dd. 9. Find the sum of 3 (a?» — y% 8 {x^ — y^), — 5 (a;2 — y^) -f 6 (a;2 ^ y2)2, and 7 (a; — y)2. Ans. 6 (a;2 _ y2) ^ 6 (a;2 + 7/2)2 + 7 (a; - ;?/)2. 10. Find the sum of a"* — 5'' + 3a: ", 205"* — 35" — x^, and or 4- 45** — a;«. Ans. 4a"* + 2a;^ — xf^. LAWS RESPECTING THE USE OF THE PARENTHESIS. 56, If a polynomial has any number of its terms enclosed by a parenthesis preceded by the sign + , the parenthesis may be omitted and the value of the polynomial will not be changed. Thus, a — b + {c — d-\-e)=a — b-\-c — d + e {^5). Cor. — The value of a polynomial will not be changed by enclosing any number of its terms by a parenthesis^ provided the parenthesis have the sign + prefixed. Thus, a — 5 + c — d-\- e=za — b ■\- c-^ {—d -\- e)=. a — d + (c -\- e—b) = a -^ {— d -\- c ■{■ e —5). SUBTRACTION. 16 SUBTRACTION. 11 5-3 6 + 3=9. 57. Subtraction is the process of finding the simplest ex^ pressioQ for the difference between two quantities. 5S, Uie Minuend is the quantity from which the sub- traction is to be made. 59. TTie Subtrahend is the quantity to be subtracted. 60. The Hefnainder is the result obtained by the sub- traction. 61. To find the diflference between two quantities. 1. Let it be required to subtract 5 — 3 from 11. Subtracting 5 from 11 we obtain 6. This result is too small by 3, for the number 5 is larger by 3 than the number which was required to be subtracted. In order, therefore, to obtain the correct remainder, 3 must be added to 6. 2. Let it be required to subtract c — d from a -\- b. Subtracting c from a + 5 we obtain a ^ h —c. This result is too small by d, for c is larger by d than the quantity which was required to be subtracted. In order, therefore, to obtain the correct remainder, d must be added to a + b — c. Hence a-{-b — {c — d) = a-{-I?—c-}-d. The same result may be obtained by adding — c + ^ to a + b (55). RULE. CJiange the sign of every term of the subtrahend from. -\- to —y or from — to •\-, and add tUe result to the minuend. Remakks— 1. Beginners may solve a few examples by actuaUy changing the sign of every term in the subtrahend. After this, it is better merely to conceive such change to be made. 2. Subtraction, in Algebra, is proved in the same manner as in Arith- metic, by adding the remainder to the subtrahend ; the sum should be equal to the minuend. a + b c — d a -\- b — c -\-d. Id FUNDAMENTAL PROCESSES. EXAMPLES, (1.) From 3a + J ^ f 3a + 5 Take a- J ^^^ ^°^^ ^^^^ *^® ^^^^^ ""^{-a^- I T^ . -, 7 ^ the subtrahend chaQged. —r rr Remainder 2a + 2J J ^ I 2a + 25 (2.) (3.) From \\a^-\-Zdb^^x\j From 5«_3J + 4c— 7rf Take 5a'^ + 4a^>— 6a;y Take 2a— 2^> + 3c— d Remainder Ga^— aj-|-2a;^. Remainder 3a— 5+ c— 6^. 4. From 7^-\-^3? — Ix^-^-'^x — \ take a:* + 2a;3 — 2^:2 + 6a; — 1. Ans, 2a^ + a;. 5. From 3a2 — 2ax-{-a^ take a^^ax+ x\ Ans. 2a^ — ax. 6. From 2a — 2J — c + ^ take a — b — 2c+ 2d, 7. From 4a + 35 — 2c + Sd take a + 25 + c — 5d J?iA'. 3a + 5 — 3c + 13d, 8. From Sar^ _ 3aa: + 5 take 5x^ + 2aa: + 5. 9. From 1!xy — 10^ + 4a; take 3a:y + 3y + 3a;. ^W5. 4:xy — ISy + a;. 10. From 3a 4- 5 + c take a — 5 — c. Ans. 2a + 2b -\- 2c. 11. From a2 -|- 2a5 + 5^ take a^ — 2a5 + 52. v4m5. 4a5. 12. From a^ + 352c + a52 — abc take a52 — fl5c + b\ 13. From 5a;2 — y^ take 4a:2 _ ^ ^ ^4 ^^^^ 2;2 — y*. 14. From 4a'"+ 2xp— x^ take a*"— 5'»+3a;i'+ 2a"»— 35"— a;**. ^W5. a"*+ 45"— a:'. LAWS RESPECTING THE USE OF THE PARENTHESIS. 63, A parenthesis preceded by the sign — may be omitted without affecting the value of the expression in which it occurs, provided the sig7i of every term within it be changed. Thus, a—{b — c^d-\-e)=^a — b-\-c-\-d — e. Cor. — The value of an expression will not be changed by en- closing any number of its terms by a parenthesis, preceded by the sign — , provided the sign of every term thus enclosed be changed. Thus, a—b-{-c-\-d—ez=za—b-\-c—{—d-\-e)=a—{b—c—d-\-e) = a + c— (5— £?+e). SUBTRACTIOK. 17 EXAMPLES. Ans. a —b + c — d. Ans. a — h -{• c -\- d* Ans. a — 7 J. Ans, 5a, Eemove the signs of aggregation from each of the following expressions : 1. a—[b—(c — d)]. % a- [(5 — c) — d]. 3. « 4- 25 — 6« — [U — (Qa — 65)]. 4. 7fl — f 3a — [4fl — (5a — 2fl)] i . 5. da—\a + h— \a^h-\-c—{a + h -\- c + d)W Ans. 2a — h^d. 6. 2x _ [3y — {4a; — (by — 6x)\]. Ans. 12a; — Sy. 7. a — {a — [a — (a — x)]\. Ans. x. 8. 4a; — I a; — [a: — (a; — 3) + 3] — 3 } — { — a; — [— a; (_ a; + 3) 4- 3] _> 3}. Ans. 4a; + 12. 63. SYNOPSIS FOR REVIEW Positive terms. Negative terms. Order of terTns. ADDmON. j When preceded by + . ( When preceded by — . SUBTBACnON. " Use op Pahknthesis. Reduction of terms. 1^^^^ P««^^^^ •^ (Rule. Sum. Mule. JUimiend. SvbtraJiend. Remainder. Rule. Toremcyte (When preceded by +. ( When preceded by — , To introduce. . . .\ ^^^^ P^^^^^^^ ^^ +* ( When preceded by — . 18 FUNDAMENTAL PROCESSES. MULTIPLICATION. 64. TJie ^Product of two quantities is a quantity which is as many times greater than one of them as the other is greater than a unit Thus, the product of 5 and 4 is 20. 65. Multiplication is the process of finding the product of two quantities. 66. The Multiplicand is the quantity which is to be multiphed. 67. The Multiplier is the quantity by which to multiply. 68. The product of three or more quantities is sometimes called a Continued Product The product of any number of factors is the same in whatever order they may be taken ; thus, abc = ach = bca. The literal factors are generally arranged in alphabetical order. 69. To find the Product of Monomials. Let it be required to find the product of taW and ha^¥c» We may indicate the multipHcation thus : laW X 6a^¥c\ and since the product is the same in whatever order the factors are taken, we have la^J^ X ba%^c = 7 x 5 x a^aW¥c = dhaHW¥c. Here a occurs as a factor five times, h occurs six times, and c once. Therefore the required product may be written thus : ZbaWc. Hence, Principles. — 1. The coefficient of the product of given mono- mials is equal to the product of their coefficients. 2. Every letter which occurs in any of the given factors must he written in the product with an exponent equal to the sum of all its exponents in the given factors. MULTIPLICATION. 19 Cor. — We may, if we please, indicate the product of the hke powers of different letters by writing them within a parenthesis and placing the exponent over the whole. Thus, aW = {aVf ; for {aUf = ahxah = aaxhh =aW. EXAMPLES. 1. Multiply ah by x. 2. Multiply ^ax by 2ay. 3. Multiply 4am by Zlc^n. 4. Multiply ba^x by Zah?. 5. Multiply Za^x"" by ^a'^oif*. Ans. abx, Ans. ^dhoy. Ans. VZdbc^mn, Ans. 15aV. Ans. 27a'"+»a;'"+'» 70. To find the Product of two Polynomials. 1. Let it be required to multiply a -f 5 by «?. The product of a and c is ac ; but this is too small by he, for it is the sum of a and h which is to be multiplied by c. Hence {a + h)c ^ ac -\- he. 2. Let it be required to multiply a — hhy c. Here the product of a and c must be diminished by the product of h and c. Hence {a — h)c = ac — he. 3. Let it be required to multiply a -{- b by c ■{■ d. The product of a -^ h and e is ac -}- he; but a -f & is to be multiplied by the sum of c and d ; hence ae 4- he is too small by the product of a -\- h and d ; that is, by ad + hd, which must, therefore, be a-{-b e-\- d a + b ae -f be. a - -b c ae- -be. ae + he + ad -\- bd. added to ae + he to produce the correct result. Hence (a -^ h) {e + d) = ac -^ be -\- ad -{- bd. *0 FUNDAMEJSTAL PROCESSES. 4. Let it be required to multiply a -^ bhy c — d. Here the product of a + b and c must be di- minished by the product ota + h and d. Hence c—d ac + bc—(ad-{-bd)=ac + bc—ad—M. {a -\-b) (c — d) = ac -{-be— {ad + bd) =zac-^bc — ad — bd, 5. Let it be required to multiply a — bhj c — d. Here the product of a — b and c, which is ac—bc, is to be dimin- ished by the product of a — b and d, which ha ad — bd. Hence a — b c—d ac—bc— {ad— M)^aC'-bc — ad-^bd. {a — b) {c — d) = ac — bc — {ad — bd) = ac — bc — ad-\- bd. In this example, we observe that corresponding to the terms — b and c, one of which occurs in the multiplicand and the other in the multiplier, there is the term — be in the product ; and corresponding to — J of the multiplicand and — d of the multi- pher, there is the term -j- bd in the product. Hence it is often stated as an independent truth, that (—b) X c= — be, and (— b) x {— d) = ■{- bd. Thus, the sign of the product is deduced from the signs of the factors by the rule. Like signs produce +, and unlike signs produce — . 6. Let it be required to multiply 4^2 — 6ab + 653 by 2a^ — Sab + 4^. 4flr2 _ 6ab + 6J2 2^2 - 3ah 4- 4^ Sa^ — lOa^ 4- 12«2J2 — 12a^ + 15a^^ — 18ab^ ' +16g2^ — 20^53+ 24^ - 8a^ — 22a^ + 43^2^ — 38ab^ + 24^. MULTIPLICATIOIT. 21 7. Let it be required to multiply 2a^ + 3ic -f 4 by 2a;2_ 3^ _l_ 4 2a;2 4. 3a: + 4 4a4 ^ 6^ ^ 8a;2 — 6a^ — 9x^^ — 12a; + Sx^ 4- 12a; + 16 4a4 ^ 7a;2 _|_ 16. MULE. Multiply every term of the multiplicand hy each term of the multiplier in succession ; if a term in the multiplicand and a term in the multiplier have like signs, prefix the sign + to their product ; if they have unlike signs, prefix the sig7i — ; then take the sum of these partial products to form the complete product, 1. Multiply 2p^qhj 2q + p. Ans. dpq + 2^2 _ 2q^, 2. Multiply a^ + Sab + 2b^ by 7a — 55. Ans. W 4- l^a% — aW' - IW, 3. Multiply a^ — ah^- W by a^ + ah — b^ Ans. a^ — aW + 2ab^ — ¥. 4. Multiply a^ — ab-^ W by a^ + «5 — W. Ans. a* — a^y^ + ^aW — 45*. 5. Multiply or2 4- 2ax 4- x^ by a^ 4- ^ax — x\ Ans. a^ + ^aH 4- ^a^x^ — a:*. 6. Multiply a2 _]_ 4^3; 4. 4a;2 by a^ — 4aa: + 4a:2 Ans. a* — 8a2a:2 4. iBrc^. 7. Multiply a2 __ 3 _ ^ ^ ^ahc. 12. Multiply a:* + 2a:3 ^ ^ _ 4a; __ 11 by a,-2 — 2a: + 3. ^ws. n« + 10a; — 33. 13. Multiply a* — 2(z3 + 3^2 - 2a + 1 by a^ + 2a3 + 3^2 -f 2a H- 1. Ans. a^ + 2a« + 3a* + 2a^ + 1. 14. Multiply together a — x, a + Xy and a^ + a:^. Ans. a* — x^. 15. Multiply together x — 3, a; — 1, a; + 1, and a; + 3. Ans. x^ — 10a:2 + 9. 16. Multiply together a^--a; + l, ar^+a; + l, and a^— a^^ + l. Ans. a:^ H- ic* + 1. 17. Multiply together a -}- x, b -^ x, and c + x. Ans. abc + (ab + be -\-ac)x -\- (a -t b -{■ c)x^ + 7?. 18. Simplify (a + b){b + c) — {c-\-d) (a + o?) — (a + c) (^>— ^). 19. Simplify (a + J+ c+ ^)2 + (a -^ & — c + ^)^+ («— ^ + c— (?)2 -f- (a + J — c — dy. Ans. 4 (a2 + ^2 + ^2 + d^). 20. Simplify (a + J + c)^ — a (J + c — a) — J (a + c — J) — c(a + Z, _ c). Ans. ^{a? + b"^ -{- c^). 21. Prove that a^ j^ f ^ [x -^ yf = 2 (x^ + xy -\- y^f + ^7?y'^{x-\-yf{3^-{-xy-\-y^). 22. Prove that 4a;2^ (ar^+ y2) = (.r2_|- xy-\-y^f— {x^-^xy^y'^f. 23. Multiply (a;2 - 3a; + 2)2 by a^^ + 6a; + 1. ^?i5. a;« — 22a;* + 60a;3 _ 55^ ^ 12a; + 4. 24. Multiply (a + «»)2 by (a - Z')^. ^ws. aS — a*5 — 2a3&2 _^ 2a2Z»3 + aJ4 __ js. 25. Prove that (a + If - (a — &)2 = 4a&. MULTIPLICATIO]^^. 23 71. The square of the sum of tico quantities is equal to the sum of their squares increased by twice their product. If we multiply by a -\-b a -\-b a^-^ab -\-ab-\-l^ Tve obtain a^ + 2ab -{■ l^ ', hence {a-\-hf=a^-\-2ab-\-J?, If we wish to obtain the square of the sum of two quantities, this theorem enables us to write the terms of the result without the necessity of performing the actual multiplication. EXAMPLES. 1. (2 + 5)2 = 4 + 20 + 25 = 49. 2. (2w + 3/2)2 _ 4^2 _|_ i2mn + 9^2. 3. (ax + byf = a^x^ + "^abxy + Wy\ 4. (c 4- 2^)2 = c2 + 4c3 ^ ^6. 7. [(a; + yY -^{x- 7/)«]2 =1 (2: + y?"^ + 2(a; + «/)"»(a; -yY + (a;-2/)2". 72, 27^6 square of the difference between two quantities is equal to the sum of their squares diminished by twice their product, K we multiply a —b by a —b a2_ ab — ab^-U^ we obtain a^ — %ab + b^\ hence {a—bY=^a^—%ab-\-b\ EXAMPLES, 1. (5 - 3)2 = 25 - 30 + 9 = 4. 2. (2a; — yf = 4a;2 — 4a;2/ + y\ 3. (3rc — 5^)2 = 9ic3 — 30a;i2 + 24: FUNDAMENTAL PROCESSES, 4. (c — 2dy = c2 - 4c^ + 4:cP. 5. (a2 _ ^2)2 = ^4 _ 2a2^ ^ j*. 6. [a + b-(c-\- d)Y = a^ + 2aJ + ^ -2 (a + 5) (c + ^) + 73. ^e product of the sum and the difference of two quan- tities is equal to the difference between their squares. K we multiply a -\-b by a -b a^ + ab -ab- -^ we obtain a^- -^; ^; hence (a-\-b) {a—b)=a^—b^. EXAMriiJSS. 1. (3 + 2) (3 - 2) = 9 - 4 = 5. 2. (3a + 2*) (3a - 2b) = 9a^ - AJP. 3. (m + 1) (m — 1) =m2 — 1. 4. (c + 2d) (c — 2d) = (^- 4^. 5. (a2 4- ^) (a2 _ ^>2) = ^ _ J4. 6. [(a + J) + c] [(a 4- *) - c] = a^ + 2aJ + ^ - c». 7. [(a + J)2 + (a: - y)T [{a + ^)2 - (a: - 2^)2] = (a + J)* - 7^. Meaning of the Sign ±. We may here indicate the meaning of the double sign ±, which is sometimes used. Since (a + b)^ = a^ + 2ab + V^, and (a - bf = a^ - 2ab + l^, we may write both formulae in the following abbreviated form : (a ± bf = a^± 2ab + 2^. By this notation we are enabled to express two different theorems by one formula. The expression a ± & is read a plus or minus b. MULTIPLICATION. 25 75. By the aid of the preceding theorems the process of mul- tiplication may often be abridged. Thus, (aj^-bj^c-^d) {a-\-b—c—d)^\{a + h)-^{c + d)] \{a-^h) — {c-\-d)'\ = {a + hf-{c + df (73) =a?+ 2ah + ^2 _ (^2 ^_ <^cd + d^) (71) =a^-\-'itah-\-V^—(?—'^cd—d?. REMARKS ON MULTIPLICATION. 76. The degree of the product of two monomials is equal to the sum of the degrees of the multiplicand and multipher, since all the factors of both monomials appear in the product (69, Prin. 2). Thus, if we multiply 'ia% which is of the third degree, by Zab\ which is of the fourth degree, we obtain Qa%\ which is of the seventh degree. Hence, if two polynomials are homogeneous, their product will be homogeneous (70, 6). 77. The number of terms in the product of two polynomials, when the partial products do not contain similar terms, is equal to the product obtained by multiplying the number of tenns in the multiplicand by the number of terms in the multiplier. Thus, if there be m terms in the multiplicand, and n terms in the mul- tiplier, the number of terms in the product will be mn. If the partial products contain similar terms , the number of terms in the product after reduction, will evidently be less than 7nn. 78. When the multiplicand and multipher are arranged in the same way, according to the powers of some common letter, if there be one, the first and last terms of the product are unlike any other terms. Thus, in the sixth example of Art 70, the mul- tiplicand and multiplier are arranged according to the descending powers of a ; the first t^rm of the product is 8a* and the last term is 24S*, and there are no other terms which are like these ; for the other terms contain a raised to some power less than the fourth, and thus differ from 8a*; and they all contain a to some power, and thus differ from 24 J*. Therefore the product of two polynomials cannot contain less than two terms. FUNDAMENTAL PROCESSES. 79. SYNOPSIS FOR REVIEW. CHAPTER 11— Continued. MULTIPLICATION. iMuUiplicand, Multiplier. Product. Factors. ^ ( Order of factors. Product of ) Monomials. ) ^^,„ - ^Coefficients. {Law of . . I Exponents. / Investigation for rule. Product op ) n^u. Theorems ^Remarks. Degree of product. Product, homogeneous when. Number of terms in product. First and last terms, different •when. DIVISION. 80. Division is the converse of Multiplication. In Mul- tiplication we determine the product of given factors. In Di- vision we have the product of two factors, and one of them given to determine the other factor. 81. Tlie Dividend is the given product. 82. The Divisor is the given factor. 83. Tlie Quotient is the factor to he determined. DIVISION. 27 84. To find the Quotient of two Monomials. . Let it be required to divide Soa^^c^ by 7a^b^c. The division may be indicated thus : naWc ' Now, since the quotient must be such a quantity that when it is multiphed by the divisor the product shall be equal to the dividend, the coefficient of the quotient multiplied by 7 must give 35 ; hence, the coefficient of the quotient is found by dividing 35 by 7. Again, the exponent of any letter in the quotient added to the exponent of the same letter in the divisor, must give the exponent of this letter in the dividend (69, Prik. 2) ; hence, the exponent of any letter in the quotient is found by subtracting its exponent in the divisor from that in the dividend. Therefore, -r-^v„— = 5a^^c. Hence, Prin"CIPLES. — 1. Tlie coefficient of the quotient of two given monomials is the quotient obtained by dividing the coefficient of the dividend by that of the divisor. 2. Every letter which occurs in the dividend must be tvritten in the quotient, ivith an exponent lohich is found by subtracting its exponent in the divisor from that in the dividend. Cor. — = «'"-'« = fl", and — = 1 : hence a" = 1. 1. Divide abx by x. Ans. ab. 2. Divide Qa^xy by Zax, Ans. 2ay. 3. Divide 12abc^mn by Sbchi. Ans. ^anu 4. Divide Iha^T^ by ^a^x\ Ans. haH. 5. Divide 27a"» + "a:^ + " by 9a"a^. Ans. Sa^'ai^. 28 FUNDAMENTAL PROCESSES. 85. It follows from Art. 84 that the exact division of mono- mials will be impossible : 1st. When the coeflBcient of the dividend is not divisible by that of the divisor. 2d. When the exponent of a letter in the divisor is greater than the exponent of the same letter in the dividend. 3d. When the divisor contains a letter that is not found in the dividend. 86. To find the Quotient of two Polynomials. 1. Let it be required to divide ab — he by 1). — 7 — = a — c ; for {a — c)h = ah — be. In this example, we observe that corresponding to the term ah in the dividend and to the divisor h, there is the term a in the quotient ; and corresponding to the term — he in the div- idend and to the divisor h, there is the term — c in the quo- tient. We have already seen that ix(-c) = - he, and (- h) (- e) = he (70). In like manner, the following statements may be admitted : — he ^ :i he , = hf and =z — h. — c —c Thus the sign of the quotient is deduced from the signs of the dividend and divisor by the rule. Like signs produee +, and unlike signs produee — . 2. Let it be required to divide ah'^ — ahe -\- ahd by ah. at^ — ahe + ahd ah — l^c-^d. We divide each term of the dividend by the divisor, then collect the partial quotients to obtain the complete quotient. Divisio]^-. 29 3. Let it be required to divide 8a^ + 43«2^»2 _ 22a^ + 245^— S8ab^ by 2a^ + U^ — 3a6. The operation may be conveniently arranged as follows : DIVIDEND. 8a^—22a^ + ^3a^b^—dSa¥ + 24.¥ 8a^—12a^-\-16a^^ DIVISOR. 2a2_3«&4-4Z>2 4a2_5a<$> + (Jj2--Quo. 1st Rem. = -10a^-\-27aW—38ab^-\-2i¥ —lOa^ + 15a^b^—20ab^ 2dEem. = 12aW—18aI)^ + 2U^ 12aW—18ab^+24:¥ Now, the term of the dividend, which contains the highest power of any letter as «, must be equal to the product arising from multiplying the term of the divisor which contains the highest power of that letter by the term of the quotient whicli contains the highest power of the same letter. Therefore, if we arrange the dividend and divisor according to the descending powers of a common letter as a, the first term of the quotient js found by dividing the first term of the dividend by the first term of the divisor. Hence, in this example the first term of the quo- tient is tt-t; = 4^2. 2a^ Again, the dividend is equal to the sum of the partial pro- ducts obtained by multiplying the divisor by each term of the quotient in succession ; and, therefore, if the product of the divi- sor by the term just found is subtracted from the dividend, the remainder must be equal to the sum of the partial products ob- tained by multiplying the divisor by the remaining terms of the quotient, and hence may be used as a new dividend to obtain the second term of the quotient. Proceeding in this manner, we find the complete quotieut to be 4^2 — 5ab + 6^. A similar course of reasoning is applicable when the dividend and divisor are arranged according to the ascending powers of a common letter. 30 FUNDAMENTAL PROCESSES. JR VLB. I. Arrange the dividend and divisor according to the poivers of some common letter. II. Divide the first term of the dividend by the first term of the divisor ; the result will be the first term of the quotient Miil- tiply the whole divisor by this term, and subtract the product from the dividend. III. Regard the remainder as a neiu dividend ; find the second term of the quotient in the same manner, and proceed with it as with t1i£ first term ; and so on. Remarks. — 1, Tlie situation of the divisor in regard to the dividend is a matter of arbitrary arrangement ; but by placing it on the right it is more easily multiplied by the several terms of the quotient as they are found. 2. Wlien there are more than two terms in the quotient, it is not neces- sary to bring down any more terms of the remainder, at each successive subtraction, than are required by the quantity to be subtracted. 3. It is evident that the exact division of one polynomial by another will be impossible, when the first term of the arranged dividend is not exactly divisible by the first term of the arranged divisor ; when the last term of arranged dividend is not divisible by the last term of the arranged divisor, or when the first term of any arranged remainder is not divisible by the first term of the divisor. EXAMPLES. 1. Divide a:^ + 1 by re + 1. Ans. x^ — x -]-l, 2. Divide 213^ + Sy^ by 3x -^ 2y^ Ans. Qx^ — 6xy + AyK 3. Divide a^ — 2a^ -{- 1)^ hj a — b. Ans. a^ -\- ab — b^. 4. Divide a^ — 2a^b — ?>al^ by a + ^>. Ans. a^ — Zab. 5. Divide 64^ — \f by 2x — y. Ans. 32a:5 _^ 16^4^ ^ %^y% 4. 4a4j^ _^ '^xy'^ + if. 6. Divide a^ -\-lr' by « + b. Ans. a* — a% -{■ aW — a¥ + b\ K. Divide a« — IQa^x^ + 64:^:6 by 4^2 4. ^^2 _ 4^ax. Ans. 16x^ + IQas^ + 12a^x^ + 4:a^x + aK Divisio:^:. 31 8. Divide 1 — ISz^ + Sl;^* hj 1 + 6z + 9z^. Ans, 1 — 62; + 9zK 9. Divide 81a^ + 16^12 _ 72a4J6 by 9^^ + 12^2^,3 ^ 4J6. ^?zs. 9^4 _ I2a2^,3 _j_ 4ja, 10. Divide a^ — x^y + a^y^ — a:2^^ + ^y^ — y^ by 0^ — y\ Ans. x^ — xy ■\- y\ 11. Divide x"^ + a? — ^x^ -\- hx — ^ hj x^ + 'Zx — 3. Ans. ic2 — a; + 1' 12. Divide a* + 2^252 _j_ 9^4 by ^2 _{_ ^ab + 3^»2. ^^5. ^2 _ 2ah + 3^>2. 13. Divide a^ — ¥ by a^ + 2a2Z> + 2«Z^2 ^ js, J?z5. a3 _ 2^25 ^ 2a52 _ J3. 14. Divide 7^ — 27?-\-l by a^ — 22; + 1. Ans. a:* + 2.r3 + 3a:2 _|. 2a; + 1. 15. Divide a^ + a^h -\- a^c — ahc — hH — hc^ by a^ — Ic. Ans. a -\-h + c. 16. Divide a^ + 1^ — (^ + dale hj a + h — c. Ans. a^ -\-b^ + d^ -\-ac-\-hc — ab. 17. Divide 1 — Qa:^ — Sa:^ by 1 + 2a; + x\ Ans. 1 — 22; + 3a;2 — 4a:3 ^ 5^4 _. ca^s ^ 7^^ _ ga;^ 18. Divide {a + b — c) {a — b + c) ib -\- c — a) hy a^ — b^ — c^ H- 2bc. Ans. b -\-c — a. 19. Divide {a^ — bcf + 8^c8 by a^ + ^ + C^). 11. ex — Zcxz -\- cx^. Ans. ex (1 — 3^ + a;). 12. Vl(^b7? — \h(?7^ — G(?:i?y. Ans. Zc^t? {4:C-b — 5cx — 2y). In resolving a polynomial into two factors, one of which shall be a monomial, it is common to divide by the greatest monomial that will exactly divide each of its terms ; but it is not necessary to do this. Thus, x^y + xy^ may be expressed under any one of the three following forms : ^y (^ + y)y ^ {^y + y% y {^^ + xy). FACTOEIKQ. 35 PRINCIPLES USED IN FACTORING BINOMIALS. 94. The difference letween the squares of two quantities is equal to the product of the sum and the difference of the quanti- ties (73). Thus, ^2 _ J2 ^ (^ + ^) (^ _ ^). 95. The difference iettueeti any tiuo like poivers of two quan- tities is divisible hj the difference between the quantities (87). Thus, 3 73 ^ ~ ' = a^-[-ab + b'^) whence, a^— b^= {a — b) {a^ + ab + b^). a — 96. The diff^erence between any two like even powers of tivo quantities is divisible by the sum of the quantities (87). Thus, ^4 _ J4 i- z=z a^— a% + ab^ — b^ ; whence, «*— J* = a -\-b {a + b){a^ — a%-\-a¥—b% 97. The sum of any tiuo like odd powers of tiuo quantities is divisible by the sum of the quantities (87). Thus, ^1±-?. = a2_ ab + b^', whence, a^ + b^= {a + b) (a^- ab + b^), a -J- u EXAMTLES. 98. Eesolve each of the following expressions into its prime factors : Ans. {a -\- c){a — c). Ans. (2x + y) {2x — y). Ans. (z -{'l)(z^ — z-\- 1). Ans. {a^-\-b^){a + b)(a-b), Ans. {x + y)(x^ — xhj + xhf — xy^ + y% Ans. (a^ + (^) {a^ + ^2) (« + c) {a — c). Ans. {x!^ + y^) (^^ + y) {^^ — y\ Ans. (1 + (^){l + c) (1 - c). Ans. (Sx + 1) (dxi -dx-i- 1). Ans. l2x - 1) (4a;2 -\- 2x i- 1). 1. a^- -c2. 2. ^x^- -f. 3. z^-[- 1. 4. a^- -¥. 5. o^ + yK 6. a^- -C8. 7. a^- -t' 8. 1- ■ (^. 9. 27a^» + 1. :o. 8x3 -1. 36 FUNDAMENTAL PEOCESSES. 99, Certain trinomials can be factored in accordance with the following principle : If two terms of a trinomial are positive squares, and the other term is twice the product of the square roots of these two, the trinomial is equal to the square of the sum, or the square of the DIFFERENCE, of these squarc roots, according as that other term is positive or negative (71-73). Thus, a^ ± 2ab -{^ = (a ± hf z=(a±h)ia± h). EXAMPLES. Resolve each of the following ten expressions into two equal fectora: 1. ir» -f 2fla; + a\ Ans. {x ■\-a){x-\- a), 2. 7n^ -f w* + 2^2/^2. Ans. (m^ + n^) (m^ + n^)- 3. 16a^b^m^ — Sa^hhn + 1. Ans. (^a^U^m. — 1) (4:a^li^m — 1). 4. 36a2 -f 12ab + R Ans. (6a + b) (6a + b). 5. c2 — lOcd + 25^. A71S. {c — 5d) (c — 5d). 6. ah^ + 2aa^y + f. Ans. (ax^ •+ y) {ax^ + y), 7. 25x^y* + 20x3^2^ + 4cZ^. Ans. (5xy^ + 2z) {5xy^ + 2z). 8. Ore* — 63^z^ + ^. ^ns. (3a^— z^) (Sx^ — z^), 9. (a + J)3_ 2(a + b){c + d) -{- (c -\- d)\ Ans. [a-^b — {c-\-d)'][a + b — (c-\- d)\ 10. a^*" 4- 2a'»5" + ll^\ Ans. («"* + 6») (a"* + ^>"). 11. Can x^ — 2xy — y^ be resolved into two equal factors? 12. Resolve ^y^(? — (^ 4- c^ — fl2)2 into its prime factors. Here we have the difference between two squares; hence, 4J2c2 _ (&2 +c2 — a2)2 = (2Jc 4. J2 _^ ^2 _ ^2) (2^^ _ ^>2 _ ^2 4. «2), But, 2Jc+J2+c2— a2=(5 + c)2— a2=(J4-c + a)(5 + c— a), and 25c — 52 _ ^ _^ ^2 = «2 _ (J2 _ 2JC + C2) = a2 _ ( J _ c)2 =: (a + 5 — c) (a — 5 + c). Therefore, 452^_(^4c2— a2)2=(54-c + a)(5 + c— a)(a + 5— c)(fl5— 5 + c). FACTOKING. 37 13. Resolve m^ + ^mn -^-n^ — a^ -\-%ab — W into its prime factors. This expression may be put under the form, But m^ + 2mn -]-n^=i{m-\- n)% and a^—2ab + l^={a — b)^; hence, m^ + 2mn +n^ — a^ + 2ab — b^ = {m -{- nf — {a — bf = {m -\- n -\- a — b)(m + n — a -}- b). 100. The following formulae may be yerified by performing the operations indicated in their second members : x^-\- {a-\-b)x + ab = {x-}- a){x-\-b) . . . (1), a^—{a + b)x-{-ab=(x — a){x — b). . . (2), a^-\- (a — b)x — ab = {x + a)(x — b) , . . (3), xi—(a — b)x — ab = {x — a)(x-\-b) . . . (4). From (1) and (2) it follows that Any trinomial of the form ofx^ + mx + n, or of the form of xi — mx -\- n, can be resolved into two binomial factors, if the coefficient of the second term is equal to the sum of two quafitities whose product is equal to the third term. From (3) and (4) it follows that Any trinomial of the form of x^ + mx — n, or of the form of x^ — mx — n, can be resolved into two binomial factors, if the coefficient of the second term is equal to the difference of two quantities whose product is equal to the third term. It will be observed that we have used the words sum and difference in their arithmetical sense. In the first form both of the terms in each binomial factor are positive. In the second form the second term of each of the binomial factors is negative. 38 FUNDAMENTAL PROCESSEft. In the third form the second terms of the binomial factors have contrary signs, the larger being positive. In the fourth form the second terms of the binomial factors have contrary signs, the larger being negative. EXJ.M1*LES. 1. Eesolve ar^ + 5rr + 6 into two binomial factors. This comes under the first form. Let us now seek two num- bers whose sum is 5 and product 6. We see that these numbers are 2 and 3 ; hence, a^ -\- ox -h 6 = {x -\- 2) (x -\- 3). 2. Resolve x^ — 9x -\- 20 into two binomial factors. This comes under the second form; and, therefore, since 4 + 5 = 9, and 4 x 5 = 20, we have a^ — 9x + 20z=z(x — 4:){x — 5). 3. Resolve a:^ + 4a; — 32 into two binomial factors. This comes under the third form; ond, therefore, since 8 — 4 = 4, and 8 X 4 = 32, we have a5 + 4a; — 32 = (a; + 8) (a; — 4). 4. Resolve x^ — ox — 6Q into two binomial factors. This comes under the fourth form; and, therefore, since 11 — G = 5, and 11 x 6 = 66, we have a^ — 5x — 66 = {x-{-6)(x — 11). Resolve each of the following ten expressions into two bino- mial factors : 5. ic2 + 8a; + 15. Ans. {x + 3) {x + 5). 6. a;2 + 8a; 4- 7. Ans. (x + 1) (x + 7). 7. a:2 _ a: _ 6. Ans. (x + 2) (a; - 3). S. a^ -{- Sx -\- 2. Ans. (x + 2) (a; + 1). Q, a^ — x — 72. Ans. (x + 8) {x — 9). 10. a:2 _ 13a; + 42. Ans. (x - 7) {x - 6). 11. x^-x — 42. A?is. {x — 7) {x + 6). FACTORING. 12. a^-x-2. 13. x^ + 2^; - 35. 14. x^ — x — 30. 39 Ans. (x 4- 1) {x — 2). ^7^5. {x — 5)(x -{- 7). A71S. {x -{■ 6) {x — 6). 101. Since a^p-{-{a-\-b)xP-{-ab=(xP-{-a){xP-{-b) . . (1), x^P—{a-\-b)xP-{'ab = {xP—a){xP — b) . . (2), x^p+{a—b)xP-ab={xP-{-a){xP—b) . . (3), and a^P—{a — b)xP—ab={xP—a){xP-\-b) . . (4), it follows that such expressions as a;^ -f- 8x^ + 15, x^ — Idx^ +42, a^ -{-Sx^ -{- 2, and a;^ — 5a;^ — 66 may be resolved into binojnial factors in the same manner as the examples of the preceding Article. Thus, a4_|_8a^ + 15 = (a:2+3)(ic2+5), afi-133^-{-4:2=(a^-'7){a>^—6), .-^8 + 30:*+ 2 = (x^-\-2)(x^-{-l), x^—6a^—66 = (x^^e)(x^—ll). MISCELIjANEOUS examtt^es, 102, Kesolve each of the following expressions into its prime factors : 1. 0!^ — X. A71S. {x — l){x-\- 1) X. 2. 3ax^ + 6axy + 3ai/^. Ans. 3a (x -{- y) (x -\- y). 3. 2cx^ — 12cx + 18c. Ans. 2c (x — 3) (x — 3). 4. 27a — ISax + 3ax^. Ans. 3a (3 — x) (3 — x). 5. 3m^n — 3mn^. Ans. 3mn{m -\- n) (m — 7i), 6. 2sfi -{-6x — 8. Ans. 2 {x + 4) (a; — 1). 7. 2a^ + ^xi — 70a;. A7is. 2x {x + 7) (a; — 5). 8. a^ — W — c^ — %bc. Ans. (a^b-^c) (a—b—c). 9. ac -{- ad + bd -{- be. Ans. a(c -^ d) -\- b {c -j- d) = (a + b) (c + d). 10. am + 2bx + 2ax + bm. Ans. a {m + 2^;) -\- b (m -{- 2x) = (a + b) (m + 2x), 11. a^ — ab^. Ans. a{a -\- b){a — b). 12. 7x^—12x-ir^. Ans. X {7x — b) — {7x — 6) = {x — l) (7x - 5). 13. x^ — x^ — 2x. Ans. x{x-{-l){x-- 2). 14. 7^ — \W + 9. A71S. {x^ - 9) (x^ -l)=.{x + 3)(x- 3) {x + 1) (^ - 1). 15. x^ — 17ji? + 16. Ans. {x + 4) (a; — 4) \x + 1) {x — 1). 40 FDIfDAMENTAL PROCESSES. 103. SYNOPSIS FOR REVIEW. RELATION TO MULTIPLICATION. TERMS USED . . . Dividend. Divisor. Quotient. M0N0M.-5-M0N0M. Law of coefficients. Law of exponents. {Like. Law of signs . . ( Unlike. POLYNOM. ^POLTNOM. TERMS USED 1. Coefficient. When impossible. ^ 2, Exponent. 3. Literal paH i Investigation for rule. Rule. Proof. When impossiblb. r Prime quantity. J Relatively prime. j Composite quantity. I Prime factor. Eh O <1 MONOMIALS— Rule. POLTNOML^ WITH MONOMLiL FACTOR— RULB. BINOMIALS— Principles. TRINOMIALS . First form. Second form. Third form. Fourth form. CHAPTER III. POSITIVE AND NEGATIVE QUANTITIES. 104. In Algebra we are sometimes led to a subtraction which cannot be performed, because the subtrahend is greater than the'^minuend. In the equation a — {h -\- c)=a — l) — c, it is implied that J + c is less than a ; but suppose that « = 7, J = 7, and c = 3 ; we shall then have 7- 10 = 7-7-3 = - 3. In writing this equation, we may be understood to make the following statement : It is impossible to take 10 from 7 ; hut if 7 he taken from 10, the remainder will he 3. 105. It might at first seem unlikely that such an expression as 7 — 10 should occur in practice ; or that if it did occur, it would only arise either from a mistake which could be instantly corrected, or from an operation being proposed which it was obvi- ously impossible to perform, and which must therefore be aban- doned. As we proceed, we shall find, however, that such expres- sions occur frequently. It might happen that a — h appeared at the beginning of a long investigation, and that it was not easy to decide, at once, whether a were greater or less than h. The object of this chapter is to show that in such a case we may proceed on the hypothesis that a is greater than h, and that if it should finally appear that a is less than h, we shall still be able to make use of our investigation. 42 POSITIVE AND NEGATIVE QUANTITIES. 106. Suppose a merchant to gain in one year a certain num- ber of dollars, and to lose a certain number of dollars in the fol- lowing year; .what change has taken place in his capital? Let a denote the number of dollars gained in the first year, and J) the number of dollai'S lost in the second year. Then if a is greater than h, the capital has been increased hj a — b dollars. If b is greater than a, the capital has been diminished by ^ — a dollars. In this latter case a — Z> is the indication of what would be pronounced, in Arithmetic, to be an impossible subtraction ; but, in Algel)rii, it is found convenient to indicate the change in the capital by a — b, whether a is greater or less than b, which we may do by means of an appropriate system of interpretation. Thus, if « = $400 and b = ^500, the merchant's capital has suf- fered a diminution of $100. The algebraist indicates this in sym- bols thus : 400 — 500 = — 100 ; and he may convert his symbols into words by saying that the capital has been increased by — $100. This language is far re- moved from that of ordinary life ; but if the algebraist under- stands it and uses it consistently, his deductions will be sound. 107. There are numerous instances in which it is convenient to be able to represent, not only the magnitude, but also what may be called the quality of the things about which we may be reasoning. In business transactions a sum of money may be gained or it may be lost ; in a question of chronology we may have to distinguish a date before a given epoch from a date after that epoch ; in a question of position we may have to distinguish a distance measured to the north of a certain point from a dis- tance measured to the south of it ; and so on. These pairs of re- lated magnitudes the algebraist distinguishes by means of the signs + and — . Thus, if the things to be distinguished are gain and loss, he may denote by -f a a gain of a dollars, and then he will denote by — « a loss of the same extent. 108. In Arithmetic we consider only the numbers repre- sented by the symbols 1, 2, 3, 4, etc., and intermediate fractions. POSITIVE AND NEGATIVE QUANTITIES. 43 In Algebra, besides these, we consider another set of symbols, — 1,-2, — 3, — 4, etc., and intermediate fractions. The relation between positive and negative quantities is ex- hibited to the eye in the following diagram, where the distance from the zero point A to any point in the indefinite line BC is considered positive or negative according as that point is on the right or on the left of A : Negative. PoBitive. I I I I I I I I I I I I I I i I I I I -9-8-7-6-5-4-3-2-1 012 3456789 109. In the preceding chapter we have given rules for the Addition, Subtraction, Multiplication, and Division of algebraic expressions. Those rules were based on arithmetical notions, and were shown to be true so long as the expressions represented pos- itive quantities. Thus, when we introduced such an expression as a — h, we supposed a and h to be positive quantities, and a to be greater than h. But as we wish hereafter to include negative quantities among the subjects of our reasoning, it becomes neces- sary to recur to the consideration of these primary operations. Now it is found convenient to have the laws of the fundamental operations the same whether the symbols denote positive or nega- tive quantities, and we may secure this convenience by suitable definitions. 110. The Absolute Value of a quantity is the number represented by that quantity taken independently of the sign which precedes it. Two quantities are equal when they have the same absolute value and are preceded by like signs. Two quan- tities may have the same absolute value and be unequal. Thus, + 7 and — "^ have the same absolute value, but they are not equal. Sucn quantities as + 7 and — 7 are sometimes said to be numerically equal. 111. In Arithmetic the object of addition is to find a number which shall contain as many units as all the given numbers taken together. This notion is not applicable to negative quantities ; that is, we have as yet no meaning for the phrase " add — 3 to + 5," or " add — 3 to — 5." We shall therefore give a meaning 44 POSITIVE AND NEGATIVE QUANTITIES. ' to the word add in such cases, and the meaning we propose is determined by the following RULES. I. To add two qiiantities tuith like signs, add their absolute values, and prefix the common sign to the sum. n. To add tivo quantities with unlike signs, subtract the less absolute value from the greater, and prefix to the remainder the sign of that quantity which has the greater absolute value. Thus, the sum of 3 and 5 is 8 ; the sum of — 3 and — 5 is — 8 ; the sum of — 3 and 5 is 2 ; and the sum of 3 and — 5 is —2. 112. That the rules of the preceding Article are not alto- gether arbitrary will appear from the following illustrations: 1. Suppose a man starts from A in the line BC (108), and travels first 3 miles toward the right, and then 5 miles further in the same direction ; his final distance from A will be 8 miles in the positive direction. This may be considered as an intei-preta- tion of the 8 obtained by adding 3 to 5. 2. Suppose a man starts from A and travels first 3 miles toward the left, and then 5 miles further in the same direction ; his final distance from A will be 8 miles in the negative direction. This may be considered as an interpretation of the — 8 obtained by adding — 3 to — 5. 3. Suppose a man starts from A and travels first 3 miles toward the left, and then turns and travels 5 miles toward the right ; his final distance from A will be 2 miles in the positive direction. This may be considered as an interpretation of the 2 obtained by adding — 3 to 5. 4. Suppose a man starts from A and travels first 3 miles toward the right, and then turns and travels 5 miles toward the left ; his final distance from A will be 2 miles in the negative direction. This may be considered as an interpretation of the — 2 obtained by adding 3 to — 5. POSITIVE AND NEGATIVE QUANTITIES. 45 113. In Algebra, addition does not necessarily imply aug- mentation in an arithmetical sense ; nevertheless, the word sum is used to denote the result. Sometimes, when there might be an uncertainty on the point, the phrase algebraic sum is used to distinguish such a result from the arithmetical sum which would be obtained by the addition of the absolute values of the terms considered. 114. In arithmetical subtraction we have to take one num- ber, which is called the subtrahend^ from another, which is called the minuend, and the result is called the remainder. The re- mainder, then, may be defined as that number which must be added to the subtrahend to produce the minuend, and the object of subtraction is to find this remainder. We shall use the same definition in algebraic subtraction; that is, we say that in subtraction, we have to find the quantity which must be added to the subtrahend to produce the minuend. B, ULE. Change the sign of every term in the subtrahend, and add the result to the minuend ; the sum thus obtained will be the remain- der required, 115. By the rule of Art. 114, the following results are obtained : 1. Subtracting 3 from 8, w^e obtain 5. 2. Subtracting 8 from 3, we obtain — 5. 3. Subtracting — 3 from — 8, we obtain — 5. 4. Subtracting — 3 from 8, we obtain 11. 5. Subtracting 8 from — 3, we obtain — 11. Let us now recur to the diagram (108) and see how these results are to be interpreted. 1. Starting from the subtrahend 3, we must move a distance of 5 toward the right — that is, in the positive direction — in order to reach the minuend 8 ; hence, the remainder is 5. 2. Starting from the subtrahend 8, we must move a distance of 5 toward the left — that is, in the negative direction — in ordei to reach the minuend 3 ; hence, the remainder is — 5. 46 POSITIVE AND NEGATIVE QUANTITIES 3. Starting from the subtrahend ~ 3, we must move a dis- tance of 5 towai'd the lefty in order to reach the minuend — 8 ; hence, the remainder is — 5. 4. Starting from the subtrahend — 3, we must move a dis- tance of 11 toward the right, in order to reach the minuend 8; hence, the remainder is 11. 5. Starting from the subtrahend 8, we must move a distance of 11 toward the lefty in order to reach the minuend — 3 ; hence, the remainder is — 11. 116. In the multiphcation of one monomial by another there are four cases to be considered. 1st. When the multiphcand and multipher are positive. 2d. When the multiphcand is negative and the multiplier positive. 3d. When the multiplicand is positive and the multiplier neg- ative. 4th. When the multiphcand and multiplier are negative. It was shown in Art. 70 that {a — h)(c^d) = ac — ad — hc + hd . . . (1) Kow, although the result was obtained on the supposition that ayh and cy d, it will be convenient to assume that (1) is true for all values of the letters. In this way uniformity of re- sults will be secured. Suppose J = 0, and J = 0; then (1) becomes (a — 0)(c~0) = flc — flxO — Oxc + OxO; that is, a X c = ac. Suppose a = 0, and ^ = ; then (1) becomes (^^h){^c) = -hc. Suppose J — 0, and c = ; then (1) becomes a{—d) = —ad. Suppose a = 0, and c = ; then (1) becomes {-h){-d)=hd. Hence, to multiply one monomial by another, we have the following POSITIVE AKD NEGATIVE QUANTITIES. 47 MULE. Multiply without considering the signs, and prefix -f or — to the 2>roducty according as the two monomials have like signs or unlike signs. 117. In division we have the product of two factors, and one of them given to find the other. Therefore, since the product of the divisor and quotient must he equal to the dividend, we have for the sign of the quotient the following MULE, When the dividend and divisor have like signs, the quotient must have the sign + ; tuhen the dividend and divisor have un- like signs, the quotient must have the sign — . 118. The words greater and less are often used in Algebra in an extended sense. We consider a greater than b, or b less than a, when a — J is a positive quantity. This is consistent with ordinary language when a and b are positive numbers, and it is found convenient to extend the meaning of the words greater and less, so that we may still consider a greater than b, when a or b is negative, or when both are negative. Thus, in algebraic lan- guage, 1 is greater than — 2, and — 2 is greater than — 3 ; for 1 -I (— 2) = + 3, and _ 2 — (- 3) = + 1 (114). In this extended or algebraic setise a negative quantity may be said to be less than zero. Thus, — 2 is algebraically less than zero; forO — (— 2)= + 2. 119. That a negative quantity is not less than zero in the arithmetical sense may be shown thus : It is evident that ^ = ^ (H'^)- Now, if — 1 is less than zero, much more will it be less than -|- 1 ; that is, the nu- merator of the fraction ^ will be greater than its denominator; hence, the numerator of the fraction ~ will be greater than its denominator ; therefore, — 1 is less than + 1 and greater than -f 1, which is absurd. CHAPTEE lY. GREATEST COMMON DIVISOR AND LEAST COMMON MULTIPLE. GREATEST COMMON DIVISOR. 130. A Common Divisor or Common Measure of two or more quantities is any quantity that will exactly divide them. Thus, a, J, and ab are common divisors of ab^ and abx. Any factor common to two or more quantities is a common divi* sor of them. 121. Commensurable Quantities are those which have a common divisor. Thus, ah^ and abx are commensurable. 122. Incom^mensurable Quantities are those which have no common divisor. Thus, ab^ and cdx are incommen- sunible. 123. The Greatest Commmi Divisor of two or more quantities is that common di\isor of them which contains the greatest number of prime factors. Thus, Qdh^ is the greatest common divisor of Via^bx^ and l%a^cxz. For brevity, Ave shall sometimes use G. C. D. for the phrase greatest common divisbVi 124. To find the G-. C. D. of two or more quantities. Since every factor of a quantity is a divisor of that quantity, it follows that all the factors common to t^vo or more quantities are all the common divisors of those quantities. Again, since the product of any number of factors of a quantity is a divisor of that quantity, it follows that the product of all the factors common to two or more quantities is a common divisor of those quantities. GREATEST COMMOlf DIVISOR. 49 Moreover, this product is the greatest common divisor, for it con- tains all the factors common to the given quantities ; therefore, if another factor were introduced into this product, the result would not divide at least one of the given quantities. Hence, when the given quantities can he resolved into prime factors by methods already explained, the G. C. D. may be found by the following RULE. Resolve each of the given quantities into its prime factors, then the product of all the prime factors which are common to those quantities will ie the 0. C. D. required. EXAMPLES. 1. What is the G. 0. D. of ^%x, Qalh?, and IWl^c^da^'i Aa%x = 2 • 2 • aal)x, Qab^a:^= 3 * 2 * abixxx, and lOaWc^dx^ = 2 • baahhhccccdxx. The common factors are 2, a, h. and x ; hence, 2dbx is the G. C. D. required. That ^ahx is the greatest C. D. is evident; for if an additional factor, as a, be introduced, the product 2a^x will not be a divisor of all the given quantities. 2. What is the G. 0. D. of ^am^ + ^bm^ and Zan + Un ? 4flm2 4- ^hm^ = 2 • 2 • mm {a -f 5), and ^an -f ^hn = 3 * 7^ (a + &). The only factor common to both the given quantities is a -\- b; hence, it is the G. 0. D. required. Remakk. — When there is only one common divisor, as in the preceding example, it would seem to be improper to speak of it as the greatest G. D. Nevertheless, since the common divisor, in such cases, is found by the gen- eral rule, we shall, for the sake of uniformity, call it the G. C. D. 3. What is the G. C. B. of a^ — y^ ond a? — y^? a?-y^={x-y) (x^ -\-xy + y^), and x^ — y^ = (x — y){x + y); hence, x — y ia the G. 0. D. 4 50 GREATEST COMMON DIVISOE. 4. What is the G. C. D. of aa^ -{- 3ax + 2a and ax^ — ax — 2a? aa^ + 3ax~{-2a = a{x + l){x+ 2), and ax^ — ax — 2a = a(x -\- 1) (a; — 2) ; hence a(x + 1) is the G. C. D. 6. What is the G. C. D. of a.-2 — 7a; + 12 and a« — Sa- + 15 ? Ans. x — 3. 6. What is the G. C. D. of a** — a;— 12 and apu^—lax + 12a? Ans. X — 4. 7. What is the G. C. D. of 2a:3 ^ g^. _ 3 and 2a:3 + 2a; — 24? Ans, 2 (a; + 4). 8. What is the G. CD. of 2ar»4- 4a;y + 2y^ and Zax^-\-^axy-\- 3a^ ? Ans, (x + y) (x -\- y). 125. It is sometimes rery difficult, if not impossible, to re- solve the given quantities into their prime factors by inspection. We shall therefore proceed to demonstrate the following rule, which is more general in its application : RULE FOR FINDING THE G. C. D. OF TWO ALGEBRAIC EXPRESSIONS. I. Let A aiid B denote the two expressmis j let them he ar- ranged according to the desceiiding powers of some common letter, and suppose the exponent of the highest power of that letter in A not less than the exponent of the highest power of the same letter in B. n. Divide A hy B ; then make the remainder a divisor and B the dividend. Again, make the new remainder a divisor and the preceding divisor the dividend. Proceed in this way until there is no remainder ; then the last divisor is the G. C. D, required. The demonstration of the preceding rule depends upon the following Lemmas: Lem. I. — If P is a divisor of A, then it will be a divisor of mK. For, since P is a divisor of A, we may suppose A = aP; then mA=i ma? (42, 4) ; but P is a divisor of maV ; therefore, since wA = maV, P is a divisor of mA. AiB pB p B ^0 ^ c D r\) r GREATEST COMMON DIVISOR. M Lem. II. — If P is a divisor of A and B, then it will be a divisor of mK ± nB. For, since P is a divisor of A and B, we may suppose A=«P, and B=Z'P; then mK±nB={ma±nb)^; hence P is a divisor of mA. ±_ wB. Let A and B denote the two expressions whose G. C. D. is to be found; let them be aiTanged according to the descending powers of some common letter, and suppose the exponent of the highest power of that letter in A not less than the exponent of the highest power of the same letter in B. Divide A by B ; let p denote the quotient, and the remainder. Divide B by ; let 5' denote the quotient, and D the remainder. Divide C by D ; let r denote the quotient, and suppose there is no remainder. Now, since the dividend is equal to the product of the divisor and quotient, increased by the remainder, we have the three fol- lowing equations : A=i?B + C . . . (1), B = ^C + D . . . (2), C = rD . . . (3). We shall first show that D is « common divisor of A and B. D is a divisor of C, since C = rD ; hence (Lem. I), D is a divi- sor of q(jy and, therefore (Lem. II), it is a divisor of qO -\-D\ that is, D is a divisor of B. Again, since D is a divisor of B and C, it is a divisor of joB + C ; that is, D is a divisor of A. Hence D is a divisor of A and B. We have thus shown that D is a conmion divisor of A and B ; we shall next show that it is their greatest common divisor. Equations (1) and (2) may be written as follows: A-pB = Q . . . (4), B-^C = D . . . (5) (42,3). Now, every common divisor of A and B is a divisor of A~joB, that is, C (Lem. II) ; hence every common divisor of A and B is a common divisor of B and C. Similarly, every common divi- 52 GREATEST COMMOJS^ DIVISOR. sor of B and C is a common divisor of C and D. "We have thus shown that D is a common divisor of A and B, and that every common divisor of A and B is a divisor of D. But no expression of a higher degree than D is a divisor of D. Therefore, D is the G. C. D. required. Cor. 1. — Every common divisor of A and B is a divisor of their G. C. D. ; and every divisor of their G. C. D. is a common divisor of A and B. Cor. 2. — Suppose we have to find the G. C. D. of A and B ; and at any stage of the process suppose we have the expressions K and R, one of which is to be a dividend and the other a divi- sor. Let R = ?wS, where m has no factor which K has ; then m may be rejected ; that is, instead of continuing the process with K and R, we may continue it with K and S. For, by what has been already shown, we know that A and B have the same com- mon divisors as K and R have. Now, any common divisor of K and S is a common divisor of K and R. Therefore, any common divisor of K and S is a common divisor of A and B. Again, any common divisor of K and R is a common divisor of K and wzS, for ??iS = R But m has no factor which K has. Therefore, any common divisor of K and R is a common divisor of K and S. Hence, A and B have the same common divisors as K and S have. Cor. 3. — A factor of a certain kind may be introduced at any stage of the process. Suppose we have to find the G. C. D. of A and B ; and at any stage of the process suppose we have the expressions K and R, one of which is to be a dividend and the other a divisor. Let L = wK, where n has no factor which R has ; then n may be introduced ; that is, instead of continuing the process with K and R, we may continue it with L and R. For A and B have the same common divisors as K and R have ; and any common divi- sor of K and R is a common di\nsor of L and R. Therefore, any common divisor of A and B is a common divisor of L and R. Again, any common divisor of L and R is a common divisor of wK and R But n has no factor that R has. Therefore, any GREATEST COMMOiT DIVISOR. 63 common divisor of L and E is a common divisor of K and E. Hence A and B have the same common divisors as L and E have. Jii USTJIJLTIONS. 1. Find the G. C. D. of x^—6x-^S and 4:X^—21x'^-\-16x+20. The operation may be arranged thus : 4a^ — 21x^-\-15x-\-20 a;2 _ 6a; 4- 8 4:^3 — 24x2+32^ 4a; + 3 3a;2— 17a; -f 20 Sx^- 18a; + 24 a;8 _ 6a; + 8 a;8— 4a; a;-4 a;-2 — 2a; + 8 — 2a; + 8 Hence a; — 4 is the G. 0. D. required. 2. Find the G. C. D. of a;2 + 5a; + 4 and a^ -{. 4:X^ ^ 6x + 2. a;2 + 5a; + 4 a^-\- 4a;2 + 5a; + 2 x^-i- 5x^ -{- Ax a;-l a;2+ a; + 2 x^ — 5a; — 4 a;8 + 5a; + 4 a^^-l- a; 6x -\- 6 4a; + 4 4a; + 4 a; 4 6"^ 6 This example introduces a new point for consideration. The last divisor here is Gx -{- 6; this, according to the rule, must be the G. 0. D. required. When a;^ + 5a; + 4 is divided by 6a; + 6, X 4 the quotient is ^ + - . If the other given expression be divided by 6x + 6, the quotient will be -^ + « + q • O /i O It may at first appear that 6a; + 6 cannot be a divisor of the two given expressions, since the quotients contain fractions. But we observe that in these quotients the letter x does not appear in 54 GREATEST COMMON DIVISOR. X 4 the denominator of any fraction. Such expressions as - + - and x^ X \ — -f 5 + ^ are said to be entire with reference to x, Xi At O When we say that 6a; -|- 6 is the G. C. D. of the two given expressions, we mean that no common divisor can be found which contains a higher power of x than 6a: + 6. Other common divi- sors may be found which differ from this so far as respects numer- ical coefficients only. Thus, 3a; + 3 and 2a; + 2 are common di\nsors. Again, a; + 1 is also a common divisor, and the corres- ponding quotients are a; 4- 4 and a:^ -f- 3a; + 2. We may then conveniently take a; -f 1 as the G. C. D., since the quotients do iiot contain fractional coefficients. We may avoid fractional coefficients by proceeding as in the following example : 3. Find the G. C. D. of Zx^ — V^t? + 15a; + 8 and a* — 2a:* — 6a;3 ^ 4^ ^ 13a; ^ 6. 3a« — 10a;s+ 15a: + 8 32« — 6a;* — 18a;8 + 12a;2^ 39a; +18 a;«— 2a:*— 6a*+4a;»+ 13a;+ 6 G^4- 8a:3_i2a4j_24a;— 10 Before proceeding to the next division, we may reject the factor 2 from every term of the new divisor (125, Cor. 2), and multiply every term of the new dividend by 3 (135, Cor. 3). We then continue the operation thus : 3a:5— 6.x-*— 18a;3 + 12ay^ + 39a; + 18 [3a;^ + 4a;^— 6a;2— 12a:— 5 3a:5_^ 4y4_ 6a;3_i2a:2— 5a: \ x ' -^lOa;*— 12a:3^24a;24.44a;^18 Rejecting the factor 2 from every term of the last remainder, and multiplying the result by 3, we have the expression, — 15a:* — 18a;3 _^ 36^^ ^ ee^: + 27. We then continue the operation thus : — 15a;* — 18a:3 ^ 36^:2 4. gga: + 27 I 3a:* + 4a:^ — 6a:^ — 12a: — 5 — 15a:* — 20y3 + 30a;g + 60a; + 25 | —5 2a;3^ 6ar5+ 6a;+ 2 GREATEST COMMOIT DIVISOR. 65 We now reject the factor 2 from every term of this remainder, and continue the operation thus : — 03^ — Idx^ — Ibx — 5 — 5a^ — 15a:=2 — 15a: — 5 Hence o:^ -\- djfi + 3x + 1 is the G. C. D. required. 126. Suggestions.— Suppose the given expressions A and B to contain a common Victor F, which is obvious on inspec- tion. Let A = aF, and B = bF. Then F will be a factor of the G. C. D. (120). We may then find the G. C. D. of a and b, and multiply it by F ; the product will be the G. 0. D. of A and B. In like manner, if at any stage of the operation we perceive that a certain factor is common to the dividend and divisor, we may omit it and continue the operation with the remaining fac- tors. The factor omitted must then be multiplied by the last divisor obtained by continuing the operation ; the product will be the G. C. D. required. 127. To find the G. C. D. of three Algebraic Ex- pressions, A, B, and C. Find the G. C. D. of two of them, as A and B. Let D denote this G. C. D.; then the G. 0. D. of C and D will be the G. C. D. of A, B, and 0. For every common divisor of and D is a com- mon divisor of A, B, and C (125, Cor. 1). Again, every com- mon divisor of A, B, and is a common divisor of C and D. Hence the G. C. D. of C and D is the G. C. D. of A, B, and C. 128. Find the G. C. D. of 1. a^ — Sx-j-2 and x^ — x — 2. Ans. x — 2. 2. a:3+3a^i+4a;4-12 and y^ + 4.x^ + 4.x + Z. Ans. a; + 3. 3. :i^j^y?j^x—^ and a?+37?-^6x-\-3. Ans. x^ f 2a; + 3. 56 LEAST COMMON MULTIPLE. 4. a^ H- 1 and a^ + fux^ + ma; + 1. Ans. x + 1. 5. 6a:* — laa^ — 20ah: and 3a^-\- ax-^ 4a\ Ans. 3x + 4«. 6. a;5 — ^ and gfl _ y2^ j^ns. X — y. 7. ^a? — Ua? + 23a; — 21 and Qa^ -\- t? — ^^ -^ 21. ^ws. 3ic — 7. 8. a;* — 3a;8 ^ 22^3 + a; — 1 and a^ — xi^%x + 2. Ans. X — 1. 9. a4 _ 7a^ 4. 8a?« + 28a; — 48 and a:8 — 8a:2 ^ 19^; _ 14 Ans. X — 2. 10. a;* — a:® + 2arJ 4- a; + 3 and a;* + 2a;8 — a; — 2. Ans. a;^ + a; 4- 1. 11. 4a;* + 9a;8 _j. 22?» — 2a; — 4 and 3a;8 + 5a?5 — a; + 2. Ans. a; + 2. 12. 2a;* — 12a;8 + 19a;8 — 6a; + 9 and 4a« — 18a;2 + 19a; — 3. Ans. a; — 3. 13. 6a;* + a;* — a; and 4a;8 — 6a;2 — 4a; + 3. Ans. 2x — 1. 14. 2a;* + lla;® — ISx^— 99a; — 45 and 2a;3 _ 7a;2_ 45^. _ 21. A71S. 2xi -\-7x-^ 3. 15. a;s — 9a;2 ^ 26a; — 24, x^ — 10a^» + 31a; — 30, and a^ — liar' + 38a; — 40. Ans. x — 2. LEAST COMMON MULTIPLE. 129. When one quantity is divisible by another, the first is called a Multiple of the other. Thus, 6 is a multiple of 2, and a J is a multiple of d. 130. A Common Multiple of two or more quantities is a quantity which is divisible by each of them. Thus, 12 is a common multiple of 2 and 3, and 20a;y is a common multiple of 2a; and 6y. 131. The Least Coimnon Multiple of two or more quantities is that common multiple of them which contains the least number of prime factors. Thus, 6 is the least common mul- tiple of 2 and 3, and 10a;y is the least common multiple of 2a; and 5y. LEAST COMMON" MULTIPLE. 57 For brevity, we shall sometimes use L. 0. M. for the phrase least common multiple. 133. To find the L. C. M. of two or more quantities. It is obvious that the L. C. M. of two or more quantities must contain all the factors of each of them, and no otlier factors. Hence, when the given quantities can be readily resolved into their prime factors, the L. 0. M. may be found by the following JRULE. I. Resolve each of the given quantities into its prime factors. II. Multiply one of the given quantities ly the product of such prime factors of the other quantities as are not found in it; the result will be the L. C. M. required. Cor. — If the given quantities are relatively prime (90), their product is their L. C. M. Thus, the L. C. M. of lab and Qcd is 4:'Zabcd. EXAMPJLES. 1. Find the L. 0. M. of ^x^y and 12xtj\ Wy =z3'3'x'X'y, and 1 2xy^ = 3 '2'2' x' y y; hence the L. C. M. is 9x^y x2'2'y = ddxy. 2. Find the L. 0. M. of 4«2^, 6a% and 10^2. 4a2J2 ^ 2 • 2a^^ 6a^ = 2 • da% and lOa^^ = 2 ' 6a^ ; hence the L. C. M. is 4^252 x 3 x 5ax^ = QOa^b^x^ It is not necessary, when the given quantities are monomials, to actually separate the literal parts into prime factors, since the exponent of any letter shows how many times it occurs as a factor. 3. Find the L. 0. M. of a^x — 2ahx -f Vhi and a^y — %. ah^—2alx-\-l^x={a—l)(a—l)x, and a^y—¥y={a + li){a--l)y\ hence the L. 0. M. is {a^x — 2ahx -f l^x) {a -\- l)y. 58 LEAST COMMON MULTIPLE. 4. Find the L. C. M. of 2aWcx, 3a^b(^x% ^acx, 9cVo, and 24fl8. Ans, naWc^x^^ 6. Find the L. C. M. of l^ax, 40^a;, and 2baWx^. Ans. 400a'Z»5^. 6. Find the L. C. M. of a:2 __ 3^; 4. 2 and a^ — 1. Ans. a? — %3? — x -{■% 7. Find the L. C. M. of a^c + ^ and a^ - 1^. Ans. ahc — ciRtx + ab^ — hhc. 8. Find the L. C. M. of a^ + 2aJ + 53 and a^ — %db + V^. Ans. (a2_^)2. 9. Find the L. C. M. of a^ + V^ and a^ — 1^, Ans. a^ — J*. 10. Find the L. C. M. of o? — x and x^ — 1. A71S. a? — X. 11. Find the L. C. M. Of xz -\- yz and x^y -\- xy\ Am. o?yz + xyH, 133. It is sometimes very difficult, if not impossible, to re- solve the given quantities into their prime factors by inspection. We shall therefore proceed to demonstrate the following rule, which is more general in its application : RULE FOR FINDING THE L. C. M. OF TWO ALGEBRAIC EXPRESSIONS. Divide the 'product of the two expressions by their G. C. D. ; n divide one of the expressions by the G. C. D. and multiply the quotient by the other expression. Let A and B denote the two expressions, and D their G. C. D. Suppose A = dD, and B = Z>D. From the nature of the G. C. D., a and b have no common factor ; hence the L. C. M. of A and B is abJ). But dbJ) — -^ =:=xB=:^xA. Cor. — If M be the L. 0. M. of A and B, it is obvious that every multiple of M is a common multiple of A and B. LEAST COMMOIT MULTIPLE. 6^ 134. Every common multiple of two algelraic expressions is a multiple of their L. C M, Let A and B denote the two expressions, M their L. C. M. , and let N denote any other common multiple. Suppose, if pos- sible, that when N is divided by M, there is a remainder, E ; let q denote the quotient. Then E = K" — gM. Now A and B are common divisors of M and N, and therefore they are divisors of E (125, Lem. II) ; that is, E is a common multiple of A and B. But E is of loiuer dimensions than M ; hence there is a common multiple of A and B of lower dimensions than their L. C. M. This is absurd ; hence, there can be no remainder ; that is, N is a multiple of M. 135. To find the L. C. M. of three Algebraic expres- sions, A, B, and C. Find the L. C. M. of two of them, as A and B. Let M denote this L. C. M. ; then the L. C. M. of M and C is the required L. C. M. of A, B, and C. For every common multiple of M and is a common multiple of A, B, and (133, Cor.). Again, every common multiple of A, B, and C is a multiple of M and C (134). Therefore, the L. 0. M. of M and C is the L. 0. M. of A, B, and C. EXAMPLES, 136. Find the L. C. M. of 1. Qx^ — x — 1 and 2x^ -f 3a: — 2. Ans. (2aj2 4. 3a; — 2) {Zx + 1). 2. 0^ — 1 and rr2 -f a; — 2. Ans. {a? — V){x + 2). 3. a;3 __ 9^ ^ 23a; — 15 and x^ — %x + 7. Ans. (^3 _ 9^ _^ 2Zx — 15) (x — 7). 4. 3a;2 __ 5a; 4. 2 and 4^ — 4:X^ — x -\- 1. Ans. (dx — 2){^x^ — Axi — x-{- 1). 5. (x + 1) {x^ — 1) and a^ — 1. Ans. (a;^ _ 1) (a; + 1)2. 6. a^-\- 'Hx^y — xy^ — 2f and a^ — 2xiy — xy^ + 2y^ Ans. (a^i — ?/2) (a;2 _ 4^2), 60 LEAST COMMOiq- MULTIPLE. 7. 2z _ 1, 4a:2 — 1, and Aa^ + 1. Ans. 16ar* — 1. 8. a^ — x, a^ — 1, and a;^ + 1. Ans, x{7^ — 1). 9. a4> _ 4^2^ (a; ^ 2«)3, and {x — 1d)\ Ans. {x^ — 4«2)3. 10. 2)3 __ 62;2 4- 11a; — 6, a;3 — 9a:2 + 26a; — 24, and a;^ _ 8a;2+ 19a;— 12. ^w«. (a; - 1) (a; — 2) (x _ 3) (a; — 4). 11. a;2 4. 7a; ^ 10, a;2 _ 2a; — 8, and a^^ 4- a; — 20. Ans, a;3 _|_ 33;2 _ isa; _ 40. 12. a2 — 3a J + 252, a^-ab — 2^, and a2 _ jz. ^ns. a3 _ 2ff2^> - «Z>2 ^ 2&8. 13. 22?* — 7a;y + 32/2, ^^ _ ^^y ^ ^y^, and a;2 — bxy + 6i/2. ^715. 2a;3 — lla;2y + 17a;^2 _ 5^3^ 137. SYNOPSIS FOR REVIEW. r G. 0. D. L. 0. M. Terms used Special Rule— Dem. General Rule. — Dem. Cor. 1, Ck)R. a. Cor. 3. Comtnon Diusor. Q. C. D. Commenaurdble. Incommenaurable. Lemma I— Dem. Lemma U—Dem. Application of Lemmas^ {Introduction of Factors, Rejection of Factors. Suggestions .... G. C. D. OP three Algebraic expressions. {Multiple. Common Multiple. L.C.M. Special Rule— Dem. — Cor. General Rule — Dem. Theorem— Dem. L. C. M. OF THREE Algebraic expressions. CHAPTER Y. -^ FEAOTIOI^S. DEFINITIONS AND FUNDAMENTAL PRINCIPLES. 138. A Fraction is a quotient expressed by placing the dividend over the divisor, with a line between them. Thus, ■J and T are fractions. 4 139. The Numerator of a fraction is the quantity above the Une, and the Denominator is the quantity below the line. The numerator and denominator of a fraction are called its Terms, 140. If the terms of a fraction are integers, we may regard the denominator as denoting the number of equal parts into which the unit (1) is divided, and the numerator as denoting how many of those parts are expressed. 141. A Fractional Unit is one of the equal parts into which the unit is divided. Thus, in the fraction t> the fractional unit is T. 142. An integer may be considered as a fraction with unity for its denominator. Thus, a = zr. 143. An Entire Quantity is one which does not con- tain a fraction. Thus, a + b -^ c is an entire quantity. 144. A Mixed Quantity is one which contains an en- 62 PEACTIONS. tire part and a fractional part. Thus, a-\-b + -^ is a mixed quantity. 145. A Simple Fraction is one whose terms are entire. Thus, :; is a simple fraction. c -\- d ^ 146. A Complex Fraction is one which has a fraction in one or both of its terms. Thus, is a complex fraction. 147. A Compound Fraction is the indicated product of two or more fractions. Thus, t X ;i X ;7; is a compound frac- tion. 148. A Froper Fraction is one whose numerator is less than its denominator. Thus, 7 is a proper fraction. 149. An Improper Fraction, is one whose numerator is equal to or greater than its denominator. Thus, - and are improper fractions. 150. The J^ecijrrocal of a quantity is the quotient ob- tained by dividing unity by that quantity. Thus, the reciprocal of a is - . a 151. To multiply a fraction by an integer. Let T be a fraction, and c an integer: then y x c = ^. ° For, in each of the fractions ^ and -7- the fractional unit (141) is t; hence, -r- is c times t (140). Again, — x c = y For, the fractional unit in t is c times DC the fractional unit in 7-. DC DEFINITIONS AND PRINCIPLES. 63 RULE. Multiply the given numerator ly the given integer, and divide the product by the given denominator, or, divide the given denomi- nator hy the given iiiteger, and divide the given numerator hy the quotient. 153. To divide a fraction by an integer. a ^ _ a^ For, T is c times ^ (151) ; hence r- is -th of t. DC ^ be c b . . ac a Again, -^c = j^ For, ^ is c times t; hence y is -th of -7-. b b' b c b MULE. Multiply the given denominator by the given integer, and di- vide the given numerator by the product, or, divide the given nu- merator by the given integer, and divide the quotient by the given denominator. 153. The value of a fraction is not changed by multiplying or dividing both of its terms by the same quantity. It is evident that if we multiply the fraction t hy c, and then divide the product hy c, the resulting fraction will he equal to the given fraction. Now ^ x c = -r- (151), and "t "^ ^ = l^ (153) ; , a ac ac a hence t = i-> or ^ = t . b be be b 64 FRACTIONS. REDUCTION OF FRACTIONS. 154. A fraction is in its Lowest Terms when its terms have no common factor. 155. To reduce a fraction to its lowest terms. RULE. Divide both terms of the fraction by their G. C. D. Or, Resolve both terms of the fraction into their prime factors, and then cancel those factors ivhich are common. ILZU8TBATION8, lOacx^ 1. Reduce ^,, ' , to its lowest terms. The G. C. D. of the terms of this fraction is 5caf, Dividing both terms by this, we "have lOaca? ^ _ 2a 15bca^ ~ 3bx' 2. Reduce r-5 — ^-^ to its lowest terms. 3a^ — 3ab 3 a^ + Sab _ da {a + b) _ a_+b da^ — 'Sab~3a{a — t>) ~~ a—b' ^ _ , 6a:2_ 7a;_20 ^ ., , , , 3. Reduce -r-5 — 7^ ? to its lowest terms. 4^ — 27a; 4- 5 Here the G. C. D. of the numerator and denommator is 2x — 5. Dividing both terms of the fraction by this, we have Qa^— 7a; — 20 _ 3a; + 4 4a:3_27a;+ 5 ~" 2a;3 _^ 5^; _ 1 ' 156. To reduce a fraction to an entire or mixed quantity. a^-\-ab ^ a^+b b , a^— b b — ■ = a -\- b, — ■ — = a -f - , and = a . a 'a a a a BEDUCTIOl^ OF FRACTION'S. 65 ^ MULE, Divide the numerator hy the denominator, expressing any term of the quotieMdn a fractional form when the divisio7i cannot de exactly performed.^ 157. To reduce an entire quantity to the form of a fraction having a given denominator. Let it be required to reduce x -\- y to the form of a fraction whose denominator shall he x — y. x^y = ^ + ^ = (^ + y) (a^ - ;/) ^ x^ - y\ ^ 1 X — y X — y ' RULE. Consider the entire quantity as a fraction whose denominator is unity ; then multiply hotli terms of this fraction iy the given denominator, 158. To reduce a fraction to an equivalent one having a given denominator. Let it be required to reduce the fraction ^ to an equiva- lent one having the denominator a^ — h\ Dividing o^ — W- by a — h, the quotient is a-^-h. Multiplying both terms of the fraction 7 by this quo- tient, we have a^-b _ {a ■\-b){a-\- h) _ {a + hf a — b" a^ — l^ - a^ — yi' RULE. Multiply both terms of the given fraction by the quotient ob- tained by dividing the denominator of the required fraction by the denominator of the given fraction, 159. To reduce a mixed quantity to the form of a fraction. Let it be required to express a + - under a fractional form. 66 FRACTIONS. h ac -\-h a; + - = . c c For, if we reduce the fraction to a mixed quantity, c we obtain a -\- - (156). c In like manner we may show that a = . •^ c c RULE. Multiply the entire part hy the denominator of the fractional part ; the7i add the numerator to the product, or subtract it from the product, according as the fraction has the sign +, or the sign —, prefixed ; the result will he the numerator, and the given denominator will be the denominator of the required fraction. 160. To reduce fractions to equivalent ones having a common denominator. a c e Let ^, -j, and ^ be the proposed fractions. If we multiply both terms of each of these fractions by the product of all the de- nominators except its own, the values of the fractions will not be changed (153). Moreover, the denominators of the new fractions will be equal, since each is the product of the denominators of the given fractions. ^, a adf c bcf , e bde RULE. Multiply both terms of each of the given fractions by the pro- duct of all the denominators except its own, 161. To reduce fractions to equivalent ones having the least common denominator. Let — , — , and — be the proposed fractions. The L. 0. M. nix my mz ^ ^ of the denominators is mxyz, Now reduce each of the given frao- COMBINATIOlfS OF FEACTIOKS. 67 tions to an e^ivalent one having mxyz for its denominator (158) ; tlie resulting fractions are ayz hxz , cxy mxyz' mxyz' mxyz' Now since mxyz is the least quantity that can be divided sep- arately by mar, my, and mz, it follows that the given fractions have been reduced to equivalent ones having the least common denominator. Divide the L. C. M. of all the denominators hy each denomina- tor separately ; then multiply both terms of each fraction by the corresponding quotient. ScH. — Before commencing the operation, each fraction must be in its lowest terms. COMBINATIONS OF FRACTIONS. 162. To find the sum of given fractions. a c 1. Let it be required to find the sum of the fractions t? t? and ^ . Here the given fractions have a common denominator. In the first fraction the fractional unit ^ is taken a times ; in the second it is taken c times ; and in the third, d times ; hence, in the snm o therefore, the snm of these fractions, t must be taken (a + c -t d) times ; a £ ^_ a -\- c -^ d h^b'^b- b ft p 2. Let it be required to find the sum of the fractions ^, ^, and ^. Here the given fractions have unequal denominators. Re- ducing them to equivalent fractions having a common denomina- tor (160), we have a c e _adf M. jl-^ — 0^^/"+^^/+ ^^^ b^d'^f-bdf'^bdf^bdf- bdf 68 FBACTIONS. n ULES, I. If the given fractions have a common denominator, form a fraction tvhose numerator is the sum of the given numerators, and whose denominator is the given common denominator j this fraction luill he the sum of the given fractions. II. If the given fractions have not a common denominator, re- duce them to equivalent ones having a common denominator ; then proceed as directed in L 163. To find the diflference between two fractions. c fl 1. Let it be required to subtract -r from t- The fractional unit T is taken a times in the minuend, and c times in the sub- . trahend ; hence, it must be taken {a — c) times in the remain- der; therefore, a c _a — c h~h~~h~ c a 2. Let it be required to subtract -^ from t. Reducing these fractions to equivalent ones having a common denominator, we have a c _ad Ic _ad —he b li~ hd hd~ hd RULES. I. If the given fractions have a common denominator, form a fraction whose numerator is the remainder ohtained hy suhtr act- ing the numerator of the suhtr ahend from that of the minuend, and luhose denominator is the given common denominator ; this fraction will he the difference required. IL If the given fractions have not a common denominator, re- duce them to equivaleiit ones having a common denominator ; then proceed as directed in L COMBINATION'S OF JRACTIOIJS. 69 164. To find the product of given fractions. CL C Let it be required to find the product of t and -^. The fol- lowing is usually given as a solution : Put Y = ^> and -^ = w. Then a = bm, and c = dn. Hence, ac = bmdn =zhd x mn; or, dividing both members dC by bd (42, 5), we have j-j = mn. This process is satisfactory when m and n are really integers, though under a fractional form, because then the word multipli' cation has its common meaning. It is also satisfactory when one of them is an integer, because we can speak of multiplying a frac- tion by an integer, as in Art. 151. But when both m and n are fractions, we cannot speak of multiplying one of them by the other without defining what we mean by the term multiplication j for, according to the ordinary meaning of this term, the multiplier must be an integer. The following definitions will show more clearly the connec- tion between the meaning of the word multiplication when ap- plied to integers, and its meaning when applied to fractions. When we multiply one integer, a, by another, b, we may describe the operation thus : What tve did with unity to obtain b, we must now do with a to obtain b times a. CL C Now, let it be required to multiply t hy ^. Adopting the definition just given, we may say that, what ive did with unity to obtain -^, we must now do with ^ to obtain the product of ^ c c and -^ . To obtain -^ from unity, we divide it into d equal parts, and multiply one of the parts by c ; therefore, to obtain the pro- duct of T and ^, we divide t into d equal parts, and multiply 70 FRACTIONS. one of them by c. Now ^ -^ J = r^ (152), and -^-7X^ = ^-7 •^ Id ^ ' bd bd (151). We may therefore give the following extended definition : Multiplication is the process of finding a quantity having the same relation to the multiplicand that the multiplier has to unity. RULE. Form a fraction whose numerator is the product of the given numerators, and whose denominator is the product of the given denominators ; this fraction will he the product required. ScH. 1. — This rule embraces all the cases in which a fraction is a factor. Thus, if it be required to multiply a fraction by an entire quantity, the latter may be considered as a fraction whose denominator is unity (142). ScH. 2. — If any factor is a mixed quantity, it is best to reduce it to the form of a fi*action before commencing the operation. ScH. 3. — if the numerator and denominator of the product have any common factor, it should be canceled. Thus, 2a2 (a-\.bf _ 2fl2(rt + *)2 _ 2a^{a + b){a-\-b) _ a + b ai — l^- 4a26 {a'^—l^)^ib U^b{a + b){a-b) 2b(a-b) (155). 165. To find the qtiotient of two fractions. a c Let it be required to divide t 1^7 3 • Denoting the quotient by X, we have a c _ V'^'d^'^' But the product of the divisor and quotient is equal to the dividend; hence, c a Multiplying both members of this equation by - (42, 4), we c have c d a d d c c COMBIITATIOKS OF FEACTIOKS. 71 Canceling common factors (155), we have a d c that is, the quotient is equal to the product obtained by multiply- ing the dividend by the divisor inverted. nULE. Multiply the dividend by the divisor inverted. Cob. 1. — The product of a quantity and its reciprocal is unity. Thus, a X - = 1. a Cob. 2. — To divide by a quantity is the same as to multiply by its reciprocal ; and, conversely, to multiply by a quantity is the same as to divide by its reciprocal. Thus, « -^ Z* = a X T J and a x J = a ~ t • 166. In the present chapter we have thus far supposed each letter to represent an integer ; but, by virtue of our extended definitions, it may be shown that all the rules and formulae given are true when any letter represents a fraction. For example, let it be required to show that t = t- when a = — , J = — , and ^ b he n q r c = -. s a m n m "~ n P mq ~ np* ac = _m ~ n r X - s mr - ns' q s qs , ac mr pr mr qs mrqs mq hence, t- = J—^= — x -^^ = — — = — ^• oc ns qs ns pr nspr np 72 FRACTIOKS. THE SIGNS OF FRACTIONS. 167. Each sign in the numerator and denominator of a frac- tion affects only the term to which it is prefixed. Thus, in the fraction ^, the sign of a is +, that of J is — , that of c is +, and that of d is — . 168. TJie dividing line of a fraction answers the purpose of a vinculum ; that is, it connects the terms which the numerator and denominator may each contain. Therefore the sign prefixed to the dividing fine affects the fraction as a luhole, 169. If the sign prefixed to the dividing line be changed, the sign of the fraction will be changed. Thus, -r- = fl^ ; but ah 170. If the sign of each term of the numerator be changed, the sign of the fraction will be changed. Thus, ^ — = a — c ; ^ , —ah -^-Ic but 7 = — a + c. 171. If the sign of each term of the denominator be changed, ah the sign of the fraction will be changed. Thus, -7- = a ; but 172. We may sum up the three preceding Articles thus: If the sign prefixed to a fraction,, or the sign of each term of the numerator, or the sign of each term of the denominator, he changed, the sign of the fraction will he changed. CoE. — If any two of these changes be made at the same time, the sign of the fraction will not be changed. 173. Tlie Apparent Sign of a fraction is the sign pre- fixed to the dividing hne of that fraction. The Meal Sign EXAMPLES. 73 of a fraction is the sign of its numerical value. Thus, the ap- parent sign of the fraction ^^— is — ; but, if « = 3, 5 = 4, and c = 5, the real sign is -f . 1*74. EXAMPLES, Simplify the following fractions, from 1 to 12, inclusive : a;2+2a; — 3 . a; + 3 ^' x^-{-6x — 1l' a; + 7 ^ X2-3X-4: a; — 5 a^-6a^ -\.llx — 6 a;2 _ 3a; + 2 * -47i5. a; — 3. a^ -{- 3a^ + Sal?^ -{- b^ ^ws. a-{-b. a;* + 10a;3 ^ 35^ ^ 50^ + 24 ^ + 9a;2 + 26a; + 24 Ans, X -\-l. 3a:3_i6^^23a;-6 • "iT^ — Wx^^Yix — ^' A 3a;— 1 Ans. ^ r. 2a; — 1 ^ 6a;3 — 5a;2 4- 4 ^x? — x?—x^%' . 3x- + 2 Ans, — i—-. X+1 2a;3 4.9^^ ^x — 3 • 3a;3^5^2_i5a;^4' . 2a; + 3 ^^^•3^-4- 1 — a;2 ^''''1+x' 5a^-\-5ax Ans. . a — x a;^ + 2a;2 + 9 • ari_4a:3^4^_9- "a;2 + 2a; + 3 ^g a;2 4- (« -f c) a; + ac * a;2 + (6 + c)x+ be' Ans.^'^l X + b 74 FKACTIONS. Perform the additions and subtractions indicated in the fol- lowing examples, from 13 to 28 inclusive : a + ^ a— J a^ — W 14. ^^ r-7 + -7 jr- AtlS. ^. 2a — 2^ 2^ —- 2a 2 ^„ 2 3 2rr — 3 ^ 9 a; 2a; — 1 4.^2— 1 * (4a;2_i)^ 17. rr — -, rrr7.. AuS. 2a Ans. — . n 18. x — l ic-f 2 (a;-f-2)2* "* (a; — 1) (a; + 2)2' 5 1 24 2(a;+l) 10(.'?; — 1) 5(2a; + 3)' . 2a: — 3 Ans. (a;2_l)(2a; + 3)' h—a a — 2b ^ Sx(a — b) . ax — 1^ ^^' x-b x-\-b^ a^-lP '^^^' w^r^' _^ 3 4- 2ar 2 — 3a; IGa; — x^, 1 20. -r jr— 1 r — . Ans. 2 — x 2 + a;' a-2 — 4 a;4-2 oi 3 7 4-20a; . ^ ^^•nr^-rT2^-4^rrT- ^^^- ^• 22.4-. + ^^--.^,. Ans. '^^ a + b ' cfi — V^ a- + V>' a* — b*' o„ 1 1 1 . x' — ixy — y- ^^- a:^ _ 2,3 + (.^ + yf (X - y)2- ^'**- (a^ - y2)» 24. i2i±^-?-*-2. Ans. "" ab(a — by b a {a — bf ^K fl , 3a 2«a: . 4a 25. \ ; -. Ans. a — x a -{- X a^ — 7?' ' a -j- x' q^ 3a — 45 2a — b — c 15a — 4c a — Ab , Sla — ib ^»*- -ST— EXAMPLES. 75 27 ^ + ^ L ^ + g , c + a {h — c){c — a) {c — a){a^ b) {a —b) (b — c)' Ans. 0. a^ — be b^ — ac & — ab (a + 6) (a + c) (^> + c) (^ + a) (c + «) (c + ^') * ^4ws. 0. 29. Multiply -5^ r-^- by -^ tt- ^w5. ^ .-(-. ^ -^ a 4- 6 ^ x{a — b) {a + b)x 30. Multiply ^i^ by _?!zilL. ^^^. -^^^^ . o. T»r IX- 1 . XI 3«a; «2_/p2 Jc + ^a; , c — a; 31. Multiply together ^, ^-^, -^^-^^, and --. 3a; ^72S. -7-. 32. Prove that 33. Multiply together y^y , ^^, and 1 + j-^. Ans. --. X 34. Multiply -^ — —- ^ by „ \^ — '—' ^ " a^ -\-2ax + x^ ^ a^— 2ax + x^ A c^x Ans. -^ -„. a^ — x^ OK o- vc a^ — b^ a^-b . a^ + b'^ 35. Smphfy ^^^-^^ X ^^^^. Ans. —-. 36. Simplify (^+l-^:zl_ JyL\ ^+1. Ans. 3. ^ '^ \x — y X -{■ y a^ — yv 2y 37. Simplify ^-^3 - ^-^ - (^^^:py,j. «2 _ ^J _|_ J2 ^''^- ^2 + «^, + J3 38. Multiply -'-- + 1 by ^ + ^ + 1. ^^z-?. ^ H- -' + 1. 76 FRACTIONS. 39. Multiply ir2 — a; +1 by 1 + - + 1. Ans. x^ + l-}- ^. 40. Simplify -^ ) ; ,( | — t X -^ Tn. Aiis, ) ^ ,( -,. 42 Divide ^(^'^^^) by -^^ Ans l^^^- 43. Divide -^^ by -f-- ^w^- zs ^^"a- 44. Divide '-^ + -^ r by ^, . a; + y a; — ^ o? — 'f '' x? — y^ Ans. ^ 45. Simplify g + ^).(j-J + i). Ans.^-±y 46. Simplify (-^ + —j) - (-^ ^—j). Ans. 1 47. Simplify (^±^ + ?) - (^-±^ ^-) . ^^.. 1 ^ -^ \x-\-y yl \ y x + yj 48. Divide x^ j by x •\- -. Ans. o^ — x A „ x^ •^ X X x^ 49. Divide 0? ^ \ + 2 by x+^, Ans. — - x^ -^ X X 50. Divide x^^l^^ by ^ - 1+ a;. ^tjs. ^^ + ^ + ^ a;-* " X X 51. Divide fl2_^_^^2Z'c by ^-±4-f^. Ans. a^ — b^ A- (^ + ^cic. 62. Dmde — ^- — ^— r — ^^— by -oV" — f— 2- « + a; • x — y 53. Diyide a^ — h^ — c^ — 2hc by EXAMPLES. 77 a -\- h -\- c a -\- b — c' A71S. a^ _ J2 _|_ ^2 _ 2ac. 54. Divide x^ - Sr- - 2a^ + ^^fV by Sx - Ga -^. a;2 + 36?a; — 2a^ Ans, X -\- 6a 5o. Divide zr^ — 4 H — ^ by ^r . Ans. 2a^ x^ -' 2a X ax a + h a — h t^n cs' ^'c^ ^ + d c — 7l . ac — M 56. Simplify 7 t* ^ns. --^-^. ^ -' a + b a — b ac -\- bd c — d c + d a ■\- X a — x 57. Simplify . Ans. —-: . ^ '' a -\- X a — x 2ax a — x a -{- X a—1 b—1 c—1 ^o o- vn ^abc a b c 58. Simplify be + ca — ab 1,1 1 be -\- ac -\- ab a b c Ans. be -^ ac — ab Ko Q- r^ (a + b a^-^b^\ (a-b a^ - b\ 59. Simplify (^-^ + -^-^^j - (— ^- - ^3-^^). a^ + aW + ^* ^^5. «^> (a - bf 60. Simplify (^-^^ - -^,-^3) ^ (--^ + -,_-^-^). bc(b-cf Ans. ei.Si.piiiy(J±|-5^:).g^^-^D. Ans. ^ 78 FRACTIONS. 62. Simplify (i:+* + ^) ^ (^ _ ^) . 63. Simplily — i ^ x ^ — =. ^ns. m, n m X -{- a x — a 64 Simplify ■ — . Ans. -, ''x — a X -\- a X -\- a x — a x^ — a x — a X -{■ a 65. Simplify ^ ^(l + — ^^— )• a b + c fifi bimplify 1 X-\- 1 ^ + 3- ±_l — X 07. Simplify h-¥ a c A Ans. d(x^l) '-J CO — bv 68. Find the value of ax + 5?/, when x = -^^ — - and «r — c» y= r^. Ans. c. ^ aq — bp 69. Find the value of ^r- -\ -rj , when x = x — 2a X — 2b' a -\- y Ans. 2. 70.If| + |=l,showthatf-|=J-J. a c^ fv^ (^ n c 71. If T + :7 = 72- ^> sliow that t > = 1. a If^ (P b d I SYNOPSIS rOR REVIEW. 79 175. SYNOPSIS FOR REVIEW. CHAPTER V. FEAOTIONS. f Denominator. ) rn, ^ , , ,. , ^ , > The terms of a fraction. Numerator. S ' Terms used . . J ^rrac. unit. Entire quantity. I Mixed quantity. Simple Frac. Proper. Improper. To REDUCE . . Complex Fractions. Compound Fractions. Value of fraction not changed, when. Fraction in its lowest terms, when. Fraction x Integer. Bute. Fraction -t- Integer. BuZe. Fraction to mixed qwintity. Rule. Entire quantity to fraction having given denominator. Rule. Fraction to fraction having given de- nominator. Rule. Mixed quantity to form of fraction. Rule. Fractionsto common denominator. Rule. Fractions to least com. denom. Rule. Addition. Investigation for rule. Rule. Subtraction. Investigation. Rule. Multiplication. Def. Investigation. Rule. Division. Investigation. Rule. Cor. 1, 3. ^gn prefixed to a term of numerator or denominator. Sign prefixed to fraction. Signs of J Methods of changing sign of fraction. Fractions. | Changes of sign not affecting sign of fraction. Apparent sign of fraction. ^ Real sign of fraction. CHAPTEE VI. DEFiraiOIfS AND GENERAL PRINCIPLES RELATING TO EQUATIONS. 176. An Equation consists of two expressions connected by the sign of equality. Thus, x -\- a^=m -\- n is an eqnarion. The First Member of an equation is the quantity on the left of the sign of equality, and the Second Member is the quantity on the right of the sign. Thus, in the equation x -\- a =z m -\- n, ic -f a is the first member, and m -\- n the second member. 177. An Identical Equation^ or An Identity , is an equation whose members are either identical, or may be made identical by performing the indicated operations. Thus, a^ — x^ , a ax ax -^ b =: ax -^ b, — = a + x, and a ^ ^ ^ ' a — x ^ ' 1 + a: 1 +0; are identities. 178. It follows, from the definition, that the members of an identity are equal for all values that may be assigned to each letter which it contains. Thus far the student has been almost entirely occupied with identities. Thus, the equations given in Articles 71, 73, and 73 are identities. 179. An Equation of Condition is one whose mem- bers are equal only for a limited number of values of each letter which it contains. Thus, x -\- 1 =z 7 is an equation of condition, because its members are not equal unless x = 6. An equation of condition is called briefly, an equation. 180. An Unknown Quantity is a letter to which a particular value or values must be given in order that the mem- bers of an equation may become identical. The equation is said DEFINITIONS. 81 to be Satisfied for such particular value or yalues. Thus, in the equation x^ — ^x = — 3, the unknown quantity is x, and when a; = 3 or 1, the equation is satisfied. 181. An Unknown Term of an equation is a term con- taining an unknown quantity. 183. A Moot of an Equation is a quantity which, w^hen substituted for the unknown quantity, satisfies the equation. Thus, 3 and 1 are the roots of the equation x^ •— 4lX=: — 3. 183. To solve an Equation is to find its roots. 184. A Niimerical Equation is one in which all the Tcnoion quantities are represented by numbers. Thus, 2a;2 -\-Zx = V)x + 15 is a numerical equation. 185. A Literal Equation is one in w^hich the known quantities are represented entirely or in part by letters. Thus, ax -\- b ^= ex -\- d and ax — 5 = 3a; — 5 are literal equations. 186. Hie I>egree of an equation is denoted by the num- ber of unknown factors in that term which contains the greatest number of such factors. Thus, ax — h=^c is of the first degree, a:^ + 2/?a; =zq is of the second degree, a?y -{- x^ — ex z=z a is of the third degree, ic'* + «a;'*-i -\-hx''~^ = c is of the n^'' degi'ee. Remark. — It should be observed that the definition implies that the equation is of such a form that no unknown quantity occurs under the radical sign, or in a denominator. 187. A Simple Equation is one of the first degree. 188. A Quadratic Equation is one of the second degree. 189. A Cubic Equation is one of the third degree. 190. A biquadratic Equation is one of the fourth degree. 191. Higher Equations are those of higher degrees than the second. 82 EQUATIONS. 192. For brevity, the following symbols are sometimes used : . • . signifies hence, therefore, or consequently. • . • signifies since, or because. TRANSFORMATION OF EQUATIONS. 193. To Transform an equation is to change its form without destroying the equality of its members. 194. Clearing of Fractions and Transposition of Terms are the principal transformations. 195. To clear an equation of fractions. Let it be required to transform the equation, ^l-Ve-' (^^' into another, all of whose terms shall be entire. Multiplying both members of (1) by al^c^ which is the pro- duct of the denominators, we obtain lex — dbx = dtt^cd (2). Instead of multiplying both members of (1) by ab% we may clear the equation of fractions by multiplying both members by dbc, which is the L. C. M. of the denominators ; we thus obtain ex — ax = abed (3). JR ULE. Multiply both memhers of the given equation by the product of all the denominators or by the L. C. M. of all the denominators, 196. To transpose a term from one member of an equation to the other. Let us consider the equation x-a = b-y . . . (1). TRAKSPOSITION. 83 Adding a to each member of (1), x — a-\-a = b — y-{-a (42, 2) ; that is, x=^d -\- a — y (2).. Subtracting h from each member of (2), x-lz=za-y (3) (43, 3). Here we see that — a has been removed from one member of the equation and appears as + a in the other ; and + l has been removed from one member and appears as — ^ in the other. ■Remove the term, which is to he transposed, from the member in which it stands, and write it, with its sign changed, in the other member. Cor. — If the sign of every term in an equation be changed, the equality still holds. Thus, if x^a=b—y, then a—x=y—b* CHAPTEE VII. SIMPLE EQUATIOI^J'S SIMPLE EQUATIONS WITH ONE UNKNOWN QUANTITY. 197. To solve a simple equation containing only- one unknown quantity. 1. Let it be required to solve the equation 3a; _ 4 = 24 — X. By transposition, 3a; + a; = 24 + 4; that is, 4a: = 28 ; 28 whence, by division, a; = — = 7. We may verify this result by substituting 7 for a;. in the given equation. The first member becomes 3x7 — 4, that is, 17; and the second member becomes 24 — 7, that is, 17. 2. Let it be required to solve the equation ^"T" 8 "^32' Multiplying both members of this equation by 96, which is the L. C. M. of the denominators, 240a; — 128a; — 1248 = 60 + 3a;. By transposition, 240a; — 128a; — 3a; = 1248 + 60 ; that is, 109a; = 1308; whence by division, x = — — - = 12. J ' 109 ONE UNKNOWl^ QUANTITY. 85 RULE. I. Clear the equation of fractions, if it has any, ■ II. Transpose every unhnown term of the second meniber to the first, and every known term of the first member to the second; and reduce each member to its simplest form. III. Divide both members by the coefficient of the unhnown quantity ; the second member of the resulting equation will be the value of the unhnown quantity. IIjL ustbations. Sometimes it is more convenient to clear of fractions par- tially, and then to effect some reductions before getting rid of the remaining fractional coefficients. For example, 1. Solve x^l 22; — 16 2 a; + 5 _ ^, 3a; + 7 "n 3~" + "T~-^* + ~i2~- Multiplying both members by 12, ^^ ^^:^ '^^ - 4 (2a: - 16) + 3 (2a; + 5) = 64 + 32; + 7; that is, ^^ (^ + '^) _ 8:c + 64 + 6a; + 15 = 64 + 3a; + 7. By transposition and reduction, 12 (a; + 7) _ 11 "" Multiplying by 11, 5a; -8. 12a; + 84 = 55a; — 88; by transposition, 12a; — 55a; = — 88 — 84; by reduction, — 43a; — — 172 ; by changing signs, 43a; = 172 ; 172 by division, x = -j-^ = 4. 86 SIMPLE EQUATIONS. 9 SoIto ^ _ ^ "" 2x-\-l'~ 5x — 8' Multiplying each member by (2a; + 1) (ox - -8), 25a; — 40 = 4a; + 2. By transposition and reduction, 21a; = 42; 42 by division, a; = — = 2. Verification, — Putting this value for x in the given equation, we have or 1=1. 3. Solve 4 + 1 10 — 8 2a; — 3 4a; — 5 3a; — 4 6a; — 7 Clearing of fractions, (2a; - 3) {Qx _ 7) = (4a; - 6) (3a; - 4) ; that is, 12a;2 _ 32a; 4- 21 = 12a?8 — 31a; + 20. Subtracting 12a;2 fpom ^Qth members, 21 — 32a; = 20 — 31a;; by transposition and reduction, — a; = — 1, or a; = 1. Verification. — Putting this value for z in the given equation, we have 2-3 4-5 3-4~6-7' or 1=1. . c! 1 a; _ 10a; 7 4. Solve -_8 = -3--3. Clearing of fractions, 3a; — 48 = 20a; — 14. By transposition and reduction, — 17a; = 34, or 17a; =—34; OifE UNKNOWN QUANTITY. 87 by division, x = — — = — 2. 2 20 7 Verification, - — 8= -^ — ^> or — 9=— 9. 5. Solve ax -\- b = ex -\- d. By transposition, ax — ex = d — b; that is, (a — c) ic = (Z — 6 ; d-b by division. a— c Verifieation. — Putting this value for x in the given equation, we have a — c a — c which is an identity, since each member reduces to . 6. Solve Ix^iain -\- x). By transposition, bx — ax = an ; by division, x = -. •^ b — a ^ .^ ^. , an i an \ abn abn Verincahon. b x 7 =^ ain + 7 ), or t =7 . •^ b — a \ b — aj b — a b — a 198. A simple equation eontaining only one unknoiun quan- tity has one root, and no more. By clearing of fractions, transposing, and reducing, if neces- sary, any simple equation containing only one unknown quantity may take the form of ax-=b (1) ; whence, x = - (2). a The value of x is verified thus : a X - = &, or b^b. a 88 SIMPLE EQUATION'S. Now, it is evident that any value for x greater than - would make the first member of (1) greater than the second, and that any value for x less than - would make the first member of (1) less than the second. Hence, there can be no root either greater or less than -; that is, x is equal to -, and to nothing else. This theorem may also be demonstrated thus : Suppose the equation ax^=h (1), has two different roots, ^j and q ; then these roots will satisfy equation (1), and give ap = 'b (2), aq = l) (3). Subtracting (3) from (2), a(p-q) = (4) (42,3); but this is impossible, for a is not zero, and, by hypothesis, p— q is not zero. EXAMPLES. 199. Find the value of x in each of the following equations: . 22: + l 7z + 5 . 1. — - — = — 5 — . Ans,x = l, 2. I _2 = ^ + ?-l. Ans. a: = 20. • 2 4 x + \ 3^-4 l_ 6^ + 7 3. -y-+ ^ +g- g . Arts. X- 6. , 5a;— 11 ir— 1 11a;— 1 . .. ^ X X X 1 A ^ ^•2 + 3-4 = 3- Ans.^^-^. 6. ^ + -+^^16-^4^. Ans. ^ = 13. Z o 4 ONE UNKNOWN QUANTITY. 89 11— a; 26 — a; 7, X -\ — = — ^ — . Ans, x = S. 1 ^5 8. 19x ^-{7x-2)=z4^ + ^. Ans. x = l. _ ic— 3 a;— 4 a;— 5 x-\-l . 10. — — = ^x — 14. Ans. x = 1. ^ ^ x—S 2x — 5 41 3a; — 8 5a; + 6 11.-^ 6^=60 + ^ 1^- ^^''^ = ^' 12. 5£^_?:^:^5^_10. ^^.. a; = 3. 13. liS-x) +a;-l| = -^-|. Ans. x = b. a;H-3 a;-2 3a; — 5 1 14.^ ^=-12- +4- ^7^..a; = 28. ,„ 3a; -1 13— a; 7a; 11 (a; + 3) ._ 5a; — 3 9— a; 5a; 19, ,, 10. — ^ ^ z= — -f — (a; — 4). ^W5. a; = 2. .„ 5a; — 1 9a; — 5 9a; — 7 17. — jt; 1 -j — = — - — . Ans. a; = 3. 7 11 ^„ 3a; + 5 2a; + 7 ,^ 3a; ^ 18. — Y ^ + 10 - y = 0. Ans. X = 10. a;-l 4a; -f 7a; - 6 _ ^ , a;-2 . 3a; - 9 » ^/i. days he can perform - ths of the work. In one day B can perform T^rth of the work, therefore in x days he can perform — ths of the work. Hence, since A and B together perform the whole work in x days, 8^ 10 By clearing of fractions, 6a; + 4a: = 40 ; by reduction, Oa- = 40 ; whence, x = -^ = 4^, 5. A laborer was employed for 20 days, on condition that for every day he worked he should receive 50 cents, and for every day he was idle he should forfeit 25 cents. At the end of the 20 days he received $4 ; how many days did he work, and how many days was he idle ? Let X denote the number of days he worked ; then he was idle 20 —X days. 50a; = wages due for work, and 25 ^0.— x) = the amount he forfeited ; .-. 50a; — 25(20 — rr) = 400; that is, 50a; — 500 + 25a; = 400 ; by transposition and reduction, 75a; = 900 ; whence, a; = 12 = the number of days he worked, and 20 — a; = 8 = the number of days he was idle. 94 SIMPLE EQUATIONS. 6. How much rye, at 54 cents a bushel, must be mixed with 50 bushels of wheat, at 72 cents a bushel, iu order that the mix- ture may be worth 60 cents a bushel ? Let X denote the number of bushels required ; then h\x is the value of the lye ; and since the value of the wheat is 3600, the value of the mixture is 54a; + 3600. The value of the mixture is also (x + 50) 60 \ (a; + 50) 60 = 54a; + 3600; that is, 60a; + 3000 = 54a; + 3600 ; by transposition and reduction, ^x = 600 ; whence, x = 100. 7. A smuggler had a quantity of brandy, which he expected would sell for ^100 ; after he had sold 10 gallons, a revenue officer seized one- third of the remainder, in consequence of which, the smuggler received only $80 for his brandy. How many gallons had he at fii-st, and what was the price per gallon ? Let X = the number of gallons ; then — is the price per X X 10 gallon, and — r — is the quantity seized, the value of which is o 100 — 80 = 20. The value of the quantity seized is also ex- ^ ^ .-c — 10 100 pressed by — ^ — x X l«xi^ = 20. 3 a; Clearing of fractions, 100 (a; -10) =60a;; that is, 100a;— 1000=60a;; by transposition and reduction, 40a; = 1000; whence, x = 25, the number of gallons ; , 100 100 , -, 11 .. . 11 and — = --^ = 4 dollars, the price per gallon. OKE UN^Kiq^OWK QUANTITY. 95 204c, rMOBJLEMS. 1. Divide 13870 between two persons, A and B, so that A eliall receive twice as mucli as B. Ans. A gets 12580, and B $1290. 2. Divide $420 between A and B, so that, for every dollar A re- ceives, B may receive $2 J. Ans. A's share = $120, B's = 1300. 3. How much money is there in a purse when the fourth part and the fifth part together amount to 145 ? A7is. $100. 4. After paying the seventh part of a bill and the fifth part, $92 were still due ; what was the amount of the bill ? Ans. $140. 5. Divide 46 into two parts, such that if one part be divided by 7 and the other by 3, the sura of the quotients shall be 10. Ans. 28 and 18. 6. A company of 266 persons consists of men, women, and children; there are four times as many men as children, and twice as many women as children. How many of each are there ? Ans. 38 children, 76 women, 152 men. 7. A person expends one-third of his income in board and lodging, one-eighth in clothing, one-tenth in charity, and saves $318. What is his income ? Ans. %^%^. 8. Three towns. A, B, C, raise a sum of $594 ; for every dollar B contributes, A contributes three-fifths of a dollar, and C seven- eighths of a dollar. What does each contribute ? Ans. A contributes $144, B $240, C $210. 9. Divide $1520 among A, B, and C, so that B shall have $100 more than A, and C $270 more than B. Ans. A gets $350, B $450, C $720. 10. A certain sum of money is to be divided among A, B, and C. A is to have $30 less than the half, B is to have $10 less than the third part, and C is to have $8 more than the fourth part. What does each receive? Ans. A $162, B $118, C $104. 11. The sum of two numbers is 5760, and their difference is equal to one-third of the greater; find the numbers. Ans, 3456 and 2304. 96 SIMPLE EQUATIONS. 12. Two casks contain equal quantities of beer ; from the first 34 quarts are di*awn, and from the secon(J 80 ; the quantity re- maininir in tiie first cask is now twice that in the second. How much did each cask originally contain ? Ans. 126 quarts. 13. A person bought a picture at a certain price, and paid the same price for a frame ; if the frame had cost $1 less, and the pic- ture $J more, the price of the fi-ame would have been only half that of the picture. What was the price of the picture ? ^ Ans. $2}. 14. A house and garden cost $850, and five times the price of the house was equal to twelve times the price of the garden ; find the price of each. • A71S, House, $600 ; garden, $250. 15. One-tenth of a rod is colored red, one- twentieth orange, one-thirtieth yellow, one-fortieth green, one-fiftieth blue, one- sixtieth indigo, and the remainder, which is 302 inches long, violet; find the length of the rod. Ans. 400 inches. 16. Two-thirds of a certain number of persons received $18 each, and one-third received $30 each. The whole sura received was $660. How many persons were there ? A7is. 30. 17. Find the number whose tliird paii; added to its seventh part gives a sum equal to 20. Ans. 42. 18. The difference between the squares of two- consecutive numbers is 15. What are the numbers ? Ans. 7 and 8. 19. A performs ^ of a piece of work in 4 days ; he then re- ceives the assistance of B, and the two together finish it in 6 days. Find the time in which each alone can do the whole work. A71S. A in 14 days; B in 21 days. 20. A bought eggs at 18 cents a dozen ; had he bought 5 more for the same money, they would have cost him 2^ cents a dozen less. How many eggs did he buy ? Atis. 31. 21. A man bought a certain number of sheep for $94 ; having lost 7 of them, he sold one-fourth of the remainder at prime cost for $20. How many sheep did he buy? Ans. 47. 22. A man leaves home in a stage which travels 12 miles an hour, and agrees to return in 2 hours. How far may he ride if he walks back at the rate of 4 miles an hour? Ans. Q miles. OIJE UNKNOWN QUANTITY. 97 23. A and B play at a game, agreeing that the loser shall always pay to the winner $1 more than half the money the loser has; they commence with equal sums of money, hut after B has lost the first game and won the second, he has twice as much as A; how much had each at the beginning? Ans. $6. 24. A person who possesses $12000 uses a portion of the money in building a house. One-third of the money which re- mains he invests at 4 per cent., and the other two- thirds at 5 per cent., and from these investments he obtains an income of $392. What was the cost of the house ? A7is. $3600. 25. A takes from a purse $2 and one-sixth of what remains ; then B takes $3 and one-sixth of what remains ; they then find that they have taken equal amounts. How many dollars were in the purse, and how many did each take ? Ans. There were $20 in the purse, and each took $5. 26. A vessel can be emptied by three taps ; by the first alone it could be emptied in 80 minutes ; by the second alone, in 200 minutes ; and by the third alone, in 5 hours. In what time will the vessel be emptied if all the taps are opened? Ans. 48 min. 27. A person buys some tea at 36 cents a pound, and some at 60 cents a pound ; he wishes to mix them, so that, by selling the mixture at 44 cents a pound, he may gain 10 per cent, on each pound sold ; find how many pounds of the inferior tea he must mix with each pound of the superior. A?is. 5. 28. A cask, A, contains 12 gallons of wine and 18 gallons of water ; another cask, B, contains 9 gallons of wine and 3 gallons of water ; how many gallons must be drawn from each cask, so as to produce, by their mixture, 7 gallons of wine and 7 gallons of water? Ans. 10 from A, and 4 from B. 29. A can dig a ditch in one-half the time that B can ; B can dig it in two-thirds of the time that C can ; all together they can dig it in 6 days ; find the time in which each alone can dig the ditch. Ans. A in 11 days, B in 22 days, and C in 33 days. 30. At what time between one o'clock and two o'clock is the minute hand exactly one minute in advance of the hour hand? Ans. 6-j^ minutes past one. 7 98 SIMPLE EQUATIONS. 31. A man leaves home in a stage which travels h miles an hour, and agrees to return in a hours. How far may he ride, if he walks back at the rate of c miles an hour ? Ans. -r miles. b-\-c ScH. — As «, h, and c may have any values whatever, the solu- tion of Problem 31 furnishes a formula which can be used for the solution of any similar problem. Thus, to obtain the answer to Problem 22, we have only to substitute 2 for a, 12 for Z>, and 4 for c, which gives 2 X 12 X 4 96 _ ^=-T2-+-r- = i6 = ^- A problem is said to be generalized when letters are used to represent its known quantities. 32. A crew, which can row a boat at the rate of 9 miles an hour in still water, finds that it takes twice as long to come up a river as to go down ; at what i-ate does the river flow ? Ans, 3 miles an hour. 33. A certain article of consumption is subject to a duty of 72 cents per cwt. ; in consequence of a reduction in the duty, the consumption increases one-half, but the revenue falls one-third. Find the duty per cwt. after the reduction. Ans. 32 cents per cwt. 34. A merchant maintained himself for 3 years at a cost of ^250 a year ; and in each of those years augmented that part of his stock which was not so expended by one-third thereof. At the end of the third year his original stock was doubled ; what was that stock ? Ans. $3700. 35. A market woman bought some eggs at 2 for a cent, and as many more at 3 for a cent ; she sold them all at the rate of 5 for 2 cents, and found she had lost 4 cents. How many did she buy ofea« ELIMINATION BY SUBSTITUTION. Let it be required to solve the equations Ax-\-3ij = 22 ... (1), bx — 7y = Q. . . . (2). From (1) we find 22 - 4a; ,„. y = — o — .... (3). Ans, x = 5, y = 4. Ans. X — G, y = 5. Ans. X = 1, 1/ = 3. Ans, x = 5, y = 2. Ans. X = .S' -\-d 2 ' s 2 d TWO UNKNOWN QUANTITIES. 103 Substituting this value for y in (2), 22 — 4a; that is, bx 5a;- 7 X 154 - 28a; 3 = 6 . = 6; . (4). Multiplying (4) by 3, 15a; — 154 4- 28a; = 18 . . (5); by transposition and reduction, 43a; = 172; whence, a; = 4. Substituting 4 for x in (3), we find y = 2. RULE. Find, from one of the give?i equations, an expression for the value of the unknown quantity to be eliminated, and substitute this value for the same unknown quantity in the other equation ; there loill thus be formed a iiew equation containing only one un- knowti quantity. EXAMPLES. Solve the following groups of simultaneous equations : 1. j 2a; + 3?/ = 33 ) ( 4a; + 5?/ = 59 ) x-{-y x—y_ 2 3 ~ x + y x—y 3 "^ 4 11 3. (3a:-2^=:l) ( 3?/ - 4a; = 1 f 4. 2^3 ^ 4. ^ 3'^4 Ans. x=i^, «/ = 7. Ans. a; = 18, ?/ = 6. Ans, x = 6, y = 'il' Ans. a? = — 6, y = 12, 104 SIMPLE EQUATIONS. 5. X y a b ! ^ , _ (ic{dn—hm) _ hd{am—cn) X y ' ~ ad — be ^ ^ ~ ad— be 'c'^d^ 210. ELIMINATION BY COMPARISON. Let it be required to solve the equations 72: + 62^ = 20 (1), 9a; — 4^ = 14 (2). From(l), y = ^^^ (3), g-P 14 and from (2), y = j (4) ; 9^—1^ 20 — 72; ,_. .-.(42,6), ___- = ___ . . . (5). Clearing of fractions, 27a; — 42 = 40 — 14ar; whence, a; = 2. Substituting 2 for x in either (3) or (4), we find y = 1. Br ULE. Find, from each of the given equations, an expression for the vahte of the unknown quantity to be eliminated, and equate the expressions thus obtained ; an equation will thus be formed con- taining only one unknown quantity. EXAMTIjES. Solve the following groups of simultaneous equations : ^ j4a; — 2v= 20) , _ ^- TWO Ui^'KITOWN ^UAifTITIES. 105 (2a;— v= 1) I 4. 5. 1/ S 7a; - 3?/ = 12 ( 2a; + 2i/ = 12 a:+ y = 21 ^x - y 3_3^ -1 2-T~'~^ a; + 2/ 2| Ans. X =z2, y = 3. Ans. x = 3, y = 3. Ans. x = 9, y = 12, Ans. a; = 3, y = 5. 211, Genekal Scholium. — In the solution of simultaneous equations, any of the preceding methods of ehmination can be used, as may be most convenient, each method having its advan- tages in particular cases. Generally, however, the equation ob- tained by using the second or third method contains fractional terms. This inconvenience is avoided if we eliminate by the first method. The second method may be preferable whenever the co- eflBcient of one of the unknown quantities in one of the given equations is unity; for then the inconvenience of which we have just spoken may be avoided. We shall sometimes have occasion to use the second and third methods, but generally the first method is preferable. EX A MP JjES Solve the following groups of simultaneous equations : 1. i^ + 2/ = in. ^^,.^^^11, 2/ =4. [x — y=l) ^ (3a;-2y = l) \3y-^x = l) (3a;-5y = 13) ( 2a; + 7y = 81 f ( 2a;-f-3y = 43) * (lOa;— y= 7) Ans. a; = 5, «/ = '^* Ans. a; = 16, y = Z Ans. x = 2, y = l3. 106 SIMPLE EQUATIONS. ( 6x- 1y= 33) ( 11a; + 127/ = 100 ) i3^-7.= 4) j 21y + 20^ = 165 ) = 295) ( 77y _ 30a; ( 5a: + 7?/ = 43) ( 11a; + 9«/ = 69 r (8a;-21y= 33) (6a; + 352/ = 177) ( 5a; -f- 7y = 41 ) j 16a; + 17y = 500 ) (17a:- 3?/ = 110) 12. 13. \ 14. 15. J 5 + 6 - ^^ ^-^-21 2 4~ - + ^-9 3 + 4-*^ ? + ^=7 4^6 2 3~ ^ + ^ = 1 3^4 a: + i2/ x—y ^ "2 3"= ^ ~3 4"- ^ ^W5. a: = 8, y = 1. Ans. x=z2, y = 6. Ans. a: = 3, y = 5. Ans. a: = 3, ?/ = 4. ^W5. x = 12, y = 3. ^?J5. a; = 4, y = d' Ans, a: = 10, ?/ = 2ft ^W5. a; = 60, ^ = 36i Ans. X = 12, ?/ = 20. Ans. x= —6, y=Vl Ans. a: = 18, y = Q. TWO UNKNOWN QUANTITIES. 16. 17. 18. 19. 11a; — oy_^x-\-y n " 16 8a; — 5^=1 3 4 + 2+^-^ 4+12 |_| + 2=^-2. + 6 x = 4.y \{%x + ly)^l=.h2x-Qy-\-l) 107 Ans, X = 11, y =zll, Ans. x = 2f y='7. Ans. a;=4, y=l. x + l(3x-y-l)^l + l{y-l) 20. 21. 5(4x + 3y)=g + 2 ' 3a;— -Sy , , 2x + y x — 2y _x y ^ ^-2 + 3 3aj ;/ 4 _ a; «/ 10 ~" 15 ~ 9 ~ 12 ~ 18 o. s _ ^ .V , H ''^ ~ 3 - 12 ~ 15 "^ 10 ^/i5. a; = 3-J, y ^ 6f . Jw5. a; = 12, y = 6, Ans. a; = 2, ^ = — 1. 22. 42; — 3.y — 7 _ 3a; 2y 5 5 ~10 15 6 ?/— 1 a; Sy_y—x a; 11 ""3" "^ 2""20~"15~ "^6"^ 10 Ans. x = 3f y=2. 23. ^ 2a; 6y 3a; y 3 12 2 3 7 23 4 2 X- 1^ _1 La; + 2/ 5 Ans. a; = 18, y = 12. 108 SIMPLE EQUATIONS. (12a;- Gv = a ) 25. X + t- 2 m 71 X U- 1 {m n Ans. X = y =:-. . 3m n Ans. a; r= -;^ , y = ^. 2 ' 26. a b 3a 6b 3 ( 7nx — ny = a ) Ans. X =:3a, y = 2b. en -\-bd _ cm — ad bm -\- an' ^ ~ bra + an ' Ans. X = r, : , y 28. - 29. - b-\-c a-\-c ax — %_1 {a- b)^c~' X a-\-b ^ a-b x-y Aab = 1 = %a Ans. x-=b -\- c, y =ia -{• c. Ans. x={a + by, y = {a — by. SIMPLE EQUATIONS WITH AXT NUMBER OF UNKNOWN QUANTITIES. 212. To solve a group of simple equations contain- ing any number of unknown quantities. Let it be required to solve the equations Sx-\-4:y-2z = 10 . . . (1), 6a; — 2?/ + 32 = 16 . . . (2), 4a; + 2^ + 2;z = 22 . . . (3). AN^T NUMBER OF UNKITOWIS" QUANTITIES. 109 < Combining (1) and (3), also (1) and (3), eliminating z in each case, we have the new group 19a; + 8?/ = 62 . . . (4), ^ 7r?: + 6?/ = 32 . . . (5). Combining (4) and (5), eliminating y, we have 29ic = 58; whence, . a; = 2. Substituting 2 for x in (5), we have 14 + 6y = 32; whence, 2/ = ^• Substituting 2 for x and 3 for y in (1), we have 6 + 12 — 2;z = 10; whence, z — L B ULE. I. Cofnhine one equation of the group with each of the others, eliminating the same unknoiun quantity in each case ; there tuill result a new group containing one equation less than the original group. II. Combine one equation of the resulting group tvith each of the others, eliminating a second unknown quantity; there ivill result a new group containing two equations less than the original group. III. Continue the operation until a single equation is found, contaiiiing only one unknoiun quantity. IV. Find the value of this unknown quantity hy the rule of Art. 197; substitute this value in either one of the group of tioo equations, and find the value of a second unknoimi quantity; then substitute the two values thus fou7ul in any one of the group of three equations, and find the value of a third unknown quan- tity ; and so on, till the values of all are found. •110 SIMPLE EQUATIONS. ScH.— When any one of the unknown quantities does not occur in all the equations, it will generally be best to eliminate that quantity first. EXAMPLES. Solve the following groups of simultaneous equations : 1. )2x — dtj+z = iy' Ans. x = S,y = 2yZ = l. (dx— y + 22 = 9 ) (dx-{'2t/-'4:Z = 15) 2. }6x — 3y-\-2z = 2S>' Ans. x=il,y = b,z = L [dy ^^— x = 2^) I a;+ y— z = l ) 3. <8x + 3y — 6z = l>' Ans. x = 2, y = 3^ z=z 4, i3z — 4^— y=l) (2x-7y + ^= 0) 4. l3x — 3y-\--z=o[' Ans. x=l, y = 2, z = 3. idx + by -{-3z = 28 ) iLx — 3y-\-2z=9\ 5. < 2x + 5y — 3z = 4 ?■ • A^is. x = 2^ y =z 3, z = 5. ( 5a: + 6?/ — 22 = 18 ) (2x-^ 4:y + 92 = 28 I 6. •] 7a; + 3y — 6z = 3>, Ans. x = 2, y = 3, z = 4:. i 2x 4- lOy — ll2 = 4 ) ( x — 2y-\-3z= 6 ^ 7. \2x-{-3y — 4zz=i20y' - Ans. x = S,y=z4,z = 2. {3x — 2y + 5z = 2Q) (4^-^3y + 2z = 4L0) 8. -j 5a; + 9?/ — 72 = 47 [■ • Ans. x = 10,y = 2j 2=3. ( 9a; + 82/ — 32 = 97 ) ( Sx-\-2y -{- z = 23) 9. I 5a; + 2z/ + 42 = 46 y . A71S. x = 4=, y = 3, z = 0, (lOa; + 5?/ + 42 = 75 ) ANY NUMBER OF UNKNOWN QUANTITIES. Ill 10. 11. 12. 13. 14. 16. 17. DX — 6y -\- 4:Z =z 15 7a; -\.4:y — dz = ld ^^ + y + Qz = ^6 x-{-y -{-z = 31)' X -j- y — z = 26 [' X — y — z= 9; x-}-y^z = 26) X — y = 4 V • X— z = 6 ) x — y — z= (j\ 3y — x — z = 12\' '^z — y — x = 24:) Ans. X =z S, y = 4:, z =z 6, Ans, ic = 20, y = S, z^3. Ans. X = 12j y = S, z = 6. Ans. X = 39, «/ = 21, z = 12. t3y-l 4 ~ 6z 5 X 2 -1 5x ^ = y + 5 6 3a: + 1 L 7 z 14 + 1 6 =1- " 10a; 4- 4y -6z = 4a; + Gy 5 9 Ans, x=2, y=3j z=l. 3z 10a; -]- 4:y — 5z = 4x -h 6y — 3z — 8 10a; -{- 4ry — 5z 4x -\- Qy — 3z x -^ y ■{■ z 10 + 3 20 46 Ans. x = e,y = —,z = Y' '7x-3y = l} llz —7u = l 4z —7y = 1 19a; — 3w = 1 "3u — 2y= 2 5x — 7z = 11 2a; + 3y = 39 I4:y -\-3z =4:1 ) Ans. x = 4, y = 0, z = IGf u = 'Z5« Ans. X = 12, y = 5, z z= 7, u zzz 4. 112 SIMPLE EQUATIONS. 18. 19. 2x 37/ -\-2z = 13 4:1/ -\-2z = 14 4:U — 2x = 30 5y + ^u = 32 J Ans. x=3, ?/=!, z=5, u=zd. Hu — 13z = 87 lOy— 3a; = 11 Sic + 14a; = 57 2x — llz = 50 >■ . Ans. x = d, y = 2, z=: — 4:, uz=6. ['^x-2z + Su = 17 ^ 4:y — 2z + ?; = 11 20. ^ 5y — 3x — 2u=z 8 % — 3?^ 4- 2v = 9 3z -^Su =33 ^W5. a; = 2, ?/ = 4, 2 = 3, ?^ = 3, ?; = 1. ' 3a; — 4?/ + 32J + 3t; — Gu = 11 ^ 3a; — 5?/ 4- 2z — 4w = 11 21. ^ lOy — 3z+3u — 2v= 2 5;z + 4w + 2v — 2a; = 3 Qu — 3v -{- 4:X — 2y = 6 ^?i5. a; = 2, 2/ = 1, 2; = 3, 2^ = — 1, ?; = — 2. X y ^ 22. ^ ? + i = 1 c (ay+I}x = c) 23. -^ ca; + a^; = 5 )■ . \ iz -\- cy =: a) Ans. X = :tt , y 2bc ^ . a h c Ans. x=z^, y--^, ZZZZ-, 2ac z = a^ + l? — (f 2ab r a; + «= ?/+ z 24. K 2/ 4- fl^ = 2a; + 25? ( z + a = 3x -]-3y Ans. X -a,y^-^a,z = -a. ANY NUMBER OF UNKNOWN QUANTITIES. 113 25. I {b-^c)x+{a + c)y+ (a + b)z i hex + acy + abz = 1 1 1 Ans. X = V y 26. 27. 28. 29. 30. (/f-a){b-cy (a—b){a—c) ax -\- hy -\- cz =: A. \ a^x 4- l^y + c^^ = A2 V . Ans. x a^x -\- l^y + c^^ = A^ ) x-\-y-\-z = a + h-\-c \ hx -\- cy -\- az ^cx -{- ay + hz\ . ex -\- ay -\- iz = a^ -}- b^ + c^ ) X — ay -^ a^z^a^ \ x—by+b^z = b^>. X — cy -]- ch = c^ ) ex -\- y -}- az = 2a ) c^x -\- y -}- ah = 2ac Y . aex — y + aez=za^-{-c^) Ans. x=z (c—a){c—b) A{A-b){A~ c) a(a — b) {a — c) Ans. x = b + e Ans. X = abc. a-\-l y=ia — e, 2; = ^ It + V -\- to -\- X -\- y ^16 V -\-w + x-^y-{-z =20 w +x -{- y -{■ z -\- u =19 X -\- y -\-z -^u -{-V =1^ y -\- z -{-ti -\-v -\-w = l'^ ^Z -\- U -\- V -{• W -\- X =z\^ ) Ans. u = l, 1; = 2, ?(; = 3, ic = 4, y = 6, z=6. 213. FROBIjEMS. 1. A and B engage in play ; in the first game A wins as much as he had and four dollars more, and finds he has twice as much as B ; in the second game B wins half as much as he had at first and one dollar more, and then it appears he has three times as much as A ; what sum had each at first ? Let X = the number of dollars which A had, and y = the number of dollars which B had; 114 SIMPLE EQUATIONS. then after the first game A has 2a; 4-4 dollars, and B has y—x—^ dollars. . • . by the first condition, 2a; -I- 4 = 2 (?^ - a; -4) . . . (1). Again, after the second game A has 2a; + 4 — *^ — 1 dollars, and Bhas?/ — a; — 4 + ^-fl dollars. . • . by the second condition, y-a;-4 + | + l=3(2.T + 4-|-l) . . .(2). By transposition and reduction, (1) and (2) become, y — 2x= 6 . . . (3), 3y - 7a; = 12 . . . (4). Multiplying (3) by 3, 3?/ — 6a? = 18 . . . (5). Subtracting (4) from (5), a; = 6. Substituting 6 for x in (3), we find 2/ = 18. 2. A sum of money was divided equally among a certain num- ber of persons ; had there been three more persons, each would have received one dollar less, and had the number of persons been two less, each would have received one dollar more than he did ; what was the number of persons, and what did each receive ? Let X = the number of persons, and y = the number of dollars each received; then xy dollars is the sum divided. By the conditions of the problem, the sum divided is also ex- pressed by (x + 3) (y - 1), or (a; - 2) (2/ + 1) ; .-. (43, 6), we have. AN^Y NUMBER OF UKKKOWN QUAi^TITIES. 115 {x-\-2){y-l)=xy . . . (1), {x-^)(y + l)=xy . . . (2). By transposition and reduction, (1) and (2) become Zy-x = Z . . . (3), x-^ = 2 . . . (4). Eliminating x from (3) and (4), ^ — '^y — 5, or «/ = 5; .-. by (4), a; rrz 2?/ + 2 = 10 + 2 = 12. 3. What fraction is that which becomes equal to J when its numerator is increased by 6, and equal to \ when its denominator is diminished by 2 ? Let a; = the numerator, and y = the denominator of the fraction ; then, by the conditions of the problem, 3 y -^ ' ' ' W> X 1 y-%-2 ' ' ' (2). Clearing of fractions, transposing and reducing, 3y — 4:X = 24: . . • (3), y-2x = 2 . . • (4). Multiplying (4) by 2, and subtracting the result from (3), we y = 20; .-. by (4), x= 9. Therefore the required fraction is ^. 4. Find two numbers whose sum is «, and whose difference is b. 116 SIMPLE EQUATIONS. Let a; = the gi*eater number, and y = the less number ; then, by the conditions of the problem, x-\-y = a . . . x — y = b . . . (1), (3); whence, a , b ^ = 2 + 2' and a b Since a and b are any numbers whatever, we have the follow- ing general principles, by means of which all similai* problems can be solved : 1. The greater of two numbers is found by adding half their differe7ice to half their sum. 2. TJie less of two numbers is found by subtracting half their difference from half their sum. 5. A and B together possess $570. If A's money were three times what it really is, and B's five times what it really is, the sum would be ^2350. How much money does each possess ? Ans. A $250, B $320. 6. Find two numbers such that if the first be added to four times the second, the sum is 29 ; and if the second be added to six times the first, the sum is 36. Ans. 5 and 6. 7. K A's money were increased by $36, he would have three times as much as B ; but if B's money were diminished by $5, he would have half as much as A. How much has each ? Ans. A f542, B |26. 8. A and B lay a wager of $10 ; if A loses, he will have $25 less than twice as much as B will then have; but if B loses, he will have five-seventeenths of what A will then have. How much money has each? Ans. A $75, B 835. 9. For $21, either 32 pounds of tea and 15 pounds of coffee, or 36 pounds of tea and 9 pounds of coffee, can be bought ; find the price per pound of each. Ayis. Tea 50 cts., coffee o^\ cts. ANY NUMBEE OF UNKNOWN QUANTITIES. 117 10. A pound of tea and three pounds of sugar cost $1.20; but 11 lea were to rise 50 per cent, and sugar 10 per cent., they would cost $1.56 ; find the price per pound of each. Ans. Tea 60 cents, sugar 20 cents. 11. A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and in 10 days ; in what time could each person alone perform the same work ? Ans. A, 14}f days; B, 17f|; C, 23^. 12. A and B together can perform a piece of work in a days, A and C together in b days, and B and C together in c days ; in what time could each person alone perform the same work ? . . 2abc , A in f 7 days, ac + oc — ao *' Ans. ^ B m -7 ^ days, ab ■\- be — ac '' ^ . 2abc , C m —, r- days. ab -\- ac ^ be '' 13. A person possesses a certain capital, which is invested at a certain rate per cent. A second person has $1000 more than the first, and investing his capital one per cent, more advantageously, has an income greater by $80. A third person has $1500 more capital than the first, and investing it two per cent, more advan- tageously, has an income greater by $150. Find the capital of each person and tlie rate at which it is invested. j Sums at interest, $3000, $4000, $4500. ( Rates of interest, 4, 5, 6 percent. 14. If there were no accidents, it would take half as long to travel the distance from A to B by railroad as by coach ; but three hours being allowed for accidental stoppages by the former, the coach will travel all the distance but fifteen miles in the same time ; if the distance were two-thirds as great as it is, and the same time allowed for railway stoppages, the coach would take exactly the same time ; find the distance from A to B. Ans. 90 miles. 15. A and B are set to a piece of work which they can finish in thirty days, working together, for which they are to receive 118 SIMPLE EQUATIONS. $64. When the work is half finished, A rests eight days and B four days, in consequence of which the work occupies five and a half days more than it would otherwise have done. How much ought each to receive? Aiis. A $22, B 142. 16. A and B run a mile. First A gives B a start of 44 yards, and beats him by 51 seconds; at the second heat A gives B a start of 1 minute and 15 seconds, and is beaten by 88 yards. In what time can each run a mile ? Ans. A in 5 minutes, B in 6 minutes. 17. A and B start together from the foot of a mountain to go to tlie summit. A would reach it half an hour before B, but, missing his way, goes a mile and back again needlessly, dur- ing which he walks at twice his former pace, and reaches the top six minutes before B. C starts twenty minutes after A and B, and, walking at the rate of two and one-seventh miles per hour, arrives at the summit ten minutes after B. Find A's and B's rates of walking, and the distance from the foot to the summit of the mountain. Ans. 2^, 2 miles per hour; distance, 5 miles. 18. A number expressed by two digits is four times the sum of the digits, and if 27 be added to the number the order of the digits will be inverted; find the number. Let X = the left digit, and y = the right digit; then, since x stainds in the place of tens, the number will be rep- resented by 10^ + y. .*. by the first condition, 10x + y = 4.{x-{-y). . . (1); and by the second condition, lOx + y -{- 211 = lOy -}-x. . , (2). Solving these equations, we find x = 3f and y = 6; . • . lOo: 4- y = 30 4- 6 = 36, the number required. 19. A number is expressed by three digits. The middle digit is equal to twice the left-hand digit, and greater by 3 than the ANY NUMBEK OF UKKNOWN QUANTITIES. 119 right-hand digit. If 99 be subtracted from the number, the order of the digits will be inverted; find the number. Ans. 241. 20. A number consisting of two digits contains the sum of its digits four times, and their product three times; find the number. Ans. 24. 21. A railway train, after travehng for one hour, meets with an accident which delays it one hour, after which it proceeds at three-fifths of its former rate, and arrives at the terminus three hours behind time; had the accident occurred 50 miles further on, the train would have arrived 1 hour and 20 minutes sooner; find the length of the line, and the original rate of the train. Ans. 100 miles; original rate, 25 miles per hour. 22. A railway train, running from London to Cambridge, meets with an accident which causes it to diminish its speed to -th n of what it was before, in consequence of which it is a. hours late. If the accident had occurred b miles nearer Cambridge, the train would have been c hours late. Find the original rate of the train. Ans. — miles per hour. a — c ^ 23. The fore-wheel of a carriage makes six revolutions more than the hind-wheel in going 120 yards. If the circumference of the fore-wheel be increased by one-fourth of its present size, and the circumference of the hind-wheel by one-fifth of its present size, the six will be changed to four. Required the circumference of each wheel. A71S. 4 yards and 5 yards. 24. A man starts p hours before a coach, and both travel uni- formly ; the latter passes the former after a certain number of hours. From this point the coach increases its speed to six-fifths of its former rate, while the man increases his to five-fourths of his former rate, and they continue at these increased rates for g hours longer than it took the coach to overtake the man. They are then 92 miles apart; but had they continued for the same length of time at their original rates, they would have been only 80 miles apart. Show^ that the original rate of the coach is twice that of the man. Also, if p + q = 1G, show that the original rate of the coach was 10 miles per hour. 120 SIMPLE EQUATIONS. 314. SYNOPSIS FOR REVIEW. CO Eh Members . Kinds First Member. Second Member. Identical Equation. Equation of Condition. Numerical Equation. Literal Equation. Klnds respecting \ *^^, ' . Degree. U^^adratw. ' Higher C Cubic. \ Biquadratic, etc. Transformations. { Clearing of Fractions. Rule. ( Transposition of Terms. Rule. Cor. Solution of Simple Equations containing only one Un- known Quantity. Solution of Pros. Simultaneous Equations. Statement. Solution of Equation. Rule. fEliminav'on by addi" tion or subtraction. Rule. Sch. 1, 2,3.4. EliminatioD by substi- tution. Rule. Elimination by com- parison. Rule. , General Sch. Rule for groups of Equations containing any number of unknown quantities. ' Solution for two un- known quantities. DISCUSSION OF PROBLEMS LEADING TO SIMPLE EQUATIONS. 215. After a problem has been solved, we may inquire what values the unknown quantities will have, when particular suppo- sitions are made with regard to the given quantities. The deter- mination of these values, and their interpretation, constitute the Discussion of the I*rohlem, DISCUSSIOK OF PROBLEMS. 121 216. IKTEEPRETATION" OF K^EGATIVE RESULTS. 1. What number must be added to a number a in order that the sum may be ^ ? Let X = the required number ; then, by the question, a + ic = $; whence. x=ih — a. This is a general solution, a and b being arbitrary quantities. If a=^ 12, and b = 25, we have a; = 25 — 12 = 13. But suppose a = 30, and b = 24=; then a; = 24 — 30 = — 6. How is this negative result to be interpreted ? If we recur to the enunciation of the problem, we see that it now reads thus : What number must be added to 30 in order that the sum may be 24 ? Here it is obvious that if the words added and sum are to re- tain their arithmetical meanings, the proposed problem is impos- sible. But we see at the same time that the following problem can be solved : What number must be taken from 30 in order that the difference may be 24 ? The answer to this problem is 6. The second enunciation differs from the first in these respects : The words added to are replaced by taken from, and the word sicm by difference. Hence we may say that, in this example, the negative result indicates that the problem, in a strictly arithmetical sense, is im- possible ; but that a new problem can be formed by appropriate changes in the original enunciation, to which the absolute value of the negative result will be the correct answer. This indicates the convenience of using the word add, in Alge- bra, in a more extensive sense than it has in Arithmetic. Let X denote a quantity which is to be added algebraically to a ; then the algebraic sum is a -{- x, whether x be positive or neg- ative. Hence, the equation a -\- x = b will be possible algebraically, whether a be greater or less than b. 122 SIMPLE EQUATIONS. 2. A's age is a years, and B's age is l years; when will A be twice as old as B? Let X = the required number of years ; then, by the question, a-\-x=:2{b -\-x); whence, x=a — 2b. K a = 40 and b = 15, then re = 40 — 30 = 10. But suppose a = 35 and h = 20, then a; = 35 — 40 = — 5. Here, as in the preceding problem, we are led to inquire into the meaning of the negative result. Now, with the assigned values of a and Z>, the equation which we have to solve becomes 35 4- a; = 40 + 2x. This equation is impossible, if a strictly arithmetical meaning is to be given to the symbols x and + , for 40 is greater than 35, and 2x is greater than x. But let us change the enunciation to the following : A's age is 35 years, and B's age is 20 years ; when tvas A twice as old as B ? Let x = the required number of years ; then, by the question, 35 — re = 2 (20 — a:) = 40 — 2rr; whence, x = 5. Here again, we may say the negative result indicates that the problem, in a strictly arithmetical sense, is impossible ; but that a new problem can be formed by appropriate changes in the original enunciation, to which the absolute value of the negative result will be the correct answer. We may observe that the equation corresponding to the new enunciation may be obtained from the original equation by changing x into — x. Suppose the problem had been originally enunciated thus: A's age is a years, and B's age is b years; find the epoch at which A's age is twice that of B. DISCUSSION or PROBLEMS. 123 We cannot tell from the enunciation of the problem whether tlie required epoch is before or after the present date. If we sup- pose the required epoch to be x years after the present date, we obtain x-=a — %l). If we suppose the required epoch to be x years lefore the present date, we obtain x:=^%b—a. If 25 is less than a, the first supposition is correct, since it leads to a positive value for x ; the second supposition is incorrect, since it leads to a negative value for a*. If V) is greater than a, the second supposition is correct, since it leads to a positive value for x\ the first supposition is incorrect, since it leads to a negative value for x. Here we may say, then, that a negative result indicates that we made the wrong choice out of two possible suppositions which the problem allowed. But it is important to notice, that when we discover that we have made the wrong choice, it is not neces- sary to go through the whole investigation again, for we can make use of the result obtained on the wrong supposition. We have only to take the absolute value of the negative result and place the epoch before the present date if we had supposed it after, or after the present date if we had supposed it before. 3. A's age is a years, and B's age is h years ; when was A twice as old as B ? Let X = the required number of years ; then a — x:=2{b — x)', whence, x = 2b — a. Now let us verify the solution. Substituting 2J — « for x, we have a — X = a — {2b — a) = 2a — 2b; and 2{b — x) = 2{b-'2b-^a) = 2a — 2b. If b is less than «, these results are positive, and there is no arithmetical difiiculty. But if b is greater than «, although the 124 SIMPLE EQUATIONS. two members are algebraically equal, yet, since they are both neg- ative quantities, we cannot say that we have arithmetically veri- fied the solution ; and when we recur to the problem, we see that it is impossible if a is less than i ; because, if at a given date A's age is less than B's, then A's age never was twice B's, and never will be. Or, without proceeding to verify the result, we may observe that if d is greater than a, then x is also greater than a, which is inadmissible. Thus it appears that a problem may be really absurd, and yet the result may not immediately present any difficulty, though when we proceed to examine or verify this result, we may discover an intimation of the absurdity. The equation a 4- a: = 2 (J + a:) may be considered as the symbolical expression of the following verbal enunciation : Suppose a and J to be two quantities ; what quantity must be added to each, so that the first sum may be twice the second ? Here the words quantity, sum, and added may all be under- stood in algebraic senses, so that x, a, and J may be positive or negative. This algebraic statement includes the arithmetical question about the ages of A and B. It appears, then, that when we translate a problem into an equation, the same equation may be the symbolical expression of a more comprehensive problem than that from which it was de- rived. When the solution of a problem leads to a negative result, and the student wishes to form an analogous problem that shall lead to the corresponding positive result, he may proceed thus: Change x into — x m the equation that has been obtained, and then, if possible, modify the verbal statement of the problem, so as to make it coincident with the new equation. We say if j^ossible, because in some cases no such verbal mod- ification seems attainable, and the problem may then be regarded ( DISCUSSION OF PROBLEMS. 125 as altogether impossible. To illustrate, take the following prob- lem: 4. A's age is 20 years, and B's age is 30 years ; when will the age of A be twice that of B ? Let X = the required number of years ; then 20 + 2; = 2 (30 + a;) = 60 + 2a;; whence, a; = — 40. This negative result shows that the epoch is not in the future. Suppose it to be in the past. Changing x into — x, the original equation becomes 20 — a: = 2 (30 — a;) ; whence, x = 40. This result seems to indicate that 40 years ago — that is, 20 years before A was bom, and 10 years before B was born— A was twice as old as B. A manifest absurdity. Hence, the problem is an impossible one. Prii^ciples — 1. A negative result may arise from the fact that the problem, contains a condition ivhich cannot he arithmet- ically satisfied ; or from the fact that, of tivo possible suppositions respecting the quality of a quantity, ive adopted the wrong one. 2. After a problem has been translated into an equation, the qualify of any quantity involved will be changed, if we change the sign of the symbol of that quantity, PJROBLJE31S. 1. A father's age is 40 years, and his son's age is 13 years ; when will the age of the father be four times that of the son ? Ans. x = — 4:. Modify the enunciation so that the result shall be + 4. 2. Find two numbers whose sum is 2 and difference 8. Ans. — 3 and -|- 5. Modify the enunciation so that the result shall be +3 and +5. 126 SIMPLE EQUATIONS. 3. The difference of two numbers is 6, and four times the exceeds five times the greater hy 12 ; find the numbers. Ans. — 42 and — 36. Modify the enunciation so that the result shall be -\- 42 and + 36. 4. Two men, A and B, began trade at the same time, A having three times as much money as B. When A had gained $400 and B 1150, A had twice as much money as B ; how much did each have at first ? Ans, A was in debt $300, and B $100. Modify as in the preceding examples. 5. There are two numbers whose difference is a ; and if three times the greater be added to five times the less, the sum will be b. What are the numbers ? . b -\- 5a , b — Sa Ans. — - — and — - — . o o Interpret this result when a = 24 and b = 48. 6. Two men were traveling on the same road toward Boston, A at the rate of a miles per hour, and B at the rate of b miles per hour. At 6 o'clock a.m. A was at a point m miles from Boston, and at 10 o'clock a.m. B was at a point n miles from Boston. When did A pass B ? A71S. 7 — hours after 6 o'clock a.m. a — b If m = 36, n = 28, a = 5, and 5 = 3, at what time did A pass B? ZERO AND INFINITY. FINITE, DETERMINATE, AND INDETER- MINATE QUANTITIES. 217. The symbol 0, called Nothing, or Zero, is used to denote the absence of value, or to represent a quantity less than any assignable value. A quantity less than any assignable value is sometimes called an Inftnitesinial, 218. The symbol 00 , called Infinity^ is used to represent a quantity greater than any assignable value. ZERO AND INFIlflTY. 127 219. A Finite Quantity is one whose absolute value is comprisea between the limits and oo . 220. A Determinate Quantity is one which has only SL finite number of values. 221. An Indeterminate Quantity is one which has an infinite number of values. A A INTERPRETATION OF THE FORMS — , — , -r-, -, oo X 0, U 00 j^ y) AND 00 — 00 . 00 222. In order to explain the meaning of these symbols, let us consider the fraction :jr-. 1. Suppose A to be finite, and to remam unchanged, while B continually decreases ; then the value of the fraction -^ will con- tinually increase. Thus: If B = A; then ^=lj IfBz=:|; then ^ = 2; IfB = A; then 1 = 10; If B = A; then ^ = 100; Hence, it is evident that, when B becomes less than any assign- e quantity, the frac able quantity ; hence, able quantity, the fraction ^ will become greater than any assign- A 2. If the denominator B is made to increase continually, while the nunjerator A remains unchanged, then the value of the frac- 128 SIMPLE EQUATIONS. tion rrj will continually decrease ; and when the denominator B becomes greater than any assignable quantity, the fraction will become less than any assignable quantity ; hence, 00 3. If the numerator A is made to decrease continually, while the denominator B remains unchanged, then the value of the fraction .^ will continually decrease ; and when A becomes less than any assignable quantity, the fraction will also become less than any assignable quantity ; hence, 1 = 0. 4. Multiplying both members of the equation ^ = by B, we have = B X 0. Dividing both members of this equation by 0, we have But B is any finite quantity ; hence tt is a Symbol of In~ det elimination (221). 6. Multiplying both members of the equation — = by c» , we have A = oo X 0. But A is any finite quantity ; hence oo x is a symbol of in- determination. 6. We may place the equation x = B under the following form: 1 INFINITESIMALS AND INPINITIES. 129 But 7: = ^ ; hence, 00 -D . therefore ^o is a symbol of indetermination. 7. In the identity ^ = — r- make o^ = and 5 = ; we -^ a b ab ^ then have 1_1_0 0"~ that is, 00 — 00 = - ; hence oo — oo is a symbol of indetermination. ORDERS OF INFINITESIMALS AND INFINITIES. 223. An Inflnitesinial of the First Order is one that is infinitely small in comparison with a finite quantity; that is, so small that it may be contained in a finite quantity an infinite number of times. An Infiiiitesifnal of the Second Order is one that is infinitely small in comparison with an infinitesimal of the first order. An Jnfinitesi' mat of the Third Ordet* is one that is infinitely small in comparison with an infinitesimal of the second order; and so on. In order to illustrate, let us consider the continued identity 1 _ X _x^ _a^ Let X be an infinitesimal of the first order ; then - = oo ; that ' X is, 1 is infinitely great in comparison with x. Again, since 1 r?/ 1 X - =: —, and - = 00 , it follows that -5 = oo ; that is, x is infinitely X Xt X X great in comparison with x^\ but x is, by hypothesis, an infinitesi- mal of the first order; therefore x^ is an infinitesimal of the sec- ond order. In like manner, it may be shown that 7? is an infi- nitesimal of the third order, and so on. 130 SIMPLE EQUATIONS. 234. Infinities are of different orders also. Let x be an infi- nitesimal of the first order, and A any finite quantity; then, AAA — = 00 ... (1), -2 = 00 ... (2), ;;5 = 00 • • • (3), and so on. Now the denominator in the first member of (1) is infinitely great in comparison with the denominator in the first member of (2) ; therefore the second member of (2) is infinitely great in com- pai'ison with the second member of (1). In hke manner it may be shown that the oo in (3) is infinitely great in comparison with the oo in (2) ; and so on. 235. PROBLEM OF THE COURIERS. The discussion of the following problem, originally proposed by Clairaut, will serve to illustrate some of the preceding prin- ciples : Two couriers, A and B, were traveling along the same road and in the same direction, namely, from C toward C ; the former going at the rate of m miles per hour and the latter at the rate of n miles per hour. At 12 o'clock, A was at P, and B was d miles in advance of A. When were the couriers together ? q: IP 12 !£ 2 We cannot teU from the enunciation whether the couriers were together before or after 12 o'clock ; but in order to effect a statement of the problem, we will suppose the required time to be after 12 o'clock. We must then regard time after 12 o'clock as positive, and time before 12 o'clock as negative. Suppose R to be the point where the couriers met, and Q to be the point where B was at 12 o'clock. Let X = the required number of hours ; then, since A traveled at the rate of m miles per hour, and B at the rate of n miles per hour, we have PR = mx^ and QR = nx. But PR = PQ + QR; mx =z d -\- 7ix', whence, x = DISCUSSION. 131 d m — n DISCUSSION. I. Suppose 771 > 71. Under this hypothesis the vakie of x will be positive, because the denominator m — 7i is positive. Now, since x is positive, we infer that the couriers were together after 12 o'clock. This conclusion is consistent with the conditions of the prob- lem. For, the supposition is that A was traveling faster than B. A would therefore gain upon B, and overtake him some time after 12 o'clock. n. Suppose m < w. Under this hypothesis the value of x will be negative, because the denominator m — n is negative. This imphes that the cou- riers were together before 12 o'clock. This interpretation, also, agrees with the conditions of the problem under the present hypothesis. For, since m <,7i,B was traveling faster than A ; and, as B was in advance of A at 12 o'clock, he must have passed A before that time. III. Suppose m = n. Under this hypothesis we shall have d . ^ = ^ = 00. This result implies that the time to elapse before the couriers are together is greater than any assignable quantity, or infinity ; therefore they can never be together. This interpretation is in accordance with the conditions of the problem under the present hypothesis. For, at 12 o'clock the couriers were d miles apart ; and, i?m = n, they were traveling at equal rates. Hence, they could continue to travel forever without meeting. IV. Suppose d = 0, and myn, or m n, or m < w, the couriers were traveling at difierent rates, and must have been either approaching or receding from each other at all times except at the moment of ^ = 0, and m = n. V. Suppose Then 0' Here the value of ic is represented by one of the symbols of in- determination. This result implies that the couriers were to- gether all the time. This conclusion is evidently confirmed by the conditions of the problem. For, if rf = 0, the couriers were together at 12 o'clock ; and, if m = n, they were traveling at equal rates, and would never separate. 236. SYNOPSIS FOR REVIEW. CHAP. Yll.— Con. DISCUSSIONS. Teems Used . Interpretation OP f Principle 1. Neg. Results. [ Principle 2. Zero. i First order. Second order. Third order, etc. Infinity. Determinate Quantity. Indeterminate Quantity. I Finite Quantity. Symbols of indetermination. i x, oo x n 5o , oo — oo A Symbols of zero, Problem of Couriers. I A' 00 CHAPTER VIII. VANISHL\G FRACTIOXS.-INDETERMIMTE EQUATIONS AW PROS- LEMS.-IXCOMPATIBLE EQUATIONS. VANISHING FRACTJONS. 227. A Vanishing Fraction is one which, on a cer- tain supposition, assumes the form of indetermination. Thus, assumes the form of - , if a; = «. X — a The vahie of a fraction sometimes reduces to the form of-, for a particular supposition, in consequence of the existence of a factor common to both terms, which factor reduces to for that suppo- sition. Thus, the fraction tt^-t A reduces to the form of -, db (a — b) if fl = 6, because the factor a — b becomes in that particular case. But if this factor be canceled, and the supposition that 2 a = ^ be made afterward, the value of the fraction will be 7:. Before deciding, therefore, upon the nature of the symbol -, we must ascertain whether it results from a factor common to both terms, which reduces to for the supposition made; if it does not, the value of the fraction is really indeterminate. MULE. I. Reduce the given fraction to its loioest terms. II. Malce tlie supposition tuhich would cause the original frac- tion to assume the form of indetermination ; the result will be the value of the fraction for that supposition. 134 INDETEBMINATE EQUATIONS. EXAMPLES. 1. Find the value of —. — —^, when x = v. x^ — f- Canceling the common factor x^ — y% we have, which, when a; = y, reduces to 'Hy^', p^^ = 2y% when x = y. This may be expressed algebraically as follows : 2. Find the value of \ ^^(^-^) \ ^^8. 14, 1(1 -\-x){x — l) I « = !. 3. Find the value of 1 1 ^^~ ^l^ I Ans, 0. (3(a3— 62) ) a = b. 4. Find the value of \ ~ J > Ans. oo . 5. Find the value of (-, -r — = — rr-^ ^l Ans. oo. w — 2aa:3 + 2a^x — ayx=a. 6. Find the value of ] x 7 — ^^^^-r^ i Ans. 1. (a — x (a + xy) x = a. INDETERMINATE EQUATIONS. 228. An Indeterminate Equation is one in which each of the unknown quantities has an infinite number of values. 229. A single equation containing tiuo or more unknown quaiitities is indeterminate. Suppose we have an equation containing two unknown quan- tities, X and y, for example, 2a; — 3?/ = 15. For every value INDETERMIJSJ-ATE EQUATIONS. 135 which we please to ascribe to one of the unknown quantities we can determine the corresponding value of the other, and thus find as many pairs of values as we please which satisfy the given equation. Thus, if y = l,2,3,4:,6 . . . .; then x=% lOJ, 12, 13^, 15 ... . Again, suppose we have an equation containing three unknown quantities, x, y, and- z ; for example, x -{- y -{- 2z =z 90. For every value which we please to ascribe to two of the unknown quanti- ties we can determine the corresponding value of the third, and thus find as many sets of values as we please which satisfy the given equation. Thus if i'^^^' 2, 3, 4, 5 , ^^"""^'^ 13^ = 0,1,2,5,8....; then X = 88, 85, 82, 77, 72 ... . A similar course of reasoning is applicable to an equation con- taining more than three unknown quantities. 330. Equations are indeterminate if the number of unknown quantities involved exceeds the number of equations. For, by eliminating, we can obtain a single equation contain- ing two or more unknown quantities, which is indeterminate (239). Thus, suppose we have the two equations x^ y-{-2z= 90 . . . (1), 6x -\- 2y — 2z = 366 . . . (2). Eliminating z, 6a; + 3?/ = 456 . . . (3), which is indeterminate. 231. An equation containing only one unhnoion quantity may he indeterminate in consequence of certain relations which subsist betiveen the known quantities. 136 INDETERMINATE EQUATIONS. If we solve the equation ax -[- b = ex -\- d d-b we obtain X = a — c Now, if d = b, and a = c, ^=0 • • (1), (2). (3); hence, under this hypothesis, the value of x is indetenninate. But, ]£ d = b, and a = c, (1) becomes ex -\- b = ex -{- b . . . (4), which is an identity, and may therefore be satisfied for any value of X (178). Here, then, we have one unknown quantity and 7io equation ; that is, no equation of condition (179). 232. Two equations involving tivo unknown quantities may be indeterminate in consequence of certain relations which sub list among the known quantities. (1), (2), (3), and V = "^ — ^ . . . (4). (5), and bcz= ad (6), If we solve the equations ax -\-by = r ex -\- dy = s obtain dr — bs ~ad—bc as — cr L y~ ad-bc Now, if dr = bs . . I be = ad . . then, by multiplying (5) by (6), member by member, and re- ducing, cr = as (7) ; .'. (3) and (4) become INDETERMIiq^ATE PEOBLEMS. 137 , x = -, and y = -. Let us now see what is implied by the relations (5) and (6). From (5) we have d = —, and from (6), c = ^ = —. These values of d and c reduce (2) to (1), and we then have only one equation containing two unknown quantities, which is inde- terminate. Cor.. — The four theorems which have just been demonstrated may be reduced to the following one : Indetermination arises if the number of unTcnown quantities exceeds the number of equations. INDETERMINATE PROBLEMS. 233. An Indeterminate JProbleni is one which ad- mits of an infinite number of solutions. We may often limit the number of solutions by imposing the condition that the values of the unknown quantities shall be pos- itive integers. When an indeterminate problem is expressed in algebraic lan- guage, it will be found that the number of unknown quantities exceeds the number of equations. PnOBZEMS. 1. A boy paid 50 cents for some apples and oranges, giving 2 cents each for apples and 10 cents each for oranges. How many of each did he buy ? Let X = the number of apples, and y = the number of oranges; then, by the question, 2x-\- 10?/ = 50; whence, a; = 25 — 5y. Now, if x and y are to be positive integers, y must be some in- teger between and 5. Let y= 1, 2, 3, 4; then X = 20, 15, 10, 5. 138 INDETERMINATE PROBLEMS. 2. Find two positive integers sucli that 12 times the one ex- ceeds 13 times the other by 9. Let X = one of the numbers, and y = the other ; then, by the question, 12a; — 13^ = 9; whence, x= ^^^ =^ + S^' t/ + 9 Since x and y are to be positive integers, ^ must b^ an integer. Let y + 9 whence^ ^ = 12?i — 9. r . 1 o o . ^i, ( 2^ = 3, 15, 27, 39 Let .. = 1, 2, 3,4....; then 1^^^^^^; 3^^ ^3^^^^ 3. A man bought 100 animals for $100 ; sheep at $3^ each^ calves at $1^, and pigs at $j^. How many did he buy of each kind ? Let X = the number of sheep, y = the number of calves, z = the number of pigs ; then, by the question, x+ y+ z = 100 . , . (1), 3^2; + li2/ + i^ = 100 . . . (2). Combining (1) and (2), eliminating z, 18a; + 5y = 300 ... (3); whence, y=60 — . . . (4). o From (1) it is evident, that if x and y are positive integers whose sum is less than 100, z will be a positive integer also. From (4) it is e\ddent that x must be a multiple of 5, and that 18a; —^ must be less than 60. o INCOMPATIBLE EQUATIOKS. 139 Let X =: 5, 10, 15 ; then y = 42, 24, 6, and z = 53, 66, 79. 4. The sum of three positive integers is 11 ; and if the first be multiplied by 3, the second by 5, and the third by 7, the sum of the products will be 57. What are the numbers ? I a; = 4, 3, 2, 1, Ans. i .y = 2, 4, 6, 8, ('z=6, 4, 3, 2. 5. Divide 200 into two parts, such that if one of them be di- vided by 6 and the other by 11, the respective remainders may be 5 and 4. Ans, 185, 15 ; 119, 81 ; 53, 147. 6. Can the equation 4a; + 6y = 27 be solved in positive in- tegers ? 7. Find the least number which, being divided by 5, leaves a remainder 3, and divided by 7 leaves a remainder 5. A^is. 33. 8. Solve the equation 8a; -f 13?/ = 159 in positive integers. . j a; = 15, 2. Ans. i _ ^. (y = 3y 11. INCOMPATIBLE EQUATIONS. 234. Incompatible liquations are those which can- not be satisfied for the same values of the unknown quantities. 235. Equations are said to be Independent when they express conditions essentially different, and Dependent when they express the same conditions under different forms. Thus, ] K _ qq f ^^® independent equations. But \ ^ J ~ ^^c are dependent equations, since the ( 2a; + 6i/ = 38 ) second may be obtained from the first by multiplying both mem- bers by 2. 236. If the number of independent equations exceeds the number of unknown quantities, these equations may be incom- patible. 140 INCOMPATIBLE EQUATIONS. Let US consider the three equations rr + y = 8 . . . (1), x-y=:2 . . . (2), ?=2 . . . (3). y Combining (1) and (2), we find a; = 5, and y = 3 ; but these values will not satisfy (3). In like manner it may be shown that the values which satisfy (2) and (3) do not satisfy (1), and that the values which satisfy (1) and (3) do not satisfy (2). 237. If the number of independent equations exceeds the number of tinknoiun quantities, such relations between the hnotun quantities can be found as icill make the equations compatible. Let us consider the equations x^y = s . . . • (1), x — y = d . . . (2). x = ay. . . (3). Combining (1) and (2), we find x = s -\- d 2 ' y = s-d 2 • Substituting these values in (3), s + d 2 ~ 'H- '-}■■ ince, s + d '' = s-d . . . (4). If the relation expressed by (4) subsists, (1), (2), and (3) will be compatible. Thus, the equations, x + y = 9, x — yz=zd, x = 2y, are compatible, for 94-3 2 = 3* SYNOPSIS FOK KEVIEW. 141 Cor. — In order that a problem may be determinate, the con- ditions must furnish as many independent equations as there are unknown quantities. ScH. 1. — When a problem contains more conditions than are necessary for determining the values of the unknown quantities, those that are unnecessary are termed redundant. ScH. 2. — A problem, from which incompatible equations are deduced, is called an impossible jwohlem. Such a problem is said to involve incompatible conditions. 338. SYNOPSIS FOR REVIEW. CHAPTER Vin. ' Vanishing Fractions. Investigation. Rule. 229. Theorems relat- ing TO Indetermi- nate Equations. 230, 231. 232. Reduction of the four theorems to ONE. Indeterminate Problems. No. solutions limited, how. i Terms j Dependent Equations. used. I Independent Equations. ^ *>36 Theorems. \ ' 237. Cor.;8ch.l,!2. CHAPTER IX, INEQUALITIES. 239. An Inequality consists of two expressions con- nected by the sign of inequality. The First Member of an inequality is the expression on the left of the sign of inequality, and the Second Member is the ex- pression on the right of the sign. 240. Two inequahties subsist in the same sense when the first member is the greater in each, or the less in each. Thus, 5 > 3 and 7 > 4 subsist in the same sense. Two inequahties subsist in a contrary sense when the first member is the greater in one, and the less in the other. Thus, 5 > 1 and 4 < 8 subsist in a contrary sense. 241. If the same quantity be added to, or subtracted from, each member of an inequality, the resulting inequality will subsist in the sams sense. For, suppose a > J ; then a — J is positive (118). Again, since a ± <^ — (& ± c) = a -— J, it follows that « ± c — (J ± ^) is positive ; Cob. 1. — The rule for the transposition of terms in equations is applicable to inequalities. Thus, if then a3 + £2 _ ^ab > )i>ab — 2ab + c^, or a2--2a& + ^>c8. INEQUALITIES. 143 Cor. 2. — If an equation be added to an inequality, member to member, or subtracted from it, member from member, the result- ing inequality will subsist in the same sense. Thus, if a>h, and ^ = 2^> then « ± ^ > ^ ± y« 242. If an inequality he subtracted from an equation^ mem- her from member, the sign of inequality will be reversed. For, suppose ^ = y, and « > ^ ; then x — a — {y — b)z=h — a, and 5 — a is negative ; x — a<^y — h, 243. If both members of an imquality be multiplied or di- vided by the same positive quantity, the resulting inequality will subsist in the same sense. For, suppose m to be positive, and a>h] then, since a — J is positive, m{a — h) and — (a — J) are pos- itive ; maymb and — > — m m 244. If both members of an inequality be multiplied or di- vided by the saine negative quantity, the resulting inequality ivill subsist in a contrary sense. i or, suppose m to be negative, and a>b', then, smce a — & is positive, m(a — b) and — (flf — J) are neg- ative ; ma<,mb and — < — m 7n 144 INEQUALITIES. Cor. — If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For changing the signs of all the terms is equivalent to multiplying each member by — 1. 245. If tivo or more inequalities subsisting in the same seiise he added, member to member, the resulting inequality luill subsist in the same sense as the given inequalities. For, if ay by a' y b', and a" > b", then a~-bi a' — b\ and a" — b" are positive ; therefore, a — b -^ a' — b' -\- a" — b'\ or a + a' + a!' — (^ + Z>' + b"), is positive ; a + a' + a" > Z> + J' + J". ScH. — If one inequality be subtracted from another subsisting in the same sense, the result will not always be an inequality sub- sisting in the same sense as the given inequalities, or an inequality ai all. Take the two inequalities 4<7, and 2 < 3. By subtraction, 2 < 4. Here the result is an inequality subsisting in the same sense as the given inequalities. But take 9 < 10, and 6 < 8. By subtraction, 3 > 2. Here the result is an inequality subsisting in a sense cont'^ary to that of the given inequalities. Again, take 9 < 10, and 6< 7. By subtraction, 3 = S.- Here the result is an equation. SOLUTIOIS^. 145 346. TJie Solution of an inequality consists in trans- forming it in such a manner that the unknown quantity may stand alone as one of its members. The other member will then denote one li7n,it of the unknown quantity. EXAMPLES, Solve the following inequalities : , X , 2x ^*^x , 9 Multiplying both members by 20, 10a;4- 8a;>15a;'4-45; transposing and reducing, 3a;>45; ence, a;>15. 2. 5:.>y+14. u47zs. a; > 4. 2x 2x 2x ^715. a; < 3. 8 ^ 4 ^ 6 ^ 12 ^WSv^< 14:. 247. If there be given an inequality and an equation, involv- ing two unknown quantities, a limit of each unknown quantity may be found by ehmination. EXAMPLES. 1. Find a limit for x and y in the following groups: (2a; + 5«^>16 . . (2a; + 2^ = 12 . . . (1), . (2). Subtracting (2) from (1), 4y>4j whence, y > 1. 10 146 INEQUALITIES. If we substitute 1 for y in (2), the first member will be made less than the second ; hence, 2rr + l<12; whence, a; < oj* j2a; + 4y>30) (3:r + 2^ = 31) (5a:-3?/<15) (9a;-f-2«/ = 46) (4a;+ 5^ = 68)' (5a: + 3y>121) (7a;+4y = 168i' 6. . 8 6 -^ 3£--24 rr~2/_-,o -4ws. a; < 8, 2/ > ^i* Ans. x<^^,y> 2f|. -47^5. a; < 13, yy ^. Ans. a: < 20, y > 7. 248. CHAPTER IX. mEQUALITIES. SYNOPSIS FOR REVIEW. Membeks. stjbsisting in the same sense. Subsisting in a contrary sense. 241 Cor. 1, 8» 242. Theorems .... J 243. 244.— C<7r. , 245.— .a^. SOLtTTION. Combination of an Equation with an Inequality. OHAPTEE X. Il^YOLUTION A]^D EYOLUTIOJSr. INVOLUTION. 249. A Potver of a quantity is the product of factors each of which is equal to that quantity. A quantity is said to be raised or involved when any power of it is found. 250. Involution is the process of raising a given quantity to any required power. 251. The Base or Hoot of a power is one of the equal factors of the power, and the Degree of a power is equal to the number of times the base is used as a factor to produce the power. Thus, a^ is the third power or cube of a, and a is the base of a\ 252. The Exponent of the Potver is the exponent which indicates the power to which the given quantity is to be raised. 253. A Perfect Power of the n^^ degree is a quantity which can be resolved into n equal factors. Thus, a^ — "Zab + W' is a perfect power of the second degree. 254. An Imperfect Power of the n*^ degree is a quan- tity which cannot be resolved into n equal factors. Thus, a^— h^ is an imperfect power. A quantity may be a perfect power of one degree, but an im- perfect power of another degree. Thus, a^ — 2al) + ^^ jg ^ perfect power of the second degree, but an imperfect power of the third degree. 148 INVOLUTION". 255. Tlie Sign of the Power, — Any power of a posi- tive quantity is positive ; for, when all the factors are positive, the product is positive. If the quantity to be involved is negative, the even powers will be positive, and the odd powers will be neg- ative. For, { — a){—a) = a\ (— a) (— a) (— a) —a^ (— «)= —a\ (—«)(—«)(—«)(—«)= — a^ (— a) = a\ and so on. 256. The 7i*^ JPotver of a Product,— It follows, from Art. 249, that {abY=ab xdbxdb . . .U) n factors=a xaxa. . . to n factors xoxbxb . . . to n factors = a"b\ In like manner, {abcY = a^'b^c", and {abc . . . ^•)'* = a"Z>"c" . , , k". . ' . The 71^ power ofjhe product of any number of factors is equal to the product of the n^^ poivers of those factors Again, (a^^f = a" x a" = a"+" = a^, (a")^ = «•» X a" X a" = «"+"+" = a**, . • . If the n^ power of a quantity be raised to the m^ poioer, the result will be the mn*^ power of tMt quantity. 257. Hie Coefficient of the Power.— Since (5aY = 5a X ba X 6a = bhi^ = 12oa% and (5a)" = S^a", it follows that The coefficient of the n^ power of a quantity is the n*^ power of the coefficieiit of that quantity. 258. To find any power of a monomial. B, ULE. Raise the numerical coefficient to the required power, and write after the result all the letters of the given monomial, giving to each an exponent equal to the product of its original exponent by the exponent of the power, 1. Find the 3d power of 2a^^c. Ans, %a%^<^. 2. Find the 6th power of the 5th power of a%(^. Ans, a^b^(^. I]S"VOLUTION. 149 3. Find the 5th power of — abx"". Ans. —a^^a^. 4. Find the 4th power of — aWc^. Ans. a%^\hl^. 5. Find the 5th power of ^a^x\ Ans. 243«%io. 6. Find the 7th power of aWc (— x^fzf^). Ans. — a^Wc^ x^yh^. 7. Find the 3d power of (— aMc^) (— a^bh). Ans. aWc^ 8. Find the 5th power of (— af (— by (— cy. Ans. —aW^c^. 9. Find the m^ power of — a^b^(^ when m is an even positive integer. Ans. a^b^'^c^"'. 10. Find the m^^ power of — (abcY when m is a positive in- teger. A71S. ± c^'^b^c^K 359, To find any power of a polynomial. RULE, Find the product of as many factors, each of ichich is equal to the given polynomial, as there are units in the exponent of the required power. 11,1. USTBATIONS. a -\-b a -\-b a^ + ab -j-ab-hb^ a^+2ab +b^ a -\-b a^+2a^-\- ab^ -\-a^ -^2ab^ + b^ a^ + Za% + ^alP +^3 a -\-b + a^b + ^aW + dab^+¥ a2 + 2ab + ^^, a^-{- Sa^ + dab^ + b% a'^ + ^a^ + Qa^^ + ^ab^ + b\ {a-\-bY = a^-\- 2ab-^b% (a + by = a3 + Sa^ + 3«^^ + b% (a + by = a^ + 4:a^b + 6^2^,2 4. 4a§3 ^ j4. In like manner it may be shown that (a — by = «2 - 2a,b + b\ (a — by = a^ — M% + 3fl^2 _ js^ \a — by = a^ — ia^b + QaW — 4ab^ + i^- 150 INVOLUTION Cor. 1. — Since [a^'Y = (a")"* (256), we may reach the same result by different processes of involution. For example, we may find the sixth power of a -\- ihj repeated multiplication by a + Z>; or we may first find the cube of a -\- b, and then square the re- sult ; or we may first find the square of a -\- b, and then cube the result. The work may sometimes be abridged by using the principle expressed by the equation «*"«"= «»'*+". Thus, we may find the fifth power of a -{- b by multiplying the cube of a + b hy the square of a + b. Cor. 2. — It may be shown by actual multiphcation, that (a-{-b-^cy=a^+2a(b + c)-\-b^+2bc-^c^, and (a + b-\-c-^dy=a^-\-2a{b-\-c + d) + b^-\-2b{c-{-d)-i-ci-{-2cd+(P. Hence, we may infer that 77ie square of any polynomial is equal to the sum of the squares of its terms, together with twice the sum of the products obtained by multiplying each term by the sum of all the terms which follow it, EXAMPLES. 1. Find the square of {a — b-\- c). Ans. a^ + l^ -\- c^ — 2ab -\- 2ac — 2bc, 2. Find the square of \ -\- x -\- ofi -^ a^. Ans. 1 + 2:r + 3a:2 + 4a;3 ^ 3^ _^ 2:^5 + 2:6. 3. Find (1 - 2a; + ^3?Y. Ans. 1 — 4a; + lOx^ — 12a:3 + ^x^. 4. Find {a -\- b — cf. Ans. a^ + b^—(^+^a\b—c) + W{a-c) + d(^{a + b)^Qabc. 5. Find (1 + 22; + x^f. Ans. 1 + 6a; + 15a;2 + 20a;3 4. 15^4 _f_ 6:^5 + a;^. 6. Find {a + bf. Ans. a^ + Wb + Iba^lr^ + 20^3^ + lha^¥ 4- Qa¥ + ¥. (I)" EVOLUTION". 151 260. To find any power of a fraction. a X a X a ... to n factors a" X T X T to 71 factors = b b b ~bxbxb...tx)n factors ~~ 6"' Hence, The n^^ power of a fraction is a fraction tuhose numerator is the n^^ poiver of the given numerator, and whose denominator is\ the n^ power of the given denominator, EXAMPZES. W - ^/ • 3. Find ^^-^-r • Ans. a^—'^ab-\-¥ 4 Find I ■ . . ^^,.o 5. Show that ( ^^ ) +i ._2J_.J=J.. EVOLUTION. 261. Let n be any positive integer ; then The n*^ Hoot of any given quantity is a quantity the n*^ power of which is equal to the given quantity. 262. Evolution^ or T/ic Extraction of Hoots, is the process of finding any root of a given quantity. Evolution is the converse of involution. 263. Tlie Sif/n of the Boot— If the i7idcx of the root to be extracted be an odd number, the sign of the root will be the same as the sign of the given quantity (255). Thus, V«^ = ay and \/—a^= — a. 152 EVOLUTION. If the index of the root to be extracted he an even number, and the given quantity be positive, the root may be either positive or negative. Thus, ^/a^ = ± a. 264. If the index of the root to be extracted be an even num- ber, and the given quantity be negative, the root cannot be ex- tracted ; because no quantity raised to an even power can produce a negative result {2>55). The indicated even root of a negative quantity is caUed an Imaginary Quantity, Thus, V— 9, V— «^, and V— {(I + if are imaginary quantities. 365. To find any root of a monomial. (3a'»Z''-)" = S^a"^^ (358); ^3«^m«^rn _ ^^m^r (261). RULE. Extract the required root of the numerical coefficient, and write after the result all the letters of the given monomial, giving to each an exponent equal to the quotient obtained hy dividing its original exponent hy the index of the root, 1. Find VSo^^^- -4w5. 2a^l^c, 2. Find V^^^^- ^ns. ± aH^c^^ 3. Find V— «^^^- ^^^s. — ahxT. 4 Find \/(^bVd^ Ans. ± aWcdK 5. Find a/«*"^^"^"*'*> when m is an even positive integer. Alls, ih a%^(^. 6. Find ^ — a''¥c^, 266. To find the square root of a polynomial. Since the square root of c? + lal + J^ ig a -^-l, we may be led to a general rule for the extraction of the square root of a EVOLUTION. 153 polynomial by observing in what manner a -{- b may be derived from a2 _|_ 2ab + bl 2a-i-b\2ab + b^ Arrange the terms according to the descending powers of a ; then the square root of the first term, a% is a, which is the first term of the required root. Subtracting its square, that is, a^^ from the given polynomial, we obtain the remainder 2ab -\- bK Dividing 2ab by 2a, we obtain b, which is the second term of the root. Multiplying 2a -{- bhj b, and subtracting the product from the first remainder, we obtain for the second remainder ; hence a + b is the required root. When the root contains three or more terms, it may be found by a similar process. Thus, fl2 + 2a5 + ^ + 2 (flj + J) c + c2 1 « + 5 + c 2a-{-b\2ab + b^ ' 2ab + b^ 2(a-{-b) + c\2{a-{-b)c-^c^ ^2{a + b)c + c^ The first term of the required root is a. Subtracting a^ from the given polynomial, we obtain the remainder 2ab + b^-{-2(a-^b)c + c2. Dividing the first tenn of this remainder by 2a, we obtain the second term of the root. Multiplying 2a -\- b by b, and sub- tracting the product from the first remainder, we obtain the sec- ond remainder, 2 {a -\- b) c -{- c^. Dividing the first term of the second remainder by 2a, we obtain tiie third term of the root. Multiplying 2 (a + Z>) + c by c, and subtracting the product from the second remainder, we obtain for the third remainder ; hence a -{-b + c is the required root. We call 2a the partial divisor, 2a -\-b the first complete divi- sor, and 2 (a + 2>) + c the second complete divisor. 154 EVOLUTION". RULE, I. Arrange the given polynomial according to the powers of one of its letters. IL Extract the square root of the first term; the result will he the first term of the required root. Subtract the square of this term from the given polynomial. III. Divide the first term of the remainder by twice the first term of the root, and annex the result to the first term of the root and also to the divisor ; then multiply the divisor thus completed by the second term of the root, and subtract the product from the first remainder. IV. Take twice the sum of the first and second terms of the root for a second divisor. Divide the first term of the second re- mainder by the first term of the second divisor, and annex the re- sult to the part of the root already found and also to the second divisor ; then multiply the divisor thus completed by the third term of the root, and subtract the product from the second re- mainder, V. If the required root contains additional terms, proceed in like manner until all the terms are found. Cor. 1. — If the first term of the arranged polynomial is not a perfect square, or if the first term of any arranged remainder is not divisible by twice the first term of the root, the exact square root cannot be found. Cor. 2. — All even roots admit of the double sign (263) ; hence the square root of a^ + 2ab + J^ jg _ (^j _^ j)^ ag ^rgu j^g a -\-b. In fact, the first term in the root, which we found by extracting the square root of a\ might have been — a ; and by using this we should have obtained — b for the second term of the root * EXAMPLES Find the square root of each of the following expressions : 1. ^x^ — 123^ + hx^ + 6a; -f 1. Ans. ± {2x^ — 3a; — 1). 2. 1 + 42: -f 10a;2 + 127? + ^x^. Ans. ± (1 + 2a; + ox^). EVOLUTION. 155 3. 9x^ + 12a;3 _|_ 22a:2 _|_ i2a; 4. 9. A71S. ± {Sx^ + 2a; + 3). 4. 9^2 -j- I2ah + 4^ 4- 6ac + 4^>c + c^ Ans. ± (3a + 25 + c). 5. 4^4 — 12a3 4- 25a2 _ 24« + 16. Ans. ± (2^2— Za + 4). 6. I62:* — Uabx^ + 16^2^2 _|_ 4^2^2 _ g^js ^ 4^4^ 7. a;(5 _ 4^ _|_ i0a;4 _ 20a;3 + 250:2 — 24.2; + 16. 8. Slx^ — 432a;3 ^ 864a:2 — 7G8a; + 256. 9. {a -by -2 (a2 4. ^) (^ _ j)2 + 2 (a* + ^4). 10. a^^M + (^-{-d^ — 2«2 (Z,2 ^ J2)_2J2(^_^2)_,_2c2(a2_^2). 267. Wlien a Trinomial is a JPerfect Square. — Since {a ± hY =z a^ + %ah + h^. it follows that a ainomial is a perfect square, if two of its terms are squares, and the other term is twice the product of the square roots of these two. When a trinomial is a perfect square, its square root may be found by extracting the square root of each of the square terms and connecting the results by the sign of the other term. 1. Extract the square root of \.x^ — Vtxy + ^y\ This is a perfect square, because 4a:2 and 9?/2 are squares, and Vlxy is equal to twice the product of the square roots of these terms. The square root of 4a;2 is 2a;, and the square root of 9^2 jg Zy. Connecting these results by the sign of the term 12a:?/, we obtain 2a; — 3y, or Zy — 2x. 2. Extract the square root of a;2 _|_ ^xy + 9^2, Ans. ± (a; 4- Zy). 3. Extract the square root of 9^2 4- 25^>2 _ 30fl;5. Ans. ± (3a — 5Z>). 4. Extract the square root of 9a!2 _ Vlab 4- 16^2. 268. An expression wJiich, in its simplest form, is a bino- mial, cannot be a perfect square. For the square of a monomial is a monomial, and the square of any polynomial contains at least three terms. 156 EVOLUTION. 269. To find the square root of a number. Pkinciples. — 1. Tlie square of a number consisting of tens and units is equal to the sum of the squares of the tens and the units increased hy tiuice their prod^ict. Thus, 782 ^ (70 + 8)2 = 702 + 2 X 70 X 8 + 82 = g084. 2. TJie square of a number expressed by a single figure con- tains no figure of a higher order than tens. For 9 is the largest number which can be expressed by a single figure, and 92 = 81. 3. The square of tens contains no significant figure of a lower order than hundreds, nor of a higher order than thousands. Thus, 102 _ 100, and 902 _ gioo. 4. Hie square of a number contains twice as many figures as the member, or twice a» many less one. Thus, 12= 1, 102= 100, 92= 81, 1002= 10000, 992 =9801. 10002 =1000000. Hence, h. If a numher be ^rvarated into periods of two figures each, beginning at units' place, fhfi member of periods tuill be eqticl to the number of figures in the square root of that number. Thus, there are two figures in the square root of 43,56. 1. Let it be required to extrsc*; the squi^re root of 4356. Let a represent the value of the first figure 43,56(60 + 6^^0^ of the root, and b that cz2— 36 00 of the second figure. 2« + J=120 + 0=i26')7^ Since 56 cannot be a 2a6 + ^2-^7 55 part of the square of the tens (Peik. 3), a must be the greatest multiple of ten whosf* Equare is less than 4300 ; this is found to be 60. Subtracting a" EVOLUTION-. 157 that is, 3600, from the given number, we find the remainder to be 756. This remainder consists of ttvice the product of the tens hy the units, and the square of the units (Prin. 1). But, since the product of tens by units cannot be of a lower order than tens, the last figure, 6, cannot be a part of twice the product of the tens by the units; this double product must therefore be found in the part 750. Now, if we double the tens and divide 750 by the result, the quotient, 6, will be the units' figure of the root, or a figure greater than the units' figure. This quotient figure cannot be too small, for the part 750 is at least equal to twice the product of the tens by the units ; but it may be too large, for 750, besides the double product of the tens by the units, may contain tens arising from the square of the units (Prin. 2). To ascertain if the quotient, 6, is correct, we add it to 120 and multiply the sum by 6. Subtracting the product from 756, we find the remainder to be ; hence 66 is the required square root. 2. If the square root contains more than two figures, it may be found by a similar process, as in the following example, where it will be seen that the partial divisor at each step is obtained by doubling that part of the root already found. The letters show how the different steps correspond to those of the algebraic process in Art. 266. a h c 18,66,24(400 + 30 + 2 =432 fl2= 160000 2a + Jr=800 + 30 = 830)26624=2ff5 + J2_|_2«c + 2Jc + c3 24900=2^^ + ^2 2(a-f J)+c=:800 + 60 + 2=862)1724=2«c + 25c + ^2 1724=2ac + 25c + c2 For the sake of brevity, the ciphers on the right are usually omitted; thus, 43,56(66 18,66,24(432 36 16 126)756 83)266 756 249^ 862)'l724 1724 loS EVOLUTIOIT. RULE, I. Separate the giveji number into periods of ttvo figures, each^ beginning at the units' place, II. Find the greatest number tchose square is contained in the period on the left; this tvill be the first figure in the root. Sub- tract the square of this figure from the period on the left, and to the retnainder annex the next j^eriod to form a dividend. III. Divide this dividend^ omitting the figure on the right, by double the part of the root already found, and annex the quotient to that part, and also to the divisor ; then multiply the divisor thus completed by the figure of the root last obtained, aiid subtract the product from the dividend. IV. If there are more periods to he brought dow7i, continue the operation in the same mariner as before. EXAMPLES. Find the square root of each of the following numbers : 1. 177241. Ans. ^9A. 2. 4334724. - ^?i5. 2082. 3. 14356521. Ans. 3789. 4. 17.338896. Ans. 4.164. 5. 2.5. Ans. 1.5811+. 270. To find the cube root of a polynomial. {a^\-b^ircY=a^■^^a^b + ^aJyi■\-b^-\-Z{a-\-bYc-{■^a■\-l)c^-\-c^. Let us now find the root a + b + c from its cube. a^J^^a:ih^^al^J^l^J^^aJ^byc+^a^\-b)(^-\-(^ a-\-b + c 3a24.3aJ + 52 dd^b + '6aU^ + b^ da% + dal^-\-¥ 3(a + 5)2 + 3(rt + % + (^ ^a + bfc-\-'d{a-\-b)c^ + (^ ^a+bfc+^a+by■\-(^ EVOLUTION. 159 The first term of the root is obtained by extracting the cube root of c^. Subtracting a^ from the given polynomial, and dividing the first term of the remainder by Za^^ we obtain the second term of the root. Multiplying Za^ + Zab + W- by l, and subtracting the product from the first remainder, we obtain the second remainder. Dividing the first term of the second remainder by ?>o?, we obtain the third term of the root. Multiplying 3 (a + J)2 + 3(« + l)c + &■ by c, and subtracting the product from the second remainder, we obtain for the remainder. I. Arrange the given polynomial according to tlie powers of one of its letters ; then the cube root of the first term will be the first term of the root. Subtract the cube of the first term of the root from the given polynomial, II. Divide the first term of the remainder by the partial divi- sor, tuhich is three times the square of the first term of the root; the quotient ivill be the second term of the root. III. To the partial divisor add three times the product of the first and second terms of the root, also the square of the second term ; the result will be the first complete divisor. IV. Multiply the complete divisor by the second term of the root, and subtract the product from the first remainder. V. Divide the first term of the second remainder by the par- tial divisor, which is three times the square of the first term of the root ; the quotient will be the third term of the root. VI. To three times the square of the sum of the first and sec- ond terms of the root, add three times the product of the sum of the first and second terms by the third, also the square of the third term ; the result will be the second complete divisor. VII. Multiply the second complete divisor by the third term of the root, and subtract the product from the second remainder. VIII. If the required root contains additional terms, proceed in like manner until all the terms are found. 160 EVOLUTION. Cor. — ^We may dispense with the complete divisor, if, after each time that we find a new term of the root, we subtract the cube of the sum of the terms already found from the given poly- nomial. jexampi.es. Find the cube root of each of the following expressions : 1. a^ + 6x^y + 12a:^2 ^ 3^. ^;^, ^ + 2y, 2. a^-\- 12a^ + 48a; + 64. Ans, a; + 4. 3. «» — 9a2 + 27a — 27. Ans, a — 3. 4. 8a3 — SQa^b + 54a^ — 27^8. Ans. 2a — dh. 6, afi + ea^ — 40a:S + 96a; — 64. Ans. a^5 + 2a; — 4. 6. a« + 6a5 4. 150* + 20a^ + ISa^ + 6a + 1- Ans. a^ -\-2a -{- 1. 7. a;« — 12a* + 54a;4 _ 112:2:3 ^ iQg^ _ ^Sx + 8. Ans. a^^ — 4a; + 2. 8. «6 _ 3^5j 4. 6a4^2 _ 7«8j3 ^ 6«2^ — 3a^ + J«. Ans. a^ — ab-{- ^. 9. ^3 — Z>3 _|. c3 — 3 (^2^ — a^c — ab^ — ac^— V^c + h(P) — Ubc. Ans, a — b -\- c. 10. 1 — 6a; + 21^-2 — 56a;3 ^ iiia4 _ 1743.5 ^ 219a;6 — 204a;74- 144a;8 _ 64^9. Ajis, 1 _ 2a; H- 3a;2 — 4a;3, 11. 8a;6 + 48ca;5 _^ eOc^a;* — 80cSa? — 90^a;2 _^ loscSo; — 27c«. ^;^5. 2a;2 ^ 4^^; _ 3^, 12. a;9 — 3a;8 + 6a;^ — 10a;«+ 12a;S— 12a;4^ 10:6^— 6a;2+ 3a; — L Ans. a;^ — a:2 -j- a; — 1. 13. a;9 + 6a;8 _ 64a;« — 96a;5 _^ 192^4 4. 512^3 — 7G8a; — 512. Ans. a;3 _|_ 2a;2 — 4a; — 8. 14. 8a8 _ V2a% + ^Qa^bc + 6a3^>2 _ 360^2^2^ _ a%^ + 5405^2^2 _p 9a2^>3c _ 27a^c2 + 27^c3. Ans. 2a — ab + 3/./;. EVOLUTION. 161 271. To find the cube root of a mimber. Pkixciples. — 1. The cube of a number contains three times as many figures as the number, or three times as many less one or two. Thus, 13= 1, 103= 1000, 33= 27, 1003= 1000000, 93= 729, 10003= 1000000000, 993 = 907299, 100003 = 1000000000000. Hence, % If a number be separated into periods of three figures each, beginning at unit^ place, the number of periods will be equal to the number of figures in the cube root of that number. Thus, there are two figures iu the cube root of 405,224. 1. Let it be required to extract the cube root of 405224. a b 405,224(70 + 4=74 a3=343 000 Sa^= "702x3= 7^x300=14700 3aJ=70 x4x3 = 7x4x 30= 840 ^= 42 = 16 62224=3«2^ + 3«J2_^j3 62224=(3a2 + 3a& + ^>2)j da^+3ab-\-l^= 15556 Denote the tens of the root by a, and the units by b ; then, since the cube of tens contains no significant figure of a lower order than thousands, a must be the greatest multiple of ten whose cube is less than 405000 ; that is, a must be 70. Subtract- ing the cube of 70 from the given number, we find the remainder to be 62224. Dividing this remainder by 3a% that is, by 14700, we obtain 4 for the value of b. Adding Sab, that is, 840, and b% that is, 16, to 14700, we find the complete divisor to be 15556. Multiplying the complete divisor by 4, and subtracting the pro- duct from 62224, we find the remaiader to be ; hence, 70 + 4, that is, 74, is the required cube root 2. If the cube root contains more than two figures, it may be found by a similar process, as in the following example, where it 162 EVOLUTION. will be seen that the partial divisor at each step is obtained by- multiplying the square of that part of the root akeady found by 3. 12,812,904(200 + 30 + 4=234 8 000 000 2002x3=120000 200x30x3= 18000 4812904 4167000 302= 900 645904 138900 645904 2302x3=158700 230x4x3= 2760 42= 16 161476 The work in the preceding example may be abridged as follows ; 12,812,904(234 8 22x300= 1200 2x3x30= 180 4812 4167 3^= 9 645904 1389 645904 232x300=158700 23x4x30= 2760 42= 16 161476 RULE. I. Separate the given number into periods of three figures each, beginning at the units' place. II. Find the greatest number whose cube is contained in the period on the left; this will be the first figure in the root. Sub- tract the cube of this figure from the period on the left, and to the remainder annex the next period to form a dividend. III. Divide the dividend by the partial divisor, which is three hundred times the square of the imrt of the root already found; the quotient will be the second figure of the root. EVOLUTION^. 163 IV. To the partial divisor add tJiirty times the product of the first and second figures of the root, also the square of the second figure ; the result will he the complete divisor, V. Multiply the complete divisor hy the second figure of the root, and subtract the product from the dividend. VI. If there are more periods to he drought down, contiime the operation in the same manner as hefore, EX A MPIjE S. Extract the cube root of each of the following numbers: 1. 9261. Ans, 21. 2. 15625. Ans. 25. 3. 12167. Ans: 23. 4. 32768. Ans. 32. 5. 103.823. Ans. 4.7. 6. 884.736. Ans, 9.6. 7. 12.812904. Ans, 2.34. 8. 8741816. Ans. 206. 9. 2.5. Ans. 1.357 +. 10. .2. Ans. .5848 + . 272. The Higher Hoots of Quantities.— W\\m the index of the required root contains no prime factor greater than 3, the root may be found by methods already explained. In order to show this, it will be necessary to prove the following principle : The mn^^ root of a quantity is equal to the m^^ root of the n^^ root of that quantity. Let |/V« = r . . . (1). Kaising both members of (1) to the m*^ power, Va = r™ . . . (2). 164 EVOLUTION. Kaising both members of (2) to the n^^ power, az=r^''' . . . (3). Extracting the mn^ root of both members of (3), V^a = r . . . (4). Comparing (1) and (4), "»"/— "* /♦» /— Va = 4/ ya. Thus, Vl6 = |/V16 = a/4 = 2, V64 = |/V64 = ^8=2, 1. Extract the fourth root of 6fl2^ + a^ — 4^85 _ 4flJ3 ^ J4. a^ ' 2a2-2flJ| _4a3d + 6fl2/^ 2a2_4«j_}.^ I 2«2Z>2_4«Z>3_^ j4 a2_2aJ+62|«_j «^ ' 2a— b\ —2ab-\-b^ We extract the square root of the given polynomial, and thus obtain a^ — 2cf J + ^^ ; we then extract the square root of this last expression, and find the root to be a — J ; this is, therefore, the fourth root of the proposed expression. 2. Extra<3t the fourth root of 81a;* — 432a:3 + 8642;2 — 768a; + 256. Ans. ± (3a;— 4). EVOLUTIOK. 165 3. Extract the sixth root of Qa^b + l^a^h^ + a« + "^^aW + Iba^b^ + ^^ + ^a¥, . Ans. ±{a + b). 4. Extract the eighth root oia^ — 16x'^ -^ lUa:^ — US:^ -{- 1120a;4 — 1792a;3 + 1792a;2 — 1024^ + 256. Ans. ± {x — 2). 273. Hoots of Fractions* — TJie n^^ root of a fraction is a fraction tohose 7iumerator is the n^^ root of the given numerator, a7id whose denominator is the n^ root of the given denominator. Thus, ^_ = _, for (-)=-, •^74. SYNOPSIS FOR REVIEW. CHAPTER X. § POWEB Degree. Exponent, Perfect. Imperfect. Sign. Powers of Products, Coefficient. Power of Monomial. Rule. Power of Polynomial. Rule. Power of Fractions. Root Index. Imaginary Quantities. Root of Monomial. Rule. Square Root. Polynomial. Rule. Cor. 1, 2, Trinomial. Binomial. Cube Root op Polynomial. Rule. Cor. Higher Roots. Roots of Fractions. CHAPTEE XL THEORY OF EXPOE'EKTS, a"* X a** = a^' > — - = a»"-", or a" 1 275. We have defined «"», where w is a positive integer, as the product of m factoi-s, each equal to a, and we have shown that and that according as m is greater or less than n. Hitherto an exponent has been regarded as Si^jositive integer j it is, however, found con- venient to use exponents which are 7iot positive integers, and we now proceed to explain the meaning of such exponents. 276. As fractional exponents and negative exponents have not yet been defined, we are at liberty to define them as we please. For the sake of uniformity, we shall give such a meaning to them as will make the relation true, whatever m and n may be. 277. Find the meaning of a* ai X «* = «! = « (276); that is, a* must be such a quantity that if it be squared, the re- sult will be a ; hence, a* = Va, THEORY OF EXPONENTS. 167 378. Find the meaning of cfi. a^ xa^ X a^ :=a^^^^^=a>=a', a^ = V«- 279. Find the meaning of a^. d^ X a* X a^ X a^ = a^; a^ = \/a^. 1 280. Find the meaning of a", where n is a ^^ositive integer. i 1 i l+1+l+....tontenna a** X a" X a" X ... to w factors = a** " ** =a^=a; a" = V«. 281. Find the meaning of a", ivhe7'e m and n are positive integers. tn tn tn tn , tn , tn — — — , ,. , — I 1 f- ... to n terms a" X a" X a" X .... to 7i lactors = ««»*» =a^; m Hence, the numerator of a positive fractional exponent denotes the power to which the quantity is to be raised, and the denomi- nator denotes the root to be extracted. 282. Find the meaning of a~\ w X a" = a** * = a. 3-2 — Dividing both members of this equation by a*, -2 « 1 a 2 r= — = -2 a^ a^ 283. Find the meaning of «"% where n is any positive num- her, integral or fractional. 168 THEORY OF EXPONElTrS. Dividing both members by a^+\ Hence a~^ is the reciprocal of a*». 284. It follows, from the meaning which has been given to a negative exponent, that — = a"*-^ when m is less than n, as well as when m is greater than n. For, suppose m less than n ; then 285. General Scholium. — It thus appears that it is not absolutely necessary to introduce fractional and negative expo- nents into Algebra, since they merely supply us with a new nota- tion for quantities which we had already the means of represent- ing. Thus, «■» = V«^, a^ = Va^f «* = a/«^ = a^, , 1 _± 1 1 -111 a-^ = —„, a « = — = -— , a * = — = — a' J Va J a^ If m is a positive integer, the expression a"^ is read tlie mP^ power of ttj OT a m^ power. But if 9n is not a positive integer, a"» should be read a exponent m. Thus, a^ is read a exponent two-thirds f not " a two-thirds power," for there is no such power. i i 1 286. To show that a" x ^" = («*)**. Let 11 ( i 1.Y / ty ( lY ic = a" X ^"; then a;« = [a'' x 2'"/ = V«"/ x W) . But Uv x\bi =ab (2S0); x^ = ab; 1 whence, x = (ah)^. THEORY OF EXPONENTS. 169 ill 11 1 Cor. a" X 5" x C^ = {obY x c" = {ahc^. In like manner it may be shown that 111 1 1 a" X ^** X c" X . . . . ^" = (abc .... k)^. Suppose now that there are w of these quantities a, b, c , , , k, and that each of them is equal to a ; then the last equation be- comes 1 ^ But {a^)^=a^ (281); \anj —a «. That is, The m^ power of the n^ root of a is equivalent to the n^ root of the mP^ power of a. 287. To show that ^=(1)". 1 1 Let x=^, then a:«= /^C =| (380-260) Jn \bn} whence. x = (D* 288. To show that \ar 1 J_ amn. Let ( ^\~- X = \a'"/ ; then X^: 1 whence, X = A. 289. To show that a" = mp Let a; = a" ; then ai^ = 0^; and of^p = a"*'; mp whence, a; = a"^. 170 THEOBY OF EXPONENTS. 290, EXAMPLBS, 1. Simplify {x^ x x^)^. Am. x^, 2. Find the product of fl*, a % a S and a~^. Ans. cT^, 3. Find the product of (f ) , (|) , and (t^ . Ans, -^Y* {Ixy 4. Multiply a^ + J* + cT^i by aF^ — a^ + h^. Ans. ah~^ 4- ah^ + a~h^. 5. Multiply ic* — xy^ + x^y — y* by x -\- x^y^ + y, Ans, x^ + x^y — xy^ — y^, 6. Multiply a^ — flS + a^ — a3 + a^ — a + a*— 1 by a^+l ^W5. a* — 1. 7. Multiply a"^ — fl^ + 1 — a~* + cT^ by a^ + 1 + a"^. ^?i5. a + d^ —1 + a~* 4- a~*. 8. Divide a;* — a:?/^ + x^y — y^ by x^ — y^, , ^W5. X -{- y, 9. Divide a:^ + x^a^ + a^ by a;** + x^a^ + a* ^Tis. x'^ — x^a^ + a* Sn _3n n _n 10. Divide a^ — a '^ by a« — a «. .4ws. «« + 1 + a"". 11. Divide a^ — o*^' + ab^ - 2«iz>2 4. jf by «* — aJ^ + ah — b^. Ans. a + aH^— l. 12. Simplify THEOKY OF EXPOiTENTS. (^ — ax^ + a^x — x^ a^ — a^x^ + Sa^x — dax^ + aJx^ — x^ a + X 171 Ans. a^ + Sax + x^ ifi x^ 2y^ x^ 13. Extract the square root of — + ^ — h ~ :; — Ans.^ + x^tj^--"^. x^ 2if 14. Extract the square root of 4« — 12aH^ + 9^* + l^a^c^ — 2^i^c^ + leA vd^s. 2Gf^ — dl)^ + 4A a 15. If a* = 1% show that (t) = «* ; and if a=z2b, show- that ^» = 2. 391. SYNOPSIS FOR REVIEW. CHAPTER XI. THEORY OF EXPONENTS. f Bam of theory (276). Meaning of a', a , a% Meaning of ««, a». Meaning of a~^, a~*. General Scholium, L L i Show that an x bn — {aby. 1 i_ Show that — = 1^1. m mp CHAPTER XII. EADIOAL QUANTITIES. DEFINITIONS. 292. A Simple Hadical Quantity is an expression of the form of a V^, or ah^. Thus, 2 a/9, 3 Vs, l\/a, and ^(a _|_ lYYc are simple radical quantities. 293. The Hadical Factor is the indicated root, and the Coefficient of the radical factor is the quantity affixed to the radical sign. Thus, in the expression 2 V^, the radical factor is Vo and 2 is its coefficient ; and in the expression [(« 4- HfYCf the radical factor is [(« + J)^] and c is its coefficient. If the coefficient of a radical factor is 1, it is usually omitted. Thus, a/3 is equivalent to 1 \/3. 294. The Degree of a simple radical quantity is denoted by the index of the radical sign, or by the denominator of the fractional exponent. Thus, h Va and a (5 + c)^ are of the second degree ; h a/« and a (J + c)"* are of the third degree ; Ja/» and a ( J + c)^ are of the fourth degree ; and so on. 295. Two or more simple radical quantities are said to be Similar if their radical factors are identical. Thus, 2 %/Z and 4 a/3 are similar. 296. A simple radical quantity is said to be in its Simplest Form when the quantity under the radical sign is entire and EEDUCTION. 173 does not contain a factor which is a perfect power corresponding to the degree of the indicated root. Thus, 3 a/5 is in its simplest form. 297. A national Quantity is one which may be ex- actly expressed without using the radical sign or a fractional ex- ponent. Thus, 5, — 3, 42, and a + b are rational. Any rational quantity may be expressed under the form of a radical quantity. Thus, 5 = a/25, — 3 = — a/9, 4?=V^% and 298. An Irrational Quantity is one which cannot be exactly expressed without using the radical sign or a fractional exponent. Thus, 3 a/8, 2\/d, and 5 a/3 are irrational. Irrational quantities are sometimes called surd quantities, or simply surds. REDUCTION OF SIMPLE RADICAL QUANTITIES. 299. TJie deduction of radical quantities consists in changing their forms without altering their values. 300. To reduce a rational quantity to a radical quantity of the n^^ degree. 3 = a/9 = V^ = a/81 = V3^; and a -{- x= V(« + xY = V{a + x)\ MULE. Raise the given quantity to the n^^ power and indicate the n*^ root of the result, EXAMPLES, 1. Reduce 2a2 to a radical quantity of the third degree. Ans, %/W. 2. Reduce a -\- x to a radical quantity of the fifth degree. Ans. ^/(a + x)^. 174 RADICAL QUAIfTITIES. 3. Reduce - to a radical quantity of the sixth degree. Ans. y —z. 4. Reduce -~ 6a^b to a radical quantity of the third degree. Ans. V— 125^6^8. 5. Reduce — {x -^ y) to a radical quantity of the fourth de- g^e®- Ans. - V(x + y)K 6. Reduce (a — hf to a radical quantity of the w^^ degree. Ans. V(a — by^, 301. To introduce the coeflacient of the radical factor under the radical sign. 4a/2 = vTg X V^ (297) = V32 (286) ; aVx = Vc^ X Vx = V^; fl/=(fly)*; X %/%a — x — V^ X V^a — x = V2aa^ — x^. Hence, denoting the degree of the radical quantity hy n, we have the following It ULE, Multiply tlie quantity under the radical sign ly the n^^ poicer of the coefficient and indicate the rtP*- root of the product. Cor. — In a similar manner any factor of the coeflBcient may be placed under the radical sign. Thus, 3x2 a/5 = 3 V^O- EXAMFJLMS. Introduce the coefficient of the radical factor, in each of the following expressions, under the radical sign : 1. 6 a/5. Ans. a/180. 2. 3a/3T Ans. V243. KEDUCTIOif. 176 3. (a + d) Va+^. Ans, V{a + by. 4. ay -. Ans. ^/ab. '■^''-'yy^^T^- Ans.V6al^. ^ a./c , a . ,/a , a^ 6. -y - + -. Ans, y - + 35. 7. 5a\/bc, Ans. ^/b^w^c. 8. 6x V25^, Ans, V5«+2a;»-2 9. 3a?» {x — y)^. Ans. {2W — 272:62^)* 1 f2 302. To remove a factor from under the radical sign to the coeflacient. The reduction is performed by reversing the process of Art. 301, Thus, 2a/8 = 2\/4xV2 = 2x2a/2, and a VWc = ab ^/c. Henee, denoting the degree of the radical quantity by n, we have the following It ULE. Divide the quantity under the radical sign by the factor to be removed; and to the indicated n^ root of the quotient prefix, as a coefficient, the product of the given coefficient and the n^ root of tliefa^ctor to be removed. JEXAMPTjES. 1. Keduce \/20 to such a form that the factor 4 shall not occur under the radical sign. Ans. 2 Vs. 176 RADICAL QUANTITIES. 2. Reduce V24 to such a form that the factor 8 shall not occur under the radical sign. 4,^,,^ 2 V^. 3. Reduce 3 'v/TS to such a form that the factor 25 shall not occur under the radical sign. j^^^g 15 a/3^ 4. Reduce (a — h) V(a + h^c to such a form that the factor {a + by shall not occur under the radical sign. Ans. (a2 — ^) a/c. 5. Reduce a (5 + c) V^ to such a form that the factor d" shall not occur under the radical sign. 303. To reduce the indicated root of a fraction to an equivalent expression in which the quantity under the radical sign shall be entire. |/^ = j/lTe = l/| X V6 (286) = I a/6 ; l/n l /nTl 1/3 '/I : »/r ,/- I3,- 'Ai "./T^TS^^ \fa^^ "/I T" "/T .__ n-l Hence, denoting the degree of the radical quantity hy n, we have the folio wiug RULES. I. If the fraction under the radical sign has a denominator which is a perfect n^^ poiver, prefix to the indicated n^^ root of the numerator the reciprocal of the n^^ root of the denominator. n. If the fraction under the radical sign has a denominator which is not a perfect n^^ power, multiply or divide loth of its REDUCTIOiT. 177 terms ly such a quantity as will reduce it to one wliose denomina- tor is a perfect n^^ power ; then substitute this fraction for the given one and jJroceed as directed in I. Cor.— If the given radical quantity has a coefficient, the re- sult obtained by the rule must be multiplied by it, in order to obtain an expression equivalent to the given one. Thus, JEXAMPLES. Reduce each of the following expressions to another in which the quantity under the radical sign shall be entire : 1. V^- A7is.^V6. 2. ||/f- Ans.lVm 3. 2|/|- Ans, a/3. 4. IVIf. Ans.\VT^. 5. -^V^- Ans.,^VlSu. 6. m{a -\- x)\ ^^~~ • Ans, m V(« — a;) (« + x)''-\ a "I" 3/ 304. To reduce a simple radical quantity to its simplest form. 1. Reduce 3 VS to its simplest form. The largest perfect square which is a factor of 8 is 4. Remov- ing this factor from the radical sign to the coefficient, we have dVS = 6V2 (303). 178 KADICAL QUAKTITIES. 2. Reduce 5 \/4:Sa^x^ to its simplest form. The largest perfect cube which is a factor of ASa^x*- is Sa^ofi, Bemoving this factor from the radical sign to the coefficient, we have 3 /g 3. Reduce 4 y ^ to its simplest form. 4 V^ = 4i/| = I V2i (303) = I X 2 V3 (302) = | Vs. Hence, denoting the degree of the radical quantity by 7i, we have the following MULBS. I. If the quantity under the radical sign is e7itire, resolve tt into two factors, one of which is the greatest n^ poiver contained in that quantity ; then remove this factor from the radical sign to the coefficient. n. If the given radical quantity contains the indicated root of a fraction, reduce it to such a form that the quantity under the radical sign shall he entire, and proceed with the result as directed in I. EXAMrms. Reduce each of the following radical quantities to its simplest form: Ans. 5a Vah. Ans. da^cVdbx, Ans. Sa^^c^V^. Ans. 3a \^. Ans. Qx \/a -f ba?. Ans. 35al}\/c. 1. V25a^. 2. V'^Wl^ch;- 3. 4. Vl92a^^cl VlOSa^. 5. 6 Vax^ 4- ^^. e. 7V625a*Z>V. REDUCTIOJS^. 179 7. 3 Va^+"^. Ans, 3a Va^'b, 8. V(a + x)H**. Ans, (a + x) V6". 9. ||/|. Ans.^Vn, 10. 6 |/|- ^W5. 2 Vis. 11. t1/t* -47i5. TiV^d, b ^ d Id 12. 2 1/ ,/^ , ' ^^& -^- VC^ + ^) «. 13. (— ) . Ans, — (a'^l^T^y^)^, \xy / XV 14 xyf xy ^^^ ^ /5(^^^^) . ^;^.. ^3V5TrT^)^. 16. (a^b)\/^—4' Ans. V^^::^, a -\- 17. (a — b) \/ ' ; Ans, ^/cim + n). 305. To reduce a radical quantity of the form of V^ to another of a lower degree. V9 = |/V^(272) =V3; V8 = /Vi='V/2; mrij — *" /n / — m /— 180 RADICAL QUANTITIES. Hence, denoting the factors of the index by m and n, we have the following RULE. Extract the n^^ root of the quantity under the radical sign, and indicate the rrif^ root of the result, EXAMJPLES. 1. Reduce V^a^ to a radical quantity of the third degree. Ans. a/^«- 2. Eeduce v64a^ to a radical quantity of the second de- gree, j^ns. V^i' 3. Reduce \/lM¥c^ to a radical quantity of the second de- gree, jins. ^/%aM, 4. Reduce V^oa^^c* to a radical quantity of the third degree. Ans. ^/hab(^. 6. Reduce \f^J^(^ to a radical quantity of the third degree. Ans, VoW. 6. Reduce Va^^ to a radical quantity of the fifth degree. Ans. V«^c3. 306. To reduce a simple radical quantity to another of a higher or lower degree. Vff = a* = a^ = a^ = «^ = a'* = v^a»; '^^« = a^=a^ = c.* = a* = a* = V«. nTJL,B. Express the given radical quantity hy means of a fractional exponent; then suMitute for this exponent any equivalent frac- tion having a denominator greater or less than that of the given fractional exponent. KEDUCTION. 181 Cor. 1.— If equal factors be introduced into the index of the root and the exponent of the quantity under the radical sign, the result will be equal to the given radical quantity. Thus, CoE. 2. — Conversely, if equal factors be canceled in the index of the root and the exponent of the quantity under the radical sign, the result will be equal to the given radical quantity. Thus, EXAMFZJES. 1. Reduce V« to a radical quantity of the 12th degree. Ans. V«®. 2. Reduce *Va to a radical quantity of the mn^^ degree. Ans. va". 3. Reduce Va — b to a radical quantity of the 20th degree. Ans. ^\/(a — by. 4. Reduce a/(« + by to a radical quantity of the 10th degree. Ans. ^^/(a + by. 5. Reduce V{a — by^ to a radical quantity of the 3d degree. Ans. V(a - by. 307. To reduce simple radical quantities having unequal indices to equivalent ones having equal in- dices. 1. Reduce Va and \/a to equivalent expressions having equal indices. Va = a^ = J= V^; and Va = a^ = a^= V^. 2. Reduce v^a and ^/a to equivalent expressions having equal indices. 182 KADICAL QUANTITIES. % = a- = a"- = T^; JRULE. Express the indicated roots by means of fractional exponents, and reduce the expressions thus obtained to equivalent ones, in which the fractional exponents shall have equal denominators, EXAMPLES, 1. Reduce v^ and 3 V3 to equivalent expressions having equal indices. j^^. »^ and 3 '^/m. 2. Reduce a/2, VS? V^, and %/b to equivalent expressions having equal indices. j^^s, ^2*, v^, ''V^% ^^/^. 3. Reduce \/2, V3, V^, and V^, to equivalent expres- sions having equal indices. Ans, v'ioge, v^729, ''V/626, '^v^sS. 4. Reduce 3* 2*^, and 5* to equivalent expressions having equal indices. ^^^_ 3^ ^-h^ 5A i 1 5. Reduce a»» and J"* to equivalent expressions having equal 6. Reduce Vox, Vxy, and Vc^ to equivalent expressions having equal indices. Ans. *v^^^^^ "v"^^; ^"a/^^^. 7. Reduce V^, V^, V?, V^, and T 6^ to equivalent expressions having equal indices. Ans. a/«, ^/h, ^~c, Vd, \/e. COMBINATIONS. 183 COMBINATIONS OF RADICAL QUANTITIES. 308. To find the STim of two or more simple radi- cal quantities. 1. Find the sum of 6 V2 and 8 V^. 6 v^ + 8 VS = (6 + 8) a/2 = 14 V^. 2. Find the sum of 2 V^i and 3 Vl92. 2V24 =2V8^3 = 4V3, and 3 VTd2 = 3 V64 x 3 = 12 V^ ; 2V24 +3Vl9^ =:16V3. 3. Find the sum of V^^, V^^, and a/«^^. V^ = a; V^, V^^ = 2a; V^, and 'v/^= aVii^; .-. V^+ V4^+ V^=^A/^+2a; V54-aV^=(3a; + a)\/i. 4. Find the sum of 2 Vi08 and 5 V24. 2Vl08 = 2V27 X 4= 6V4, and 5V24 =5V8ir3 =10V3; 2V1O8 + 5V24 = 6V4 + IOV3. In this example the radical quantities cannot be made similar; hence the addition can only be indicated. 5. Find the sum of 2 V36 and 3 a/6. 3 a/6 = 3 V36 (306, Cor. 1). .-. 2 V36 + 3 a/6 = 2 a/36 + 3 a/36 = 5 V36 = 6 V^ (305). 184 BADICAL QUANTITIES. RULES. I. If the given radical quantities are similar, prefix the sum of the coefficients to the common radical factor. II. If the given radical quantities are of the same degree, but not similar, reduce them, if possible, to equivalent similar ones by the rule of Art. 304, and proceed with the results as directed in I. If they cannot be so reduced, indicate their sum. in. If the given radical quantities are of different degrees, reduce them to equivalent ones of the same degree, and proceed with the results as directed i7i IL EXAMPZE8. 1. Find the sum of 7 VlO and 2 \/90. Ans, 13 a/To. 2. Find the sum of VsOO and V256. Ans. 9 Vi 3. Find the sum of 4 -^^500 and 3 v'lOS. Ans. 29 Vi. 4. Find the sum of |/ ^ > r 5» ^^^ V Tq' Ans. \/2. Z V lo 6. Find the sum of kV qj 7 r -g"? 8,nd gV 05* 6. Find the sum of Va^x and V^ Ans. (a -\- b) ^/x. 7. Find the sum of Vu, 2 \/72, and a V^. Ans. 2 ( Ve + 6 a/2) + ax a/6. 3 / o>v„^^ 3 8. Find the sum of y -^r- and y —r . 3^4-1 3 Ans, --^ A/4a%. COMBINATIOIfS. 185 9. Find the sum of V(l + a)-\ Va^ (1 + a)~\ and a V(l + «)(! + ay. A71S. {a^ -{- a + 1) Vl -\- a. 10. Find the sum of 3 VlQa^^ and 5 V^ab?. Ajis, (3c -\- 5c2) ^tabc. 11. Find the sum of ^/%a:j?—^ax-\-%a and ^/%a7?-\-^ax-\-%a, Ans, 2xV^ci- 12. Find the sum of V^a'^^^b% Vl^a^'~^b% V2a4"'+», and ^^^' Ans. (Sa^ + ^ + a-»+3 + c) ^2^, 13. Find the sum of a V^ and c Vb^- Ans. ahV^+cI^Vb. 14. Find the sum of « (l + ^) and b(l + ^) . ' Ans.[(a^ + b^)T 309. To find the diflference of two simple radical quantities. 1. Subtract 6 V2 from 8 V2. SV2 — 6V2 = {8-6)V2 = 2V2. 2. Subtract 2 V^i from 3 Vl92. 3 a/T92 = 3 V64 X 3 = 12 V3, and 2V24 =2V8~>r3 = 4V3; 3V192 — 2V24 = 8V3. 3. Subtract V^ from V^- a/4^ = 2xVx, and V? = xVx; 186 RADICAL QUANTITIES. 4. Subtract 2 \/l"08 from 5 V^i. sV^i =5V8^3 =10V3, and 2Vi08 = 2V27~xl= 6V4; 5V24 — 2V108 =ioV3 — eVi. In this example the radical quantities cannot be made similar; hence the subtraction can only be indicated. 6. Subtract 2 V36 from 3 \/6. 3^/6 = 3^/30(306,00^1); ... 3a/6 — 2 V36 = 3V36-2V36 = V36 = >v/6 (305). MULES. I. If the given radical quantities are similar, subtract the co- efficient of the radical factor ^71 the subtrahend from that of the radical factor in the minuend, and prefix the remainder to the common radical factor, II. If the given radical quantities are of the same degree, but not similar, reduce them, if possible, to equivalent similar ones by the rule of Art, 304, and proceed with the results as directed in I, If they cannot be so reduced, ijidicate the subtraction. in. If the given radical quantities are of different degrees, reduce them to equivalent ones of the same degree, and proceed with the results as directed in 11. EXAMPLES, 1. Subtract ^/ba from V^a. Ans. 2 ^/Ea. 2. Subtract ^/U from ^^^192. Ans. 2 V3. 3. Subtract ^V^W from 3a Vb, Ans. 2a Vb. 4. Subtract 3 V^ from 6 V^. Ans. 3 Va. COMBII^^ATIONS. 187 6. Subtract (a - x) |/^-±^ from {a - x) Va^ - x\ a — x Ans. (a — x — l) ya^ — x^. 6. Subtract i/ ^ ., , . ,p from y -^ — ^ . , .o - 7. Subtract V32« from 2 V40«. Jw5. 4 VSa — 2 V2a. 8. Subtract 2 V54 fi-om 6 V320. Ans. 24 Vs - 6 V^- 310. To find the product of two or more simple radical quantities. 1. Multiply 6 \/54 by 3 ^2. 6 \/54 X 3 V2 = 6 X 3 V54 x \/2 = 18 V54 x 2 (286) = 18 VT08 = 108 Vs. 2. Multiply 3 ^/U by 2 V3«. 3 V^ x 2 V3« = 3x2 ^/^ x V3^ = 6 V(2^ X V(3^ (306, Cor. 1) = 6 V(2«)M3«)^ = 6 V72^^. RULES. I. i/* /^e ^iVe» radical quantities are of the same degree, find the product of the radical factors ty the principle of Art. 286, and to the result prefix the product of their coefficients. Express the final result in its simplest form. II. If the given radical quantities are of different degrees, re- duce them to equivalent ones of the same degree, and proceed with the results as directed in I. Cor. 1. — If the roots are indicated by fractional exponents, the product may be found by the principles of Ari 69. Cor. 2. — The product of two or more simple radical quantities can always be reduced to a simple radical quantity. 188 RADICAL QUANTITIES. 1. Multiply i a/6 by -ft V9. Ans. -^ VEi = ^ \/6. 2. Multiply 4y I by 3|/|. Ans. 4^/15. 3. Multiply 3 -v^ by 4 V3. Ans. 12 Vi32. 4. Multiply a/24^ by Vl2x. Ans. 12a x^2. 5. Multiply d \/ax by c Vxy. Ans. be Va^y^ 6. Multiply 3 V^ by 4 ^a. Ans. 12 '\/^. 7. Multiply (rt + i)^ by (a + ^>)i. Ans. a -\- b. 8. Find the product of V^bc~\ V«, VS"^*^^, and V«~i. 9. Multiply (a + 5)* by {a — V)^. Ans. (a^ — b^)^. 10. Multiply aVi by b^^/y. Ans. aZ> V^aj^'y^ 311. To find the product of polynomial radical quantities. The product of two polynomial radical quantities is found by combining the rules of Art. 310 with that of Art. 70. If frac- tional exponents are used to indicate the roots, the rule of Art. 70 is sufficient. EXAMPLES. by 2 — Vs. 3+ \/5 2— V5 6 + 2\/5 - 3 V5 - 5 Product, 1 — a/5. COMBINATIOKS. 189 2. Multiply x + 2Vy + sVz hy x — 2Vy-\- Z\fi, ic + 2 V^ + 3 v^ X— 2 v^ + 3 V^ —2x Vy —^y — 6 Vy x \/z ■i-dxVz + QVy xVz + ^V^ Product, aj2_4y j^exVz-\-9 V?. 3. Multiply J + a^b^ + ah + ^'^ by d^ — bi a^ + a^b^ 4- ah + b^ a + aJ^>i" + ah + a^Z^^ _ ahi — ah — ah^ — &2 Product, a — b"^. 4. Multiply V8 + \/3 by a/8 — V3. ^W5. 6. 5. Multiply V^+ V^ by Va — 2^/l^, Ans. a^ + Va^* — 2 V^ — 2 V^ 6. Multiply Va + Vb -\- x by V^ — ^b + a;. ^Tis. a — b — x, 7. Multiply a^ - 2a^^ + ^a^^'"* — 8a5 4- IGa^Z** — 32Z>'^ by fli -f 2J^. ^W5. «8 — 642>2. 8. Multiply a^ + a^b^ + Z>^ by a*" — Z** ^W5. a — J. 9. Multiply x^y + y^ by a;* — y ». ^W5. x^y — y^. 312. To find the quotient of two simple radical quantities. 1. Divide 6 ^54 by 3 \/2. 3v^ V^ 190 EADICAL QUANTITIES. 2. Divide 8 A/2a by 2 Va. RULES. I. If the given radical quantities are of the same degree, divide the radical factor in the dividend hy that in the divisor, hy the principle of Art. 387, and to the result prefix the quotie)it obtained by dividing the coefficieiit in the dividend by that in the divisor. Express the final result in its simplest form. II. If the given radical quantities are of different degrees, reduce them to equivalent ones of the same degree, and proceed with the results as directed in I. Cor. 1. — If the roots are indicated by fractional exponents, the division may be performed by the principles of Art 84. Cob. 2. — The quotient of two simple radical quantities can always be reduced to a simple radical quantity. 1. Divide 8 \/l08 by Vq. Ans. 24 V^. 2. Divide V512 by 4 V2. Ans. VI. 3. Divide 12 V54 by 3 V2. Ans. 12. 4. Divide 4 Vi2 by 2 V3. Ans. | Vl6 x 3s. 5. Divide a by Va. Ans. Va. 6. Divide Va by V^. Ans. yVab'^K be 7. Divide 2aWby 4vWc^. Ans. ^.^/V'cd^. 2d COMBIlfATIONS. 1£ 8. Diyide |/| by j/|. a 9. Divide cfi by a*. Ans, d^. 10. Divide a» by «g. mq-np Ans, a nq . 313. To find the quotient of two polynomial radical quantities. The quotient of two polynomial radical quantities is found by combining the rules of Art. 313 with that of Art. 86. If frac- tional exponents are used to indicate the roots, the rule of Art. 86 is sufficient. EX AMPLE 8. 1. Divide V^ — a/«^ — Va + V« by \/a — 1. \/a^ — "s/a — \/a^ + \/a \ \/a — 1 3. Divide a^ + ^ah^ - 4.Jb'^ - Sh^ by J— 4.A gi _ 4at^,i J+2b^ 2a^b^- Sb^ 2ah^- Sb^ 3. Divide a^ + aVb — 6b hy a — 2Vb. Ans. a + SVb. 4. Divide a — 41 V« - 120 by V^ + 4 V« + 5. Ans. V^-4V^+llV«-24. 5. Divide x^ — a^x^ — 4«a: ^+ 6a^x — 2a^xi by x^ — 4^:?;^ + 2a*. ^ws. ic — «'^a:». 192 RADICAL QUANTITIES. 6. Divide x^ — y* by 2;^ — y*. Ans. x^ + yi. 7. Divide x^ — %x*y* + 1/^ hy x* — y'^. Ans, a;* — y^. INVOLUTION OF RADICAL QUANTITIES. 314. To find any power of the indicated n^ root of a Quantity. ( Vs)' = V3 X V3 = V32 (310) ; (V3)'=V32x V3= V38 = 3; (Va)"* = V^ (286, Cor.). Hence, denoting the index of the given indicated root by n, we have the foUowing RULE. Raise the quantity under the radical sign to the required power, and indicate the n^ root of the result. Cor. 1. — If the index of the given indicated root is equal to the exponent of the power to which that root is to be raised, the required power may be obtained by simply removing the radical sign. Thus, (V«) = a, ( V«) = «j and (Va) = a. Cor. 2. — If the index of the given indicated root and the ex- ponent of the required power contain a common factor, the result obtained by the rule may be reduced to a radical quantity of a lower degi-ee. Thus, {V~af = V^ = Va^ (306, Cor. 2). CoR. 3. — If the root is indicated by a fractional exponent, the ruleof Art. 258issuflacient. Thus, [a^) =a^, (a") =a\ 315. To find any power of a simple radical quan- tity. (5 V3)^ = 6V3x5V3 = 5x5xV3xV3 = b'^V^', {aVby = a^Vb^. INVOLUTION. 193 Hence, denoting the exponent of the required power by m, we have the following nULE, Raise the given radical factor to the m^^ power (314), and to the result prefix the m^ power of the given coefficient. CoR. — If the root in the given expression is indicated by a fractional exponent, the rule of Art. 2>5S is sufficient Thus, (2a*) = 23a^ = 8a* EXAMPLES, 1. Find the square of 5 V«- -^ns. 25 \/a\ 2. Find the third power of ha %/x. Ans, 125a%. 3. Find the square of a^ V^- ^ns, a* V36. 4. Find the 3d power of | Vd, Ans, | a/3. 5. Find the 4th power of — \/a^, Ans. a^ \/~o^. 6. Find the 75th power of x ^Vy- Ans. x^ ^/y\ 7. Find the square of x^y. Ans. x^\/~y. 8. Find the n^^ power of x^'s/y, Ans. uf^^/y. 9. Find the 4th power of ^a . Ans. -^cb^- 10. Find the 6th power of a (Z> + c)^. Ans. a^ (^^+2dc4-c2). 316. To find any power of a polynomial radical quantity. Any power of a polynomial radical quantity is found by com bining the rule of Art. 315 with that of Art. 359. If fractional exponents are used to indicate the roots, the rule of Art. 359 is sufficient. 194 RADICAL QUANTITIES. EXAMPLES. 1. Find the square of Vs + a a/2. Ans. 3 + 2a Ve + 2a^ 2. Find the third power of 3 + V5. Ans. 72 + 32 \/5. 3. Find the square of a* + b^- ^ns. a + Za^b^ + b^. 4 Find the 4th power of a* — b^. Ans. a2 _ 4:a^l>i 4- 6aJ — 4ai7>^ + P, EVOLUTION OF RADICAL QUANTITIES. 317. To find any root of the indicated root of a quantity. v¥ = V32 (272) = V3 (306, Cob. 2) ; m /n /- mnr- Hence, denoting the index of the given indicated root hy w, and that of the required root by m, we have the following L If the quantity tinder the given radical sign is a perfect m^ power, extract the m^^ root of it, indicate the n^^ root of the re- suit, and, if possible, reduce to a lower degree. IT. If the qicantity under the given radical sign is not a per- fect m^^ power, indicate the mn^^ root of it, and, if possible, re- duce the result to a lower degree. Cor. — If the given indicated root is expressed by a fractional exponent, the rule of Art. 265 is sufficient. Thus, l/J = «^ EVOLUTION. 195 318. To find any root of a simple radical quantity. |/ 25 V32 = 5 1/ V32 = 5 V32 = 5 V3 ; 1/5V9 = \/yVZb X 9 (301) = ''v/125 X 9 = '^^1125 ; yl)Va = y'^^alt = *a^«Z>". Hence, denoting the index of the required root by m, we have the following I. If the given coefficient is a perfect wP^ power, prefix the mP^ root of it to the m^ root of the given radical factor. IT. If the given coefficient is not a perfect nnP^ power, introduce it under the given radical sign, and find the m^ root of the result (317). Cofi. — If the radical factor is expressed by means of a frac- tional exponent, the rule of Art. 265 is suJGficient. Thus, |/5x 9^ = 5* X 9* EXAMPLES. 1. Find the square root of 9 V3. Ans. 3 V3. 2. Find the square root of 3 V^. Am^^ Vl35. 3. Find the cube root of |y | Ans, ^V^- 4^/4 1 3 / — 4. Find the fourth root of ^ V g. ^ns. ^ Vl^- 5. Find the sixth root of a^ V?- ^^^- «^ ^^' 6. Find the fourth root of a^s V^- ^^«- «* V^. 196 RADICAL QUANTITIES. 7. Find the fifteenth root of VJaTW^- Ans. V(« + b)^ 8. Find the 7i*^ root of c^^/a\ A7is. a Va. 319. To find the square root or the cube root of a polynomial radical quantity. The square root or the cube root of a polynomial radical quan- tity is found by combining the rules of Art. 318 with those of Articles 266 and 370. If the roots in the given expression are indicated by fractional exponents, the rules of Articles 266 and 270 are sufl&cient. 1. Find the square root of Vx + 2 \/xyi + Vy. Vx-\-2 \fxy + v^ I V^+ Vy 2Va;-i- Vy | ^Vxy^-y/y 2\/xy -\- Vy 2. Find the square root of x^ — 2x^y^ -f ?/* ^ __ '■tx^y^ -f y^ \x^ — y^ 2a;^ — y^ — 2icy + y^ — 2x^y'^ + y^ 3. Fmd the square root of 4 Vc^ -f 12 Vab + 9 %/¥, Ans. 2 V« + 3 Vb. 4. Find the cube root of « + 3 VM + 3 V«^ + b- Ans. %/a + \f'b. REDUCTIOIJ". 197 5. Find the square root of a -{- 2a^b^ + ^ + 2aM + 'Zbh^ -\- c. Ans. «i -I- Z>i + c^. 6. Find the cube root of 8a + dQah^ + 54a^<5>* + 27h, Ans. 2a^ + 3^^ REDUCTION OF FRACTIONS HAVING SURD DENOMINATORS TO EQUIVALENT ONES HAVING RATIONAL DENOMINATORS. 330. A Simple Surd is a surd of the form a\/b or ab^. Thus, 2 V3 is a simple surd. 321. A Polynomial Surd is a surd having two or more terms. Thus, 2 a/3 + 3 \/5 — 6 \/7 is a polynomial surd. 322. To reduce a fraction whose denominator is a simple surd to an equivalent one having a rational denominator. 2 _ 2 V3 _ 2\/3 _ 2 a/3 . 5 a/3 "" 5 a/3 X a/3 "" 5 a/32 ~" 15 ' 2 a/5 _ 2 a/5 X V3 _ 2 a/125 x a/9 _ 2 a/1125 ^ 3V9~3V9xV3~ 3V27 ~' 9 ' V^ a/^xV^^ V^xa/^"^ a/^"^" Hence, denoting the degree of the denominator of the given fraction by n, we have the following RULE. Divide some perfect n^^ power which is a multiple of the quan- tity under the radical sign in the given denomi^iator by that quantity, and multiply both terms of the given fraction by the indicated n^ root of the quotient. 198 BADICAL QUANTITIES. HXAXPZES. Reduce each of the following fractions to an equivalent one having a rational denominator : Ans. Ans. 3 2V9 2 vr 2 w 2a/6 4 2 6Va^ Ans. 5. ^=r. -4n«. 2V3a; 8. 10. 11. V2 V9' a/2 V3 V2" a/2 vr a/2* V! V3' ^W5. 2Va 5a2 ' flA/6 2 ■ V72 ^W5. V2. JW5. V41472 2 Arts. Ans. a/25 X 812 3 • *a/288 2 • Ans. '\/8 X 9^ 13. 13. Vb' Vb Vb 15. 16. 2! J*' 1 a** -J" 17- 4- 18. 19. 20. 21. a^/b b b n n 7' aW 22. ^%. V. Ans. aVb , aVW Ans. — 5—. b ' Ans, Ans. ah^ b ' 1 m-l a^'b"' b ' mn-^ Ans, ab " Ans. — , a Ans. Ans. a Vb a Ans. Ans, ^/~ac KEDUCTION-. 199 323. To reduce a fraction whose denominator is a binomial surd to an equivalent one having a rational denominator. 1. Reduce to an equivalent fraction having a 2 V3 - V2 rational denominator. 6 5 _ 5(VT2 4-V2) 2a/3 — V2 VT2-V2 (a/12 — V2)(Vl2 + a/2) _ 5(a/12 + a/2) _ a/12 + \/2 ^ "" 12 — 2 ~ 2 We obtain the multiplier VT2 -f a/2 by dividing 12 — 2 by \/T2 — a/2. 5 2. Reduce = — to an equivalent fraction having a 2 a/3 + a/2 rational denominator. 5 5 5(a/T2— a/2) 2 a/3 + a/2 a/12 + a/2 (vl2 + a/2) (vT2 — a/2) _5 (vl2 — a/2 )__ a/12 — a/2 "~ 12 — 2 ~" 2 3. Reduce 5-^^ g— = to an equivalent fraction having a ra- A/a — V^ tional denominator. Dividing a — J by V« — V^j we obtain V^ + a/«^ + A^ Multiplying both terms of the given fraction by this quotient, we have c(a/^+a/^+ Vb^) a — b c 4. Reduce 57= 5/= to an equivalent fraction having a ra* tional denominator. Dividing a -f 5 by V« + V^, we obtain V? — Vad + \/^\ 200 RADICAL QUANTITIES. 5. Keduce ^— : ~ to an equivalent fraction having a ro" V « — yb tional denominator. c __ c _ c{y^-^Vab + lVa + V¥) _ We reduce the given fraction to an equivalent one, in which the simple surds in the denominator are of the same degree, and then multiply both terms of the resulting fraction by the quotient obtained by dividing a — ^ by V^ — V^. MULES. I. If the given denominator is of the form of V« — *Vb, divide the indicated difference of the quantities under its radical signs by the denominator, and multiply both terms of the givm fraction by the quotient, II. If the given denominator is of the form of%^ + V^? f^nd its indices ere even, proceed as directed in L III. If the given denominator is of the forjn of ^fa + a/^, and its indices are odd, divide the indicated sum of the quantities un- der its radical signs by the denominator, and multiply both terms of the given fraction by the quotient. IV. If the given denominator is not of the form of V^ — V^, nor of the form of \fa -\- Wb, reduce the given fraction to an equiv- alent one having a denominator of one or the other of these forms (306), and proceed with the result as directed in the rule which corresponds to the form of its denominator. REDUCTION. 201 Cor. 1. — If we multiply both terms of the fraction — =. a± Vb by a =F V^j the resulting fraction will have a rational denomina- tor; for a±Vl = V^ ± Vh, and a^ — b-r- {V^ ± Vb) = « =F V^ Cor. 2. — A fraction whose denominator is a trinomial of the form of Va ± Vb ± a/c may be reduced to an equivalent one having a rational denominator by two multiplications. Thus, d _ d{Va±^TVc) ____ d{Va±VhTV~c) _ d{Va±VbTVc){a±i—cTWab) __ a±b—c±2Vah ~ {a±b—c±2Vab) {a±b—c^^2Vab)~~ d{Va±VbTVc)(a±b—cT2Vab) ^ ia±b—cy—^ab EXAMPLES, Reduce each of the following fractions to an equivalent one having a rational denominator: 1. — ^— . Ans !M±:v^) A/5-V2 3 V5-A/2 ^ „ a—Vl . a^+b—2aVb 3. — Ans, 5 — 7 — . a-^Vb ^'-^ ^a-\-^/b \ a + b + 2Vab 5. lr±^. Ans. '-±fi. 3(3— 3 V2) ^ 202 RADICAL QUANTITIES. 6. r-/-T7- ^ns. 5 (V9 + V6 + Vi). Ans, d (a^W-aWbc + acVb-V^) dVb+Vc ' a^b—c^ ^ Va-\-x-[-Va—x . a-\-Va^—^ 9. —-=^ » -4ws. . Va+a;— ya— a; ^ 10. ^^ — ■ ■ — ^— — ^^ ^. Ans. 5^ — j ^. (x^+x-^iy-^(xi-x-iy ^'^^ 324. Utility of tJie Two Preceding Transforma- tions, — The two preceding transformations enable us, in many cases, to abridge the computation of the approximate value of a numerical fraction whose denominator is a surd. JXi USTJRATIONS. >v/5 4-V^ _ V40+V24 _ 2(\/l0 + V6) _ VTo+Vq V8 " 8 - 8 " 4 • 2. i.^%, = TiVE-WlS) ^ 248024 ^ ^^^^^^ 3^ V6 ^ V42-V18 ^ 2^ ^ ^^^^^^^ V7 + V3 ^ 4 . 9 + 2V1O 2(9 + 2vloy 242 + 72a/10 . __ 18-4\/l0 164 164 2^/^ _ 2V3(V25-V30 + V36) _ 2(V75-V90 + Vl08) ^.^_ lERATIONAL QUANTITIES. 203 PROPOSITIONS RELATING TO IRRATIONAL QUANTITIES. 325. An irrational quantity cannot de expressed by a rational fraction. This follows from the definition of an irrational quantity (398). 326. A simple quadratic surd cannot le equal to the sum of a rational quantity and a simple quadratic surd. For, if possible, suppose V^ =z « -f- Vm . . . (1), in which ^/n and Vm are surds. Squaring both members of (1), 7i = ^2 _|_ 2a ^m + m : y2 m whence, ^m=i ^r . . . (2); <»a that is, we have Vm, an irrational quantity, equal to a rational fraction, which is impossible (325) ; hence (1) cannot be true. 327. Tlie product of two simple quadratic surds, which are not similar, and which cannot be made similar, is irrational. Let Vm and Vn be two such surds; then, if possible, sup- pose Vm^ =.an . . . (1). Squaring both members of (1), mn = d^v? \ whence, \/m = a Vn . . . (2) ; that is, Vm and Vn may be so reduced as to be similar. But this is contrary to the hypothesis ; hence (1) cannot be true. 204 RADICAL QUANTITIES. 328. Tlie quotient of itvo simple quadratic surds, which are not similar, and lohich cannot he made similar, is irrational. Let Vrn and 's/n be two such surds ; then, if possible, sup- pose _ 4/f = a» . . . (1). Squaring both members of (1), in = a*w 2«3; •whence, ^/mz=an's/n . . . (2); that is, "s/m, and Vn may be made similar. But this is con- trary to the hypothesis; hence (1) cannot be true. 329. Tlie sum, or difference of tivo simple quadratic surds, which are not similar, and which cannot he made similar, cannot he equal to a simple quadratic surd. Let Vrn and Vn be two such surds; then, if possible, suppose Vrn ± Vn = Va . . . (1). Squaring both members of (1), m ± 2 Vnm -{- nz=a; u . / — a — m — n ... whence, ± yinn = . . . (2). But Vmn is irrational (327) ; hence we have an irrational quantity equal to a rational fraction, which is impossible (325) ; hence (1) cannot be trua 330. In an equation, of which each memher is the sum or difference of a rational quantity and a simple quadratic surd, the rational quantities of the two memhers are equal, and also the irrational quantities. Suppose X ± Vy = a ± Vh . . . (1), lERATIONAL QUANTITIES. 205 in which ^/y and V^ are irrational ; then will x-=za and V^ = V^. For suppose ic = « ± w . . . (3) ; then (1) becomes whence, n ±, ^fy = ± Vb . . . (3). But (3) is impossible (326) ; hence (2) cannot be true. Therefore x =za, and consequently Vy = V^. 331. If i/a -{- Vb =zx -{• A^y, in which ^/b and V^ are irrational, then ya — V^ = a; — Vy. For since |/a + V^ = a; + a/^ . . . (1), we have by squaring, a + Vb = x^^-2xVy-\-y . . . (3); .-. « = 2:2 + y . . . (3), and V^ = 2a; a/^ . . . (4) (330). Subtracting (4) from (3), a — Vh =zx^ — 2x Vy + y . . . (5) ; whence, y a — Vb =:x — Vy . . . (6). 333. 7/" y a 4-^/3 = a/^ + Vy , in which Vb, ^/x, and Vy are irrational, then y a— Vb = Vx— Vy- For since |/a + V^ = Va; + Vy • • • (1)> 206 EADICAL QUANTITIES. we have by squaring, a-\- Vl = x-{-2 Vxy H- y . . . (2); .-. a = a;4-y ... (3), and Vb=2Vxy, ..(4) (330). Subtracting (4) from (3), a—Vb = x — 2Vxy-^y . . . (5); whence, ya — Vl = Vx—Vy . . . (6). SIMPLIFICATION OF COMPLEX RADICAL QUANTITIES. 333. A Conijilex Radical Quantity is an expres- eion in which one radical sign includes one or more others. Thus, yVS, |/9 4-3v^, and \ 0A/hy~Q are complex radical quan- tities. 334. The complex radical quantity y a ± ^/h may le sim- plified if b is a perfect square, or if a^ — h is a perfect square. 1. Suppose that 5 is a perfect square; then y d±:Vb may- be reduced to Va±Cy in which c is the square root of h. Thus, |/5±\/9 = V'5±3. 2. Suppose that "s/b is a surd, and that a^ — h is a perfect square; then ya-\-Vb may be reduced to ± y ^"^^ J_ i A~^ j and \/a-Wb may be reduced to ± i/^±£ T j/^""^ r 2 ^ V 2 '^ 2 ' in which c is the square root of a^ — b. Assume Vx -\- Vy=A/a-{-Vb . . . (1). In this equation one or both of the terms in the first member must be irrational, because the second member is a surd ; SIMPLIFICATION. 207 V^-^Vy=\/a-Vb . . . (2) (331-332). Multiplying (1) by (2), x-y=V'^^b . . . (3). Squaring both members of (1), a; + 2 Vx^ + ^ = « + VJ . . . (4). In this equation 2 Vxy is a surd (327) ; x-\-yz=a . . . (5). Combining (3) and (5), we find 2 (6), and Vy= ± j/ a—Va^—b r,^>. (8), /^Zvi=±v ^+^~^'-^ :fV^ ^^'-^ , , . But Va^— b=c by hypothesis; |/a+V^=±/^±y^ . . . (10), and ^a^Vb=± |/^ ip >j/^ . . . (11). EXAMPIjES. 1. Simplify |/3 + 2\/2. |/3 + 2V2=|/3 + V8; 208 RADICAL QUANTITIES. a = 3, ^> = 8, and c = a/9— 8 = 1. Substituting in (10), i/3 + 2a/2 = ±|/S ± |/?Z^ = ± V2 ± 1. 2. Simplify |/7-2VlO. |/7-2a/10 = |/7-\/40; a = 7, J = 40, and c = V49-40 = 3. Substituting in (11), -j/7-2Vl0 =:±V6TV2. Simplify each of the following expressions : 3. |/i1 + 6a/2. Ans. ±3±\/2. 4. y/7_4V^. Ans. ±2^V3, 5. 1/94+42 a/5. ^w5. ±7±3a/5. 6. |/ll + 6 V'2 + |/7-2a/10. ^7i5. ±3±a/5. 7. 4/Jc + 2Z>a/^c— R ^,i5. ±5±a/&c— Z>2. 8. |/(a + J)2— 4(«— J) a/«^. ^^. ±(a— &)=F2A/a?. 335. Tlie complex radical quantity y a\fc-^Vi^ in which aVc and a/^ are supposed to le surds, may he simplified^ if fl2 {g d perfect square. SIMPLIFICATION^. 200 aVc ± Vb = Vela ± y -jf '''^aVc±^b = VV~((a±\/-]= ^W a±\/\ (302). This expression may now be simplified by the method of Art. 334 when c^ is a perfect square. EXAMPLJES, 1. Simplify |/V324VM. a/32 + V30 = V2 (4 + a/15) ; |/a/32 + V'SO z= V2>(/4+ Vl5. But /4+A/r5=±/|±|/^. Simplify each of the following expressions. 2. |/a/27 + Vl5. ^ W5. V3 f ± ;^ ± |/i j. 3. |/5a/2 + 4a/3. ^ws. V2(±a/3±a/2). 4. |/8a/3~2a/45. ^^. a/3(±a/5 =F V3). " / ^ 336. Complex radical quantities of the form of y a\/b may ie simplified by the rules of Art, 318. Thus, "sa/B = yV^ = V45. 14 210 RADICAL QUANTITIES. 337. Complex radical quantities of the form of 3 iUayi) ± c^d ± etc., or of the form of y a\/b ± c \/5 ± etc., may, in some cases, be simplified by the method of Art, 319. Thus, i/ V5 + 2 Vl5 + a/3 = V5 H- V3. IMAGINARY QUANTITIES. 338. An Imaginary Quantity is one which, when in its simplest form, contains an indicated even root of a negative quantity. Thus, 2 V^, 5 V— 10, and (a + i) V^ are imaginary. 339. The term Meal is applied to all quantities that are not imaginary. Thus, 5, — 3, VS, and V— 27 are real 340. Imaginary quantities are classified in the same way as other surd quantities. Thus, 2 V— 3 is simple and of the second degree, 4/3 + 2 \/^ is complex, and 8 + 3 V^ + 5 V^i— 7 V — 1> considered as a single expression, is a compound or polynomial imaginary quantity. An imaginary quantity usually consists of a real and an imag- inary part. Thus, 2 + 3 V— 1 consists of the real part 2 and the imaginary part 3 V— 1. The whole expression is considered as an imaginary quantity on account of the presence of the imagi- nary part. COMBDfATIONS OF SIMPLE IMAGINARY QUANTITIES. 341. To find the sum of simple imaginary quan- tities. EXJLMrLES. 1. Find the sum of V-^ and V— 16. >v/I~9 + V-IQ = VQ (- 1) + V16(-1) = 3 \/^ IMAGIKARY QUANTITIES. 211 2. Find the sum of 3 V— 81 and 2 V^16. 3 V^ZM + 2 V-16 = 3 V81(-l) + 2 Vl6(-1) = 9 v^^ + 4 V^i = 13 V^. 3. Find the sum of V— 50 and V— 18. ^W5. 8 V^^. 4. Find the sum oi ^/ —a and ^ —h. Ans. (Va + Vb) a/^I. 342. To find the diflference of two simple imag- inary quantities. EXAMPLES, 1. Subtract 2 V^^ from 9 V^I 9 V^^ - 2 V^^ = 9 V^l — 2 V4(— 1) = 9 V^Ti— 4 V^i = 5 a/^i. 2. Subtract 2 V^^ from 9 9 v^- 2-2 V— 3 = 9 V2(-l) - 2 a/3(— 1) = 9 \/2 V^iri _ 2 a/3 a/^I = (9 a/2 — 2 a/s) a/^I. 3. Subtract V^^ from V— 8^. Ans. V-^- 4. Subtract V— b from V~— a, Ans, (Va — Vi) V— 1. 343. To find the product of two simple imaginary quantities of the second degree. £!XAMPZES. 1. Find the product of b V— ci and c V— «• It is evident that b ^/—a x c \/~^a = bc{—a) (314, CoE. 1) = — abc. It is also evident that b V — « X c a/— a=.bc Va^; 212 RADICAL QUANTITIES. for if this be not true, the rule for the sign of a product is not general ; it therefore follows that, in this case, V«^ = — a. But it may be said that V«^ = ^y and therefore a = — a. This reasoning is erroneous, for it is not true that v^ =i -\- a and — a at the same time (74). We are enabled to remove the ambiguity with regard to the sign of a/«^ by knowing that a^ resulted from the involution of — a. If we did not know in what way a^ was produced, that is, whether a^ represented {-\- of or (— a)^ then the sign of V^ would be ambiguous. 2. Find the product of a/— a and ^/—b, a/^a = ^/a{—l) = Va ^/~— 1, and V^^ = V^* (— 1) = V^\/^^; .•. a/^^ X V— * = Va ^f^^ X ^/h V— 1 = VabW— l) = — a/«^. The ambiguity with regard to the sign of the product may therefore be removed, if we reduce each of the imaginary factors to the form of a V— 1? and remember that V— Ix V— 1 or / / \^ • \V — 1) IS equal to — 1. 3. Multiply 4 ^^l by 3 a/^^. Ans. — 12 Vs. 4. Multiply — 5 a/^2 by — 3 a/^^. Ans. — 15 a/10. 5. Multiply a/— ^«^ by ^T^W. Ans. — ah. 344. To find the quotient of two simple imaginary quantities of the same degree. JEXAMTZJES. I. Divide \^—a by V— h. V—a _ VaV^l^ _ Va _ /^ IMAGIJS^AR'Y QUANTITIES. %13 2. Divide V~~ a hj — V— i h a/_ h V h' 3. Divide a/— a by — V— b. — ^—i — v^V-1 v& =-1^' The ambiguity with regard to the sign of the quotient of two imaginary quantities is removed, therefore, by reducing each of - 1, and observing that — ^=r = 1. V-i Ans, ^\/3. Ans, eVs them to the form of « '^ 4. Divide 6 V-^ by 2 v^^^. 5. Di^dde — V^^ by — 6 V~—3, 345. To find all the powers of V— 1. (V:riy = v--^, (V:ri)'^_a, (V--ri/ = (-if^i. If we multiply these powers, in their order, by the 4th, we shall obtain the 5th, 6th, 7th, and 8th ; (V--ri)'^_i, / / \8 214 RADICAL QUANTITIES. Therefore all the powers of ^— 1, arranged in order, be- ginning with the lowest, form a repeating cycle of the following terms: V— 1, —1, — V— 1? audi. 346. MISCELLANEOUS EXAMPLES IN IMAGINARY QUANTITIES. If the student will observe the directions given in Articles 343 and 344, and remember that imaginary quantities are surds, he will have no difficulty in solving the following problems : 1. Multiply 4 + \/^^ by V^^- Ans, 4 V— 5 — a/15. 2. Multiply 3 + V^^ by 2 — ^T^^. Ans. 6 + 2 V--2 — 3 V'^^ + VS. 3. Multiply 1 + V^^ by I — V^^. Ans. 2. 4. Multiply a-\-h V^^ hy a — b V^^. Ans. a^ + h\ 5. Divide (\/^i)* by V^^. Ans. — V^^. 6. Divide 4 + V"^ by 2 — V^^. ^W5. 1 + ^/^2. 7. Reduce to an equivalent fraction having a ra- tional denominator. Ans. V^l. 8. Simplify y^7 + 30 V— 2. ^W5. ± 5 ± 3 ^^^2. 9. Simplify /j/si + 12 V^^ + |/- 1 + 4 V- 5. ^W5. ±8±2V^5. 10. Simplify y^a^ _ 2a^ + 2 (a — h) V— ^. Jws. ± (« — i) ± J a/^^. 11. Find the 3d power of a a/— • 1. -4;i«. — fl^A/--^i. RADICAL EQUATIONS. 215 12. Find the 3d power of « — Z> V'^- Ans. «3 4- J3 >v/:^i _ Sab{h + a \^-^). 13. Find the 4th power of a + V^b. Ans. a* — Qa^b -\- b^ + (4«8 _ 4«5) V^- 14. Find the values of x and «/ in the equation ^ + «/ + ^ V^5 = 5 + a; + 2/ V^^- x = 2 + VlO, y = 6 -\- Via RADICAL EQUATIONS. 347. A Madical Equation is one which involves one or more radical quantities. 348. To free a radical equation from radical quan- tities. EXAMPLES, 1. Free the equation 0/^-^3=2 ... (1) from radical quantities. Transposing V3 to the second member, and squaring the re- sulting equation, a; = 44.4-v/3 + 3 = 7 + 4^3 . . . (2). Transposing 7 in (2) to the first member, and squaring the re- sulting equation, je2_i4^4.49-48; whence, dfi — 14a; = — 1 . . . (3). 2. Free the equation Va; + ll + \/^"^^ = 5 ... (1) from radical quantities, and find the value of x. 216 EADICAL QUANTITIES. Transposing Vrc — 4 to the second member, and squaring the resulting equation, a; + 11 = 25 — 10 Va; — 4 + a; — 4 ; whence, V^c — 4 = 1 . . . (2). Squaring (2), a; — 4 = 1 ; whence, x = 6. 3. Free the equation /y/i _ Va; — 6 _ 4a; — 35 , . Vx + Va; — 5 ^ from radical quantities, and find the value of x. V x—V x — 5 _ 2a; — 5 — 2 Va^ — 5x ,323) . Vi + Vx — 5 hence (1) becomes 2a; — 5 — 2 ViB^ — 5a; = 4a; — 35; (1) whence, v^ — 5a; = 15 — a; . . . (2). Squaring (2), a;^ _ 5^; :_ 225 — 30a; + a?*; whence, a; = 9. 4. Free the equation c m ^/a ^ V^+ V^ a; — a—Vi— Va from radical quantities, and find the value of x. Multiplying both members of (1) by x — a, c {Vx — ^/a) + m Va = m ( Vi + v^) > whence, {c — m) Vx = c Va . . . (2). Squaring (2), (c ■— m)H = ac* ; whence, x = -, r^ . RADICAL EQUATIONS. 217 Find the value of x in each of the following equations : 5. V^ + 7 + v^ = 7. Ans, a: = 9. G. x + Z = Vx^ — 4:X + 59. Ans. x = 5. 7. \/Vx 4- 48 — a/^ = V^. -4/iS. a; = 16. a^—4a 4 8. a/x + 2 a/« + a; = ya — V« + ^z^- -4/is. a: 9. — + — = 4/ -. -4/15. a; = c (Vm — «). 10. H- , = ^ Ans.x = -. Vi-{-x Vi—3?^ Vi—s^ ^ 11. V ^ + 2'' = , ^ws. a; = — tr — . c2 — a2 12. g + ^/& — (ix= -— . Ans,x = \(^ — ax ^ 1^- S + 5 = ^ 25 + 5^5 + ^- ^^^.^ = 20. 14. Va — a: = V« +^' ^^i^* a; = — ^ — . ^2-a/S V4 + a; 13 16. V5 + ^ + Vs"^^ = VlO. Ans. x = 5. 17. yo; + ya + a; =1 - . Ans. x = -. Va -\- X ^ 18 . X -\- a = jJ €? 4- a; V^ + a?^. -4ws. a; = ^-4^2 4a • 218 RADICAL QUANTITIES. 19. V^x + 'Z 4^/6^+6* 4 + x 20. V64 H- a:^ - 8a; = ? 21. V5 + » + VS V4 + a;' 15 VS 4-aJ ^W5. a; = 6. ^W5. ir = 3. ^W5. iC = 4. !. ^x+Vi- ^x-^x = K^rp^) . An,. xJ±. ^/ax — h 3Vax — 2h A6. — -=. = -:s=. . \ax -\-h 3 wax 4- 5 J 24. V4a: + 1 + V4^ ^g^ 25. V4x H- 1 — \/4^ 3V^^4 _ 3V^-f.l5 V^4-2"" V« + 40 9^2 a x = Ans. a; = 4. 349. EH i Definitions . . SYNOPSIS FOR REVIEW. ' Simple radical quantity. Radical factor and its coefficient. Degree of dmple radical quantity. Similar radical quantities. Simplest form of radical quantity. Rational quantity. L Irrational quantity. Reduction. . ' To reduce rational quantity to radical quantity of n'* degree. Rule. To introdu^ coefficient of radical factor under radical sign. Rule. To remme a factor from binder the radical sign to the coefficient. Rule. To reduce the indicated root of a fraction to an equivalent ea-pressioji in which the quantity under the radical sign sMU he entire. Rules. Cor. SYNOPSIS FOR REVIEW. 219 SYNOPSIS FOR REVIEW— Continued. Reduction— (7(cmf(f. Combinations . . Involution op Rad- ical Quantities. Evolution op Rad- ical Quantities. Reduction of pkac- tions having surd denominators to equivalent ones HAVING rational denominators. To reduce simple radical quantity to sim- plest form. Rules. To reduce radical quantity of the form of ^^a* to another of lower degree. Rule. To reduce simple radical quantity to another of higher or lower degree. Rule. Cor. 1, 3. To reduce simple radical quantities liaving un- equal indices to equivalent ones having equal indices. Rule. To find the sum of sim,ple radical quantities. Rules. To find the difference of two simple radical quantities. Rules. To find the product of two or more simple rad- ical quantities. Rules. Cor. 1, 2. To find the product of polynomial radical quantities. To find the quotient of two simple radical quantities. Rules. Cor. 1, 2. To find the quotient of polynomial radical quantities. " To raise the indicated nP" root of a quantity to any power. Rule. Cor. 1, 2, 3. To raise a simple radical quantity to any power. Rule. Cor. To raise a polynomial radical quantity to any power. To find any root of the indicated root of a quantity. Rules. Cor. To find any root of a simple radical quantity. Rules. Cor. To find the square root or cube root of a poly- nomial radical quantity. A simple surd. A polynomial surd. To reduce a fraction whose denominator is a simple surd to an equicalent one having a rational denominator. Rule. To reduce a fraction whose denominator is a binomial surd to an equivalent one having a rational denominator. Rules. Cor, 1, 2. Utility of preceding transformations. 220 a I 5 RADICAL QUANTITIES. SYNOPSIS FOR B^YJBW— Continued, Prop, relating TO Irrational < Quantities. 325. 326. 327. 32§. 329. 330. 331. L332. ' A complex radical quantity. Simplipica'n op Complex Radi- cal Quan, To simplify Imaginary Quantities. \/a±Vb, ya V^ ± cVd ± etc. /j/a V^ ± c Vd ± etc. ' An imaginary quantity. A real quantity. Classification ofimxiginary quantities. To find the sum. To find the difference. To find the product. To find the quotient. . To find all the powers of y — 1. Combinations of imagiTiary quan- tities. CHAPTER XIII. QUADRATIC EQUATIONS WITH ONE UNKNOWN QUANTITL- QUADRATIC EXPRESSIONS. DEFINITIONS AND PRINCIPLES. 350. An equation which contains only one unknown quan- tity as X, and whose members are entire and rational with refer- ence to X, is of the Second Degree when it contains a^ and does not contain a higher power of x. Thus, 7x^ = dx + 160 is an equation of the second degree. 351. A Quadratic Equation is an equation of the second degree. 352. An equation of the second degree containing only one unknown quantity as x, when expressed in such a form that its members are entire and rational with reference to x, cannot have more than three kinds of terms, namely : terms which contain the square of x, terms which contain its first power, and known terms, that is, terms independent of x. Therefore, by transposing and uniting terms, the equation can be made to take the form of aoi? •\-hx=:c'', a, h, and c being given quantities, which may be either positive or negative. For example, the equation can be transformed successively into the following equations : 222 QUADRATIC EQUATIONS. 6«2 _ 30^:2 ^ 135^ + 78a; = 360 + 18, — 25a^ H- 213a; = 378, 25a:2 _ 213a: = - 378. We may consider a in the general equation as positive ; for, if it is negative, we may make it positive by changing the signs of all the terms of the equation, as in the preceding example. 353. A Comjyfete Equation of the Second De- gree is one which can be expressed in the form of ax^ -\- bx = c, in which neither b nor c is zero. Thus, a;^ -f 5a; = 24 and 2a:2 — 3a; = 2a; + 12 are complete equations of the second degree. The coefficient a cannot be zero ; for then the equation would cease to be of the second degree. A complete equation of the second degree is sometimes called an Affected Quadratic Equation, 354. If J or c is zero, the equation takes one of the forms ax^ = c, aa^ -\- hx =^0. In either case the equation is said to be Incomplete, Thus, dx? = 27 and 2x^ — 6a; = are incomplete equations of the second degree. An incomplete equation of the second degree of the form of ax*' = c is sometimes called a Pure Quadratic Equation. INCOMPLETE EQUATIONS OF THE SECOND DEGREE. 355. To solve an equation of the form of ax? = c. Dividing both members of the equation by a, and extracting the square root of both members of the resulting equation, we find •1 INCOMPLETE EQUATIONS. 223 R ULE. Find the value of the square of the unlcnown quantity ly the rule for solving a simple equation ; the result will be an equation of the form of x^ = q ; then extract the square root of both mem- bers of this equation. Cor.— The two roots of a pure quadratic equation have equals ahsolute values, but contrary signs. EXAMPLES, Solve the following equations : 1. 8 3-39 5^. -4/15. a; = ± 3. 2. { = U-3.». ^^5. a; = i: 2. 3. ^.-^-'X 16. Ans. a; = ± 3. 4. (a; + 2)2 = 4:c + 5. ^ws. a; = ± !• 5. 1 -h a; ' 1 — a; -4^15. X=i ±-, 6. 3 1 7 42^2 6a;2~3* Ans. ^ = ± 2' 7. o , 7 65a; ^ws. a; = ± 2f 8. >y/flr2 ^ x^ -\-X h ^a^ j^x^ — x c . a(b — c) Ans. x= ± —^ — — ^. 2Vbc Ans. x= ± -4 '-. V2n-1 9 ^,1 ^„2 1 ^2_ ^^ Va2 + x^ Va2-a;2_ V'52 4.a;2 c Va3 — a?J+ V^'^+ic^ ^ Ans.x-±y 2i(^+W) 224 QUADRATIC EQUATIONS. The foUowiDg equations have such a form that they may be solved by a method similar to that employed in the solution of equations of the form of a^ = c : 28a: _ 63 (g + 18) .. a; 4- 18 ~ ^x ' ' ' ^ ^' ^ Clearing of fractions, (1) becomes 112ic2 = 63 (a: + 18)2 . . . (2). Dividing both members of (2) by 7, 16a:8 =9(x-\- 18)2 . . . (3). Extracting the square root of both members of (3), 4a:=±3(a:+18) . . . (4); whence, a: = 54 or — 74- 12. (x — ay = h. Ans. x = a ± Vb, 356. To solve an equation of the form of ax^ + ix = 0. This equation may be expressed thus : x{ax-\-b) = . . . (1). Now, in order that the product of x and ax -]- b may be equal to zero, we must have either x = . . . (2), or era; -f- J = . . . (3). From (3), x = _b a' Hence, an equation of the form of ax^ -\- bx = has two roots, one of which is zero. EXAMPLES. Solve the following equations: 1. 2a^ - ^ = S^a:. Ans. x = or 9. PROBLEMS. 225 2. ^35 ^-r = 2^- Ans. x = or 4. 3. 2 + 4-? = 0. 4. a;(2a; + 5)=a;(3a; — 9). 357, PROBLEMS, 1. Find two numbers, one of which is four times the other, and the sum of whose squares is 153. Ans, ± 3 and ± 12. 2. Find two numbers, one of which is three times the other, and the difference of whose squares is 32. A7is. ± 2 and ± 6. 3. Find two numbers, one of which is three times as great as the other, and whose product is 75. Ans, ±5 and ± 15. 4. A merchant bought two pieces of cloth, which together measured 36 yards. Each piece cost as many dimes a yard as there were yards in the piece, and the entire cost of one piece was four times that of the other. How many yards were there in each piece ? Ans. 24 yds. in one, and 12 yds. in the other. The negative numbers are not given because they do not sat- isfy the question in its arithmetical sense. 5. Two persons, A and B, set out from different places to meet each other. They started at the same time, and traveled on the direct road between the two places. On meeting, it appeared that A had traveled 18 miles more than B ; and that A could have traveled B's distance in 15|- days, but that B would have been 28 days in travehng A's distance. Find the distance between the two places. Ans. 126 miles. 6. The product of the sum and difference of two numbers is 8, and the product of the sum of their squares and the difference of their squares is 80. What are the numbers ? Ans. ± 1 and ± 3. 16 226 QUADRATIC EQUATIONS. 7. The product of the sum and difference of two numbers is a, and the product of the sum of their squares and the difference of their squares is ma. What are the numbers ? Ans. ±y—^- and ± |/ _— — • 8. Two workmen, A and B, were engaged to work for a certain number of days at different wages. At the end of the time, A, who had been idle a of those days, received m dollars, and B, who had been idle b of those days, received w dollars. Now, if B had been idle a days, and A had been idle b days, they would have re- ceived equal amounts. For how many days were they engaged ? b Vm — a Vn , Ans. — -=z — days. ym — Vn COMPLETE EQUATIONS OF THE SECOND DEGREE. 358. To solve a complete equation of the second degree. Let us consider the complete equation acp^ + bxz=zc . . . (1). Dividing both members of (1) by a, x^ + h = - . . . (2). a a ^ ^ b c Let o = - , and o = -', then (2) becomes ■^ a ^ a ^ ' x^^px = q . . . (3). Adding ^ ^ ^^^^ members of (3), ^^-px^^^q-V^ . . . (4). The first member of (4) is the square of (^ + f ) ; hence, ex- tracting the square root of both members, COMPLETE EQUATIONS. '+1 = ±l/?+^ . . Transposing |, x.= - -|±l/7^ • . 237 (5). The given equation, therefore, has two roots, namely : -i+/,.A The operation of transforming (3) into (4) is called Completing the Square, RULE, I. Reduce the given equation to the form of x^ -\-px^ q. II. Add to loth memhers of this equation the square of half the coefficient of x. III. Extract the square root of both members of the equation thus obtained ; the result will be an equation of the form of X + -^z=z ±m, from which the values ofx may be found by trans- position, BXAMPZJES, Solve the following equations : 1 — 3x2 + 36a; = 105. Dividing both members by — 3, ic2 _ 12a; = — 35. Completing the square, ai2— 12a;+36=:-35 + 36=l. 228 QUADRATIC EQUATIONS. Extracting the square root of both members, x-6=±l; x = 6±l; that is, ic = 7 or 5. Verification. - 3 x 7^ 4- 36 x 7 = - 147 + 252 = 105; - 3 X 52 + 36 X 5 = — 75 + 180 = 105. 2. ar» — 4a; + 3 = 0. Ans.x = l or 3. 3. 6a:2 _ 13a: = — 6. * Ans. x = ^ or -, 4. a;2 _ 52; _^ 4 = 0. Ans. x =zl or 4. 6. 3ar5 — 7a; = 20. Jt^s. a; = 4 or — -. o 6. 2r5 — 7a: + 3 = 0. Ans. x = 3 or -. 7. 3arJ — 53a; + 34 = 0. Ans. a: = 17 or | o 8. a^ -{- 10a; + 24 = 0. ^W5. a; = — 4 or — 6. 9. {x — l)(x — 2) = 6. Ans. a; = 4 or — 1. 10. (3a; — 5) (2a; — 5) = (a; + 3) (a; - 1). Ans. a; = 4 or -. 5 11. (2a; — 3)2 = 8a;. Ans. ^ = 1 ^^ l- 1^- ^ = ^-^- ^.... = 51 or 5. 13. V(2a; + 7) + ^{3x - 18) = ^{7x + 1). A71S. x=z9 or — 3|. 14. aa?*— ac = ex — 03^. Ans. x = -^= -— — ! — . 2{a-\-b) 15. a2+Z>2_2Ja; + a;2 = ^. n^ Ans. X = -J— — ^^{bn ± ^/ahr? + Ithn?' — a^n%. COMPLETE EQUATlOi^S. 229 16. 3x-^ 4- 2x-^ = 1. Ans. a; = 3 or — 1. 17. 4o-4i = 2a;0. Ans. x=l ov -% X ^ x^ 5 18. 3x-^2Vx= 16. A71S. x = '7^ ov 4. 12 19. Va; + 5 = -—r . ^7^5. ic = 4 or — 21. Vx + 12 21. V^ 4- m — a/^+ /^ = \/2i. Ans. x= — ± - V2m^ + 2nK 22. (X - c) (ab)^^ ^^ = 0. Ans. ^ = f or ^'. {cx)-^ * ^ 23. ^ + 1 ^ 12a (g + x)~^ {a-hx)~^ (a-x)'^ ^ J 4:a Sa Ans. x = -- or -=-. 5 5 a:- V^+ l 5 ^ Q 8 x-^Vx + 1 11 9 25. — : + hx-^ = — , • Ans. x = — =^-^^t 359. When a complete equation of the second degree is pro- posed for solution, instead of going through the process of com- pleting the square, we may use the formula x= — ^±y$'+^. For example, take the equation — 3a^ + 36a; = 105. Dividing by — 3, a^« — 12a; = — 35. In this case, p = —12, and q =: —35; hence, by the for- mula, x= ±y — 35 + ^^ — j—^ = 7 or 5. 230 QUADEATIC EQUATIONS. EXAMPLES. Solve the following equations by using the formula 1. a;2 — 6ir=7. 2. a? -{-l^ = 95. 3. x^—2x = ^, 4. a? +10x= — 9. 6. 0^2 _ 14^ — 120. 6. a:2 _^ 32a; = 320. 7. 0^2 ^ lOOrr = 1100. 8. x'-x = \. 4 9. a;2 ^ 3^^: _ 19. 13 5 ''*i- Ans . X ;=:7 or — 1. Ans. X: = 5 or - -19. Ans . X : =r 4 or -2. Ans. X = — 1 or -9. Ans. X: = 20 or -6. Ans. X : = 8 or - -40. Ans. X = 10 or - 110. Ans. X : = li or 1 2* Ans. X : = 3 or - -6i. 1 Q 10. xi + ^x = 14u Ans. ic = 7| or — 10. 11. 2iC = 4 4- -. Ans. a; = 3. or — 1. X x^ — S 1 12. X ^— — r = 2. Ans. xz=:2 OT -. x^ -{- 6 2 a:2 X m^ — 4a' 13. 3w — 2a 2 4a — 6m ^W5. a; = m — 2a or ^m -\- a. 14. ^_^ = a. ^««. rr = 5(2±A/^+l). x—d x-\-6 a^ 15. mrc^ a; = 1. Ans. x=z— or =. mn n m^ 16. ^[a^5 + a(a4.J)]+^da: = ^a;(20a4-73). Ans. a; = 2a or 2 (a + ^). COMPLETE EQUATION'S. 231 360. Solving the equation ax^ -^ bx = c in the usual way, we find _ ^b±Vb^ -{- 4:ac ""- 2a To solve an equation of the second degree by means of this formula, it is only necessary to reduce it to the form of ax^-\-bx=Cf and then make the proper substitutions. For example, take the equation —3cc^-{- 36x = 105. In this example, a = — 3, ^ = 36, and c = 105 ; hence, by the formula, _ — 36± V362+4(— 3)105 _ — 36 J: Vl296 — 1260 -^^±^--^^±^ = 6Tl = 5or7. — 6 —6 Let the student solve some of the equations of Articles 358 and 359 in this way. 361. To complete the square by the Hindoo Method. Take the equation aa? -{-hx = c . . . (1). Multiplying both members by 4«, ^ah? + A:abx = ^ac . . , (2). Adding l^ to both members of (2), ^aho^ + ^abx -\- If^ = 4.ac -\- b^ . . . (3). Extracting the square root of both members of (3), 2ax + h= ±V4:ac + b^ . . . (4); whence, ^^-•h ±^^ac + b^ . . . ^^y Solve by this method the following equation : bx^ — 3x = 224. 232 QUADRATIC EQUATIONS. Multiplying both members by 4 x 5, the given equation be- comes lOOa^ — 60a; = 4480. Adding 3^ to both members of this equation, 100^:2 _ 60a; + 9 = 4489. Extracting the square root, 10a; — 3 = _^ u7 ; 3 4- 67 whence, x = ^ = 7 or — 6^. Let the student solve some other equations in this way. 363. To cause the term containing the first powei of the unknown quantity to disappear. If, in the equation we substitute z — ^ for a;, we obtain (._|)V,(._|).,, that is, z2_^ ^'-■J = r whence, zz=±)/q-\-^; Solve by this method the following equation : a^5 — 11a; = — 18. Substituting z -\- — for x, this equation becomes PEOBLEMS. 233 I whence, ^ = -7-, and ^"==^2' 11 7 =. = ^±3 = 9 or 2. 363. pmobijEms. 1. Find a number, such that the square of one- tenth of it shall be equal to the remainder* obtained by subtracting 24 from the number. Let X = the number ; then, by the problem, whence, x = 60 or 40. 2. Divide the number 10 into two parts, such that their pro- duct shall be 24, Let X = one part ; then will 10 — a; = the other part Hence, by the problem, x{10 — x) = 24:; whence, a; = 4 or 6 ; therefore 10 — a; = 6 or 4. Here, although x may have either of two values, yet there is only one answer to the problem ; one part must be 4 and the other 6. 3. A person bought a certain number of oxen for $400. If he had bought 4 more for the same sum, each ox would have cost $5 less. How many did he buy ? 234 QUADRATIC EQUATIONS. Let X = the number of oxen ; then will — = the cost of each in dollars. X If he had bought 4 more for the same sum, the cost of each would have been a; + 4' 400 400 ^ -: — 5; a; 4- 4 a; whence, x=16 or — 20. Only the positive value of x is admissible ; hence, the number of oxen is 16. In solving problems by algebra, results will sometimes be ob- tained which do not apply to the question actually proposed. The reason is that the algebraic language is more general than ordi- nary language, and thus the equation, which is a proper expression of the conditions of the problem, is also appUcable to other con- ditions. It is sometimes possible, by making suitable changes in the enunciation of the original problem, to form a new problem, corresponding to any result which was inapplicable to the original 400 problem. If we change the sign of x in the equation X-\-4: 400 , .^, 400 400 , 400 400 ^ 5, it becomes -. = 5, or = f- 5. X 4: — X —X X — 4: X This equation is the algebraic statement of the following prob- lem : A person bought a certain number of oxen for $400. If he had bought 4 less for the same sum, each ox would have cost $5 more. How many did he buy ? o 1 • XI. J.' 400 400 ^ Solving the equation j = 1- 5, we find a; = 20 or X — TC X — 16. In this connection the student should review Art. 216, 4. Find two numbers whose difference is 8 and whose product is 240. Ans. 12 and 20. 6. Find two numbers whose difference is 2a and whose product is h, Ans. a ± V«^ -|- h and — a i ^/d? -f h. PROBLEMS. 235 6. The remainder obtained by subtracting a certain number from 10 is equal to the quotient obtained by dividing 25 by that number. What is the number ? Ans, 6, 7. Divide the number 40 into two such parts that their pro- duct shall be equal to 15 times their difference. Ans. 60 and — 20, or 10 and 30. The numbers 60 and — 20 satisfy the problem in the algebraic sense, but not in the arithmetical sense. 8. Divide a into two such parts that their product shall be equal to m times their difference. a — 2m± Va^ + ^^^ . « + 2m =f Va^ + 4m2 A71S. ==-^ and ^ 9. Di\4de 100 into two such parts that the sum of their square roots shall be 14. Ans. 64 and 36. 10. Divide a into two such parts that the sum of their square roots shall be s. a + V2as^ — s* j « — V2as^ — s* Ans, 2 ^^^ 2 * 11. A and B start at the same time from different places and travel toward each other. At the end of 14 hours they meet, when it appears that A has traveled 10 miles more than B, and that their rates of travel are such that B requires half an hour more than A to travel 20 miles. Find B's rate of travel. Ans. 5 miles. The negative result is rejected, because it does not satisfy the problem in its arithmetical sense. 12. A and B start at the same time from different places and travel toward each other. At the end of m hours they meet, when it appears that A has traveled a miles more than B, and that their rates of travel are such that B requires n hours more than A to travel b miles. Find B's rate of travel. A a ^ / ab ^ a^ Ans. —^ hi/ h T— 5' 2m ^ m?i 4m2 236 QUADRATIC EQUATIONS. 13. A started from C toward D, and traveled at the rate of 10 miles an hour. When he was 9 miles from C, B started from D toward C, and went every hour one-twentieth of the distance from D to 0. When B had traveled as many hours as he went miles in one hour, he met A. Find the distance from C to D. Ans, 180 miles or 20 miles. 14. A went from C to D, traveling a miles an hour. When he was b miles from C, B started from D toward C, and went every hour - th of the distance from D to C. When B had traveled as n many hours as he went miles in one hour, he met A. Find the distance from to D. Tn — a , /'/n^^^^ ~~\ 15. A and B were traveling on the same road, and at the same rate, from Columbia to St. Louis. At the 50th mile-stone from St Louis, A overtook a flock of geese which were traveling at the rate of three miles in two hours, and two hours afterward met a wagon which was moving at the rate of nine miles in four hours. B overtook the same flock of geese at the 45th mile-stone, and met the same wagon 40 minutes before he reached the 31st mile- stone. Where was B when A reached St. Louis ? Ans. 25 miles from St. Louis. THEORY OF QUADRATIC EQUATIONS WITH ONE UNKNOWN QUANTITY. 364, Every equation of the second degree containing only one unknown quaiitity has two roots, and only two. Every equation of the second degree containing only one un- known quantity can be reduced to the form of x^ -\- px=z q. Solving this equation, we find -|±/^+f has THEOKEMS. 237 Hence x two values, namely: — "| + y S' + x and Denoting we have thef irst of these values by a;' and the second by x"y .■=-t+^,+t, ^--i-/,.?- The equation ot^ -^ px = q cannot have more than two roots. If possible, let a, b, and c be three different roots of this equa- tion ; then will these roots satisfy the equation. a^ -\- pa = q . . . (1), l^+pb = q . . . . (2), c^ -]- pc = q . . . , (3). Subtracting (2) from (1), a^ — ^2^^(^a-b)=0 , . . . Subtracting (3) from (1), a2 — c^^p(^a — c) = . . . . (4). (5). Dividing both members of (4) by a — b, which is, by hypoth- esis, not zero, we obtain a-{-b-{-p=0 . . . (6). Dividing both members of (5) by a — c, a + c+p = . . . (7). Subtracting (7) from (6), b — c = 0; whence, 5 = c ; that is, two of the supposed roots are equal to each other ; there- fore the equation x^ -\- px = q cannot have three different roots. 238 QUADRATIC EQUATIONS. 365. The sum of tlie roots of an equation of the form of Qi? -\- px=z q is equal to the coefficient of the second term taken with the contrary sign. Solving the equation a^ + px=.q, we obtain ^■=-| + iVf^ • • • «' XX and -„"=_|_yVh| • • • (2)- whence, by addition, X -f x" = —p. Thus, the roots of the equation a? —10xz=z -^1^ areS and 2, and their sum is 10. 366. The product of the roots of an equatio7i of the form of x^ -\- px = q is equal to the second member taken with the coti- trary sign. From (1) and (2) of Art. 365 we obtain, by multiphcation, (-| + i^)(-|-v^)=?-(«+?) Thus, the roots of the equation a?— 10x= — 16 are 8 and 2, and their product is 16. Cor. — The independent term q is divisible by each of the roots. 367. Every equation of the second degree containing onlp one unknown quantity can be reduced to the form of (x - x') (x - x") = 0. Denoting the roots of the equation a^-\-px = q . . . (1), by a' and x", we have p=-{x'-\-x"\ and q = -- x'x" ; APPLICATION OF THEOREMS. 239 hence (1) becomes if2 _ (^x' + x") x=— x'x" . . . (2). By transposition and factoring, (2) becomes (x — x') (x — x") = . . . (3). Cor. — Hence a^ -\- px — q ^ {x — x') (x — x") ; therefore the first member of the equation x^ -\- px — q = is divisible by X — x' and by a; — x", 368. EXAMPI.E8. 1. Find the equation whose roots are 2 and 3. 1st /SoZw^iow.— Substituting —(2 + 3) for ^ (365), and —2x3 for q (366), the general equation 7? -\- px =z q becomes x^-hx= — Q. 2d Solution. — Substituting 2 for x and 3 for x", (3) of Art. 367 becomes (a;_2)(ic-3) =0; that is, a:2 _ 5;^; + 6 = 0. 2. Resolve the first menaber of the equation nf -^ 6a; -f 8 = into two binomial factors. Solving this equation, we find a:' = — 2, a;" ir: ~ 4; hence the given equation may be wrtteo in the form [z-(-2)][a;-(^4)]= 0(367), that is, {x + 2) (a; + 4) = 0. 3. Find the equation whose roots are 5 and 2. Ans. x^—'ixz=z -10- 4. Find the equation whose roots are 3 and 3. Ans. a:2 — 6a; = — 9. 44 6. Find the equation whose roots are 10 and — —, o Ans. 3? ■\- -3--'^* = -g-- 240 QUADRATIC EQUATIONS. n -^ :. 1.^. i_- 1- X 13+V85 ^ 13 — \/85 6. Find the equabon whose roots are and — ^ ^ 13 3 Ans. a^ — ;z-x= — j^ 7 7 7. YmA the equation whose roots are 5 + V— 1 and 5— V— 1. Ans. 2:2 — 10a; = — 26. 8. Resolve the first member of the equation Sa;^— 10a;— 25=0 into three factors, . _ , r\ / . 5\ „ Ans, d{x — 5)lx-\--j = 0. 9. Resolve the first member of the equation 3^-\-'ildx-\-780=z0 into two binomial factors. Ans. {x -f 60) {x + 13) = 0. 10. Resolve the first member of the equation 2a;2+ a: — 6 = into three factors. . « / . «x / 3\ Ans. 2 (a; + 2) (a; — -I = 0. 11. Resolve the first member of the equation a;2— 88a; + 1612 = into two binomial factors. Ans. {x — 62) (x — 26) == 0. 12. Resolve the first member of the equation x^ + a^ = into two binomial factors. Ans. (x — a V^^) (a; + « a/^^) = 0. DISCUSSION OF THE EQUATION a;2 -f ^a; = q. 369. The Discussion of an equation consists in making every possible supposition with regard to the arbitrary quantities contained in it, and interpreting the results. The arbitrary quantities in the equation ^ ^ px^q are ^ and q, 370. We shall first make every possible supposition in rela- tion to the signs of 'p and q. Suppose, /rs^, that j9 and q are positive; second, thatjt? is neg- ative and q positive ; tliird, that p is positive and $- negative ; fourth, that p and q are negative. We shall thus have DISCUSSION. The Four Forms x^-\-px=z q . . • m, Q^—px— q . . (2), a? -\- px=: — q . . (3), a? — px= — q . , (4). 241 371. In the first form one root is positive, the other negative, and the negative root is numerically the greater. Since q is positive, the product of the roots is negative (366) ; • hence one root is positive and the other negative. Again, since p is positive, the sum of the roots is negative (365) ; hence the negative root is numerically the greater. Illustration. — The roots of the equation a? -\- x=iQ are 2 and — 3. 372. In the second form one root is positive, the other nega- tive, and the positive root is numerically the greater. Since q is positive, the product of the roots is negative ; hence one root is positive and the other negative. Again, since p is negative, the sum of the roots is positive ; hence the positive root is numerically the greater. Illustration. — The roots of the equation a? — x=z 210 are 15 and — - 14. 373. In the third form both roots are negative. Since q is negative, the product of the roots is positive ; hence they have like signs ; and since p is positive, the sum of the roots is negative ; hence both roots are negative. Illustration. — The roots of the equation x^ -^ ^x =i — 12 are — 4 and —3. 374. In the fourth form both roots are positive. Since q and p are negative, the product and the sum of the roots are positive ; hence both roots are positive. Illustration. — The roots of the equation a? — "^x^^ — 1% are 4 and 3. . " 16 243 QUADRATIC EQUATIOJN^S. 375, For convenient reference, the four fonns and their cor- responding roots are here given. ^ -\- px^q . . . (1) ; whence Q?-^px=:q . . . (2); whence -'•=i+i/?+f =^"=i-i/.+^- oi^j^ px:^^q . . . (3); whence ci^^px^^q . . . (4); whence ^-hV^ 376. Unequal Moots,— The roots of an equation of the first or of the second form are unequal, whatever the relative values of j9 and q may be (371-373). The roots of an equation of the third or of the fourth form are unequal if -^ is greater or less than q. 377. Equal Hoots, — The roots of an equation of the third or of the fourth form are equal if ^— is equal to q. Illustration,— ^oWing the equation a;^ -|- 6a; = — 9, we find a;' = — 3 and x" = — 3. Solving the equation a;^ — Ga; = — 9, we find x' = 3 and x" = 3. 378. Heal JSoo^«.— The roots of an equation of the first DISCUSSION. 243 or of the second form are real, whatever the relative values of p «2 . . . and q may be, for in these forms -r ■\- q '^^ positive. The roots of an equation of the third or of the fourth form are real if -r is not less than a, 4 Remabk. — The quantities jP and q are liere supposed to be real. 379. Imaginary Moots, — The roots of an equation of the third or of the fourth form are imaginary if —■ is less than q'y for the radical part of each of the roots, in this case, is the square root of a negative quantity. Illustration. — The roots of the equation a;^ + 6a; = — 10 are __ 3 ^ V-- i and — 3 — V— 1 ; and the roots of the equation 0^2 _ 6a; = — 10 are 3 + V^^ and 3 — V~-^. 380. Imaginary roots indicate incompatible conditions. The demonstration depends upon the following Lemma. — 77ie greatest product wJiich can be obtained by sep- arating a given number into ttvo parts and multiplying one by the other is the square of half that number. Let p be the given number, and d the difference of the parts into which it is separated. 7) d Then -^ + - = the greater part, 10 d and -g — - - = the less part (213, 4). Denoting the product of the parts by P, we have 4 4' Now, since jO is a given number, it is evident that P will in- 244 QUADKATIC EQUATIONS. crease as d diminishes, and will be the greatest possible when d = 0\ that is, P, when greatest, is equal to (^ . Illustration. 8 = 1 + 7; 7x1= 7. . 8 = 2 + 6; 6 X 2 = 12. 8 = 3 + 5; 5x3 = 15. 8 = 4 + 4; 4x4=16. In the first form the sum of the roots is — ^ (365), and their product is — q (366) ; hence (Lem.), in this form, — q can- not be greater than -^. In the second form the sum of the roots is p, and their pro- duct is — 5' ; hence, in this form, —q cannot be greater than '-j-- In the third form the sum of the roots is — jo, and their pro- duct is q; hence, in this form, q cannot be greater than ^. In the fourth form the sum of the roots is p, and their product is q ; hence, in this form, q cannot be greater than ■^. In the first and second forms q is positive; hence an equation p2 in which — q is greater than ■'— can never occur in either of these forms ; for ^— being positive is greater than any negative quantity. But in the third and fourth forms an equation may «2 occur in which + 5' is greater than •'—. Thus, in the equations x^ + 6x= —10 and a?^—(},x=— 10, the independent term, taken with the contrary sign, is greater than the square of half the coejQ&cient of the second term ; therefore the roots are imag- inary. Hence, if- any problem furnishes an equation of the third or of the fourth form, in which the independent term, taken with the contrary sign, is greater than the square of half the coefficient of PEOBLEM OF THE LIGHTS. 245 the second term, we infer that the problem contains incompatihle conditions. For example, let it be required to find two numbers whose sum shall be 6 and product 10. Let X = one of the numbers ; then will 6 — a; = the other. By the second condition of the problem, x((o — x)=ilO', that is, Qx — x^ = 10, or, by changing signs, t? — 6.r = — 10 ; whence, a; = 3 ± V— 1. The imaginary roots indicate that there are no real numbers whose sum is 6 and product 10. The greatest product which can be formed by separating 6 into two parts and multiplying one by the other is 9. PROBLEM OP THE LIGHTS. 381. To find, on the straight line joining tivo lights, the points which are equally illuminated hy those lights. ! I [ I I P'' A P B P' Let A and B be the two lights. Denote the intensity of the light A at a unit's distance by a, the intensity of the light B at a unit's distance by h, and the distance between the lights by d. Let P be a point equally illuminated by the two lights, and let ic = AP; then will d — x = BP. One of the laws of light is, that the intensity of a light at any distance as x, is equal to the quotient obtained by dividing its in- tensity at a unit's distance by x^\ hence -^ = the intensity of the light A at P, and -7-^ i-g = the intensity of the light B at P. yCL — Xj 246 QUADRATIC EQUATIONS. But P is to be equally illuminated by the two lights; ±- ^ , . . (1). Clearing this equation of fractions, a{d-x)^ = bx^ . . . (2). Extracting the square root of both members of (2), (d-x)Va= ±xVb . . . (3) ; Wa ± VbJ whence, Separating the values of x, Wa + Vb/ and a^''=d(-J^-\ \Va - Vb' From the nature of the problem, a and b are positive ; hence the values of x are real ; therefore there are two points of equal illumination, and only two, on the line of the lights. Six different suppositious can be made upon the arbitrary quantities a, b, and d^ namely : 1. ayb and dyO. 4:, ay b and d = 0. 2. a = b and ^ > 0. 5. a = b and d = 0. 3. a <.b and c? > 0. 6. a 0. In this case — = — is a proper fraction ; that is, it is less V« + Vb than 1 ; and since the denominator is less than twice the numera- tor, the fraction is greater than ^. PROBLEM OF THE LIGHTS. 247 < d and > -d. The point P is therefore between the two Hghts and nearer the weaker one. The fraction ^^ - > 1 ; ivBvh' V« — vh) The second point of equal illumination is, therefore, at some point P' on the right of B. 2. a=zb and e^ > 0. In this case x' = - and x" = -—— = oo (222, 1) ; that is, the first point of equal illumination is at the middle point of AB, and the second is at an infinite distance to the right of A. The symbol oo indicates impossihility ; that is, it shows that there is no second point of equal illumination. 3. a<,b and d>0. In this case —-= — < ^ and -— — is negative ; Va + V 6 ^ Va — yb x' 5 and d = 0. In this case x' = and x" = 0. How are these results to be interpreted ? They seem to indi- 248 QUADKATIC EXPEESSIONS. cate that the point at which the lights are placed is equally illu- minated by them ; but this is not true, as we shall see by consid- ering equation (1). Under the hypothesis that d=zO, (1) becomes ^2 a;2 • But this is not an equation in fact, for a'>l), and the de- nominators are equal. It would not be an equation if a; = 0, for then -^ and -^ become unequal infinities. There is, then, in this case, no eqiiation, and hence no point of equal illumination. 5. az=l) and J = 0. In this case a;' = and x" = -. The first value of x indicates that the point at which the two equal lights are placed is equally illuminated by them, and the second value of x indicates that a7iy point on the line of the lights is equally illuminated by them (322, 4). As the lights are now at the same point, the line of the lights may be drawn in any direction in space ; hence, in this case, any point in space will be equally illuminated by the two lights. 6. a 2. Subtracting (4) from (3), 2t/= ± V«* + 5^^ =F V«2 — 2^; whence, 2^ = ± i V^ + 2^ =F i Va2 — 2^. 3. Solve the equations x^^f = a^ . . . (1), xy = lfi . . . (2). Adding four times the square of (2) to the square of (1), X^ ■}- 2a^f -^ y* = a* -{- ^b* . . . (3) ; whence, 3^ + y^= ± Va* + U* . . . (4). Adding (1) to (4), 2xi=za^± V«* + 4^; whence, x=± ^cfi±V^+W^ Subtracting (1) from (4), 2f=z ^a^±Va* + ^b^; , . i/ - a" ± Va' + 4^ whence, y=:±y — -! . 4. Solve the equations 2/2-^:2^16 . . . (1), 2y^ - 4a;y + 3a;2 = 17 . . . (2). Assume y = vx; then (1) and (2) become v2a>J — a;2 = 16 . . . (3), PABTICULAR SYSTEMS. 263 2v2a^ _ 4vx^ _|. sx^ = 17 . . . (4). 16 From (3), x^ = 1' and from (4), ^2^__i|__; 17 16 (5). 2v^-.^v-\-3 v^ — 1 ' Clearing (5) of fractions, transposing, and reducing, Uvu — 642; = — 65 . . . (6) ; 6 13 whence, ^~q ^^ ~^* 5 16 Substituting - for v in the two equations x^ = -^ — - and yz=vx, we have x? = ^, and yz=-x', m x= ±3 and ^ = ± 5. 13 Substituting -^ for v in the same equations, we find a^=±3 and y=±Y' The artifice here used may be adopted whenever all the un- known terms in both equations are of the second degree with ref- erence to the unknown quantities. 5. Solve the equations gfi ^ xy — 6y^ = 24: . . . (1), x^ -{- 3xy - lOy^ = 32 . . . (2). Assuming y = vx, (1) and (2) become x^ 4- vx^ — 6vW = 24 . . . (3), ayi ^3v(x^ --lOv^x^ = 32 . . . (4). 264 SIMULTANEOUS EQUATIONS. 24 From (3), x^ = 1 4- V — 6t^^ 32 and from (4), ^,2,^ _____; 24 32 1 4- v — 6y2 1+ 3i; — 10t;2 whence, ^^ = o ^^ q * 2 3 (5); 1 24 Substituting ^ for v in the two equations x^ — ^ and y = vx, we have 3?^ = cx^, and y = -x; a; =±00 and y=:t^» Suostituting ^ for v in the same equations, ^ a; = ± 6 and ^ = ± 2. 6. Solve the equations a^ + 2xy-\-y^-{^ax + ay = b . . . (I), xy + y^ = c . . . (2). The first equation may be written thus : {x + yY-{-a{x + y)=zb; whence, x -\- y = .• • • (3)- From (2), x-hy = - . . • (4). Combining (3) and (4), we find a2^2b — 2cTaV¥TU x = — a± Va^+ 46 2c a ± ^Ai^ 4- 42> PARTICULAR SYSTEMS. 265 7. Solve the equations a? — 2rr^ -\- y^ — ax ■]- ay = b . . . (1), xy — y^ = c . . . (2). Equation (1) may be written thus: {x — yY — a(x — y)=h', whence, x — y=z • • • V^)* Prom (3), ''-y = l • • • W- Combining (3) and (4), we find y _a^ + 2b-\-2c±a Va^ + ^b ~~ a± Va^ + 46 2c a ± Va^ H- 45 8. Solve the equations xiy-\.xy^ = 30 . . . (1), 1+1=1 ■ • • (^)- Equation (1) may be reduced to the form xy{x-\-y)=SO , . . (3), and (2) may be reduced to the form 6(x + y)=6xy . . . (4). Dividing (3) by (4), xy _ 30 . 6 ~ 6xy' whence, 6x^y^ = 180, or, a;y = 36; whence, xy= ±6 . . . (5), 266 SIMULTANEOUS EQUATIOKSw Combining (3) and (5), iP + y=±5 . . . (6). Combining (5) and (6), we find x = S, 2, 1, or —6, y = 2, 3, —6, or L 9, Solve the equations X -hy =a . . . (1), af^Jff = h^ . . . (2). Dividing (2) by (1), ic* — a^Sy + x^y^ — xy^ -\- y^ = -', ct which may be placed under the form ^^-y''-xy(xi^y^)-\-x^y'^ = ^ . . . (3). Squaring (1) and transposing the term 2xy, a^-i-y^ =ai — 2xy . . . (4). Squaring (4), X^-\-2x^f-\-y*z=za*^^^y-\-4xiy^ . . . (5); whence, x* -{- y^= a* — 4:a^xy + 2x^y^ . . . (6). Equation (3) may therefore be placed under the form a^'^^aHy ■\'2o?y'^ — xy{a^ —2xy) + x^^-^ that is, hx^y'^-^aHy — ^'^a?' ... (7); Cb a^±y 4:b^ + a^ whence, xy = —^ — . . . (8). The values of x and y may now be found by combining (1) and (8). PARTICULAK SYSTEMS. Solve the following pairs of equations : 267 10. i^ + ^^ = 65) 11. Ans. )^=±^or ±4, ± 4 or ±7. \y Ans. (^=±3 or qp 8, 12. j x^ + Zxy= 54 ) \xy + 4.y^ =115 4" x= ±d or ±36, Ans. \ 23 «/ = ± 5 or =F y . 13. (^2 + ^^ = 15) ^^^^ , a; = ± 3 or ± -—, ( V2 y=±2 or ± V2 I 3a^ 4- 8?/2 = 14 ) a;= ±2or ±\/% 15. i^+*y = 12l. j„,. ( a:?/ — 2?/* = 1 ) (i/2_2a:?/ + 15= 0) ( a:^ + 2w2 = 3 i W5. -4W5. 18. a; y 5 rrj^ OK 180 a; = ± 3 or ± ^ = ± 1 or ± 8 1 V6 a: = ± 4 or ±3^/3, 2/ = ± 5 or ± Vs. 15 x=± V21 3 or i: 00, ^ =^ V21 Ans, i^ = 30or6 ( 2/ =i 6 or 30. 268 SIMULTANEOUS EQUATIOlfS. 19. 20. 21. 22. ( ic2 + 1/2 = 45 i ^ + ^ = ^[. Ans, iSy — x = y^) . Ans. ]' U x=±6, ±3 or :F3. ^ + 3^^=o^2' x — y xy x = 0, 2, or ± a/2, 3^ = 0,2, or 2T A/^. . ( a; = 0, 4, or — 2, ( y = 0, 2, or — 4. (3?^y^^x + y=\^) X xy == Q) A71S, ( a; = 3, 2, or — (y = 2,3, or - -3±V3, 3:fV3. 23. \3^ + y'-x-y = 32) ' \x -\-y -\-xy =29) ^W5. rr = 5, 4, or — 5 ± V— 14, y = 4:, 5, or — 5 =F V— 14. 24. -1^+.^= n ( a:3 + y3 ,^3 65 f . ( a; = 4 or 1, Ans, \ . / \y =il or 4. ( 2/ = or — 2. In some of the following answers the roots are not all given 26. {xy{x+y)=30\ {7? + if = 35 ) 2:5 __^=: 3093 28 i ^-2^ = 3093) ■ix —y — 3 ) Ans. \y = x = 3 or 2, 2 or 3. 27. ■i!, + ^=i[- ^^^. (a;4 + «* = 82i a:=3, 1, or 2±5\/— 1, 2/=l, 3, or 2T5a/^1. ^/i.. i^ = 5or -2, (3/ = 2 or —5. PAETICULAR SYSTEMS. 269 29. j^^-^y+2/'=in. J, ( X — xy -\- y ^= 4 ) 30. 32. 33 \x^ + xY-\-y^ = ^^^ ^' 31 U2 + 2/2 + :r^ := 49 ) ix^ + y^ + xY=931 ) (a4_^^_y4_^2^84) ( 2;2 + a^i/2 + «/2 = 49 f • J (2;2 — y^){x —y) = 16a:y ( (V _ ^) (^2 _ y2) = UOx^y^ 34 \(f<^ = y{s-y) I, I y^+{s — yY = x^) Ans. Ans, Ans. Ans. 2/=J(9TV73). {x= ±3 or ±2, U=±2 or ±3. X=: ±5 or ±3, y=±3 or ±5. x= ±3 or ±3, y=±^ or ±3. Ans, \ ^=^ ^^ ^' ( «/=3 or 9. y = 35. J a;2^i + 2/^^1 = 9 ) ^^5. ic=:4 or 2, 2/ = 2 or 4. PAIRS OF EQUATIONS INVOLVING RADICAL QUANTITIES. 393, EXAMPLES. 1. Solve the equations V^ + V^ = 5 . . . (1), x + y = 13 . . . (2). Squaring (1) and subtracting (2) from the result, 2 Vxy = 12 ; whence, ^y = 144 . . . (3). 270 SIMULTANEOUS EQUATIONS. Squaring (2) and subtracting (3) from the result, a^2-2a-^ + 2/2 = 25; whence, a: — y = ± 5 . ^ . (4). ix — 9 or 4 y = 4: or 9! 2. Solve the equations Jx — y 20 .-V --y + V^| = ^T^ • • • «> 0^5 + 2/^ = 34 . . . (2). From (1), by transposition and reduction, ^,y2^20 ^ /x-y _ ,3x x^-y ^ x + y Dividing the denominators of (3) by Vx-\-yy Vx-\-y = — Vx — y; whence, a^ - y"^ -20 = - Vx'-y^; or, x^-y^-\- Vx^-y^ = 20 . . . (4). Assuming \/Q? — y^ = z, (4) becomes «a + ^ = 20 . . . (5); whence, 2; = 4 or —5; that is, jr2-y2=i6 or 25 . . . (6)/ fa; = ± 5 or ± 3 5;9 2' INVOLVING RADICAL QUANTITIES. 271 3. Solve the equations a;J + /=:35 . . . (1), xi + y^= 5 . . . (2). Assuming a;* = v and y^ = 2, (1) and (2) become ^3 + ;23 = 35 . . . (3), V -{- z =6 ... (4). These equations may now be solved as Ex. 24, Art 39^. 4. Solve the equations ^i*^i=:k*' • • . (1), V^-'^ + V^^ = 78 . . • (2). Clearing (1) of fractions, it becomes x + y = 61 + \/xy . . . (3). Equation (2) may be written V^(V^ + V^) = 78 . . . (4). * Assume x + y = v , . . (5), and \/xy = z . , , (6). Multiplying (6) by 2 and adding the result to (5), X -\-2 Vxy -\- y = v + 2. > whence, ^/x -f V^ = ± ^v + 'Zz • (7)- Hence (3) and (4) become t;=61 + 2j . . . (8), ^/z{± Vv + 22) = 78 • (9). The values of v and z are easily found from (8) and (9), and then (5) and (6) will furnish the values of x and y. 272 SIMULTANEOUS EQUATIONS. Solve the following pairs of equations : '-{'. -h y + Vxy= 7 = 21) + r + xy + ?/ —VI ^^f^xy =84 ^ U + ?/ — Vxy = 6 . ( a; = 4 or 1, ( ^ = 1 or 4. \y — % or 8. (a:+ Va^_3^2 = 8) ^' \x-y =1) a; = 13 or 6, 12 or 4. a; + 2^ = 10 • I a; 4- 2/ = 20 i (y = 2 or 8. -4/15. a;=10±4V6 or 10 T ^Vl5, y=10:F4\/6 or 10 ± Vl5. 10. y J . Ans. a^c — 2c±:ac Va^ — 4 y=± ±2V^'^^ c 11. a:y — (a; + 2^) = 54 ^ ^TZS Vfl2-4 (2/= 6 or —4 J, 12 or —9. 12 iC = ^2 ^ 2 a: = 6 or 3, 2 or 1. (2/ = MORE THAif TWO Ui^KNOWJ^ QUANTITY. 273 13. )^^^y^ 14. j ^ + y^ An8 = 189 { X -i- y + ^x + ?/ = 12 15. :^^y — 's/y — xz=^ Va — x 16. s^ Jx-\-y 6 a: + w — i/ — i-^ = ^ X — y X — y iC2 4. ^2 _ 41 -4?^s. J a; = 81 or 16, • ( ?/ = 8 or 27. ic = 5 or 4, ^W5. or 5. Ans. 4 y = -^a. ±5 or ±3l/|, y = ± 4 or ± ^1 2 ' GROUPS WITH MORE THAN TWO UNKNOWN QUANTITIES. 394, EXAMPLES. 1 Solve the equations ^ + ^^ + 2/'-^ = 37 . . . (1), x^-^xz + z^z^^^ . . . (2), y^ + yz + z^=:l^ . . . (3). Subtracting (2) from (1), {y^z)x + y^^z^ = ^', whence by factoring, (y-z){x + y + z)z:^% . . . (4). Subtracting (3) from (2) and factoring the result, {x-y){x-{-y -^z) = ^ (5). 18 274 SIMULTANEOUS EQUATIONS. Combining (4) and (5), whence, x -{■ z = 'Zy . . . (6). Combining (5) and (6), (x-y)^y = ^', whence, a; = - + ^ . . . (7). if OomlDining (1) and (7), (? + y)'+ 3 + 3^ + ^2 = 37; whence, ^ = ± 3 or ± o "V^- Substituting these values in (7), x=±i or ±yV3. Substituting for x and y their values, we find from (6) 2 = ± 2 or =F I V3. 2. Solve the equations x + y + z—a . . . (1), a?+y^+z^=l^ . . . (2), xy=cz . . . (3). Transposing z in (1), x-\-y = a — z; tyhence, x^ + 2xy + y^ = a^ ^ 2az + 2:2 . . . (4^^ Transposing z^ in (2), a^^y2=:l2 — zZ . . . (5). Subtracting (5) from (4), 2xy = a^ + 2z^^ 2az — i^ . . . (6). MORE THAN TWO UNKNOWN QUANTITIES. 275 Combining (3) and (6), 2cz = a2 ^ 2z^ — 2az — h^ (7); whence, a + c ± V(a 4- c)2 — 2 {a^ — b^) 2 Substituting these values for 2; in (1) and (3), a + c±V(a + c)^—2(a^—d^) ^ + y + — —^^ ^ = a (8), xy _ac-\-c^±cV(a + cy—2{a^—b^) (9). From (8) and (9) the values of x and y may be found. Solve the following groups of equations : ' xy^s^ = 4725 yz^ 45 3. \T ~Y j___3 x^y ~ 245 X -\-y -^ z =13 X'i -f- ^2 ^ ;g2 _ 61 2yz z= X {z -\- y) Ans. 5. X a + b" 1 X z - + i a c y^ h c z . 1 1 1 n - H- - + - = 9 X y z X y (^ = 7, Ans. Uj=:b, (z =3, * a; = 9 or 4, •. \ y = 2± V — 14 or 3, \z=2'=f V-14 or 6. ix=:0 or 2a, Ans, \y = b or —b, z •=. c or — c. r 1 6 ^=2^^2G' Ans, - 1 15 y = 3or^, 1 15 ^ = 7 or 77. L 4 44 y -{• z = z + X X -\-y = SIMULTANEOUS EQUATIONS. 1 X 1 y 1 z J Arts, x = y z=zz=i ±ro* x^ +xy -{- xz = 27 ^y + y^ +yz = is xz -\- yz -^ z^ =^ 36 U = ±4. 10. y = z * "a; 2; = V v = a ' X [bx= ^vyJ ^xyz = 6 xyv = 8 xzv = 12 L yzv = 24 J ^W5. ± Va .^* y = V*, 2; = ± a/«, ^v = ± Va Vb. r x = i, Ans. - y = 3. 0=3, w = 4. 395. PROBLEMS. 1. The sum of two numbers is equal to nine times their differ- ence ; and if the greater be subtracted from their product, the re- mainder will be equal to twelve times the quotient obtained by dividing the greater by the less: find the numbers. Ans, 5 and 4. 2. The sum of two numbers is equal to a times their differ- ence; and if the greater be subtracted from their product, the re- mainder will be equal to h times the quotient obtained by dividing the greater by the less: find the numbers. -j^(i±^ and U/^M- PROBLEMS. 27? 3. Find two numbers whose difference is equal to two-ninths of the greater, and the difference of whose squares is 128. Ans. 18 and 14. 4. Find two numbers whose difference is equal to - of the greater, and the difference of whose squares is a. 5. The sum of two numbers is 16, and the quotient obtained by dividing the greater by the less is 2J times the quotient ob- tained by dividing the less by the greater : find the numbers. Ans, 10 and 6. 6. The sum of two numbers is «, and the quotient obtained by dividing the greater by the less is l times the quotient obtained by dividing the less by the greater : find the numbers. — 1 — \ 7. The difference of two numbers is 15, and half their product is equal to the cube of the smaller number : find the numbers. Ans. 18 and 3. 8. The difference of two numbers is d, and half their product is equal to the cube of the smaller number: find the numbers. . 1 + 4^ ± VsTTTi . 1 ± V8^ + i Ans. ^ ■ — and —^ — ^ ■ — 4 4 9. The product of two numbers is 24, and the product of their sum and their difference is 20 : find the numbers. Ans. 6 and 4. 10. The product of two numbers is «, and the product of their sum and their difference is h'. find the numbers. Ans, ± y b ± V^a^ + ^ and ± ^ -h ± Vla^ -^l^^ 2 2 278 SIMULTANEOUS EQUATIONS. 11. The product of two numbers is 18 times their difference, and the sum of their squares is 117 : find the numbers. Ans. 9 and 6. 13. The product of two numbers is ?;i times their difference, and the sum of their squares is a : find the numbers. Ans, c ± Vc (2?n + c) and — c ± Vc (2//i + c), where c = " "^ ± f "+^ . 13. Two persons, A and B, bought a farm containing 600 acres, for which they paid $600, each paying 1300. A paid 75 cents more per acre than B in order to be permitted to take his share from the best land. How many acres did each get, and at what price per acre ? A7is. A 200 acres at $1.50, B 400 acres at $0.75. 14. A certain number of workmen in 8 hours carried a pile of stones from one place to another. Had there been 8 more work- men, and had each cari'ied each time 5 pounds less, the pile would have been removed in 7 hours ; but if there had been 8 workmen less, and had each carried each time 11 pounds more, it would have required 9 hours to remove the pile. How many workmen were employed, and how many pounds did each carry at a time ? Ans, 28 workmen, and each carried 45. pounds; or 36 workmen, and each carried 77 pounds. 15. The fore- wheel of a carriage makes 6 revolutions more than the hind-wheel in going 120 yards ; but if the circumference of each wheel be increased one yard, the fore-wheel will make only 4 revolutions more than the hind-wheel in going the same distance. What is the circumference of each ? Ans, Fore-wheel, 4 yds.; hind-wheel, 5 yds. 16. What number is that, which being divided by the product of its two digits gives 2 for the quotient, and if 27 be added to it the order of the digits will be inverted ? Ans. 36. 17. A sets off from London to York, and B at the same time from York to London, each traveling at a uniform rate. A reaches York 16 hours, and B reaches London 36 hours after they have met on the road. Find in what time each has performed the journey. A7is, A in 40 hours, B in 60 hours. SYNOPSIS FOR REVIEW. 279 18. A man had $1300, which he divided into two parts, and placed at interest at such rates that the incomes from them were equal. If he had put out the first part at the same rate as the second, he would have drawn for this part 136 interest; and if he had put out the second part at the same rate as the first, he would have drawn for it $49 interest. Find the rates of interest. Ans. 6 per cent, for the larger part, and 7 per cent, for the smaller. 19. A and B engage to reap a field for $24; and as A alone could reap it in 9 days they promise to complete it in 5 days. They found, however, that they were obliged to call in to assist them 2 days in order to complete the work in the stipulated time, in consequence of which B received $1 less than he would have done if he and A, \nthout the assistance of C, had continued until they completed the work. In what time could B or C alone reap the field? Ans. B in 15 days, in 18 days. 20. A number consists of three digits. The sum of the squares of the digits is 104; the square of the middle digit exceeds twice the product of the other two by 4 ; and if 594 be subtracted from the number the order of the digits will be inverted: find the number. Ans. 862. 396. SYNOPSIS FOR REVIEW. O t— I EH .< >^ ^ H S5 ' Groups and pairs. Degree of an equation. General form of an equation of the second degree in- volving two unknown quantities. Pairs consisting of one equation of the first degree AND one of the SECOND DEGREE. Particular systems. Pairs of equations involving radicals. Groups involving more than two unknown quantities and one or more equations of a higher degree than the first. OHAPTEE XVL EATIO, PEOPORTIOl^, AJSTD YAEIATION. RATIO. 397. Hie Hatio of two quantities is the quotient arising from dividing the first by the second. Thus, the ratio of 6 to 3 is 2, and the ratio of a to J is t.* 398. The Sign of ratio is the colon. Thus, a:b ia read the ratio of a to b, 399. The Terms of a ratio are the quantities com- pared. Thus, in the expression a : b, the terms of the ratio are a and b. 400. The Antecedent of a ratio is its first term, and the Consequent of a ratio is its second term. The antecedent and consequent of a ratio together form a Couplet, 401. A Simple Hatio is one whose terms are entire. Thus, a:b is a simple ratio. * Ratio as thus defined is sometimes called geometrical ratio or ratio hy quotient, to distinguisli it from arithmetical ratio or ratio hy difference. The arithmetical ratio of a to & is a — &. The sign of arithmetical ratio is •• . Thus, a --h is read the arithmetical ratio of a to h. When the word ratio is used without modification it is understood to mean geometrical rcuio. EATIO. 281 403. A Complex Matio is one in which at least one of the terms involves a fraction. Thus, - : b and - : - are complex c c a ^ ratios. 403. A CoTVipound Hatio is the ratio of the products of the coiTesponding terms of two or more ratios. Thus, the ratio compounded of a:b and c:d is ac: bd. A compound ratio does not differ in its nature from any other ratio. The term is used to denote the origin of the ratio. 404. The Duplicate Matio of two quantities is the ratio of their squares. Thus, a^:l^ is the duplicate ratio of a to b, 405. The Triplicate Matio of two quantities is the ratio of their cubes. Thus, a^ : b^ is the triplicate ratio of a to b. 406. Tlie Subduplicate Matio of two quantities is the ratio of their square roots. Thus, ^/a : Vb is the subduplicate ratio of a to b. 407. The Subtriplicate Matio of two quantities is the ratio of their cube roots. Thus, ^s/a : Vb is the subtriplicate ratio of a to b. 408. T7ie Direct Matio of two quantities is the quo- tient arising from dividing the antecedent by the consequent. Thus, the direct ratio of a to 5 is t- 409. Tlie Inverse or Meciprocal Matio of two quantities is the direct ratio of their reciprocals, or the quotient arising from dividing the consequent by the antecedent. Thus, the inverse ratio of a to 5 is - : ^ or -. a b a 410. A ratio is called a ratio of Greater Inequality, of TjCSH Inequality 9 or of Equality, according as the antecedent is greater than, less than, or equal to, the consequent. 282 PEOPOETIOK. 411. EXAMPLES, 1. Find the ratio of a^ — h^ to a + h. a -\- 2. Find the inverse ratio of a^ — h^ to a ■\- b. Ans. t. a — h 3. Find the ratio which is compounded of 3 : 5 and 7 : 9. Ans, 15 4. Find the subduplicate ratio of 100 to 144. Ans. -. 6 6. Show that a ; ^ is the duplicate of a + c : J + c if c^ = a5. PROPORTION. 413. A Proportion is an equation in which each member is a ratio, both terms of which are expressed.* The equality of the two ratios may be indicated either by the sign =, or by the double colon : :. Thus, we may indicate that the ratio of 8 to 4 is equal to that of 6 to 3 in any of the following ways: r8:4 = 6:3 8:4::6:3 . 8_6 4~"3 8-T-4 = 6-^3^ This proportion, in any of its forms, is read tlie ratio of 8 to U is equal to the ratio of 6 to 3, or, 8 is to Jf as 6 is to 3. * A geometrical proportion is one in which the ratios are geometrical. An arithmetical proportion is one in which the ratios are arithmetical. Thus, 6 •• 5 — 9 •• 8 is an arithmetical proportion. When the word proportion is used without modification, it is understood to mean geometrical proportion. PEOPORTIOK. 283 413. Tlie Terms of a proportion are the four quantities which are compared. The first and second terms form the first couplet ; the third and fourth, the second couplet. 414. The Antecedents in a proportion are the first and third terms. 415. Tlie Consequents in a proportion are the second SLud fourth terms. 416. The Extremes in a proportion are the first and fourth terms. 417. The Means in a proportion are the second and third terms. ■ 418. If four quantities a, h, c, and d are so related that a'.h ^=c'.d, d is said to be a Fourth Proportional to a, h, and c, 419. If three quantities «, i, and c are so related that a-.h = h:Cf c is said to be a Third JProjwrtional to a and b. 420. K three quantities «, J, and c are so related that a:b ^b:c, b is said to be a Mean Proportional between a and c, 431. A Continued Proportion is a continued equa- tion in which each member is a ratio, both terms of which are expressed. Thus, a : b = c : d = e :f := g : h is a continued pro- portion. 422. If four quantities a, b, c, and d are so related that a\b — ~.-y 1.1 c'd' they are said to be Inversely or Iteciprocally Propor- tional, 284 PEOPOKTION. 433. JEquitnultiples of two or more quantities are the products obtained by multiplying each of them by the same quan- tity. Thus, ma and mb are equimultiples of a and i. 424. A proportion is taken by Alternation when the means or the extremes are made to exchange places. Thus, if a : b=c : d, we have by alternation, either a : c^=b :d,or d'. bz^c-.a. 425. A proportion is taken by Inversion when the terms of each couplet are made to exchange places. Thus, if a : 5 = c : df we have by inversion, b:a=. d'.c. 426. A proportion is taken by Composition when the sum of the terms of each couplet is compared with either term of that couplet, the same order being observed in the two couplets. Thus, if « : 5 = c : i/, we have by composition, either a -{- b'.a =z c + d:c, OT a -\- b:b = c -\- did. 427. A proportion is taken by Division when the differ- ence of the terms of each couplet is compared with either term of that couplet, the same order being observed in the two couplets. Thus, if a:b = c:df we have by division, either a—b:a=c—d:Cj or a — b'.b ^ c — d:d. 428. In every proportion the product of the extremes is equal to the product of the means. Let a:b = c:d; . then |:=.|(412); whence, ad = be. Cor. — If the means are equal, as in the proportion a:b = b:c, we have b"^ = ac, whence b = Vnc; that is, a mean propor- tional between two quantities is equal to the square root of their product. 429. If the product oftivo quantities is equal to the product of two others, either two may be made the extremes, and the other two the means, of a proportion. Let PROPOBTION. ad=zlc . . . (1) ; 285 then, dividing both members by cd, -=— , or a\c=z o:d . . . c d In like manner it may be shown that a\h =^ c \d . . . (3), (2). c \a-=. d'.h c \d=. a'.h d'.h z= c\a (4), (5), (6), and so on. CoE. 1. — Any one of these proportions may be inferred from any other. Thus, from (2), ad = be, from which any one of the proportions may be derived. Cor. 2. — Since (3) may be derived from (2), it follows that If four quantities are in proportion, they will he in proportion hy alternation. Cor. 3. — Since (4) may be derived from (2), it follows that If four quantities are in proportion, they will he in proportion hy inversion. 430. Equimultiples of two quantities are in the same ratio as the quantities themselves. ma mh a ma:mh = a: h. Cor.— K m = l± P we have that is, a + -a'.h ±-b = a:h; 286 PROPORTION. If tico quatitities he increased or diminislied hy like parts of each, the results will he in the same ratio as the quatitities themselves. « 431. Any equimultiples of one couplet of a proportion are in the same ratio as any equimultiples of the other couplet. Let a\h = c'.d\ then whence, a V c -d' ma mb nc ma : mh = nc : nd. I and n = l±K CoE. — ^If m = l ±- and w = 1 ± =^, we have ^ <1 a±^a:b±^b=ic±^c:d±^d; , . 9 9 9 9 that IS, If tJie terms of the first couplet of a proportion he increased or diminished hy like parts of each, and if the terms of the second couplet he increased or diminished hy any other like parts of each, the results will be in proportion. 433, Any equimultiples of the antecedents of a proportion are in the same ratio as any equimultiples of the consequents. Let a'.h=zc:d . . . (1); then %^% . . . b a (2). Multiplying (2) by ^, ma mc nh "" nd' whence, ma : nh = mc : 7id, which by alternation becomes ma:mG=:^nb: nd. PROPOKTioi^r. 287 433. Axiom. — Ratios that are equal to the same or equal ratios are equal to each other. Thus, if Q,'*l) ^= c: d, e\f—g\h, and a\b =.e\f', then c I d =g\h. 434. If the ratio of the antecedents of one proportion is equal to the ratio of the antecedents of another proportion, the ratio of the consequents of the one will he equal to the ratio of the conse- quents of the other, and conversely. Let a\l =^ c d . . . (1), e'.f=g:h . . . (2), and a'.c=e'.g . . . (3) ; then will h : d =/ : h. For, from (1), a\c — h\d (429, Cor. 2), and from (2), e : g =/: h ; b:d=f:h {'^SS). In hke manner it may be shown that, if a'.h=:c\d, e\f — g'.h, and 'b\d^=.f'.h\ then will a'.c = e:g. 435. If four quantities are in proportion, they will he in proportion hy composition or division. Let a'.h = c'.d . . . (1), then 1 = 1 •• • (2)5. whence, | ± 1 = | ± 1, 288 PROPORTION". a±h c±d b = d- ' ' . 0^); whence, a ^^h '.h = c ±d: d . . . (4), Dividing (3) by (2), a±l} c±d a c . (5); whence, a±d:a = c±6?:o . . . (6). Cob.— Separating (6), a -^ b '. a=^c -{- d: Cj and a — b'.a=:ic — d'.c; a-\-b',a — b=iC + c? : c — a? (434); that is, If four quantifies are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference. 436. Tlie sum of any number of the an tecedents of a continued proportion is to the sum of the corresponding consequents as any antecedent is to its consequent. Let . a\b =: c \ d =. e \ f z:^ g \h ^ &c., and let r denote the ratio ; then a c e a o^ whence, a = br, c = dr, e = fr, g = hr, &c. ; whence, by addition, a + c + e+^ + &c. = (J+ d ■\-f+h-\- &c.)r; whence a±cj±_e_^r_g_±&^ _ ^ _ ^ _ £ _ &_ .-. a + c + e+^ + &c. iS + c^+Z + A + Ac. =za'.b =c:d = &Q. PKOPORTION. 289 437. The products of the corresponding terms of two or more proportions are in proportion. Let a\h =^c'.d, e\f =g\hy and m\n =p:q; then a c b-d' « _^ f-h' and m p n-q' whence, by multiplication. aem cgp ^ bfn ~ dhq' aem : bfn = cgp : dhq. 438. Tlie quotients of the corresponding terms of two propor- tions are in proportion. Let a:b =ic:d. and e'.f = g\h\ then ' a c l~d' and -f-v whence, af ch te~dg' or, a d _c b ^ e h g f « a b c d 19 290 proportio:n^. 439. Like powers or like roots of the terms of a proportion are in proportion. Let a:i = c:d . . . (1), then l='a ■ ■ ■ (')■ Kaising (2) to the n^ power, a" : i" = c" : """l^^^ is the value of 2J when .t = ?/ =2? ^?Z5. 4. 4. If zee px + ?/, and if ;$; = 3 when a; = 1 and y = 2, and 2=5 when a; = 2 and y = 3, what is the value of p ? ^ws. 1. 5. If x^ccy^ and a; = 2 when y = 3, what is the value of y in terms of a; ? 3 -^ ^W5. 2/ = 3 |/ — • G. If ^ oca; 1 X , and, if 1/ = 4 when a: = 1, and y=5 when a; = 2, what is the value of y in terms of a; ? 2 A ns. y = 2x + -. X 7. If xccy when 2; is constant, and if xazz when y is con- stant, how does x vary when both y and ;? are variable ? ^W5. a: X ^jj. i>SS VARIATIOIT. 458. SYNOPSIS FOR REVIEW. ' EATIO PEOPORTION. L VAEIATION. I Terms. — Antecedent. — Consequent.— Couplet. Simple. — Complex. — Compound. Duplicate. — Triplicate. — Sub-dup. — Sub-trip. Direct. — Inverse or Reciprocal. — Of greater IN EQUAL. — Of less inequal. — Of equality. Geometrical. — Arithmetical. — How indi- cated? How READ? Terms — Antecedent. — Consequent. — Ex- tremes. — Means. Fourth proportional. — Third proportional. — Mean proportional. Continued proportion. Inverse or reciprocal proportion. Equimultiples. — Alternation. — Inversion. — Composition. — Division. Cor. Cor. 1, Cor. Cor. Cob. 42§. 429. Cor. 1, 2, 3. 4:iO, 431. 4:{2, Propositions. ^ 433, 434. 4:{5. 436. 437. 438. 439. ' Direct variation. Sign of variation. A variation. Inverse variation. Joint variation. Direct and inverse variation combined. 447. 44§. 449. 450. 451. 452. 453. Cob. 454. 455. 450. Propositions. CHAPTER XYII. MATHEMATICAL Ilv^DUCTIOK 459. Mathematical Induction or Demonstra^ tive Induction may be thus described : We prove that if a theorem is true in one case, whatever that case may be, it is true in another case which we may call the 7iext case ; we prove by trial that the theorem is true in a certain case ; hence it is true in the next case, and hence in the next to that, and so on ; hence it must be true in every case after that with which we began. This method of reasoning is exemphfied in the demonstration of the following theorems : 460. The sum of n consecutive integers hegitining with 1 is n {n 4- 1) 2 We see that this theorem is true in some cases ; for example, 1 + 2 = ^ti), 1 + 3 + 3 = iii+il ; we wish, however, to show that the theorem is true universally. Suppose the theorem were known to be true for a certain value of n ; that is, suppose for this value of n that 1 + 2 + 3 + 4+.. ..+« = ^"^^ . . . (1). Adding n -\-\ to both members of (1), fi (n 4- 1 ^ 1 + 2 + 3 + 4+.. .+7j + 7i-fl = -^^-^ + n H- 1 = (. + i)[l!L±|I±i] . . . (3). 300 MATHEMATICAL INDUCTION. Therefore, if the sum of n consecutive integers beghming with ^i (fi _i_ 1) 1 is ^ — -, the sum of w + 1 such numbers will bo Z (/I _[- 1) ^^ .J • Ii^ other words, if the theorem is true when 71 is a certain number, whatever that number may be, it is true when we increase that number by 1. But we have seen by trial that the theorem is true when 7i = 3 ; it is therefore true when « = 4 ; it is therefore true when ji = 5; and so on. Hence the theorem must be universally true. 461. Tlie difference between the like poiuers of any tivo quan- tities is divisible by the difference between the quantities. Let a and b denote any two quantities, and let n be any posi- tive integer ; then will «" — b^ be divisible by a — b. « ^ an-1 ^ b^±—-J—l . . . (1); a — d a — b hence a" — b^ is divisible by a — b, if a"~^ — Z*""^ is divisible by it. Now a — b\s divisible by a — b\ therefore a^ — 1? is divisible by a — b\ therefore, again, a^ — - b^ is divisible by a—b, and so on ; hence a" — i" is divisible by a — b, if n is a posi- tive integer. Performing the division indicated in (1), we obtain Cor. 1. — The number of terms in the quotient is n. Cor. 2. — If b = a, each term of the quotient becomes equal to a**"!; hence, (?^l.=-- • • • (^)- Cob. 3. — Substituting (^ for a and c^ for b in (2), we have C2» ^n ^ _^3 = c2«-24-c2/i-4^_j. .... 4. c2^»-4^^«-2 . . . (4). SYi?^OPSIS FOR EEVIEW. 301 hence, Tlie difference hetween the like evenpoivers of any tivo quan- tities is divisible by the difference between the squares of the quantities, OoB. 4. — Multiplying both members of (4) by c — dy we obtain ^~f^ ={c — d) (c2^-2 + c^-^d^ 4- . . . . _j_ c'dJ^-i -I- d^-^) ; c ~\~ a hence, TJie difference hetween the like even powers of any two quan- tities is divisible by the sum of the quantities. Cor. 5. — Substituting c^ for a and d"^ for b in (2), we have pinn flrnn Qin _ dm hence, The difference between the like powers of any two quantities is divisible by the difference betiveen any other like powers of the two quantities, if the exponent in the first set of powers is divisible by that in the second set. CoE. 6. — When 7i is odd, we have a — (—b) ~ a + b' Now by the theorem, a" — (— i)" is divisible by a — {—b); hence a" -|- 5" is divisible hy a -{- b when n is odd ; that is, T/ie sum of the like odd powers of any two quantities is divisible by the sum of the quantities. Performing the division indicated, we obtain a ■\- 462. SYNOPSIS FOR REVIEW. CHAPTER XVII. ( (§60. MATHEMATICAL INDUOTION. ( ^^^^^^^^- \ 461. Cor. 1, 2, 3, 4, 5, 6 CHAPTEE XVIII. PERMDTATmS-COMBIXATIOXS-BINOMIALFOMUU-EXTRACTIO.N' OF HIGHER ROOTS. PERMUTATIONS. 463. Tlie Per^nutations of n things, taken r at a time, ai-e the results obtained by arranging the things in every possible order in groups of r each. Thus, the permutations of the letters a, b, Cf taken two at a time, are ab, ba, aCf ca, bCy cb. The permutations of the same letters taken all at a time, are abCf acb, bac, bca, cab, cba. It is evident that r cannot be greater than n. 464. To find the number of permutations of n things, taken r at a time. Suppose the ti things to be n letters, a,b, c, d . .' , , The number of permutations of n letters, taken one at a time, is n. In order to form all the permutations of n letters, taken two at a time, we must annex to each letter each of the n — 1 other letters. We thus obtain n {n — 1) permutations. In order to form all the permutations of n letters, taken three at a time, we must annex to each of the permutations, taken two at a time, each of the n — ^ other letters. We thus obtain n(n — l) {}i — 2) permutations. PERMUTATIOIS-S. 803 In the same manner it may be shown that the number of per- mutations of n letters, taken 4 at a time, is 7i (^^— 1) (/^— 2) (?^— 3). From these cases it might be conjectured that the number of permutations of n letters, taken r at a time, is n{n — l){n — ^){n — d) . . . . {n — r + 1). To show that this is true, we employ the method of mathemat- ical induction. Denote the number of permutations of n letters, taken r at a time, by P^, and suppose for a certain value of r that P, = ^ (7^ - 1) (?i - 2) (;i - 3) . . . . (?^ - r + 1) . . . (A). Now, in order to form all the permutations of n letters, taken r -f 1 at a time, we must annex to each of the P^ permutations each of the n — r other letters. We thus form nin — l){n — 2)(n — Z).,.. {n — r+l){n — r) permutations. Hence, denoting the number of permutations of n letters, taken r + 1 at a time, by P^+i, we have Fr+i = n{n — l){n — 2)(n — d).... {n — r + 1) (n — r), which may be written Fr^, = n{n-^l){n-2)(n-.3) {n-r + l)[n-{r^l)-\-l]. This equation is of the same form as (A) ; that is, it may be derived from (A) by simply substituting r + 1 for r. If then (A) is true when r is a certain number, it is true when we increase that number by one. But (A) has been shown to be true when r = 3 ; it is therefore true when r == 4 ; it is there- fore true when r = 5 ; and so on. Hence the formula must be universally true. Cor. — K r = n, (A) becomes 'P^ = n{n---l){n-2){n-^d) . . , ^1 . . . (B). - That is, T^ie nnmher of permutations of n things, taken n at a time, is equal to the product of the consecutive numbers from 1 to n in- clusive. 304 PERMUTATIONS. For brevity, n(n — 1) {n — 2) {7i — 3) . . . . 1 is often de- noted by the symbol \n, which is resid, factorial n. ^ 465. To find the number of permutations of n things, taken n at a time, when some of the things are identical. Suppose the n things to be w letters ; and suppose p of them to be a, q of them to be J, r of them to be c, and the others to be uuhkc. Denote the required number of permutations by N. If in any one of these N permutations thej3 letters a were changed into p new letters different from each other, and also different from all the other letters contained in the permutation, then, without changing the situation of the other letters, we could from the single permutation form [^ different permutations; therefore the whole number of permutations would be N x [^ In like manner, if the q letters b were changed into q new lettei-s differ- ent from each other, and also different from all the other letter's contained in the N x \p permutations, we could form N x |j'' X 1^ permutations; and if the r letters c were also changed \\\ the same way, the number of permutations would be N x |/ X [^ X [r. But this number must be ecjual to the number d permutations of n dissimilar things, taken w at a time; hence. Nx[£x[£x[r = l£; \n whence, ^ = 1 1= — r • • • (^« [£ X [£ x|r ^ ' 466. PB0BZEM8. 1. How many different permutations may be formed of 8 let- ters, taken 5 at a time ? A7is. 6720. 2. How many different permutations may be formed of all the letters of the alphabet, taken 4 at a time ? Ans. 358800. 3. How many different permutations may be made of 6 things, taken 6 at a time ? Ans. 720. COMBIXATION-S. 305 4. now many different numbers may be formed with the five figures, 5, 4, 3, 2, 1, each figure occurring once, and only once, in each number? Ans. 120. 5. How many different permutations may be made of the letters in the word Longitude, taken all together ? Ans. 362880. 6. How many different permutations may be made of the let- ters in the word Caraccas, taken all together? A^is. 1120. 7. How many different permutations may be made of the let- ters in the word HeliopGlis, taken all together ? Ans. 453G0O. 8. How many different permutations may be made of the let- ters in the word Ecclesiastical, taken all together ? ^r^5. .45-1053600. 9. What value must 7i have in order that the number of per- mutations of n things, taken 4 at a time, may be equal to 12 times the number of permutations of n things, taken 2 at a time ? Ans. n = 0. COMBINATIONS. 467. T7ie Combinations of n things, taken r at a time, are the results obtained by arranging the things in as many differ- ent groups of r each as possible, without regarding the order in which the things are placed. Thus, the combinations of the let- ters a, h, c, taken two at a time, are ah, ac, he. It will be observed that if the letters be regarded as factors, the combinations which may be formed by taking r at a time will constitute all the different products of the r^^ degree, of which the letters are capable. 468. To find the number of combinations of n things, taken r at a time. Denote the number of combinations of n things, taken r at a time, by C^, and the number of permutations of n things, taken r at a time, by P,- 20 306 COMBINATIONS. It is evident that all of the P^. permutations can be fonned by gubjectiug the r letters of each of the C^ combinations to all the per uuitat 10)18 of which these letters are susceptible, when taken r at a time. Now, the number of permutations of r letters, taken r at a time, is |r (464, Cob.) ; therefore the number of permuta- tions of n letters, taken r at a time, is 0^ [r ; hence, C, xlr = P,; p whence, C,. = -p. But P^=n(»-l)(»-2)(«-3) .... (w-r+1) (464); _ n(n-l)(;i-2)(;?-3) . . .'. (n-r +l) . . w- ^ . . . (1^}. 469. The number of combinations of n things, taken r at a time, is equal to the number of combinations of n things, taken n — r at a time. Denote the number of combinations of n things, taken n — r at a time, by C„_r ; then by (D), \, _ 7i{n-l){n-2){n-3) [n-{n-r) + l] _ n(n-l)(n-2)(7i-3) (r+l) ,,, - \n-r ' ' ' ^^^' Multiplying both terms of the second member of (1) by |r, c^=tir|E:r • • • ('>• Multiplying both terms of the second member of (D) by \n Gr = C^ . . . (4). THE BIXOMIAL FORMULA. 307 470. PnOBLEMS. 1. Find the number of different products that can be formed with the numbers 1, 2, 3, 4, 5, taken 2 at a time. Ans. 10. 2. What vahie must n have in order that the number of per- mutations of n things, taken 5 at a time, may be equal to 120 times the number of combinations of n things, taken 3 at a time ? Am. 8. ^ 3. When n is eren, what value must r have, in order that Gr may be the greatest possible ? j _ ^ Z 4. When n is odd, what value must r have, in order that C, may be the greatest possible ? Ans. r z=z 5. From a company of soldiers numbering 96 men a picket of 10 is to be selected ; in how many ways can it be done so as always to include a particulai* man ? |95 6. From a company of soldiers numbering 96 men, a picket of 10 is to be selected ; in how many ways can it be done so as always to exclude a particular man ? |95 ^"^^ [10xi85 THE BINOMIAL FORMULA. x-\- ah. 471. (x + a)(a; + ^=^ + « {x -\- a) {x -\- h) {x -\- c) =0? -{- a {x-\-a){X'\-b){x-\-c){X'\-d)=7^-\-a 7? ■\- dbx-^- abc. + ac + be a^-\- ah a^-{- abc + ac ■\-abd -\-ad + acd 4- be -{-bed -\-bd -\-cd x + abcd. 308 THE BINOMIAL FORMULA. In each of these identities we observe the following laws : 1. The number of terms in the second memher is one greater than the number of binomial factors in the first memher. 2. The exponent of x in the first term of the seco7id member is equal to the number of binomial factors, and in each of the suc- ceeding terms the exponent of x is one less than in the preceding term. 3. Tlie coefficient of the first term of the second member is unity ^ the coefficient of the second term is the sum of the second terms of the binomial factors j the coefficient of the third term is the sum of all the products of the second terms of the binomial factor s, taken two at a time ; the coefficient of the fourth term, is the sum of all the products of the second terms of the binomial factors, taken three at a time ; and so on : the last term is the product of all the second terms of the binomial factors. 4!li2i» That the laws stated in the preceding Article are gen- eral may be shown as follows : Suppose the laws to be true in the case of n binomials, x -f a, X -\- by X -\- c . . . . X -{- k'y that is, suppose {x + a)(x-{-b)(x-\-c) .... (a;-f /^•)=^" + Pi^""^ + P2?^""HP3a^""^ + Pn . . . (1), in which Pi = the sum of the terms a, b, c . . . . 7c, Pg = the sum of the products of these tei-ms, taken two at a time, Pg = the sum of the products of these terms, taken three at a time, P„ = the product of all these terms. Multiplying both members of (1) by x + I, (x-^a){x + b){x+ c) . . . . {x + Jc) {x + l) = a;«+i -}- Pj a;" + Pg 3;^-^ + P3 ^'"-^ .... +PJ . . • (2). + 1 + Pi^ + P2^ THE BINOMIAL FORMULA. 309 Now Pi+Z = a + Z> + c....+^' + ? = the sum of all the terms a, 5, c, . . . . Ic, I; P2 + Pi^ = P3 + (a + Z^ + c -^k)! = the sum of the products of all the terms a,b,c,....k, I, taken two at a time ; P3 + Pg? = P3 + {ab + ad -\- ac + he ■\- hd + cd . . . .)l = the sum of the products of all the terms a, bf c, . . . . h If taken three at a time . PJ = the product of all the terms a^b, c, . . . . k, I The law of the exponents in (2) is also the same as in (1). Hence, if the laws are true when n factors are used, they will be true when 7^ -f 1 factors are used. But they have been shown to be true when ?j = 4; therefore they are true when oi=:b; and so on. Hence the laws must be true universally. 473. The number of terms in P^ is obviously n; the number of terms in P2 is equal to the number of combinations of n thiugs, 41 ( ji -_ 1 ^ taken two at a time, that is, — ^j^ — - (468) ; the number of terms in P3 is equal to the number of combinations of n things, taken three at a time, that is, — ^^ — ; and so on. Now suppose a=:b = c=.,..k; then Pj = na, _n{n — l)^ p _ n {n — l){n— 2 ) Under this hypothesis, (1) of Art. 473 becomes (X + a)-=x- + 7iaa^-' + n{n-l) ^^^^_^ _^ n{n-l){n-2) ^^^^_^ ^ If This is the Binomial Formula, The second member of this formula is called the Expansion or Developtnent of {x + a)^, and when we put this expansion or development in the place of {x + a)^ we are said to expand or develop (x -f- a)". P^ = ^^Z^ a2, P3 ^ 'lK!i_tJS-^±l „s, and so on. 310 THE BINOMIAL FORMULA. 474. The coefficient of the product of the powers of a and x in any term of the expansion of {x + a)« is called the coefficient of that term. Thus, the coefficient of the thu'd term of the « / N • ^ (^ — 1) expansion of (x + aY is tt . 475. The first letter in an expression of the form of {x + «)** is called the leading letter, 476. Another Proof of the Binomial Formula,— We can verify the Binomial Formula by trial for small values of ^ as 2, 3, 4. Assume If l£ + +«" . . . (1). Multiplying both members of (1) hj x ■\- a and reducing, +^fc^a»."-H . . . . +«»- . . . (^); that is, the expansion of {x + fl)"+* is of the same form as that of {x H- a)". Hence, if the Binomial Formula is true for any ex- ponent, it is true when the exponent is increased by unity. But the formula is true when w = 4; it is therefore true when w=5; it is therefore true when w = 6 ; and so on. Hence the Bino- mial Fonnula is true for any positive integral exponent. Cor. 1. — If we multiply the coefficient of any term in the ex- pansion of {x + aY hy the exponent of x in that term, and di- vide the 'product hy the number obtained by adding 1 to the expo- nent of a in the same term, the quotient will be the coefficient of the succeeding term. Cor. 2. — Tlie sum of the exponents of a and x in any term of the expansion of {x + aY is equal to n. THE BINOMIAL FORMULA. 311 477. To find the stun of the coefficients in the ex- pansion of {x 4- a)\ The formula (J^l .2.n-2 j_ nin-l){n-2) . _ {x 4- ay = a;" + naaf"-^ -\ — ^-uT-- «^^"~^ + -^ ^ d^^ + . . . . H-a» is true for all values of x and «, and the coefficients are indepen- dent of x and a. Suppose a; = c = 1 ; we then have n{n — V) n(n — 1) (7i — 2) . , ^ 2»= 1 + w + -^^ — ^ + -^ ^ + +1. That is, the sum of the coeflScients in the expansion of {x-\-aY is2». 478. To find the r^^ term of the expansion of (x 4- ay. The exponent of a in any term of the expansion of {x + aY is one less than the number of that term ; hence the exponent of a in the r^ tenn is r — 1. The sum of the exponents of a and x in any term is n ; hence the exponent of x in the r^^ term is n — r-\-l. The coefficient of the second term is equal to the number of combinations of n things, taken one at a time ; the co- efficient of the third tenn is equal to the number of combinations of n things, taken Uvo at a time ; and so on ; hence the coefficient of the r'* term is equal to the number of combinations of n things taken r — 1 at a time ; that is, n(n-l){n-%){n-^). . , , (^ - r + 2) ^^^^^^ |r — 1 Therefore, denoting the r^ term by T„ we have *■ |r — 1 It should be observed that r cannot be greater than w -|- 1. 312 THE BIN-QMIAL FORMULA. 479. In ilic expansion of (x + «)'» Me coeficicnf of the r^^ term from the heyinning is equal to the coefficient of the r^ term from the end. The coefficient of the r^ term from the beginning is n(n-l){n^2){n-^3) , , , . (n - r -i- 2) ^^^^^^ Multiplying both numerator and denominator by \n — r 4 1, this becomes |r — 1 X \n — r-i- 1' The r^ tenn from the end is the {71 — r -{- 2y^ term from the beginning, and its coefficient is n{n-'i){n-2)(n-3) [n-(n-r + 2)-{-2] \n — r -\- 1 ^ '* which becomes by reduction n{n^l){n—2){n—Z) . . . . r N-r + l Multiplying both numerator and denominator by \r — 1 , this becomes \r — 1 X \n — r-\- 1' 480. To obtain the Expansion of (x — «)», it is suffi- cient to put — a in the place of -\- a in the expansion of (x + a)« The tenns which contain the odd powers of — a will be negative, and the terms which contain the even powers of — a will be positive. Hence, If E , "(«-i)(«-2)f'i:z^^4^„-4 . _ . ^ a\ a^, a\ a\ a\ cfi; V, b\ V, b% ¥, ¥; 1, 5, 10, 10, 5, 1 (4'76,CoR.l), THE BINOMIAL FORMULA, 313 481. EXAMPZES. 1. Expand (a 4- ^)^- First Solution. — In the expansion of {a+by the powers of a are the powers of b are and the coeflBcients are (a + Z>)5 = flS ^_ 5^4^ ^ i0a3^2 _f_ 10^2^3 + 5^^ _|_ js. Second Solution. — The Hteral parts of the terms of the expan- sion of (a + by are a^ a% aW, aW, a¥, b\ and the coefficients are 1, 5, 10, 10, 5, 1 ; (a H- 5)5 = ^5 + ba'^b + 10^3^^ + 10^2^,3 4. ^a¥ + 5^ Third Solution. — Substituting a for x, b for a, and 5 for n, in the Binomial Formula, we have {a-\-by = a^ + 5a*b + lOaW + lOa^h^ + 6aM + b^. 2. Expand {2x — Say. Powers of 22-, 16a:*, Sa^, Aa^, 2x, (2a:)0; Powers of —da, (— Say, — Sa, 9a\ — 27a^, 81a*; Coefficients, 1, 4, 6, 4, 1; .-. (2x — Say = Ux^ — ^^ax? + 216«2|;2 _ 2l6aSa; + 81a*. 3. Expand (a -\- b -^ c -\- dy. Put ic = a + i and y=zc-^d\ then (fl + 5 + c + t?)3 = (a; + «/)3 = a;3 ^ 3^4}^ _f_ 33.^2 ^ ^^s Substituting for x and y their vakies, (a + & + c+(^)3=(a4-5)3 + 3(fl + ^')2(c4-6?)+3(a4-5)(c + 6?)2-f(c + c?)3 =a3 ^ 30^2^ ^ 3^j2 + 53 ^ 3^2^ _}. 6ar^,c + 3^2^ + 3^2^ + ^abd + SbH + 3ac2 + Gac^+3a6?2 4-35c2+65crf + 3i(^ + c3 + 3c2(/ + Scd'^ + <^. 814 THE BINOMIAL FORMULA. 4. Find the 5th term of the expansion of {a + ly^. Substituting a for x, b for a, 15 for ??, and 5 for r in the formula of Art. 4T8, we have ^ 15 X 14 X 13 X 12 y. .. -ion-ii. 11 ^ 1x2x3x4 5. Expand {a — hf. Ans. flS — 5^45 ^ X0a3i2 _ i^^aW ^ 5^j4 _ js. 6. Expand (1 + cf. Ans, 1 + 6c + 15c2 + 20^8 + 15c* + Sc^ + c«. 7. Expand {x + y)". A 718. x' + 7a:«?/ + 21x^if + 35ar*j^3 ^ 35^:8^ + 212^2^8 + 73:^6 ^ ^7, 8. Expand (a^ _ 1)8. Ans. ai«-8fliH28ai2_56aio+70a8-56a6 + 28fl4-8a2 + l. 9. Expand (a — c)^. — 36flV + 9ac8 — c*. 10. Expand (1 -\-axy. Ans. 1 + 5fl«; + lOa^x^ + lOa^s^ 4- 50*^ + a^^^ 11. Expand (3«-f-2c)^ A ns. 243^5 + SlOa^c + 1080^3^2 + 720^^ + 240ac* + 32^5. 12. Expand (o-l)^. >! -.rc^o^ 010- .3125„ 625-125, 5^', x^ Ans. 15620-3120.^ + -^:^-— ^+^3^^-*-^ + -4^. 13. Expand (^2 — ah -\- lf)K Ans. «» — ^a?h + lOa^Z^ — Ua^h^ + 19a*Z»* — IGa^Z^s 4. i^a^^ — 4m1P + ¥. 14. Find the middle term of the expansion of {a + xy^. Ans. '2l2a^x\ (3a 4r\* 16. Find the 2001»' term of the expansion of («^ + x'^J* Ans. 2003001a^^a;«». THE 'nP^ KOOT OF QUANTITIES. 315 THE n^^ ROOT OF QUANTITIES. 433. To find the n^^ root of a polynomial. Find the vJ'^ root ofa;" + y^Q^"~^ + ^^ ^^,1" aH''-'^ .... + a". if x^ + waa^-i 4- ^1.7" Q^^^""^ . . . . + a'^Iic + a if Arrange the terms according to the descending powers of x. The n^^^ root of the first term, a;", is x, which is the first term of the required root. The second term of the root may be found by dividing the second term of the given polynomial by wa;«~^ If there were more terms in the root, we should proceed with X -\- a as we did with x, RULE. I. Arrajige the given polynomial according to the powers of one of its letters. II. Extract the n^^ root of the first term; the result will be the first term of the required root. Subtract the n^ power of this term from the given polynomial. III. Divide the first term of the remainder by n times the (n — 1)^ poiver of the first term of the root ; the quotient will be the second term of the root. Subtract the n^^ power of the sum of the first and second terms of the root from the given polynomial. IV. Take n times the {n — 1)<* potver of the sum of the first and second terms of the root for a second divisor. Divide the first term of the second remainder by the first term of the second divisor ; the quotient will be the third term of the root. Subtract the n^^ power of the sum of the first, second, and third terms of the root from the given polynomial. V. Proceed in this manner until all the terms of the root have been found. 316 THE W'^ ROOT OF QUANTITIES. Con. — If the root contains only two terms, it may be obtained by extracting the n^ root of the extreme terms of the arranged polynomial, and placing the proper sign between the results. Thus, the cube root of a^ ^- Za^h + oaW' ■\-'b^ is a-^l, and the cube root of a^ — Zd^h + ZaU^ — W is a — h. EXAMPLES. 1. Find the fourth root of 16a* — 96a^ H- 216fl2ic3 + Sla;^ — 216aa:8, j^^g^ 2a — Zx. 2. Find the fifth root of SOa^ + 32^5 _ SOa* — 40^2 + 10a— 1. Ans. 2a — 1. 3. Find the fourth root of 336^5 + Sia^ — 21Ga^ — 56a* + 16 — 224a3 + 64a. A ns, 3a^ — 2a — 2. 4. Find the fourth root of a* — ^^b + Qa?b^ — 4aJ3 -|- l\ Ans. a — h, 5. Find the fifth root of cfi+ba^h + lOaW+10aW-\-oa¥-\-l)\ Ans. a + h. 6. Find the sixth root of a^— 6a5 4-15a*— ^Oa^ + lSa^— 6a + l. Ans. a — 1. 7. Find the seventh root of a7+7a«+21a«+35a*+35a3+21a2 + 7a + l. Ans. a -\-\. 8. Find the eighth root of 3?—%x'^-\-2^ofi—hQj^+l^x^-bQ7? + 28arJ— 8a; + l. Ans. x — 1. 483. To find the 71^^ root of a number. For a reason similar to that given in Art. 271, we separate the given number into periods of 71 figures each, beginning with units. The 71^ root of the greatest n^^ power contained in the period on the left will be the first figure of the root. If we sub- tract the 71^^ power of the first figure from the given number, and divide the remainder by n times the (71 — 1)^^ power of that figure, regarding its local value, the quotient will be the second figure of the root, or a figure too large. The result may be tested by subtracting the 71^^ power of the number represented by the SYNOPSIS FOR EEVIEW. 317 first and second figures of the root from the given number. If there are additional figures in the root, they may be found in the same manner. :exampIjJes. 1. Find the fifth root of 33554432. 335,54432(30 + 2 = 32 305 _ 243 00000 5 X 304 = 4050000) 92 54432 325 ^ 33554432 2. Find the fourth root of 79502005521. A7is. 531. 3. Find the fourth root of 75450765.3376. Ans. 93.2. 4. Find the fourth root of 2526.88187761. Ans. 7.09. 5. Find the fifth root of 418227202051. Ans. 211. 6. Find the seventh root of 34359738368. Ans. 32. 484. SYNOPSIS FOR REVIEW. PEEMUTATIONS OOMBINATIONS, BINOMIAL FOKMULA. EXTRACTION OP HIGHEE. KOOTS. n THINGS TAKEN r AT A TIME. n THINGS TAKEN 71 AT A TIME. n THINGS TAKEN 71 AT A TIME, WHEN SOME ARE IDENTICAL. 7? THINGS TAKEN T AT A TIME. Interpret the equation C^ = C„--r. Product of n binomials whose first terms are identical and whose second terms are different. General laws. 1, 2, 3. Binomial Formula. Coefficient of a term of the ex- pansion OP (a: + a)". Leading letter. Another proof op Binomial For- mula. Cor. 1, 2. To find the sum op coefficients in THE expansion OF {x + O,)^. To FIND THE r'* TERM OF THE EXPAN- SION OF {x -\- ay. Expansion of {x — «)". To FIND THE n'^ root OF A POLYNO- MIAL. Rule. Cor. To FIND THE 7l'* ROOT OF A NUMBER. CHAPTEE XIX. IDEIfTIOAL EQUATIOI^fS, PROPERTIES OF IDENTICAL EQUATIONS. 485. Jf the equation A + B.r + Cu^5 + D^+ etc. = A' + B'a; + C'a;2+D'^H etc., in tuliich A, B, C, D, etc., A', B', C, D', etc., are finite quantities independent of x, is an identity y the coefficients of the like powers of X are equal to each other. Since this equation ig tnie for every value that may be assigned to X (178), it must be true when a; =: 0. But when x = 0, all the terms disappear except A and A', and the equation becomes A = A'. Dropping A from one member of the given equation, and A' from the other, B^; + Ca;'2 + Bx^ + etc. = B'x -f- C'x^ + B'x^ + etc. Dividing both members of this equation by a:, B -{• Cx -{- Da;2 -\- etc. = B' + C'x + D'x^ + etc Making a; = 0, as before, this equation becomes B = B'. In like manner it may be shown that C = C', D = D', etc. DECOMPOSITION OF RATIONAL FEACTIONS. 319 486. If the equation A + Bir + Ca;2 + Da:3 _^ etc. = 15 an identity^ each of the coefficients A, B, 0, D^ etc., is equal to zero. Since this equation is true for every value that may be assigned to X, it must be true when a; = 0. But when a; = 0, the given equation becomes A = 0. Dropping A from the given equation, and dividing the rqsult by a;, B + Ca; + D:c2 _|. etc. = 0. Making a; = 0, as before, this equation becomes B = 0. In like manner it may be shown that • C = 0, D = 0, etc. 487. Undetermined Coefficients are such as are un- known in an assumed identity. Thus, if we assume {x + aY = Ax'^ + Ba:2 j^ Qrf. _i_ J) to be identically true. A, B, C, and D are undetermined coefficients. DECOMPOSITION OF RATIONAL FRACTIONS. 488. To Decomimse a Mat ion al Fraction is to separate it into fractions whose sum is equal to the given fraction and the product of whose denominators is equal to the given de- nominator. The parts into which the given fraction is separated are called Partial Fractions. EXAMPZES. 1. Separate -^ — -r into partial fractions. X ~— iX ~p xo The factors of the denominator are x — b and rr — 2 : hence 320 IDENTICAL EQUATIONS. the denominator of one of the partial fractions is x — 6, and that of the other is a; — 2. . 8a: -31 A . B ... Assume -r— -, — —r?. = ^ H :t • • • (I)' 2^2 _ 7a: + 10 a; — 5 a: — 2 ^ ^ Since the first member is the sum of the two fractions in the second member, this equation is an identity. Clearing (1) of fractions and uniting similar terms, 8a:-31^(A + B)a;-(2A + 5B) . . . (2). (2A4-5B=:31) ^ ^' whence, A = 3 and B = 5. Substituting 3 for A and 5 for B in (1), 8a: — 31 3 5 a:* — 7a: + 10 x — 5 a: — 2 72^ 4- X 2. {Separate -. -r-rr -^ into partial fractions. ^ (ar-|.l)(2ar — 1) . W + x A , B .^. Assume ^^ ^ ^^ ^^^ _ ^^ ^ —^ + ^— -^ ... . (1). Clearing of fractions and uniting similar terms, 7ar» + a: = (2A + B) a: + B — A . . . (2). The coefficient of a^ in the second member of (2) is ; 7 = (485), which is absurd. Hence the given fraction cannot be separatee! into partial fractions having numerators independent of x. . 7a;2-|-a: Aa: , Ba: ,. Assume -. —-rz -r = + 7i t • • • {pJ- (a; + 1) (2a: — 1) a: + 1 2a: — 1 ^ ^ Clearing of fractions and uniting similar terms, 7a?J-ha:=(2A + B)a:2+(B-A)a: . . . (4). SYNOPSIS FOR REVIEW. 321 Equating the coefficients of like powers of x in (4), ( 2A + B = 7 ) ( B-A=:lP whence, A = 2 and B = 3. Substituting 2 for A and 3 for B in (3), 72;2 + a; 2^- _ Zx {x 4- 1) (2a; — 1) x^\ ' 'Zx—\ From this example we learn that if we assume an impossible form for the partial fractions, the fact will be made apparent by some absurdity in the equations of condition (179). Separate each of the following fractions into its partial fractions : 7a; -24 5,2 20^:4-2 . 8 6 2.6-2-1- 3a; — 20* ^^* 2a; — 5 "^ a; + 4* 6arJ-22a; + 18 . 1 , 2 , 3 (a;— l)(a;2 — 5a; + 6)* 'a;- l^a; — 2 ' .^• — 3' 1 3 ^ a; +2 .2,22 6. , . Ans, — — r H r . a; -[- 1 a; — la; 1111 .2 2 3^3 7. —, T^-^ rrrr. Arts. — -— — - — — r + y?- -x' 10 a4 _ xZx^ ^ 36 3:^3 -}- 5a;2 — 2a; a;-}- 2 X — 2 a;-|-3 a; — 3 8. ^ -^ . An8. + a;2 — 1 a; — l'a; + l 489. SYNOPSIS FOR REVIEW. t— ( n H EH w p fi ex Properties of Identical Equations. \ * Undetermined Coefficients. Decomposition of Rational Fractions. Partial Fractions. 21 CHAPTER XX. SERIES. GENERAL DEFINITIONS. 490, A Series is a succession of quantities, each of which, except the first, or a certain number of the first, may be derived from the preceding one, or a certain number of the preceding ones, by a fixed law called the Law of the Series, Thus, 1, 2, 3, 4, 6, 6, 7, 8, 9, is a series, the law of which is that each quantity, except the first, is derived from the preceding one by adding unity to it. 491, Tlie Terms of a series are the quantities of which the series consists. 492. A Finite Series is one which, by its law of forma- tion, can have only Q.Ji7iUe number of terms. Such a series is said to terminate. Thus, the expansion of {x -f a)», when n is a pos- itive integer, is a finite scries. 493. An Infinite Series is one which, by its law of formation, may have an infinite number of tenns. Such a series is said not to terminate. Thus, 1 1 1 -L _L JL * 2' 4' 8' 16' 32' . 64' is an infinite series. 494. A Converging Series is an infinite series, the sum of the first n terms of which cannot numerically exceed some finite quantity, however great n may be. Thus, ARITHMETICAL PROGRESSION. 32S 1 1 1 J_ ^' 2' 4' 8' !♦>' is a converging series, for the sum of its first n terms cannot exceed 2, however great n may be. 495. A Divevffing Series is an infinite series, the sum of the first n terms of which can be made numerically greater tlian any finite quantity by taking n sufiiciently great. Thus, 1, 2, 3, 4, 5, 6, 7, 8, is a diverging series. ARITHMETICAL PROGRESSION. 496. An Arithinetical JProgression , or a PrO" ffvession by Ulfference, is a series in which the difference between the first and second terms is equal to the difference between any other two consecutive terms. Thus, 1, 3, 5, 7, 9 is an arithmetical progression. An arithmetical progression is sometimes called an Avith- metical Series. For brevity we shall sometimes use A. P. for the phrase arithmetical progressio7i. 497. The Ertrenies of an A. P. are the first term and the last term ; the other terms are the Means, 498. The Common Difference of an A. P. is the remainder obtained by subtracting any term from the one which follows it. Thus, in the progression 1, 3, 5, 7, 9, the common difference is 2. 499. An Increasinf/ A, JP. is one in which the com- mon difference is positive. Thus, 1, 2, 3, 4^ 5 is an increasing A. P. 500. A Decreasing A. P.'i^ one in which the common difference is negative. Thus, 9, 7, 5, 3, 1 is a decreasing A. P. 324 SERIES. , 501. Notation. — In treating arithmetical progressions we shall use the following notation : a = the first term of the progression, I = the last or n^^ term, d = the common difierence, w ^ the number of terms, s = the sum of all the terms. Thus, in the A. P. 1, 3, 5, 7, . 9, a — I, 1 = 9, d = 2, w = 5, s = 25. 502. To find I when a, d, and n are given. The first term is «, the second term is a + d, the third term is a + 2eZ, the fourth term is a + M, and so on; hence the n^ term is a ■}- {n — 1) d; that is, l = a -\- (n — 1)^. 503. To find s when a, I, and n are given. sz=za-\-{a-^d)-\-{a + 2d)+{a-\-M)+ . . . . +1 . . . (1). Inverting the order of terms in the second member of (1), s = l^{l-d) + {l—2d) + {l^Zd)-\-,..,-\-a . . . (2). Adding (1) and (2), 25=(rt + + (« + + (« + 0+---- +(« + 0=^(«^ + . . . (3); whence, « = o (^ + • • • (^)- 504. In an A. P. the sum of any ttvo terms equidistant from the extremes is equal to the sum of the extremes. Let X denote a term which has m terms before it, and y a term which has m terms after it ; then -whence, x -\- y = a -\- 1 AKITHMETICAL PROGRESSIOIT. 325 505. To insert any number of arithmetical means between two given quantities. Let a and b be the given quantities, and let it be required to insert m arithmetical means between them; that is, let it be required to form an A. P. whose extremes are a and b and the number of whose terms is w -{- 2. Substituting b for I and m + 2 for n in the formula of Art. 503, we have b = a + {m -{-l)d; b — a whence, d = m+ 1 By adding the common difference to a we obtain the second term ; by adding it to the second term we obtain the third ; and so on. Example. — ^Insert 10 arithmetical means between 5 and 38. 38-5_ hence the required progression is 6, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38. 506. To find any two of the quantities a, Z, J, n, and 5, when the three others are given. ilz=a-\-{n-l)d\ The group j,= |(« + o [ contains the five quantities a, I, d, n, and s ; hence any two of them may be found when the three others are given. The ten cases are given in the following table as an exercise for the student. Each case is an example of two simultaneous equations with two unknown quantities. 326 SERIES. OITEN. TO FIND. a, d, n ly S a, dy I W, 5 a, d, s Hy I rt, w, I d, S a, 71, s dy I a, I, s dy n d, n, I a, 8 dy fly 8 ay I d, h s a, n I, fly S ayd BESULTING FOKMDUB. d—%a± V(2a — d)'^ + 8^6- 2d 10. J I — a W/ . n d = ^--^, l^^-^-a. n (n — 1) n . ^ — a^ 28 d = -; :, 71 = 1. 2s — a — l a -i- I a=l-{7i^l)dy s=^[2I-(7i-l)d], _ 2s—7i{n—\)d j_ 2s-\-n{7i—l)d ""- 271 ' ^- 2n • ^-i{d± V(z l + df - iidsjy _2l-\-d± V(2l + d)^- Sds 2d 2s 2 (In — s) n 71 {n — 1) n = 507. PJtOBLEMS, 1. The first term of an A. P. is 5, the common difference is 3, and the number of terms is 24. Find the last term and the sum of all the terms. AVe have given a = 5, c? = 3, n = 24 ; ? = 5 + (24-1)3 = 74, and 24 s = y [10 + (24 -1)3] = 948 (506, 1). ARITHMETICAL PROGRESSION. 32? After finding the value of Z, we might have found the value of s from the formula s = - (a + /). Thus, 04. s = ^ (5 + 74) = 948. 2. Given « = 15, tZ = — 2, and 5 = 60, to find I and n, ? = - ^- ± 1/2 (- 2) 60 + (15 - ^-/ = 5 or -3, and Both values of n are possible ; for there are two progressions which satisfy the given conditions, one having 6 terms, the other 10 ; these progressions are 15, 13, 11, 9, 7, 5, and 15, 13, 11, 9, 7, 5, 3, 1, —1, —3. Another Solution, — Substituting the given values in the group ilz=za + {n — l)cl\ we obtain the group i Z=15-2(7^-l) ^ J60 = |(15 + )' whence, Z = 5 or — 3, and n=zQ oy 10. 3. Given a = 275, I = 5, and n = 46, to find d and s. j^=-6, ^'''' \s= 6440. 4 Given d=z6, n = S, and s = 156, to find a and I. ^"^^ I "=3?: 5. Form an A. P. of 6 terms whose extremes shall be 7 and 37. Ans. 7, 13, 19, 25, 31, 37. Ans. {"J 328 SERIES. 6. Given a = 3, n = CO, and s = 3720, to jQnd d and I = 2, 121. 7. "What is the sum of the terms of an A. P. formed by insert- ing 9 arithmetical means between 9 and 109 ? Aiis, 649. 8. Find the sum of the first n terms of the progression 1, 2, 3, 4, 5, 6, ... . . w /I , \ ' ' A71S. S = -(l 4- 71). 9. Find the sum of the first n terms of the progression 1, 3, 5, 7, 9, ... . Ans. s = w^. 10. Sum to 30 terms the progression 116, 108, 100, ... . Ans. 5 = 0. 11. Sum to n terms the progression 9, 11, 13, 15, . . . . Ans. s =n(8 -^ n). 12. Are the squares of a^ — 2x — 1, x^ + 1, and x^ -\- 2x — 1 in A. P. ? 13. A sets out from a place and travels 1 mile the first day, 2 the second, 3 the third, and so on. Five days later B sets out from the same place and travels 12 miles a day in the same direc- tion as A. How long will A travel before he is overtaken by B ? A US. 8 or 15 days. 14. A sets out from a place and travels 1 mile the first day, 2 the second, 3 the third, and so on. B sets out a days later from the same place and travels b miles a day in the same direction as A. How long will A travel before he is overtaken by B ? Show that B will never overtake A if a > ^^ — qT~^' 15. A sets out from a place and travels 1 mile the first day, 3 the second, 5 the third, and so on. B sets out three days later from the same place and in the same direction as A, and travels 12 miles the first day, 13 the second, 14 the third, and so on. How long will A travel before lie is overtaken by B ? Ans. 6 or 12 days. ARITHMETICAL PKOGRESSIOIs\ 329 16. The distance from P to Q is 165 miles. A sets out from P toward Q and travels 1 mile the first day, 2 the second, 3 the third, and so on. At the same time B sets out from Q toward P, and travels 20 miles the first day, 18 the second, 16 the third, and so on. When will they meet ^ A?is. At the end of 10 or 33 days. Do A and B meet twice ? ARITHMETICAL MEAK. 508. T7ie Arithmetical Mean of two or more quan- tities is the quotient obtained by dividing their sum by their number. Thus, the arithmetical mean of a and Z> is — - — , and 2 the arithmetical mean of 1, 7, 11, and 5 is 6. 509. To find the arithmetical mean of the terms of an A. P. Denoting the arithmetical mean by M, we have, by definition, M = i. n But 't / -TV M = a -f I 510. To find a and I when M, d, and n are given. s {n-l)d a = s (n^l)d ^-^+ 2 (506, 8). Substituting M for -, we have a = M- l = M + (n — l)d 2 {n — l)d 330 SERIES. 511. PBOBZEMS. 1. Find five numbers in A. P. whose sum is 25, and whose con- tinued product is 945. Denote the arithmetical mean by M and the common differ- ence by X ; then and M =-^ = 5 the first term = 5 — 2a; ^ the second term =z 5 — x the third term = 5 the fourth term = 5 + x the fifth term = 5 + 2a; (510); .-. (5 - 2a;) (5 - a;) 5 (5 + x) (5 + 2a;) = 3125 - 625a;3 + 20a;* = 945; 1 whence, The required numbers are therefore 1, 3, 5, 7, 9, or 5 — a/109, 5 — I VT09, 5, 5 + | vT09, 5 + vT09. 2. Find four numbers in A. P. whose sum is 32, and the sum of whose squares is 276. Ans. 5, 7, 9, 11. 3. Find three numbers in A. P., the sum of whose squares is 1232, and the square of whose arithmetical mean exceeds the product of the extremes by 16. Ans. 16, 20, 24. 4. Find four numbers in A. P. whose sum is 28, and whose continued product is 585. Ans. 1, 5, 9, 13. 5. The sum of the squares of the first and last of four numbers in A. P. is 200, and the sum of the squares of the second and third is 136: find the numbers. Ans. 2, 6, 10, 14. 6. Find the first term and the common difierence in an A. P. of eighteen terms, in which the sum of any two terms equidistant from the extremes is 31J, and the product of the extremes is 85 J-. A J « =3, ^^^- \d=li. GEOMETRICAL PEOGRESSION. 331 GEOMETRICAL PROGRESSION. 512. A Geotiietrical Progression, or a JPrO" gression by Quotient^ is a series in which the quotient obtained by dividing the second term by the first is equal to the quotient obtained by dividing any other term by the preceding one. Thus, 1, 3, 9, 27, 81 is a geometrical progression. A geometrical progression is sometimes called a Geometric cal Series. For brevity we shall sometimes use G. P. for the phrase geometrical progression. 513. TJie Extremes of a G. P. are the first term and the last term ; the other terms are the Means, 514. The Hatio of a G. P. is the quotient obtained by dividing any term by the one which precedes it. Thus, in the progression 1, 3, 9, 27, 81 the ratio is 3. 515. An Increasing G, JP» is one in which the ratio is gi-eater than 1. Thus, 1, 2, 4, 8, 16 is an increasing G. P. 516. A Decreasing G, JP, is one in which the ratio is less than 1. Thus, 64, 16, 4, 1, ^ is a decreasing G. P. 51*7. An Infinite Decreasing G» JP. is one in which the ratio is less than 1, and the number of terms is infinite. 518. dotation. In treating geometrical progressions we shall use the following notation : a = the first term of the progression, I = the last or n^ term, r = the ratio, n = the number of terms, 8 = the sum of all the terms. Thus, in the G. P. 1, 3, 9, 27, 81, a = l, Z = 81, r = 3, w = 5, s = 121. 519. To find I when a, r, and n are given. The first term is «, the second term is ai^^ the third term is ar% 832 SEBIES. the fourth term is ar^, and so on; hence the n^ term is ar»-i; that is, I = ar""i = - . r*». 7' Cor. — If n = ao and r < 1 : then l = - x = 0; that is, r The last term of an infinite decreasing O. P. is 0. 530. To find s when a, I, and r are given. 5 = rt + ar + a?-2 + ar^ + . . . . + ar"~^ . . . (1). Multiplying (1) by r, r5 = ar + ar^ 4- ar^ + ffr* 4- . . . . -f ar^ . . . (2). Subtracting (1) from (2), rs — s = ar^ — a ... (3) ; whence, 5 = -r- . . . (4). r — 1 ^ Substituting /r for ar^ (519), (4) becomes Ir — a ... r — 1 ^ ' Cob. — K 7i = oo and r < 1 ; then Z = (519, Cor.), and (5) becomes s = ——— . . . (G) ; that is, TJie sum of the terms of an infinite decreasing O. P. is equal to the quotient obtained by dividing the first term by 1 minus the ratio. 531. In a G. P. the product of any two terms equidistant from the extremes is equal to the product of the extremes. Let X denote a term which has m terms before it, and y a term which has m terms after it ; then (519); whence, xy = al. rx = ar"^ "I GEOMETRICAL PROGRESSION. 333 522. To insert any number of geometrical means between two given quantities. Let a and h be the given quantities, and let it be required to insert m geometrical means between them; tbat is, let it be required to form a G. P. whose extremes are a and b and the num- ber of whose terms is m + 2. Substituting h for I and m -\- 2 for n in the formula of Art. 519, we liave whence, r = y -. By multiplying a by the ratio we obtain the second term ; by multiplying the second term by the ratio we obtain the thu'd term ; and so on. Example. — Insert three geometrical means between 7 and 112. hence the required progression is 7, 14, 28, 56, 112. 523. To find the continued product of the terms of a G-. P. Denoting the required product by P, we have P = « X ar X «r2 X ar^ X . . . .? . . . (1). Inverting the order of the factors in the second member of (1), we have P = ? X . . . . «r3 X ar^ X «r X a . . • (2). Multiplying (1) by (2), P2 = (al) (al) {al) . . . . (al) = {aiy .... (3) (521) ; whence, P = V(«0" • • • (^)- 524. To find any two of the quantities «, I, r, and s when the three others are given. i I — ar""-^ ) The group j _ Ir-^a > 334 SERIES. contains the five quantities a, I, r, n, and s ; hence any two of them may be found when the three others are given. Tlie ten cases are given in the following table as an exercise for the student. The value of n in the last four cases cannot be found without a knowledge of the properties of logarithms. This part of the work must therefore be deferred until Chapter XXI shall have been read. NO. eiviui Tonin> BEStTLTING rORMUL^. 1 a,r,w l,s l-ar-^ «_«^'"-« i-ctr ,o_ ^._^ . 2 I, r,n a,s I I (r- - 1) 3 n,r, s a, I s(r-l) , (r_l).9r»»-i ^ - >« _ 1 ' ^ - r" - 1 • 4 a, I, n r, s 1 " *» 5 a, w, s r,l ar» — rs = a — s, I (.9— Z)"-i= n{s—a)''-K I, n, s a,r a(s- «)»-! = l(s — 0"-S (5-.Z)r»— 5r"-i 7 a, r, I s, n Zr — « log Z — log « , ^ r — 1 ' log r 8 a, Z, s r, n 5 — a log ? — log « ^^-5- r "-log(5-«)-log(s-Z) ' 9 a,r, s Z, n « + 5(r-l) log [a + 5(r-l)]-log«. log r 10 I, r, s a, n a = ?r — s (r — 1), logi-log[Zr-.(r-l)] ^"- logr ^^- GEOMETRICAL PKOGnESSIOlT. 335 525, pjtoiiLEJis. 1. The first term of a G. P. is 3, the ratio is 3, and the number of terms is 12. Find the last term and the sum of the terms. We have given a = 3, r = 2, n = 12; ? = 3 X 2^^ = 6144, and s = ^ ^^^~^ = 12285 (524, 1). 2. Given s = 1820, w = 6, and r = 3, to find a and I 1820(3-1) ^ ^ = - 36-1 =^> . 1820(3-^1)35 ^^_ __. ^. and I = 36iri~ — ^^^^ (^^*> ^)- After finding the value of «, we might have found the value of I from the formula I = ar""^. 3. Find the sum of an infinite decreasing G. P. of which the first term is 1, and the ratio -. s = ^-^ = 3 (520, COE.). 4. Given a = 1, / — 512, and s = 1023, to find r. Ans. r = 2. 5. Insert two geometrical means between 24 and 192. Ans. 24, 48, 96, 192. 3 9 27 1 6. Multiply l+^ + jg + ^+....to infinity by - — i + l55-6i+----*^^^^^^*y- ^^''''i 7. Find the value of x in the equation l-\-x-{'a^-\-0!^ + 9:^-{-a^-\- ....to infinity = 2. Ans» X = s- 836 SERIES. 8. Find the ratio of an infinite decreasing G. P. of which the 5 1 first term is 1, and the sum of the terms -. Ans, r = -. 4 5 9. Find the fii'st term of an infinite decreasing G. P. of which the ratio is — , and the sum of the terms -— . Ans. a = l. m m — 1 10. Find the sum of the first n terms of the G. P. whose m^ tennis (— l)'»fl*^. . «* r/ i\n i» -n ^ ^ Ans,s = — -[(— l)»a^— 11. 11. Find the ratio of an infinite decreasing G. P., in which each term is ten times the sum of all the terms which follow it. . 1 Ans. r = —. 12. Find the sum of the first n terms of a G. P. whose first term is a, and third term c. ^ / c^ a' Ans, s = fi- GEOMETRICAL MEAN". 526. Hie Geometrical Mean of n quantities is the n^ root of their product. Thus, the geometrical mean of a and b is Vab, and the geometrical mean of 1, 3, 6, and 72 is 6. 527. To find the geometrical mean of the terms of a a. p. Denoting the geometrical mean by M, and the product of the terms by P, we have, by definition, m = Vp. But P = V{al)^ (523) ; M = Vol. GEOMETRICAL PROGEESSIOX. 337 538. To find a and I when M, r, and n are given. I al a = ar n-l I ^^n-1 _ alr*"-^ (519). Substituting M^ for al, we have a = whence, a = I M U = M V^""^ J Cor. 1. — If M = ^/~xy and r = ttHj, all the terms of the progression are of the first degi'ee and rational if n is even. The sura of the exponents in the ratio TT'^y is ; hence all the tenns are of the same degree. The ratio x~^y is rational; hence the terms are either all rational or all irrational. It is sufficient, therefore, to show that the first term is of the first degree and rational. M n 2— n which is of the first degree and rational when n is even. Cor. 2. — If M = icz^ and r = ar^y, all the terms of the pro- gression are of the second degree and rational if n is odd. a=z M x^y xy ___ U|/ a^;/2 Xi--nyn-l = V a:"+y~" n+l ?— n w+l n— i = x ^ y ^ =^X ^ y '^ , which is of the second degree and rational when n is odd, 22 838 SERIES. 539. rjtoBZEMs. 1. Find the first term of a G. P. of three terms whose geo- y metrical mean is xy and ratio ^. Substituting 3 for n, xy for M, and - for r in the formula M , xy xy . a = , we have a = — -== = -^=a*. 2. Write a G. P. of three terms whose geometrical mean is xy and ratio -. Ans. os^, xy, y\ X 3. Write a G. P. of four terms whose geometrical mean is ni Qfyi nut \xy and ratio -. Ans. — , x, y, ^. X y X 4. Write a G. P. of five terms whose geometrical mean is xy and ratio -. Ans, — , x^, xy, y\ '—. X y ^ ^ X 5. Write a G. P. of six terms whose geometrical mean is Vxy ... V A ^ ^ y^ y^ and ratio ^. - Ans, -5, — , x, y, ^, ^. X y^ y ^ x' x^ 6. The sum of three numbers in G. P. is 26, and the sum of their squares is 364. What are the numbers ? Denote the geometrical mean by xy and the ratio by - ; then by the conditions of the problem, a^^xy +y^= 2Q . , , (1), and x^ -}- xy -\- y* =z 364c . . . (2). Transposing xy in (1) and squaring the result, x^ -{- 2x^y^ -\- y* = 67Q — 62xy -{- xiy^ . . . (3) ; whence, x* -}- x^y^ -\- y^ =z 676 ^ 52xy . . . (4). GEOMETRICAL PROGRESSION-. 339 Since the first members of (2) and (4) are identical, 364 = 676 — 520:^ . . . (5) ; whence, y =- . . • (6). Substituting this value of y in (1) and solving the resulting equation, we find a:^ _ ig or 2. Squaring (6) and substituting for x^ its value, we find y^=2 or 18. From (6) xy = 6. Hence the numbers are 18, 6, and 2. 7. The sum of four numbers in G. P. is 15, and the sum of their squares is 85. What are the numbers ? Denote the geometrical mean by V^cy and the ratio by - ; then by the conditions of the problem, .15 . . . (1), and ^+a;2H.«/2 + ^ = 85 . . . (2). Assume x + y =z z, and xy =zp; then x^ -\. y'i z=i z^ — 2/7, and a? -^ y^ :=^ z^ — 3zp. Substituting 2; for x -^ y in (1) and z^ — 2p for x^ -^ y^ in y ■\-x -\-y + t X + a;2 + «/2 + (2), |+. + f = 15 . . . (3), and ^ + ,»_2;, + g = 85. . . . (4). Transposing z in (3) and ^ — %p in (4), ^ + -^ = 15-« . . . (5), and J + g=:85-«« + 2^ . . . (6). 540 SERIES. Squaring (5) and transposing 2xy or 2/7, p+|^ = (15-^)^-2i' . . . (7). Since the first members of (6) and (7) are identical, (15 — zy^%pz=^b — z^-\.2p . . . (8) ; whence, 2z^ — 30^; — 4p = — 140 . . . (9). Clearing (5) of fractions, 7? ■\- if = {lb — z) xy = 15/7 —pz . . . (10). Substituting z^ — 3zp for ofi + y% (10) becomes z^ — 3zp = 15p—pz . . . (11); z^ whence, p = — — . . . (12). -^10 + 2z ^ ' Substituting this value of/? in (9), clearing of fractions, trans- posing and reducing, we obtain 15^3 + 85;? = 1050 . . . (13) ; 35 whence, 2; = 6 or —, o Substituting 6 for z in (12), we find ^ = 8. We then have the equations ic + y = 6 . . . (14), and xy=% . . . (15) ; whence, a; = 4 or 2, and y = 2 or 4. The raquired numbers are therefore 1, 2, 4, 8. The second value of z leads to imaginary results. In the solution of such problems as this and the preceding one, the terms of the progression may be represented by X, xy, xy^, xif, a;^, . . . . but the notation we have used is generally preferable. THE DIFFEKEKTIAL METHOD. 341 8. The sum of three numbers in G. P. is 210, and the last ex- ceeds the first by 90. What are the numbers ? Ans. 30, GO, 120. 9. The continued product of three numbers in G. P. is 216, and the sum of the squares of the extremes is 328. What are the numbers? A^is. 2, 6, 18. 10. The continued product of three numbers in G. P. is G4, and the sum of their cubes is 584. What are the numbers ? Ans. 2, 4, 8. 11. The sum of 120 dollars was divided among four persons in such a manner that the shares were in A. P. If the second and third had each received 12 dollars less, and the fourth 24 dollars more, the shares would have been in G. P. Find the shares. Ans. $3, $21, $39, 857. 12. The sum of six numbers in G. P. is 189, and the sum of the third and fourth is 36. What are the numbers ? Ans, 3, 6, 12, 24, 48, 96. TREATMENT OF SERIES BY THE DIFFERENTIAL METHOD. 530. The First Order of I>ifferences of a series is the series obtained by subtracting each term of the given series from the following term ; the Second Order of l>fffer~ ences is the series obtained by subtracting each term of the first order of difierences from the following term ; tlie Third Order of Differences is obtained from the second in the same way as the second is from the first ; and so on. Thus, If the given series be 1, 4, 9, 16, 25, .... The 1st order of differences is 3, 5, 7, 9, . . . . The 2d order of differences is 2, 2, 2, . . . . The 3d order of differences is 0, 0, ... . 531. The Differential 31ethod is the process of find- ing any term of a series, or the sum of any number of its terms, by means of the successive orders of differences. 342 SEKIES. 532. To find the n^^ term of a series. Let a, h, Cf d, 6) , , , , be the proposed series. The 1st order of differences is J— a, c—h, d—c, e^d, .... The 2d order of differences is c — 2b + a, d—2c + 5, e— 2c?4- c, . . . . The 3d order of differencesis d—dc-^Sb—ay e—dd-\-3c—bf . . . . The 4th order of differences is e — 4e? + 6c — 45 + «, . . . . Denote the first term of the first order of differences by di, the first term of the second order of differences by d^^ the first teim of the third order of differences by d^, and so on ; then c?i =: J — a di-=c —2b -\- a d3 = d — Sc + db — a di = e — M -\- 6c — 4:b -{- a I. • . (1); whence, ' b z= a + di c =z a -}- 2di -i- di d = a -\-3di-\- dd^ + ds e = a -\- 4:di -\' 6di -^ ^dz + d^, (2). The coefficients in the value of c, the tliird term of the pro- posed series, are 1, 2, 1, which are the coefficients of the expan- sion of {x + of ; the coefficients in the value of d, the foiirth term, are 1, 3, 3, 1, which are the coefficients of the expansion of {x + of ; the coefficients in the value of e, the fifth term, are 1, 4, 6, 4, 1, which are the coefficients of the expansion of {x-\-aY\ and so on. Hence the coefficients in the value of the n^^ term are the coefficients of the expansion of {x + aY~\ Therefore, denot- ing the n*^ term of the series by T„, T„= flf + (w— 1)^1 + in- l)(n-2)^^ ^ (n-l)(n-2)(^ 3) la ^ ^3+.... (A). THE DIFFERENTIAL METHOD. 343 EXAMTIjES. 1. Find the 12th term of the series 1, 4, 9, 16, 25, ... . In this example a = 1, J, = 3, tZg = 2, d^ = 0, and w = 12. Substituting these values in (A), we obtain T. = l + 11x3 + 11^115^ =144. 2. Find the 9th term of the series 1, 4, 8, 13, 19, ... . Alls, 53. 3. Find the 15th term of the series 1, 4, 10, 20, 35, .... Ans, 680. 4. Find the 8th term of the series 1, 6, 21, 56, 126, 251, . Ans, 771. 5. Find the 20th term of the series 1, 8, 27, 64, 125, Ans. 8000. 6. Find the n^ term of the series 1, 3, 6, 10, 15, 21, .... Ans, ^J^, 1% n{n-\- 1) divisible by 2 ? Why ? 7. Find the n^ term of the series 1, 4, 10, 20, 35, ... . Ans. "(« + l)(» + ^). 6 Is w (n + 1) {n + 2) divisible by 6 ? Why ? 8. Find the n^ term of the series 1, 5, 15, 35, 70, 126, . . . . Ar,. n{n + l){n + %)(n^^) Is w (71 + 1) {n + 2) (;i 4- 3) divisible by 24 ? Why ? 533. To find the siim of n terms of a series. Let a, h, c, d, e, . . . (1) DC the proposed series, and denote the sum of n terms of it by S, Let us assume the series 0, a, a -\- h, a -{• l + c, a + 1) + c + dy . . . (2). 344 SERIES. Now it is evident that the sum of n terms of (1) is equal to the {n 4- 1)^ term of (2). Denoting the {n -[- 1)^ term of (2) by T^+i, the first term of the first order of differences of (2) by D„ the first tenn of the sec- ond order of differences by Dg, and so on, we have by (A), T.,.=0+«D,+fcllD,+^i^%.+ (3). But T,+i=S„, D, = «, J)i= b — a=id^, D3= c— 25 + « = f?2, and so on. Hence, by substitution, (3) becomes S,=yzfl+ ^ .,^ V i4-— ^ -d,+ (B). EXAMPLES, 1. Find the sum of 10 terms of the series 1, 4, 9, 16, 25, . . . . In this example cr = 1, ^, = 3, c?, = 2, dz = 0, and w = 10. Substituting these values in (B), we obtain 10 X 9 X 3 10 X 9 X 8x2 2 "^ 6 2. Find the sum of 20 terms^of the series 1, 3, 6, 10, 15, 21, ... . Ans. 1540. 3. Find the sum of 12 terms of the series 1, 5, 14, 30, 55, 91, Ans. 23G6. 4. Find the sum of 10 terms of the series 1, 4, 13, 37, 85, 1 GG, ... . Ans. 2755. 6. Find the sum of n terms of the series 1, 3, 6, 10, 15, 21, ... . Ans. M!L±iHfi±2)_ 6 INTERPOLATION. 534. Interpolation is the process of inserting between two consecutive terms of a given series a term or terms which shall conform to the law of that series. 535. The Formula for Interpolation is that given for finding the n^^ term of a series by the differential method. S,„ = 10 + ^" .^^ "" + " r = 385. THE DIFFERENTIAL METHOD. EXAMPLES. 345 ^ V2I = 2.758924 ' V22 = 2.802039 Given \ y^ V24 2.843867 2.884499 L V25 = 2.924018 to find the cube root of any in- termediate number by the differ- ential method. 1. Find the cube root of 21.75. The operation of finding the orders of differences may be con- veniently arranged as follows : NO. CUBK ROOTS. d. d. d. d. 21 2.758924 22 2.802039 + .043115 23 2.8438G7 + .041828 -.001287 24 2.884499 + .040C32 -.001196 + .000091 25 2.924018 + .039519 —.001113 + .000083 —.000008 The distance between any two consecutive terms of the given series is 1 ; hence the value of n which corresponds to the required term is If; that is, the required term is J of the way from the first to the second term. Substituting in (A) 1| for n, 2.758924 for a, .043115 for d^, — .001287 for ^g, .000091 for d^, — .000008 for d^, and reducing, we find Ti| = V2r75 = 2.791385. 2. Find the cube root of 21.325. 3. Find the cube root of 21.875. 4. Find the cube root of 21.4568. 6. Find the cube root of 22.25. 6. Find the cube root of 22.684:. 7. Find the cube root of 22}. Arts. 2.773083. Ans. 2.796722. Ans. 2.778785. Ans. 2.812613. Ans. 2.830783. Ans. 2.833525. 346 SERIES. DEVELOPMENT OF EXPRESSIONS INTO SERIES. 536. To Develop or Expand aa expression is to con- vert it into a series. 537. To develop a fraction into a series by division. EXAMPLES, 1. Convert 3rii 1—X mto an inuniie Beries. 1 1—X 1 — x l + a; + ic2 + a^ + etc. X x — a? a? a?-x^ a^ 1 +x -\-x^ r^x^ + x^ -{■ etc. to infinity (1) If a; = |, (1) becomes 2 = l + | + ^ + i + etc. ... (2) If x = l, (1) becomes oo = l + l4-l-f-l4- etc. ... (3) If x = 2j (1) becomes — 1 = 1 + 2 + 4 + 8 + etc. ... (4) How is this result to be explained ? Convert each of the following fractions into an infinite series ^ a , ^ X x^ x^ X* 2. — --. Ans. 1 + -2 -A + ~i— ••• ri -i- ^ a a^ a"^ a^ . ^ X x^ x^ X* 3. . Ans, 1 + - + -o + -3 + -. + ... 4. ^— ^-^. A71S. l-\-2x + 2x^-\-2a^-\-2x^+ , . . ■ 1 X^ , X^ X^ X? Ans, « + — , 7 + -9 — • • • a a^ a'' a' a^ 6. -. Ans, l + a—a^—a^^-a^-\-a^—a^—a>^+ .. . 1 — a + a^ a + X a a — X l-{-x 1 — x a -\- X a^ + x?' 1 DEVELOPMENT OF EXPRESSIONS INTO SERIES. 347 538. To develop an expression of the form of Vm ±,n by extracting the indicated root. EXAMPLES. 1. Convert a/1 + ^ into an infinite series. l + ^l + 3-^ + ^-j28 + etc. ^ + 5 X 2 4-a; 8/ 4 a;2 ic3 ^4 ~T~"8 "^64 2+^~l+fi) "8 ~6i "8 "*" 16 ~ 64 "^ 256 ""64 "^ 64 "" 256 X ^ ^ hxS ^^^^+^=^ + 2^8+16-128+ •'•• Convert each of the foUowing expressions into an infinite series : 2. \/a — X, i( x_ x^ 3a:3 3-5a^ \ ^'^^- "^ r~'2a-2-4a2 2-4-6a3 2-4-6-8a* /* 3. V«M^. Z»2 ^4 3^_ 3-5^8 4. Vl - X. X x^ _ a^ _ 5x* _ Ix^ Ans. 1------ ^^g ^^g 348 SERIES. 539. To develop an expression by means of unde- termined coefficients. 1 4- 2a: 1. Convert jr- into an infinite series. 1 — 3a; 1 -h2x Assume = A + Ba; + Cie2 + DaJ* + Ea:* + . . . (1). 1 — 3a; Multiplying (1) by 1 — 3a;, we obtain 1 + 2a; = A 4- B\x + C — 3a! -3B a--2+ D -3C -3D a;* + . . . (2). Equating the coefficients of like powers of x in the two mem- bers of (2), A = l 1 B - 3A = 2 C - 3B = D — 3C = E — 3D = whence, A = 1, B = 5, C = 15, D = 45, E = 135, Substituting these values in (1), 1 + 2a; 1 — 3a; 2. Convert Assume = 1 + 5a; + 15a;2 ^ 4^^^ ^ 135^:1 + . . into an infinite series. 3x-j^ 1 = A + Ba; 4- Ca;2 + I)a:3 ^ Ea;4 + . . . (ij. 3x — a;2 Multiplying (1) by 3a; — x^, we obtain 1 = 3Aa; + 3Bb2 + 3c - Al - B a;3 + 3D - C ^ + . . . (2); whence, 1 = 0, which is absurd ; hence the second member of (1) is not of the proper form. Assume DEVELOPMENT dF EXPKESSIONS INTO SERIES. 1 349 'dx — x^ Multiplying (3) by Zx — x\ 1 = 3A + 3B|a; + 3C - aI - B = Aic-i + Ba:0+Ca;+Da;8+Ea;3^. . . (3). a;2 + 3D - C 7^-\-. . . (4). Equating the coefficients of like powers of x in the two mem- bers of (4), 3A = 1 ' 3B — A = 3C — B z= 3D — C = whence, A B Substituting these values in (3), ■" 3a; "^ 9 "^ 27 "^ 81 "^' • • • The proper form of the second member of the assumed identity may be detemained in each case by observing what the given expression becomes when the variable is supposed to be zero. If the given expression becomes a finite quantity, the first term of the series will not contain the variable ; if it becomes zero, the first term of the series will contain the variable ; and if it becomes infinity the first term of the series will be of the form Ax"^. Convert each of the following expressions into an infinite scries : 1 o^ 3. 5^. ^ws. l+a;+3a;2+9a;3 + 37a^+8l2:54..... 1 — 6x 1 4- 2r 4. = — .. Ans.l + Zx+4.x^-\-W-\-llx^-\-l^x^-\-..>. 5. l-^x — x^ \-^x l_3a;_2a;2 Ans. l + 2a; + 8a;2-f 28rr3 + 100a;^ + 356a;5^. 350 SERIES. {L 4X) 2 A . ^ . ^''^ . ^^^ 3^^ ^- 3^32^^- ^''^•32;'^9 + 27'^"8r + 2i3+*--- 8. 1 1 + 22^^4- 3a;* Ans, l—23^+x^+4:3f^—lla^-{-10x^-\-ldx^— RECURRING SERIES. 540. A Recurring Series is one which may be pro- duced by expanding some rational fraction. Thus, is a recurring series, because it is the expansion of the fraction :j (537, 1). In this series all the terms after the first two recur according to a definite law. 541. The Generating Fraction of a recurring series is the fraction which can be converted into the given series. Thus, the generating fiuction of the series l-\-x-\-3x^-\-%a?-^ 27ic* 4- . . . . 1 — 2a; IS l-3a;' 543. In the series given in Art. 540, each term after the first may be obtained by multiplying the preceding term by x ; and in the series 1 + 42: + lla^^ + 34a:3 j^ loia:* + . . . . the sum of the products obtained by multiplying the first of any two consecutive terms by 3x^ and the second by 2a; is equal to the next succeeding term. The expression by means of which any term of a series may be found when the preceding terms are known is called the Scale of Relation, Thus, the scale of relation of the series l-^x-{-x^-\-3?-i^x^-\-.,,, is a;, and the scale of relation of the series 1 + 4a; + 1 la;^ + 34a:^ + . . . . is 3x^-\-2x. 543. A recurring series is said to be of the n^ order when the number of terms in its scale of relation is n. Thus, the series 1 + 4a; + lla;3 _|_ 34^ + . . . . is of the second order. RECUEEING SEEIES. 351 544. To find the scale of relation in a recurring series. Let a-{-b-{-c-{-d-\-e-\-,... be the proposed series. 1st. Suppose that the series is of the first order. Let m denote the scale of relation ; then b = ma ; whence, b m = -. a 2d. Suppose the series to be of the second order. Let m + 71 denote the scale of relation ; then i^=:::::i(^*^-^3); mb -f- 7ic , c^ — bd , ad —be whence, m = j-„, and w = rr-. ac — IF ac — b^ 3d. Suppose the series to be of the third order. Let m -\-n -{• r denote the scale of relation ; then i d z= ma -{- nb -\- re \ < e ■=. mb -\- nc -{- rd > , \ f =zmc -\- nd -\- re ) From this group of equations m, n, and r may be found. If the series is of the fourth order, fifth order, sixth order, &c., the scale of relation may be found in a similar manner. CoE. 1. — If the proposed series is of the n^ order, 2n con- secutive terms must be given to enable us to find the scale of relation. CoE. 2. — If we assume any proposed series to be of a higher order than it really is, one or more of the terms of the scale of relation will be found to be equal to zero. If we assume any proposed series to be of a lower order than it really is, or if we attempt to find the scale of relation of a series which is not recurring, the error will appear if we attempt to apply the scale. SERIES. EXAMPLES. Find the scale of relation in each of the following series: 1. 1 + 4a; + IQx"' + 223^ -\- . . . . Assume the scale of relation to he m + n; then J 10a;2 = fn-\- ^nx ) ^ whence, m = — 2a^, and n = 32?. Therefore the scale of rela- tion is — 2j?5 + 3a;. 2. l4-6a; + 12arJ + 48a;3 + 120a:*+ 3. l4-2a; + 3a;H4a:3_f_5a4^ . . . 4. l + 2a;+8ir3+28j^+100a;*-h . ^ws. 6a;2_|_^, Ans, —x^-\-2x, Ans. 2xi-^dx. Ans. 2x''' + 2x. 54:5. To find the generating fraction of a recurring series. Let a-\-b-{-c-^d + e + .... be the proposed series. 1st. Suppose the series to be of the first order. Let m denote the scale of relation ; then b = ma c = mb d= mc e = md whence, bi-c + d+e-{- . . . . =ni(a + b + c + d+e-^ , . . .). Hence, denoting the generating fraction or the sum of the series by Fj, we have whence, ^'-r:^ m (0,). EECURBINQ SERIES. 353 2d. Suppose the series to be of the second order. Let m -{- n denote the scale of relation ; then c = ma -f nb d=:mb -i- no e = mc + nd f = md -f ne whence, Hence, denoting the generating fraction by Fg, we have Fg -{a + l) = wFg + n(F^ - a) ; ^ a + 1) — an whence. 1 — 7/^ (0»). 3d. Suppose the series to be of the third order. Let m •{■ n -{• r denote the scale of relation ; then d = ma -^ nh -\- re e =imh ■\- nc -\- rd f ^rnc 4- nd + re g = md ■\-ne -^r rf whence, d+e+f-\-g-\-. . . .=m {a-\-h-\ c-\-d+e+f+g-\-. . . .) Hence, denoting the generating fraction by Fg, we have F3 -- {a + h + c) = m¥^ H -^ (F3 -a) + r[F,-{a + b)] ; whence, F, a -{- b -{■ c — an — (a + b) r 1 — m — n — r ■ (0,). If the series is of the fourth order, fifth order, sixth order, etc., the generating fraction may be found in a similar manner. ScH. — The formulae fOj), (O^), (O3), etc., have been obtained on the hypothesis that the given series is infinite and converging. 354 SERIES. EXAMPIiES, Find the generating fraction of each of the following series : 1. 1 + 4a; +10a?J+ 322:34- 46a:* -f .... In this series the scale of relation is —^x^-\- 3x (544, 1) ; hence (Og) is appUcable. Substituting — 23^ for m, 3x for w, 1 for a, and 4:X for b, we have -, l+4a;— 3a; l-fic ^ ^ 1 + 23^ — 3x~ 1 -]- 23^ — 3x 1 + 2a; 2. l + 3a;-f4a^5+7a;3^11a4^__ j;^^. 3. l + 6a;+12a;2^48a:3^120a:4-}-.... A7is, 4. l + 2a:— 5ar^ + 26a;3— 119a:*+.... Ans. l — a;— a:2" 1 + 5a; l_a;__6a;2" 1 + 6a; 1 + 4a; — Sxi' 6. l + 4a;+3a;2_2a;3+4a4^X7a;5_j.3a4_.._ H-3a; + arJ Ans. l—x-\-2xi—3a^' 1-^.x 6. lH-3a;4-5a;8+7a;8^9a4^ ,.., Ans,, .^. (1 — a;) REVERSION OF SERIES. 546. To Hevert a Series containing an unknown quan- tity is to express the value of that unknown quantity in terms of the sum of the given series. Thus, to revert the series in the sec- ond member of the equation y z= ax ■{■ hx^ -\- CT^ •\- dx^ + ea;^ + .... is to find the value of x in terms of y. EXAMPLES. 1. Revert the series in the equation y=zx-[-x^-\-x^-\-a^-\- .... X This is a recurring series whose generating fraction is ; X. •"" X REVEKSION OF SEEIES. 355 henco y = 1-x' whence, x = y ^ y — y^+y^—y^+ Q? iC^ iC^ 2. Revert the series in the equation y=ix——-\r t — -k-\- 2x This is a recurring series whose generating fraction is — - — /v -p X hence y = 2x ._ ^y _, y^ , y^ 2 H- .;' ^^^'^''' ^=21:^=2/ + Y + t + t + A recurring series cannot be reverted by this method when the equation, formed by placing y, the sum of the given series, equal to the generating fraction, cannot be solved. The method used in the following example is applicable to any series. 3. Revert the series in the equation yz=ax-^la^ + ca^-\-dx^-^ .... Assume x=Ay-[-By^-\-Cy^ + T)y^-{- .... in which the coefi&cients A, B, C, D, . . . , Substituting for y its value, (2) becomes • • • (!)• . . . (2), are undetermined. X = aAjx + bA + a2B ^2 + cA\ + 2a^B + a^C\ + dA + hm -f 2acB + Sa^C + a^D x^-^ (3). r«A = i bA -\- a^B = cA + 2abB -\- a^C = dA + ^B + 2acB + da^C -f a^D = whence. a 0^ a? Substituting these values in (2), 1 b ^ 2W'-ac ^ = -^y--^y^-^ — :;5— ^ aH—6abc-\-bl^ a^d—5abc-\-5b^ y+. (4). 366 SERIES. 4. Eevert the series in the equation i/ = x-\-2x^ + 4^-\-Sx^-\- . . . . In this series « = 1, 5 = 2, c = 4, J = 8, . . . . Substituting these values in (4) of the preceding example, we have 6. Eevert the series in the equation 1_ 4xi 63^ Sx^ -_Zx--^+ ^ - ^ + and find the value of x. Substituting j for y, 2 for «, — q ^or b, - for c, — - for «/, . . .. 4 o o 7 in (4) of example 3, we have 1\* \ & '■'& »©' + .000013 + . . . . = .135993 +. 6. Eevert the series in the equation y = x-\-dx^'^5a^ +»7a.-4 ^ Q3^ ^ , , , , Ans. x = ij^dy^ + 13/ _ e7y^ + 381?/» — . . . . 7. Eevert the series in the equation a^ x^ x^ a^ y2 jiS jA ^ 8. Eevert the series in the equation y = X-\-7? -^-X? -^X^ ^-X^ -{■ .... Ans. x=z y — y^ -\- 2y^ — by' + 14/ — .... 9. Find the value of x in the equation I = 5a; — 20a;2 + SOa:^ _ 3202?^ + I280a:5 _ . . . . Ans. 2: = .117047+. BIKOMIAL FORMULA. 357 THE BINOMIAL FORMULA FOR ANT EXPONENT. 547, It has been shown that when nis q> positive integer 71/ ( 11 — 1 ^ {x + «)« = x^ + nax''-'^ + — ^- — ' a2^«-2 + . . . . "We now proceed to show that this formula is true, whether n \% positive or negative, entire or fractional, 548. Lem. — Hie value of —, when y = x, is nx'^~\ X — y whether n {^positive or negative, entire or fractional. 1. When w is a positive integer. The proposition has been shown to be true for this case (461, Cor. 2). 2. When nis a. positive fraction. Let n = -, in which j!? and q are supposed to be positive in- ( 2 P) tegers. We are to show that 1 — \ z=:^x9 . ^ i ^ — y }y=x q ( -Y ( -Y p XQ p { ^\p ( iV 1 1 — yq_\XQ/ —Kyi) a:? — ;/g ""^ \xl) — Ky'n) \x'q) — \yl) 1 1 x'd — yi But, by the first case, i \x J + 3D 22 + 4E2: 3+....=w4. nBz + B + 2C + 3D B = n 2C + B = nB •• 3D + 2C = nQ 4E + 3D = nD (8). Y, 360 SERIES. whence, B = n n(n — l) n(n—l){n—2) n{n—l)(n—2){n— 3) C = D = E = Substituting these values in (2), we have Substituting for z its value, -, (9) becomes (9). (10). Multiplying (10) by a", 1* . 12 (11). Cor. — If n is not a positive integer, the expansion of {z + a)" is an infinite series; for no one of the factors in the coefficients can be equal to zero under this hypothesis. j:xampz:es. Expand each of the following expressions : 1 a + 5 2. Vl + a or a^ a^ Ans-l+la-la^ + ^^a" SYKOPSIS FOR REVIEW. 361 3. (a-x)i. ^ns, ai(l-^-^,-^-^^,-. .. ), 4. (l-x)^. ^^'' ^-3-3:3- 3^^^-"- 5. (a + J)t ^^^^'- «^ (1 + 3^ -3-7(3^ +3T6-.-9^3--- •)• 1 , 1 b b^ b^ ¥ a — b a a^ a^ a* w ». ^« 6j. ^ws. « \^i g^^ ^.^^2 3.6'9a3 •••7 9. (1— a)-3. Ans. l + 3a + 6a2 + 10«8+15a4+21tt«+.... ^ . , a;2 6a;3 6-11:^4 ^^' Vf^' ^^'•^+5+2^+2^3T«+--" 550. SYNOPSIS FOR REVIEW. CHAP. XX. SERIES. General Defini- tions. Amthmetical Pro- gression. 2'erms. Finite Series. — Infinite Series. ^ Converging Series. — JDicerg'g Series. " Extremes. — Common difference. Increasing A. P. — Decreasing A. P, Notation. To find I, when a, d, and n are given. To find s, when a, I, and n are given. Sum of two terms equidistant from extremes. To insert any number of arithmeti- cal means between two given quantities. To find any two of the quantities a, I, n, d, s, when the three others are given. To find arith. mean. Arith. Mean, i To find a and I when M,c?, 71, are given. SERIES. SYNOPSIS FOR BEVIEW^Cmiinved. CHAP. XX. SERIES. Continued. Geometrical Pro- gression. Treatm't of Series BY THE DlFFEREN- TiAii Method. Development of Ex- press'n into Series. Extremes. — Ratio. Increasing G.P.—Decreamng O. P. — Infinite Decreasing G. P. Notation. To find I, icTien a, r, and n are given. Cor. To find s, when a, I, and r are given. Cor. Product of two terms equidistant from extremes. To insert any number of geometric means 'between two given quan. To find the continued product of the terms of a O. P. To find any two of the quantities a, I, n, r, and s, when the three others are given. Geom ^ '^o find geometrical mean. Mean. 1 "^^ ^^ ^ ^^'^ ^ when M, r, ' and » are given. Cor. 1,2. ' Orders of differences. Differential method. To find the n"" term of a series. To find the sum of n terms of a scries. Interpolation.— FcfTmu\& for Inter- polation. ' Development of fractions by division. Development of expressions of the n by extract- form of "s/m ± ing the indicated root. Development of expressions by means of Undetermined Coefficients. Generating Fraction. Scale of Relation. Order of Recurring Series. To find the Scale of Relation in a Recurring Series. Cor. 1, 2. To find the Generating Fraction, Sch. Reversion op Series. . B1NOML&.L Formula for any Exponent, -j -5^^^^- { Cor. Recurring Series . . OHAPTEE XXI. LOGARITHMS AND EXPONENTIAL EQUATIONS. LOGARITHMS, 551. The Logarithm of a number is the exponent by which some fixed number must be affected in order to produce the given number. The fixed number is called the Hase of the System, Thus, in the equation a'' = ?^, x is the logarithm of n to the base a. For brevity, the expression log^w is sometimes used to denote the logarithm of ?^ to the base a» Thus, x = logaW expresses the same relation as «* = n. 552. Any number except + 1 and -— 1 may be used as the base ; hence there may be an infinite number of systems of logarithms. There are only two systems, however, in general use, namely: Briggs' system, the base of which is 10, and Napier's system, the base of which is 2.718 -f . Briggs' system of logarithms is used more than that of Napier, and is hence called the common system. 553. If in the equation a* = 7i we suppose n to represent a perfect power of «, then x will be some integer; but if n is not a perfect power of a, then x will be a mixed number or i\, fraction. 554. The Characteristic of a logarithm is the integral part of it, and the 3Iantissa is the fractional part. Thus, the characteristic of logg243 is 2, and the mantissa is .5; for 92.8 _ 9^^ 35 _ 243. 364 LOGARITHMS. GENERAL PROPERTIES OF LOGARITHMS. 555. In any system the logarithm of 1 is 0. For a* = 1 when x = (84, CoE.). 55^. In any system the logarithm of the base is 1. For a'^ =i a when a; = 1. 551, In a system whose base is greater than 1, the logarithm ofOis — 00 . For «~* = -— , and — z = when « > 1. 558. In a system whose base is less than 1, the logarithm of is + 00 . For a* =: when a < 1. 559. In a system whose base is positive , a negative quantity has no real logarithm. For, if a is positive, a* is positive, whether x is positive or neg- ative. Thus, 102 = 100, and 10-2 = -L = ^. 10^ 100 560. The logarithm of a product is equal to the sum of the logarithms of its factors. Let X = logflW, and y = log^w ; then m = a^, and 7i=za^; whence, mn = a^^a^ = a'^K Therefore, ]og„7nn = x ■{- y (551) = log^wi + hg^n. 561. TTie logarithm of a quotient is equal to the remainder obtained by subtracting the logarithn of the divisor from that of the dividend. Dividing m = a' by n^^aM^ n hence logo —= x — y = hg^m — log^w. LOGAKITHMS. 365 563. The logarithm of any poiuer of a numher is equal to the product of the exjjonent of the power and the logarithm of the number. Raising both members of the equation m = a* to the r^^ power, loga(m^) = rX:=r loga77Z. 563. TIte logarithm of any root of a numher is equal to the quotient obtained by dividing the logarithm of the number by the index of the root. Extracting the r^^ root of both members of the equation m = a*, , {r /—\ X logam 5(54, EXAMPLES. Prove each of the following statements: 1. log (abc) = log a + log b + logc. 2. log \^J = log a + log 5 4- log c — log d. 3. log {aWc^) = 2 log a + 3 log J + 4 log c. 4. log(^'-) = 2]og« + 31og5 + 41ogc- 5log^. 5. log Vabc = ^ (log a + log J + log c). 6. log Va^ — ^ = I [log (a + b) + log {a — b)]. S66 LOCAEITHMS. THE COMMON SYSTEM. 565. To find the characteristic of a logarithm in the common system. In this system, log 100 = log 1 = 0, log 10-1 _ log .1 = — 1, log 101 ^ log 10 = 1, log 10-2 = log .01 = — 2, log 102 = log 100 = 2, log 10-3 = log .001 = — 3, log 103 = log 1000 = 3, log 10-4 _ log .0001 = — 4, log 10^ = log 10000 = 4, log 10-s = log .00001 = — 5, Hence, supposing n to be a positive integer, 1st. The logarithm of a number between 10** and 10"+^ is greater than n and less than n -\- 1\ its characteristic, therefore, is n. Now, the number of figures in the iutegi'al part of a num- ber between 10" and 10"+* is n + 1. Hence, The characteristic of the common logarithm of an integer, or of a number composed of an integer and a decimal fraction^ is pos- itive and one less than the number of figures in the integral part of that number. Thus, the characteristic of logio 258.045 is 2. 10-(»+i)j that is, between -— and -ttz-^x, is some negative number 2d. The logarithm of a decimal fraction between 10"" and 10"» ^^^ W between — n and — (n ■\-\)\ hence, if we agree that the man- tissa shall in all cases be positive, the characteristic will be — (7i 4- 1). Now, the number of ciphers preceding the first sig- nificant figure in a decimal fraction between -— and — r^, is n. Hence, Tlie characteristic of the common logarithm of a decimal frac- tion is negative and numerically one greater than the number of ciphers preceding the first significant figure in that fraction. Thus, the characteristic of logio .0546 is — 2. LOGARITHMS. 367 566. If the ratio of two numbers is any perfect power of 10, the mantissas of their logarithms in the common system will he the same. This follows from Art. 561. Thus, denoting the mantissa of logio 5468 by m, log 5468 = 3 + m, log 546.8 = log (^) = log 5468 — loglO = 3 -f m —1 —%+m, log 54.68 = log (^^) = log 5468 — log 100 = 3 +m-2=l+m, log 5.468 = log(^) = log 5468 - log 1000=3 + m-3=:0 + m, log .05468 = log (^-^^) = log 5468 - log 100000 = 3+^-5 = — 2 + m. COMPUTATION OF LOGARITHMS. 567. To express the logarithm of a number in terms of that number and the base of the system. Let X be the logarithm of n to the base a ; then a' = n , . . (1). Assume a = 1 -f- t??, and n^l -{- p\ then (1) becomeg {l+mY = l-^p . . . (2); whence, (1 -f m)*y=(lH-^)y . . . (3). Expanding both members of (3) by the Binomial Formula, =i+yp+y^-^f+y^y^-:^f+ (4). 368 LOGARITHMS. Dropping 1 from both members of (4) and dividing the result x(xy — 1) - , ir (xy — 1) (xy — 2) _ . If ^ l£ =^ + (j^-iV + (y-iKv-^V + (5). Making y = 0, (5) becomes m^ m^ 7n* w^ whence, ^ 2 + 3 4+5 ^^^' ^ _ ^ a. ^ _ -^ a. ^ _ ^ 2+3 4 + 5 • • • • ,^, ^ = 2 ^ 2 i^ • • • (7). m^ m^ m* m^ ' But X = logaW =: loga (1 ■\- 'p)\ hcncc, if we put ^^ 1 rn? m^ m^ m^ ' '"-y + y-T + T-----. we have x = log„(l+^)=M(;,-^ + |!-^V^-....)...(L). The second member of (L) consists of two factors, tiamely : the series within the parenthesis, which depends only upon the number, and the quantity M, which depends only upon the base of the system. The factor M, which depends only upon the base, is called the modulus of the system. The series in (L) is called the Loffarithmic Series, 568. To find the Base of Napier's System. Baron Napier arbitrarily assumed the modulus of his system to be unity. Making M = 1, and denoting the Base of Napier's System by e, (L) becomes LOGARITHMS. 369 ^ = log,(l+p)=p-f + f-^ + f- (1). Eeverting the series in the second member of (1), we obtain X^ CC^ X^ , 0^ , Ofi , /Qx ^=^+l-+l+l+l+f + (^^• But 6^ = 1 +i?; x^ a^ x^ x^ x^ ,o\ Making a; = 1 in (3), we have 1111.1. ^A\ ^ = ^ + l2+f + f + f+^+ (^)- Summing the series in (4) to nine terms, we find e = 2.718282. 569. T7ie logarithm of a number in any system is equal to the product obtained by multiplying the modulus of that system by the Napierian logarithm of the same number. For log„(l +^) = M (^ - I + I- 1^ + ^-. . . .)(567), and \os,(i+p)=p-^^l._t + ^_ .... (568); C4.1A«J. iV^g ^J. T-i^y-y^ 2^3 4^5 .*. loga(l+i?)=Mloge(l+i?), Cor.— If l^p = a, we have \ogaa = Clogs', but loga^ = 1 (556) ; •*• 1 = M logeffi ; whence, Hence, l0ge« The modulus of any system is equal to the reciprocal of the Napieria7i logarithm of the base of the' system. U 370 LOGARITHMS. 570. To transform the Logarithmic Series into a Converging Series. The formula log„(l+i,) = M(;,_^ + ^-^ + |^-....) . . . (1) cannot be used for the computation of logarithms when ^ > 1, because the series in its second member does not converge. Substituting —p for jt? in (1), we have log.(l-,) = M(-,-|-|_^-^-. ...)...(.). Subtracting (2) from (1), observing that loga(l-fj9) — log«(l-i<)=log,{i±-|)(661), ^^^•([^|) = ^^K^+l + l + ^+--) • • • (^)- . 1 ., 1+jy z + 1 Assume p = t: — -— r ; then -■ = . ^ 2^; + 1 1 —p z Substituting these values in (3), l0g« y-^) — ^^oa (2; + 1) — l0ga2; = ^^^(rin[ + 3(2^3 + 5(^^ • • • (^>- For Napier's System (4) becomes loge (z + 1) — logeZ = ^ (2^^ "^ 3 (2^ + 1)3 + 5 (2;^ + If + ) • • • • (^)' whence, by transposition, log,(^ + l) = ^^^^ + ^27^1 + 3^2^3+57^^ • • • W- LOGARITHMS. 371 571. To compute a Table of Napierian Logarithms. log,0 = - 00 (551), log,l = (555), log,2 = log.l + 2(3U3i^3+^ + ^+....) = 0.693147 (570), log.3=.log.24-2(l-|-3L_^+^ + y^+....) = 1.098612, log,4 = log,22 = 2 log,2 = 1.386294, log,5 = log,4 + 2g + 3i^3 + ^, + ^+ ....) = 1.609438, log,6 = log,2 + log,3 = 1.791759, log.7 = log/. + 2(l+3434-^3+ ....) = 1.945910, log,8 = loge23 = 3 loge2 = 2.079442, log,9 = log,32 = 2 log,3 = 2.197225, logelO = log,2 4- logeS = 2.302585, 572. To find the modulus of the common system. Denoting the modulus of the common system by M, we have 573. To compute a table of common logarithms. If we multiply the Napierian logarithm of a number by the modulus of the common system, the product will be the common logarithm of the same number. Thus, logio5 = 1.609438 x .434294 = 0.698970. 373 LOGAKITHMS. TABLE OF COMMON" LOGARITHMS FROM 1 TO 100. N. Loo. N. Loo. N. Loo. N. Loo. 1 0.000000 26 1.414973 51 1.707570 76 1.880814 2 0.301030 27 1.431364 52 1.716003 77 1.886491 3 0.477121 28 1.447158 53 1.724276 78 1.892095 4 0.602060 29 1.462398 54 1.732394 79 1.897627 5 0.698970 30 1.477121 55 1.740363 80 1.903090 G 0.778151 31 1.491362 56 1.748188 81 1.908485 7 0.845098 32 1.505150 57 1.755875 82 1.913814 8 0.903090 33 1.518514 58 1.763428 83 1.919078 9 0.954243 34 1.531479 59 1.770852 84 1.924279 10 1.000000 35 1.544068 60 1.778151 85 1.929419 11 1.041393 36 1.556303 61 1.785330 86 1.934498 12 1.079181 37 1.568202 62 1.792392 87 1.939519 13 1.113943 38 1.579784 63 1.799341 88 1.944483 14 1.146128 39 1.591065 64 1.806180 89 1.949390 15 1.176091 40 1.602060 65 1.812913 90 1.954243 16 1.204120 41 1.612784 66 1.819544 91 1.959041 17 1.230449 42 1.623249 67 1.826075 92 1.963788 18 1.255273 43 1.633468 68 1.832509 93 1.968483 19 1.278754 44 1.643453 69 1.838849 94 1.973128 20 1.301030 45 1.653213 70 1.845098 95 1.977724 21 1.322219 46 1.662758 71 1.851258 96 1.982271 22 1.342423 47 1.672098 72 1.857333 97 1.986772 23 1.361728 48 1.681241 73 1.863323 98 1.991226 24 1.380211 49 1.690196 74 1.869232 99 1.995035 25 1.397940 50 1.698970 75 J 1 1.875061 100 2.000000 574, EXAMrT.ES. 1. Find the product of 9 and 7 by means of logarithms. Log (9 X 7)=:log9+log 7 (560) =0.954243 + 0.845098=1.799341. The number corresponding to this logarithm is 63 (573). EXPONENTIAL EQUATIONS. 373 2. Divide 210 by 7 by means of logarithms. Log (^i^)=log210-log 7=2.322219-0.845098=1.477121 = log 30. 3. Find the square of 9 by means of logarithms. 4. Find the fourth root of 625 by means of logarithms. 5. Find the logarithm of 33^. 6. Find the logarithm of 6^ x 7^ x 8^. EXPONENTIAL EQUATIONS. 575. An Exj^onential Equation is one in which the unknown quantity occurs as an exponent. Thus, a^ = n is an exponential equation. 576. To solve the exponential equation a^ = n. Taking the logarithm of each member of this equation, we have xloga = log 71 (562) ; whence, x = ——. log a EXAMPTjES. Solve each of the following equations : 1. 3^ = 27. Ans. x = 3. 2. 5^ = 100. Ans. x = 2.861. Ans. 1.5. Ans.x = ^.^^^P^. A log J Ans. x— ^ 6. a^ — 2pa^ = I, Ans. x 3. 3 2^ = = 4. 4. ah^ = n. 5. 1 = n. log n — log a \og{p±Vb+ p^) log a 374 LOQABITHMS ±ND EXPONENTIAL EQUATIONS. 5111. SYNOPSIS FOR REVIEW. LOGARITHMS. Log. of a number. Base of a system. Characteristic. — Mantissa. General Properties. Common System. Computation. . Log. 1. Log. Base. When Base >• 1. WTien Base < 1. When Base is positive. Log. of a Product. Log. of a Quotient. Log. of a Power. L Log. of a Root. To find th^ characteristic. Mantissas of log. of two ?ium- bers whose ratio is a perfect power of 10. "" To express log. of a number in terms of that number and the base of the system. Modulus. — Logarithmic series. To find the base of Napier's sys- tem. Log. number = Modulus x Na- pier's Log. same number. Modulus of any system = recip- rocal of the Napi.rian log. of la83 of the system. To transform log. scries into converging series. To compute table of Napierian log. ^ To compute table of common log. I EXPONENTIAL EQUATIONS. CHAPTEE XXII. COMPOUND INTEREST AND ANNUITIES. COMPOUND INTEREST, 578. To find the amount of p dollars at compound interest for n years at r per cent, per annum. At tlie end of the 1st year the amount will be p -irpr=p{l + r); at the end of the 2d year the amount will be p{l + r) + p{l + r)r =p{l + r )2; at the end of the 3d year the amount will be ^ (1 + r)2 +i? (1 + rfr =p (1 + rf, and so on. Hence, denoting the requhed amount by A, A=i?(l+r)» . . . (1). Any one of the four quantities, A, p, n, and r may be found from this equation when the three others are given. The compu- tation is most readily performed by means of logarithms. Taking the logarithm of each member of (1), log A = log;? + ^ log (1 + r) . . . (2) (560-562) ; whence, log;; = log A — w log (1 + r) . . . (3), log(l + r)z. ^"g^-'°g-P . . n (4), _logA — log;? and n = -:p—j——~ . . . (5). log (1 + r) ^ ^ 37G ANNUITIES. Example. — How much will $500 amount to in five years at 6 per cent, compound interest? Given jl^gl-^^ =0.025306) ( log 609.10 = 2.825491 5 * Substituting 500 for j9, .06 for r, and 5 for n in (2), we have log A = log 500 + 5 log 1.06 = 2.698970 + 5 X .025306 = 2.825500. Since log 669.10 = 2.825491, it fx)llows that A = $669.10. ANNUITIES. 579. An Annuity is a sum of money which is payable annually. The term is also applied to a sum of money payable at any equal intervals of time. 580. To find the amonnt of an annuity of a dollars for n years at r per cent, per annum, when the inter- est is compounded every year. The first payment a becomes due at the end of the first year, and m n — 1 years this will amount to « (1 -f r)"~i (578) ; the second payment a becomes due at the end of the second year, and in n — 2 years this will amount to « (1 + r)«-2j the third pay- ment will amount to a (1 + r)"-^ in n — Z years ; and so on. Hence, denoting the amount of the annuity by A, A = a (1 + r)"-i + « (1 + r)"-2 + a (1 -f r)"-3 + . . . . + « (1 + r) + « . . . (1). By reversing the order of the terms in the second member of (1). A=a + f((l + r) + a(l + r)2+.--- +«(! + '•)"-' ■ • • (3); whence, A ^ (1 + ,.) _ T- "= " ' r " ' ' ^^^- SYNOPSIS FOR REVIEW. 377 581. To find the present value of an annuity of a dollars for n years, at f per cent, per annum, the in- terest being compounded every year. Denoting the present value of the annuity by P, (1 4- r)» - 1 P(l+r)' whence, P=: a (1 + r)« — 1 (1 + tY Cor. — If ?^ = 00 , (2) becomes P = ^. (1) (578-580) ; (2). ^^2,. SYNOPSIS FOR REVIEW. r ooMPouro INTEREST CHAP. XXII. M To FIND THE AMOrifT OP ^ DOLLARS AT COMPOUND INTEREST FOR 71 YEARS AT r PER CENT. PER ANNUM. AOTTJITIES. ^ To FIND THE AMOUNT OF AN ANNUITY OF a DOLLARS FOR 71 YEARS AT T PER CENT. PER ANNUM, WHEN THE INTER- EST IS COMPOUNDED EVERY YEAR. To FIND THE PRESENT VALUE OF AN ANNUITY OF a DOLLARS FOR 7^ YEARS AT r PER CENT. PER ANNUM, WHEN THE INTEREST IS COMPOUNDED EVERY YEAR. Cor. CHAPTEE XXIII. THEORY OF EQUATIOIsTS. DEFINITIONS. 583. Every equation of the n^ degree containing only one unknown quantity may be written under the form of a;" + Ao:"-! + Bs^-^ + ....+ Ka; + L = 0. This equation is called the general equation of the n^^ degree. The term L, which is called the absolute or independent term, may be considered as the coefficient of aP. 584. A Function of a quantity is an expression contain- ing that quantity. Thus, aoi? -{- bx is a function of x. For brevity we shall sometimes use the symbol /(ic) to denote & function of x. If f(x) is entire and rational with reference to a;, it is called a rational integral function of x. In the present Chapter, when f{x) is used without modifica- tion, it is understood to denote a rational integral function of x. 585. Any quantity, which substituted for x in f (x) causes f(x) to vanish, is a Moot of the equation f{x) = 0. GENERAL PROPERTIES. 586. If f{x) vanishes when x=^r, the function is divisi- ble by X — r. Suppose f{x) to be divided by x — r, and the operation continued until a remainder is obtained which is independent of x. GEN-ERAL PROPERTIES. 379 Denote the quotient by Q and the remainder, if there be one, by R; we then have the identity f(x)=Q{x-r)-{-K By hypothesis, / (x) vanishes when x = r; and since Q is a rational integral function of x, it cannot become infinite when x = r; hence Q{x — r) vanishes when x=r. Therefore E vanishes when x = r. But R does not contain x', hence it van ishes without regard to the value of x. ^SH. If f{x) is divisible hy x — r, then r is a root of the equation f{x) = 0. Let Q denote the quotient obtained by dividing f{x) by x — r; we shall then have the identity f{x) = (i{x-r). Now Q{x — r) vanishes when x = r; hence f(x) vanishes when X = r. It therefore follows that r is a root of the equation f{x)=0 {5S5). 588. If the equation f{x) = is of the n^ degree, it has n roots, and no more. Let a represent a root of the equation f{x)=0 . . . (1); then f{x) is divisible hjx — a (586). The quotient obtained by dividing f(x) hy x — a will be of the (n — iy^ degree. Denoting the quotient by fi(x)y (1) may be written {x-a)f,(x)=0 . . . (2). Again, let b represent a root of the equation Mx)=0 . . . (3), which is of the (n — iy^ degree; then f^ix) is divisible by x — b. The quotient obtained by dividing /i (a;) hj x — b will be of the {n — 2y^ degree. Denoting this quotient by f^ix), (2) may be written {x-a){x-h)f,{x) = . . . (4). 380 THEORY OF equatio:n"S. By continuing this process, f{x) will ultimately be resolved into n binomial factors, x—a,x—h, x—CyX—d, . . . ., (x—k), x—l. r. f(x) = {x-a) {x-b) (x-c) (x-d) .... (x-k) (x-l) . . . (5). Now f{x) vanishes when x is equal to any one of the n quan- tities a, b, c, df . . . . Tc, l\ hence f{x) = has n roots. This equation has no more than n roots, for if we ascribe to a; a value m which is not one of the n values a, b, c, d, . . , . h, I, the value of / {x) becomes {m —a){m— b) {m ~c)(m—d).,. (m — k) {m — /) , which is not zero, because each factor is different from zero. Cor. — K a is a root of the equation f(x) = 0, then f{x)=z{x — (f)ft{^), where ft{x) is one degree lower than f{x) ; hence the remaining roots of the equation f{x) = can be found if we can solve the equation f^ (x) = 0. In like man- ner, if a and b are roots of the equation / {x) = 0, then f{x) = [x — a) (x — b)f^{x) ; hence the remaining roots of the equation f{x) =0 can be found if we can solve the equation /jj {x) = 0, which is . two degrees lower than the equation /{x) = 0. 589. To find an eqiiation when its roots are given. Let a, b, c, df . . . . k be the n roots of an equation ; then (x—a) (x—b) (x—c) (x—d) .... (x—k) = will be the equation required; for each of the n quantities, a, bf c,d,.... k is a root of this equation, and it has no other roots. MULTIPLICATION" BY DETACHED COEFFICIENTS. 590. To multiply a rational integral function of x by X ±_a, by means of detached coefficients. JEXAMPZBS. 1. Multiply a^ -\- ox^ — Gx + 4: hj X — d. Since the coefficients of the product do not depend upon x, the product may be found as follows : GEXEBAL PROPERTIES. 381 1 + 5 — 6 + 4 Detached coefficients of multiplicand. 1 — 3 " « « multiplier. 1 + 5 — 6 + 4 Detached coefficients of 1st partial product. _ 3 _ 15 -^- 18 — 12 « " " 2d " " 1+2 — 21 + 22 — 12 Detached coefficients of product. Hence the product is a^-\-2a:^ — 21x^ + 22^; — 12. Since the coefficients of the first partial product are identical with those of the multiplicand, this operation may be still further abridged as follows : 1 + 5— 6+ 4 Detached coefficients of multiplicand. _3_15 + 18_12 1 + 2—21 + 22—12 « " "product. Multiplying 1, the coefficient of the first term of the multi- plicand, by — 3, and adding the product to 5, we obtain 2; multi- plying 5 by — 3, and adding the product to — 6, we obtain — 21 ; multiplying — 6 by — 3, and adding the product to 4, we obtain 22 ; and multiplying 4 by — 3, we obtain — 12. When multipUcation is performed in this way, the terms should be arranged according to the powers of x ; and if a term is want- ing, its place should be ffiled with a cipher. 2. Multiply 7^-\-Qx^ ^ 5a; — 10 by a; — 5. 5 1 + + 6+ 5 — 10 __ 5 _. — 30 — 25 + 50 1 _ 5 + 6 — 25 — 35 + 50 Hence the product is a;^ — 5a;^ + Q>x^ — %bx^ — 35a; + 50. 3. Multiply 2:5 _ 4^ ^ 6a:2 — 8a; + 15 ^y a: + 8. 8 1 + 0-4+6—8 + 15 + 8 + — 32 + 48 — 64 + 120 1 + 8 — 4 — 26 + 40 — 49 + 120 Product, a;6 + 8a;5 — 4a;4 — 26a;3 _|_ 40a;2 _ 4Qx + 120. 4. Multiply a;7 — 4ar3 + 6a; — 7 by a; + 3. Ans. x^ + 3a;7 — 4a;* — 12a;3 + 6a?J + 11a; — 21. 382 THEORY OF EQUATIOi^S. DIVISION BY DETACHED COEFFICIENTS. 591. To divide a rational integral function of x by X ± «j ^1 means of detached coefficients. EXAMPIjEa. 1. Divide a^s — 9a:2 ^ 26a: — 24 by x — 4. Since the coefficients of the quotient do not depend upon Xy the quotient may be found as follows: Coeflfic'ts of dividend, 1—9 + 26— 24[1— 4 Coefficients of divisor. 1—4 1—5 + 6 " "quotient. —5 + 26 —5 + 20 6—24 6-24 Hence the quotient is ^ — hx-\-^. This operation may be still further abridged as follows : 1-9 + 26-24 -4 . . . (A). _ 4 + 20 — 24 1 — 5 4- G + The coefficient of the first term of the quotient is evidently 1 Multiplying 1, the first coefficient in the dividend, by — 4, and subtracting the product from — 9, we obtain — 5, which is the second coefficient in the quotient ; multiplying — 5 by — 4, and subtracting the product from 26, we obtain 6, which is the third coefficient in the quotient ; and multiplying 6 by — 4, and sub- tracting the product from — 24, we obtain 0. "We may substitute addition for subtraction in (A), if we mul- tiply by + 4 ; thus, l_-94.26-24 4 . . . (B). 1_5-|. 6+ GEN-ERAL PROPERTIES. 383 When division is performed by means of detached coefficients, the terms should be arranged according to the powers of x ; and if a term is wanting, its place should be filled with a cipher. The process used in (B) is called Synthetic Division. 2. Divide x^ — ^a? — Ibx^ + 49a; — 12 by sr — 5. l_3_15_j.49— 12 4.54-10—25 + 120 1 + 2— 5 + 24 + 108 Hence the quotient is a:^ 4. ^^^ __ 5^; 4. 24, and the remain- der is 108. 3. Divide x* — Sx^ — \lx^ + 198a; — 360 by cc — 7. 1 _ 8 — 11 + 198 — 360 1 7 4 7 — 7 — 126 + 504 1 1 _ 1 _ 18 + 72 + 144 Hence the quotient is o? — a? — IHx -\- T2, and the remain- der is 144. 4. Divide o^ + bx^ + ^x — % by a; + 4. 1 + 5 + 2 — 8 _4_4+8 — 4 1 +l_2+0 Hence the quotient is a;2 + a; — 2. 593. GENEBAZ EXAMPLES, 1. Show that 1 is a root of the equation a;3 4. 3a:2 _ iq^ + 12 = 0. That the first member of this equation is divisible by a; — 1 may be proved as follows : 1 + 3-16 + 12 414 4-12 1(591); 1 4 4 — 12 + hence 1 is a root (587). 384 THEOEY OF EQUATIONS. 2. Show that 3 is a root of the equation t^ — lOa^ + 35a.'2 — 50a; + 24 = 0. 3. Show that + 12a; + 35 = 0. 7 is a root of the equation a;* + 22;^ — Z\x^ 4. Show that — 1 and — 2 are roots of the equation a;^ — 4a;^ + 22a;3 _ 25a; - 42 = 0. 5. Show that 1 + V— 5 and 1 — \/- equation a;* _ 2a;8 ^ ^^ ^ 10a; — 30 = 0. 5 are roots of the 6. One root of the equation a;^ + 5a;2 _|_ 2a; _ 8 = is 1 ; what are the other roots? 1+5+2-8 +1+6+8 1-5 — 1 _ 7-1-29 + 30 + 6 + 1—30 1 — 6 — 2 _ 1 + 30 + + 16-30 1+6+8+0 a^J + Ga; + 8 = {^^^^ CoR.) ; whence, a; = — 2 or — 4. 7. Two roots of the equation a;*— 5a;3— 7a;2^29a;+30=0 are — 1 and — 2 : what are the other roots ? — 1 — 2 1_8 +15+ a;2 _ 8a; + 15 = ; whence, a; = 3 or 5. 8. Three roots of the equation a:5_4^^22a;2— 25a;— 42 =0 are — 1, — 2, 3 ; what are the other roots ? Ans. 2 + V^^, 2 — V"^^. 9. Two roots of the equation 7^ — Zx^ — \x^ -\- 30a; — 36 = are 2 and — 3 ; what are the other roots ? Ans. 2 + V^^, 2 — V^^. 10. One root of the equation 7? — \ = ^ is 1 ; what are the other roots? Ans. i(— 1 ± a/^^). 2 1 —1 + 2 -2 4 1 +1 + 4 — 2 + 4 -8 GEITERAL PKOPERTIES. 385 11. Find the equation whose roots are 1, — 2, — 4. The required equation is {x^l)[x-{-2)][x-{-4.)] = {x-l){x + 2){x + ^)=:0. The indicated multiplication may be performed as follows: (590); 1+5 -1-2—8 hence a^ -\- bx"^ -\- 2x — S = is the required equation in its simplest form. 12. Find the eouation whose roots are 3, — 2, — 1, 5. Ans. x^ — 6x^ — W + 2dx-{-dO= 0. 13. Find the equation whose roots are 1 + V"^, 1 — V— 5, //5, __ //s. Ans. x^—2a^-{-x^-{- 10a; — 30 = 0. 14. Find the equation whose roots are —1, —2, 3, 2 + V— 3, 2 _ V^Is. Ans. x^ — 4:X^ + 22:^2 _ 25a; — 42 = 0. 15. Find the equation whose roots are a, b, c. Ans. 7^ — {a ■\- b -\- c) x^ -\- (ab + ac -{- bc)x— abc = 0. 16. Find the equation whose roots are a, b, c, d. Ans. x!^—(a-\-b-\-c + d)x^-\-{ab-\-ac-\-ad-\-bc-\-bd-\-cd)a? ■^ (abc -^abd-\-acd-\-bcd)x-\- abed :=0. 593. To find the relation between the coefficients off{x) and the roots of the equation / (x) =0. Suppose the terms of f(x) to be arranged according to the descending powers of x and that the coefficient of the first term is 1; then 1. The coefficient of the second term with its sign changed is equal to the sum of the roots (593, 15, 16) ; 2. The coefficient of the third term is equal to the sum of the products of the roots, taken two and two j 25 886 THEORY OF EQUATIONS. 3. TJie coefficient of the fourth term with its sign changed is equal to the sum of the products of the rootSytahen three and three; and 80 on. 4. If the degree of the equation is even^ the absolute term is equal to the product of all the roots. If the degree of the cquaimi is odd, the absolute term with its sign changed is equal to the product of all the roots. By a method similar to that employed in Art. 472 it may be proved that these laws are true universally. Cor. 1. — If the roots of f(x) = are all negative, each term of f{x) is positive. Cob. 2. — If the roots of f(x) = are all positive, the signs of the terras of /(a;) will be alternately + and — . Cor. 3. — If the second term of f{x) does not appear, the sum of the roots of the equation / (x) = is equal to zero. Thus, the sum of the roots of tlie equation a;^ — 2a; -f 4 = is zero. Cor. 4. — K f(x) has no absolute term, at least one of the roots of f(x) =0 is zero. Thus, one root of the equation a^ — 2ci^ + 3x = 0ia0. Cor. 5. — The absolute term of f{x) is divisible by each root of the equation f{x) = 0. Cor. 6. — Let a, b, c, d, . . . . I denote the roots of the equa- tion a^+Aa;"-i + Ba:«-?+ 4-Ka: + L = 0; then — A = « + Z> + c + 6/+ + I, and B^ab -\- ac + ....-{- bd + be -^ . . . . ; whence, A^ — 2B = a^ -}- b^ -{- c^ + d^ -\- + P; that is, A2 — 2B is equal to the sum of the squares of the roots of the proposed equation. Hence, if A^ — 2B is negative, the roots of the equation cannot be all real. Thus, the roots of the equation a^ — 4:X^ -\- 22a^ — 25x — 42 = are not all real, for (—4)2 — 2x22 is negative. GEKEEAL PROPERTIES. '38? 594. An equation ivhose coefficie7its are integers, that of its first term being unity, cannot have a root which is a rational fraction. Let the equation be x^ + A:k^-i + Ba:«-2 +....4-Ka; + L = . . . (1), in which the coefficients A, B, . . . . K, L are supposed to be in- tegers. Suppose, if possible, that (1) has a rational fractional root which in its lowest terms is expressed by t- Substituting t for x in (1), and multiplying the resulting equation by h"~\ wq obtain J + Aa«-i + Ba«-2^> + .... 4- KaZ>"-2 + L2»«-i = . . . (2) ; whence, j = - (Aa"-i + B«"-2J +....+ K«5«-2 + LJ^-i) . . . (3). The second member of (3) is an integer, and its first member is an irreducible fraction. Hence j- cannot be a root of the pro- posed equation. 595. If a-^bV^ 1 is a root of an equation whose coeffi- cients are real, then loill a — bs/—\ be a root of that equation. Let a -\- b^/ —1 be a root of the equation ^n _|. ^^n-1 j^ Ba;"-2 + . . . . H- Krc + L = . . . (1), in which the coefficients are supposed to be real, then will rt — Z> V— 1 be a root of that equation. Since a -{• b ^/ — I is a root of (1), {a JrbV-'lT+ ^{a + b A/-"ir' + B(« 4- h ^-1)"''+ + K(« + Z>V^1) + L = . . . (2). If we expand those terms of (2) which contain a -\- b a/— 1, the resulting equation will contain some terms which are real and 388 THEOEY OF EQUATIONS. 6ome wliich are imaginary. Since the coefficients A, B, C, . . . , and the even powers of ^V— 1 are real, it follows that V— 1 will occur only in connection with the odd powers of h. Denoting the snm of the real terms by P, and the sum of the imaginary terms by Q V— !> we have P + QV^=1 = . . . (3); whence, P = — Q \/^^ . . . (4). To satisfy (4) we must have P = and Q = 0, for a real quantity cannot be equal to an imaginary quantity. Now if a — bv^—1 be substituted for x in (1), its first mem- ber, when expanded, will differ from the result obtained by ex- panding the first member of (2) only in the sign of the odd powers of b V— 1 ; that is, the first member of (1) may be represented by P — QV— 1 when a — JV— 1 is substituted for a:. But P = and Q = 0; P-QV~1 = . . . (5). Therefore a — bV— I is a root of (1). CoE. 1. — An equation of an odd degree whose coeflBcients are real has at least one real root. CoE. 2. — The product of the two roots a -{- b V~— 1 and a — b V^l is a^ + W, which is a real positive quantity ; hence, an equation of an even degree whose coefficients are real, and whose absolute term is negative, must have at least two real roots. CoE. 3.— The product o{ x—{a-\-bV^^^ and x—{a—bV-i) is (x — ay + Z^, which is a rational quadratic expression, and positive for all real yalues of x. Cor. 4. — If a + v^, in which V^ is a simple quadratic surd, is a root of an equation whose coefficients are rational, then will a — Vb be a root of that equation. TEANSFOEMATION OF EQUATIONS. 389 EXAMPLES, 1. 1 _ 2 V^^i is a root of the equation a:^— :r'^ + 3:c + 5=0; what are the other roots? Ans. — 1 and 1 + 2 ^/ — 1. 2. V— 1 is a root of the equation a;* + 4a:3H-ea;'^ + 4a; + 5 = 0; wiiat are the other roots ? 3. 3 + a/— 2 is a root of the equation x^-{-a^—2bx^-\-A:lx-{-QQ =0; what are the other roots? 4. V^ is a root of the equation a;^ + 2a;3— 4ic2_4a; + 4=0 ; what are the other roots ? 5. 2 + V3 is a root of the equation a:*— 2a;3— 5.^2_6:c + 2=0 ; what are the other roots ? 6. a/3 and 1 — 2 a/— 1 are roots of the equation a^—a^-\- 8a;2— 9x— 15=0; what are the other roots? 7. Has the equation a;^— 2a; 4-4=0 a real root? Why? 8. Has the equation a."*— 4a;2-f 4.c— 1=0 any real roots? Why? TRANSFORMATION OF EQUATIONS. 596. To transform an equation into another, the roots of which shall be those of the proposed equar tion with contrary signs. Let r represent a root of the equation a;" + Aa;"-i + B.^"-2 + Qx""-^ +....= . . . (1) ; then r» + Ar«-i + Br«-2 + Cr"-^ -f ....== . . . (2). Changing the signs of (2), __ ,.n _ Ar"-i — Br"-2 — Cr"-^ . . . . = . . . (3). Changing the signs of the alternate terms of (1), ^« _ A^«-l _|_ I3a,n-2 _ Q^n-Z _|_ _ _ ,33 . . . (4). 390 THEORY OF EQUATIONS. Substituting — r for Xy the first member of (4) becomes /•« 4- Ar"-i + Br" -2 _|_ Cr"-3 +...., or — r« — Ar"-i — B?'«-2 — C?'»-3 _ . . . . according as n is ere;i or orftZ. But, by (2) and (3), each of these expressions is equal to zero ; hence — r is a root of (4). Since — r is a root of (4), it is a root of the equation obtained by changing all the signs of (4) ; that is, — r is a root of the equation — a;" + Aa;"-i — Ba;«-2 + Ca;~-3 —....= . . . (5). Hence, If the signs of the alternate terms of a complete equation be changed, the signs of all the roots will he changed. An incomplete equation may be rendered complete by insert- ing the missing temis, with zero for the coefficient of each of them. Thus, by inserting Ooc/^ and Oai^, the equation a^ -\-Zx^ — 4^3 ^ 4a; + 7 = becomes 3^ + ^3^-\-0x^—^-\-0x^—^x-\-^=Q' EXAMPLES, 1. The roots of the equation a^ — Ix^ -\- I'dx — 3 = are 3, 2 + ^/3, and 2 — V3; find the equation whose roots are _ 3, _ 2 — V3, and — 2 + V3. Ans. a^ + 7a;2 4- 13a; + 3 = 0. 2. The roots of the equation a;^ _ 3^ ^ 3^2 _^ 17^ _ 18 == are 1, — 2, 2 + V— 5, and 2 — a/— 5 ; what are the roots of the equation x^ -\- ^3^ -\- Zx^ — llx — 1% = {)'i Ans. — 1, 2, — 2 — V~-^, — 2 + V^^. 3. The roots of the equation a;* + 4a;3 — a;^ — 16a; — 12 = are 2, — 1, — 2, and — 3 ; what are the roots of the equation — a;* + 4a;3 ^ ^2 _ 16a; + 12 = ? Ans. — 2, 1, 2, 3. 4. The roots of the equation a;^ — 1 = are 1, i(— 1 -f V— 3), and ■|-(— 1 — V— 3); what are the roots of the equation Ans, -1, -^(-1 + V^, -iC-l-V^^^^). TKANSFOEMATION OF EQUATIOlirS. 391 597. To transform an equation containing frac- tional coefficients into another in -which the coefficients shall be integers, that of the first term being unity. Let the proposed equation be a;« 4- A^«-i + B:c«-2 + + Kx-{-h:-0 . . . (1), in wliicli some or all of the coefficxnts A, B, C, . . . . are sup- posed to be fractional. Assume y — kx, or :?; = "-. Substituting | for x in (1) and multiplying the resulting equation by ^", ?/« 4- A/j?/"-! + B7cY~^ + + K^-"-i?/ + W = , . . (2). Now, since 7c is arbitrary, we may give it such a value as will make the coefficients Ak, Bk\ .... Kk''-\ Lk"' integers. EXAiUrLES. Transform each of the following equations into another in whicli the coefficients shall be entire, that of the first term being unity: 1. ^ + |.. + |,. + ^,.Hu.| = o . . . (1). Substituting j for x in (1) and multiplying ^.he resulting equation by h\ ah , ck^ - ck^ qlc^ ^ . ,^. y' + Tf + -ay+-fy+-h=^ ■ ■ ■ (^)- Assuming h = bdfh, (2) becomes yi + adfhif + cWphhf-^eW?ph^y+gVd^fh^=0 . . . (3). 2. ^+"^V- + - = . . . (1). pm m p ^ Substituting j for x in (1) and multiplying the resulting equation by k^, y. + ^y^ + ^^i + ?E = o . . . (2). ^ pm m p ^ ^ 392 THEORY OF EQUATIONS. Assuming Jc=pn, the L. C. M. of the denommators, (2) becomes ys + o?/2 4- Ip^my + cj^^ ~Q . . , (3). 3. ^_§^ + ^.,__J_.__H_ = o . . . (1). Substituting y^ for a; in (1) and multiplying the resulting equation by k^, ^--6-2^+12^-150^-9000 = ^ • • • (^)- Kesolving the denominators in (2) into prime factors, we have 6=2x3, 12=22x3, 150=2x3x52, 9000=23x32x53. Assuming ^' = 2 x 3 x 5, (2) becomes 5-2'3-5 3 5-22-32-52 ^ 7-2«'38-53 13-2^-3^ -5^ ^ 2-3 ^ "^ 22-3 ^ 2-3-52 ^ 23-32-5~3 = . . . (3). Canceling common factors in (3), j^_5-5z^-5-3-52y2- 7-22-32-5y-13-2-32-5 =0; that is, 7/4— 25y3^375y2_i260^— 1170=0 ... (4). If we had assumed k = 9000, which is the L. C. M. of the denominators of the given equation, the coefficients in the trans- formed equation would have been much larger than those in (4). ^ ^-35-^2450"^ 68G0O~ Ans. if — 6?/2 + 26y — 85 = 0. K n 13^ . 21 , 32 2 43 1 . ^' ^-12^ + 40^-225^'-600^-800 = ^- ^725. 2/5— 65y* + 1890y3_30720/-928800^— 972000=0. = 0. Ans, if — 14^2 ^ \\y _ 75 _ 0. '- ^-V^l^-% = '- TRANSFOEMATIOK OF EQUATIOITS. 393 598. To transform an equation into another, the roots of which shall differ from those of the given equation by a given quantity. Let the proposed equation be .T" + A:r"-i 4- Bx""-^ + 4-Ka; + L = . . . (1). Substituting y -\- h for x in (1), we have (y4_7,)n + A(?/ + /i)"-i + B(2/H-/i)"-2 + ... + K(?/ + 70 + L=:O...(2). Expanding and reducing, (2) becomes y^ + nh r~' + 2 yn-^^ + h'' + (^^ — 1) A^ + Ah""-^ -f B + B/i«-2 I = ... (3) + ... + L The roots of (3) differ from those of (1) by h, forx = y + h. Denoting the coefficient of r^-^ by A', that of ^""^ by B', .... , and the independent term by L', (3) becomes ^n^Ay-i + By'-2 + ....+jy + K'2^ + L' = o . . . (4). We now propose to show that (4) may be deduced from (1) by Synthetic Divismi. Restoring the vahie of ?/, (4) becomes + L' = . . . (5). Now the first member of (5) is identical with the first member of (I); for, in deducing (5) from (4) we merely retraced the steps by which (4) was deduced from (1). Hence the equation a;«H- Aa:«-i+ B.c«-2-f-....+J.'c2 4.Ka; + L== (a:— 70~ + A'(a:—7i)«-i + B\x-UY-^-\- . , . , +;5'(x-hf-\-^{x-U) + U • • • (6) is an identity. 394 THEORY OF EQUATIONS. Dividing the second member of (6) by a; — //, we obtain the remainder L'; dividing the quotient by x —h, we obtain the re- mainder K'; dividing the second quotient by a: — //, we obtain the remainder J' ; and so on ; hence if we treat the first member of (G) or the first member of (1) in the same way, vv^e shall obtain the same remainders. But these successive remainders are the coefiBcients of (4). Hence the coefficients of (4) may be obtained from (1) by the following R ULE. Divide the first member of (1) by x — h, continuing the oper- ation until a remainder is obtained luhich is independent of x ; then divide the quotient by the same divisor, and so on, until n divisions have been performed : the successive remainders will be the coefficients of (4). EXAMPLES. 1. Find an equation whose roots are less by 2 than those of the equation a:* — 4^:3 — 8a; + 32 = 0. Substituting y + 2 for a; in this equation, we obtain y^ + 4^ — 24?/ = 0, which is the equation required. The same result may be obtained by Synthetic Divisiouy as follows : l_4+0— 8 + 32|2 + 2-4- 8 — 32 — 2-4- 16+0 1st rem. + 2 + 0- 8 + 0-4- 24 2d rem. + 2 + 4 + 2 + 3d rem. + 2 + 4 4th rem. Hence the required equation is ?/^+ 4cy^-\- Oy^— 24y+ = 0. 2. Find an equation whose roots are greater by 3 than those of the equation x* + IGa^ + 99x^ + 228a: + 144 = 0, TRANSFOBMATIOISr OF EQUATIONS. 395 Substituting ?/ — 3 for x in this equation, we obtain y^ + 14^3 _^ Qy2 _ 42^ = 0. The same result may be obtained by Syn- thetic Division, as follows : 1 + 10 + 90 + 228 + 144 I - 3 _ 3 — 39 — 180 — 144 1 + 13 + CO + 48 + 1st rem. _ 3 — 30— 90 4- 10 + 30 — 42 2d rem. — 3-21 + 7 H- 9 3d rem. — _3 -f- 4 4th rem. Hence the required equation is y^ + 42/^ + 9y^ — 42?/ = 0. 3. Find an equation whose roots are greater by 2 than those of the equation x^ + 4cX^—24:X = 0. Ans. y*—iy^^Sy-\-32 = 0. 4. Find an equation whose roots are less by 3 than those of the equation a;* — 12:c3 + lUx^ _ 9a: + 7 == 0. Ajis. y^ - 37?/2 _ 123?/ — 110 = 0. 599. To transform an equation into another in which the second or third term shall not appear. Since h in equation (3) of Art. 598 is arbitrary, we may give to it such a value as will cause the second term of that equation to vanish. A Assume nh-{-A = 0; then h= . Substituting this value for h in (3), we obtain an equation of the form of yn ^ B'2/"-2 -f- Cy-3 + . . . . + K>y + L' = 0. If we assume ''(''- '^)^'' -\- (n - 1) Ah + B = 0, the third term of (3) will disappear. Cor.— The value of h which makes the second term disappear may cause the disappearance of the third or some other term. 396 THEORY OF EQUATIONS. In order that the third term may disappear at tho same time with the second, it is necessary that the value of h wliich satis- fies the equation nh + A = shall also satisfy the equation n (n - 1) h^ _^ (^ _ 1) A7i + B z= 0. Substituting - - for h . X1-. X- 1 w(/i — 1) A2 , ^,A2 ^ ^ in this equation, we have ,, • -^— (w — 1) t-B = 0; whence A^ = -. This equation expresses the relation which must subsist between the coefficients A and B in order that the third term may disappear with the second. EXAMPLES. 1. Transform the equation x^ — Q>x^ + Sx — 2 =i into an- other wanting the second term. Ans. y^ — 4ty — 2 = 0. 2. Transform the equation cc* — 12a;3 + IW — 9a: + 7 = into another wanting the second term. Ans. y^ — 311 y^ — 123y — 110 = 0. 3. Transform the equation a:^ — Gic^ -|- 13a; — 12 = into an- other wanting the second term. Aiis. y^ + y — 2 = 0. 4. Transform the equation 3^-\-bx^-\-^x — 1=0 into two others, each wanting the third term. 139 Ans. ?/3 _ 2^2 _ 5 _ and y^ + y^ ^r^ = 0. 5. Can the equation x^ + Ga:^ -f 12a; — oG = be transformed into another wanting the second and third terms ? THEOREM OP DESCARTES. 600. In any series of quantities a pair of consecutive like signs is called a JPertnaneuce of signs, and a pair of consecu- tive unhke signs is called a Variation of signs. Thus, in the expression x^ — S-c^ _ 4a:« + 7a;5 _j. 3^^ 2:x^— x^— a; + 1, there are four permanences and four variations. 601. If the equation f{x)=0 is complete, the sum of the num- ber of permanences and the number of variations in the signs of the terms of /(a;) is equal to the greatest exponent of a; in the equation. THEOREM OF DESCAETES. 397 602. TJieorem of Descartes, — 77ie mimber of real pos- itive roots of the equation f(x) = cannot exceed the numher of variations in the signs of its terms ; and, if the equation f{x) — {) is complete, the number of real negative roots cannot exceed the numher of permanences in the signs of its terms. Eepresent the real positive roots of the equation f{x)=Q ... (1) by «, 5, c . . . . , and suppose (1) to be divided by the product of all the factors x — a, x — h, x — c, . . . . corresponding to the real positive roots (586). Eepresent the resulting equation by /,(i) = . . . (2). This equation has no real positive roots. We shall now show that if (2) be multiplied by the factor X — a corresponding to a real positive root, the number of varia- tions of the resulting equation will be at least one greater than in (2). I. Suppose (2) to be complete, and let the signs of its terms be The signs of the multiplier are The signs of the product are + ± — T T H . A double sign is placed where the sign of any term in the product is ambiguous. Now, taking the ambiguous signs as we please, the number of variations in the product is greater than in the multiplicand; and this is still true if we suppose some or all of the terms having ambiguous signs to vanish. 11. If (2) is incomplete, reduce it to a complete form by in- serting the missing terms with zero for the coefficient of each ; the resulting equation will contain at least as many variations as (2). Multiplying the completed equation by x — a, the number of variations in the product will be greater than in the multipli- cand (I). But the product thus obtained is the same as the pro- + + - + -. +• + + - — +. + + + -. 398 THEORY OF EQUATIONS. duct of /j (x) and x — a; hence, the number of variations in the product of f^{x) and a; — a is greater than in fi{x). We have thus shown that when the factor x — a is introduced into (2), the resulting equation contains at least one more varia- tion than (2). In like manner it may be shown that when the factor a; — ^ is introduced into the resulting equation, at least one more variation is introduced ; and so on. Hence the number of real positive roots of the equation f(x) = cannot exceed the number of variations in the signs of its terms. We prove the second part of the theorem as follows : Suppose (1) to be complete, and let the signs of its alternate terms be changed ; then the signs of the roots will be changed (596), the permanences will become variations, and the varia- tions will become permanences. But the number of real positive roots of the resulting equation cannot exceed the number of vari- ations in the signs of its terms ; hence the number of real negative roots of the given equation cannot exceed the number of perma- nences in the signs of its terms. Cob. 1. — Whether the equation f{x) =0 is complete or not, its roots are numerically equal to those of the equation/(— a:) =0 ; but the signs of the two sets of roots are opposite. Hence the number of real negative roots of the equation /{x) = is equal to the number of real positive roots of the equation f (— x) = 0. But the number of real positive roots of the equation /(— x)^0 cannot exceed the number of variations in the signs of its terms. We may therefore state the theorem of Descartes as follows • The number of real positive roots of the equation f(x)=zO camiot exceed the nuniber of variations in the signs of f {x), and the number of its real negative roots cannot exceed the number of variations in the signs of f{— x). Elustratio7i.—The equation a^ -\- 3a^ -\- 5x — 7 = has only one variation of signs ; therefore it cannot have more than one real positive root. By putting — x in the place of x, we obtain the equation x^ -{- 3x^ — 5x — If = 0. This equation has only one variation of signs; therefore it cannot have more than one THEOREM OF DESCAETES. 399 real positive root ; lience the original equation cannot have more than one real negative root. Cor. 2.— If the equation f{x) = is complete, and all its roots are real, the number of positive roots is equal to the number of variations in the signs of its terms, and the number of negative roots is equal to the number of permanences in the signs of its terms. Denoting the number of permanences by p, the number of va- riations by V, the number of positive roots by P, the number of negative roots by N, and the highest exponent of x in f{x) by n, we have v -\- p = n and P + N = ?i ; hence v + j9 = P + N. Now P cannot exceed v, and N cannot exceed p ; hence P = v, and N ^= p. Cor. 3. — By means of the theorem of Descartes we can sometimes detect the presence of imaginary roots in an equa- tion. Illustration. — The equation a;^ + 16 = has no variation of signs ; therefore it has no real positive root. By putting — a; in the place of x, we obtain the equation x^ -f 16 = 0. This equa- tion has no variation of signs ; therefore it has no real positive root; hence the original equation has no real negative root. Therefore the roots of the equation a;^ -f 16 = are imaginary. EXAMPLES. 1. Show that the equation x^ -{- 5x -{- IS = has only one real root. 2. All the roots of the equation x^ -\- 5x^ -{- 2x — S = are real; how many of them are negative? Ans. Two. 3. All the roots of the equation x^ — ^x'^—llx^ + 29x +30=0 are real ; how many of them are positive ? Ans. Two. 4. All the roots of the equation x^— 3x^^—50^ -{-16x^-}-^x— 12 = are real ; how many of them are positive ? Ans. Three. 400 THEORY OF EQUATIONS. DERIVED FUNCTIONS. 603. Substituting x -{- h for x in the identity f{x) = a;» + A2;"-i + Ba:'»-2 + + Kic + L . (1), and arranging the result according to the ascending powers of /^, we obtain f{x + h) = x" n3f*~^ + K h H- n{n—l)x**~^ + (n— l)(?i— 2)Aa:«-3 (2). 7^2 Denoting the coefficient of h by /'(:z^), that of --— by f"{x)j and so on, (2) may be written f{x + h)^f{x)+f(x)h+r{x)'-^+f'\x) ^+ . . . . (3). The expression f{x) is called the primitive fimction, the expressiony'(.r) is called the Jirst derived function, or simply the first derivative, i\\Q, expression /'(a:) is called the second deriva- tive, and so on. The first derivative may be obtained from the primitive func- tion by multipl}ing each of its terms by the exponent of x in that term and dividing the result by a;; the second derivative may be obtained from the first in the same way that the first is obtained from the primitive function ; and so on. EXAMPLES. 1. Find the derivatives of a;^ _ q^z ^ Sa; — 2. ( 1st. Sx^ — Ux-}- 8, Ans. I 2d. (jx — 12, ( 3d. 6. DERIVED FUNCTIONS. 4:01 2. Transfoiin the equation a^ — Gx^ -\- Sx — 2 =zO into another wanting the second term. Substituting x-\-2 for x in the identity/ (2-) =:a;2 — G.^2_|-8:c— 2, we have 22 23 f{x-^2) = x^-6a^-\-Sx—2-^{3xi—12x + 8y2-{-(6x—12)^-\-6~ — a^ — 4:X — 2; hence the required equation is a^ — ix — 2 = 0. 3. Find an equation whose roots are less by 1 than those of the equation x^ — 2^:2 _|. 3^ — 4 = 0. Substituting x-^1 for x in the identity/(a:)=a:3 — 20^24- 3a; — 4, we have /(rr + 1) =a;3_22;2-f3a;—4 + (3a;2—4a; + 3)l + (6a;— 4)^ + 6^ = x^ -{- x^ -\-2x — 2; hence the required equation is xi-^x^-\-2x — 2 = 0. Let the student solve all the examples of Art. 599 by the method of derived functions. 604. TJie first derivative of the product of tivo ftmctions of the same quantity is equal to the sum of the products obtained hy multiplying each hy the first derivative of the other. Substituting x -\- h for x in the two expressions f{x) and /i (^)> w® obtain f(x-\-h)= f{a)-\-f{x)h+ (1) (603), and fr{^+h)=f^{x)-{-f^{x)h-\- (2). Multiplying (1) by (2), f{x + h)f^{x+n) =f{x)f,{x)+f,(x)f\x)h+f{x)f^{x)h + .... . . . (3). The coefficient of h in (3) is the sum of the products obtained .by multiplying /^ {x) by the first derivative of / {x) and / {x) by the first derivative 0^ f^(x) and this coefficient is the first deriva- tive of /(2;)/i (2:) (603). 26 402 THEORY OF EQUATIONS. Cor. — In like manner it may be shown that the first deriva- tive of the product of three or more functions of the same quan- tity is equal to the sum of the products obtained by multiplying the first derivative of each by the product of the other functions. EXAMrZES. Find the first derivative of each of the following expressions : 1. x^ (x — ay. Ans. 2x (x — af -\- 3{x — a)- x\ 2. (a -^ x){l) + x). Ans. J)-^x+a-\-x = a-\-h + 'ilx, 3. {x-aY (x-by. Ans. 2{x-a){x-bY-i-.S(x—by {x-ay. 4. {x — ay{x — by(x — cy. Ans. 3(x—ay (x^bf (x—cy+4:(x—bf {x—ay {x—cf + b{x—cy(x—ay {x—by. 5. {x — ay {x — b)"". Ans, n (x—ay~^ {x—b)^ + m(x-'by"~^ {x—ay. ROOTS COMMON TO TWO EQUATIONS. 605. If a is a root of the equation f{x) = 0, f{x) is divisible by X — a, and if a is a root of the equation f^ix) = 0, fi{x) is divisible by x — a; hence the roots of the equation obtained by putting the G. C. D. of f(x) and f^ix) equal to zero will be the roots common to the two equations f{x) = and fi{x) = 0. EXAMPLJES. 1. Find the root which is common to the two equations X^—2x^—7x^ + 20x—V2=0 and A3^—6x^—Ux + 20=0. The G. C. D. of the first members of these equations is x—2; hence 2 is a root common to the given equations. 2. Find the roots common to the two equations a;4— 22:3— llic2 4-122; + 36=0 and ^a^—6x^—22x-\-12=0. Ans. 3 and —2. EQUAL ROOTS. 403 3. Find the roots common to the two equations x'^—Sx^-\-c(P—4^- + 12x—4:=zO and 2x^—Qx^ + 3x^—3x-^l=0. Ans. ^r . 4. How many roots are common to the two equations a^—4,Q:^^e^x^-\-10a^—26x—4:=0 and 2x^—18x^-\-d9x^—2bx^ + a; + 1 = ? Ans. Three. EQUAL ROOTS. 606. The equation f{x)=0 is called the Primitive Equa- tion, and the equation f'{z) = 0, which is obtained by putting the first derivative of f{x) equal to zero, is called the First Derived Equation, 607. If a root occurs n times in the equation f{x) = 0, it will occur n — 1 times i7i the equation f(x) = 0. Let the proposed equation be f(x) = {x-ay{x-b)(x-c),,,.=0 . . . (1), in which a occurs as a root n times. The first derivative of each of the factors x — b, x — c, . , , , is 1 ; hence the first derived equation is f'{x) =n{x — «)"~^ {x — i) (x — c) . . . . -\- {x — aY{x — c) . . . . ^ {x-aY{x-b) .... -\- = Q . . . (2), in which a occurs as a root n — 1 times (587). Cor. — A root which occurs only once in (1) does not occur in (2) ; hence any root which is common to (1) and (2) is one of the equal roots of (1). Find all the roots of each of the following equations : 1. f(x)—x^—W-\-Ux — VZ = 0. f\x) = 3xi — Ux + 16 = 0. The G. C. D. of f(x) and f'{x) is X — 2. Putting this equal to zero, we have x — 2 nr ; whence x = 2. The given equation, therefore, has two roots equal to 2. The remaining root of the given equation may be found by the principle of Art. 588^ Cor. 404 THEORY OF EQUATIOKS. 2. x^- ll3^ + Ux^ - 76a; + 48 = 0. Ans, 2, 2, 3, 4. 3. 2x^^123^ -\-ldxi—Gx+ 9=0. Ans. 3, 3, ± \/ - )- 2 4. 7^ — 2x^ + ^3?— Ix^ -\-^x—^ = 0. Ans. 1, 1, 1, -i±|A/^^ni. 6. /(a;)=a;7— 9a:5^C2:4 + 15a;8— 12a?2-7a;+6=0 . . . (1). The first derived equation is f'{x) = W-^ 45a:* + 24a;8 + 45a:3 _ 24a; — 7 = . . . (2). The G. C. D. of f{x) and f'{x) is a^-x^-x + \. Equat- ing this with zero, we have a^-x^^x-\-\=0 . . . (3). The G. C. D. of the first member of (3) and its first derivative is a; — 1. Equating this with zero, we find :?; = 1 ; hence (3) has two roots equal to 1. The remaining root of (3) is — 1 (593). Now, since (3) has two roots equal to 1, and one root equal to — 1, (1) must have three roots equal to 1 and two roots equal to — 1. Dividing f(x) by {x — 1)8 (a; + 1)^ we obtain x^-\-x—(j. Equating this with zero, we have 7? + x — 6 = 0; whence, a; = 2 or — 3. The roots of (1) are therefore 1, 1, 1, — 1, — 1, 2, —3. 6. 0^ — 2x^ — 23? + 4.7? ■]-x — 2 = 0, 7. a;^ — 6a;* + 4a;3 + ^x^ — 12a; + 4 = 0. LIMITS OF THE ROOTS OF AN EQUATION. 608. If the coefficients of f{x) are real, and the results oh- tained Inj snhstituting p and qfor x in f{x) have Wee signs, the equation f(x) = has either no root or an even number of roots lying heticeen jp and q ; hut if the results have contrary signs, the equation has an odd number of roots lying between p and q. LIMITS or THE ROOTS OF AX EQUATIOIsT. 405 Let the real roots of the equation f{x) = 0.,.(l) be de- noted by «, by c, . . . . Jc, and let the quotient obtained by divid- ing /(re) by {x — a){x — b)(x — c) , . . . (x — k) he denoted by /iW; then f(x) = {x-a)ix-i){x-c) {x-k)Mx) . . . (2). Now, since /^ (x) is the product of all the factors correspond- ing to the imaginary roots of (1), and since the number of these imaginary roots is even, it follows that /^ (x) is positive for all real values of x (595, Cor. 3). Substituting jj and q, in succession, for x in (2), we have f(p) = {p-a){p-b){p-c) . . . . (p-Ic)Mp) . . . (3), /(?) = (?-«) (?-*)(?-<■•) ( t-/ > y ^ that is, a; > th • Li Hence p is an inferior limit of the positive roots of the given equation. EXAXPZES. Find an inferior limit of the positive roots in each of the fol- lowing equations : 1. a;5+ 5a;* + 2a:3_i4a;2_26a;+ 10 = 0. Substituting - for x and reducing, we have - 26 . 14 3 2 2 5 1 . 2^-l0 2^-10^ +10 2^' + io^+Io = ^- A superior limit of the positive roots of this equation is 3.6; hence ^- is an inferior limit of the positive roots of the given equation. stusm's theorem. 409 o 2H-V40 3. a:« — 5^:5 _|. ^4 + 12a:3 _ i^x^ 4-1 = 0. 4. 3^ — ^3^ + 12a;2 + IQx — 39 = 0. 611. To find the limits of the negative roots of an equation. Substitute — x for x in the given equation, and find the limits of the positive roots of the resulting equation. By changing the signs of these Hmits we obtain the limits of the negative roots of the given equation (602, Coe. 1). EXAMPLES. Find the limits of the negative roots in each of the following equations : 1. 3^ — ^X^ + bX-{-l = 0. ^W5. — (l + V7), — :^. 2. ic* — 15a?^ — lOz 4- 24 = 0. 3. 2;6__3a;5_|.2a;4 4_27a:3 — 4a;2— 1 = 0. STURM'S THEOREM. 612. If the coefficients of / {x) are real and the equation f{x) = has no equal roots, then, if x is made to assume, in suc- cession, all real values from — oo to + oo, the sign of/ (a;) will change as often as x passes a real root of the equation (608, Cor. 3). Sturm's Theorem enables us to determine the number of such changes of sign. 613. Stiirtn^s Functions.— h^i f{x)z=zO be an equa- tion whose coefficients are real, and which is freed from equal roots (607) ; and \etf{x) be the first derivative offix). We now apply to f(x) and f'{x) the process of finding their G. C. D. (125), with this modification, namely: 1. When a remainder is found which is of a lower degree than the correspond- ing dividend and divisor, ice change its sign and use the result for the next divisor. 2. We neither introduce nor reject a nega- tive factor in preparing for division. 410 THEORY OF EQUATIONS. We continue the operation until a remainder is obtained which is independent of a:, and change the sign of that remaiader also. Let /i(:r), f^(x), f^(x), f^ix) denote the series of modi- fied remainders thus obtained. The functions f{x), f\x), f,{x), f,(x), f,{x), . . . .^x) are called Sturm^s Function^. The functions f'(x), /^(.r), f^{x), f^(x), f^{x) are called Auxiliary Functions, 614. Stiirm^s Theorem, — If x he conceived to assume, in succession, all real values from — - oo to -f oo , there will be no change in the number of variations in the signs of the scries of functions f {x), f{x), fiix),f^(x), f^(x), f„(x)y except lohen X passes through a real root of the equation /(.r) = ; and when X passes through such a root, there will he a loss of only one varia- tion. I. fn{x) is not zero; for, by hypothesis, it is independent of a;; hence, if it were zero, / (x) and f'{x) would have a common di- visor, and the equation. /(a:) = w^ould have equal roots (607) ; but this is contrary to the hypothesis. II. Two consecutive functions cannot vanish for the same value of X, Let q^, q^, q^, . . . , qn denote the successive quotients ob- tained by performing the operations described in Art. 613 ; then, by the principles of division, /(x)=(7,.f(.r)-/,(x) /i(a^)=?3/s(^)-/3(*) f.-,(x)=q,f,-,{x)-f,{x) (2), (3), («). Now suppose f'{x) and fi{x) to vanish at the same time; then by (2) we shall have f2(x) = ; hence by (3), f^ix) = ; and so on ; that is, if two consecutive functions vanish at the same time, all the succeeding functions^ including f„{x) would vanish ; but this is impossible (I). 411 III. When any auxiliary function vanishes, the two adjacent functions have contrary signs. Thus, if f^^^) ■= 0, we have by (3),/.(.^) = -/3W- IV. No change can be made in the sign of any one of Sturm's functions, except when x passes through a vahie which causes that function to vanish (608, Cor. 3). V. Sturm's functions neither gain nor lose a variation of signs when X passes through a value which causes one or more of the aux- ihary functions to vanish, but which does not cause /(a;) to vanish. 1. Suppose /i {x) vanishes when x-=.c, and that no other function vanishes for this value of x. Let h be a positive quantity so small that no one of Sturm's functions except /^ {x) vanishes while X is passing from c — li to c •\- h. When x = c, f'{x) and /g {x) have contrary signs (III) ; hence they have contrary signs all the time that x is passing from c — h to c -\-h (IV). Now at the instant x becomes equal to c, f^(x) changes its sign (608, Cor. 3) ; hence, before the change, its sign is like that of one of the adjacent functions, and after the change it is like that of the other. But no change in the number of variations of signs in a row of signs can be made by simply changing a sign situated between two adjacent contrary signs. Thus, in the row of signs -\ 1 1 1 there are seven variations; and if we change the fourth sign there are still seven variations. Hence Sturm's functions neither gain nor lose a variation of signs while x is passing from c — h to c -\- h. 2. Suppose that when /j (x) vanishes, other auxiliary functions vanish. The vanishing functions cannot be consecutive (II) ; the functions adjacent to each vanishing function have contrary signs while X is passing from c — h to c -\- h; and each vanishing func- tion changes its sign at the instant x becomes equal to c. But, as we have just shown, this change of sign does not change the num- ber of variations in the row of signs. VI. Sturm's functions lose one variation of signs, and only one, each time x passes through a real root of the equation f(x) = 0. Let a be a real root of the equation f{x)z=0; thep 'f{a)=0. 412 THEORY OF EQUATIONS. Substituting a + h for x in f{x) and f'{x), and developing by Art. 603, we bave /(a + /0=/^(/>)+/>)|-+/>)|+ ••'•) • • • (1)' 7,2 f\a + li)= na)+f'\a)h^f"'{a)^^ (2). Now assume tbe absolute value of h to be so small that the first term in each of these developments shall be numerically greater than tlie sum of the other terms; then the sign of f{a 4- h) will be the same as that of hf\ci), and the sign of f\a + 1i) will be the same as that of f\a). Hence f{x) and f'(x) will have contrary signs when h is negative, and like signs when h is positive. But when h is negative, x is less than a, and when /* is positive, x is greater than a ; hence when x passes a real root of the equation f{x) = 0, a variation is changed into a per- manence. Now it is evident, from (2), that f\x) cannot vanish as long as h has such a value that f\a) is numerically greater than f"{a)h +/'"(«) uT + • • • • Some of the auxiliary func- tions lying between f'{x) and fj^x) may, however, vanish and change signs while x is passing through the root a ; but the change of a sign lying between two adjacent contrary signs (III) does not change the number of variations in the row of signs (V, 1). Therefore, when x passes through the root a, Sturm's functions lose one variation of signs, and only one. In the same way it may be shown that when x passes through any other real root of the equation f{x) = 0, Sturm's functions lose another variation of signs. Cor. 1. — The number of real roots of the equation f(x) = is equal to the number of variations of signs lost by Sturm's func- tions while X is passing from — oo to + oo ; the number of real negative roots is equal to the number of variations of signs lost while X is passing from — oo to 0; and the number of real posi- tive roots is equal to the number of variations of signs lost while X is passing from to -f oo . 413 CoK. 2. — Let a be the smallest real root of the equation f{x) = 0, ^ the next greater, c the next, and so on. Just after x passes through the root a, f{x) and f'(x) have like signs; and just before x passes through the root h, f{x) and f'{x) have contrary signs (VI). But f{x) does not change its sign while x is passing from a to h', hence f'{^) must change its sign. Therefore the equation /'(a;)=0 has one real root between a and h. In the same way it may be shown that the equation f'{x) =z has one real root between b and c. Therefore, between any two consecutive real roots of the equation f(x)=:0 there is one real root of the equation f'{x) = 0. ScH. — The sign of each remainder is changed in order that there may be neither a gain nor a loss in the number of variations in the row of signs, except when x passes through a real root of the equation (III-V). Find the number and situation of the real roots of the follow- ing equations : 1. a^ — 3a^ — ^x -^13 = 0. f{x)=a^ — 3x^ — 4:X + 13, fix) = 3x^ — 6X—4. (603), f^(x) = 2x-6 (613), Mx)=-{-l. AS8T7HED VALTTES OP X. FUNCTIONS AND THEIR SIGNS. NtmBEB OP TAKIATIONS. fix), fix), f,ix), f,ix). — 00 — + — + 3 + — — + 2 1 + — — + 2 2 + — •— + 2 3 + + + + + 00 + + + + Hence all the roots of the equation are real; two of them are positive and the other negative; and the two positive roots are situated between 2 and 3. When x = -'3 the signs of the functions are — + — +? and when x= —% the signs of the functions are + + — f ; 414 THEORY OF EQUATIONS. hence the negative root is between — 2 and — 3. To separate the two roots which he between 2 and 3 we must substitute for x some number or numbers lying between 2 and 3. AVhen x = 2^ the signs of the functions are — — d: +• Here we have only one variation whether we consider the vanishing function f^ (x) to be positive or negative; hence one of the positive roots lies between 2 and 2J, and the other between 2^ and 3. 2. 2^3 _ 3.^ _ 12.^^ _f_ 24 = 0. Ans. Three ; one between 1 and 2, one between 4 and 5, and one between — 3 and — 4. 3. a:3 ^ 6r^ + 10.T — 1 = 0. 4. x^ — Qx^ -{- Sx -\- ^0 = 0. 5. a:^ + 4 = 0. 6. a^ — 2x^-{-3a^ — W-\-Sx — d = 0. 7. x-i — 92.-« + 6a:* + 15a^^ _ 12ar5 - 7a; + 6 8. cc4 _|_ a;3 _ a4j _ 2x-\- 4 = 0. HORNER'S METHOD OF APPROXIMATION. 615. Let it be required to find a root of the equation a:» + Aa:«-i + Bx"-2-f 4-Ka; + L = . . . (1). Suppose a to be the integral part of the root required, and r, s, t, . . . , , taken in order, to be the digits of the fractional part. Let a be found by trial (608) or by Sturm's Theorem ; then find an equation whose roots shall be less by a than those of (1) (598). Let y« + Ay-i + By-2 + . . . -f K> + U = . . . (2) be that equation. In this equation the value of y is less than 1 ; hence the terms containing the higher powers of y are comparatively small; neg- lecting these, we have, approximately, 'K'y -f L' = 0, whence y ^ —=-,, The first figure in the value of y is r. Xow find an equation whose roots shall be less by r than those of (2). Let ^«+A";z»-i+ B"z^-^+ .... +K";2+L"=0. . .(3) be that equation. HORKER'S METHOD OF APPROXIMATION. 415 In tliis equation the value of z is less than .1 ; hence, we have, L" approximately, K";2 + L" = 0; whence 2; = — =7-,. This pro- cess may be continued to any desired extent, and we shall have finally x^a-\-r-^s-\-t-\-,,.. RULE. I. Find the integral part of the root hy Sturm's Theorem or otherwise. II. Find an equation whose roots shall he less than those of the given equation hy the integral part of the required root III. Divide the independent term of the transformed equation hy the coefficient of the adjacent term^ change the sign of the quo- tient and write the first figure of the result as the first figure of the fractional part of the root, IV. Find an equation whose roots shall he less than those of the second equation by the first figure in the fractional part of the required root, V. Divide the independent term of this transformed equation hj the coefficient of the adjacent term, change the sign of the quo- tient, and write the first figure of the result as the second figure of the fractional part of the required root. VI. Continue this process until tlw root is obtained to the required degree of accuracy. ScH. 1. — To obtain the negative roots it is best to change the signs of the alternate terms of the given equation, and then find the positive roots of the result; changing the signs of these, we obtain the negative roots required. ScH. 2. — If a trial figure of the root, obtained by any division, causes the two last terms of the succeeding equation to have the same sign, that figure is not the correct one and must be changed. ScH. 3. — K K' should reduce to zero in the operation, then / T"' we should have, approximately, J'^2 + L'=0 ; whence yz=^y —^* 416 THEORY OF EQUATIONS. JEXAJirPLES, 1. Find one root of the equation a:® — 2a^ — 20.?- — 40 = 0. By Sturm's Theorem we find that this equation has only one real root, and that the integral part of this root is 6. We now find two figures of the fractional part as follows : 1-2 — 20 — 40 6.23 + 6 + 24 + 24 + 4 + 4 — 16^i> -f 6 + 60 + 10 + 64 + 6 + 16^*> 1(1) _,_ 16(1) + 64 - 16 + 0.2 + 3.24 + 13.448 + 16.2 + 67.24 — 2.552^*> + 0.2 + 3.28 + 16.4 4- 70.52^«> + 0.2 + 16.2^2] 1<8> + 16.6^2> + 70.52^2> — 2.552^«> We find the coeflBcients of an equation whose roots are less hy 6 than those of the given equation, using the method explained in Art. 598. These coeflBcients are 1, 16, 64, and — 16, marked (1) in the operation. Dividing 16 by 64, we obtain .2, which is the second figure of the root. We next find the coefficients of an equation whose roots are less by .2 than those of the second equa- tion. These coeflBcients are marked (2) in the operation. Divid- ing 2.552 by 70.52, we obtain .03, which is the third figure of the root! This process may be continued until the root is obtained to any required degree of accuracy. 2. Find one root of the equation x^-{-a^—d0x^~20x—20=0. By Sturm's Theorem, we find the integral parts of the two real roots to be 5 and — 5. Changing the signs of the alternate terms of the equation, we find the fractional part of the negative root as follows : HOENER'S METHOD OF APPROXIMATION-. 417 1 — 1 + 5 — 30 + 20 + 20 —50 - 20 1 5.73 — 150 + 4 + 5 - 10 + 45 —30 + 175 — 170^i> + 9 + 5 + 35 4- 70 + 145^i> + 14 + 5 + 105^i> 19(i> l(i)_^19(i) + 0.7 + 105^i> + 13.79 + 145(i> + 83.153 — 170^i> + 159.7071 + 19.7 + .7 + 118.79 + 14.28 + 20.4 + .7 + 133.07 + 14.77 + 228.153 + 93.149 + 321.302^2) 10.2929(»> + 21.1 + .7 21.8^2) + 147.84^«> 1(«> + 21.8<2> +147.84(2> +321.302^2) _ 10.2929(8> Hence the negative root of the given equation is — 5.73 +. Find the real roots of the following equations : 3. a^^2Ti—23x—70c=0, Ans. 5.1345. 4. a:3_ic2 4-70a:-300=0. Ans. 3.7387. 5. a^4_a;2_500=0. Ans. 7.6172. ( 3.3792, 6. a^_a;2_40a; + 108=0. Ans. I 4.5875, —6.9667. ( 1.7191, 7. a:3_4c2_24a;+48=0. Ans. \ 6.5461, ( -4.2652. 8. .^_l_a;3_|_^_a;_500=0. . j 4.4604, ^^^- \ -4.9296: 9. 24_9^_lla^_20a; + 4=0. 27 Ans \ '^^^^^ ^'''- ] 10.2586. 418 THEORY OF EQUATION'S. Definitions CQ . o X < XI P a EH OENERAIi PbOPERTIES. ^ 616. SYNOPSIS FOR REVIEW. The general equation oftJiG v}^- degree. The absolute or independent term. A function of a quantity. A rational integral function of x. < A root of the equation f {x) =^ 0. When f(x) is divisible by X — r. When Visa root off(x) = 0. Number of roots of f {x) = 0. Tofiiid an equation whose roots are given. Bdatim between the coefficients of i\. fix) and the roots of f{x) =0. j 3. Cor. i, 2, 3, 4, 5, 6, ' 3. When f{x) =0 cannot have a root which is a rational fraction. Boots of the form of a -^bV—l and a — b V— 1. Cor. i, 2, 3, 4. To change the signs of the roots of an equation. To transform an equation containing fractional coefficients into another in irhich the coefficients are integers, that of the first term being unity. To transf/rm an equation into another, the roots of which differ from those of the given equation byagicen quantity. Rule. To cause the second or third term of an equation to disappear. Cor. Cor. 1, 2, 3. Primitive function. First deHvative, Second derivative, etc. First derivative of product of functions. Roots common to two equations. Equal Roots. r Number of roots of f{x) = lying be- Limits of the Roots I tween p and q. Cor. 1, 2^ 3. OF AN Equation. j Limits of positive roots. I Limits of negative roots, Sturm's Theorem. . . 3 ^' ^^' ^^^' ^^' ^' ^^' \ Sch. Hornek's Method of Approximation. Mule. Transformation Equations. of Theorem op Descartes. Derived Functions WEBSTER'S DICTIONARIES. UNABRIDGED QUARTO, NEW EDITION. 1928 Pages, 8000 En- gravings. Over 4600 New Words and Meanings, Biographical Dictionary of over 9700 Names. NATIONAL PICTORIAL, OCTAVO. 1040 Pages, 600 Illustrations. COUNTING-HOUSE DICTIONARY. With Illustrations. ACADEMIC QUARTO. 834 I llustrations. 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First Lessons in our Country's History, bringing out the salient points, and aiming to combine simplicity with sense. I vol. square, fully illustra' ed. Condensed Seliool History of U. S. A Conden-ed Schol History of the United State ;, constructed for definite results in Re^ itation, and con- taining a new method of Topical Reviews. New edition, brou'jh. down to the present time. Illustrated with Maps, many of wnich are colored, Portraits and Illustrations. 1 vol. cloth. 300 pages. Outlines of the World's Ifl«tory. Ancient, Mediaeval and Modern, with special reference to the History of Mankind. A most excellen: work for the proper introduction of youth into the study of General History. 1 vol . With numerous maps and illustrations. 500 pages, lamo. S '.yiNTON'S GFIOGIl APHICAIi COURSE. The famous "two book series," the freshest, best graded, most beautilul and cheapest Geographi- cal Course ever published. Of the lar.e cities that have adopted Swinton's Geographies, we mention Washington',' D. C, Rochester, N. Y., Troy, N. V., Brooklyn, N. Y., New York; City, Ki.ngston, N. Y., Augusta, Me., Charles- ton, S. C, Lancaster, Pa., VN'illiamsport, Pa., Macon, Ga. I iround numbeis, they have been adopted in more than l.OOO Cities and Towns in all parts of the countrv, and h.ive, with inarkjd P'e/irence,hcQn made the bas s oi Pro- fessional Training in the Leading Normal Schools of the United States. Elementary Course in Oeojjraphy. Desifjned as a class-book for primary and intermediate grades ; ancf as a complete Shorter Course for un- graded schools. 123 pages, 8vo. Complete Course In (ieosrapliy. Physical, Industrial and Political; with a special Geography for each State in the Union. Designed as a class- book far intermediate and grammar grades. 136 pages, 4to. The Maps in both books possess novel features of tne highest practical value in education. S\ri\TON'S RAHIBIiES AI?IONG WORDS: Their Poetry, History and Wisdom. A Standard Work to all who love tne riches of the English Language. By William Swin ton, M.A. Hands mely bound in flexible cloth and marbled edges. 302 pages. *»* The above may be had, as a rule, from any bookseller ; but when not thus obtainable, we will supply them, transportation paid, at liberal rates. De- scriptive Circulars and Price Lists will be sent on applicatton. Very liberal terms for introduction^ exchange and examination. IVISON, BLAKEMAN, TAYLOR & CO., Publishers, NEW YORK and CHICAGO. Approved Text-Books FOR HIQH SCHOOLS. THERE are no text-books that require in tlieir preparation so much prac- tical scliolarship, combined with the teacher's experience, as those com- piled for use in High Schools, Seminaries and Colleges. The treatment must be succinct yet thorough ; accuracy of statement, clearness of expression, and scientific grad;ition are indispensable. We have no special claims to make for our list on the score of the Ancient Classics, but in the modern languages, French, German and Spanish, in Botany, Geology, Chemistry and Astronomy, and the Higher Mathematics, Moral and Mental Science, etc., etc., we chal- lenge comparison with any competing books. Many of them are known to all scholars, and are reprinted abroad, while others have enjoyed a National repu- tation for many years. WOODBURY'S GERMAN COURSE. Comprising a full series, from "The Easy Lessons" to the most advanced manuals. FASQUEf.LE'S FRENCH COURSE. On the plan of Woodbury's Method ; also a complete series. MIXER'S MANUAL OF FRENCH POETRY. LANGUELLIER db MONSANTO' S FRENCH GRAMMAR. UENNEQUIN'S FRENCH VERBS. MONSANTO' S FRENCH STUDENT'S ASSIS7ANT. MONSANTO it LANOUELLIER'S SPANISH GRAMMAR. GRA Y'S BOTANICAL SERIES DANA'S WORKS ON GEOLOGY. ELIOT db STORER'S CHEMISTRY. ROBINSON'S HIGHER MATHEV-i TICS. SWINTON'S OUTLINES OF HISTORY. WILLSON'S OUTLINES OF HISTORY. CATHCARTS LITERARY READER. HICKOK'S WORKS ON METAPHYSICS. HUNTS LITERATURE OF THE ENGLISH LANGUAGE. WELLS' WORKS ON NATURAL SCIENCE. KERVS COMPREHENSIVE ENGLISH GRAMMAR. WEBSTER'S ACADEMIC DICTIONARY. TOWNSEND'S CIVIL GOVERNMENT. WHITE'S DRA WING. TAYLOR-K'UHNER'S GREEK GRAMMAR. KIDDLE'S ASTRONOMY. SWINTON'S COMPLETE GEOGRAPHY. Etc. E*c. **♦ Descriptive Circulars and Price Lists will be sent to Teachers and Edu- cationists on application. Liberal terms will be made for introduction, exchange and examination. IVISON, BLAKEMAN, Taylor & Co., Publishers, NEW YORK and CHICAOO. FOURTEEN DAY USE RETURN TO DESK FROM WHICH BORROWED This book is due on the last date stamped below, or on the date to which renewed. Renewed books are subject to immediate recall. 8Feb'56K0j[ i^AiM c, o njsiv wv 2lMar'57Ls RcUl) - isMayeojo RECD LD MAYl 1960 ?5^r!;.'63 •'■;''■ --■ MAU 2 6 1963 tB^al;^???,!'^^ ""'-ilgS'-- LIBRARY USE RETURN TO DESK FROM WHICH BORKOWED LOAN DEPT. THIS BOOK IS DUE BEFORE CLOSriMr Tuji: ON LAST DATE If AMPED slfoV™^ LI8RARY US^ nanriyBTT Titt^"^ f^"') iiMARYUSi #a2I3I_ • .A « e27'67 -e^ LD 62A-50m-7,'65 (F5756sl0)94l2A General Library University of California Berkeley