GIFT OF - J^e ^H /^e - iV'^ ' - ^7 ^' % >^ Digitized by the Internet Archive in 2008 with funding from . IVIicrosoft Corporation http://www.archive.org/details/essentialsofalgeOOstonrich ESSENTIALS OF ALGEBRA COMPLETE COURSE (AN ADEQUATE PREPARATION FOR THE COLLEGE OR TECHNICAL SCHOOL) FOR SECONDARY. SCHOOLS BY JOHN C. STONE, A.M., MICHIGAN STATE NORMAL COLLEGE, CO-AUTHOR OF THE SOUTHWORTH-STONE ARITHMETICS JAMES F. MILLIS, A.M., THE SHORTRIDGE HIGH SCHOOL, INDIANAPOLIS, INDIANA OV TTOAA' ttA.Aa TToXv BENJ. H. SANBBrN & CO. BOSTON NEW YORK CHICAGO Wv^V J^/^ S7^ Copyright, 1905, Br JOHN O. STONE and JAMES F. MILLI8. U '■ PREFACE. That there is a place for a new Algebra recent corre- spondence has abundantly proven. This book contains features which will not only arouse and sustain interest in the subject, but are now demanded in an elementary course in algebra. It is believed that the book is modern and progressive, yet free from fads, and in no sense extreme. It is simple in style and rigorous in treatment. In order to make this possible, some of the topics commonly treated in elementary algebra, difficult for the beginner, and of comparatively little value to him, have been postponed or omitted. Highest common factors by divis- ion, and square and cube roots of polynomials and of arith- metical numbers by formulse, have been put into the Appendix. These may be read whenever the teacher thinks advisable, or they may be omitted without detriment to the subsequent work. The fundamental laws of numbers have been explained and carefully illustrated when introduced, but rigorous proofs of these laws have been put into the Appendix, where they may be read when the pupil has become sufficiently familiar with algebraic processes. The beginning pupil should not be over- burdened with the proof of certain simple principles ; yet he must see, before he leaves the subject, that there is a demon- stration for every principle. The following are some of the special features of the book. iii 1 '4732 iv PREFACE Extension of number. The notion of a general number, rep- resented by some letter of the alphabet, is introduced by many illustrations. The pupil is shown that all of the new kinds of numbers, — negative numbers, surds, imaginary numbers — like fractions, arise logicaMy from an attempt to make the funda- mental processes universal. Checks. The pupil is taught from the beginning how to check his work — multiplication, factoring, etc. — by substitut- ing particular values for the general numbers in the identities which he obtains. This gives him constant practice in the in- terpretation of the symbols of algebra, broadens his grasp of the significance of general number, and trains him to be ac- curate and self-reliant. Factors. The subject of rational factors is thoroughly dis- cussed, and made the basis of subsequent work whenever possible. The pupil is taught to reduce all expressions to be factored, when possible, to one of the fundamental type- forms ; otherwise to try the remainder theorem. Abundant miscellaneous exercises are given. The equation. Special care has been taken in the treatment of equations. The pupil is required to see that every step in the solution of an equation is the application of a fundamental principle. " Transpose " and " Clear of fractions " are terms }vhich the pupil is allowed to use only after he knows the meanings of the processes which they represent. Correlation with science. The formula. Any letter is used as an unknown number and formulae are solved for any letter in them. For tliis purpose many fundamental formulge are borrowed from science. Consequently, the algebra which PREFACE y the pupil will meet in his subsequent science work will be familiar to him. Expression of laws by equations. It is shown that the equa- tion may be used to express a law of science. Practical appli- cation is thus made of the theory of variation. Exercise is given in interpreting the law expressed by an equation, and in writing the equation which expresses a stated law. The sim- ple laws and formulae of science are used in the work. The graph. The graph is used in the treatment of indeter- minate equations, both linear and quadratic, as a means of illustration. It is used to illustrate the different kinds of sys- tems of equations, such as impossible systems, etc. The graphic solution of a system is shown, and the solution of a single equation in a single unknown is found by solving a system. Exercises. The exercises are numerous, new, and well graded. They are sufficiently difficult to call for effort on the part of the pupil. Reviews. Miscellaneous review exercises are distrilmted throughout the book. These reviews consist of many exer- cises and questions intended to bring out in review the funda- mental principles of algebra. The Brief Gourse consists of the first twenty-one chapters of the Complete Course and is intended for classes that go only through quadratics, inequalities, ratio and proportion, and the theory of exponents. The Complete Course furnishes an ade- quate preparation for the College or the Technical School. The authors take pleasure in acknowledging their indebt- edness for many valuable suggestions to Professor E. A. Vi PREFACE Lyman and to Dr. N. A. Harvey of the Michigan State Normal College, who have read both the manuscript and the proof; to Professor J. L. Love of Harvard University, and John A. Avery, of the English High School, Somerville, Mass- achusetts, who have carefully read all of the manuscript ; and to E. Harry English, Supervisor of Mathematics in the Wash- ington (D. C.) High Schools, who has carefully read the first proof, and made helpful suggestions. J. C. S. • June, 1905. J. F. M. PUBLISHER'S NOTE Before accepting the manuscript of this Algebra it was placed in the hands of Professor James Lee Love of the Law- rence Scientific School of Harvard University, who gave it his" cordial approval and endorsement. The Southworth-Stone Arithmetics, published in January, 1904, have in one year been introduced in more schools, ex- ceeded in sale any competitors, and received more general commendation from the leading superintendents and teachers than any other Arithmetics ever published in this country. 'Many of the features that have made these Arithmetics so deservedly popular have been introduced in this Algebra, and the book, with its large clear type and open page, will be cor- dially welcomed by the educational public. CONTENTS. CHAPTER I. PAGE Introduction 1 Definitions 1 Signs of Grouping " . . 6 General Number 8 CHAPTER II. Fundamental Laws of Numbers 16 Fundamental Laws 16 The Equation 19 Axioms 21 CHAPTER III. Negative Number 30 Definitions 30 Addition of Algebraic Numbers 35 Subtraction of Algebraic Numbers 37 Multiplication of Algebraic Numbers 39 Division of Algebraic Numbers 41 Exercises for Review I 43 CHAPTER IV. Addition and Subtraction of Literal Expressions 46 Addition of Monomials 46 Addition of Polynomials 47 Subtraction 49 Removing Signs of Grouping , . 53 Inserting Signs of Grouping 58 vii Viii CONTENTS CHAPTER V. PAGE Multiplication of Literal Expressions 54 Laws of Exponents 54 Multiplication of Monomials 54 Multiplication of Polynomials by Monomials 56 Multiplication of one Polynomial by Another 60 CHAPTER VI. Division of Literal Expressions 64 Law of Exponents 64 Meaning of a", .... 64 Division of Monomials 64 Division of a Poljaiomial by a Monomial 65 Division of one Polynomial by Another 66 The Fraction 71 CHAPTER VII. Powers and Roots 75 Powers 75 Roots :.) 83 Roots of Monomials 86 Roots of Polynomials by Inspection 90 CHAPTER VIII. Special Products and Quotients 92 Products 92 Quotients 96 Exercises for Review II 100 CHAPTER IX. Factors 103 Factors— Type Forms 103 The Remainder Theorem 123 Synthetic Division 125 CONTENTS ix CHAPTER X. PAGE Common Factors and MuLTiPiiES 130 Definitions 130 Highest Common Factor 131 Lowest Common Multiple 135 CHAPTER XI. Fractions 139 Definitions, Laws, etc 139 Addition and Subtraction of Fractions 146 Multiplication of Fractions 150 Division of Fractions , 153 Complex Fractions 155 Exercises for Review III 157 CHAPTER XII. Linear Equations — One Unknown 161 Definitions , 161 The Solution of a Linear Equation 165 Fractional Equations 167 The Formula 170 CHAPTER XIII. Linear Equations— Systems 180 Definitions 180 Elimination 183 Systems of Fractional Equations 190 The Graph of an Equation 193 Graph of a System 197 Exercises for Review IV 209 CHAPTER XIV. Surds and Imaginary Numbers 211 Definitions, Reductions, etc 211 Addition and Subtraction of Surds 215 X CONTENTS PAGH Rationalization ^ , 220 Powers and Roots of Surds 221 Imaginary Numbers. ... 223 Geometric Representation of Imaginary Numbers 228 CHAPTER XV. Quadratic Equations— One Unknown 230 Definitions 230 Solution of Pure Quadratics 231 Solution of Complete Quadratics 233 The Discriminant 241 Solving a Quadratic Formula 244 CHAPTER XVI. Higher Equations— Equations involving Surds— One Unknown 253 Solution of Equations of Higher Degree 253 Solution of Equations involving Surds 256 Introduction of New Solutions 257 CHAPTER XVII. Systems involving Quadratic and Higher Equations 261 Solution of Systems 261 Special Devices 270 Graphs of Systems of Quadratics 273 Exercises for Review V 285 CHAPTER XVIII. Inequalities 291 Definitions 291 Principles 292 CHAPTER XIX. Ratio and Proportion 298 Definitions 298 Principles , 300 CONTENTS Xi CHAPTER XX. PAGK Variation. Algebraic Expression of Law 310 Definitions, etc 310 Law expressed by an Equation 314 Limits of Variables '. 318 Properties of Zero 321 ~ CHAPTER XXL Fractional and Negative Exponents 324 Definitions and Interpretations 324 Exercises for Review VI 334 CHAPTER XXII. Permutations and Combinations 337 Permutations, — Definitions, etc 337 Combinations, — Definitions, etc 342 CHAPTER XXIII. The Binomial Theorem 346 Proof — The Exponent a Positive Integer 346 Exponent Negative or Fractional 350 Extraction of Roots by use of Binomial Theorem 352 CHAPTER XXIV. Progressions. 354 Series, — Definitions, etc 354. Arithmetical Progression 355 Geometrical Progression ". 362 Harmonical Progression 369 Exercises for Review VII 371 CHAPTER XXV. Undetermined Coefficients 375 Tlieorem of Undetermined Coefficients 376 Xii CONTENTS PAGE Expansion of Fractions into Series 378 Expansion of Surds into Series ... 380 Reversion of Series 382 Partial Fractions 383 CHAPTER XXVI. Logarithms 388 Exponential Equations, — Definitions, etc 388 Fundamental Principles of Logarithms 390 Exponential Equations 406 Table of Mantissas t 411 Exercises for Review VIII 407 APPENDIX. Square and Cube root by the formula 1 The Highest Common Factor and Lowest Common Multiple. 16 The proof of the Fundamental Laws : 23 ESSENTIALS OF ALGEBRA CHAPTER I. INTRODUCTION. 1. Algebra. No clear line of distinction will be found between algebra and arithmetic. Algebra deals with number and with the same fundamental processes which are dealt with in arithmetic. The processes of arithmetic, however, are extended in algebra ; and the meaning which is attached to number in arithmetic is also extended in algebra. Algebra deals particularly with the equation. 2. Signs. The signs +, — , X, -^, and =, which are used in arithmetic, are used with the same meanings in algebra. Other signs used in algebra will be introduced later, when they are needed. 3. Fundamental processes. The fundamental processes of algebra, as in arithmetic, are addition, subtraction, multipli- cation, and division. a. Addition. The addition of two or more numbers is the process of uniting them into oiie whole. As in arithmetic, two numbers to be added are called addends, and the result is called the sum. To indicate addition the sign + is placed between the two numbers to be added. Thus, 7 + 3 indicates that 3 is to be added to 7, The result, which is 10, is called the sum. 2 ALGEBRA The symbol =, placed between two numbers, indicates that they have the same value. Thus, 7 + 3=10. This expression is called an equation, and is read "seven plus three equals ten." We shall have more to do with equations later. Three or more numbers may be added by adding two at a time. Thus, to add 3, 6, 5, and 2; adding 6 to 3 gives 9; adding 5 to 9 gives 14; adding 2 to 14 gives 16, the sum. From the foregoing it is evident that to find the sum in a series of additions we may proceed from left to right, uniting two numbers at a time. Thus to find the sum of 4 + 6 + 2 + 9^; we have 4 + 6=10; 10*f2=:12; 12 + 9|=21^. b. Subtraction. Subtraction is the inverse of addition. Given one of two numbers and their sum, subtraction is the process of finding the other number. Subtraction is expressed by placing the sign — between the two given numbers. When^ so used, it means that the second number is to be subtracted from the first. Thus, 9 — 5 indicates that 5 is to be subtracted from 9. As in arithmetic, the number obtained by subtracting one number from another is called the difference, or remainder ; the number to be subtracted is called the subtrahend ; and the given sum, or the number from which the subtrahend is to be subtracted, is called the minuend. From the definition of subtraction, it follows that to sub- tract the second of two numbers from the first is to find a third number such that if it be added to the second, the sum will equal the first. Hence the following important principle : subtrahend + remainder = minuend, INTRODUCTION 3 A series of two or more subtractions may be performed by proceeding from left to right, subtracting one number at a time. Thus, to find the value of 26 - 12^ - 3| ; 26 - 12| = 13^ ; 13^ — 31 = 10, the value. Likewise, a series of additions and subtractions may be performed by proceeding from left to right, performing one operation at a time. Thus, to find the value of 100 - 8 + 2| + 10 - 25 ; we have 100-8 = 92 ; 92 + 2| = 94| ; 941 + 10 = 104^ ; 104^— 25=79|. c. Multiplication. Multiplication has been defined in arith- metic as the process of taking one number, called the multiplicand, as many times as there are units in the other, called the multiplier. It is evident that this definition holds only when the multiplier is a whole number, and fails when it is a fraction. Thus, to multiply 7 by 2| would mean to take 7 as many times as there are units in 2|, that is, 2| times. This is im- possible. One cannot do a thing 2| times. Instead of this definition, which is not sufficiently general, we shall use the following : To rmdtiply one number^ called the midtiplicand^ by another^ called the nudtiplier^ is to use the multiplicand as we tnust use unity to obtain the multiplier. The result is called the product. For example, let us multiply 3 by 4. To obtain 4 we take 1 + 1 + 1 + 1=4. Hence, to obtain the product, we take 3 + 3 + 3 + 3=12. Here we have done to 3 what we did to 1 to obtain 4. ^-^2 GEBRA Again, to multiply 2 by 4| ; we take Hence the product i^jf^J^'J^^'^' ^ y^ ^-^^ ::- /J'/i^ _ 2 + 2 + 2 + 2 + I + |=8 + ^=9|. ^'^ This definition evidently holds when the multiplier is either an integer or a fraction, hence we see that it is iiion geoieral than the old one. The sign commonly used in arithmetic to indicate mul- tiplication is the oblique cross X? placed between the num- bers. When an integral multiplier is written first, the sign is read '• times ; " and when the second number is consi« vired the multiplier, the sign is read " multiplied by." Thus, 8x7=56, is read " 8 times 7 equals 56" or " 8 multi- plied by 7 equals 56." In this book the second number will be considered the multiplier, thus, 5x2| is read "5 multiplied by 2|." In algebra the dot ( ; ) also indicates multiplication when placed between two numbers, in a position above that which would be occupied by the decimal point. Thus, 8x7 may be written 8 • 7. A series of two or more multiplications may be performed by proceeding from left to right, performing one multiplication at a time. Thus, to find the value of 7 x 8 x | x 2 ; we have 7x8=56 ; 56 x 1= 84 ; 84 X 2 = 168. The result obtained by a series of two or more multiplica- tions is called the product of all of the given numbers. In the above example 168 is the product of 7, 8, | and 2. d. Division. Division is the inverse of multiplication. Given one of two numbers -and their product, division is the process of finding the other number. The given product, or INTRODUCTION 5 number to be divided, is called the dividend ; the given number, or number by which the dividend is divided, is called the divisor ; and the result, or number found, is called the quotient. Division is indicated by placing the sign -=- between the two given numbers. When so used, the second number is the divisor. Thus, for "divide 10 by 2," we write 10-^2. In algebra division is often indicated by use of the horizon- tal line — , oblique line /, or colon : , placed between the dividend and divisor. Thus, 10^2 may be written >/, 10/2, or 10 : 2 ; read " 10 divided by 2." From the definition of division we have the following impor- tant principle : quotient x divisor = dividend. Thus, since 20^4=5, then 5x4=20. A series of two or more divisions may be performed by proceeding from left to right, performing one division at a time. To find the value of 256^8^16 ; we have 256^8=32 ; 32^16=2. Likewise, a series of multiplications and divisions may be performed by proceeding from left to right, performing one operation at a time. To obtain 28-^4x^x2^3; we have 28-=-4=7 ; 7x9=63; 63x2 =126 ; 126^3=42. 4. Expressions. Any combination of number-symbols indica- ting one or more processes such as addition, multiplication, etc., is called an expression. Thus, 2x6 — 3 + 1/5 is an expression. 6 ALGEBRA A rational expression is an expression that contains only indi- cated additions, subtractions, multiplications, divisions, and powers of numbers. See §13. 5 . Signs of grouping. An expression is often used as a si7igle number. To indicate that it is to be so used the expression is usually enclosed within a sign of grouping, or sign of aggrega- tion. We shall use four different signs of grouping to enclose ex- pressions. They are the parentheses ( ), the brackets [ ], the braces { }, and the vinculum , the last being placed above the expression. They all have the same meaning, and the different signs are used in different places merely for con- venience. They indicate that the expressions enclosed by them are to be used collectively, i.e., used as one number. Thus, 9 — (7—2) indicates that 7—2, as one number, is to be subtracted from 9. We have 7—2=5 ; 9 — 5=4. The student will note carefully the difference between 9 — (7— 2) and 9—7—2. The expression 9-(7-2)=9-[7-2]=9- {7-2; =9-Tr2. Again, 6 x (5 + 7) indicates that 6 is to be multiplied by 5 + 7 as one number. Thus 5 + 7=12 ; and 6 x 12=72. 24 4 In the expression ^ — 5— the horizontal line serves three pur- poses at once. It is a vinculum under the 24—4, a vinculum over the 2 + 8, and a sign of division. Thus ^:ii=?^ =2. It follows from the foregoing that in expressions containing signs of grouping., the operations indicated within the signs of grouping must be perf or n^ed first. 6. Evaluation of expressions. The value of an expression is the result obtained by performing all of the operations in- dicated within it. Tlius, the value of 38-4 + 12-3 -^ (2 + 3 + 4)- ^^ is 34. INTRODUCTION 7 MathematiciaiTs have agreed that in expressions containing additions, subtractions, multiplications, and divisions, the multiplications and divisions shall take precedence over the additions and subtractions.* Hence the following rule is to be observed in the evaluation of all expressions : If the expression contains no signs of grouping : (1) all operations of multiplication and division should be perfor7ned in the order in lohich they are written from left to rights before any of those of addition and subtraction are performed ; (2) in the resulting expression which vnll contain only additions and subtractions^ these should then be performed in order from left to right. If the expression contains signs of grouping^ all operations indicated within these must 'first be performed according to the foregoing rule. - \f^ >2^^ ^ ^ ^ Example 1. Evaluate 98-(10-2 + 7) + (12-2) -^ (1 + 4). Performing the operations witfiin the parentheses, we have 10—2 + 7=15; 12—2=10; 1 + 4=5. Hence the expression becomes 98—15 + 10^5. Performing the division, we have 98 — 15 + 2. Then performing the subtraction and addition, we have 98-15 + 2=85. Example 2. Evaluate ?-±i_-i^ + 7(2 + 3-1). We have ^_^ + 7 (2 + 3-1) =-|-4 +7^4 =3-2 + 28 =29. EXERCISE 1. 1. Find, by using the definition of multiplication, the product 7x5- 2. Find, in the same way, the product 5 X ^^' * The reason for this will appear later. Members connected by X or -r form a single term. § 15. 8 ALGEBRA 3. If the subtrahend is 28 and the remainder 15, what is the minuend ? 4. If the divisor is 8 and the quotient 12, what is the dividend ? 5. In the product 9x7 what is the multiplier ? 6. What is the difference between 8x5 and 5x8? 7. What conclusion may you draw from Problem 6 as to the relation between the multiplicand and multiplier in any problem ? Evaluate the following expressions : 8. 12-7-3 + 6. 17. (6-2)X(8 + 3). 9. 16 + 4-3-10 + 8. ■ 18. (5-3)- (5 + 3). (8-7). 10. (16-4)-(8 + 2). 19. (9 + l)-(3 + 2)X(ll-8). 11. 18 + 3 X (7-2). ^Q 7 + 19 12. 6-(3 + 9) ^(6-4). ' '^-^' 13. 4x(7 + 8-3). 21. ^5^X3. 14. 5x(3-2) ^ (6 + 4). ^^ 2 + 12^8-(10-4) 3 ^^ 15. 8 + 7x9-8-2x7-5. "'"" 7 ' 12 16. 3X7X2---6--7X4. . 23. jl5 + 3]-|15-3 24. 2 + 3^. (6-4)--(l + 5) 2 .- ^ 2. 1 + 2 -^^.J 26. (6 + 5) • (8 -6) -^7^=^. 4 26. 10-4--[13 + 2-7] 15--9r=^ GENERAL NUMBER. 7. In the following paragraphs we shall discuss a new kind of number. The student understands that the character or figure 3 is not a number, but that it is a symbol which represents a num- ber. All of the numerals 2, 7, 102, |, 5J, etc., and the Roman characters I, V, X, etc., represent numbers with which the INTRODUCTION 9 student is familiar. We shall see that there are' other symbols which may represent number. 8. Definite number. Each of the symbols 1, 2, 3, 4, etc., represents a number with one definite value. For example, 3 represents the single number-idea which we have learned to call three. The numbers which these symbols represent have the same values in all problems in which they may occur. Hence they are called definite, particular, or single-valued numbers. 9. General number. On the other hand, it is sometimes con- venient to use number symbols that can represent more than one value. Such a number evidently can not be represented by one of the symbols 1, 2, 3, 4, 5, etc. It is usually repre- sented by some letter of the alphabet., as a or x. Note carefully the following illustrations. (1) To get the cost of a number of similar articles at a certain price per article, we always multiply the price of one article by the number of articles. In other words we always use the rule : cost of all articles ={pribe of one article) x {number of articles). Thus, to get the cost of 7 books at $1.25 per book, we have cost of 7 books =$1.25x7 =$8.75. A-lso, to get the cost of 12 pencils at 5 cents a pencil, we have cost of 12 pencils=5c x 12 =60 cents. Now, if in place of the statement ^'■cost of all articles,^' which represents a number, we put c, the first letter of "cost" ; in place of ''price of one article,'' we put p, the first letter of '' price" ; and in place of ''number of articles,'' we put n, the first letter in " number " ; the above rule becomes c=p x n. This rule, c=pxn, in which c, p, and n represent immbers, ap- 10 ALGEBRA plies to all particular problems of the above sort. The letters c, p, and w, represent different particular values in different partic- ular problems. Thus, in the book problem above, c=$8.75, p= $1.25, n=7 ; in the pencil problem, c=60c, p=:5c, and n=12. (2) Simple interest on money is always obtained by the rule interest = principal x rate x time. Thus, $200 at 6^, for 3 years, would give interest = $200 x. 06x3 =$36. Now, if in the rule interest = principal x rate x time, we replace each word by its first letter, we get i = p X r X t^ in which i, p, r, and f, represent numbers. These letters stand for different particular numbers in different problems. Note. It is understood, of course, that " time " refers to the ab- stract number representing the number of years. (3) The distance an object, as a train, has moved in a given time is found by the following rule : distance = rate x time. In the same manner as above, this may be written d = r X t. d, r, and t represent different values in different particular problems. 7 2 7 + 2 (4) The statement Trr+-=r() ~~iT)" i^ ^ P^^'ti^^^^^ ^^^^ of the following principle in addition of fractions : first numerator second numerator _ first num. -\- Sftcmid num. common denom. common denom. ~ common denom. This may also be written which expresses m a way the rule for adding fractions. Here, c c c INTRODUCTION H /, s, c, represent numbers which, in one set of fractions, will have certain particular values, and in another set of fractions, will have other particular values. . It must now be evident that numbers may sometimes be represented by letters of the alphabet ; * and that a number represented by a letter may have any particular value v^hatever. Thus, a may equal 1, 2, 3, 4, |, ||, or ayiy other definite or par- ticular number. Hence, numbers represented \>j letters are called many- valued, or general numbers. General numbers are largely em- ployed in the problems of algebra. 10. The same laws which apply to particular or definite numbers must evidently apply also to general numbers. Gen- eral 7iumbers may be added, multiplied, or subjected to any other operation which may be performed upon definite numbers. For example, a + 6, a — 6, a x b^ a-i-b, represent respectively the sum, difference, product, and quotient of the general num- bers represented by a and b. When a = 8 and 6 = 4, what are the values of a + 6, a X 6, and a-^b'i Hereafter, instead of speaking of the " numbers represented by a, 5, etc.," we shall speak of " the numbers a, ^, etc." 11. Some special notation in the nmltiplication of general numbers may now be considered. In addition to the use of the two signs of multiplication dis- cussed in § 3, multiplication may sometimes be indicated by the absence of any sic/n between the given numbers. * By attaching superscripts and subscripts to the letters of our al- phabet, other symbols for representing numbers may be obtained. Thus a', a", a'", etc., and oci, a2, cca, tti, etc. The letters of the Greek alphabet, a, /3, y, S, etc., are also much used for representing numbers. 12 ALGEBRA Thus, a X 6, ab, and a&, all indicate the product of a and b. The product of 4, a, and x may be written 4ax. The product of a + £c and b + y may be written (a+a?) (b + y)^ as well as (a + a?) x {b-\-y), or (a+a^)-(6 + 2/). It is clear that on account of the place value feature of our notation for definite number by which a figure represents both an intrinsic value and a local value,, the absence of a sign may not be used to indicate the product of two definite numbers. Thus 34 does not mean 3x4, the value of which is 12, but means 3x10 + 4. 12. Factors. Coefficients. Numbers which are multiplied so as to form a product are called factors of the product. Thus, in 5aa?, 5, a, and x are called the factors of ^ax. Any factor of a product, or the product of any two or more factors, is called the coefficient of the product of the remaining factors. For example, in 5aa?, 5 is the coefficient of ax, 5a of x, a of 5a?, etc. If a coefficient is a definite number, it is called a numerical coefficient. For example, in 5aa?, 5 is the numerical coefficient. If no numerical coefficient is written, 1 is understood. Thus, abc is the same as la6c. 13. Powers. If all of the factors of a product are equal, the jjroduct is called a power of one of the factors. It is usually written in an abbreviated form. Thus, aaaaaa is written a^, and is called a power of a. Similarly xxxx is written x*. In the power a^, a is called the base and 6 the exponent. The exponent is a number written at the right of, and above the INTRODUCTION 13 base, to show the number of times the base is to be used as a factor to form tlie power. Thus, a2, read " a square," or " a second power," denotes aa; a^, read " a cube," or "a third power," denotes, aaa; a*, read "a fourth power," denotes aaaa; a^, read "a fifth power," denotes aaaaa ; a"\ read "a exponent ?i2." or "a mth power", denotes aaa ... to the product of m factors.- If no exponent is written, tlie^rs^ power is understood. Thus, a is the same as a^. 14. Literal expressions. An expression containing one or more general numbers is called a general or literal expression. For example, 2a + bG—x is a literal expi'ession. A literal expression may liave any definite, or particular value, which depends upon the values of the general numbers involved in it. A definite value of a literal expression may be found wlien a definite value is assigned to eacli general number involved in it. For example, when a ='10 and a; = 6, 2a— x = 14 ; 2ax = 120 ; 3a^x = 5. To find the value of a literal expression use tlie method of § 6. Note carefully the following examples. Example 1. Find the value of a + d—c + b, when a=l, 6=2, c=3, d=4. The expression becomes 1 + 4—3 + 2. Performing additions and subtraction, 1 + 4 — 3 + 2=4. Hence a+d— c + 6=4. Example 2. Find the value of 3a^ 6^ when a=4, 6=2. We have 3a^ ¥ =^aa666=3-4-4-2-2-2=384. Example 3. Find the value of a + 3S— 2^ + (i-f-2, when a=l, 6=2, c=3, d=4. 14 ALGEBRA We have a + 36—2c + -3-/a 8. 6.5£c + 3.25 = 15.75-6£c. 16. 5A;-5=X; + 3. 26. The equation may be easily used to solve certain kinds of arithmetical problems. To solve such problems, the values of certain unknown nuni- bers are to be found. If some of these unknown numbers are represented by letters of the alphabet, the conditions of the problem will lead to one or more equations containing the un- known numbers. In some problems there will be but one THE EQUATION 25 equation containing one tinknoimi number. By solving this equation the value of the unknown number is found. It is evident that the unknoton numbers in these problems become the general numbers in the equations. Consequently, the general numbers of any equation are sometimes called unknown numl)ers. The following examples will show in detail how the equation may be used to solve arithmetical problems.* Example 1. Of two unequal numbers the greater is twice the less, and the two together equal 99. Find the numbers. Let i»?=the less. Then 2ic=zthe greater^ for " the greater is twice the less." Hence, iZJ + 2a?=99, for " the two together equal 99." Therefore 3x=99. § 21. a?=33. Axiom 4. And %x=m. Axiom 3. Hence the less number is 33, and the greater is ^^. Observe, that when x was taken to represent one of the num- bers, the next two statements followed from the conditions stated in the problem. Example 2. John, Henry and James have among them 30 marbles. John has twice as many as Henry, and James has as many as John and Henry together. How many has each ? Let a= the number of marbles Henry has. Then 2a = the number of marbles John has. Why ? And 3a = the number of marbles James has. Why ? Therefore a + 2a + 3a = 30 . Why ? Adding hke terms, 6a =30. Law of Dis. * Some authors prefer to represent unknown numbers by only the last few letters of the alpliabet. Tliere is no good reason for this. On tlie other hand the student sliould be able to consider the number represented by any letter in the equation as the unknown number, and to solve for its value. 26 ALGEBRA Dividing by 6, a=5. ^ Axiom 4. Hence 2a =10. ' Axiom 3. And Sa=15. Axiom 3. Therefore Henry has 5, John 10, and James 15. The student should see that these numbers satisfy the conditions stated in the problem. Example 3. The difference between the ages of two persons is 10 years, and the sum of their ages is 60 years. What are their ages ? Let n= age of younger. Why? Then n + 10= age of other. Why? Hence n-\-n + 10=Q0. Why? Then n+n=50. Why? 2n=50. Why? n=25. Why? 71 + 10=35. Why? What are their ages ? EXERCISE 6. 1. In a certain algebra class there are 24 pupils, and there are twice as many girls as boys. How many boys are there ? 2. I paid $7 for a pair of shoes and a hat. The hat cost three-fourths as much as the shoes. AVhat did the hat costp^ Suggestion. If you let 4ic represent the cost of the shoes, what will represent the cost of the hat ? If you let x represent the cost of the shoes, what then must represent the cost of the hat ? Which is preferable ? Why ? 3. A certain man's age is 8 times that of his son, and in ten years it will be twice as great as the son's age will then be. What are their present ages ? Suggestion. If you let x represent the son's age now, what must represent the age of the father now ? What will represent the ages of each ten years hence ? What equation follows ? 7 THE EQUATION 27 4. Divide 125 into two ^arts one of which is 35 greater than the other. Suggestion. If a? = the smaller part, what must equal the larger part ? When x = the larger part, what must equal the smaller part ? 5. Find two numbers whose difference is 32, and one of which is 3 times the other. 6. The sum of the ages of two boys is 23 years, and one is 7 years older than the other. What are their ages ? * ^ 7. A, B, and C buy a piece of property for $1550. A invests -^ twice as much as B, and C invests $50 more than A and B ^ ^ together. How much does each person invest? 8. A farmer sold a number of cows at $45 each, and three times as many hogs as cows at $13 each. If all sold for $336, how many cows and how many hogs were sold ? 'Z ♦ 1 1— . 9. A rectangular field is twice as long as it is wide, and the distance around it is 672 yards. What are its dimensions ? 10. Two men, starting from points 40 miles apart, walk toward each other, one at the rate of 2 miles an hour, and the other at the rate of 3 miles an hour. In how many hours will they meet? 11. If a man can row 2 miles an hour in still water, what / must be the speed of the current of a stream, if by its aid he ^ can row down the stream 10 i miles in 3 hours ? 12. What number increased by f of itself will make 112? l {llint. Let 4£c = the number. Why ? Also solve by letting x= the number.) /13. What number must be added to 48 in order that twice the sum shall be 114 ? 14. If to a certain weight you add its half, its third, and 8 28 ALGEBRA pounds more, the sum will be twice the weight. What is the weight ? 15. Divide 63 into two parts whose difference is 15. 16. The difference between two numbers is 7, and their sum is 137. Find the numbers. 17. The sum of two numbers is 80 and the greater one exceeds the smaller one by 16. What are the numbers? J 18. Find three consecutive whole numbers whose sum is ^174. ' 19. Two. men were employed to dig a ditch 630 feet long. One of them dug an average of 45 feet a day, and the other 60 feet a day. How long was required for them to dig the ditch ? ^;^20. A man started on a bicycle to a town 25 miles away. He rode at the rate of 6 miles an hour. On the way the bicycle broke, and he walked the remaining distance at the rate of ^\ 2 miles an hour. It required 6i hours to make the whole trip. How far was he from home when the bicycle broke ? —21. A started from a place and traveled at the rate of 3 miles an hour, and 3 hours later B • was sent from the same place to overtake A. B traveled at the rate of 4 miles an hour. How long was it from the time that A started until B overtook him ? ^ 22. A tank holding 300 gallons of water has two pipes. One lets in 15 gallons a minute, and the otiier draws out 12 gallons a minute. If both pipes are running, how long will it require to fill the tank ? 23. A chain which contains 60 links is divided into three segments, whose lengths are as 3, 4, and 5. How many links in each segment? Suggestion, Let 3a; represent the number in the shortest THE EQUATION 29 segment. What then will represent the number in each of the other two ? 24. One number is twice as large as another. If I take 4 from the smaller and 16 from the greater, the remainders are equal. What are the numbers ? 26. Four men planned to form a partnership and to buy a piece of property, but, one man dying, each of the others had to invest $2000 more than he had planned to invest. What was the cost of the property. 26. Find the number whose double exceeds 12 by as much as 9 exceeds the number. ' 27. A man is now seven times as old as his son, and in five years he will be only four times as old as his son will then be. W^hat are their present ages ? 28. How far can I drive into the country at the rate of 4 miles an hour, in order that by walking back at the rate of 2 miles an hour, 1 may return in 9 hours from the time that I started? ' 29. A boy bought equal amounts of two kinds of candy, one kind at 10 cents a pound, the other at 20 cents a pound. It all cost him 90 cents. How much of each kind^did he get? 30. A solves a certain number of the problems in this exer- cise. B solves all of those which A cannot solve and twenty- two of those which A solves. B solves two more problems than A. How many does each solve ? CHAPTER III. NEGATIVE NUMBER. 27. Let us attempt to solve the equation Subtracting 2a; from both members, 03+7 = 4. Axiom 2. Subtracting 7 from both members, x = 4:—7. Axiom 2. What is 4 — 7? Evidently this equation has no solution, unless we may subtract 7 from 4. This leads us to the follow- ing considerations. 28. Extension of fundamental processes. Counting gives rise to the series of arithmetical whole numbers 1, 2, 3^ 4, 5, 6, 7, 8, 9, 10, 11, 12, etc. The sum of any two numbers in this series is another one of these numbers. Thus 3 + 7== 10. The product of any two numbers of this series is also another one of these numbers. Thus 2x6 = 12. But when we attempt to divide, the quotient of two numbers of this series is sometimes another one of these numbers, and sometimes it is not. Thus 12-^-3 = 4; but 10^7 gives no one of these numbers. However, problems which arise have de- manded that we be able to divide any number in this series by any other of the numbers. Hence, since the whole num- 30 NEGATIVE NUMBER 31 bers are insufficient to express all quotients, we indicate the quotients, and thus obtain a neio kind of 7iimibei\ not found in the above series of whole numbers. These new numbers in arithmetic have been called fractions. Thus, 10-r-7=-i^, a fraction. The series of numbers in arithmetic has, therefore, been extended to include fractions as icell as lohoU niimhers^ and this has followed from the necessity of making division always Likewise, when we attempt to subtract, the difference be- tween two numbers of the above series is sometimes another number of the series, and sometimes it is not. Thus, 8 — 5 = 3 ; but, 4 — 7 gives no one of these numbers, and we say 7 cannot he subtracted from 4- In general, we say that a number can not be subtracted from a smaller number. However, problems which arise have demanded again that we be able to subtract any number from any other number. Hence, v^hen the sub- trahend is greater than the yninuend^ we indicate the remainder, and thus obtain another new kind of number., not found in the old series of arithmetical numbers. These new numbers are called minus numbers, or negative numbers. Thus, 4—7 gives a negative number. • The series of numbers has, therefore, been extended in algebra to include negative numbers ; and this has followed from the necessity of making subtraction always possible. 29. Negative and positive numbers. One number may be subtracted from another by separating the subtrahend into parts and subtracting the parts one at a time. It follows that we may write 4— 7 = 4— 4— 3 = 0— 3. Hence it appears that a negative number may be written to indicate the subtraction of a, number from. zero. 32 ALGEBRA Dropping the in — 3, we have — 3=— 3. Likewise, 4-5 = 4-4-1 = 0-1 = -!; 4-6 = 4-4-2 = 0-2 = -2; 4-8 = 4-4-4 = 0-4= -4; 4-9=4-4-5 = 0-5=-5; etc. It is clear that a negative number, such as —3, indicates a reserved subtraction, there being nothing, when it stands alqne, from wliicli to subtract it. It is in nature always a subtra- hend. Negative numbers may be added, subtracted, or used in any- other operation in which other numbers may be used. For the sake of distinction, ordinary or arithmetical numbers in algebra are sometimes called plus or positive numbers. Also for distinction in writing positive and negative numbers, the positive or ordinary numbers are often preceded by the sign + when standing alone. Thus, 6 is written + 6 ; a is written + a. When clearness would not be sacrificed, however, the sign + may be omitted from the positive numbers. The sign --^^jnust never be omitted. When standing alone the positive numbers 1, 2, 3, etc., or + 1, +2, +3, etc., are read either « plus 1," " plus 2," " plus 3," etc., or " positive 1," " positive 2," " positive 3," etc. And the negative numbers —1, —2, —3, etc., are read either " minus 1," " minus 2," "minus 3," etc., or " negative 1," " negative 2," " negative 3," etc. The positive and negative numbers of algebra are called algebraic numbers. The signs + and — written before numbers are called the signs of the numbers or signs of quality. They may always be considered as signs of operation i7i algebra; and when con- venient, as signs of quality^ or distinction. Two numbers which differ only in their signs, such as + 8 NEGATIVE NUMBER 33 and —8, are said to have the same arithmetical, or absolute value. 30. Opposite numbers. Since a negative number, such as — 5, always implies a reserved subtraction, when it is combined with an arithmetical or positive number, it tends to destroy of the positive ntmiber, a part equal to itself. Thus, when 9 and —5 are combined, —5 destroys 5 of the 9, leaving 4. Accordingly, positive and negative numbers are sometimes called opposite numbers. In subtraction, when the minuend and subtrahend are equal, the remainder is zero. This is equivalent to saying that when two opposite numbers with equal absolute values are combined, the value of the result is zero. Thus, 3 and —3 give ; 25 and —25 give ; +a and —a giveO. 31. Opposite concrete magnitudes. Many concrete magni- tudes are capable of existing in opposite states^ such that one tends to destroy an equal amount of the other ; and hence their numerical values may be represented by positive and negative numbers. A few are here suggested. The student should study these examples carefully. (1) Debt and credit. If I have $20 in a bank, and owe the bank $20, paying the debt will leave me nothing. The indebt- edness destroys the credit. Hence debt and credit are opposite magnitudes. If I have $10 in my pocket, and make a purchase amounting to $15,1 will have left $10 — $15, or — $5. The — $5 means that I not only have no money left, but am $5 in debt. In this sort of problem, then, a negative number means indebtedness. If we call indebtedness negative^ credit will \)Q positive. Indebt- edness will destroy credit. 34 ALGEBRA (2) Forces in opposite directions. If two persons pnll in oppo- site directions on cords attached to the sanie object, each with a force of 100 lbs., the object will not move. The two forces have destroyed each other. If one pulls 100 lbs., and the other l^ulls only 80 lbs., the -effect upon the object will be the same as if the first person alone had pulled with a force of 20 lbs. If the force exerted by one person be represented by a^9ost7iy6 number, the other force will be represented by a negative num- ber. (3) Motions and distances in opposite directions. If a person walk 100 ft. east, then turn about and walk 100 ft. west, he will arrive at his starting point. The result is the same as if he had not moved at all. The second motion destroyed the effect of the first. If the motion in the first direction be called positive, the second motion will be called negative. Accordingly, distances moved through, or distances measured in, opposite directions are called opposite distances. Thus if distances measured to the right of the point A are called posi- A 1 . - + tive, distances measured to the left of A are called negative. Hence in representing distances, the signs + and — serve to indicate the directions in which they are measured. (4) Positive and negative temperature. If temperature above zero is called positive, temperature below zero will be negative. + 10° means 10° above zero. —10° means 10° below zero. From the above illustrations it is seen that a negative num- ber has the property of oppositeness. Thus both positive and negative numbers may be represented by distances along a straight line measured from a common starting point. If dis- tances to the right represent positive numbers, then negative NEGATIVE NUMBER 35 numbers must be represented by distances in the opposite direction —5 ~4 --3 -2 -1 o 1 2 3 4 5 To every point to tlie right there corresponds a similar one to the left. + 3 is represented by a point 3 units to the right of some point, as o, on a straight line. —3 is represented by a point 3 units in the opposite direction. If we measure 3 units in one direction, then from this point, measure 3 units in the opposite direction, we return to the starting point, ^^e., 3 and -3 = 0. ' . 32. Addition of algebraic numbers. Addition was defined in § 3 as the process of combining two or more numbers into one. We shall indicate the addition of algebraic numbers by writing them in succession with their signs. Thus, to indicate the addition of +6, +2, —4, +3, and —5, we write +6 + 2—4 + 3 — 5. The first term in the expression being positive, we may omit the sign, giving 6 + 2—4 + 3—5. Algebraic numbers to be added are sometimes placed in a column with their signs. Thus, -3 + 2 -6 Evidently two positive numbers, since each is an ordinary arithmetical number, ina\j he combined by adding their absolute values ; and their sum will be a positive number. Thus, 5 + 9 = 14. Since the negative number —5 indicates a reserved subtrac- tion, then to add — 5 to any positive number means to combine it by subtraction; that is, to subtract 5 from the number. (See § 30.) The same reasoning would apply to any other negative number. 36 ALGEBRA Thus, the sum of 7 and —5 is 7—5=2. And the sum of 3 and — 5 is 3 — 5=— 2. Remember that when the subtrahend is greater than the minuend the remainder is negative. Also, to subtract each of two numbers in succession is equivalent to subtracting their sum. Hence two negative numbers, since each is a subtracted number, inai/ be combined by adding their absolute values ; and the sum will be negative. Thus, to add —7 and —5 we have —7—5= — 12. From the above considerations we evidently have the follow- ing rules of addition of algebraic numbers : (i) To add two numbers with like signs, find the sum of their absolute values., arid %>refix the common sign. Thus, 7 + 15=22; -7-15=-22. (^) To add t\no numbers with iinlike signs.^ firul the difference between their absolute values, and attach the sign of the arithmet- ically greater. Thus, -7 + 15 = 8; 7-15 = -8. {3) To add three or more algebraic numbers, 2^'^oceed from left to right, performing each step according to (1) or (2). Thus, to find the sum of 6, —9, 3, and —7, we write 6 — 9 + 3—7; 6-9 = -3; -3 + 3=0; 0-7=-7. Another rule is sometimes to be preferred to Rule (3). It is illustrated in the following example : 7-3-9 + 6-5 + 21 = -3-9-5 + 7 + 6 + 21 Law of order. = ( — 3-9 — 5) + (7 + 6 + 21) Law of grouping. = _17 + 34. Rule (1). = 17. Rule (2). The same method is applicable to any problem. Hence we have the following rule : (4) To add three or more algebraic numbers, find the sum of NEGATIVE NUMBER 37 the positive numbers and the sum of the negative numbers hy {1); then find the sum of the sums hy {2). Thusin 21-6-3 + 5 + 1; 21 + 5 + 1=27; -6-3=-9; 27-9 = 18. 33. Subtraction of algebraic numbers. Since subtraction is the inverse of addition, the rule for subtraction of algebraic numbers may be obtained from addition. In § 3 we established the principle : subtrahend + remainder = minuend. Hence if s stand for the subtrahend in any problem, and r for the remainder, the minuend will be r + s. Now (r + s) — s represents the sum of the minuend and the subtrahend with its sign changed. But {r-\-s) — s = r-\'{s — s) by Law of Order for Addition. By § 82, s-s = 0. Hence r+(s— s) = r. That is, the sum of the minuend and the subtrahend with its sign changed equals the remainder. Hence the rule : . To subtract one algebraic number from another^ change the sign of the subtrahend^ then proceed as in addition. Thus, to subtract —5 from 10, we write 10+5 = 15, remainder. Again, 6 from —7 gives —7—6 = — 13. The latter might have been written —7 6 -13 38 ALGEBRA EXERCISE 7. Find the sum of 1, 7, -2, 3, -5. 9. From 6 take —3. 2. 6, 5, -3, -1. 10. From -6 take 3. 3. 3, -15, -12, 1, -9, 10. 11. From -7 take -5. 4. -6, -5, -12, -8, 4. 12. From 3 take 8. 5. 12 5 Ql 2'> — ¥» — IT' ^2' 13. 6-(-8)=? 6. A. -f -h 14. -3-(-7)=? 7. sV' T6' — 1» — i- 15. 5-(-3)-(-6)=? 8. 5.25,-2.5, -7.34, 4.35. 16. -10-(-4)-(-3)=? 17. Since the sign — always indicates a subtraction, what name may be given to any expression witliin a sign of group- ing which is preceded by tlie sign — ? 18. From 7-3-6 + 12 take -6 + 14 + 3-7. 19. From the sum of -3-5 + 2 and 12-6-1 take the sum of 26 + 111-9 and 3i + 7-l. 20. Is the absolute value of an algebraic number always diminished by subtraction ? Illustrate. 21. Do you ever add arithmetically when you are subtract- ing algebraically ? Illustrate. 22. Is the absolute value of an algebraic number ever de- creased by addition ? Illustrate. 23. A balloon lifts up with a force of 200 lbs. A weight of 150 lbs. is attached to it. What is the effect? If the weight is represented by a positive number, what will represent the lifting force of the balloon ? What will represent the result when the weight is attached ? 24. A freight car is running at the rate of 20 feet per sec- ond. If a man walks toward the front on the roof of the car at the rate of 5 feet per second, how fast will he be moving NEGATIVE NUMBER 39 relative to the ground ? If the speed of the car is called posi- tive, what algebraic number will represent the speed of his Avalking ? What algebraic number will represent his speed re- lative to the ground ? If he walks toward the rear of the roof of the car at the same rate, what will represent the rate of his walking? What will be his speed relative to the ground? What will represent it ? 25. I deposit in the bank at different dates $100, $175, and $95. I check out at different times $25, $10, $87.25, and $37.75. If my account is then balanced, how much will I have in the bank ? If the deposits are called positive, what will represent the checks ? Show how thus to find the balance. 26. On the earth's surface longitude west is called positive longitude ; longitude east, negative longitude ; latitude north, positive latitude ; and latitude south, negative latitude. How could you describe the position of a town whose longitude is 112° east, and latitude 75° north? In what country Avould a city be whose longitude was +90° and latitude +40° ? Longi- tude -90° and latitude +40° ? 34. Multiplication of Algebraic Numbers. The laws of signs for multiplication of algebraic numbers come directly from the definition of multiplication given in § 3 ; viz, to obtain the prod- uct of two numbers we must use the multiplicand as we m,ust use 1 (unitf/) to obtain the multiplier. Suppose the multiplier to be +3. To obtain +3 we must add three I's. Thus, +3 = 1 + 1 + 1. Hence to obtain the product ( + 5) ( + 3), we must add three +5's. Hence ( + 5)( + 3) = + 5 + 5 + 5=+15. To obtain the product ( — 5) ( + 3), we must add three — 5's. 40 ALGEBRA Thus (-5) ( + 3) = -5-5-5= -15. Suppose the multiplier to be —3. To obtain —3 we must subtract three I's from ; that is, change their signs and add them. Thus -3=-l-l-l. Hence to obtain the product ( + 5)( — 3), we must subtract three +5's from ; that is, change their signs and add them. Thus ( + 5)(-3)- -5-5-5= -15. To obtain the product (— 5)( — 3), we must subtract three — 5's from ; that is, change their signs and add them. Thus (-5)(-3)= +5 + 5 + 5= + 15. The preceding reasoning will evidently apply to any multi- plier and any multiplicand. Hence the following laws : (1) The product of tvio numbers loith like signs is positive. Thus ( + 8)( + 9) = + 72, and (-8)(-9)=: + 72. Note that all of the signs + could have been omitted. (2) The product of tico numbers vnth unlike signs is negative. Thus .( + 8)(-9) = -72, and (-8)( + 9) = -72. 35. A product consisting of an odd number of negative factors is negative ; and a product consisting of an even number of negative factors is positive. This follows easily from § 34. For ( — 1)( — 1)= + 1 ; Even number, hence (-1)(-1)(-1) = ( + 1)(-1) = -1 ; Odd number, and (-1)(-1)(-1)(-1) = (-1)(-1) = + 1; Evennumber. and(-l)(-l)(-l)(-l)(-l) = ( + l)(-l) = -l; Odd number, and so on. Thus (-6)(-5)(-2)=-60; (-3)(-4)(-5)(-6)= +360, NEGATIVE NUMBER 4| Since the product of two positive factors is positive, it is easily shown by reasoning similar to the foregoing that a pro- duct consisting of any number of positive factors is j^ositive. Thus, ( + 6)( + 5)( + 4) = +120, and ( + 3)( + 7)(+ 1)( + 2) = +42. 36. Signs of Powers. If the factors of a product are all equal, the product becomes a power. See § 13. Thus, (-3)(-3)(-3)(-3)=(-3)*; read "the fourth power of -3." Hence from § 35 we get the following laws : (1) Any power of a positive number is positive. Thus ( + 2y* =( + 2) ( + 2)( + 2)=+8; ( + 4)-^ =( + 4) ( + 4)=16 ; etc. {2) Any odd j)ower of a negative number is negative ; any even power of a iiegative nu^nber is positive. Thus, (-2)^ =(-2) (-2) (-2) = -8 ; (-3)^ =(-3) (-3) (-3) (-3) (-3) = -243. And(-2)*=(-2)(-2)(-2)(-2) = + 16 ; (-5f =(-5) (-5) = + 25. To find the value of a^6^ when a=3, 6=— 2; we have a^ b^ = 32(-2)^ =9 (-8)^-72. If a=-3, 6=2, aW={-3yx2^=9x8=72. 37. Division of Algebraic Numbers. The laws of signs in division of algebraic numbers are easily deduced from the cor- responding laws in multiplication. In § 3 it was shown that divisor x quotient = dividend. Hence, since- ( + 5) ( + 3)= +15, then ( + 15)^( + 3)= +5 ; since (-5) ( + 3) = -15, then (-15)--( + 3) = -5 ; 42 ALGEBRA since ( + 5) (-3)--15, then (-15)--(-3)= +5 ; since (-5) (-3)= +15, then ( + 15)---(-3) = — 5. Manifestly the same laws apply here which apply to multi- plication. The reasoning in the particular cases above will evidently apply to any dividend and any divisor. Therefore we have the following laws : (1) If the dividend and divisor have like signs^ the quotient will he positive. Thus,(-18)-(-2)= + 9;^^=+2;^- +4; etc. {2) If the dividend and divisor have unlike signs^ the quotient will he negative. 30 Thus, (-21)-(+3) = -7;--g=-5; (- |)- | = - |; etc. EXERCISE 8. Find the product of 1. 2 and -7. 7. 9, -2, G, -5. 2. -2 and 7. 8. 5, -7, +4, -2, -1. 3. 2 and 7. 9. -7, -1, -1, -1, -1. 4. -2 and -7. 10. -1, -1, -1, -1, -1, -1. 6. -3, 2 and -5. 11. -4, 3, -2, 0. 6. -4, —3 and -1. Find the value of 12. ( + 2f. 14. (-4)1 16. (-2)^ (-3)^ 13. (-3)*. 15. (-If. 17. (-ly (-2)*(-8)». If a=2, ^>= — 3, c^ —4, find the value of 18. a'h\ 19. ahc. 20. h'c, 21. a'h'c'. 22. (-2)(-3)i^. NEGATIVE NUMBER 4.3 Divide : 23. -27 by 3. ^ ^ ^ 27. 6.25 by -2.5. 24. -27 by -3. ^ ; 28. 8 by - |. 25. 21 by -7. 29. -2| by 17. 26. - 1 by f . 30. +216 by +12. If a=— 2, h = Z, c=- -4, find the value of 31. "^',-5. 32. (v'~-G'yJ)\ 34. 3c^--2^. -J^ 4 36. 3 . 33. c^--6--«^ 36. d'^&-^h\ EXERCISES FOR REVIEW (I). 1. How does algebra differ from arithmetic ? 2. How is the addition of three or more numbers indi- cated? 3. What are the signs of multiplication used in algebra ? Why may ah indicate the product of a and h but 87 not the product of 3 and 7 ? 4. What is the use of the " signs of grouping " ? Illus- trate. 6. IIow would you evaluate the following expression? (l + 2)(9-4)-8-=-(10-6 + 2) + 3. State a rule for evaluating expressions. 6. What is a general number ? Illustrate. How is it rep- resented in algebra ? 7. What do we mean by factors of an expression ? What are the factors of lahc ? 8. What is a ponder of a number? What does a? mean? What is the 3 called? When a = 2 what is the value of a*? of a*? oia'^i 44 ALGEBRA 9. What is the coefficient of an expression ? What is the numerical coefficieyit of lab'^ ? What is the coefficient of b^ in the expression 7a//? of a? of 7b'? 10. What is a /*7eraZ expression ? Illustrate. ^o 11. When a =10, 6=4, c = 2, find the value of -U ' (a.) {2a-Zb){4a-3c)~{Sa-Sc-b). ** (6.) {a + br-^^^ + c(a + 2b). 12. If a = 4, 6 = 8, c = 3, r?=2, and x = b^ find the value of (a.) 3 aa5 + 5M-2c'a; + 2cf?^-M\ Aa' + ^ b' Sd'{2G'-^ db' + 2x) d^x ^^■^ a'b + a {b'-c') b ' 13. What are similar terms ? Illustrate. 14. State t\iQ ftmdamental laios of nmnbers. Write them in symbols. Illustrate each law with definite numbers. 15. State the laws used in the following identities: 2a^-56^-7c2 = 2-5-7a^-6^c''=(2-5-7)(a26V) = 70a^6V. 16. What law is involved in 5a; + 2a; + 12£c = (5 + 2 + 12)£c = 19a;? 17. State the axioms given in this book. Of what use are they? 18. What is an equation? Distinguish between a condi tional equation and an identity. Illustrate each. 19. What is meant by a root^ or a solution^ of an equation? 20. State the steps in the solution of the following, giving your authority for each step. l + 4a; = 2(a! + l)+3. 21. A tree 60 feet high was broken at such a point that the part broken off was 3 times the length of the part left stand- ing ; required the length of each part. ^ko^-, ^'^ **' -^ j^ NEGATIVE NUMBER 45 22. The greater of two numbers is 5 times the less, and their sum is 126 ; required the numbers. 23. What is a negative number ? A positive number ? How did they originate ? Illustrate. 24. Why are positive and negative numbers sometimes called opposite numbers ? 25. If a boy weighing 75 pounds, is holding a toy balloon, pulling upward with a force of 15 pounds, how may these numbers be represented by positive and negative numbers ? 26. In exercise 25, if 75 is called +75 what is the 15? What force will be required to lift the boy and balloon ? 27. If I have $1000 and owe 11500, by what number may my financial condition be expressed ? 28. How else may 50° above zero and 15° below zero be ex- pressed ? 29. What is the difference between algebraic numbers and arithmetical numbers f 30. How do we add algebraic numbers ? Find the algebraic sum of -10, +5, -7, -3, +4, and +8. 31. How do you subtract algebraic numbers? Illustrate. Upon what fundamental principle is the proof based ? 32. From the sum of 6, 5, —3, —1, take the sum of ~15, -12, 1, -9, 10. 33. State the laios of signs in multiplication. Upon what important definition is the proof of these laws based ? By the use of this definition prove that (—2) (—4)= + 8. 34. Give the values of (-2)^ (-3)^; {\f\ (-i)*- 35. Give the laws of signs in division. Upon what import- ant principle is the proof of these laws based ? 36. Divide 8 by -2 ; -8 by -2 ; -81 by 3 ; -f by -|. CHAPTER IV. ADDITION AND SUBTRACTION OP LITERAL EXPRESSIONS. 38. Addition of monomials. It was shown in § 21 that similar terms could be added by use of the distributive law ax-\'hx-\-cx={a-\-h + c)x. From this law we have the following rule : To add similar terms^ add their coefficients and affix to this sum the common letters with their exponents. Thus, to add 4x^2/^ — 2a?^2/^ and —z>x^y^^ we have, writing them in succession with their signs, A.x'y^ - 23(^1/' - ^x^f = (4 - 2 - 5) x^y"" = - Sx'y^ To add dissimilar terms, write them i/t succession with their signs. Thus, to add 2ah, —^ax^ 122/^ and— 32;^ we have 2a6-3aa?4-122/'-32;l The addition of terms similar with respect to certain letters may be indicated by grouping the coefficients of the common letters, and affixing to this group the common letters with their exponents. Thus, to add 2ax^y^ —hx^y, and Scoc^y, we write 2ax^y—bx^y + Scx^y={2a—b + 3c)x^y. 46 ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 47 EXERCISE 9. Find the sum of 1. 2x 2. Za'b 3. 7c' 4. Qabcd 5. ax^ -3a; 4a'b -be' — 21abcd ^ax' 5x - a'b Qc' — Sabcd - ax' — X -lOa'b -12c' Uahcd 7 ax' 6. bA, 12A, -3.4, -7A. 9. ^pq, 4pq, -V^pq. 7. 16P§, -10P(2, 4P§. 10. 100ylC;-14.4(7, |^(7. 8. 4^"^ 2^^ -B\ -^B\ 11. ccy, -9a;V', 4ajV', -a^y. 12 p5'r, — 10^5'r, ^pqr^ —\pqr. 13. ^mhi^^ 6^iW, — 32/?^^^^ — mW. Simplify 14. ^x—^x—bx^-^x—x. 16. aWc—2a''¥c^ Q^a^'^c. 15. -12y^-3y^ + 4/-72/^ 17. 4a^»c-2cia-7^>ac. 18. |2/^-3y2;-|2/s + |2/2. Simplify by adding similar terms 19. 4a + 6a-12a; + 2a + 3a5. ^ Solution. 4a + 6a-12ic + 2a + 3ic=4a + 6a + 2a-12.r + 3£C Why ? = (4 + 6 + 2) a+ (-12 + 3).T = 12a-9x. 20. xy — ab + l(iab—\^xij — '^ab. 21. ^abc''\-2a''bc-babc''^7ah^c-\-7a''bc. 22. ^a'' + W-ba'-b\ Indicate the sum of 23. Zax^j bbx\ —7cx^. 24. 2xi/z, axijz, -bxyz. 25. -7cij\ 2y\ Zay\ 39. Addition of polynomials. If all of the terms of a poly- nomial are added to an expression, the polynomial is added to the expression. This follows from the lavi of grouping. 48 ALGEBRA Thus, if a, 6, and c are added to a?, we have x-\-a+b-\-c—x+ (a+b+c). But x+{a+b+c) indicates that the polynomial a+b + c itself is added to x. Hence, to add two or more polynomials^ write down all of the terms in successioii vnth their signs / then combine the similar terms^ if any. Example 1. Add 2a + 3&— 4c, —4a— 6 + 5c, and 5a + 6— 2c. We write 2a+36— 4c— 4a— 6+5c+5a+6— 2c=3a+3?)— c. The work may often be more conveniently performed by Avriting the similar terms in vertical columns, then adding the terms in the resulting columns. The above example might be written 2a+36-4c —4a— 6+5c 5a+ b—%c 3a + 36— c Checks. In much of the work of algebra the student can easily verify or chech his results ; i. e. , he may perform other operations that tend to show that the first result is correct. This is called checking the work. A short method of checking addition of polynomials consists of assigning particular values to all of the general numbers in- volved in the polynomials and in the sum, and seeing if the sum of the values thus obtained for the polynomials is equal to the value of the sum of the polynomials. This is illustrated in the following example. Example 2. Add 6a— 56-f-3c, 7a+106— 6c, and 8a— 96- 10c. Work. Check. 6a- 56+ 3c 6- 5+ 3= 4 7a + 106- 6c 7 + 10- 6= 11 8a- 96-lOc 8- 9-10=-ll 21a- 46-13C 21- 4-13= when a=l, 6=1, c=l. ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 49 EXERCISE 10. Add and check : 1. x-{'y-\~z^ x—2i/-\-Sz, —t)x—4ii/ + z. 2. Sa—b + 2G, ba + c—'2b, —a + Sb—4:C. 3. 2P + 4§ + i?-7.S; -6P-§ + 3i? + 2>S'. 4. 7ac-\-Sxi/, 2ac—lxy. 5. 20ji?— <7 + r, 2/) + 5^— 7r, — 7p + 2y + 3r. 6. 2a6 — 3*c + 5ac, 7^c— 2ac + 6a*, — 3ac + 2Z>c. 7. a;=' + i«' + a; + l, x'-x' + aj-l. 8. a^ + 2a^ + ^>^ a'-2ab + h\ 2a' -W. 9. a;^— y^ a;^ + 6£c^y, —^xy'^—y^. 10. |6«-^^-i-c, fa-J^ a + |$-3c. 11. 3£c-'-2£c' + a;- -4, x' + 4aj^ + l, 3a;^ -2a; + l, x^-x\ 12. -12a;* + 2a;- -1, Zx' '-2a;^ + 3a^, , a;^ + 2a; + 5, Zx' + A.x\ 40. Subtraction. The reasoning in § 33 evidently holds for algebraic expressions in general, since any expression is itself a number. Hence, to subtract one expression front miother^ change the sign of the subtrahend and add the result to the minuend. The sign of an expression is changed by changing the sign of each of its terms. This follows directly from Rule 4 in addition, since an expression is but the algebraic sum of its terms. Thus, changing signs in 7—3 we have —7+3. Now 7—3 and —7 + 3 are opposite in sign, but have the same absolute value, 4. Therefore, for subtraction we have the following rule : Add to the minuend the subtrahend loith the sign before each of its terms changed. Example 1. Subtract — 4a& from — 2a6. Changing the sign of the subtrahend and adding gives -2a6+4a6=2a6. 50 ALGEBRA Example 2. From 2a— 36 + 5c take 3a— 2b— 2c. Changing the sign before each term in the subtrahend arid adding, we have Work. Check. 2a-3h + 5c =4] -3a + 2b + 2c =1 I when a=l, 6=1, c=l. — a— 6 + 7c = 5 J The change of signs need only be made mentally, the writ- ten signs of the subtrahend remaining unaltered. This is illus- trated in the following examples. Example 3. Subtract —2x^ — 4:y^+15a from 7x^—2y^. Work. Check. lx^-2y' = 5 1 -2a;2- 41/2 + 1 5a = 9 I when x*=l, i/=l, a=l. 9xH2i/2-15a = -4 J Example 4. From 2a^ + 4^* — 3a7 + 7 take a^ — 3ar* + 2x^ — x. Work. Check. 2x^ + 4cc* -3a.'+7 = 10] x^ -Sx'-{-2x-- X ^- 1 lwhena;=l. af' + 4x' + 3x'-2x'-2x + 7 = 11 EXERCISE 11. 1. From 2 a'b' take —SaW. 4. From —7abc take ^abc. 2. From — £cy take 5a»/. 5. hhy'^—{—%hif)='i 3. From -^aa;Uake -^ax\ 6. 21a;y2-(-3a;y^)=? 7. From Zx—2y-\-lz take x^ 6y — Sz. 8. From 7a + 2a^— 2c take 8a— 12a^ + 5c. 9. From 2x—7y take 3y— 5a;. 10. From x^—x^+x""—! take 2x^—x^^x^—l. ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 5I n. From a^^-Za'b + ^ab'-^b^ take a'-a^h-\-ah'-h\ 12. From lAB^^lxy — ZPQ take 'lxy-V\Pq. 13. 3ic2-5a; + 9-(2ec2 + 6£c-4)=? - 14. The subtrahend is x^^-x'-^x'-^-x-^ the remainder is x^^x^ + £c"^ + l. Find the minuend. 15. Subtract — 4c^s^ + «*— r^ from 2rt^ + 3i^'^— cV. 16. Subtract \—x^^x^ from a?^ ; from a\ from 0. 17. From the sum of cv' + a'b'—d'b'-\'If and ^aW—2b'—a' take a^— 5^ + 2£c. 18. From the sum of ^x'—^^x' + l and ^x' + p' + ^x take ia;^' -x' + 2. 19. From 1.5a-7.2ic'-3.25m=' take the sum of .4a;^-7.5a + 5m' and l-.125a + 3m^ 20. From the sum of a^^— 1, ce^+aj + l, and x^—1 take the sum of x^ + a?^ + 2 and x^ — 1 . 21. What operations are indicated by 3«^ + 2«— (a^— a^ + «— 1) ? 22. Simplify ^x^ -\-Sx^+1- (2x^ -3 + 0^). 23. From 7a^ + 3^2-2 take a' + Sa + b. If ^ = 2a;^ + a;-3, J?=a;='-a;'^ + 2a! + 2, C=Sx'-6x + l, find the value of 24. A-B+O. 26. -A + B+O. 2b. A + B-a 27:^ A-B- a If a= -2, b=~l, c=3, f?=2, find the value of 28. 2abc + ^a'cl 33. 3a -2/; + 4c. 29. b'c + adc. 34. -2ac-2bcl 30. a^-5^ . 35. a' + b' + c' + dK 31. 2^>c + 3«(?. 36. 4a'^' + 2aiV-^>cW. 32. a + Sd-2c + b. 37. after?- 2a^ + cr. 52 ALGEBRA 41. Removal of signs of grouping. The negative sign (— ) always indicates that the number following it is a subtracted number. Hence, an expression inclosed within a sign of grouping which is preceded by the negative sign is a subtrahend. Thus, in 3a — (5a— 26), the expression (5a— 26) is a subtrahend. But subtraction is performed by changing the sign of each term of the subtrahend and adding the resulting expression to tlie min'uend. Hence, a sign of grouping ^yreceded by the negative sign may be rernooed if the sign before each tertn inclosed is changed. Thus, 3a-(5a-26)=3a-5a + 26 ; -{-Qx-{-5x') = 6x—5x\ The positive sign ( + ), preceding an expression inclosed within a sign of grouping, either indicates an addition or serves as a sign of distinction. Hence, a sign of grouping preceded by the jyositive sign may be removed vnthout changing the sign of any of the terms inclosed. Thus, 4x + 7.r+(3a?— 2j?+1)=4:c + 7x + 3x— 2j?+1. Sometimes signs of grouping are inclosed within other signs of grouping. In such cases the use of different kinds of signs is advantageous. Thus, a-{2a-(a— 26)} ; x-^y—{2x—y) + \;Zx-{^y—x)]. In expressions of this kind, it is best for the beginner to remove the innermost sign first ,' then the innermost remaining sign ; etc. Thus, a-h-{-a-{-h-a-b)) =a— 6— .{a— (— 6— a + 6)}, removing vinculum, =a — 6— {a + 6 + a— 6}, removing parentheses, ' =»a— 6— a— 6— a4-6, removing braces, = — a— 6, adding like terms. Again, Ix-Zy- {(4a— 6)— [5«— 6— (3a:— 2^)— 36] j ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 5^ =7x—:^^J—{4:a—b—[5a—b—3x + 2y—Sb]} =7x—3t/—{4.a—b—5a + b + Sx—2y + Sb} =7x—3i/—4a-\-b + 5a—b—Sx-h2y—3b =4x — y + a—3b. Some Avork may be saved by adding the like terms as each sign is removed. EXERCISE 12. Simplify by removing signs of grouping and adding like terms : 1. 2a + Sb + (a-4:b.) 3. 2B'-{B'—4.AC). 2. {lx + 2y) + (Zy-2x). 4. « + [2a-(26-r0]. 5. x' + x'-{x-'lx^)+{x'-{%x''-l'Mx'-x)\, 6. a;V-(2.^'^-3.^//)-(2.^^V- {Zxy^-^^f-l^-f^}). 7. —(a— {« — [«— a— 26]}) 8. -(^^ + 2iC"-a5)+(3a;-a;3+l)_(2cc2_^8.T + 5). 9. 6a5-- (2y2-4aj^) -7y' + (3a;y-2y^) -(3a3'^-4/). 10. -[-{-(-,7=^)}]. 11. 10rt-(3^»-4a)-{2rt-(35 + «)}-{3/>-(2« + 6)}. 12. 2.«-{a?-(-y-a^^^)}. 13. 7- {8-[3 + (6-2^=a;)]}. . 14. —\_x— {x+ {a—x) — {x— a) —x) —a~\—x. 15. ^(-.(^(^(^(-.1))))). 16. 10-[16-(14-{12-2}-4)-10]-2. 17. Solve the equation 8i«-(5-2^+{3 + 4)=«-(2a;-10). 42. Insertion of signs of grouping. It follows immediately from § 41 that terms of a pohpioniial may he inclosed within a sign of (frouping^ ichen this sign of grouping is preceded by the positive sign, icithout changing the signs of the terms ; and. 54 ' ALGEBRA when preceded by the negative sign., by changing the signs of the terms. Thus, to inclose the last three terms of 3a?-'' + 2a:^— 4a; +1 in brack- ets preceded by the sign +, we have 3a?^ + [2x^—40? + 1]. If pre- ceded by the sign — , it becomes 3^— [— 2a?2 + 4a?— IJ. This principle is of use In combining the terms of a poly- nomial which are similar with respect to some general num- ber. Thus, combining the terms having the same powers of x, a£C* + &ir^ + 3Ce»+5x*-3iC=*-x + 4=(a + 5)x* + (6-3)^+(3c-l)x + 4. Again, 4-5^ + 3ca^-ai;c-|-6aa^ + 3.r-7ii;2^4-(a-3)x-|-(3c-7)ar' -(5-6a)ar\ EXERCISE 13. Without changing the values of the expressions, inclose the last three terms of the following expressions in signs of group- ing preceded by the sign — : 1. ^—x'-^x^-x'+x-l. 3. a + 2^>-3c + 4(^. 2. ax—by — cz-Vdw. f L(/^»Aj^. Sd-lOe^bf-g. In the following expressions combine the terms having the same powers of x, so as to have the sign + before each group : 5 . 2x^ — Sx^ + aoi? + b^ + hx — ex. 6. 7 + dx'-^x'—2ax—4ax' + Qbx^ + ^x. 7. ax*—l + 2x^ — dx* + x'^ + ax^ — cx + dx — bx^. In the following expressions combine the terms having the same powers of y, so as to have the sign — before each group : 8. -y + b + 2y' + ay-by\ 9. py-qy + ry- sy* + 2py' - Sqy\ 10. -Qxif + xY-dx'y-2y'-Sy' + by. ADDITION AND SUBTRACTION OF LITERAL EXPRESSIONS 55 In the following expressions remove all signs of grouping, and then combine the terms having like powers of x : 11. 2x-(ax'-bx)+{cx-(2x'-10)\. 12. ax'-('2x-6x'-^x). 13. (x'-x+l)-{2x-(Sx'-2)-x'}. Add the following expressions, combining like powers of x : 14. a^—1, ax^ + bx, ax^ — bx^ + cx, x-\-b. 15. ax^ + a^x\ 2ax—Sx\ ax^ — %^. 16. x^'^—ax^^b, bx' + c, 2x>—d. 17. From a^ + bx + c take bd^ + gx— d. 18. From 2£c^ — 3a; + 5 take a—bx^-\^cx. 1 9. From 2^^^ ~ 2^ take px^ -{-rx—q. CHAPTER Y. MULTIPLICATION OF LITERAL EXPRESSIONS. 43. Law of exponents. The Lxav of exponents in multiplica- tion is derived immediately from the meaning of an exponent. It is understood here that Ave are dealing only with exponents whidi are positive integers. The law expressed in symbols is flr™-a" = a"'+". WehnYG a"' = aaaa torn factors;* and . a" — aaaaa to n factors. IIe«ice a"'-a" — aaaaa to 7n-j-n factors. m+u = a Thus, d'^a*=aaaaaaa=a' ; tA/ tA/ — eA-/«A/eAytA-/«A/ tAytAy tAj > and y^-y^=i/+^ = y^. By similar reasoning this law may be extended to any num- ber of factors. Hence, fl'"-a"-a;'=fl'«+«+/-; etc. That is, the product of two or more poicers of the same base is equal to that base tcith an exponent equal to the sum of the exponents of the given poicers. * Tlie sign • is called the sign of continuation, and means " and so on," or " and so forth." Thus, 1, 3, 3, 4 is read " 1, 3, 3, 4, and so on." And aaaa is read "aaaa and so on." 56 MULTIPLICATION OF LITERAL EXPRESSIONS 57 44. Multiplication of monomials. By use of the law of order, the law of grouping, the law of signs, and the law of exponents, the product of two or more monomials may now be found. Thus, to find the product of 2i)c^y, — 3ic^2/^ ^"d 7xy^^ we have {2oc^y){—3x^y%7xy^) = —2-3-7x^a^xyy^y^, law of order and signs, = — {2-S-7){oc^3cr^x){yy^y^)^ law of grouping, =—42x®2/*', law of exponents. By these laws we have the following rule for the multi- plication of monomials : Find the j^roduct of the numerical coefficients^ using the law of signs ; and affix, to this the products of the literal factors^ using the law of exponents. Example. The product of — 3a&^ 7a^x^, and — 2&V is ( - 3a62) (Ta^x^) ( - 2h'x^) = ^2aWx\ Note. — The student must be careful not to combine the exponents of different kinds of bases, as of a and 5. It must be remembered also that if no exponent is written above a base, the exponent 1 is under- stood. EXERCISE 14. Find the product of : 1. aW, and -a'b'\ 2. Qa'b, -baWand -2a'b\ 3. Sx% — 2a;y, —bxg and 7x\ . 4. —iJfy^g'^z^, —^axg^ and ^a^. 5. \d'h'c'd^ and -Ua''hc\ 6. -'Ia-b\ -4a=^^^c and l^c. 7. 1^, -lA'B and ^A'B\ 8. F\ -P'QsLudP'Q'. 9. — ^mW and J^mn^. 68 ALGEBRA 10. 2.bx\ S.2bxy and l.lbxi/. 11. 12pqr, — 4/9^$'r^, jo^^V and —r^. 12. B''^, -lOi^/S'^ and -^BS. 13. (a + by and 2(a + b)\ 14. 3«2(^> + c) and -ba{b + cy. 16. What is the meaning of x-? Of »;'»+«? Of j^"-*? Of 16. What is the product of x"^", £c"+»' and x*"-'"? 17. What is the product of a^+S —2a^" and --3«? 18. What is the meaning of {a'f? Of (ay? Of (a;'')''? 19. Find the product of (aj\ (a'f and {a')\ 45. Multiplication of a polynomial by a monomial. The rule for multiplying a polynomial by a monomial is obtained directly from the law of distribution. Stated in symbols the law is (fl + 6 + cH- ) x = ax-\-bx-\-cx-\- If a + ^ + c+ represents the polynomial, and x the monomial, then ax-^bx-\-cx-^ is the product. Hence, the rule : The product of a polynomial midtiplied by a monomial is the si(,m of the products obtained by multiplyi?iy each term of the polynom,ial by the monom^ial. Example 1. Multiply 3ic*— 2a^+6a?— 5 by 4a^. (3x*— 2x2 + 6x— 5)4.r^=3x*-4£e— 2x2-4aT^ + 6x-4a^— 5-4a?* = 12x' - 8^ + 24a?* - 20x^ The work is conveniently arranged thus : 3ic*-2^2 + 6ic-5 4ic^ 12x' - 8x^ + 24x* - 20itr» MULTIPLICATION OF LITERAL EXPRESSIONS 59 Check. When ir=l, multiplicand=3, multiplier =4, product = 8, as it should. Observe that since any power of 1 is 1, substituting 1 for x does not check the exponents 7, 5, 4 and 3, but merely the coefficients. It is always more convenient to perform the multiplication from left to right. EXERCISE 16. Multiply and check : 1. a'-'lab + h'^hy a'h\ 2. cc^— i// + a;— 1 by £c*. 3. 6«'-5«2^>+25^by -«^6^ 4. - 2a;* - ZxY + 5y * by - Ix^y"-, 5. x^y^—m^z^ + x^w by Sxyzw. ^6. 2pq—Sqr-\~4:rphjbpqr. 7. A'-B' by AB. 8. 20x'-dx' + 2hy—4x\ ^- .9. i a^y + 1 xyz-^ £cV by -xh/z'. 10. —ha^h'^&^'^abc—ax—hy—cz by —Zabcx^y^. 11. I a;y^— i «icy + J f^ — f aV by —'la^xyK 12. Is there any difference between 4(a— ^) and («— ^)4? Why? Remove the signs of grouping and simplify : 13. x{a—h) — (a^h)x. Note. — In these expressions, products preceded by the sign — are sub- trahends. Hence, when the signs of grouping in products preceded by the sign — are removed by multiplication, the signs of all terms arising from such products must be changed. Thus, — (a +6) a? is a sub- trahend. Hence x (a — 5) — (a + h) x =^ax — hx — ax — bx 2bx. 14. A(2x-Sy)-2(4:X + y). 16. 10{x-2y)-(2x-y)S. 16. {a-b)c-(a + b)c. —17. -Sy(xy-x') + 2x(x'-y'). 60 ALGEBRA 18. -^cJ?h-a¥)-{ah^'-a'b). 19. -2(— 2j«* + 3a;V-y') + 3(^* + 2£cV-2y*)- 20. 3[2a;y-4fc{y-2(a^-2/)}]. 46. Product of polynomials. The product of two polyno- mials is also obtained from the law of distribution. Thus, (ct + 6 + c)(£c+y + ^) = «(^+y + 2!) + i(£c+y + 2;) + c(£c + 2/ + s) = ax-\' ay -}- az-{-hx-{-by -\'hz-\- cx-\- cy -{- cz. Here the product shows that each term of the multiplicand has been multiplied by each term of the multiplier ; and the product is the sum of all the products thus formed. The same method will evidently hold for any two polynomials. - Hence the rule : To obtain the product of ttoo polynomials^ multiply each term of the multiplicand by each term of the multiplier^ and take the sum of the residting products. Example 1. Multiply 2x'^—^xy-\-y'^ by ^x^—xy. Write the multiplier below the multiphcand ; then multiply each term of the multiplicand by 3^"^ ; then multiply each term of the multiplicand hj —xy. The similar terms obtained by multiplying should be arranged in columns and added. Thus 2aj2- Sxy + y^ Sx^— xy (2a^-Sxy + y^){Sx^)= 6x'- dx^'y + Sx'y' {2x^—3xy + y^){—xy)= — 2x^y + 3x''y^—xy^ 6x* — 1 IxTf + 6x^2/'' — ^if Check. When x=2 ; i/=l ; multiplicand =3 ; multiplier =10 ; product =30. Note. — A polynomial is said to be arranged according to the powei^s of some letter when the exponents of that letter either increase or decrease in the successive terms as we pass from left to right. MULTIPLICATION OF LITERAL EXPRESSIONS. 61 Thus, x*—2oiy^+x^+x~5, is arranged according to the descending powers of x ; while ^if+xy^+x^y'^+a^y+x^ is arranged according to the ascending powers of x. The student will find it an advantage in multiplication of poly- nomials to arrange, if possible, both multiplicand and multiplier according to the powers of some letter. Example 2. Multiply x^—2-\-2xhjx — ^ + x^. Arranging both trinomials according to the descending powers of x^ we have x''-\-2x -2 x^-^ X —6 a?* + 2ar^- 2x' ie + 2x'- - 2x — ex'- -12;r+12 x'* + 3x^- 6x'—Ux+12 Check. When x=3 ; multiplicand =13 ; multiplier =6 ; pro- duct=78, as it should. The product of three or more polynomials may evidently be obtained by multiplying the product of any two by a third, and so on. EXERCISE 16. Multiply and check : 1. a + b by a + b. 9. 2a;-5 by 2£c + 7. 2. a + b by a—b. 10. 4a+7 by 4«-7. 3. 2x-ij by Sx + 2i/. 11. -^-10 by -a^ + lO. 4. 2x' + by' by x—Sij. 12. Scd+Qx^/ by 2xy—cd. 5. x^-\ by 3£c^ + 4. 13. \x^-\ by fcc^-i 6. 4m^-5/^^ by 3?^^ + 2m^ 14. |a + |^ by \a-\b. 7. X^pq^^pr by 2pq—'lpr. 15. |£c'4-|y' by \x—\xj. 8. a; + 3 by £c + 2. 16, 2.25a; + 7.5y by -1.5y + 2.5a;. 62 ALGEBRA 17. .2a-lM by 4.7a + 2M. 20. a + b + c hy x + y-\-z. 18. ax + b}/ + czhj ax-hy. 21. x^—'lx-^^ by 2a;^+5ic-4. 19. pq-Vqr—pr hj pq + qr. 22. £c^ — 1 by a;^ + ic + l. 23. 4£c=^-2a;-^ + 3ix;-5by a;' + aj^ + l. 24. £c* + a;^ + l by £c*-a;^ + l. 25. a^-}-a^x—ax'^—x^ by «^— aa; + £c^ / 26. 2a'^ ba'b' - 36* by a' - ^a'b' + 26*. 27. a'-^ab + b'hj a'-ab + b\ 28. x' + x'-i-x' + x + lhy x-1. 29. J?^-4^C^by J^^-4i?(7. 30. pc+qd—rehy Sqd—2pc-^re. 31. iK^ + 2aa;^-£c* by 2x'-Sa + ax\ 32. «-^-3a'6 + 3a6^-6=^ by a— 6. 33. a'x'-2ax' + Sa'x-x' by aa;-a;^ + 6e^ 34. ab + bc—cd—bdhjab — bc + cd+bd. 35. a''—aW + a'b^ by a*-6l 39. ic^'^+^ + y"-^ by a;^-2/^ 36. aj'^ + y'^by aj + y. 40. x''—x''-' + x''-''hjx''-i-l. 37. X"— y" by x^ + y"". 41. |t«'— ^ 6«6 + i6'' by ^a—^b. 38. ^=^'' + 6^'' by a^^' + J^^ 42. ^x'-^a' by ia^^ + ^a^ 43. ^x'-^x^^ by 2£t— 1. 44. i«=^-^a;2 + ia;— ^ by 3.c^ + 9£c-27. 45. f^^ + |^-iby|a.^-|;^-i. 46. 1.4i«^-2.7a; + 3.2 by 2»5''- 1.4a; -3.2. 47. x''—y^\)yx^—y'. 48. £c»+y'^ by a;™ + y'". Find the product of 49. x-1, iB + 2, aj + 1. 52. x' + x-^l, x-1, a;M-l. 50. x^-1, a;2 + l, aj* + l. 53. a^-6', a^-\-b\ a^ + b\ 51. 2a-36, 3a + 26, a + b. 54. ic + 1, aj + 2, £c + 3, aj+4. MULTIPLICATION OF LITERAL EXPRESSIONS 63 56. ia^ + iy, ix + y, i^ + iy- 6 6. £c» — y % ic« + y ", x"' + y "'. 57. f/ -h ab -}- b'^, a — b, a^ — ab + b'^, a-\-b. Remove signs of grouping and simplify : 58. {a-{'b){a + b)-(a-b){a-b). - 59. 2(Qi'-Sx+l)-(x + A)(x-l). 60. (x + y)(x-2ij) + 2(x'-y') — {x—y)(x+2i/). 61. 2Xx+l)(x-l)(x + 2)-4x(l-x)(x + S). 62. -2a^{a^-3a(a-6)} +(a' + a^>-^>')(«'-6' + a^»')- CHAPTER VI. DIVISION OP LITERAL EXPRESSIONS. 47. Law of exponents. The law of exponents for division is obtained directly from the corresponding law for multiplica- tion, by means of the principle quotient x divisor = dividend. Expressed in symbols the law is : This follows from the preceding principle, for the quotient «'"-", multiplied by the divisor a" gives «"*-"«'", or a% the dividend. Thus, a^-i-a^=a^-^=a\ This agrees with the above principle, for a'^-a^=a^. 48. Meaning of a". By § 47 we have But «"^a'' = l, for any quantity divided by itself gives 1. Hence fl°=l. Axiom 7. That is, am/ base loith the exponent zero equals 1 . Thus, a^=l ; 2"=1 ; 10''=1 ; 45"=1; x^-^x'=3(P=\. It is therefore evident that if a base appears to the same power in both dividend and divisor, it will have the exponent zero, and hence gives the value 1, in the quotient. 49. Division of monomials. Since division is the inverse of multiplication, i. e^ since quotient X (Umsor = dividend^ then from § 44, the rule for multiplying monomials, we obtain the following rule : 64 DIVISION OF LITERAL EXPRESSIONS (;5 To divide one monomial by another^ divide the numerical co- efficient of the dividend by that of the divisor^ using the law of signs ; then divide the literal ^factors by subtracting the ex- ponents of the bases in the divisor from the exponents of the like bases in the dividend to obtain the exponents of these bases in the quotient. Thus, since (4a'&V)(-3a*60=:-12a«6V, ( - 1 2a«65c*) ^ ( _ 3a*6c'0 = 4.a'b*c\ which may evidently be obtained by the above rule. Likewise, ( - IGx'y'z') -- {Sx^y^z') = - 2o^z'. EXERCISE 17. Divide : 1. a^ by a^. 2. a'b' by a'b. 3. -QaW by 3a' 4. lSx^y''z* by 6. 4«^6''c^ by a^bcK 6. 2Sa*b'c' by -7ab\ 7. -lOOyyby -2bp'r/. 8. -bOx'a' by 4x'a\ 9. 42aWc by QaWc. 10. \m^ii^p^ by — \rnn^p^. 11. r'sH' by — 3rV^. 13. -5a;™+i by -ic». 13. by a" 14. — 12£C'»+"2/™+"by Sy'^aj" 15. x^'-^'y by of-'^y. 16. -12s¥by ^st\ 17. 18y^"-^ by 2y"-l 18. IZz'^-^y"' by 62"-"^'". 19. 7i5"'5'" by 3^"'5". 20. lls^+V-^ by Qs-'-^r'-^' 50. Division of a polynomial by a monomial. If a polynomial be divided by a monomial, the quotient multiplied by the divisor must equal the dividend. Hence, the quotient must be such an expression that the product of its terms by the divisor will give the terms of the dividend. Therefore, the terms of the quotient must be obtained by dividing the terms of the dividend by the divisor. Hence the rule : 5 QQ ALGEBRA To climde a polynomial hy a tnonomial^ divide each term of the dividend hy the divisor, and take the sum of the residting quotients. Example. Divide a?^ 4-40?^— 5x* by ic^ {x^ + 4:Qd' — ^x^) -^x^=x^ + 4a^ — ^x^. The work might be written x^ 4- 4x'-5x' ^^, 4x'-5x;\ x^ Another form often used is x^ )a^ + 4x^-5x^ x* + 4:X^ — 5x^ ' EXERCISE 18. Divide : 1. x' + x' by x\ 5. -S0a'b'-27a'b' by -SaW. 2. xy-xy + 4:S(^y by xy. 6. Sa'-6a*b + 9d'b' by Sd\ 3. x''—bx*^Sx' by —x\ 7. a'-d'b-d'c by -a' 4. ^Ix^-l^x' by -9x\ 8. 1x'''-\^x'y^ by ^x\ 9. 4a;y + 8a;y-12j«yby2£cy. 10. 3£c2— |a;y + | a^y by fa^. 11. — «+5— cby— 1. 12. «=^ + a6 + ^'by aW. 1 3. ^ficy + 5£cV' - ^a^y by — liaj^'. 14. 3.25,73' -5.2a;« + 9.75x^ by .25i«l 15. 10a;'»+='-4a;»+3 by 2a!l 17. y»+V"+^ + y'"+V'+^ by a;»+y+». 16. a^'^—a"^^ by a». 18. x"- + x''+'' -V x'^^" by a;'. 5 1 . Division of one polynomial by another. The process of dividing one polynomial by another is based upon the principle that the quotient multiplied by the divisor gives the dividend. The process is best explained by taking an example as follows, DIVISION OF LITERAL EXPRESSIONS 67 First arrange both dividend and divisor according to the descend- ing powers of x. See § 46. The work may be indicated as below. Dividend x*+ a^ + 7x^—&x + 8 a^ + 2x-\-8 Divisor. {x'^-\-2x + 8)x^= x^ + 2x^ + 8x^ of— x+1 Quotient. —x'— af—Qx + S {x^-\-2x + 8)-{—x)= —af'—2od'—8x x^ + 2x + 8 (x^4-2x + 8)-l= ' a;^ + 2a?+8 Now the product of the term of highest power in x in the quotient and the highest term in the divisor must give the highest term in the dividend. Hence, the highest term in the quotient is the quotient obtained by dividing the highest term of the dividend, X*, by the highest term of the divisor, x^. This gives x^^ the first term of the quotient. Multiply the whole divisor by the term of the quotient just found. This gives x* + 2x^ + 8x'^, which is placed below the dividend. The dividend is the product of the divisor by the whole quo- tient. And X* + 2x^ 4- 8x^ is the product of the divisor by the term of the quotient found. Hence, subtracting this from the dividend, the remainder —x^—x^—Qx-\-8 must bo the product of the divisor and the part of the quotient to be found. Therefore the product of the next highest term of the quotient by the highest term of the divisor must equal the highest term of the remainder. Hence, dividing —x^ of the remainder by x'^ of the divisor gives — x, the second term of the quotient. Multiply the whole divisor by the new term, —x ; subtract the product from the remainder. This leaves oc^ + 2x+8. Evidently the third term of the quotient will be obtained from this second remainder just as the second term was obtained from the first remainder. By continuing this process, all of the terms of the quotient may be found. 68 ALGEBRA The above reasoning will evidently apply to any dividend, divisor, and their quotient. If the divisor is an exact divisor of the dividend, the work may be carried on until a remainder zero is found. Otherwise, the process may be continued until a remainder is obtained in which the highest term is of lower power than the highest term of the divisor. This is a true remainder. It is evident that, in the latter case, the dividend is com- posed of the remainder and the product of divisor and quo- tient. That is, dividend = quotient x divisor + remainder. If, in the preceding example, we had arranged both dividend and divisor according to the ascending powers of x^ we would have obtained the same result, except that the order of the terms would have been reversed. We have, therefore, the following rule for dividing one poly- nomial by another : Arrange both the dividend and divisor according to the de- scending or ascending powers of some letter. Divide the first term of the dividend by the first term of the divisor to. obtain the first term of the quotient. Multiply the lohole divisor by this term of the quotient^ and subtract the result from the dividend. Treat the remainder as a neio dividend, {being careful to arrange the terms as before) and repeat the process, continuing until either the ronainder zero, or a true remainder, is found. Example 1. Divide a^— 11a + 30 by a— 5. a^- 11a + 30 a^— 5a — 6a + 30 — 6a + 30 a — 5 a—^ Quotient. DIVISION OF LITERAL EXPRESSIONS 09 Check. When a=l ; dividend =20; divisor =—4; quotient =—5; as it should. Example 2. Divide 789ify' + 45xy^ + Uy* + Ux* + 45a^y by 2x' + 7y^ + 5xy. Uoc* + 45a?*2/ + 78ic22/2 + 4:5xy^ + Uy* 14x* + 35^ + 49.rV 2xM- 5xy + 72/^ 7^^;=^ + 5xy + 2i/'' lO.r'2/ + 29xY^ + 45it?2/"^ + 14y* 10x'y + 25x'y' + S5xy' 4:X^y'^ + 10xy^-{-14y^ 4;rV + 10a^y' + 14 x'^z + xyz—xz^ xy^+ xyz + xz^-Vy^ ^ xy"" -vy'-y^z xyz + xz"^ +y''z—z^ . xyz -vy^z-yz^ xz^ +yz'^—2^ [^ 3dz^ +yz'^—z^ EXERCISE 19. Divide and check : 1. Grt^-7a-3 by 2a-3. 3. 17.y + 2y^ + 21 by 3 + 2y. 2. lQx + bx' + ^ by £c + 3. 4. 5a-^ + lla + 2 by a + 2. 5. 3a^-7a-2-2a^ by 1 + a. 6. a;*-2a;y + y*by ic'-2a;y + 2/l 7. c^-lOc + 24 by c-6. 11. aj^' + y^^ by x + y. 8. a' — (^' by a + b. 1 2. ic' + y ^ by aj + y. 9. a'-h' by a-^>. 13. ««-^»« by ci'-h\ 10. a;*-.y* by x-y. 14. a« + ^»« by a=' + 5^ 15. 144x^-1 by 12a; + l. 16. a;^ + 3a;2 + 3a;+lbya;^ + 2ic + l. 17. 03*— 2ic^y + 2a!y— y* by a;^— 2/1 18. a;^ + a!* + l by aj^-a^^ + l. 19. aj^-1 by a?-l. 20. 7aV-3a*-5aV + 3aaj«-2a;« by a;? + 2aaj^— aj*. 21. 2a*-9a^ + 17a^-14abya'-2a. 22. x^-y^hy ^-y\ DIVISION OF LITERAL EXPRESSIONS 71 23. iz;*-12£c=^ + 54£c^-108a;+81 by x^-Qx+9. 24. x'+x' + l by x* + x' + l. 25. 15/ + 13y-17/-3 by 6/ + 3-4y. 26. x'-a^hj x' + 2x'a + 2xa' + a;\ 27. l-x-Sx'-x' by l + 2a; + £fl 28. x' + 2a;=^y2 + 9y* by x' - 2xi/ + 3yl 29. a3* + 81 + 9£c^ by Sx-x'-9. ZO. a'-Sb'-l-Qabhya-l-2b. 31. a^"^— 41a^-120 by x'-h^x + b. 32. ^—\f^2yz—z^ by x V y—z. 33. x^—x^y—xy^-\-y^ by x- Ty^ — 2xy. 34. a-' -243 by a- 3. 35. 13«^^ + 71t«-70a^-20 + 6(«'by 3a'^ + 4-7a. 36. i6''-'-y' by cc^'— y. 37. a* + 2«^-8a + 12-7a^ by a^ + 2-3«. 38. £c^ + y^ + 2=* — 3£cys by £c + y + ^. 39. ic=^ + 3xV + ^xy^-{' y' + 2' by a^ + y + s. 40. \a''\-^^ab' + ^^h' by ia + i^. 41. ^i^a^»-32^^ by \c(}-2b. ' ^% ^-^x'-\-\xY + y'hy ^x^ + ^xy^f. 43. \xY-\-Th^\^l\x^ + \xy. ^4. 6«"'"* + 3a^"*6*" + a'"^"''*— ^=^'" by (r + ^'". 45. £c*"— y*" by x'+y". 46. 12,x''*+^ + 8£c'^— 45iK"-^ + 25£c»-^ by 6a;- 5. • 47. -^\x^^ + ly^'' by ^aj^u _|_ i.yu^ 48. 4a2 + 4a^ + <^^-12ac-6^>c + 9c2 by 2a + ^-3c. 52. The fraction. An indicated quotient is called a fraction. The terms used in arithmetic are also applied to 72 ALGEBRA algebraic fractions. In a fraction, the dividend is called tlie numerator, and the divisor, the denominator. The numerator and denominator are called the terms of the fraction. A fraction may be expressed by any of the signs used to ex- press division. a , x^ + 1 Thus, ,— a/6, a-v-6, a : 6, -' are fractions. ' h, ' ' x+1 Any laws that apply to quotients must evidently apply to fractions. In the following sections a few principles are established concerning fractions that will be needed in the subsequent work. For the full treatment of fractions see Chapter XI. 53. Since quotient X divisor = dividend, it follows from the preceding definitions that fraction X denominator = numerator ; a that is, J-- b = a. This is a useful principle. 54. The product of tioo or more fractions equals the product of the 7iumerators divided by the product of the deno)ninators ; a b c abc that is, , — = . X y z xyz To establish this, call the product p ; i. e., let abc -.-.-=p. xyz Multiply both members of this assumed equation by x, y and z in turn. DIVISION OF LITERAL EXPRESSIONS Multiplying both members by x we have Then or Then a X- • X b y c z =px. ' a b y c z ~ =px h y c z • a = =px. bers of this equa h c z a = =pxi/. z • a = =2)xi/ or a b=pxy. ^3 Axiom 3. § 63. Law of order, y, we have Why? Why? Why? Then multiplying both members by z^ we have cab=pxyz^ or abc=pxyz. Now dividing both members by xyz^ gives abc xyz abc abc X y z xyz since each member equals ^x This reasoning can be extended to any number of fractions P- Therefore, Axiom 4. Axiom 7. Example 1. — — k-^ • -^ = ^ ,. ., — ■. 5 2a^ a* 5-2a^ a Example 2. 2^2 3x^ 3z/ 42/' -2/' 32/-42/^-(-2/3) 60^ 12/- 55. To dlolde any number by a is equwaUnt to muUiplying the number by the fraction — ; that is, rV ^4: ALGEBliA n 1 ~ = n. — a a For, since quotient X divisor = dividend^ to multiply - by a gives ~a = n. And to multiply n- by a gives n—a = n- { --a] =nl^n, for the the same reason. Therefore - and ?i-must each be a quotient obtained by dividing n by «, and hence must be equal. Thus, 184=J/=3. 56. The law of distribution holds also for di vision ; that is, a-\-b + c _a be X XXX For, " + ^' + '' = (« + j + c)i, by § 55, = a- + /'>>- + c-,by law of distribution, XXX a h c = - + - + -, by §55. XXX CHAPTER VII. POWERS AND ROOTS. 57. Involution, The definition of a power of a number was given in § 18, and the laios of signs ofpoioers were established in § 36. Tlie student sliould now reread those two sections. The process of raising a base to any power is called involution. The following laws of exponents will now be established for involution, wliere the exponents are assumed, of course, to be positive integers. 58. Power of a power. (a")"' = fl"™. That is, the mth povter of the nth poioer of any number equals the nmth povner of that number. For, by definition of an exponent, («»)'" = a" • a" -te" to m factors, :^^«+n+n+ m terms, ^^^ ^f CXpOUeUtS, § 43. Thus, {a^)^=a^-a^-a^-a^=a''^ ; (x^)^=aj^-'=a;2^ 59. Power of a product. {aby = a'b'. That is, the nth poioer of the product of tioo numbers equals the product of the nth powers of those numbers. For, by the meaning of an exponent, * {abY = ababab to n factors, 75 76 ALGEBRA = {aaa to /i factors) (^^^ to ?* factors). Laws of order and grouping. By similar reasoning the law can be shown to hold for any number of factors. Thus, {xyzwy=xYz'iv'; {2aby=2^aW=8aW; {-3aWy={-3yia'y{b'y=81aW\ by § 58. Combining the laws of §58 and §59, we get the following rule : To raise a monomial to a required poioer^ raise the numerical coefficient to the required poicer^ using the laws of si(/?is; then inultiply the exponent of each literal factor h\j the exponent in- dicating the required power y then indicate the product of the Example. Raise — Sa^V^^^ to the third power. We have (-5a.-V^2)'=(-5)'^*V'^'''=-125^^2^V. 60. Power of a fraction. \b] "F That is, the nth power of a fraction equals the nth poicer of the numerator dioided by the nth poioer of the denominator. For, It) = ----- to n factors, by def. of a power, \oJ h b h a aa • • to 7i factors e c^ § 54, Thus, -hbh to n factors' a^ -3xVV 2d'b' J (-3^V)' 81^1/1^ POWERS AND ROOTS 77 EXERCISE 20. Raise to the indicated powers : 1. (ay. 13. (-ba'xyy 2. (ay. 14. (Smhy. 3. (-aWy. 15. (-2d'x'y. 4. (-xyy. .a^ 21 22. u-^^: 5. (xYzy. *"• V^V 23. (ay. 6. (axy)«. ^^ /^'Y 24. (a^hy\ ». ( ^^2/). 18. ^— ^^). 26. (-x^Y'^^K 9. (7a^yni 10. (2.y)l 19. (-1^). 27. (^). 12. (-xyy. 2«- ^7-wV- 28. (^) . 6 1 . Square of a binomial. (a + 6)^ = a^ + 2a6 + 6^ That is, the square of a blnotnial. equals the square of the first term^ plus two times the product of the tico terms^ plus the square of the second term. For, (a-^hy means (a+^)(« + ^). By actual multiplication this becomes a'^ + 2ai + ^^ Note. — It is understood here that the symbols a and h represent any terms whatever. Either term may be positive or negative. Thus, Zx^—2y is of the form a + &, where a stands for 3a?^ and 6 stands for —2y. Example 1. Square 2x-{-3y. {2x+Syy={2xy + 2{2x){3y) + {Syy=4x'-{-12xy + Qy\ 78 ALGEBRA Check. When x=2 and y=l; base =7; po\Yer =49, as it should. Example 2. Square 5x*— 2?/^ {5o(f-2y'y={5x'y + 2(5sc'){-2y')-h{-2yY=25af-20x'y' + 4y'. Check. When x=2 and y=S ; base =22 ; power =484, as it should. Examples. Square —2aH 66. (-2aH66)2=(-2a=^)2 + 2(-2a'')(66) + (66)' =4a'-24a^b + 3Qb\ Check. When a=l, h=2; base =10; power =100, as it should. Example 4. {(a + 6) + l}'=(a + 6y' + 2(a + 6) + l =a' + 2a6 + ^'-^ + 2a + 26 4- 1. It is observed that since the square of any number is positive, the terms obtained by squaring the terms of the given binomial are always positive. The other term is positive^ if the terms of the given binomial have like sigtis, and negative., if they have unlike signs, EXERCISE 21. Write out the following squares by the above rule : 1. {x + y)\ 11. {^x'-^y. 21. {m'~^)\ 2. (2a + J)^ 12. {x-Viy. 22. {mn-2y)\ 3. i'Za^Uy. 13. (2i« + 3)l 23. (46^-5)1 4. {2x^-\-yy. 14. (3ic=^ + 4)l 24. {x'-l)\ 5. {a-bf. 15. (5«^' + l)l 26. {d'-l)\ 6. {x-y)\ 16. (a* + 10)l 26. {x'-yy. 7. {^x-Zy)\ 17. {la^-^-lxf. 27. {x' + xy. 8. {a'-by. 18. (Sa^' + Sa^^)^ 28. (^y-7)^ 9. (2a;^-2/'/. 19. (a;y + 2a)l 29. (£c'-10)^ 10. (aj*-5/)l 20. (m-3)l 30. {x'-by. POWERS AND ROOTS 79 50. U + 36. (2x-Shjy. 37. (2ab-{-4:bcy. 31. (a;*-a;^)l 41. (x^'-iy. 32. (2a;*-' + 0^)^ 42. (x" + iy. """ V* ' ^/ 33. {6x'-xy. 43. (cc"-l)l 51^ (-,-~X 34. (7a + 2^.)l 44. (^._2^»)2. ' \*' 2/7- 36. (3aV + ^T- 46. (2.» + 3a»)l ^2. (^l + l^. 46. (a^+'-n'^-^y. 53. {(2a + ^) + 6-} I 38. (-2^xyy. ^7- i^^^^' + ^-^^y- 54. {3a + (5-c)}- 39. (-a^i/^ + a^V)^ ^^' («"*"-!)'• 56. {4-(2a + ^.)}^ 40. (.t'"+1)1 49. (a"5»-^-a"-'^«)'- 56. {(4-3/>)-3c}^- 62. Square of a polynomial. (a + 6 + c)^=fl^ + 6' + c'+2fl6 + 2flc + 26c. By actual multiplication, it will be found that (a + 5 + c)' = a^ + ^' + c\+ 2ab + 2ac + 2bc ; also {a + b-\-c + dy = a' + b' + c' + d' + 2ab + 2ac + 2acl -^2bc + 2bd+2Gd; and so on, for any number of terms. That is, the square of any polynomial equals the sum of the squares of all of its terms, plus two times the product of each term into all of the terms following it. Example 1 . Square 2x^ + 3a? + 5. '" (2a?^ + 3.^+ 5)2= (2ic'0=^ + (3a^)2 + (5)2 + ^{^x'^Zx) + 2{2x^){S>) + 2(3if)(5). = 4a?^ + 9a;2 ^ 25 + 12x^ + %W + 30^. =4x^12x^ + 29x^ + 30.^ + 25. Check. When x=l;* base=10; power=100. To check ex- ponents also let X equal some other number than 1. Check when x=2. go ALGEBRA Example 2. Square a*—2a^ + 3a^—^a. {a*-2a' + 3a'-4ay={ay + {-2a'r + {Sa'y + (-4.ay + 2{a')(-2a^) + 2{a'){Sa')+2(a*){-4a) + 2{-2d'){3a')-{- 2(-2a^)(-4a) + 2(3a'-^)(-4a) =aH4a«4-9a*4-16a2-4aH6a«-8a5-12a^ + 16a*-24a3 =a«-4a^ + 10a«-20a5 + 25a*-24a^ + 16a^ ' Check. When a=l; base=— 2; power=4, as it should. Let a=2 and check. Note. — The student should learn to write out these values without indicating the work as in the first step. He should always check his work. Example 3. Write out (S—2x + x''y. {S-2x + x''y=9-\-4x^ + x*-12x + 6x''-4x^ =9-12x+10x^-4x^-{-x\ Check. Whena?=2; base=3; power=9. EXERCISE 22. Write out the squares of the following polynomials : 1. l+x + x\ 10. 4x'-a' + Sax. 2. a* + a' + 2. 11. Sx'-^y"-] bb-a\ 3. ^x + Sx' + 4x\ 12. b'-4ac+10. 4. x'+x' + x + l. 13. ab-hc\cd-ad. 5. 2iK*-3a3^ + 4. 14. a'-^¥-&-d\ 6. a-2J + 3c-4f7 H5e. 15. 2.^•^ - 5a; + 7 - 3al 7. l_a;-|-a;2_|_a;3 4-a;4. 16. \-x-Vx^—x^-^x'-x\ 8. 5a;•'-2 + 7«^ 17. ic" + y« + ^". '^,"x-\-y—z-\-w. 18. a?" — y"+^4-2«+^ 19. Show that the rule of § 61 is a special case of § 62. 20. Show also how § 62 could have been obtained from § 61 by so grouping as to form a binomial. POWERS AND ROOTS 81 63. Any power of a binomial. By actual multiplication it is found that (a-{-by = a'+3a'b + Sab' + b'; (a + by=a' + ^a'b + Qa'b' + ^ab'-}-b'; (a-^b)'=a'-^5a'b + 10a'b' + 10a'b'-{-5ab'-hb'; (a-^by=a'-}-Qalb + 15a'b' + 20a'b'-\-16a'b'-\-Qab'-\-b'; and so on. Now by comparing these few values of the different powers of a+b, it is found that they all may be written out by the following laws : (i) The first term in each case is a vnth an exponent equal to the exponent of the binomial ; the last term is b with the same exponent. (^) The expo7ient of a in each term after the first is less by 1 than its exponent in the preceding term, b appears to the first poy>er in the second ter^n., and its exponent in any term, after the second is greater by 1 than its exponent in the preceding term. The sum of the exponents of a and b in any term is the same for all terms., and equals the exponent of the binomial. {S) The coefficient in the second term equals the exponent of the first term, 'jind the coefficient of any term is obtained from the preceding term by multiplying the coefficient of term by the exponent of a and dividing the product by a number greater by 1 than the exponent of b in the term. {4) The number of terms is always greater by 1 than the ex- ponent of the binomial. Note.— These laws constitute what is known as the Binomial Theorem. This theorem was first estabhshed about the year 1665 by the great mathematician, Sir Isaac Newton. 6 82 ALGEBRA In chapter XXITI this theorem will be proved to hold for any positwe integral exponent, and its application to fractional and negative exponents will be shown. Example 1. Write out the value of {x-\-yf. By (1) and (2), the terms without the coefficients will be ixf" x'y ixfy'^ scr'if x^y^ x^y^ x^y^ xy^ y^ By (3), the coefficients will be 1 8 28 56 70 56 28 8 1 Hence, {x + yf=j(^ + ^x^y + 2%x^y'' + mx>y^ + H)x'y^ + 56xV'^ + 28ic'7/« + ^xy'' + if. Check. When x=\, y=l ; base=2 ; power=:256. Example 2. Write out the vahie of {x—yf. The exponents and coefficients may be calculated at once. {x-yf=x' + ^x\-y) + ^x{-yy + {-yf = x^ — 3x^y + 3xy^ — y^ . Check. When x=3, y=l ; base=2 ; power=8. Example 3. Write out the value of (2x^—3y-^y. {2x'-3fy=(2xy + 4{2xy{-:iy') + ()(2xy(-3yy + 4(2x')(-:iy'y + (-3yy =16x^-96x^y^ + 216xY-21Qx'y^ + SU/\ Check. When x=2, y=l ; base=5 ; power=625. EXERCISE 23. Write.out the values of the following powers : 1. (x + yy. 8. (Ax'-Syy. 15. (f x— fy^. 2. (x~yy, 9. (x + iy. 16. (ix' + Wy- 3. (x-ay. 10. (1 + ay. ^^ /a c 4. (x + yY\ 11. (x-iy. ' \^ ^^ 5. (^x + 2yy. 12. {x-2y. 18. (^-2 6. (^x^ + yr- 13. (1+yy. /I^ , M' 7. (x'-yy. 14. (x + ^yy, ^' \a'^ bj' POWERS AND ROOTS 83 20. (2x-yy. 22. (2a-^y. 24. (f«-f^»)«. 21. {x + i^y. 23. (i-2a)«. " 25,{ix-yy ROOTS. 64. If all of the factors in a product are equal, one of the factors is called a root of the product. Thus, a is a root of aaaa, or a\ The nth root of an expression is one of its 7i equal factors. Thus, the square root of an expression is one of its two equal factors. The cube root of an expression is one of its three equal factors. The fourth root of an expression is one of its four equal factors. The fifth root of an expression is one of its five equal factors ; and so on. To indicate a root of an expression the radical sign (|/ ) is used, ^/x represents the fourth root of x. Here 4, the num- ber placed above the radical sign, is called the index of the root. The index of a root of an expression indicates what root it is, or the number of equal factors in the expression. The index of a square root is usually not written. Thus, f/x represents the cube root of x; i.e., one of the three equal factors of x. ya represents the fifth root of a. ya is the same as j/d. y 81 represents one of the four equal factors of 81 ; i.e., y 81=3. A root is called an even root if its index is an even number, and an odd root if its index is an odd number. Thus, yd represents an even root ; yd an odd root. It follows from the above definition that to find the nth root cf (i gimn number is to find a second number whose nth power equals the given number ; that is, {y^ay = a. 84: ALGEBRA. Thus, since 5^^=25, therefore |/25=5 ; since {a}f=a^, therefore The process of obtaining a root of an expression is called evolution. 6 5 . Laws of Signs. The laws of signs of roots are obtained from the laws of signs of powers. The following principles are true : (J?) A 2^ositwe number has at least two even roots which differ only in signs. For, if two numbers have the same absolute value, but differ in sign, their like even powers are equal and positive. Thus, since ( + 3)^=9 and (-3)^=9, therefore v'^9=+3 or -3. Also, since ( + 2a^)*=16a^2 and (— 2a=')*=:16a'^ therefore y'T^'^ + 2a^ or -2a^ The two even roots of a positive number are often written together, by use of the double sign ± . Thus ]/25a*= ± 5a^, means ]/25a*= + 5a^ or — 5a^ {2) Any jjositive nnmher has at least one odd root., which is also positive / and any negative nnmher has at least one odd root, which is also negative. For any number has the same sign as any odd power of itself. Thus, since ( + 3)=^ = + 27 and (-3)=^= -27, therefore ^27= + 3 and ^^^=-3. Since ( + 2)^= + 32 and (-2)'^=-32, therefore |/32=+2 and 5/ V -32=-2. (3) A negative number has no even root that can be expressed as a positive or negative number. For any even power of a positive immber, or of a negative number, is positive ; i. 6., no positive or negative number, raised to an even power, can give a negative number. POWERS AND ROOTS " 85 Thus, y/ —9 is neither +3 nor -3, for ( + 3)'= + 9 and (-3)" = + 9. The indicated even root of a negative number is called an imaginary number, and does not belong to the series of num- bers with which we are now acquainted. Imaginary numbers will be discussed in Chapter XIV. 66. Root of a power. A root of a j^otrer of a base equals that poioer of the base vnhose exponent ^.9 tJie quotieiit obtaiued by dioidlng the given exponent by the index of the root. That is, n — 77 ya"' = a"'^". For, by § 58, (a'"-5-") " = a"'-5-''x» = a"'. m Hence, i/o^^^"'^'* or <^"- Thus, j'/^o :^ ct2o - 4 ^ ^5 . ^^ = x^ = x' ; f{x-yf=(x-yy^-= {x-yf. 67. Root of a product. The nth root of a product equals the product of the nth roots of its factors ; that is, y ab — yay b. For, by § 59, ^yay'br = {{raY-{yby=-ah. Hence, ^} ^ == ^^ ~ . ^« ^. Thus, ^^^ = -,^^l>^; -i>^Y^= |/^|/^=a^Y; y" - 32a?^ = y^^y'¥^=-2-x' or -2x\ 68. Root of a fraction. The nth root of a fraction equals the nth root of the numerator divided by the nth root of the denomi- nator ; that is, V\ S6 ALGEBRA For, by § 60, Hence, ^ * yb Thus, |/g = ^'' y 2436^^ ^'2436^'^ 2a« 36=^' 69. If the value of the mdicated root of a rational expres- sion can not be exactly obtained, the indicated root is called a surd. An expression which contains one or more surds is called a surd expression, or an irrational expression. Thus, |/3, ]/ c?, -y/a + b, are sia^ds ; Vx—\/y'^ 24-i/5, ixvQ irra- tional expressions, yl + y^ is not a surd since 1 + ^/3 is not a rational expression. A perfect ntla. power is an expression whose ni\\ root can be obtained ; i. e., the ?^th root of which is a rational expression. Thus, since (it^— 4)2=x*-8a^' + lG, then yiJc^—%x'-\-U=x'-4.. Hence, a?*— 8a?^ + 16 is Si perfect square. 70. Roots of monomials. Roots of monomials may be extracted by means of § 65, § 66, § 67, § 68. Example 1. Find the square root of 25a^y*. We have |/25aV= V25\/a^i/y* § 67. = ±.5aV. §65, §6G. Example 2. Find the fifth root of ~B2x'Y^. We have ^-32x'Y'= \/~^^ y'x^^Vy^ % 67. = -2icy. § 65, § 66. si 125a;V Example 3. Find the value of \~ 21 6a'>&' >' ./ I2r»y_ rm£i- ^^ j„ 4 216a»&8 f 216 a«6« POWERS AND ROOTS 87 §67. -"""^2* §65, §66. J^Y5- Example 4. Find the value of i/f+ Adding tr.c«on.. l/ff^.y/S = / 16^2 25 , =±i|/2, surds. EXERCISE 24. Find the value of Al ^9Z^5^ 11. tM6^5«. J^^ i/J^, 2. |/T6-^y. ^12. ^>8k,Tvy^. ' /^l^l 4. v/225m'»«'. /^U. ^"=«Tpy?^. 23. r5?^«. ^24. j;/i?5^ ^ ^5. if 27^^«. 15. i:^32^ 6. f-^xY\ '16. ^/-243«^^y. *^^"'** ^^ "^ 7. r^^^. n. i^-,^V^a^6-A26. V^T^. 8. f/-125m^Vi«. ,18. i^e^^^s. \ 26. i/TT^T. 9. iTGlSy: 19. ^-Vl^aH'^ \ 27. ^/fT]. 10. i^ip^l /20. v'afy^ \28. Vf^. 71. Square roots of trinomials by inspection. In §61 it was shown that the square of a binomial was a triuoraial. I'lie square root of a trinomial that is the square of a binomial can be found by inspection. 88 ALGEBRA Note. — It is not always possible to extract any root of any expres- sion. See § 69. Since the square root of the trinomial is to be a binomial, it must take the form a + h. Plence, the trinomial must be of the form {a-^b)\ov ce \'lah-\^h\ From the form of a' + 2ah-^b\ we have the following : A trinomial is a perfect square if tioo of its terms are perfect squares (a^ and 6^), and the other term is tvnce the product of t?wir square roots (2a 6). Thus, a*— 6a2 + 9 is a perfect square ; for |/c? is either a^ or —a^, |/ 9 is either 3 or —3, and twice the product of two of these values, a^ and —3, or —a^ and 3, gives the other term — 6a^ In fact a*— 6a''' + 9 is the square of a^— 3, or of — a^ + 3. Why not select a'^^- 3 or -a='-3 ? From the type form, fl^^-2a6^-6^ = (a^-6)^ we have the following rule for obtaining the square root of the trinomial : Write the sum of the square roots of the terms that are p>er- feet squares^ using such signs that twice the product of the result- ing terms will gim the other term of the trinomial. Example 1. Find the square root of a;^— 14a? + 49. We have j/p=±ic,|/49=±7'. ^ Since the product — 14.:r^ is negative, the terms must have unlike signs. Hence we use ^x, and —7 or —x and +7. Therefore y x' — \^.x\ ^^—x—1 or— ^ + 7. See §65, (1). Check. When x=\ ; base=36 ; root= — 6, or 6. Example 2. Find the value of |/9x* + 30x'-^?/ + 25 z/^. Here |/ 9x*= ± 3a?', |/25p= ± 5?/ Since + ^^x^y is positive, we must use like signs. Hence |/9^*T30^pT252/^=3a?H5?/ or -3ic2-52/. Check, When a?=l, 2/=l ; base^^Oi ; root=8 or —8. POWERS AND ROOTS g^ EXERCISE 25. Determine which of the following expressions are perfect squares : 1. 4a' + 4ab + b\ 8. a'b'-2a'bc' + c\ 2. a'-^ab + 9b\ 9. {x' + x' + l. 3. lQx' + 24x}/ + 9if. 10. l--ia + ^i-a2 4. x + 2xij-\ri/, 11. 25xY- + 20axij + 4a\ 5. 9r«^ + 24a&-16^>l 12. 121aj«-20ic^ + l. 6. m^-10m?2 + 25?zl 13. «* + 50a"' + 625. 7. // + 16^yV + 64c\ 14. a^-4a/y^ + 4^>l 15. IQx'-Ux + SQ. Find by inspection the square roots of the following trino- mials : 16. ii;' + 10a; + 25. 30. 25a'' + 10a'x + x\ 17. x' + 12x + SQ. 31. a'x' + 2axy + 7/\ 18. a;^ + 16a; + 64. 32. 9.^y-24«^Z.% + 16rt*6«. 19. CC2 + 18.T + 81. 33. l-6a; + 9a;l 20. a3--20i^ + 100. 34. 1^.2 + ^^^^ + ^^2^ 21. ic^-30ic + 225. 22. a^ + 50« + 625. 23. 4a;^ + 28a!-f49. 24. 9^^ -30a; + 25. 25. l6a2-48« + 36. 26. 81a;^ + 36£c^ + 4. 27. 121aj«-22a;M-l. 28. lQ9a' + U2ab + 4db\ 29. 9a;*-12ajy + 4y. — 40. -^ 144 35. 9^ -v^v '+¥2/*- 36. a' %•- 2/^' 37. 2- + 1. 2/ 38. 16a;^ 49 -2 + 49 39. 4 +* + «* 2 + ^^^. ^ 1 iC« 90 ALGEBRA 72. Roots of polynomials by inspection. Since the process of finding the ?^th root of an expression consists of finding a second expression whose nth power is the first expression, the roots of some polynomials may be found by the aid of § 63. Note. — A general process of finding the square roots and cube roots of polynomials, and of arithmetical numbers, will be found in the Appendix. These methods by inspection will suffice for the present. Example 1. Find the cube root of x"* — 3x'^y + 3xy^ — y^. Here the first and last terms are perfect cubes, and there are four terms. This suggests that the given expression may be the cube of a binomial. See § 63, (1) and (4). Taking the cube roots of the two terms which are perfect cubes, we get x and —y. Their sum, x—y, is the cube root required; for if we cube x—y by § 63, we get xr'-Sx'y + Sxy^—yK Example 2. Find the fourth roots of 16x^—96x^y + 21QxY— 216xV + 8l2/^ This expression has five terms, and the first and last terms are perfect fourth powers. Taking the fourth roots of these two terms, we get ^^320; and J^Sz/, respectively. Hence the fourth root will be either 2x^-{-^y^ 2x^—3y, —2x^ + 3^, or —2x^—3y. Of these, the fourth powers of 2x^—3y or -2x^ + Sy will give 16x«-96^«i/ + 216xY-216xV' + 8l2/*. Hence, the fourth roots are 2x^—3^ and —2x^ + 3y. From these examples it is seen that If a perfect cube contains just four tenns^ arranged according to the poicers of some letter^ the cube root is the sum of the cube roots of its first and last terms y and if a perfect fourth poicer contains just five terms^ arranged according to the powers of some letter^ its fourth root is the sum of the fourth roots of its first and last terms ; and so on for other poicers. The terms of the nth root of an expression must always be given such signs that when the root is raised to the iith power by the method of § 63, the result will be the given expression. POWERS AND ROOTS 91 Tlie work should be checked by seeing if the root will produce the given power. EXERCISE 26. Find the cube roots of 1. x' + V2x' + 4Sx + Q'^. 3. 8a;^ + 36a;- + 54.^ + 27. 2, x'-Ux' + 7bx-12^. 4. 27cf;'-10Sa'b + lUab'-Ub\ 5. 64^«-144e^y + 108a^y-27/. Find the fourth roots of 6. a'-4:'ab^Qa'b'-4ab' + b\ 7. x' + 20x'-\ 150^^ + 500a; + 625. 8. Sla''-4^2a'P'}'SQ4a'b'-lQSc(;'b' + 2^Qb'\ Find the fifth roots of 9. aj5-10a;* + 40£c-^-80a;^ + 80a;-32. 10. S2x' + 240x\i/ + 720£cy + 1080£cy + 810a^y* + 243y^ Find the sixth roots of 11. 729a«-1458tr^^^ + 1215«V-540a^5« I Uba'b'-Uab'' + b'\ 12. x'' + 12x''a + QOx'W + 160icV + 240icV + 192^3^^ _|_ q^^b CHAPTER VIII. SPECIAL PRODUCTS AND QUOTIENTS. 73. There are some especially important products and quo- tients which it is essential that the student should master before proceeding. They are fundamental forms that are often met in algebra. The student should learn to write out the pro- ducts or quotients that come under these forms by using the rules or formulae without performing the actual multiplications and divisions. PRODUCTS. 74. Product of the sum and the difference of two terms. By actual multiplication, {a + b){a-b)=a'-b\ when the symbols a and h stand for any terms ichatever. That is, the p)roduct of the sum and the difference of the same two expressions equals the diff^erence between their squares. Example 1. Find the product of x-\-2 and x—2. {x + 2){x-2)=x'-2'' =x^-4.. Check. When x=4: ; factors are 6 and 2 ; prodnct=12. Example 2. Find the value of (2^^+ 6a') {2x'-5a'). {2x' + 5a^) (2x'' - 5a^) = {2x'f - {^a^ =4x*—2oa\ Check. When x=l, a=l ; factors are 7 and— 3 ; product= —21. Example 3. Find the product otx + y + 5 and ,r + ?/ — 5. 92 SPECIAL PRODUCTS AND QUOTIENTS 93 By grouping terms, these trinomials can be written in the type form of the binomials a-^h and a—h. We have x-\-y->t-^= (-r + i/)+ 5 ; x-\-y — ^= (x^y) — 5. Hence ix + y + 5){x + y—5)=[(x + y) + 5][(x+y) — b] = {x + i/Y-5' =x'^ + 2xy + y^—2o. Check. When x=l, y=l\ factors are 7 and— 3; product =—21. Example 4. Write out the product {a + h-\-c) (a—b—c). Grouping terms, a-\-b'\-c=a + {b-\-c) ; a—b—c=a—{b-\-c). Hence {a-\-b + c)(a—b—c) = {a + b-\-c){a—b-\-c). =a^-{b-\-cy =a'-(jb^ + 2bc + c') =a^-b''-2bc-c\ Check. Whena=4, 6=2, c=l; factors are 7 and 1; product ;=7. EXERCISE 27. Write out the following products without performing the actual multiplication : 1. {x + y){x-y). ^^ 10. 0«^ + y^)(a3^-y^). 2. {m-n){m-Vn). 11. {a''-b'){d' + J)'). 3. (a— 5) (a-f-5). 12. (m«— /i'')(m*^ + n*'). 4. (a + 10)(a-10). 13. {xy-\-ab){xy-ah). 5. (y-3)(y + 3). 14. (xY-z>){xY-Vz'). 6. {a + x){a-x). 15. {'lx'-hy'){^x'-\-by^). 7. (2a + 3)(2a-3). 16. (3a^-75^)(3a^ + 7^0- 8. (3a;-2y)(3a5 + 2y). 17. (4aic-^ + 5%)(4aar'-5%). 9. (5m-4?z)(5m + 47z). 18. (\x'^-ly'){\x'-lf). 19. (2.5a^-1.7^)(2.5a^ + 1.76). 20. {^a'x' - 88iy) (3|a V + 33 ly*) . 21. (a!H-y + 2)(a! + y-2). 22. (a^-y-8)(a!-y + 8). 94: ALGEBRA 23. (x + a + b)(x-a-b). 25. (r'-r + l)(r' + r-^l). 24. f2a;-3y + 4)(2a; + 3y-4). 26. (x-l)(x^l)(x' + l). 27. (s' + s + l){s'-s + l)(s'-s'-{-l). 28. (x*-4:)(x' + A)(x' + lQ). 29. {x + y-i-z)(x'{-y—z)(x—i/-{-z)(—x+y+z). 30. (10a;" + a») (a"-10«"). 31. (af+' + y-')(x-+'-y--'), 33. ^-i + J-VJi-l-Y \4x' 2i/y\4x' 2yV 75. Product of two binomials having a common term bs (x + a) (jr + b). By multiplication, (x + a)(x + b) =x'^ + ax + bx + ab. Adding like terms, this becomes x^-^{a-^b)x-hab. Hence, (x-^a)(x-{-b) = x~ + (a-^b)x-\-ab. That is, the j^^'odiict of tvno binomials having a common term equals the square of the common term^ plus the product of the common, term atid the sum of the other terms^ plus the product of the other terms. Example 1. (x' + 2)(e;i; + 3)=x2 + (2 + 3)x + 2-3 Check. When a?=l ; factors are 3 and 4 ; product =12. Example 2. {x-4) {x + 2)=x'+ (-4 + 2)a? + (-4-2). =a?'-2x-8. Check- Let x=l \ then (-3-3) = l-2-8, or-9=-9. SPECIAL PRODUCTS AND QUOTIENTS 95 Example 3. {5x'-{-2y''){5x''-7y') = {5a^Y-\-{2y'-7y'')5x'-\-2y'{-7if) =2^3C^—2^xhf-l^y\ Check. Let x=2^ and 2/=3. (Left to the pupil). Example 4. (a + fo + 5)(a + 6— 2) = (a + 6 + 5)(a + 6— 2) = (a + 6)' + (5-2)(tt + 5)-(5-2) =a^ + 2a& + 62 + 3a + 36-10. Check. Let a=l and h=2. (Left to the pupil). Note. — In many of the examples worked out in this book hereafter the process of checking will he omitted, in order to save space. .But the student is advised to always check his work. The liabit of check- ing cultivates the indispensable habit of accuracy. EXERCISE 28. Write out the following products ; 1. (^ + 3)(^ + 4). 15. {A^-^){A'-1), 2. (a; + 7)(i»-3). 16. (2+i>)(jt>-5). 3. {h-Q){h-b). 17. (mV + 8)(6 + mW). 4 (ic-10)(i« + 2). 18. (6-a;)(12-a;). 5. (a-8)(a + 6). 19. (3-a)(10-a). 6. (m-ll)(m-2). 20. (-c + 5)(-c-7). 7. {x-b){x-^). 21. (a;» + 3)(£c'' + 7). 8. (s-10)(«-3). 22. (a;" -3) (a;" -5). 9. (jt)^ + 12)(jt>^ + 10). 23. (««+^-6)(«"+^ + 5). 10. (a;=^ + 6)(a^^ + 8). 24. {B'-4.AC){B'-QACy 11. (r^-5)(r^-3). 25. (« + 5 + 3)(a + ^>-2). 12. (a;=' + 12)(a;2-4). 26. (aj-y + 7)(«-y+ 3). 13. (a;^H-12)(a;^-3). 27. (i? + ^ + 10)(jt> + ^-16). 14. («;y^-4)(l+a;2/-^). 28. (a;^ + a; + 6)(a;^ + a;-3). 96 ALGEBRA 29. (««-«« + 10)(a«-«^-20). 32. (x'-l + x)(x'-2-^x), 30. (xy + xy + xi/) (xy + xy 33. (x' - a') (x' - 3a^). — 3a;y). 34. (aj« + 3y»)(a;" — Ty"). . 31. {z + b-b)(z^7-b). 35. (iC«+^-22/"-^)(3y»-^ + a;"+0. QUOTIENTS. Since division is the inverse of multiplication, from the type forms of multiplication, certain type forms of division follow. .76. Difference of two squares. Since, by § 65, (a + b)(a—b)=a~ — b% then a'-b a-^b; _ . L =a — b- fl-6"~""' a + b That is, the quotient obtained by dimding the difference be- tween the squares of tioo expressio?is by the diff^'erence between those expressions equals the su7n of the expressio9is. And the quotient obtained by dividing the difference betioeen the squares of two expressions by the su7n of those expressio7is equals the difference between the expressions. EXAMPLE 1. ^^ = (£!rr:(5)^=^. + 5. XT— 5 X*— 5 Check. Whena?=l: divisor= — 4; dividend =—24; quotient=6. E^^^^^^- 2- 9a^a^ + 2y^ ^ 9a-.^+2^- "^^ ^^"^^ • EXERCISE 29. Perform the indicated divisions : 1 x'-9 o «*— 25 K 4ic' 3 'a'- 5 2x-l a; +3* ' ^=^ + 4 3a +1* 1 — lax 1- -IQx^ 1- -4.x* ' r- -1 SPECIAL PRODUCTS AND QUOTIENTS 97 y 2^a'^l .. 64x*-Sly' 2ba'-22^b' ' ^ab+1 ' ' Sx'-Qf ' ^^' ba-lbb' ' ' ^ " • lla^«+10a * a«-6«* 1 — Kymrw 10 ^'"-^ 14 169a;y- 144a^ - (« + ^)2_^2 • 1 + ^^* ■ V6xy'^Vla' ' -^^5-— • a + h—x—y ' ' {a + by-\-{x—yf 77. Sum and difference of like powers of two expressions. By actual division, ^■^—^ = a' + ab + b'; a —0 ' ^lz^' = c(?-\-ceb-Va¥-VW', CI/ d'^ — b^ is not divisible by a + J; ^^J^ = d' - cr5 + «6^ - h' ; d^^-b^ is wo^ diinsible by a— ^ ; a* + ^* is 7/^0^ diuisible by a—b\ d^ + b^ 2 7 I Z.9 z=a' — ab + b^\ a -{-b a^ + 5Ms ?io^ divisible by 6« + 5. These examples illustrate the following principles.* (a) a" — b* is always divisible by a — b. The quotient is a''-^ + a''-264-a"~^5^+ +6"~^ Thus, a'^—b^, a^—¥, a*—b*, etc., are divisible by a—b. * The general proof of the divisibility in these various cases is left for the student in Exercise 43. 7 98 ALGEBRA (b) a" — b" is divisible by a-\-b only 'when n is even. The quotient is a"-^ — W'-'^b + a''-W— — 6«-i. Thus, a^—b^, a*—b\ a^—lf^ etc., are divisible by a + 6 ; whil^ a^— 6^*, a^—b^,a'—b\ etc., are not divisible by a + b. (c) a''-^b'' is never divisible by a — b. Thus, a^ + 6^ a^ + 6% a* + b\ are not divisible by a—b. (d) fl" + 6" is divisible by a^b only when n is odd. The quotient is a"-'^—a>'-^b + a''-W— +6""^ Thus, a^4-6^ a^ + 6^, a^ + 6^ etc., are divisible by a + &, while a^ + b^, a* + b\ a^ + b^^ are not divisible by a + b. From the preceding, observe that : W7ien the divisor is a — b, the signs of all terms of the quo- tient are +. When the divisor is a + 6, the signs of the terms of the quo- tient are alternately -\- and —. The exponent of a in the first term of the quotient is less by 1 than the exponent of a in the dividend., and decreases by 1 in the successive terms. The exponent of b is 1 in the second term of the quotient ., and increases by 1 in the successive terms. Example 1. Divide a^—Whj a—b. ^^ =a' + a'b + a'¥ + a'b'-\-ab* + b\ Check. When a=2, b=l ; dividend=63 ; divisor=l ; quo- tient =63. Example 2. Divide 64^2/' + i25 by 4xy + 5. Uoc^y^ + 125 ^ (4xyf + 5^ 4:Xy-^5 ~ Axy + 5 = (4xyy-(4:xy) (5) + 5' =^lQx'y'-20xy + 25. SPECIAL PRODUCTS AND QUOTIENTS 99 Example 3. Divide 32a^-2436^" by 2a-Sb\ 32a^ - 2436^ ^ (2af - {Sb'f 2a-3b' 2a-W = {2ay + (2a)^(36*0 + {2ay{^by + {2a) {Wf + {W)' = 16a* + 24a=' h' + 36a' b' + 54a6« + 816«. EXERCISE 30. 1. Can a*— 1 be divided by a-\-l ? 2. Can cc'—af be divided by a—x? 3- Can a'—x^ be divided hj a + x? 4. Can a' + i' be divided by a + ^* ? By Avhat expressions can the following be divided ? 6. ay' + S. 6. l-£c\ 7. a;^-32. 8. a= + ^>l 9. xy-l. Determine which of the following indicated divisions are possible, and write out all possible quotients : 10. m—n 11. 1-/ l+y- 12. a^ + 1 a-1* 13. aM-1 a + 1* 14. x'-l x-1' 15. a'-b' IB 16a«-81J* 18. 19. 20. 21. x'^'y" x^-\-y^ 'xTy" a* + b* a—b' y x-y 22.^' x+y 23.^ 'Id' + Zb' xy'-a' «• + ! 100 ALGEBRA EXERCISES FOR REVIEW (II). 1. What is the rule for adding similar terms ? From what law does it come? Add ^xihj^ —\Qi'}y^ |a?^y, — 6a;-y, —\xhj. 2. How do you add dissimilar terms ? Add 2a^, — 3a;, — 2c, ^x\ 3. Simplify 3.T^- 22/ -7y + 6£c^ + 2«. 4. In what letter are the terms 3icy^, 2aa?, Zcxy similar ? Add them. 6. How do you add polynomials ? Add 6a;^— 2£c + 5, —^x^^ 4i«-l, 7a;^-5£c-2. 6. What is meant by checking work in algebra ? How would you check the result of exercise 5 ? Check the work. 7. Add d^—y^^ ^a^y—xy'^^ x^ + ^xy'^—y^. Check the work. 8. How do you subtract polynomials ? From 3c«"' — 2«^ + a— 4 take a? + 4a^ + 1. Check the result. 9. What are the laws of signs to be observed in removing signs of grouping ? 10. What is indicated by 6£c-4y — (3a?— 2y) + (5£c + ?/)? From what fundamental processes do the laws of grouping follow? 11. Simplify a-[3^>+ {3c-(c-5) + «} -2a]. 12. What laws must be observed when inserting signs of grouping ? 13. Group like terms in x so as to have the sign + before each sign of grouping : Ix^ — 3c^£c — ax^ + 5ic + Ix^ — ahx". 14. Group like terms in x so as to have the sign— before each sign of grouping : lyx — ax^ — hx" + 3a;^ — ^ax — ^bx? + 2ic^ — ex?. 15. Add by combining like powers of x\ «^— 2a;, aa;^ + 5, ax'-Zx^^-hx^^ EXERCISES FOR REViEW 101 16. What is the law of exponents in multiplication ? Find the value of a^a^ ; a-a}^-(]^\ x^-x'-x^. 17. How do you multiply monomials ? Find the product of — 2£cy, Za^xif^ —ax^, and — |^.y. 18. What is the meaning of x' ? Of x" ? Of (xy ? 19. From what law do we obtain the rule for multiplying a polynomial by a monomial? State the rule. Multiply x'-'2x' + Sx-bhj 2x\ 20. How can you check your work in multiplication? Check the work in the preceding multiplication. 21. From what do we obtain the rule for multiplying a poly- nomial by a polynomial? Multiply 2a^^ — 3a'' + 26' by d^~ab -\-b\ and check the work. 22. Simplify ^[^ab-'2a{b-4(a-b)}']. 23. What is the law of exponents in division ? Find the values of «'"^-a^ ; a^-^a^ ; a^~a\ a}-^a^. 24. What is the meaning of a^ ? How is it shown ? 25. How do you divide a polynomial by a monomial ? Divide a'-'2a;'b' + b' by a'-2ab + b\ 26. What is the relation between the dividend, divisor, quo- tient and remainder ? 27. What is a fraction ? - X 28. Prove that ~y=x. 29. How do you find the product of two or more fractions ? TV ^ ^^ ^ y 21a; 1 30. {ay='i («»)"•=? {by{my='i (3^)*=? State the law. 31. {xyy='i {xYy=? {4d'b'y=? State the law that you used. 102 ALGEBRA 32. What are the laws of signs in involution? { — 2x^yh^y^ = '> %7 • 33. What is the square of a-\-b? Show that your result gives a rule for squaring any binomial. (2x^ — Sy*y-=? {la' + Shy = ? {x'-2y=? 34. Square « + ^ + c and show that the result gives a rule for squaring trinomials. Square ^x—Sy + zhy the formula you have just derived. 35. What is a root of a number ? What is the index f 36. What are the laws of signs in evolution? ]/4iK^=? f/27a^^-? f'-32a'"^>'^=? fl^hf=^. |/'=4=? 37. What is a perfect nth power ? Illustrate. 38. i/16£c*-8£c^a + a^ = ? |/81 + 25/iM-90??;^ = ? f/8a;'^-36ajy + 54£cy-27y'* = ? 39. Find the product 0,1 a— h and a-\-h and show that this gives a rule for finding the product of two binomials. (4£c' + 3y-)(4a;^-32/=^)-=? (i-6cc-^)fi + 6£c^)-? 40. Find the product of a^ + a and x^h and show this gives a rule for finding the product of two binomials which have a common term. (a;^ + 6)(a;^-4) =? i^lah + 3)(2a^> + 7) =? (4c.^-3)(4c£c + ll)=? 41 ^'-y'_^ ^.B_y6_^ a^«+y«_, X —y ' x^-y^ ' sc^+y' d' + l 42. AVhen is a" — b" divisible by a + b? 43. When is a'^ + b" divisible by a + 6? CHAPTER IX. FACTORS. Definitions and type forms. 78. Factors were defined in § 12. It follows from the prin- ciple quotient X diinsor = dimdend^ that a factor of an expres- sion is an exact divisor of it. The process of obtaining the factors of a given expression is called factoring. Hence factor- ing, like division, is the inverse of multiplication and depends upon certain type forms established by multiplication. Thus, since (a + 6) (a— 6)=a^— 6^ the factors of a"-'— 6'' are a-\-h and a—h. A common factor of two or more expressions is an exact dioi- sor of each of those expressions. Thus, a is a common factor of a&, ac, and ax+ay. It is understood that in this chapter only rational factors of an expression will be considered. 79. Any monomial expression can be factored. Thus, 6x^y^=2Sxxyyy. Hence its factors are 2, 3, x, x, 2/, 2/, y- 80. Not all polynomials can be factored into rational factors. But there are certain types of polynomials which can be factored. These types will be discussed in this chapter. 8 1 . Monomial factors. A polynomial containing a monomial factor may be factored by aid of the distributive law : ffjr+6jr+cjr+ =(a + 6 + c+ )x 103 104 ALGEBRA This identity shows that x, which is a factor of ever i/ term of the polytiomial ax + bx + cx-\- • • • • , is a factor of the poly- 7iomial itself And the other factor^ « + 6 + c+ ^may he obtained by dimding the given polynomial by the monomial factor. Hence the rule : Find., by inspection,, a monomial vihich ivill divide every term of the polynomial. Divide the given p)olynomial by this mono- mial. The divisor and quotient are the monomial and polyno- mial factors., respectively., of the given jyolynomial. Example 1 . Factor 2ax'^ —4ay'^ + 6az\ By inspection, 2a is seen to be a factor of each term. Hence 2a is the monomial factor. Dividing by 2a, we get x'^—2y^-\-oz^, the polynomial factor. The factor 2a may itself be factored into 2 and a. Therefore all of the factors of 2ax^ — 'iay^ + ^az^ are 2, a, and x^—2y^-{-2>z^. The given polynomial may be written 2a{x^—2y^ + Sz^). Note. — The factors 2, a, and x^ — 2y^+^z'^, no one of which can be factored, are sometimes called the prime factors. 2a is called a composite factor. Example 2. Factor 4:xy^—2x'^y^+x^y. Each term may be divided by xy. Dividing by xy^ gives 4y^—2xy-{-x\ Hence the required factors are a?, y^ and 4:y^—2xy + x^ ; and 4xy^ — 2x'^y^ -\- x^y = xy{'iy'^ — 2xy -f- ^0- EXERCISE 31. Factor : 1. xy'^—xy-\-x. 5. xY-\-xY- 2. a5='y + 3£cy-5a;y^ 6. bd'-l^a'b. 3. a;'— 3a;. 7. 2^x' + lxY, 4. a;' + 5a;^ 8. 18a;^-9aJ^ FACTORS X05 9. a''-^cv>h + 2a*b\ 15. 24a;y-12ar'y + 42a!y. 0^, 6a;* + 9icy + 3a;y ^6. 27wVi + 36mW + 81wi/il 11. 'Ua'b''^ZMb\ ^7. 56ay-14ay + 28ay. 12. 4x' + 4:x\ 18. ic" + a£c«. ^3. 8a'* + 4a^^> + 2a«^>^ l9. a^?r'-a\ 14. «^^>V4-«'^>'c''' + «'^^V. />^0. 5a«y«-^ + 10a"-V. 82. Polynomials that are powers of binomials. Polynomials that are poioers of binomials may be factored by aid of § 71 and § 72. In any polynomial the monomial factor, if one exists, should first be discovered and divided out. Example 1. Factor a?*— 2ay*2/ + ^*V- By inspection x'^ is seen to be a mono7nial factor. Dividing by a?^, we get x^—2xy + y'^. This is the square oix—y. Hence, x*—23c'y + x'^y'^=x^{x'^—2xy-\-y'^) =x~{x—yY. The factors are a?, x^ x—y^ ^c—y. Check. Whena?=2, i/.=l; polynomial=4; factors are 2, 2, 1, 1. Example 2. Factor ^^xHj-2Ux^y' + ^2^x'y^-1^2xyK Qxy is a monomial factor. Dividing by Qxy^ we have 8x^ — 3Qx^y + 54:Xy^—27y^. This quotient is the cube of 2x—3y. Hence, /"T^ . 4Sx*y—21Qafy'' + S24x^y''-162xy*=6xy{8x'-(^^y + Mxii^-27y") • =6xy{2x-3tjf. ' The factors are 6, a?, ?/, 2x—Sy, 2x—3y, 2x—3y. Check, When x=l, y=^\ polynomial =—768; factors are 6, 1, 2, -4, -4, -4. Example 3. Factor —x^ + 2xy^ — y*. Taking out the factor —1, we have a perfect square. Hence, —x' + 2xy'-y*=-l{x^-2xy^-\-y*) = -l{x-yy = -{x-yy. 106 ALGEBRA EXERCISE 32. Factor : 1. bx'-40x + S0. My. -a' + Sa'-16. 2. 4:a'-Sab + 4:b\ 8. 2ba'b'-10ab'x'y-{-b'xY. )^3. ax' + Qax' + 9a. \9. Sx'(/-Sxy + ^xf. 4. 4x'f-^Sbxij + 4b\ 10. 2SSa' + 4S0a'b + 200b\ J>(p. 20a'-60a' + 4ba. >L1. 7x' + 21xh/ + 21xi/' + 7i/\ 6. Qx-9x'-l. Vl2. ba'-lbd'b + lba'b'-^ab\ M^. 2x'^-{-QxY + QxY + 2xy\ >(i4. -^a'x-\-4:^a'bx-lSba'b'x+U^a¥x. VI 5. 32a* - 64a=^^ + 48a^^>^ - IQab' + 2/A Vl6. Sa'-na'b + SOa*b'-nOaW + na'b'-Sab\ 8 3 . Trinomials of the form x'-{^ax + b. By § 75, {x + m){x-i-n) = x^ + (m-\-n)x-\-mn. Now a;^ + (^ + ^?')£c + m/i is of the form x'^ + ax + b, where m-\-7i = a and 7n?i = b. But the factors of x'^-{- (m + n)x + mn are cc + m and x + ii. Hence, if x^-hax + b has rational factors, they vnll consist of two binomials, like a^ + m andx-\-n, having the common term x, and the other terms such that their sum is a and iwoduct b. Thus, since (a? + 3) (a? + 4)=x' + 7a7+12, the factors oix^-\-lx-\-\2 are a? + 3 and x-\-4. Here 7=3 + 4, and 12=3x4. Example 1 . Factor qc}^^x^ 18 . Here the factors must have the common term x, and the other terms of the two factors must be such that their sum is 9 and product 18. Two numbers whose su7n is 9 and product 18 are 3 and 6. Hence x'' + 9x + 18={x + 3)(x + Q). Example 2. Factor a?2—2x— 35. FACTORS 107 Evidently, here we seek two numbers whose product is a nega- tive number^ —35, hence the numbers must have unlike signs. And since their sum is —2, the one having the greater absolute value must be negative. Hence, they are 5 and —7. Therefore, x^—2x—i^5 = {x—7){x+5). Example 3. Factor f -25^ + 150. The common term here is t. Since the product of the other two terms is +150, they must have like signs. And since their sum is —25, they must both be negative. Hence these terms are —10 and —15. Therefore, t'-25t + loO=(t-10){t-15). Example 4. Factor 3—x^—2x. This can be thrown into the type form, x'^-\-ax+b, by taking out the factor —1. Then 3-x'-2x=-l {x' + 2x-S). But x'' + 2x—3=(x+S)(x-l). Hence, 3—x'—2x= — lix-\-S){x—l), or multiplying first and last factors, this may be written {x + S)(l-x). Example 5 . Factor a'^x* + 5a V + 6 . This is of the form x^ + ax-\-b, which may be more easily seen by writing a*x^ + 5a^x^ + 6 = (a'^x'^y + 5 (a V) + 6 . The X of the standard form is a^x^ in this exercise, hence the common term of the factors will be a^x"^. Therefore, {a'xy + 5 (a^x^) + 6 = {a'x'' + 3) (a V + 2) . Example 6, Factor x^ + ^xy + 14^/^ If we write this x^ + {^y)x-\-lAy'^, it is seen to be of the given type form. The common term is x and we have now to find two expressions whose sum is 9^/ and whose product i^ 142/^. It is easily seen that these two expressions are 2y and 7y. Hence, x^->r^xy-irl4.y^={y-\-2y){x + 7y). 108 ALGEBRA Factor : 1. x' + lSx + 42. 8. 2. x' + 2x-4S. 9. 3. £c^-9a^ + 20. >\^0. 4. x'-Sx-2S. 11. 5. a;^ + 17^ + 72. ^12. 6. x' + bx-bO. 13. /,.-^2. x'-^bx' + Q. 23. aj*-7£c^ + 10. 26. fc« + 3£cM-2. 27. a^^'-Sa^^+e. 28. a;^« + lla;-'-26. 29. a;'^-2a;«-224. 30. a'x' + 9ax+-U, 31. bhf-lhy-2,0. 32. a^^^ + 30a^> + 200. 33. £cy-28a!y + 160. 34. mV + 4m?i-60. Y35. «^^V + 13r^^6— 30. ^36. a;ys^-19£cy^ + 90. 37. xy-^xY + 2. 38. a;y + 14£cy + 33. 39. a;y-5ajy-126. ^40. «'"^>^-2«-^^>-35. EXERCISE 33. a;^-12a; + 32. 15. c^417c-84. a^ + 3a-180. ^\l6. cP-M-bb. m'-m- 240. 17. 2-{-Sr + r\ f-t-420. M8. 24:-2s-s\ a2+3« + ^. 19. 15 + 2y-yl 3c-70. ^JO. 2-rt-a^ 5''' + 20Z>+84. 21. z'-z'-2. 41 a;y + 7£cy^— 44. ]s>^42. pY-Sp'q' + 2. 43. a;=^ + 3a!y + 2yl * 44. £c^ — 5^cy + 6y^ ^5. a3^ + 17a;y + 70y^ ^"^6. a'^10ab-S9b\ 47. a'-lSab-40b\ \s, -^a^-\-2ab + b\ i)*9. -«^-5a^» + 104^1 lO. a^y — babxy + 6a^^^ >/. ^61. xY^^Uibx}f-V^ceb\ 62. £cy-3c.^?/-10cl 53. a!y + 9axy + 14al \ 64. x^if — la^bx^y^ -f 1 2«:«'^6l 65. c«^m^ + llac^w' + 30c^ C^6. l-3a + 2a^ 57. \^-Qx-21x\ 58. (a + ^)^ + 7(a + ^») + 10. V59. {x-yy-^x-y)-40. FACTORS 109 60. (x + yy + U(x+>/y + SS. 66. 2£c^-34a;-400. 61. (a + by-S(a + b)(x+i/)+ ,66. ax' + bax-Ua. 2{x-{-yy. 67. a'x' + 2a''x-Sba\ 62. 2ic^-10x-168. 68. x' + ax'-42a'x. (First remove tlie monomial factor. ) 69. dx^ — QOx^i/ — 288£cy^ 63. Sx'^Sx-lS. ^70. xy + 9xy + UxY. 64. 5ic^ + 45ic+100. 71. 260a; + 62«V + 2a;y. 72. 220-2a;-2ajl 84. Trinomials of the type form ax'^-\-bx+c. There are different methods of factoring the general quad- ratic trinomial of the type form ax'^ + bx-\-c when it has rational factors. Such a form may be factored by first changing it to the form discussed in § 83. Thus, multiplying by a, and at the same time indicating the division by a, in order not to change the value of the expression, we have ax^^bx +c= aV- abx a (axy + b(ax) + ac ~ a This numerator is now a quadratic trinomial in («.x'), the form, discussed in § 83 v^hen x is replaced by ax. Hence, we have the common ter7n ax^ and the other terms such that their product equals ac and their sum equals b ; then finally dividing by a, we get the required factors of ax^-{-bx-\-c. Example 1. Factor 2xH8a?+l. 4x^ + 6^^ + 2 2ic-^ + 3x+l = 2 {2xf + 2>{2x)-^2 2 (2.r + 2)(2a!+l) 2 110 ALGEBRA Dividing the first factor of the numerator by 2, we have (x + 1] (2ic+l). Check. When x=l ; trinomial=6; factors are 2 and 3. Example. 2. Factor Qx^ + llx + S. 6 _ i6x + 9)iQx + 2) = {2x + 3)(3x + l). Here, to divide the product of the two factors by 6, divide the first one by 3 and the second by 2. Example 3. Fsictor ax'^+{a-\-b)x + b. a = (ax + a){ax + b) a =(a? + l)(ax + 6). Example 4. Factor 1 2a?'— 23a?// + 10?/^ 1/* _ {12x-8y){n x-15y) 12 = {3x—2y){4x-5y). A second method may be obtained as follows : Since the factors are buiomlals of the form (mx + ri) and {rx \-s) whose product is rnix^^{rn + sm)x + s7i^ therefore, ax^ -\-hx + c= rmx^ + {rn + sm)x -[- sn = (mx + ?i) (rx + s). From this it is observed that if a general trinomial of this form can be factored, the ^ first and last terms of the trinomial must he so factored as to give the terms of the hinomial factors^ and the sum of the products of the first term of each hinomial factor by the second term of the other factor must give the middle term of the trinomiO/h FACTORS 111 Example 1. Factor 2x^ + 5x+2. Now the first terms of the binomial factors are factors of 2x^, and the second terms are factors of 2. The sum of the products of the first term of each by the second term of the other, called cross-products, is ^x. The possible pairs of factors may be conveniently arranged as follows : 2x+2. 2a?+l x+1 _ x^-2 from which we may easily select the pair that gives the proper cross-product. The factors then are 2a? +1 and x-\-2. Example 2. Factor 3ir^ + £c— 10. Four of the possible pairs of factors are x—2 x + 2 x—5 x-\-5 3x + 5 3j?— 5 3x+2 3a;— 2 from which it is seen that the second set is the correct one. Each set should he tested as vnritten. If this is done., it will usually be unnecessary to write all possible sets. The simpler ex- pressions of this type can be factored by inspection. Note. — The student should always look for monomial factors first, and remove any that are found. iTse the method that seems best. EXERCISE 34. Factor : 1. 2£c^ + 5a; + 2. 7. 12x' + bx-2. ]2< Sx' + 7x-\-2. 8. 6x'-llx + b. 3. 2x'-h9x-\-10. 9. 2t'-bt-l. 4. 2x' + bx-S. 10. 8a;' + 15a;-2. 5. 2x'-^x + ^. 11. 12c^-25c+12. 6. bx'-9x-2. 12. bb'-79b-U. 112 ALGEBRA 13. 15. 16. Vl7. tA 21. 23. fv 24. 26. 7«- + 36a + 5. 10/ + 21y-27. 2r* 6. Ua' + ba'-l. 6aV + 13aV + 6. 5a'^>V + 19a^>c— 4. 2icy — x^i/ — 15. Qa'b'-2Sa'b' + 20. 12x' + 21x-Q. Qax'^ + lbax + 9a. 8Qx'i/ + U2xj/-Ui/. Qa'b-4:Qab-72b. 2x' + Sxi/ + y\ 26. 12x'-7xy+f. S 27. 2x'-Sxi/-2i/\ K28. bx'-19xi/-'^tf. \29. 12ni'-lQa77i-Sa\ 30. 2£cy + «y-15. 'Nl. 3a;^ + 10a;y-8/. V 32. 21ax^—hlaxy^^ay^. • 33. 6a^^»^ + 2«Z>c—4cl 34. ax^-\-{b-a)x-b. (X35. 2y^ + (4a + % + 2«^. 36. 2;2^-(2a + ^)2; + a*. ' 37. ax'-^{ab^-\)x-^b. 38. 2a£c'''+(2a^-2«^»)a;-2«^^>. 85. Binomials of the type form a^ — b^. Binomials of the type form a^—V^ may be factored by aid of § 74. Since (a + 6)(fl-6)-a^-6% the factors of a^ — b- are a-^b and a—b. Example 1. Factor 4x^—25. = (2ic+5)(2a!-5). Note. — We might also write = (-2a^-5)(-2if+5). Usually, however, we have no use for this second set of factors. It is easily seen in general tiiat if an expression has two factors, tlie expressions obtained by changing the signs of those factors will also be factors. Thus, the factors of ah are a and 6, or — a and — 6. FACTORS 113 Example 2. Factor 27a?^— 12a?. Kemoving the monomial factor 3a;,leaves 9a?^— 4. = (3x-2)(3.r + 2). Hence 27x'-12^=3x'(3ic-2)(3x + 2). Note. — We need not write down the second step of the above solu- tion, but write merely 9^2-4=(3j;-2)(3a;+~2). ' EXERCISE 35. Factor : 1, x'-m. 8. x^-n\ ^^62bx'- -2256^ 2. x'-l^jd. 9. 4x'—a\ 16. 9a V - 1. 3. 0^—49. 4. x'-U. 10. 11. 9a3^-a^ 17. i6a^y- -9. ■25. 5. a^^-144. 6. 03^-196. 12. 13. lQx'-2ba\ 19. 25£cV- lOOy^-495^ "^20. lOOx'y -16. ^-815^, 7. x'-a\ 14. Sla' — 64b\ 21. IG^cV- -b'c\ ^ 22. imxYz'-^ha'h y%Z. \x'-\y\ 24. 1 a2_ 1 52 ^cl 36a.V^ -, 2^^- 25a^5-^ 1- 33 1 fiu2 2 5 16 K26. T-V^«'6'-^V«^. „. 4x' 81?/ V^- Sly^ 4x^ ' 27. /T-IKy'- \28. 1— Vra'*'«'- 35. 1 ^, . " 29. Ti^-T-v*y. 36. (a + by-1. 30. 2-:_«i „, a' 25(8' ^7, {a+by-(c+dy. 38. (x + yy-(a + by. . 39. (a^-y)'^-(«-^')^ 114 ALGEBRA 40. 4(x+(/y-9{a + by. -o (a + by (c + d y {a-by {c-dy (2a+by 2Hx+j/y_ w(x-j/y ^ ib{x Zy). to. ^Q^^^j^^y 8i(^^_^)2- 86. Polynomials which can be written in the type form a^—b^. Some polynomials, by grouping terms, can be put into the tjpe form a^—b^^ and can then be factored by the method of ■§ 85.' .• Example 1. Factor aH2a& + 6='- c^ Grouping terms, o? + 2a6 + 6' - & = {a' + 'itah + h^) - & =(a + 6 + c)(a + 6— c). Example 2. Factor a^—Jy"— 2bc — c\ Grouping terms, =a^-{b + cy = {a + (b + c)} {a-{b + c)} = {a + b + c) (a—b—c). Learn to omit the second and third steps and to lorite out the icork as follows : Example 3. Factor x" + 6x— a' + 4a&— 46' + 9. Grouping terms, x^* + 6a? - a' + 4a6 — 46H 9 = (x^ + 6.r + 9) - (a' - 4a6 + 4&') = (a; + 3 + a-26)(a74-3-a + 26). EXERCISE 36. \Factor : 1. a^—lab^^—x^. 4. a' + 6a5+9^>'— 4c^ 2. \-\-'lx-Vx^-if. ^b. l-x'-Sxy-Uf. ^, x^—1xy\y''—\, 6. a;'+4a£c-y' + 4a^ I^ ACTORS 115 \y, a'-l + 10ab + 2^b\ 10. l + Qax-a'-9x\ 8. -4-2ab + a' + b\ "^ 9x'-ia'-9c' + 12ac. K^ a'-^c'-Sab+lQb'. 12. a' + 2ab + b^ + 2cd-c'-d\ 14. a' + 12m7i—10ab-4m' + 2bb'-9n^. 15. ic2-12aa!-4^•y + 36a'-4?/2-6^ X XFactor, and simplify the factors : Al6. (x + 2i/y-(2x-^yy 19.- (5a-3^)^ + 12«5-^>2-36a2. 17. (2x-Si/y-(^j-xy. \20. a^^-8a;?/ + 16/-(£c + 4y)^ 3^8. (.^-5y)^ + 24£cy-9£c--16/.ai. W0a' + 20ab + b'-(a-2by, 87. Special trinomials of the type form x*+axy^+y. It often happens that trinomials ol the form x*-\ ax'^y'^+y* may be written in the form a^—¥ by the addition and subtrac- tion of a term. The addition and subtraction will not, of course, change the value of the given expression. Example. 1. Factor aj^ + o^y + ^/^- Adding and subtracting ar^i/^, we have x^ + x^y^ + y*=x^ + 2x^y^ + y*—x^y^ = (x'' + yy-x'y' = {x'' + y^ + xy) {x'' + y^—xy). Example 2 . Factor a?* + 2x'^y'' + 9y\ Adding and subtracting 4x^y^, we have x* + 2xY + ^y*=x' + Qx'y' + ^y*-4xY = {x' + 3yy-4xY = (s(f + 3y' + 2xy){o(f + Sy''-2xy), Examples. Factor a?*— a? V + 2/*- Adding and subtracting Sx^y^, x*-x^y^ + t/=x* + 2xY + y'-Sx^y' ={x' + yy-BxY ={a^ + y'+xyy'S){oc' + y'-xyi/S). 116 ALGEBRA Note. — These factors are rational in x and y, but not in the coeffi- cients. For this reason the expression is not considered factorable in the sense in which the term is usually taken. Example 4. Factor 4ic^—16a?*^* + 92/^ Adding and subtracting 4:X*y*, 4a?«— 16ic*2/* + ^y^=^o(f-12xY + 9i/«-4^Y = (2x*-3yy-4xY = {2x' - By' + 2xY) {2x* - 3y' - 2x'y') . EXERCISE 37. 4 ^ Factor : 1. x'-^-x'-i-l. ^9. 36a*-76ay + 25i/. ^^. l + Sx' + 4x*. 10. 49a'-^19a'b'-\^4b\ 3. x' + x* + l. J^^- 4iz;'-16i«y + 25yi / ^ 4. a* + 2a'b' + 9b*. 12. 04.Ty + 12a;y+l. 5. a'-Sa' + 9. ^ }^- «'-22aVy + 9a.'y- \«. ,4£c* + 16a!y + 25y*. ^4. 4m*n«-45a^^^mV + 25a*i«. " lyi^ 4a' + 4a'b' + 2^b\ />\^ 16. x''-^x' + U. V 8. a;*-15a!y + 9/. vV 16. 25a* + 16a'^»V + 4^»V 88. Binomials of the type form flf'4 6", when /7 is odd. By § 77, a" + b" is divisible by a^ b mhen n is odd. Hence, the factors of a" + b'\ when n is odd^ are a + b and the quotient obtained by dividing a" + 6" by « I ^. By division, a'+b'={a^-b){a'-ab -\-b')', a'-\-b'=(a-^b){a'-d'b + a'b'-ab^ +6*) ; a''\rb''=:{a^b){a'-a'b + a'b'-a'b'-\^a'b'-ab' + b% Example 1. Factor x? + Sy^. X? + ^y^ may be written a?^ + {2yf^ and therefore may be divided by x-[-2y. FACTORS 11' Hence, o(?-^Sy^={x-\-2y){3(?—2xy + 4ty^), Example 2. Factor 2>2w' + 2436^ 32a5 + 24365=(2a)H (36)5 = (2a + 36) j (2a)* - {2a)\Zb) + (2a)'^(36)^ - (2a) (36)» + (36)* = (2a + 36)(16a*-24a^6 + 36a'^62-54a6H816*). Example 3. Factor a^ + 6^ a» + 6»=(a»)3 + (63)» = (a« + 6^) {{ay-{a^){h^) + (6^)^} =(a3 + 6^)(a«-a='6H6«). But aH6=^=(a + 6)(a2-a6 + 62). Hence, a^ + lf={a + h){o}-ah + h''){a^-aW + W). <, Factor : 1. x^-^y\ 2. Q^-\-yK /y^Z, x' + y\ 4. a?+y\ ^b. x'' + y'\ 6. 27x^+a\ /X7. Ux' + 27y'. 8. 125a^ + 216^>=^ 9. 40a' + Ubb\ 10. 432 + 2a;^ EXERCISE 38. 0^11. 81a;^ + 3. ^2. ^' + 1. 13. l + x\ h4:. S2 + x\ 15. l + 3125a^ \6. «W + 1. ,'\17. «V + 32. 18. i25a;y. 19. J + 1. 1 7,5 /^20. ?" - 1 2/' • 21. ^i^^' + ^V. ^^2. 1024a'' 23.- 12Sx^+^i^y\ 24. i^ + 2/. 26. ic^ + l. 58. l + a'^». 89. Binomials of the type form a'*— 6", when n is odd. By § 77, fl" — 6" *s divisible by a — b when n is odd. Hence, the factors of w—b'\ when n is odd^ are a—b and the quotient obtained by dividing a"—b" by a—b. By actual division, X18 ALGEBRA a^-b^={a-b){a'-\-ab^-b') ; a'-b' = {a-b){a'^-a'b^-a'b'^-ab'^-b')\ a''-b'=^{a-b){a' + a'b^a*b' + d'b'+a'b'^ab' + b'). Example 1. Factor 27x^—8?/^ 27jL^—^y^ may be written {^xf—i^yf, and consequently may be divided by Zx—2y. Hence, 27x'-8y'={Sx-2y) {{'^x)^ + {3x){2y) + {2yY} = (Sx-2y){9x' + Qxy + ^y'). Example 2. Factor x^^—y^. x^^—y^={x^)^—y^ ={x'-y){ix'y + {x'fy + {xYy' + {oc')y' + y*} = {x^ — y) {x^ + x^y + x*y'^ + x^y^ + y*) . EXERCISE 39. Factor : 1. x^—y^. 11. 1— i^c^ \ gi ^^ V^ >"2. a^-h\ \'\2. x'-\. ' ^~^' V3. x'-y\ \ 13. 32-y\ \ 22. «;^_'i'. ^ 5. .-2/^ 15. .-32.^ )^^^ -^,_ N^ 7. 8a^-2/^ Vl7. l-a\ , ^^^, '^ 8. 27a;^-64al 18. 128-a3y. ^^6. 1^ — i. ^ 9. 125-34aal > 19. 8l£c^-3. i 27 )l 90. Special Binomials of the type form a" + 6", when n is even. When n is divisible hy Jf.^ hinomials of the form, a" + b'' may be factored as in § 87. FACTORS 119 Example 1 . Factor x^ + a". Adding and subtracting 2a*x\ ix^ + a^=3(^ + 2a* X* + a^— 2a*x* ={x* + a*y-2a*x* = {x* + a* + a^a?^/ 2)(.r* + a*-a''x'\/2). Note. — These factors are rational with respect ioX\\e general numbers X and a but not with respect to their coefficients. Factors of this kind are sometimes useful. Example 2. Factor 81a'Hl. = (9a«)2 + 2(9a«) + l-2(9a«) = {9a^ + iy-2{9a^) = (9a« + l + 3a"^|/2)(9aHl-3a=^y'2). When the exponent n has an odd factor^ binomials of the form a" + b''mai/ be loritten as the sum of two odd powers^ and fac- tored as in § 88. Example 1. Factor if-\-lf. ={y'^h') {{yy-{y') {h') + m'} :={if+h') {y'-h'y'+h*). Example 2. Factor x^" + 1024. £c" + 1024=(ir=')^ + 45 ={x^+4){(xy-{xy-4.+{xy-^^-{x:'ye+4*} = (a?2 + 4)(.T«-4a?«4-16iC*-64a72 + 256). \ EXERCISE 40. Factor : *^1. a*-^b\ \ ■'x 7. x'*-\-l. \ l\12. l + 81a;^ Y2. a' + b\ 8. x* + l. 13. 64a^« + 729/. A 3. a'-^b\ 4. a'^'^b'^ 9. 03^°+ 1. 14. X* y* y' ^'' ^ 5. a^^-\-b'\ 'Via l+ar^ ,„ 1 6. a«+l. 11. 16a;« + l. 15. -'"+5r»- 120 ALGEBRA 91. Binomials of the form a"—b'\ when n is even. Binomials of the form w—b''^ when n is even, should always be written in the form a}—h\ and factored as in § 85. Note. — Factors should themselves be factored when possible. Example 1 . Factor a?* — ?/*• Writing in the form a^—lf^ we have Factoring x^—if, =('«^ + 2/^)(<^ + 2/)('^— 2/)' Example 2. Factor 729a«-64. Writing in the form a^—b^, we have 729a*^-64=(27a=')^-8"'' = (27aH8)(27a=^-8). '. But 27aH8=(3a)H2^ = (3a + 2){(3a)2-(3a)-2 + 2=} =(3a + 2)(9a2-6a + 4). And 27a3-8=(3a)^-2=' = (3a -2) {(3a)- -f (3a) -2 + 2--^} = (3a-2)(9a2 + 6a + 4). Hence 729a«-64=(3a + 2)(3a-2)(9a2-6a4-4)(9a2 + 6a + 4). EXERCISE 41. V) Factor : 1. a*-b\ 8. a'-b\ 15. l-x'\ 2. a'-b\ 9. a''-b\ 16. 4-a;*. 3. a'-b\ 10. a''-b\ 17. 16-a^^ 4. a''-b'\ 11. x'-l. 18. 81-a;«. 5. a''-b'\ 12. x'-l. 19. 100-a^«. 6. a''-b'\ 13. l-x\ 20. 256a«-l. 7. a'-b\ 14. l-a;^". 21. 81a'^-16a3'^ FACTORS l^l 22. l-256a^^ 26. 1024-a'». 29. x^'-a"'. X /2«+4 23. 6V-64a;«. 27. -,-1. ^0. x'^'-^-if 24. xY-a'b\ ^ ^^4^, 31. (a + by-c\ 26. 16-81«^^>*. ^^- ^'~T6* 32. (x-yy-(a-b)\ 33. a^ + 4a=^i + 6a-^»^ + 4«^»^ + ^»*-c\ 34. {2x-i/y-{a + Uy. 92. Polynomials made to show a common factor by grouping terms. The terms of a polynomial may often be so grouped as to show a common factor^ then factored as in § 81. Example 1 . Factor ax +ay + hx + by. Grouping those terms which show a common monomial factor, we have ax+ay + bx+by=a{x + y) + h(x + y). . This is a binoinial whose terms are a{x-\-y}^nd b{x+y), and shows a common factor x + y. Hence a{xA-y) + b{x+y) = {x+y){a + b). Therefore ax + ay + bx+by=(x-\- y) {a-{-b). Example 2 . Factor x^ + x'^—4x—4. x^+x'—4:X—x^x'^(x + l)—4ix + l) = {x + l){x'-4:) ={x + l){x + 2)(:x-2). Example 3. Factor a^—¥—(a— bf. a-^-b'-{a-by=(a''-¥)-ia-bf = {a-b){{a + b)-{a-b)} = ia—b){a + b—a + b) =2b{a-b). Example 4. Factor a^—b^ + ax— bx. a^—b^-\-ax—bx={a'—b^) + (ax—bx) =(a + 6)(a-^&) + x(a— 5) ={a—b){a + b + x). ALGEBRA EXERCISE 42. \ % ^' / l^ Factor : 1. ac-\-bc—ad—bd. 2. ax—bx—ay + bj/. X3. ax-^bx + 2a-^2b. 4. ax-\-4:X—ay—4:y. >\5. a^b-^b-Va'c-^^c. 6. ax^'^-ax-\-bx^-^bx- k 14. £c— «+(»;— a)'. 15. a^x-\-a^y-^x + y. vVl6. x^a-Vx'b-a-b. 17. j?/ + 3ic*-a;-3. 18. x^—x—y'^^y. 19. ax^—a—bx^ + b. 7. 2a-2^• + aJ-^>6^+2c + c(7.^^ 20. aW-a*-** + l. 8. ax—bx—a + b. Jl\9. xy—x—y^\. 10. a^>-3a-25 + 6. ^11. a;^ + aa; + ^i« + «5. 12. a;^ + aa;-2a;-2a. 13. a;^— a^ + C£c— «c. 27. aV >^^1. ax^ + bx—ax—b. K23. a'-b'-{a'-by. 24. mnpq + 2+pq-^2mn. '^' 25. 4:ax—2ay—2bx + by. 26. a^^>^-6^c^^-aW^ + cW^ 1-a^-a;^ — V^8. aaj,?f 3air+5^2 + 15a; + 2a + 10. 29. a;^4-4a!^ + 4ic— aj^y— 4ajy— 4?/. ioO. TTi^n*^ — my — q'^n^ + (^y . 31. a'x'-ay-b'x'-i-by, 36. a(a + c)-^>(^> + c). 32. x'a*-y'a*-b*x^ + by. 37. (a + ^>)2 + 5c(a + ^>) +6c'. 33. «V-a2 + a^>i«=^-«7>. 38. b'-c' + a(a-2b). 34. a;^ ■2/^ ■2^?^ 2(xz-yw). 39. aj*+£c^ + y'-y*. 35. 4a;y-(a;' + 2/'-2')'. 40. x' + x'z + xyz + y'z-y\ 41. 2(l-a!)(l+a!)^+(l + a3'). 42. a^x + abx + ac + b'^y + aby + bc. 93. The methods of the foregoing sections will enable the student to factor almost any expression. Another good method, however, based upon what is known as the remainder FACTORS 123 theorem is discussed in the following sections. This method is very convenient when the given expression cannot readily be thrown into one of the familiar type forms. 94. The remainder theorem. The remainder obtained by dividing a polynomial in x by x — a is the expression that would be obtained by replacing x by a in the dividend. Thus, dividing x^-^-Zx—^ by x—a, ic^' + Sr-S x^—ax x—a (divisor) a? + (3-|-a ) (quotient) (3 + «)a?— 5 (3-|ra),r— 3a— g-^ a'^' + Sa- 5, the remainder, which is the same as the dividend when x is replaced by a. Dividing i)(f—x^ + dx+2 by x—2^ we have x^- x^ + 3x+ 2 x-2 x^-2x' x' + x + 5 x^ + Bx+ 2 x'-2x 5x+ 2 5£C-10 12, the remainder Now if, in the dividend, x is replaced by 2, the dividend be- comes i^, which is the reinainder. To show that this principle is true in general, let any divi- dend be represented by the general expression, Ax- + Bx'^-'~\'Cx^^-^+ -VMx + JSr. Let the quotient and remainder, when this is divided by .c— a, be represented by Q and B^ respectively, k Then, by § 51, M^ « \m3){x^-2x+4). Note. — In attempting to find a value for x that will reduce the given expression to zero, it is wise to begin by trying 1 or — 1 ; then try larger possible numbers until the required number is found. Example 4. Factor o(?-\-2x^y—xy'^—2y^. This expression becomes zero when x=y. Hence, x—y is a factor. Similarly a? + 2/ and a? + 22/ are factors. Hence, a^ + 2x''y-xy^-2y^={x-y)(x + y){x+2y). 126 ALGEBRA Example 5. Show that a + 6 is a factor of a" + 6" only when n is odd, and thus prove {d) of § 77. Putting —b for a in a" + 6", we get (—6)" + 6". If n is odd, (— 6)" + 6"=— 6" + 6"=0. If w is even, (— 6)" + 6''=6'^H-6''=:26». Hence, a + & is a factor of a" + 6" when n is odd and not when m is even. When a factor of the form x—a has been found, the work of dividing it out may be abridged. The method may be derived from the following example : Divide ar*— 5a;^ + 5x— 7 by £C— 3. ar»— Sar' + Sic— 7 x-S a^-3x^ a;'— 2a;— 1 -2o(f -2x' + 6x — X —x-\- 3 10, the remainder. Observe : (l) that the coefficient of the first term of the quotient is the coefficient of the first term of the dividend, and the succeeding coefficients are the coefficients of the first terms of the succeeding remainders. {2) That since the first term of each partial product caficels the term of the dividend immediately above it, the first term of each partial product need not have been written. (3) That if the sign of each term of the partial products had been changed, this term micfht have been added to the corre- sponding term of the divide^id. (This can be done by changing the sign of each term of the divisor.) (^) That if the dividend is written in descending powers of the first term of the divisor, this term of the divisor need not be written. Omitting all work that is unnecessary, writing the coefficients FACTORS X27 only, and raising each number of the several remainders into the same line, the work may be written as follows : 1-5 + 5-7 |3_ +3-6-3 1-2-1,-10 where 1, —2, and —1 are the coefficients of the quotient, and — 10 is the remainder. This abreviated form of division is called synthetic division. EX4.MPLE 1. Divide 2x^—4x^ + 7x+23hy x + S. 2- 4+ 7 + 23 1-3 - 6 + 30 -111 2-10 + 37,- 88 Hence the quotient is 2xf—10x+S7 and the remainder is —88. Observe that the first coefficient of the dividend is brought doivn for the first coefficient of the quotient. This number is multiplied by the number in the divisor and added to the next term of the dividend for the second coefficient of the quotient, and so on to the last term. Example 2. Divide 2a?*— 10x^-13 by x—2. 2 + 0-10 + 0-13 |2 + 4+ 8-4- 8 "^ 2 + 4-2-4, -21 Hence, the quotient is 2xr^-\-4iXi^—2x—4: with a remainder of— 21. Observe that where any power of x is ivanting its place must be supplied by a zero. Formulate a rule for the process. Perform the following divisions by the synthetic process : 1. x^ + Zx'-lx-Z by a;-2. 6. x'-la?-\-2x-12 by x-\-2. 2. 3a;^ + 7a;^-3a; + 15 by aj+3. 7. x' + Zx' + x—l by ic-3. 3. x'-^x'^lx-l by,£c-2. 8. 2a;''+jc=^-7 by a; + 3. 4. 2iK*— 9a;' + 3a;— 5 by cc— 4. 9. x'-i-hx'^-x—^ by a; + 4. 5. 3a;*-7a;^ + 13 by cc + 2. 10. a;* + 3aj^-9 by x-2. The pupil should now be able to factor any factorable ex- pression that is likely to occur in elementary algebra. 128 ALGEBRA The following general suggestions may be useful : 1. First remove all monomial factors. 2. Try to bring the resulting polynomial under some one of the binomial or trinomial types. 3. If imsuccessful^ try the remainder theorem. Jf.. Be sure that all factors are prime. ^ s \^' EXEBCISE 43. vi By the use of the remainder theorem, factor the following : 1. Ix^-x-l. 6. 'M- -9.^ + 9. 9. x'-^x'-Sx-2. 2. 3^^ + 5cc + 2. 6. 3a3^ + 14a!+15. 10. x'-{-7x'i-7x-16. 3. 2£c2-5ic + 2. 7. Zx'- -13^-10. 11. x* + x'-2x-2. 4. ^x^-^lx-1. 8. x'- -Qx' + Ux- -6. 12. x^ + Sx'-i. 13. 2a;-"' + 3i«'-50i« + 24. 19. a'- -a' + a'-l. 14. ^x'-1x^'\-x- -2. 20. 2x' -Sxy + y\ Jl5. a' + 3a' + 3a- + 2. 21. ^x' + 2xy-y\ 16. rt^ + 2a-'-4a- -3. 22. x'-- h 2x'^i/ —xii^ — 2?/. 17. a^-21a-20. 23. x^-2x^y-^xy''-^\%y\ 18. a* + 7a=» + 13a^ + 7a + 12. 24. a^^^a^b^ab'-W. 25. Show that w—b" is akoays divisible by a—b, and thus prove (a) of § 77. 26. Show that «"—?>>" is divisible by « + ^ only when oi is even^ and thus prove {b) of § 77- 27. Show that a" + ^" is never divisible by a—b, and thus prove (c) of § 77. By any method of this chapter, factor the following : 28. ax'-Q>^a\ 31. 4a^ + 32«^. + 39Z>l 29. x'+--2. 32. {x-\-iy-bx-2^. 30. ^ahx^ + 2abx-ab. 33. ax'-Wx^-haK FACTORS 129 34 ^ + ?? + l ^^* 4«* + 3a;y + 9y*. ' a' a ' ,60. a'b-bc' + a'c-c\ 35. xy^-z'-xz-yz. ' ^^ a.3_^2_4^_g^ 36. x'+^x'^-Qx+n. 62. a'-Va?-la-^. 37. 4(a-6y-(y + 2)l 63. x^+x'-VJx^U, 38. 9a;^-27a;y + 20y^ 64. a;^-ll£c^ + 31a;-21. 39. a' + aW-{-b\ 65. £c=^-6£c'' + lla;-6. 40. cc*— ISicy + y*. 66. a;=' + 2a3'-9a)-18. 41. a;6 + a;='-42. 67. 2/='-92^ + 25a;'^-10a;y. 42. ^*+2xy-35/. 68. Qx^'^ + x^r-lbf^ 1 9 69. 3(a-^)'-14(a-^>) + 8. ^^* ^'"^a*"!* 70. 7(a + ^)'-llc(a + ^>)-6c2. 44. x'^-^x'-x'-^x. 71. £c*-aj^^-17x^ + 5a; + 60. 45. l + ^»y-(a;' + Hi/^ 72. a^-a'b-a^h. 46. (aj^'+y'-s'O'-^a^y- 73. x'-%x-ax-V^a. 47. x'-2xy'-y'-\-^x^y. 74. aa;^-3aV + 2a'a;. 48. a;/ + 7a;y-30a;. sy75. 2£c^-3x^-9a; + 14. 49. ac-\-cd—ab—bd. 76. 35a;'— 74a; + 35. 50. iK*-(a;-6)l 77. a;^ + a;*-56a-^ 51. 72«' + 41^-45. 78. x'-^bx-^^-W. 52. a'b'-a'-b'-\-l. 79. a;^ + 7x^-5a;'-35. t^ 53. 2s' + 5s«!-12^^ 80. a;^ + 4£c'4-a;-6. 54. 6a='a;-aV-9a*. 81. £c=^-3a;' + 7£c-21. 55. bcx-\-acx^-\-bx^-\-ax'^. 82. a;3 4-25a;2_[-8a;-16. 56. a*-2(^»' + c')a' + (52 + c')l 83. 10 {x-\-yy + lz{x^y)-^z\ 57. a;'-(4a=' + ^>')a'£c + 4a''6l 84. 24(a+&)2-5c(a + 6)— SGc^. 58. a;' + (a-%-2a(a+5). 85. a;'+4ic+4-4a'^+4ay-yl 9 CHAPTER X. COMMON FACTORS AND MULTIPLES. 96. Integral and fractional terms. A fractional term is a term which contains one or more general numbers in the divisor. Otherwise, the term is integral. Thus, r, ^^7 2a ^ ^^^ ct-^{^—y), ^Y^f^CLctional terms, for there are general numbers in the divisor. And a^, :^x^y^ ^^- , and — |a&, are integral terms, for the divisors are not general numbers. A fractional term may become a whole number, and an integral term a numerical fraction, when definite values are assigned to the general numbers involved. Hence, the classifica- tion of terms into integral and fractional terms has no reference to arithmetical whole numbers and fractions. 97. An integral expression is an expression whose terms are all integral. Thus, ^a^—§ab + b^ and Soc^ + 2x'^ + x + 1 are both integral ex- pressions. Observe that the ^ and | do not make the terms fractional. The nature of a term depends upon its general numbers. A fractional expression is an expression whose terms are all fractional. 2«]t^ 6?y^ z^ abc x^ Thus, — — -J- + — and n^- 4- rr are fractional expressions. An expression may be integral with respect to certain general numbers and fractional with respect to others. That is, certain 130 COMMON FACTORS AND MULTIPLES 131 general numbers appear in the dividend only, while others appear in the divisor. Thus, ^—2^ + -^ is integral with respect to a, hwi fractional with respect to y. A mixed expression is an expression some of whose terms are integral and some fractional. 3c^ a^ a Thus, — + a and 2ab + ^ "^ 3x ^^^ wi«a?ed expressions. The sec- ond, however, is integral with respect to a. 98. The degree of a rational integral term is the number of literal factors in it. This amounts to the sum of the exponents of all letters in the term. Thus, the degree of Qa^b is 3. The degree of loc^y^z^ is 6. And 4abcde is of the fifth degree. The term 5a^a^ is of the third degree in x. And ab^x*y^ is of the seventh degree in x and y. The degn^-ee of a rational integral expression is the same as the degree of its ternj of higliest degree. Thus, 5x^—x-{-4: is an expression of the third degree. And 2x^y^—Sxy-{-l is of the fourth degree. This expression is of the second degree in x. The expression is of what degree in y ? An expression whose terms are all of the same degree is called a homogeneous expression. Thus, 5x'^—2xy + y^ is a homogeneous expression. 99. The highest common factor of two or more expressions is the expression with greatest numerical coefficient and of highest degree that will exactly divide each of them. Thus, the highest common factor of a^6V and aWc^ is a^b^c^. And the highest common factor of lOor^T/z* and Sx^y^z^ is 2o(?yz^, It follows from the definition above that the numerical co- efficient of the highest common factor is the greatest number 132 ALGEBRA that will exactly divide the numerical coefficient of each ex- pression ; i. e., the greatest common divisor of the numerical coefficients. By the coefficient of a polynomial is meant its numerical factor. Thus, 2 is the numerical coefficient of 2a;2-6i« + 4 or of 2(£c^-3a^ + 2). Also the highest power of any literal factor that will exactly divide each expression is the lowest power of that factor found in any one of the expressions. A factor which does not appear in every one of the expressions can not appear in the highest common factor. Hence the rule :* To form the highest common factor of ttoo or more m^onomials^ take the product of the greatest common divisor of their numerical coefficients^ and each letter raised to the lowest power to which it appears in any of the expressions. In polynomials^ factor each expressio7i qnd proceed as with monomials^ treating each factor as yon woidd treat a single letter. Example 1. Find the highest common factor of 12a?^i/^ and The greatest number that will exactly divide 12 and 16 is 4, their greatest common divisor. The highest power of x that will divide both a^ and x^ is a?% the lowest power present. And the highest power of y that will divide both y^ and y^ is ?/^, the lowest power present. Since z does not appear in both expressions, it can not appear in the highest common factor. Hence, the highest common factor is the product Ax^y^. * There is a method of obtaining the liighest common factor of two expressions without resolving them into their factors. It is similar to the EucUdean, or long division, process sometimes employed in arith- metic of finding the greatest common divisor of two numbers. The discussion of this method is found in the Appendix. This method is seldom used in practice. The factoring process here discussed is sufficient for our purpose. COMMON FACTORS AND MULTIPLES 133 Example 2. Find the highest common factor of SOa'b^c, 12a^¥c\ and 18d'b*c\ The greatest number that will divide 30, 12, and 18, is 6. The lowest powers present of a, 6, and c, are a^, 6^ and c, respectively. Hence, the highest common factor is 6a'*6^c. Example 3. Find the highest common factor of iK^— 6^— 27 andx' + 6a?+9. x^-ex-27={x + 3){x-9); x'' + 6x + 9={x-\-3Y. The lowest power of ic + 3 present is the first power, x + S. And x—9 is not a com- mon factor. Hence, the highest common factor is a? + 3. Example 4. Find the highest common factor of a^x—a^bx—Qab^x^ a^bx^—4ab^x'^ + S¥x^, and a^x'^—2a^bx^—Sab'^x^, We have a^x—(i%x—Qa¥x=ax{a—3b){a + 2b); a''bx'-4ab^x'' + 3b^x''=bx-{a-Sb)ia -b); a^x^-2a%x^-Sab''x^=ax\a-3b){a + b). Hence, highest common factor =x{a — ^b) =ax—Sbx. In finding the If. C.F. of two or more expressions when hut one of the expressions is easily factored by inspection, we may use the most likely factors of the factored expression as divisors of the other expressions. Example 5. Find the highest common factor oix^—Zx-\r2 and ar* — 4x^ + 4a?— 1. Qi?—Zx-^2—{x—\){x—2). Now since —2 of the second factor is not a factor of —1 of the second expression, the "most likely factor" is x—\ which by trial is found to be a factor of x'-^x'^^x-X. Hence, x-1 is the H.C.F. Example 6. Find the highest common factor of 2x!^+Sx—2 Sind4x^-\-lQx'-19x + 5. 2x^ + 3x—2={x + 2){2x—l). Herethemost likely factor is 2a?— 1. Why ? By trial 2x—l is found to be a factor of the second ex- pression and hence is the H.C.F. 134: ALGEBRA EXERCISE 44 Find the highest common factor of : ' 1. ^a'b\ Mb\ 13. SGa^Sys, 72xyz, lSOxy'z\ 2. 9a*!}'c\ 12a'b'c\ ^14. 17baWc\ 70a'b% 10ba*bcP. 3. UxY, SOxyz\ 15. 24m'na\ 42ma\ ISm'a'b. 4. Sa^yV, 9Qxy\ ^16. 15aV?i^ 40«Vwi, 35aV//2^ />t. a'b\ 10a'bc\ 1 17. ^ba'b\ bQa'b\ 9SaW. 6. 2a; V^', y^^- ^18. -98^;^', 21a;y^, -28a;yV. ^7. axy, Sbxy. - 19. acc^y^, bx*y\ cxy, 2ic^y^ j8. 49a^>Vy, 21a'¥cx\ ^20. 2(a + ^>)^ 4(rt+^»)\ ^9. ^^xy, -bQxi/z\ 21. 14(a;-y)«, 21(i«-y)*. i>^0, QSab'c^d\ -2>d'bd\ 22. 16(a + a.')VS6(a + a3)'(^> + .y)\ 11. ^xYz,_^xYz\ 2xy^z\ 23. (cc-1)^ {x-l){x-\-1). v^l2. 2Sa''b'c'd, Qa'b\ lQa'b'c\ 24. 10(a; + l)^ 5(a!-l)2(ic + l). ^5. 39(a;-l)-Xa3+l)^ 26(a;+l)(a^-l)*. j 26. bQ{x-l)(x + 2)(x + S), (x + 2y{x-^). "?^27. ic^-1, a;^-l. ■ 28. 8iB=' + l, (2x + l)(x-S). 29. 4(a; + l)', 20(a;-l)(a;+l), 36(ic-l)^ yio, x'-l,x'-dx + 2. Z4:. 2x\x'-Sx,x' + 4x. 31. x'-l, x' + bx-Q. \}^5. x'-Sx, 9-x\ x'-^x + Q, 32. a'-b\ a'-b\ 36. a^^ + l, a;=^ + l, tc + l. ^3. a'-2ab + b\ (a-b)(a + 2b). 37. 2a;'^ + a;-6, 6if^-7ic-3. 38. 3aa;' - Sa\ 4a V + 2a V - Qa'x. • 39. a^ - ab\ a' + a^^* + a^> + b\ ^^40. a■^ + 3a2a; + 2aa;^ a* + 6a^ic + 8a V. 41. a^-9a=' + 26a-24, a''-12a2 + 27a-60. ^ 42. 7m^-2m2-5, 7m=' + 12mM lOm + 5. COMMON FACTORS AND MULTIPLES 135 43. a' + ^ab + 2b\a' + ^ab + 4:b\ a}-^ab~U\ / 44. 1— £c^, 1— ic^, x—x". 46. Vla? — l^ab^Zb\ ^w'-^a^b^-'lab^—W, 46. m'^ + m— 6, m='— 2m'— m + 2, m=' + 3m'— 6m— 8. 47. ic='-3a;y'-2y\ o?-x'y—^y\ JC^S. a* + b\ a^-b\ a' + aby^+b\ 49. a=^-a'-5a— 3, a^-4a'^-lla-6. 60. l-x% l-2£c=* + a;«, l + a; + £c^ 61. a'—bax + ^x^ d^—a^x + Zax'^—Sx^. 62. ic'-7a; + 10, 4£c='-25a;' + 20a; + 25. 63. 6x'(/ + ^xf-2i/\ Sx' + AxSj-4x}/\ i 100. A common multiple of two or more expressions is an expression which is exactly divisible by each of them. Two or more expressions may have an indefinitely great number of common multiples. Thus, a few common multiples of 2o(^ and 3xy^ are 607^1/^, 6x^1/^^ ex*y\ 6x^y\ Qx'^if, 12x^y\ 18x^y\ 24x'''y\ etc. Each of these expressions is divisible by both 2x^ and 3xy^. The lowest common multiple of two or more expressions is the expression with the least numerical coefficient and of lowest degree that can be exactly divided by each of them. It follows that the numerical coefficient of the lowest com- mon multiple is the least number that can be exactly divided by the numerical coefficient of each expression ; i. e., the least common multiple of the numerical coefficients. And the lowest power of any literal factor that can be divided by each expres- sion is the highest power of that factor found in any one of the expressions. A factor appearing in any one of the expressions 136 ALGEBRA must appear in the lowest common multiple. Hence the rule : * To form the longest common multiple oftico or more monom,ials^ take the product of the least common multiple of the numerical coefficients^ and each letter raised to the highest power to v^hich it appears in any one of the expressions. In polynomials^ factor each expression and proceed as vnth monomials^ treating each factor as you icould treat a single letter. Note. — To find the least common multiple of two or more numbers, as in arithmetic, separate them into their prime factors, and take each prime factor the greatest number of times that it occurs in any one of the numbers. Thus for 72 and 96, we have 72=2-2-2-3-3, and 96=2-2-2-2-2-3. Hence, the least common multiple of 72 and 96 is 2-2-2-2-2-3-3^288. Example 1. Find the lowest common multiple of 2^aWc^ and SOa'b'c. The least number divisible by 24 and 30 is 120. And the highest powers of a, 6, and c, in the two expressions, are (i\ b^, and c*, respectively. Hence, the lowest common multiple is 120a*6V. Example 2. Find the lowest common multiple of 15aV, 21y^z, 3&a''z\ Here the least common multiple of 15, 21, and 36 is 1260. The highest powers of a, y, and z are a\ 2/^, and s^, respectively. Hence, the least common multiple is 1260a*2/"^^ ^Example 8. Find the lowest common multiple of 2x'^—4x + 2, \^' + x-2, and 5x'-^10x. * It is easily shown that the product of the highest common factor and the lowest common multiple of two expressions equals the product of the expressions. Hence the. lowest common multiple of two expres- sions may be obtained by dividing the product of the expressions by their highest common factor. Since this method could be of value only in case the factors of the expressions could not be found, it is not used in this book. ' COMMON FACTORS AND MULTIPLES 137 Here 2x'-4x + 2=2{x~iy; x' + x-2={x + 2){x—l); 5x'' + 10x=5x{x-\-2). The least common multiple of 2 and 5 is 10. The highest power of the factor xia x; of x—1 is(a?— 1)^; and of a? + 2 is a? + 2. Hence, the lowest common multiple is 10a7(a?— l)^(a? + 2). / EXERCISE 45. Find the lowest common multiple of : 1. ^a'b, Sab\ 7. 2bm'n\ 4:cmi7i\ 10a'm7i\ 2. Qa'b\ 10ab\ 8. 15^//, 6pY, Aj^Y- ' 3. M'bc\ ^ab\ y 9. SQa'% 9a'b% lQd'b\ y4. Sla!^', l^x^f. 10. 27x''i/, QxY, 4xi/\ ^xHj\ 6. lxy\ ^x% GccV. 11. {x-\-y)\ 2{x + y)\ b^. 2la'bc\ 4a'bc\ Qab\ 12. (a-b)\ S{a-b)(a + by. > 13. x{x—l)(x + l), 2x'(a; + l). 14. x'(l-{-xy(l-x), (l + a;)(l-a;). ^\6. 10(a + ^»)'(c-J), 15(c-c7)^(a + J). ^%6. 25rt^(a-^>)(2a + ^)^ 45aa!(2a + ^)l 17. aj'^-l, x'-2x + l. : 23. 2a;^ + 3a;-2, 3£c^ + 7a! + 2. 18. x'-l, x'-x-2. 24. 2iK2-aj-10, 2x' + x-S. y 19. l-a;^a;+a;^-2ar^ 2,5. 2a;=^ + 2a;, aj=' + 5a; + 6. ]M0. a'-b\ 2a' + ^ab + b\ 26. a;^-l, x'-x, 2x\ 21. a2 + 5a + 6, «^ + 7« + 12. 27. x'-l, a;^ + 3a; + 2, a;^ + a;-2 c/22. x' + x-SO, a;' + 5£c-6. 28. l-x\ l-x\ 1-x. 29. x'-fii^, 2x'-4x, x'+ 1. ' 30. a;^ + 5a;-14, 4a;^-16a;^ + 16a;, ^x\ 131. a'-b\ a^-b\ a'-b\ ^^32. l-x\ l-x\ l-a;«. 138 ALGEBRA 33. x + l,x^ + l,x'-l. 34. x' + 2x'-Sx, 2x' + Qx'' + 2x + 6, Sx^—^x\ 35. a*-b\ a* + 2aW + b\ a'-2aW + b\ 36. 27a=^-8, 9a'-4, 9a'-12a + 4. 37. x-x\ lOx + dx'-x^ x-x'-{-x^-x*, 38. 05^-1, x' + x-2, x' + bx + Q. 39. x'-l, x' + x' + x, Sx\ ^40. a'-l>'\ a' + a'b-ab'-b\ a'-a'b-ab' + b\ lAi, ic2-15a; + 36, aj='-3aj2-2a3 + 6. 42. a^-a^ + a + S, a^ + a'-Sa^-a + S. 43. 6a^ + a'^-5a-2, 6a^ + 5a^-3a-2. 44. ^x^ — lx'y—2xij\ ^x^ + xy—4y\ 45. a?*-13a;' + 36, aj*-a3='-7a;' + a; + 6. ^46. 633^-03-1, 2a;M-3a^-2. 47. 4a;^-4£c4-l, 4a;2-l, 4a;2 + 4a; + l. I 48. ax—ay — bx + by, x'^—2x7/ + y^. I 49. 6a;^ + 13aj-28, 12a^2-3l£c + 20. I 60. 8a;^ + 30a; + 7, 12a;^-29a;-8. CHAPTER XI. FRACTIONS. 101. In § 52, § 53, § 54, § 55, some principles of fractions were established which the student should now review. Other principles of fractions in common use in the treatment of algebraic expressions and equations are here given. 102. The algebraic signs of a fraction. In every fraction there are three signs to consider ; the sign before the fraction, the sign of the numerator, and the sign of the denominator. Thus, +-rk'-' ~+7' ~^' ^^^' ^hen a fraction is standing alone, the sign + may be omitted ; as — ^, ^, etc. — 7 o Since the names, numerator, denominator and fraction, are merely other names for dividend, divisor and quotient, respect- ively, the laws of signs for division must hold for fractions. From division we have the following principles : I. If the signs of the numerator and denominator of a frac- tion are both changed, the sign of the fraction is unchanged. II. If the sign of either the numerator or denominator is changed, the sign of the fraction is changed. + 2 —2 —2 -1-2 Thus, ■— o and -^ are both positive ; 370 and 30 are both negative. Since the sign before a fraction may always be considered as indicating whether it is to be added or subtracted, and since 139 140 ALGEBRA the subtraction of a term is the same as the addition of the term with its sign changed ; the sign before a fraction may be changed if the sign of either its numerator or denominator is changed. Therefore^ any two of the three signs of a fractioii may he changed without affecting the expressio?i as a whole. Thus, j.=zrh~ h~~~ '—b' ^^^^ being positive ; and -a a a —a , , • ■jr =^= — j = — -ziy ^^^" bemg negative. If the numerator or denominator be a polynomial, by § 40 its sign will be changed by changing the sign of each of its terms. Thus, ■x^-\-3x—l 'of-Sx+l x^—3x+l 103. Multiplying or dividing both terms of a fraction by the same expression does not change the value of the fraction. For, since -=1, by § 54 we get a _a X _ax , . a _ax ax a l)-\x~b^' ^^'^^ '^' l}~~b^' ^"^ Tx~b' Example 1. Multiplying both terms of —77- by a + &, we get a—b _ a^—b^ a+b~{a + bf 'L2a^b^c Example 2. Dividing both terms of 10^41^2^2 by 6aWc, we get 12aWc _ 26^ 18a*b'c' ~ 3ac The processes of multiplying and of dividing both terms of a fraction by the same expression are called reducing the frac- tion to higher tsrms and reducing the fraction to lower terms, respectively. Example 2. Reduce — — 13^2 to its lowest terms. FRACTIONS 141 104. Reducing a fraction to its lowest terms. A fraction is said to be in its lowest terms when its numerator and denom- inator have no common factor. To reduce a fraction to its longest terms, divide both its numerator and denominator by all of their common factors^ or by their highest common factor. This follows directly from the definition of highest common factor, § 99, and from § 103. Example 1. Reduce ^a^^Lyi to its lowest terms. Dividing the numerator and denominator by their highest common factor, 4,ix?yz, S6x*yz^~9xz' ocfy^z^ ■x*y*z'^ Changing both negative signs to positive, and dividing both terms of the fraction by their highest common factor, x*^V, Qc^y^z^ _oc^y^z^ _x^ ' - ~ —x*y*z^ ~x*y*z^ ~ y^' x^ -\- 2x 3 Example 3. Reduce ' ., ^ r. to its lowest terms. Factoring, and dividing both terms by their highest common laetor, a; + 3, x^-{-2x-S _ ix-l)(x-\-S) _ x-l x'' + 5x+6~{x + 2){x + S)~x+2' 1 -i^^ Example 4. Reduce , „ = to its lowst terms. Changing signs, and writing both terms of the fraction in descending powers of a?, 1-^ ^_ af-1 ^_ ix+l)(x-l) __x-l_l-x x'' + 8x + 7~ x'^ + Scc + T" {x+l)lx + 7) x + 7 x-\-7' Note. — The process of dividing the numerator and denominator by a common factor is sometimes called cancellation. 142 ALGEBRA EXERCISE 46. Reduce to lowest terms : ^ -SaW 6. 8. -256a^^V^ — 4:9x^1/^ w*' 3a.y-2/ 28 a;^-9a; + 20 i 10 + 3£c-£c^' 29. x^-h2x'y-2xf-y'' x'—'6x^y—2xy'^^\.]f' 30 }f-2hc-\-&-a} 31. 32. 33. ?yr 10m 4 16 m'^ + m— /2 ac—bc—ad^-bd ac^ad—bc — bd' x'^x'.-^x-^ x'-4x' + 2x-V^' 105. Reduction of a fraction to an integral or mixed expression. A fraction whose numerator is of a degree equal to, or higher than, the degree of the denominator may be reduced to an integral or mixed expression by division. FRACTIONS 143 Example 1. Reduce ^ — -— to a mixed expression. B S 56 Sx'-Qx + 2 _3x^_6x 2 oX oX oX oX =x-2 + §~. Sx Example 2. Reduce to an integral expression. Performing the indicated division, X -d^^x-\-l. Example 3. Reduce ^ to a mixed expression. Since the denominator will divide a^ + l, add and subtract 1 in the' numerator. Then, ^=^^!±i=2=a!±l__2^=„,_„ + l. S a +1 a+1 a +1 a+1 a+1 Example 4. Reduce — ^r^^^ to a mixed expression. x—2 ^ Dividing, we get the quotient Sor+l, and the remainder 8. Hence, 3^JzS|+6=3x+l+ « ' X—2 x—2 The division has the effect of breaking the numerator into two parts, such as in the preceding examples, one part of which is divisible by the denominator, and the other part not. EXERCISE 47. Reduce to an integral or a mixed expression : x^ ' / ' a; +1* ' X —y' 2a;' + 4a;4-l ^ - a;^ + 3a;^ 4- 3a; + 1 a;^-3a;'> + 2a; ALGEBRA 4x' + 2x'-Sx + l Q x' + lQ 2x'-l ' ' x + 'I' 11. 12. x* + U X -2 " x' + y' 9a' + Sab + 4:b' x + y' ■^"* Sa-2b ' x'-y' x-^y ' 144 8. 106. Fractions reduced to their lowest common denominator. Fractions may be reduced to equivalent fractions having a common, or the same, denominator by § 103. Since the lowest common denominator must be obtained by multiplying each denominator by some expression, therefore the lowest common denominator must be the lowest common multiple of the denominators. The expression by Avhich any one denominator must be multiplied to obtain the common denominator must be the quotient obtained by dividing the common denominator by the given denominator of the fraction. Hence the rule : To reduce two or more fractions to equivalent fractions having the lowest common denominator, first find the lowest common multiple of all of the denominators ; divide this hy the denomina- tor of each fraction in turn, and multiijly both terms of the corresponding fractions by the quotients thus obtained. Example 1. Eeduce -t' #, and — to equivalent fractions having the lowest common denominator. The lowest common multiple of ab, be, ac, is abc. Dividing OC CSC this by ab gives c. Multiplying the terms of -v by c gives —r- . Dividing abc by be gives a. Multiplying the terms of ^ by a au gives -#• Dividing abc by ac gives b. Multiplying both terms of — by & gives —^. Hence the required fractions are ex ay bz abc abc abc FRACTIONS 145 Example 2. Reduce ^2_|.4^^3 ^ s^^' ^^^ ^Hs *o equivalent fractions having the lowest common denominator. x+2 x + 2 a?— 1 x—1 X X ie + 4a;+3"'(it'+l)(a?+3)' a?''^-9~(£i? + 3)(x— 3/ x—?r'x^Z Hence the lowest common denominator is (a? + 3)(u7— 3)(a7 + l). Dividing by (ic+l)(x+3) gives ic— 3. Multiplying the terms of .T + 2 , _ . (a; + 2)(a;-3) ix^-x-^ (^•+l)(x + 3) '^y ^ '^ gi^'^^ (ic+l)(a!+3)(£C-3)' ^^af* + ar^-9x-9* Similarly, x-1 _ (..-l)(x-fl) _ 0.^-1 . and (;:c + 3)(x-3) (x + l)(ic + 3)(if-3) ar^ + x^-Ox-Q' x _ a;(a? + 3)(.r+l) __^+JteM^^^ a?-3~(^+T)(x+3)(a?— 3)"~xM^'-9a?-9* EXERCISE 48. Reduce to equivalent fractions having the lowest common denominator : 2 15 « 2a^ W ^' Zx' 'Ix'' Qx ^' Wc' Ic^c' Act'b'' ^' W 6^^' ia^' ^* xy' yz' xz 3 5 1 y j^ z X y \x^ 5. 6, £C+1' X-V X^-\ 3 5 a;^ + 3a; + 2' 2a;^ + 5a; + 2' 2a;=' + 3a; + l 2cc + l \—x X 10 (^+iy' (^+iy' («;+ir ^/ X '^ 2 a; 2a;^ ^^^^_ 146 ALGEBRA iQ ^ •'^ ^y J y' '-^y—y^' x'—y^' ii _^ y__ ^' y^ x-y' 2y-'lx' ^{x'-yy ^y' - x')' 12. _^, y -1-. a— a y—o z—G 2a 4:c 6«-6' 1 13. ^14. ^^' (c«-5)(a=^' {b-c)(b^^y (c-a)(c^' •|/-g y -\-z z + x x + y ^ ' (y-^)(^-»^)' (y-^)(y-^)' (^-^K^-y)' 107. Addition and subtraction of fractions. By § 56, a+b+c_a b c X XXX Hence, °+* + f =^±*±£.. jr * jr JT >r Therefore^ fractions hamng a common denominator may be added by adding the numerators for the numerator of the sum^ and using the common denominator for the denominator of the sum. A fraction may be subtracted by chayiging its sign and then proceeding as in addition. Fractions which are to be added or subtracted must first be reduced to equivalent fractions having a common denominator. As was pointed out in § 5, if either the numerator or denom- inator of a fraction is a polynomial, the dividing line also serves as a sign of grouping. Consequentlj^, in such cases, the signs of the terms in the numerator of a fraction which is FRACTIONS 147 preceded by a negative sign must all be changed when the numerators are added. Example 1. Simplify ^J+|^+^. Reducing to a common denominator, x+1 3 iX^-l __6 x' + Qx 9x' 2x'-2 of 2x SX" Qx' iixr' ijx' _ ex'' + Qx + dx^ + 2x^-2 ijx' _ 17x' + ex-2 Example 2. SimpHfy a^Ja+^ - a^-L+3 -a^-L^2 Here ^ 3 1 _ ' a^— 5a + 6 a^— 4a + 3 a^—i^a + 2 2 3 1 (a-3)(a-2) (a^3)(a-l) {a-2){a-'i) 2a-2 3a-6 a-3 (a-3)(a-2)(a-l) (a-3)(a-2)(a-l) (a-3)(a-2)(a-l) 2a— g— 3a + 6-2 + 3 (a-3)(a-2)(a-l) 7-2a (a-3)(a-2)(a-l)" Examples. By addition reduce x + y + ~r-^. to a fractional form. Here the integral part may be considered as a fraction whose denominator is 1. Hence, i»+2/ + -^=^^ + x—y 1 x—y _ x^-y^ ^ y^ x-y x-y x—y ^ 148 ALGEBRA Note. — It is best first to arrange the denominators of all fractions according to the powers of some letter, making use of § 102 if necessary. Example 4. Simplify >-l_ + _^-^— -J- ^ "^ x—1 1 — X^ 1 + '. + x Arranging all denominators in descending powers of x, and changing signs in the second fraction, we have, 1 a- 1 1 cc 1 1 1—x^ 1 + x x—1 x'^—l x + 1 _ x 4-1 X x — 1 ~xi^—l~x^—l~~x^ — l _ x + \—x—x-\-l ~ x'—l 2 —X V EXERCISE 49. Simplify : 1. h^-^'+' ■ 9. 1 + ^. X X^ ^X X 2. y-H 5. 10. X-\-zr-7—' DC ac ab 1 + a; _ a;4-l , 1 ,2^-1 ^^ , ^ x^ 3. —-2- +-r-:+ -.A ■> • 11. a? + l- x^ hx lOa?'^ ' x — \ 4. 1 + i + l. 'l%l-.' ^^^ £«?/ yz xz \Zy \^rx \—x u. x-\ a; + l' ^ 6. 3 2 4 a^ — £C ic^+aj iK' 7. 2a;+l a; + 2 1 a^-2 2£C-1 X 8. 3a ^ 1 ^< a ^ax 14. 15. a-\-b a — b a—b a + b' 1 1 16. o:-2 2aj^ + 7£c-4 3ic^ + 13a;+4 FRACTIONS 149 17 _« L L. 21 -^ ^-1 18 ^.+JiL_J^. V 22.^^ 1 I 2y \-\-x x—\ 1—x^ x^-\-xi/ ^—xy x^—if-' x^ 'A—y 2/ —4 ^ + 2/ x-\-A x—i 25 1 1 26. ab—ac — b'^ + be bc — ab—&-\^aG 3a + l 2^>-l 46—1 6(^ + 1 12a m ^ 16c 24c? * 27. . ,A, . + 1 1 {a—b){b—c) {b — a)(a — c) {c—a){c—b)' '*0' ;;;3:;,— ;:r3:7,+;:;23:::72- V«>"' 7777-7;;.+ or^-::.+i;F £c— 2/ aJ + 2/ x^-Vy'^ ^ ' 2b— x 2b-]-x x^—A^b OQ _^^ 2a; 1 2a; So; '*^- a;=^-l a;^ + a; + l"^a;-l' *^^- ^ ^^a;"^-l x'-^V 32.-^-^+. 1 1 a;— y x + y x—'ly a; + 2y Note. — In certain cases, like this last exercise, it is best to add only certain ones of the fractions at a time. It saves long multiplications. Here, add the first and second fractions ; next add the third j)nd fourth ; then add the sums thus obtained. 33.4-,+4-o-A-^o. 34. A— ^.4 ^ ^ cc + 1 £C + 3 x—\ a?— 3* * a — b a^b c—d c-\-o 2,3 2 3 35. a;— 5 a; + l a; + 5 a;— 1* i - 1 36. - a;'-^ — 1 a;''^ + l x^—a^ x^ + a^ a;* — 1 x*-\-x^ x*—x^ d'—b^ 150 ALGEBRA 39. ^^~-x\ 41. a+ ^,_^, +k ■ ^ x'-hy' , x'-7/ ^^- S + 2x 2^Sx , 16.x— jc^ 40. -+x ~. 42. -o —-on 2 — A~' x—y x-\-y 2—x 2 + £c £c^— 4 108. Multiplication of fractions. Fractions may be multiplied by the rule established in § 54. In case the given fractions are such that their product may be reduced to lower terms, the process of multiplication and reduction may be shortened by first cancelling any factor of any numerator hy an equal factor of any deno7ninator. This is evident, since sucli factors may be cancelled after the multiplication. See § 104, note. Example 1. Find the product of i|^, i^, ^ 12a^ Uab 5bc 'l2a^Uab-5bc W 9a-c-* 2a~" 76'^-9aV-2a 840a362c \2^a?b''& Reducing, =g^ Example 2. Simplify ( ^~^ ) x^ + 2a?, / ^^-4 \ / .T='-9 \ __ {x-2){x\2) (x + 3)(a;-3) \x^-^x)\x'^2xj~ x{x-Z) ^ x{x-\-2) ... „ ^ ix -2){x + 3) ^^, x' + x-e Cancellmg common factors, = -^ -, oi — ^^ Note. — The cancellation may be indicated by drawing lines through the common factors. {x-2)Cs»^ (x + ^){s»^^ (x-2)ix + :^) x'-^x-6 -^^^^^' xi,^^ a-iae-t^- ~ x-x ~ x' o a- ^^f a'- lSa + SO ^ a'-6a-7 ^a + 5 Example 3. Simplify ^..g^.go x a^-i5a + 56 ^ ^=:i* FRACTIONS 151 a'^-Sa-SO 'a='-15a + 56'a-l~ (ii^(a,;s*4tJ) (^^^(^.*^ (a-1) g + l -a-l' 109. In multiplication, any mixed expression should first be reduced to a fraction. Expressions free of fractions may l)e treated as fractions with the denominator 1. EXAMPLE 1. Simplify (. ^£^) (y- ^ \ i^-yj\ ^+yj \3c-yJ\3c+yj x^—y"^' Example 2. Multiply ^.-^^^^ by a^^ 4- 4x- 21. x^2 x+2 (a? + 7)(^-3) ■x^ + 12x-v^h^'^ '^ ^^^-{x+^){x + 7)'^ 1 (x+2)(x—2>) x^—x—6 x + 5 ' ^' x + 5 Note. — It is clear that the product of a fraction and an integral ex- pression, as in tlie preceding example, may be obtained by merely multiplying the numerator of the fraction by the integral expressioiij and placing the product above the denominator. Thus, in general, a an EXERCISE 50. Simplify : 4,w2 152 ALGEBRA x—y x^ Ay 20.-1 2y + 3y + 3 ^' 7+3 ^"^^^^2a;-l* J - (-S)(.-^)- J te^-l a;'' + 3a;4 2 X 4-2^a;' + 2a3 + l' x^-1 -^X ,2 4y2 9. ^Z^ x^ - ' £C + 2y x^—y^ a + 3 ^ a + 4 * 11 ^'-y' v ^'~y' 19. (a; i 14 15 16 ■( ab ^ b' y a'-^b d^_ _ y' ^)\ 3£c _27a^ / 1 L+iV J 20 xy—y 'x + y x^—i/ b-Sa 18. ^a^-b^^(^^^^y ({x±yY\ r^X 21 23 X a?^+y^ i /a^-hab\ / g ^\ 22 /^ yy. a.^ + y^ \ Vy a;A (^+y)7' /x^—y'^—z^ — 2yz \ x^—xy—x: 2z x + y+z ) 24. (^^-a^y + 2/0x5 25. 26 27 28 2a;+l X, ^^=T6^2^M^5^+2 .2V / 4^2 5c 1. (3a+jy3. / x'-Sx+ 2 \ / a;^-7a; + 12 \ / a;-^-5a;^ V£c'^-8x + 15y \x'-bx+ 4/ V a;-^-4 / '• (a? x^—4x bx + QjW + 2x Sx m- FRACTIONS 153 Multiply the following by such expressions as will make the products integral expressions : Suggestion : Multiply by the lowest common denominator. a 1 30. 31. ci'-b' a + b' 2ic + l , 3 x"^ — 1 x^ — 1 x^ + x+1 32. ^— ^ + 110. Division of Fractions. Since the divide7id = quotient X divisor, .- a; a y ^ then -= -2^ + 1 1 x—yx-\-y 9x^ — 4y^ o. 5— 7. ; 11. "o Ti — • 1 « ^~y ^ + 2/ 6x—ly X ^"t-y ^"~y x^—y '12. 7- l+a3^^ a'^-a'h'^b' ^ x Suggestion : Begin with the last complex fraction and simplify step by step. A fraction of this kind is called a continued fraction. 13.^ 14. a ^ x-V a + A- 1 + '' a-l " ' 3-a; FRACTIONS 157 EXERCISES FOR REVIEW (III). 1. What is meant by the factors of an expression? Illus- trate. 2. Name the typefomis that have been factored in this book. 3. Of what type form are the following? Factor them. / {a) Qx'-^x'y^'ixxf. (d) a'b-a'b' + aW-ab'. ,{b) ^a'b-'iaW + d'b. ^''a3^ - %^ (^0 25£c' + 70a;?/2 + 49yV. Ao) 2a;' + 5£c-12. (o) £cy + 12a7y=' + 27. (a;) ac-bd-ad+bc. 10. , Factor: "^{a) 9«« 6a^ + l. /O*) Qx'-lSxyi-Qy\ (b) 12a' + 7ab-10b\ ^ (k) 7x'-h2bxy-12f, Uc) a'-^a'b' + 4:b\ {I) 16«'^-48a + 35. * {d) 4a;* - 4a;V' + 9y*. (/?i) a"'^ - ^'". {e) IQa' + lQab-W. '' {n) a' + 216. (^) a;^« + 2a;« + l. (/>) a' + a' + l. (h) a'-b'-a-b. (q) Sx'y + Sxy-}-SxY. (^) x* + xy+y\ (r) a^—a—a'^b—ab. f^RACTIONS 150 (s) l + a—b—ab. (v) ahx^-\^{a'^^-h-)xy-\-aby'^. (t) 4:a'b'-(a' + b'-cy. (w) a'b' + a'x'-b'x'-x*. (u) Sbc-4ad+Qac-2bd. (x) Qa' + 9ab-3b-2a. 11. Find the 11. C. F. of : (a) x'-S, x'-2x'-{-x-2. (b) x'-V2x + S^, x'-2x'-19x + 20. (d) x' + Sx' + 4x + 12, x' + 4x' + 4x + ^. 12. Find the L. C. M. of : (a) Qx\ x'-2x, 3£c='-12a; + 12. (b) 2a%ba'-^ab,2ab + 2b. (c) x^—x^ — Qx, x^ + 2x\ x^ — Qx^ + dx. (d) Gx' + 7x-^,2x'-x' + Sx—4. 13. Explain the process by which algebraic fractions are added. 14. What law of signs must be observed when a fraction is subtracted? 15. Simplify : / X 3 . 2 1 (a) x'-l ' x'-Sx-4: x'-x-2' (1\ 9^ + 17 , 2x-l 2x-\-l x'-2x-4S ' 2a;+12 2a;-16 17. By what must a fractional expression be .multiplied in order to obtain an integral expression ? 18. Multiply by such an expression as will make integral: 1 QO ALGEBRA J. 1 3 2_ 3(^-1) 3a; 9^ ^ ^ i«-2 a; + 2 a; + l* 19. Sin,plify (-^-l)^(3.,-J^) 20. Simplify ^V'^'^V^ ^ x'y^—x '^ x^-\-xy-\-y^ ' y^—x^ 21. Simplify 1 1 X^3ipj3- b a 22. Divide ^ +lz:?by y^ -III? and express the quo- J. ~T~ XX X "I X X tient in its simplest form. 23. Simplify: W \^, «; • \a b)^ a + b ^b\a-by ^^ \bTc~^^rFcJ\^Tb'^^^^b~' a'-b' J' /2(^)_«+.wi ly ^ ^ \ a-^x a—xj \a xj ^ ^ Va;'-y=^ x+yj \ 2zy J x-y ab—b '^ a^-b'' a^—ab ab 1 CHAPTER XII. LINEAR EQUATIONS— ONE UNKNOWN NUMBER. 112. We showed in Chapter II the meaning of an equation and how, by the use of axioms, to solve the simplest kind of equations containing one unknown number. Now that positive and negative numbers and fractions have been discussed, we return to the discussion of equations. 113. An equation is integral with respect to its unknown numbers when both of its members are integral with respect to those numbers. Otherwise it is a fractional equation. Thus, x-\-§=Qx-\-5 and 2^^ + 3.x^=8 are integral equations in x. Sx^—xy + Si/=7 is integral with respect to both a? and y. The equation - + -=10 is fractional with respect to both x and y. X y A rational equation is one in which both members are rational expressions Avith respect to the unknown numbers. Otherwise it is called irrational. The equation j/a;— i/a;— 1 = 2 is irra- tional. All equations here to be considered are rational equa- tions. 114. The degree of a rational integral equation is the degree of its term of highest degree with respect to the unknown numbers. Thus, x + 3=4x and 3ic— 2i/=10 are of the first degree. x^—5x+Q=0 and x^—2xy + y=9 are of the second degree. a^ + a?"— x=0 is ot the third degree. a?* — 1/^ = 6 is of the fourth degree. 161 162 ALGEBRA An equation of the first degree is also called a linear equation.* One of the second degree is called a quadratic equation. One of the third degree is called a cubic equation. One of the fourth degree is called a biquadratic equation. 115. Equivalent equations. Equations which have the same solutions are called equivalent equations. Thus, 6a?— l==4a? + 7 and 2a?=7+l are equivalent, for each is satisfied by a?=4, and by no other solution. In Chapter II linear equations were solved by the use of axioms. Thus to solve 6x— 5=4a? + l. (1) Adding 5 to each member, 6x=4:X + 6. Axiom 1 (2) Subtracting 4x from each member, 2x=6. Axiom 2 (3) Dividing each member by 2, x=3. Axiom 4 (4) It will be seen that this work consists of deducing in the successive steps, successive equations, each one equwalent to the preceding one. Thus, equations (1), (2), (3), (4), in the above example are all satisfied by a3 = 3, and by no other value of X ; hence they are all equivalent. Therefore, the solution of the last equation is the required solution of the given equation. 116. Observe that to change the given equation of § 115 to an equivalent one whose first term consisted of a multiple of the unknown number, and whose second term was a known number, i. e. to reduce the given equation to the form ax = b, we proceeded as follows : I. The known member in the first member^ with its sign changed^ was added to both members to free the first member from known numbers. * The name linear equation is derived from the fact that the equation of the first degree with two unknown numbers has a pecuhar relation to a straight line. See § 134. LINEAR EQUATIONS— ONE UNKNOWN NUMBER 163 II. The term iiwolmng an unknown number in the second member urns added vnth its sign changed to both members to free the second member of unknown numbers. This process gave a new equation in which certain terms that appeared in a member of the old equation appeared in the opposite member of the new equation with their signs changed. Tills result is equivalent to transferring terms from one mernber to another and changing the signs of the transferred terms. This process is called transposition. The term is said to be transposed. Note. The pupils should use the correct phraseology of -'adding equals to both sides of tlie equation," until the thing actually done is firmly fixed in mind. When this is thoroughly understood the briefer form, "transpose", may be used if desired. Care should be taken^ however, that the pupil does not say "transpose", and mechanically perform tlie process, forgetting what he has really done. 117. To solve a linear equation for one unknown number. The following general method may be used in solving any linear equation for one unknown number : (1) Remove all signs of grouping., if any exist. {2) Transform, the given equation into an equivalent one having all the unknown numbers in the first member, and all terms free of the tinknoion number in the second member. (3) Unite like terms. {J/) Divide both members by the coefficient of the unknown number. To check the transformed equation, see if the terms that were cancelled in any member of the given equation reappear in the other member of the new equation with signs changed. Example 1. Solve 7ic+15=4a? + 3. 164 ALGEBRA 1. Adding -4a?-15, 7^-4^=3-15. (Ax. 1) 2. Uniting like terms, 3x=— 12. 3. Dividing by 3, x=—4, the solution. (Ax. 4) Check. Substituting for a? its value —4, the given equation be- comes —28 + 15= — 16 + 3, an identity. Example 2. Solve 3(a + l) = 12 + 4(a-l). 1 . Removing signs of grouping, 3a + 3=12 + 4a— 4. 2. Adding -4a-3, 3a-4a=12-4-3. (Ax. 1) 3. Uniting like terms, — a=5. 4. Dividing by —1, a= — 5, the solution. (Ax. 4) Check. Substituting for a its value —5, the given equation becomes 3(-5 + l) = 12 + 4(-5-l), or —12=— 12, an identity. Example 3. Solve (5— a-)(l + a?) = (2— .T)(4 + a?). 1. Removing signs of grouping, ^ + 4:X—x'^=S—2x—x'^. 2. Adding x^ + 2x- 5, 4x-x' + 2x-\-x'=8-5. (Ax. 1) 3. Uniting like terms, 6x=3, 4. Dividing by 6, ^=h ^^^^ solution. (Ax. 4) Check. Substituting for x its value |, the given equation be- comes (5-^)(l + J) = (2-i)(4 + |), or 4^1|=l|-4^, an identity. EXERCISE 53. Solve the following equations for x : 1. 2ic + 7 = 14-5i«. 4. 12-Ux=Q—9x. 2. 30a; + 24 = 60 + 48a5. 5. 8a;-7 = 3a; + 3. 3. 19i«-22 = 8ic-17. 6. 14aj + 20-12= -20a; + 35i«. LINEAR EQUATIONS— ONE UNKNOWN NUMBER 165 7. 2Sx-^b = 21x-10x-U7. 9. ^{2-4x) = 4{l-dx). 8. S{x-2)=2(x-S). 10. x-(Q-2x) = 9(x-l). 11. 2(a;-l) + 3(a;-2) + 4(ic-3) = 0. 12. 2x-b(20-x)-i) = 0. 13. 5(2a; + l)-7 = 3(2aj-7) + 51. 14. bx-Q(S-4x)=^x-7(4 + x). 15. 2(a;+12)-(a;-3)=0. 16. (a;-2)(a3-3) = (a;-4)(a;-5). 17. S(x + 4)(x-2)-^ = S(x + b)(x-S)+x. 18. (a; + 2)^-a;='=a;-5. Solve the following equations for a : 19. l + (a-^y = (a + iy-4. 20. («-2)(a-5) + (a-3)(a-4)-2(a-4)(a-5). 21. 2.5a-6.75-1.25a-3. 22. 0.75a + 2(l-1.2«)=0. 23. 2-6.9a(l-2a)-2(6.9a'^-3). 24. \a-\a = 2. 26. Ja = 4-lia. 118. Some fractional equations may be changed to equivalent linear equations by multiplying both members by such an ex- pression as will destroy all fractions (Axiom 3). The neces- sary multiplier will evidently be the lowest common denominator of all the fractions. This process is called clearing of fractions.* The pupil should see that the real process is multiplying both members by the same number. *Like the term "transpose," the term "clearing of fractions" is often used by pupils without their knowing the real process involved and the authority for it. A pupil sometimes thinks that clearing an equation as ^^-—j=- of fractions consists in multiplying the first member by 1 + a7 and the second by 2x-\-4 rather than each member by (l+a;)(2a;-h4). 56 ALGEBRA Thus, solve 1_ X 2 . ~3x' 5 6' Multiplying both members by 6x, 6- -4 = :5X. Adding —5x + 4- -6, 5x= =4-6. Uniting terms, 5x= r-2. Dividing by —5, X- =|. (Ax. 3) (Ax. 1) (Ax. 4) If a multiplier be used whose degree in the unknov/n number Is higher than that of the lowest common denominator, the resulting equation will usually not be equivalent to the given equation. (See § 173). Hence the rule : To clear an equation of fractions 7nulti2^ly all terms in both members by the lowest common denominator of all the fractions in the equation. It should be noticed in the following examples that the easiest way to multiply a fraction by the lowest common denominator is first to divide the lowest common denominator by the denomi- nator of the fraction, then multiply the numerator by this quo- 2a tient. For example, to multiply ^— by eer*, we divide ^x" by 3a? oX obtaining 2a?\ then multiply 2a by 2x^ obtaining 4aa?^, the product. If a fraction is preceded by the negative sign, the sign of every term of the numerator must be changed when the multiplication is performed. See § 5. Example 1. Solve -^,_-^=-A_ x—2 x+2 x^—4: The lowest common denominator is a?^— 4. Multiplying by x{x + 2)—x(x—2)=4. (Ax. 3) Removing signs of grouping, x^ + 2x—x'^ + 2a?=4. Adding, 4a7=4. Dividing by 4, x=l, the solution. (Ax. 4) a?— 3 a?— 4 a?— 6 x—7 Example 2. Solve .T— 4 x—5 x—7 x—8' LINEAR EQUATIONS— ONE UNKNOWN NUMBER 167 Here there will be an advantage in adding the fractions in each of the members before clearing of fractions. 1- Adding, ^__zl_^=___l_^. 2. Multiplying by (^—4)(a?—5)(a?—7)(x— 8), -l{x-7)ix-8) = -l(x-4){x-5). (Ax. 3) 3. Dividing by -1, {x-7){x-8) = (x-4){x-5). (Ax. 4) 4. Removing signs of grouping, x''-15x+5Q=x''-9x + 20. 5. Adding, — x'^ + 9x— 56, x^-15x—x'' + 9x=20-5Q. (Ax.l) 6. Uniting terms, — 6x=— 36. 7. Dividing by —6, a?=6, solution. (Ax. 4) EXERCISE 54. Solve the following for x : . Q 3 CC + 1 , £C^ ^' ;T-ri-:;r—i+:;j- 1. 3 1 2a5 4* 2. j« + 3 o «.-3-^- 3. »^ + l 2 1 a;' '^Sx X 4. 2a^+l 1 a; 4 05-1 2' 5. 1 2 3 2x + l x + l 'Ix 6. ^-2^03 + 1 ^• 7. cc + 1 a^-l l-f2a; 3a;^l 3.r+l W-l' R 1,1-1 £c + l a; — 1 x^ — 1 10. ?Lz| + ?i=f=,. a;— 7 a;— o £C — 2 05 + 2 ^2 x — '^x — t) /13.-A^l^-:i^^ + 7. £C + 1 aJ— 1 14. 2 + -?^=^^. 05 + 3 a;+7 15 3a; + 7 _6cc-2 4aj + 8 8a.— 5* 1R 8 a;-3_a3 + l 2a; + 3 ' 2£c-3 4a;^-9' a5"^a; + 3 cc-1" 168 ALGEBRA 17. 2a;-3 ^_ a; + 5 11 ' ' ^ jjO. ^^ 2£c-4 "^ dx-Q 2* ' aj' + 5a; + 6 a^^2 a3 + 3 j^g a? a;^ — 5a;_2 21 ^ ^ ^ ^ 3 3a;— 7 3* x—^ x—1 x—4: x—2 X ^g ^ 3_ 4 x-\-l Qo 3a; 2a; _2a;^— 5 «„ 3a; + 5 3a;'' + 5a;— 4 24. 4a;-3 4a;^-3a; + 2* a;— 7 a;— 9 _a?— 13 a;— 15 a;— 9 a; — ll~~a;— r5 a;— 17* OK a;— 5 , a;— 7 a;— 4 , x »o. — =4- 26. 27. 28. a;— 7 a;— 9 a;— 6 a;— 10 a; a; + l a;— 8 a;— 9 a;— 2 a;— 1 a;— 6 a;— 7 13 2 a; + 3 9— a;' 3— a; 3(a;-l) 3^ _ 9 x—2 a; + 2 a; + l* og 5(a; + 6) 2(5a; + 2) _^ a; + 10 2a;-l a;— 5 7— a; 2a; — 15 30. „i 9a; + 17 , 2a;- 1 2a; + l ^A. -5 — rr— r-ro + 32. a;'-2a; + 48 ' 2a; + 12 2a;-16 5 3a; + 5 _ 8 + 3a; l-9a;^ + 3a;-l~l + 3a;* 119. The formula. An equation often contains more than one general number. In that case it may be solved for the value of any one of these general numbers. It is clear that in such cases the value found for one of the general numbers may be an expression containing the others, and hence, the solu- LINEAR EQUATIONS— ONE UNKNOWN NUMBER 109 tion may not be a single- valued or definite number. Such an equation is sometimes called a formula.* Thus, ax + b=ac may be solved for a, x, 6, or c ; but the value thus found for either will consist of an expression containing the others. If the formula be linear with respect to a certain general number, it may be solved for that number by the method of the preceding section. Example 1. Solve ax-\-b=ac tor x. 1. Adding — &, ax=ac—h. (Ax. 1) 2. Dividing by a, x=^^:^^' (Ax. 4) Example 2. Solve ax-]rh=aG for &. 1. Adding — aa?, b=ac—ax. (Ax. 1) Examples. Solve ax + h=ac fore. 1. Adding —ac—ax—b, ~ac=—b—ax. (Ax. 1) 2. Dividing by —a, c= ~ ~ (Ax. 4) Or, changing signs, c= (§102) Example 4. Solve ax-\-b=ac for a. 1. Adding —ac—b, ax—ac~—b. (Ax. 1) 2. Uniting terms, a{x—c) = —b. 3. Dividing by x—c^ a= (Ax. 4) X — c Or, changing signs, a=— — -• (§ 102) Note. — The student will find it of great value to be able to solve a formula in taking up the study of Geometry and Physics. * A formula expresses a law in mathematical symbols. The type forms of multiplication or division are really formulas. When a formula is expressed in words it is a principle. When expressed as a direction how to do a thing it is a rule. Thus, the formula of § 110 was expressed as a rule. 170 ALGEBRA EXERCISE 55. 1. ^ax—h=cy. Solve for x. 2. xy-Vx=y^~y—^. Solve for x. 3. a{x — l)-{^a=x. Solve for a. 4. «(.T + 3) + 5(£c-3) = c(£c-l). Solve for h. ^6. {a— «;)(« + 2£c) =a^-\- x^. Solve for a. 6. (« + 5a^) {h + a^c) - «5(a^^ + 1) = «' + ^'. Solve for x. 7. 2(2^-l) + 3 = «(^ + 2). Solve for t. 8. 3(^ + « + a;) + 2(^ + a— £K)=-a7. Solve for if. 9. Solve Ex. 1 for 2/. 12. Solve Ex. 4 for c. 10. Solve Ex. 3 for x. 13. Solve Ex. 7 for a. 11. Solve :Ex. 4 for a;. 14. Solve Ex. 8 for a. The following formulae express important laws in Physics. lb, s=vt. Solve for?;. l^. E=\Mv\ Solve for J/: 16. v = at. Solve for t. J/y2 , ,, , 20. i^=— -. Solve for r. 17. s=\at\ Solve for a. ^' 18. TT^i^s. Solve for i^. 21. 6^=i«(2^-l). Solve for ?;. 22. PD= Tr-7>'. Solve for P. (7 i? ic 23.---. Solve for 7?. 26. 7>=-^i-,. Solve for ^^j. c r V!) — v") 24. ^,=^' Solve for P. 27. J^-|-(7+32. Solve for C, 25. (7=:f. Solve for P. 28.4=-+^- Solve tor 7?. 11 Br r' 120. Problems solved by use of linear equations with one un- known. ' In § 26 we showed how problems could be solved by the use of the equation. The important steps in the process of solving LINEAR EQUATIONS— ONE UNKNOWN NUMBER 171 such problems by use of the linear equation with one unknown number are as follows : (1) Fvrst represent by some, letter one of the iinknovm numbers mentioned in the prohle'm. {2) Then, from conditions stated in the problem, form ex- pressions containing this assumed letter xnhich icill represent the values of any other luiknown numbers mentioned in the problem. {3) By means of some other statement in the problem it should then be possible to form an equation betioeen these expressions. {4) Solue this equation and interpret the result. Example 1. The difference between the squares of two con- secutive whole numbers is 121. Find the numbers. Let x= the less number. Then, it' + l= the greater number, for the numbers are con- secutive. Therefore, x^ and {x-\- Vf will represent the squares of the two numbers. Hence, (^ + 1)^— x^=121, for the difference between their squares is 121. Removing the signs of grouping, x'^-\-2x-\-l—x^=121. Whence, 2.r=120. Therefore, a?=60, the less number, and 07 4-1=61, the greater number. Check. (61)^-(60)2=3721-3600=121. Example 2. The length of a room exceeds its breadth by 4 feet; and if each had been increased by 4 feet, the area would have been increased by 128 square feet. Find the dimensions of the room. Let x= number of feet in the breadth. Then,. a? + 4= number of feet in the length. Hence, x{x+^)= number of square feet in the area. Rule for area of a rectangle. 172 ALGEBRA If the breadth and length were each increased by 4 feet, they would be a:+4 and x+8, respectively. And the area would be (x + 4)(a? + 8). From the condition of the problem, (x -{-4){x + 8) —x{x + 4) = 128. (State the condition that gives this equation.) Removing ( ), x' + 12x + d2-x'-4.x=128. Whence, 8a;=96. a?=12, the breadth. x+4=16, the length. Example 3. The sum of two numbers is 21, and the quotient of the less divided by the greater is f . What are the numbers ? Let x= the less number. Then 21— £c= the greater number. (Why?) Hence, since their quotient is |, we have x 2 21-i»~5' Clearing of fractions, 5x=^2—2x. (What multiplier is used ?) Whence, 7x=42, and, • x=Q, the less number. 21—07=15, the greater number. Example 4. A tank can be filled by one pipe in 24 minutes, by a second pipe in 32 minutes, and by a third pipe in 40 minutes. If all three pipes run at once, how long will it take to fill the tank ? Since the first pipe alone could fill the tank in 24 minutes, in one minute it could fill ^^ of it. Likewise, in one minute the second pipe alone could fill ^\ of it; and the third pipe alone could fill jV of it. Hence, in one minute the three together could fiH^V+sV + ^V of it. Let X = number of minutes required for the three pipes to- gether to fill the tank. Then in one minute they could fill - of it. X LINEAR EQUATIONS— ONE UNKNOWN NUMBER 173 Clearing of fractions, 20x+ 15a? + 12ic=480. (What multiplier was used ?) Uniting terms, 47a? =480. a[;=10{f, number of minutes. EXERCISE 56. 1. A has 170, and B has 110. How much must A give to B in order that he may then have just three times as much as B? Suggestion. — If x representF- the required amount given by A, what will each then have? What condition of the problem, then, gives an equation ? 2. Divide 50 into two parts whose difference is 26. 3. Find two numbers whose sum is 1 and whose difference is 15. 4. Find two numbers whose difference is 15, and whose sum is f of their difference. Suggestion. — If x=: the smaller, what must equal the larger? What equation follows ? 5. Find the number the sum of whose half, third part, and fourth part is 26. 6. Find two numbers whose* sum is 36, one of which is | of the other. 7. Find the number such that \ of it shall exceed i of it by 2. 8. Find two numbers whose sum is 28, and such that one exceeds 7 times the other by 4. Suggestion. — What two conditions in the problem ? If one number is X and their sum 28, what must the other number be? What equa- tion follows from tlie second condition ? 174 ALGEBRA 9. Find two numbers whose sum is Gl and difference 11. Suggestion, — When x= one number, either of two conditions will give the other. What are the conditions? Show that you may use either condition to get the expression for the second number and the other condition to get an equation. 10. What number increased by i of itself and 20 is 2 more than double itself? 11. Eight times the difference between one-fourth and one- third of a number is 32 less than the number. What is the number ? 12. Find the number that exceeds 20 by as much as i of the number exceeds 7. 13. Find two consecutive Avhole numbers whose sum ex- ceeds 25 by as much as 25 exceeds 15. 14. There is a certain fraction whose value is J ; and if its numerator Avere greater by 2, and its denominator less by 2, its value Avould be i. What is the fraction ? 15. What number added to the numerator and to the denominator of j\ will give a fraction equal to | ? 16. John is six years older than James ; and in five years John will be 3 times as old as James was 3 years ago. What are their ages ? 17. A father's age now is 4 times as great as that of his son ; and 4 years ago it was 6 times as great. What are their ages ? » 18. A horse sold for $132.50, which was 6 % of the cost more than the cost to the original owner. What did it cost ? Suggestion. — 6^ means i^--^. If ic = the cost, then -^^ a? = the gain and |^^a?= the selling price. 19. A man invests i of his capital at 5% and the remainder at 6%. Ilis total income is $4080, What is his capital ? LINEAR EQUATIONS— ONE UNKNOWN NUMBER 175 ' 20. 16 is changed into 51 coins. If each coin is either a quarter or a dime, how many of each are there ? • 21. A train leaves a station and travels at the rate of 40 miles an hour. Two hours later a second train leaves the station, and travels in the same direction at the rate of 55 miles an hour. Where will the second train pass the first ? • 22. A tank is fitted with two pipes. One can empty the tank in 30 minutes; the other, in 25 minutes. If the tank is two- thirds full, and both pipes are opened, in what time will it be emptied ? < 23. A laborer was hired for 60 days. Each day that he worked he was to receive 12.25 and board ; and each day that he was idle he was to receive no pay, but was to be charged 60 cents for his board. At the end of 60 days he received 1106.50. How many days did he work? , 24. A rectangular field is 6 rods longer than it is wide ; and if the length and breadth were each 4 rods more, the area would be 120 square rods more than it is. Find the dimen- sions of the field. 25. What number mvist be subtracted from each of the four numbers, 12, 14, 18 and 10, so that the product of the first two remainders shall equal the product of the last two ? ^^ 26. A man rows down a stream at the rate of 5 miles an hour, and returns at the rate of 2 miles an hour. He returns to his starting point in 7 hours. At Avhat rate does the stream flow? How far down the stream does he go? How fast can he row in still water ? 27. Find a number such that i of it shall exceed f of it by 9. 28. B has $40 more than A, C has i as much as B, and be- tween them they have $360. How much has each ? 29. The difterence between the squares of two consecutive whole numbers is 23. What are the numbers ? 176 ALGEBRA 30. If a certain number be added to 8 and to 11, and the first sum be divided by the second sum, the quotient will be |. What is the number ? 31. What sum at 6% simple interest will amount to 1413 in 3 years ? 32. At what rate simple interest will $265 amount to $807.40 in two years ? 33. If linen costs i as much as silk, and I spend $19.25 in buying 10 yards of silk and 15 yards of linen, find the cost of each per yard. 34. A room is 2 feet longer than it is wide, and if its length were increased by 4 feet and width diminished by 3 feet, its area would not be changed. What are its dimensions ? 35. Two pedestrians started at the same time from points 44| miles apart, one traveling at the rate of 2i miles an hour and the other at the rate of 2| miles an hour. When and where did they meet ? / 36. Find the time between 4 and 5 o'clock when the hands of a clock are together. Suggestion. — Let x represent the number of minute spaces which the minute hand lias traveled from 4 o'clock on until it overtook the hour hand. Then j^ will be the number of spaces which tlie hour hand has traveled meanwhile. The difference is 20. Why? 37. Find the time between 4 and 5 o'clock when the hands of a clock are directly opposite each other. 38. John could remove the snow from a walk in 30 minutes. James could do it in 20 minutes. John began the work, but later James took his place, and the snow was all removed in 25 minutes from the beginning. How long did John work ? 39. A could dig a trench in 15 days, and B could dig it in 2( i' LINEAR EQUATIONS— ONE UNKNOWN NUMBER 177 days. If they worked together, how long would be required to dig it? ^ « 40. A can do a piece of woi^t in ten days ; but after he has worked two days, B comes to help him, and together they finish it in three days. In how many days could B alone have done the work ? 41. A solved 90% of the problems in an exercise, and B solved I as many as A. If B had solved 6 more, he would have solved 70% of all the problems in the exercise. How many problems in the exercise ? 42. A man made two investments amounting together to $6250. On the first he gained 6%, and on the last he lost 3%. His net gain was $150. What was the amount of each investment ? 43. If 12 lbs. of iron weigh 11 lbs. in water, and 20 lbs. of lead weigh 19 lbs. in water, find the amounts of iron and lead in a mass which weighs 72 lbs. in air and 68 lbs. in water. 44. In an alloy of gold and silver Aveighing 60 oz., there is 5 oz. of gold. How much silver must be added in order that 10 oz. of the new alloy shall contain but \^ oz. of gold ? • 45. There is a number of three digits, each less by two than the one to its right. If the order of the digits is reversed, a new number is obtained whose value exceeds that of the given number by 83 times the sum of its digits. Find the number. Note. — In algebra two general numbers written side by side, as ah, in- dicate the product of the factors a and b. In the decimal system of writing definite numbers two numbets so written do not represent a product. By the place value principle of the decimal notation the value represented by a figure not only depends upon its shape but upon the place it occupies. Thus, 37 means 3tens + 7ones, or 3x10+7. If tens' digit is represented by x and ones' digit by ?/? th^ value of the number which they represent is lOx+y, CHAPTER XIII. LINEAR EQUATIONS-MORE THAN ONE UNKNOWN NUMBER. SYSTEMS. 121. Indeterminate equations. An equation which contains two or more general numbers, or unknowns, will be satisfied by an indefinitely great number of sets of values of these general numbers. Such an equation is called an indeterminate equation. Consider the equation x + y=10. Solving it for a?, we have x=10—y. Now if different values are assigned to y, as many corresponding values are obtained for x. Thus, when 1/= 0, a?= 10 ; when y=l^ x=Q\ when y=2, x=S; when i/= — 4, x=14; when 2/=12, x=—2; etc. Since the number of values we may thus assign to y is indefinitely great, the number of sets of values of x and y which satisfy the equation will be indefinitely great. 122. Solutions. In an equation containing two or more un- known numbers, i. e., an indeterminate equation, the sets of values of the unknown numbers which satisfy the equation are called solutions of the equation. Thus, one solution of y—^x=2 isa?=l, y=5; because for these values of x and y the equation becomes 5—3=2, an identity. 123. Common solutions of two linear equations with two un- knowns. Two linear equations which involve the same two unknown numbers will, in general, have one, and only one, solution common — i. e., one set of values of the unknown num- bers which will satisfy both equations. 178 MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 179 This may be illustrated by the following example. Consider the equations x + y=G and x—y=2. Some of the solutions are : for 0^ + 2/^6, ^^g^, ^^,^, ^^4^, ^^3^, ^^g[, ^^^ x=6 ) ic= — 7 ) £c= 8 I fnr^ «-2 ^= ) a?- 1 ) x=2 \ x=3 I jr = 4 | ic=5 it-6 ) ir=7 1 x=8 I It is seen that of all of the solutions here calculated there is only one common to both equations, a?=4, y=2. That this principle is true in general will be shown in Example 3, § 127. Two or more linear equations with two unknowns may have (1) all of their solutions co7nmon ; (2) just one soliftton com- mon ; or (3) no solution common. Two such equations having all solutions common are called equivalent equations. One may be derived from the other by the use of axioms. Thus, 2x—y=3 and 4:X=Q + 2y have all solutions common, and hence are equivalent. The second may be derived from the first by multiplying both members of the first by 2, then addmg 2y to both members of this equation. Two linear equations having no solution common are called inconsistent equations. Thus, a? + 22/ =8 and3a?+6«/=5 are inconsistent equations. No solution of either equation will satisfy the other. Two linear equations having just one solution common are called independent equations.^ * Two or more equations which liave common solutions are some- times called simultaneous equations. i 180 ALGEBRA 124. Systems of linear equations with two unknowns. A sys- tem of linear equations with two unknowns is a group of two or more equations which contain these unknowns. Thus, Qx + y=5 and x—5y=4: constitute a system. A solution of a system, such as defined above, consists of a sohition common to all of the equations in the system. Thus, a solution of the system < , ^ . is x—4:, ?/=0. Since a system of two independent linear equations with two unknown numbers has one solution, such a system is called a consistent or determinate system. Three or more linear equations which contain the same two unknown numbers have, in general, no common solution. A system of such equations is called an impossible system. i2x+y=l{), (1) Thus, in the system < 3a? — y= 5, (2) the only solution common ( x + y= 2, (3) to (1) and (2) is x=^^ y—^\ and this solution will not satisfj^ (3). 125. Equivalent systems. Two systems of equations which have the same solutions are called equivalent systems. Since two equivalent equations have all of their solutions common, it follows that two systems are equivalent if the equations of one system are equivalent to the equations of the other system. In general, two systems of linear equations will be equivalent if the equations of one are derived from the equations of the other. For example, the only solution of the system ■! U J^g^^s' is a?=2, 2/ =3. Adding the corresponding members of (1) and (2), we get a new equation (3) 3a? +32/= 15. And subtracting the members of (2) from the corresponding members of (1), we get another new equation (4) x—y— — l. Equations (3) and (4) form a new system \ ^Z^,f^},\' which is ( X y — 1, MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 18l equivalent to the old system, because its only solution is also It follows that a system of equations may be solved by solving an equivalent system. 126. Elimination. A system of two linear equations with two unknowns is solved by a process called elimination. This process consists of combining the two equations of the system so as to obtain a new equation which contains but one unknown number. Tliere are three principal methods of elimination : (1) by addi'ion or subtraction, ('2) by comparison, (3) by substitution. We now proceed to discuss these three methods of elimina- tion in solving systems of linear equations with two unknowns. 127. Elimination by addition or subtraction. Example 1. Solve the system j l^^^^^^l' ^11 Let us first eliminate y. Multiplying (2) by 4, 12x-{-4y = 36. (3) Subtracting (1) from (3), 7x=U. Hence, x=2. The value of y may now be found in like manner by eliminating X between equations (1) and (2). Or, replacing x by its value 2 in equation (1), we have 10 + 42/=22. Hence, 2/=3. Let the student see if the solution x=2, y=3 satisfies both equations of the given system. Notice that the system (1) and (2) was solved by solving the system (1) and (3). See §125. This example illustrates elimination by subtraction. Example 2. Solve the system ] 4x + 3w= — 3 (2) We may first eliminate either x or y. Let us eliminate y. 182 ALGEBRA Multiplying (1) by 3, 21x-6?/=93. (3) Multiplying (2) by 2, 8x + 6y=-Q. (4) Adding (3) and (4), 29^=87. Hence, • x=^. The value of y may now be found by eliminating x between equations (1) and (2). Or, replacing x by its value 3 in equation (2), we have 12 + 3y=-S. Hence, y= — 5. The system (1) and (2) was solved by solving the equivalent system (3) and (4). This example illustrates elimination by addition. The method used in the above examples would apply to any system of linear equations. Hence we have the following rule for elimination by addition or subtraction : 3fulti2:)ly the memhers of each equation, if necessary, by such a iiumher as loill make the absolute value of the numerical co- efficients of one of the unknown numbers the same in both of the resulting equations. Add or subtract the corresponding members of the resulting equations. Note. — In any method of elimination we are concerned witli only the common solutions of tlie equations. Hence, in the addition or sub- traction, which is involved in tliis first method of elimination, it is as- sumed that the unknown numbers have the same values in each equation of the S3^stem. Eor example, x stands for the same value in each equation. Otherwise the addition or subtraction would not be allowable. The same facts apply to every method of elimination. Example 3. Show that the system of general equations in X and 7/, ] ^^i^^ZS' has, in general, one and but one solution. Given \ ax + hy=c, (1) •. ^'""^^ \dx + ey=f. .(2) Multiplying (1) by d^adx + bdy=cd. (3) Multiplying (2) by a, adx + aey=af. (4) Subtracting (4) from (3), bdy—aey=cd—af. (5) MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 183 Now (5) is a linear equation in the one unknown number y. We learned in Chapter XII that, in general, such an equation has one, and only one, solution. Solving (5) we get y=M=^e Similarly, by eliminating ?/, we get aex—bdx=ce—bf. (6) /^g Tf-f Solving (6) we get ^=^^3^^- Hence, in general, the system has one and but one solution : ^_ ce-bf cd-af ae—bd' bd—ae ^ Note. — It will be found that if certain relations exist between the general co efficients, a, b, c, d, e, and/, this conclusion fails ; i. e., the equations are equivalent or inconsistent. EXERCISE 57. Eliminate by addition or subtraction, and solve the following systems of equations : l-2a3-f2/=6. ''• \'lxVy=\. "■ l8«=5^-ll. (2£c + 5?/=15, ^ (4a-36 = l, « ( 8a=a; + 34, (3a;-4y-ll. ^' (3a-4^ = 6. ^' (6« + 8a;=53. 10. j7y^^ = 42, (3y-^-8 = 0. .. (5a-2^-35 = 0, /^^' t^> + 4a-25 = 0. 12 n«^+iy = 30, -, (1.5a^-3.7.y = 5.4, 184 ALGEBRA J. (3^-2^ + 4-0, .Q (3^ + 2y=4, ^^- t4y-3a; + l=0. 16 f2a3 + y==35, ^^' |5a;-3y = 27. on f -4a; + 3y = 45, ' ^^- \ 2y + 6^- 4. .^ (5y-5a^ = 15, ^'- |3£c4-5//-71. 21.^ 1 + ^-35, ^i^4 ^' 18. ^ ^ ^ 4 + 6-^3- [| + .==45. 22 n0a^ = 2 + 2y, Solve for x and i/ : 24 ( aa; + ^>y = a^ + ^^ * Xbx + ai/ = 2ab. 25. I ax + ai/ = a^ -{- b. 128. Elimination by comparison. EXAMPLE]. Solve |t^:^=?^6: Let us first eliminate x. (1) (3) Solving (1) for x, ^-34-2/ ^ 4 • (3) Solving (2) for x, a?=16-42/. (4)- Hence, comparing the two values of x given in (3) and (4), by Axiom 7, 3^;?/-16 42/. (5) Solving (5) for y, • 2/-3. Replacing y in (2) by 2, 8 + a?=16, whence, x=8. (6) The above example reveals the following rule for elimina- tion by comparison : jSolve each of the two equations for the value of one of the unknown niimbers^in terms of tJte other ^ and equate the resulting valim. MORE THAN ONE UNKNOWN NUMBER. SYSTEMS. 185 Solve for the unknown that gives the simplest expression from each of the equations. EXERCISE 58. Eliminate l)y comparison, and solve the following systems of equations. ^' \x-y = ^. L 3 4 3£c-y = l, r3m-4/2=10, 2^ + 5y^41. 13. Y'\^'\^n = \l, 1 (^ + 2y-4, (22/-£c + 12 \^x + y = ll. ^ (3.75.^ + 2.5y = 10.25, 22/-£c + 12 = 0. .- (£c + 2//-3 = 0, 2a-^> = 5, { 15. 1 I 7a + ^> = 265. 16. 2^3 '' + 26-25. l^,y 3"^2 jj j 10.T + 3« = 174, °- 1 3a3+10«-125. 1^ (14a+66=0, , . Lo 1 1 6^-46 = 46. g (5a + 2y-l, ^ ( 13« + 8y = ll. Solve for x and y : 10. J \2x—my=n. L7^2 + ^^^- in C«a^+y==^, 20. C ax + y {bx + y- = 30, ^' ] 3a-2^=25. 5 (3.T + 22/ = 26, „ (3a-7a;-40 = 0, '*• |4a-3.T=9. g (7a;-9y = 13, ^' |5a; + 2y = 10. 3 (4a.— 5y = 26, ^' |3.T-6y-15. 7 U^+iy=i3, - (3m + 7;?. = 16, *; |2m + 5w = 13. g ( 7a^ + 4y = l, ^' t9^ + 4y = 3. 9. (3ic- (19a; 11^ = 0, -19y = 8. 10. 11. MORE THA.N ONE UNKNOWN NUMBER. SYSTEMS 187 ^^- t-4«c + 6y=10. .« (55a;-15y = 270, ^^' (3l£c+19y = 262. 3 2 .^ j22a;-5y = 213, 2a-7_ 13-.y ^'* |6£c-22/ = 51. j3 (4.. + 3y=22, 19. i^ + y^ff. .- (5aj+ll2/=102, „ (3^ + 2.y-42, ^*- \x-Si/ + lQ = 0. '*"• |13^ + 23y = 225. EXERCISE 60. In solving the following systems choose the method of elimi- nation that seems to you best adapted to each particular system. (x+i/ = 10, ^ (8a;+6y=10, (ar-y = 4. **• |5£c + 2y = l. (8a; + i/ = 60, n (5a3-2y/-63, ^- |7a3-10y/:-9. ^* t2aj + y:-18. 3 (7^+y = 42, 10. -[r^^^^n' ., (i8.-20y^i, 11. {1:^=11^ (1^/— 1^ = 2 l5«-2y = 14. f7^-3y = 26, „ (21ffl + 86=-66, '■''■ l2x-2y = ^. "• t.49a-15S=-53. { 7x-4.i/ = 12, 14. -^ 4 2~^' Sx-bi/=0. ( 2x-2i/ = W. 188 ALGEBRA. 16. J -g-+y=18. Solve for x and y : ^g f«a5 + 6y = l, |&a?4-ay=l. jQ S hx-ay = h\ \ax—hy = a^. 20. i«f-%=«'-*', \^cry—ox=\j. 16 j2y + 79=5.x, 17. 22. 2^ {2px + ^y = 4:p' + q\ \x~'Zy = 1p'-q. [^3"^4' 6. 130. Systems of fractional equations. Certain systems of fractional equations may be solved by the methods of this chapter. Such equations should not be cleared of fractions, for the resulting equations would not be linear. Also clearing of fractions would hitroduce new solutions, ?*.e., would give solu- tions which would not satisfy the original system. ExAiMPLE 1. Solve Multiplying (1) by 2, 18_8 a h '2fi Subtracting (3) from (2), -^=6. Multiplying by 5, 26 = 66, whence. =4. (1) (2) (3) 81 36 18. Multiplying (1) by 9, Multiplying (2) by 2, f + f =30. (4) (5) Adding (4) and (5), 117 =38, MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 189 Ml] iltiplying by a, 117= 38a, whence, 117 ^-38- Example 2. Solve 3x + 22/~^' ^-?-3 (1) (2)' 9 Multiplying (1) by ^ 3 27 45 2x^Sy~ 4' (3) Subtracting (2) from 27 2 45 Multiplying by 242/, whence. Similarly, 81 + 16 = 270?/- 722/, EXERCISE 61. Solve : '• i i J-i -« y 4. ^ « + 6-B' ^ 15 4_ 7. ^ r 1 , 7 _5 4^ + %"8' 1 3 5 ^'Ix y'^28^ = 0. 1 3^6 1 6. J •'■ y Ix y 8. . 57. + 2Z."^' * 3. - ^x y 6. - X y ' Lk y 9. . fl-^ =4 10. < f-l 4- ^--1 l^x'V+y ' Ll + ^ l+y .2- 11. . ^ 4 X- 1 3 9 -1-8- 131. Systems involving three or more unknowns. It is easily shown by elimination, as in Example 3, § 127, that in general 190 ALGEBRA three linear equations containing the same three unknowns, or four linear equations containing the same four unknowns, etc., have one solution common, i. 6., one set of values for all of the unknowns which will satisfy all the equations at once. Hence, in general, a system of linear equations, in which there are just as many unknowns as there are equations, will be consistent or determinate. If there are more unknowns than equations in the system, in general the number of solutions of the system is indefinitely great. Hence, such a system is called an indeterminate system. C rjf _i_ fj t 2! -—- R Thus, ) 2x—y + z=^ ^^ indeterminate. Some of its solutions are ^=1, y=2^ z=3; x=3, y=S^ z=0; a?= — 1, y=l, z=6, etc. And if there are more equations than unknowns in the system, in general no solution common to all of the equations will exist. Hence, the system is an impossible system. To solve a system of three equations containing three un- known numbers, such as x, y, and z, we may (1) eliminate any one of the numbers from any two of the equations ; then (2) eliminate the same number from one of these equations and the equation not used. This will give rise to two new equa- tions which contain only two unknown numbers. These two equations may then be solved as a new system by the methods of the preceding sections. From a system of four equations with four unknown numbers we can, in like manner, derive a new system of three equations with three unknown numbers ; and so on. 1 x + 2y + 2z=ll, (1) Example 1. Solve ■<2x + y + z=7, (2) { 3x + 4y + z=U. (3) By subtraction eliminate x between (1) and (2). We get Sy + Sz=15. , (4) By subtraction eliminate x between (1) and (3). MORE THAN ONE UNKNOWN NUMBER. SYSTEMS 191 We get 2y + 5z=19. (5) Now by subtraction eliminate y between (4) and (5). We get 9^=27, whence, z=3. Substituting this value of z in equation (5), we get 2i/ + 15=19, whence, 2/=^. Now substituting the values of both y and z in equation (1), we get x + 4 + 6 = 11, whence, x=l. EXERCISE 62. ^7 Solve the following systems : (x+y + z=9, 1. ■< 2ic+y— 2=0, ( 3ic— y + s=5; ( x—2y + z=Q, 2. ■}x + Sy + 2z = U, (2x-y + z=lS. 3. } Sx + 4y + 6z^7, (ic + 2y + 62=4. (bx + 6y + 7z = S, 4. •} 10a;-i2y + 2l2;=3, ( 15.93— 6y + 14^ = 4. ( Sx + y-z=2, 6. -\x-2y-Sz = Q, (y + z+l = 0. (x + y = l, 6. ]y+z=9, (a; + s=— 6. C2x + y-z = 7, 7. }y-x=l, (z-y = l. ^ + ^ + ^=19 8. ^ ^_l-+^ = 6 10 10^6 ^' L4 + 5 r5""^- 9. ^ 10;« + -| + s = 7, a;.+y-2s = 16, X y (2p-2^ + 3r = 10, 10. \ ^} + q-r=b, ' (p-q + 2r=7. 11. 12. x+y+z + w = 10, X — y — z-\-w = 0, 2ic + 2/4-3^—10 = 9, Sx — y-\-z—2w= —4. ^2/9— + 5'— r— s=— 1, jL>— (Z + r— s=— 5, ^Sp + 2q~r + 2s=-17. 192 ALGEBRA 132. Geometric picture or graph of the equation. We have seen that an indeterminate equation has an indefinitely large number of solutions. The relation between these solutions, which is expressed by the equation, may be more vividly re- presented by means of the graph of the equation, discussed in the following sections. 133. Coordinates. The position of a city on the earth's surface is determined by its longitude and latitude ; i. e., one can locate the position of the Y city, if its longitude and lati- tude are both known. Now by the method of § 33, longitude east may be called negative , longitude, and longitude west may be called positive longi- tude. Also' latitude south may be called negative latitude, and ^ latitude north called positive ^^' • latitude. Thus, we speak of a certain city as being —120° longitude and +30° latitude. In like manner, the exact position of a point P (see Fig. 1), in the plane of this j^age may be determined, if we know its distances and directions from two straight lines JCX and YY\ drawn at right angles to each other and meeting at O. We shall represent the distance from P to the line YV i. e., iVP, by x, called the abscissa of the point P/ and the distance from P to the line A'X^, i. e., MP, we shall represent by y, called the ordinate of the point P. The abscissa x and ordinate y are called the coordinates of the point, and the lines XJT and YY are called the axes of coordinates. A'X' is the jr-axis and YY is the /-axis. The point of intersection is called the origin of coordinates. M MORE THAN ONE UNKNOWN NUMBER. SYSTEMS I93 The student will see that the axes XX' and YY' correspond to the equator and the prime meridian on the earth's surface, and that the coordinates x aiid y correspond to the longitude and latitude, respectively, of a place on the earth's surface. If an abscissa drawn to the rig/U of the y-axis ( YY') is called positive, and one drawn to the left is called negative ; and if an ordinate drawn above the £c-axis {XX') is called positive, and one drawn helow it is called negative ; then the exact position of the point is known when its coordinates are known. Thus, to locate a point P (See Fig. 2) whose abscissa is + 1 and ordinate +2, measure 1 unit to the right of O, then 2 units up from XX' . This point is usually re- presented by the symbol (1, 2), or P(l, 2). Similarly to locate the point Q{—\, 1), measure \ unit to the left of O and 1 unit up from XX'. To locate P(— 1, —2), measure 1 unit to the left of O, then 2 units down from XX' . To locate >S(1, —1), measure 1 unit to the right of O, then 1 unit down from XX'. Let the student draw a figure and locate the following points : (—1, 2); (3, -2); (1,3); (-2, -3). Q(-M H Rf-A-^J y' a^j X' s(''-o Y Fig. 2. 134. The graph. The graph, or geometric picture of an equation, is the line upon which are situated all of those points whose co- ordinates, represented by x and y, satisfy the equation. Consider the equation x—y=2. By assigning values to y and computing the corresponding values of x, we find a few solutions as follows : x=2 I x=S I x=4t I x=^ \ x=Q } y=0\ y=l\ y=2\ y = 3\ y=4.\ x= 1) x= 0) x*=-l ) x=—2) y=.-l\ 2/=-2i y=^-d\ V = -^V 13 194: ALGEBRA Locating the points whose coordinates are these sohitions, we get a series of points as in Fig. 3. It is seen that these points are not scattered at random over the page, but all appear to lie upon one straight line, MN. Hence, the straight line MNis the geo- metric picture or graph of the equation x—y=2. r N / A ^v^ J ^.»/ ^ \, (-X2) f^.^f \ /jj) X ^ ii. /C"/ X' ]^ ^ /-n ^ ^ (^.-J/ M -0 I / M / A V It can be shown, by the aid of geometry, that the graph of every linear equation with two unknown numbers is a straight line. We shall assume this principle in our work. Since a straight line can be drawn when two points on it are known, to draw the graph of a linear equation, we need to find only two solutions, and locate the two points whose coordinates are these solutions. Then draw a straight line through these two points. Thus, to draw the graph of x + 2y=l, we find two solu- MORE THAN ONE UNKNOWN DUMBER. SYSTEMS 195 tions to be a?=4, 2/=— |, and x= — S, y=2. Locating the cor- responding points (4, — I) and (—3, 2), and joining them by a straight line, we have the graph PQ^ Fig. 3. Any other point whose coordinates consist of a solution of this equation will be found to lie upon the line PQ. If, upon the graph P §, we locate the point A, and measure its coordinates, the coordinates Avill be found to be x= — b, y='^. These coordinates satisfy the equation a? + 2y = l. Like- wise, the coordinates of a second point B of the graph P Q are seen to be 93 = 6, 2/ =— 21. These coordinates also satisfy the equation x + '2i/ = l. Now in like manner it will be found that the coordinates of any point whatever of the line P Q will satisfy the equation x^-2y=l of which P§ is the graph. It thus appears that not only do all of the points whose co- ordhiates satisfy a linear equation lie upon a straight line^ called the graph of the equation ; hut also the coordinates of all points y^hich lie upon the graph of the equation satisfy the equation. Note. This principle will be found true of the graph of an equation of any degree. Accordingly, the graph of an equation is sometimes called the locus of all points whose coordinates satisfy the equation. In analytical geometry this principle is a fundamental notion. 135. The graph of a consistent system of equations. The solutions of each of the two equations of a system in two unknown numbers are represented by the coordinates of the points which are situated upon their graphs. Hence their common solution, the solution of the system, is represented by the coordinates of a point which is on both graphs ; i. c, the point where they intersect. Consider the system \ J^T^—a ( 2^ 3^ s /^y^ 10. 1*16327 20. i4. 2//]/ 1?~- SURDS AND IMAGINARY NUMBERS 215 28-Vi7^. 2y. |/(«_ic)^ 30. 31. iA~-'/. |/ 4«£c^^ 4 Vlax'y + 12cea;y=^ + 4ay\ 32. 2i/ 1 ,. '«• (/j+i- 35 ■Vl-l^^. 142. Addition and subtraction of surds. Only surds which, when reduced to the simplest form, have the same surd factor can be added or subtracted. Otherwise the addition or sub- traction can only be indicated. Just as 3<:< + 2a==5«, so Z\/x-\-^^/x-~b\/x. Hence, surds vnth a common surd factor are added or sub- tracted by adding or subtracting the coefficients of the surd fac- tors and affixing to the residt the common surd factor. Example 1. Find the value of i/32 + 1/50. 1/32 + 1/50=41/2 + 5.,l/2=(4 + 5) l/2=9l/2. Example 2. Simplify Va'bc-^Vab''c-\-Vahc'. Va?bc — vab^c + V abc'^ = a^ V ahc — V V ahc + c^ yabc = (a''-b'-}-c')yabG. EXERCISE 66. Simplify the following : 1. 1/T8-V8 + 1/32: 5. 2v 3 + 1/81-V3. 2. 1/75-1/3-1/12: 6. i/40-i>320-2i/5. 3. i/20 + 1/45-i/M: 7. 4i>}-y}. 4. 1/6 -v"294 + 1/486 + 1/24: 8. 2i/^ + 3i/'^-l/^ 7;s 216 ALGEBRA 9. yT^^-2i/2^' + i/W. 10. i/81a'-4i/192a^ + 3i;648al 12. 2|/V^+3v/aFc— i/VFc. 13. i/^-i/^* + V^. 17. l/F+^l/7S-2l/}- 14. 1 T + 3^|_^3. 18. I'S-V i+vV?. 15. i/27i-i/8S+l/64^^ 19. i /20 + 3i/4-2r4. 16. 3|/48-y^r_2|/|. 20. vM + i/f+i/S". 143. Change of order. A surd of a given order may be changed to an equivalent surd of a different order by the principle, ]' 0""= \ a". That is, the value of a surd is not changed by multiphjing or dimdmg the index of the root and tlie exponent of the poicer by the same number. This principle may be established as follows : Let Then Hence, or Hence, Therefore, Example 1. Express Va^ as a sixth root. Multiplying index and exponent by 2, we get \/a^=y'a^\ mx y a"' = r. a"^=r"^^ §64. y a''^=yr'"-% Axiom 6. a"=r'"- l/a«=l/r"', Axiom 6. = r. 7/6^=1/ «^. Axiom 7. Example 2. Reduce yaWc to a tenth root. SURDS AND IMAGINARY NUMBERS 217 Multiplying index and exponent by 5, we have ya^b*c= y {d'h'cf=Va'Wc\ ElXAMPLE 3. Reduce yx^y^z^ to the lowest order. Dividing index and exponent by 2, their greatest common divi- sor, gives y^x'y^^'=y^3(^y^z. 144. Reduction of surds to the same order. Surds of different orders may be reduced to equivalent surds of the same order by the principle of § 143. Manifestly the simplest common order to which surds may be reduced is the least common multiple of their orders. Example 1. Reduce Va^b, v'a■'&^ Va^U" to the same order. The least common multiple of the indices is 12. Reducing each surd to an equivalent surd of the twelfth order, ya'b=Va^¥; Vd'b^=ya''b^- ya'b'=Va^b'\ The values of surds may be compared by first reducing them to equivalent surds of the same order, and comparing the resulting numbers under the radical signs. Example 2. Compare the values of i/l5 and y Qo. 1/15 = 1/ W=y 3375 ; V 60= ;/ 60^= 1/ 3600. Since 3600 is greater than 3375, y 60 is greater thani/15. 145. Multiplication of surds. The product of surds which are of the same order may be found by the principle established in § 67; that is, y^aiyj^yab'. Surds of different orders whose product is to be found must first be reduced to surds of the same order. 218 ALGEBRA Example 1. Find the product of y'2a^, i 3a^ and F 4a. V2a^=l 8a«; ySa'=V9a*; V 4a=V 16a^ Hence, V2d'V 'Sa^ ■V4a=\/8a'' y 9a* Vl^a^ l/Sa:>-9a*lQa' = V1152a'^ =2aU/lSa\ Example 2. Multiply 21^5 + 5]/2 by 3V5-i\/2. 2VE+5V2 3|/5-4]/2 6-5 + 15l/l0 - 8t/10-20-2 30+ 71/10-40, or 7l/l0-10. 3 - Example 3. Express as an entire surd 3y 4. The coefficient may be expressed in the form of a surd of the same order as the surd factor. We have 3 = V 3-^ = 1/27. Hence, 31^4= \^21^/~4= 1/^27x4= i/^I08. EXERCISE 67. Express as surds of the same order : 1- l/a, \^a\ 2. yx\ yx, V'x\ 3. i/5, 1^2, 1^3. 4. ]>7, //I2; 1/2. 5. l/^\ 1^^, l/V^ Which is the greater : 6. 1/5 or v^lT? 7. V 2210, l'/320, or 1^48? SURDS AND IMAGINARY NUMBERS 219 Express as entire surds : 8. 21^5. 10- V7. 12. i^^. 9. 8i/8. 11. ||/3. 13. «T/X 3^ "• **'. ^y YT- 27a;' ^^- Fl/«- 1^' 2^V 27^ Simplify : 16. i/2v^3v 4. 18. i/61/'4l>'2. 20. i/^T^by/^Tl. 17. 2l^2-3l/5. 19. 1/^51/101/5. 21. i a^i/^^^l 22. i?^«^ + 2a^> + ^2v'^Hl>. 23. (^/2+i/3)(^2-i/3). 24. (l + i/2-i/3)(l + i/2+i/3). 25. (|/e + i/2)(i/6-i/2). 26. (^3 + i^2)(i>'9-i^6 + f'4). 27. (1 + 1/2 + 1/3)^ Simplify by reducing to lower order : 28. y'lM'. 31. jyg^^v". 33 VWy 29. ^2W. ./-. 'V^^^ _____ 32 i/^ 34 ,«/l^^«"^'' 30. i/Sia^'^^v^ * K F ' V ^uy^' 35. i/(^. 36. ^s/z?:. 146. Two binomials which contain surds of the second order and which differ only in the signs of the surd terms are called conjugate surds. Thus 1/2 + 1/5 and i/2— 1/5 are conjugate expressions. Since conjugate surd expressions are of the form a + 6 and a— 6, the product of a pair of conjugate expressions is a rational expression. Thus, (a + y'h) x (a— i /&} =«^— 6, 220 ALGEBRA By grouping terms, polynomials involving surds of the second order may often be expressed as conjugate expressions. Thus, l/2 + V^S+V5 and i/2-i/3+i/5 may be written (l/2+ 1/5) + VS and (V2 + |/5)-i/3. 147. Division of surds. The quotient of one surd by an- other surd of the same order may be found by using the prin- ciple established in § 68 ; that i ;, 1/6 ^ b But the quotient, where the divisor is a surd expression, is reduced to. its simplest form by first multiplying the dividend and divisor hy such an exjyression as will make the divisor rational. This process is called rationalizing the divisor. The mul- tiplier used is called the rationalizing factor. Example 1. Divide 61 2 by v 3. Multiplying both dividend and divisor by \ 3, we have 6v /2_ 6i/2i/3 ^6| /6_3 ^g 1/3 1 3F3 3 Whe7i the divisor is a binomial involving surds of the second order ^ the simplest rationalizing factor is its conjugate. Why? (See § 146.) Example 2. Simplify 1^g+l^^_ . 2i/3-|-3i/2 The conjugate of 2i/3 + 3l/2 is 2|/3-3l/2. Multiplying by this, we have l/2 + i/3^ _ (i/2-n/3)(2i,/3-3i/2)_ __:zi^__ 1 ^/e 2i/3-H3l/2 (2l/3-J-3l/2)(2i/3-3l/2) ~ -6 "^ SURDS AND IMAGINARY NUMBERS 221 When the divisor is a polynomial which involves surds of the second order, group terms so as to form a binomial, and then multiply by its conjugate. This result may, likewise, be ex- pressed as a binomial and multipliedby its conjugate; and so on until the divisor is rationalized. Example 3. Divide 2 by 1/5 + t/2— 1/3. 2 _2 V5 + V2- V 3 ~ (1/ 5 + V 2) - 1/3 2(1/5 + 1/2 + 1/3") (1/5 + 1/2-1/ 3)(l/5 + 1/2 + 1/3) _ 2(l/5 + i/2j-y3) ^j, l/5 + l/2V|/3 4 + 21/10 ' Z+VlO _ (l/5 + 1/2 + l/3)(2- l/lO) (2 + l/l0)(2-l/10) _ 2 1/3 -3 1/2 -1/ 30 -6 EXERCISE 68. Simplify : 1 i 6 ^ 11 l/"T+^^ +^ ' 1/2* • 1/-2 + 1/3' y(l+x')-x 2 V^ 1/3-1/5 1/^+6 + i/^F^ * 1/3* ' 1/3-1/2* i/^T7>-v/a=^ _ l/'2-i/3 „ 2|/7 + 3i/2 ,„ i/^Ta-i/^ «• ■ ::: • O. =: —- !«• 1/2 + 1/3 1/3-1/5 1/3-1/2 2|/7 + 3i/2 3v/2 + 2i/3 3i/5-5i/3 3|/5 + 5i/3 a'' 1/7 _ ■ • 3v/2 + 2i/3 |/ic + 3 + i/a; 4 A. o 3v/5-5i/3 14. ^t^^^ . T/2 ^' 3|/5 + 5v/r 1 + 1/2 + 1/3 ^ 12 a' ,. 1/5-1/3 5. — ^' 10 . 10. — = =: — • v3 *+l/(6^-a^) v/5-1/3-1 222 ALGEBRA 16. 17. y'2 + yS + y7 T/2 18. 19. yx+y Va y- -yz h l/a+|/ b- -yc y2 + y3 + yb 148. Powers of surds. Powers of surds are obtained by the use of the following principle : (,'/iy=l'/V. To establish this principle, we have {l^ciy=\ya-{/a-i/a to r factors = Paaa to r factors § 67. 3> Example 1. Raise 2ay2a^b^ to the fourth power. (2aV2a'b'Y=16a*y {2a'by=16a'y IQa'b"^, 149. Roots of surds. Roots of surds maybe found by the principle, y{\/a)=ya. This law follows from the fact that to take the mth root of the nth root of a is to take one of the m equal factors of one of the 71 equal factors of a, which is one of the mn equal factors of a. Example 1. Find the fifth root of y^a^if. Example 2. Simplify v^{2a''bySa¥), First express 2a^b^/3ab^ as an entire surd. y'{2a'by'3a^== V^{l/T2a^') = y'na'b^ Simplify : 1. (1/2)1 SURDS AND IMAGINARY NUMBERS EXERCISE 69. ^23 2. iY'lx'yy. 3. (5a;'^i/2^')l 4. {-'lah^/^hy. 6. (^Wby. 7. {Aayya^y. 8. (3«yv^^*)l 12. i7(^/T6). 13. i/'(|^625<). 9. {4:\/d'-by. 10. i^(i/5). 11. iV{i>(^2)}. 14. y\i/21xY)' 15. \/{VM). 16. 1/^(8^4^). 17. 1/ /(a^-^>'^)^ 18. ^ ;>8^. 19. i/ 49icV«'|?'8a^ 20. {^{^(f «"')}• 150. Imaginary numbers.* In § 65, even roots of negative numbers were called imaginary nmnhers^ and it was shown that these numbers did not belong to the series of numbers with iMch ice were then acquainted. They constitute a new series of numbers with some pecidiar properties. By means of more advanced mathematics any imaginary num- ber may be expressed in terms of quadratic imaginary numbers, i. e., square roots of negative numbers. Hence, in this chapter all of the language shall have reference only to quadratic imaginary numbers. For the sake of distinction, all numbers heretofore considered are called real numbers. Numbers of the form i — ^, where b is positive, are sometimes called pure imaginary numbers. Expressions of the form a+|/^^ when a is real and y^^b a pure imaginary number, are called complex expressions. * We have seen that the first numbers known were whole numbers ; that by extending the process of division fractions were conceived ; that by extending the process of subtraction negative numbers were conceived ; and tliat by extending the process of evo- lution surds and finally imaginary numbers were conceived. In the investigations of higher mathematics the imaginary number plays an important part. 224 ALGEBRA 151. If a=0, the complex expression a+j/^—b is a pure imaginary number i^ — b; but if b = instead, the expression becomes a real number, a. It thus appears that real numbers and imaginary numbers are both particular kinds of complex expressions. The squares of all real numbers are real positive numbers^ and the squares of all pure imaginary n^imbers are real negative numbers. Thus, ( + 6)2= + 3G; (-4)2= + 16; ( + l/2)2=+2; (-l/2)2= + 1^4. And (l/"^2^-3; (-|/^)2=-5; (-4l/^)2=-32. 152. Typical form of imaginary numbers. By the principle ]/«3=]«]/ 5 any imaginary number can be expressed in the form c]/^^, called the typical form. Thus, i/-4=i/4(-l) = i/4i/-l=2|/-l. And -|/-10=-i/10-(-1) = -i/10t/'^1. Hence, the properties of any imaginary number may be studied by studying the properties of v^^^, usually denoted by *, which is taken as the imaginary unit. Note. — The principle \/ah—\/a\/b does not hold when \/ a and \/b are both imaginary, i. e., when a and h are both negative. Thus, l/^}/— 9 does not equal i/36, or 6, but =i/a6(i/=4)2, or -6. 153. Powers of ] -1 or /. I ^or the s we have i- = y - "1; By § 64, r' = -{V- — 1 \2 = = -1; i^-- = {V~ — ~1V = =i-i'=- -*. i'- = {V~ — 1V = =i-e=- ~f- -+1. i' = = {V~ ~1.\5- =i-i'=i. t^-- = (V~ ITi \fi = =i-i'=i- ^ = ■■i'=- %'-- =(y~ ^1)'^ =i-f>=- -i. i^ = -(!'- 3~1 \H - ^i-i'=- -i'- = 4-1. i^-- -iv~ -1V = = i-i^ = i. SURDS AND IMAGINARY NUMBERS 225 Hence, the successim powers of i have only the four different values: i — i, — i, — \ — i, + 1^ rej^eating in regular order. Since *"^=— 1, and since ( — 1)^ or to any even power, is 1, then e-+^ = i*n.f =-1; Thus, *25^t*«+' = i; ^l2^^='•*=l; iP=i}-'+^=-i. 154. Addition and subtraction of imaginary numbers. Imaginary numbers to be added or subtracted must first be reduced to the tupical fonn^ and then combined as in the addi- tion and subtraction of surds. Example 1 . Simphf y 1/ - 4 + V -9 + i/-25-i/ — 16. = (2 + 3 + 5-4)j = 6^. EXERCISE 70. Reduce to the typical form : 1. V~=m. 5. y-^^. 9. y~=^K 2. V^=ms. 6. i/:=r|f. 10. i/_i6a;y. 3. |/^=^25. 7. 1/^=^7. _ 11 ,/ 9^"^^ 4. 1/-^. 8. |/-98. ' K 49a;y^* 12. |/-81(£c-//)^ 13. i/-a^-2a^-6^ 14. |/24a;y-9aj-'-16yl Simplify : 15. yZIg-^ 114+^/^307 16. 1/-^+]/ =^100"- 1/^=49: 15 226 ALGEBRA 17. i/^:^-|/~4-i/^^6. 18. i/— 8 + ^rrio + |/~7. 19. a|/^=^2+-i/^^ + 3i/^:^. 20. 2xi/—xY-i'Si/y-4x'i/'. 21. Add x + 2i/^/^=ri and 2x~i/i/^^l. 22. Add 10-|/^:^9 and 6 + i/^=^. 23. Add a + b+ ^/-^h and a^-h- V~h. 24. From a + \/^-b take a — \/^^b. 26. From 3aj — 2y\/ — 1 take a; + \/ — 9yl 155. Multiplication of imaginary numbers. Imaginary num- bers to be multiplied should first be reduced to the typical form. Then proceed as in tlie multiplication of surds, ob- taining the product of the factors V^-^\ by use of § 153. Example 1. Multiply v— 4 by ]/ — 9. Example 2. Simplify V -16V —25V —i'^. l/^;^l/^^l/-49=4l/^-5l/^.7l/^ = 140i-^=-140j. ^ Example 3. Multiply i/ — 3 + i/ — 5 by i/— 2— i/— 4. l/~i + V^=iV^+iV5 ; 1/^-1/^=^/2-2*. ^V34-^l/5 ^/2-2i i' 1/6 + 1' j/ 10 -2P }/S-2i' 1/ 5 = - 1/6 - 1/10 + 2i/3 + 2i/5. 156. Two complex expressions which differ only in the signs of the imaginary terms are said to be conjugate. Thus, a? + 7/i/— 1 and x—yi/—l are conjugate expressions. The product of tioo conjugate complex expressions is real. For {x-\-y\/^-V){x-y\/^=\)=x^—y\-V)==x^-Vy\ SURDS AND IMAGINARY NUMBERS 227 15 7. Division of imaginary numbers. Imaginary numbers in division should first be reduced to the typical form. Then, in general, multiply both dividend and divisor by such an expres- sion as will make the divisor real and rational. Example 1. Divide 6 by l/^^. Example 2. Divide ]/ —9 by y —25. 9 31/ -1 1-25 51-1 Dividing both terms by 1/ —1, =^ Example 3. Divide 2 + 1^^ by 3 - V-Tx. 2 + l/^ _ 2 + /1/5 ^ ( 2 + n/5)(3 + i]/5 ) 3-l/^~3-^l/5~(3-^l/5)(3^-^V5) _6 + 5/l/5 + 5/^ l + 5zi/5 1 5iVy - 9_5i-^ - 14 -14"^ 14 • EXERCISE 71 Simplify : 1. i/=25i/^^. 3. |/^=T00i/-49. 2. i/^=:9]/^=l6. 4. y^^^i/^^i-Oe. 5. i/^=25i/^4i/^^36i/^=49. 6. i/^=^l6i/"=^|/-25i/-81i/-4. 7. -i/=7^V^=^^^ ^' (l + l/^=4)(l-l/^. 8. -i/ir^^(_v/^. 10. (3 + i/"=2)(2-i/^=^). 11. (1/^=2 + 1 /=3)(l/'=^-l/^). 12. (2i/^=^ + 3|/^(3i/^=^-2i/^^). 228 ALGEBRA 13. (^y^^ + S]/^=l.)(2i/^^b-i'^). 14. (1/^=^ + 1/^(1/ -^-l/^. 15. {xy'^-yy^)(x}/^=^+(/i/^^). — b}/—x^ 2i/ —a; 2 ^^- 1 + 1/^2' l + i/^^F 30. ^, -==y 1-V -2 3-|/==~5 ^^•sTi;^- 16. 5 1/-1 3 17. 2i/-l 18. 10 1/-4 19. 21 l/=9* 20. a; 1/-X 91 4,/-l 22. 23. 24. 1/-16 1/-9 1/-100 i/'^2r — X y'-9x' 25. 26. 2i/-25x* 3i/ — 16a;^ 1/-16 1/-25 07 l/-25a;« 32. 6 + 1/- -3 3-1/= 1 =^ 33. x+^ 5 + 2i/ / -y -5 ■ |/-4 '^" i/ — 49a;' 34. ' ^ ' -^ ' 35. 1/-5-1/-7 .T-i -2/ 158. Geometric representation of imaginary numbers. Ac- cording to the method of Chapter XIII all real Clumbers may be represented by distances measured along the line XX' from the point 0, positive numbers by distances to the right, and negative numbers by distances to the left. Let us consider any positive number, say 2, represented by OA. By revolving OA counter clockwise about the point it may be made to take the position of OB^ C, OD^ or any other line drawn through O. Now, multiplying 2 by y^A gives 2y ^^1. Multiplying again by v— 1 gives —2, which is represented by OC in SURDS AND IMAGINARY NUMBERS 229 the figure. Hence, multiplying 2 twice by |/^^1 gives the same result as revolving OA into the position OC ; i. e., reverses the direction of the line ' which represents the mul- tiplicand 2. Since mul- tiplying twice by i/^ revolves the line OA into the position C\ multiply- j^' ing once by |/ — 1 may be interpreted as revolving OA through only half of the distance, i. 6., into, the position OB. Mul- tiplying three times would revolve it into the posi- tion OD. And multiply- ing four times would re- volve it into the position OA again. But multiplying once gives 2|/ — 1 ; multiplying twice gives —2 ; multiplying three times gives — 2]/ — 1; multiplying four times gives 2. Hence, 2|/ — 1 is represented by OB^ and — 2]/^^l is repre- sented by OD. The same interpretation would hold in case of any other real number as well as in the case of 2. Hence the following principle : If all real numbers are represented hy distances measured along the line XX' from, 0, positive numbers to the right and negative numbers to the left,, then all imaginary numbers will be represented by distances 7neasured from along the line YY' perpendicular to XX\ those with positive coefficients above 0, and those with negative coefficients below. CHAPTER XV. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER. 159. Quadratic equations. Any quadratic equation^ or equation of the second degree^ in an unknown number x must have terms in x'^, and may liave terms in x, and terms free ofx, but 710 other kinds of terms. A quadratic equation that contains a term of t\\Q first degree in the unknown number is called a CDinplete quadratic equation. A quadratic equation that does not contain a term of the first degree in the unknown number is called a pure quadratic equa- tion. Thus, 3x^ — 5x=6 is a complete quadratic equation. And 5x^—4=0 is apui^e quadratic equation. By grouping terms, any complete quadratic equation can be written in the type form in which x is the tmknown 7ium,her, and A, 7?, and C are hnovin numbers having any finite values. Any pure quadratic equation can be written in the type form Ax'^B = 0, in ichich x is the iinhnovin 7ium,her. 160. To solve a pure quadratic equation. Any pure quadratic equation can be solved like the following : Example 1. Solve 23(^ — 12=2>Q-x\ Adding a?^ + 12, 2x^-\- x' = 36 + 12. (Authority ?) Uniting terms, 3a!^=48. 230 QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 231 Dividing by 3, x^=16. (Authority ?) Hence, x must be a number whose square equals 16 ; i. e., x=i/T6 =4 or —4. This may be written x= ± 4. 2 EXx\MPLE 2. Solve <^ + l=~zrTv Multiplying both members by 'a — I , a'^ — l=2. Adding 1, 0^=2 + 1. Uniting terms, a^=3. Hence a must be a number whose square equals 3 ; i. e., a= ± I 3 The solutions, or roots, are yS and — ]/3. These examples illustrate the following rule : (1) Clear the equation of fractions^ if necessary, and rentoce all signs of grouping. (2) Transform the equation into an equivalent equation having all the unknown numbers in the first member , and all terms free of the unknoimi yiumber in the second 'member. (3) Unite like tertns. (4) Divide both members by the coefficient of the unJcnown number .^ thus giving an equation of the form x'^=A, where x is the unknoimi number. {5) Extract the square root of both ynembers of tJie resulting equation.^ attaching the double sign, ±, to the second member. EXERCISE 73. Solve for the general number : 1. 7a.'^-28 = 0. ■ 3. r^ = 845-4rl 2, 10.r^-150-4a;l 4. «(rt + 4) = 4a + 49. 232 ALGEBRA e. Sx' = 2x{x-b) + 10{x + 2).^ ' ^ ^ ^ _^ 16. ^-^ + ^J^1. 7. 125 X 3^4 17 3a;^-7 a;-l_a; + l 8. I g^- ' CC^— 1 £C+1 £C— 1 9 ^ -''/-^ 18. ^^-4l=l. ^- ^rn — 6^" 6 "+1 10 ?^=i. 19. r^.-^^r • 8 6s+l 2 + 11 »^' + 8 o on 4a3 + l_8^-19 11- -2""^^- ^"' 3^=4"7^=24- .. 9x^-l_3 21 % + ^- 1^ t/l-i. — 3 ^. -61. 2 + y 2y + 3' /l3. -^^4. 22. 4^+i = l^. ^)^ — 1 5 3a;— 2 J^_l_9 3s + 6 2s + 5 /^ 2/ y^ 4* ""• 22-6 32-6' ^ ' a; + 2 25. 4£z!=^. 27. (y + i)(y-4) = 2. 5£e- -4 4x- -5' 2a; 1 + x' 3 5"~ 2x- f3" -5" X^ + X-^1 £C- — £C+1 x—l x+l 161. Ill the following sections three methods of solving a complete quadratic equation are discussed : (1) by factoring ; (2) by completing the square ; (3) by the use of a formula. Note. — The factoring method, it will be found, may be used to solve an equation of any degree higher than the first in whicli rational fac- tors can be found. But, since comparatively few expressions have rational factors, the method of solving equations by f^,ctonng may be used in only a limited number of cases. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 233 162. To solve complete quadratic equations by factoring. This method is based upon the principle, that, in general^ a product is zero lohen one^ or tnore^ of its factors is zero^ and not otherwise. Thus, in general, ah is if a is 0, or if h is 0. Also if ah is 0, then either a or 6 must be 0. Note. — This principle fails if one of the numbers is indefinitely great at the same time that another factor is 0. See (4) § 219. For example, the product 8 (^-^>{^) Now when x=2 the product is equal to |, or 2, and not to 0, although the factor x — 2 becames 0. Example 1. Solve x^ + Q = 5x. Subtracting 5x, x^—5x + Q=0. Factoring, {x—S){x—2)=0. This equation is satisfied by a value of x that Avill make either factor of the first member 0. Such values of x are obtained by equating each factor to 0, and solving the resulting equations. Equating each factor to 0, we have a*— 3=0 ; whence, x=S. 0?— 2=0 ; whence, x=2. Check. 3' + 6 = 5 • 3 and 2' + 6 = 5 • 2, identities. Example 2. Solve Qx^=x+2. Adding -x-2, Qx^-x-2=0. Factoring, (3j;-2)(2ir 4- 1)=0. Equating each factor to 0, 3x—2=0 ; whence, a?=|; 2^^+1=0 ; whence, x=—}. Let the student check the two solutions. 231 ALGEBRA To solve an equation by factoring we evidently have the following rule : (1) Transform the equation into an equivalent equation having all of its terms in the first member. (2) Then find the rational factors of the first member. (3) Equate each of these factors to 0^ and solve the residting equations. EXERCISE 73. Solve by the factoring method : 1. x'-1x^Vl = ^. ^- xll -- r:_4 , 33 2. i«M-3£c-10 = 0. A 1 X ■ XX' /3. a3^ + 8.T + 7 = 0. j3 14-??=^:. ^^' «' + i«=¥- x" x' 22 22. 1+--Gy = 0. 6. 2a;^-5£c + 2 = 0. 14. l5a; + 4=-. ^ X 23 6f^2 = 11^ + 7 7. W + 12.^24. 15- -+1^1-22.. 24.17.^^70.-8. ^ 8. hx^=l^Ax. ^^' 6^^ + 7^=-2- ;J5. 12/-7y + l = 0. * 9. 2Lt^ = 10 + 29«!. ^'''- 4^-^l'=-77. 26. 96 + 22/=.3yl 10. 6.«'''-ll^=2. 18. P^-llP--30.27. 10^-3 = 3^^ 11. 3.^^ + 5i« = 2. 19. i«='-3£c-10 = 0. 28. lO-s^-Ss. 29. ;«^- 23a; + 132 = 0. ^. 32^-4 ^ ,, ^ 34. ^ , ,) -=2g4l. 30.6-^^4^ = 0. t' 21 y+3 '^ 31. 5r— 8= — . o 1A , ^oA r „^ 3ic— 10 £c + 120 OD. =-i = . 32. 2/^ = 13y-36. 14 x «o , , 10 , 1 ^ OP, 1 ^-7 , a; + 24 _ 33. .^+-. + -=0. 37. 2-:z2-,rxi + 5^^^=^- 38. -^,-^0=2. QUADRATIC EQUATIONS-ONE UNKNOWN NUMBER 2S5 163. Completing the sq[iiare. From the identity a?^ -\-^ax-V a? = {x-{- a) ^, we have x^^-^ax-^-a^ as the general form of the perfect square of a binomial. Hence, if we have given the binomial x^ + 2a£c, to make it a perfect square, we must add a^^ ^^e., the square of half the coefficient of x. The process of adding a term to a binomial such as a^^ + 2a.T, in order to form a perfect square^ is called completing the square. Example 1. Form the perfect square whose first two terms are x^ — V)x. /10\ ^ Here we addf -^\ , ov 25. This gives a?^— 10^ + 25. Example 2. Complete the square of which two terms are x} 4- 3a?. Here half the coefficient of x is |. The square of | is \. Add- ing I, we have x'^ + 3a? + 1 . a?' + 3a? + 1 = (,r + 1)^ Note. — It would be well here for the student to review § 61. 164. Solving the complete quadratic by completing the square. Any complete quadratic equation whatever can be solved by use of the principle of § 163. Example 1. Solve a?^ + 3x— 10=0. Adding 10, x"- + 3x- 10. Adding Q)^ to both members, £C^ + 3a? + |=— . Extracting. the square root, a? + |=±2- Solving for a?, '■^=—1 ± I- Using the + sign, a?= — 1 + |, or 2. Using the — sign, a?=— |— |, or —5, Hence the two roots are 2 and —5. Let the student check the results. 236 ALGEBRA Example 2. Solve 2x^ — 5x=7. Dividing by 2, x^—§x=l. Adding the square of half of |, a?^ — |x + f |=| + ||,=|^. Extracting the square root, x—^= ± |. When £c— 1=1, x=l. When ic— f^— |, x= — l. Check the solution. Examples. Solve x^—4:X + 5=0. Adding —5, x^ — 4a?= — 5. Adding square of half 4, x^ — 4x-\-4^= — l. Extracting the square root, .r — 2 = ± y/ — 1 . When ir— 2=1 ^, ir=24- 1/^. Whenir-2=-i ^, x=2-l ^. These may be written x=2± ]/ — 1. Here the two roots^ or solutions, are complex numbers. Check the solution. Evidently to solve any quadratic equation we have the fol- lowing rule : (/) Reduce the equation to the form Qt?-\-px = q. (2) Add to each member the square of half the coefficient of x. (3) Extract the square root of both members of the restdting equation., attaching the double sign ± to the resulting second member. (4) Solve the tico restdting linear equations. Observe that the double sign might have been placed before the resulting first member as well. The reason for not doing so may be seen from the following example. Let x^=a^. Extracting the square root of both members, ±_x=±_a, that is x=-\-a or —a and — a?= +a or —a. Now if in this second equation we divide by — 1, we have x= —a, or + a, the values in the first. That is, .t= ± a is the same as —x= ±a; hence we use the double sign before the second member only. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 237 EXERCISE 74. 3. x' 4. x' Solve by completing the square : 1. x}-%x + ^ = (). 2. x'- + 4:X-=V2,. 7. x'-^x^^. 8. a;^-9£c=22. ^ 4 + a; 9. x'-lbx^U. 31. -Sit-^ 10 . X' 12a; + 20 = 0. 11. 2a;2 + i«=l. 12. "Ix'-bx^^. 13. 6a;^ + 6 = 13a;. 14. 4a;2-lla;=3. 16. 10a;2-29£c + 10 = 0. 16. 3a;^ = 17a; + 28. 17. 2a;^ + 3a;=5. 18. f>x'-x=^l. 19. 9ic^ + 3£c + 18 = 0. 20. 12£c^-14a; + 3 = 0. 21. 0^=^ + 80; + 21 = 0. 22. aj^ + 6aj+ll = 0. 23. ^x^-'^Q = Qx, 24. i«^ + 6a; + 25 = 0. 25. 16a;^-96«=1792. 26. ic^-8a;=-15. 27. x' + Ux=-^h. 28. 39icH96 = 51a;='-96. 1 2 18 29. ic4 1 1— JC 4ic— r 14a; + 40=-0. 5. a;^-lla; + 30 = 0. 6. £c- f 5ic = 14. 5 _8 4:-x 3" ic=6. 30 36.r-105 = 0. 32. 20, 42|. 33. 6.3+?^:::^=44. 34. 7a.-=7(fc + 3) + 4. X—^ X+4: 36. ^5-,^7-2i. 36. 37. , 1 £C+1 X—1 2 = 1. 38. 3 2a!-3 ' a; 1 + 2 = 0. 39. 40. icH 2 a;-2' £C — 2 37 — 3 JK-l a; + 2 4-£c 7 0. ic-1 2aj 3 41. (a;-5)^ + (ic-10)^ = 37. 42. ^^ + ^ =0. 43. if + 9 ' a;-l • + x'-\ ' 2a;-2 4' or THi ^ 238 ALGEBRA 44 __i L = _l 45 _A . _1=11 **• £c + 2 x-'l l-x ^''' x-l^x-3 2* 46. {4:i^xy + (x-4y = Q(x-4:)(x + 4). 2.;^-l a;— i Sa;— 10 a;— 4 _ a— 4 11 *'• "^^^£^-2"" a;- 3' *^' 2a3-12 3.t-16~^' 165. Solving a quadratic equation by the formula. Since any complete quadratic equation may be thrown into the type form the solutions of this equation will give 7i formula by which the solutions of any particular quadratic equation may be written down at once. Solve Ax^ + Bx + C= by completing the square. Dividing by A^ -'+a-+3=«- Adding -^, Completing the square, ^., , ^ ^ B'- B^ a ^+yl^ + 4.P 4.4^ A' ,^B , B' B'-4A0 Extractmg square root, ^ + w~i= ±|/ - — j-p Solvnig for x, ^^~~2A^ 2A — ' / -B±yB'-4A(y or ^J x= 271 -• .p, ,. w- • -i?+i/i5^-4^6^ 1 hat IS, one solution is 2~4~~ — ~ ' ifi.fl- -B-rB'-4A0 and the other is 0-3 • QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 239 Now, by replacing A, B, and C in this formula by the par- ticular values which they have in any given equation, the solutions of any quadratic may be written out. Thus, the long process of completing the square, etc., in every equation may be avoided. Hence, the student should master this formula and tise it in all future VDorh where the quadratic eannot he factored at sight. He should be able, hoAvever, to solve any quadratic by completing the square, and in this way to derive the formida. Example 1. Solve Sa?^— 4ic=15. Written in the form Ax'^ + Bx-\-C=0, this becomes Here A=3, 5= -4, C=-15. „ 4 ±1/16 + 180 Hence, x= = — ^ ^4±14 6 • ' 18 Using the + sign, ^~"~6' ^^ ^' Usmg the — sign, x= g, or — o- Let the student check the result. Example 2. Solve 4ic2 + 2=— Sa:*. Adding 3x, we have 4^?^ + 3a? +2=0. Here, A=4, 5=3, C=2. -3 ±1 /9-32 Hence, x= ^-g _-3±j/^ ~ 8 -3 + 1-23 -3-1/-23 = '-^ or ' Let the student check the result. 8 ""^ 8 240 ALGEBRA EXERCISE 75. Solve by use of the formula : 1. a;''-3a;-10. 8. Qx' + Q = Ux. J,5, 2b'-7b + Q = 0. 2. aj'' + 10ic + 21 = 0. 9. 10x' + U = ^9x. 16. W = Qk-b. 3. 2x'-i^x=Q. 10. 8£c=^-65£c+8 = 0. ^^ 21b'-b = 4: 4. ^x' + bx=2} 11. 2«^ + 3a + 4 = 0. 1ft q^ o'2z=fi 5. 6i«^ + a;-5 = 0. 12. ba'-2a + 7 = 0. / ' ' 6. 7a;^=50a!-7. 13. 15y^-y + 3 = 0. l^. 12 + a; = lla;l 7. 4x'-}:x=b. 14. y'-8y + 4 = 0. 20. l-a^-Sa^O. 21 ?^^±I y 2 22. — ^~^ =0. 1-t 3 23. i— i =2. 24 ^-l_.9-+l_i 26. l+v = li 26. r + 3 = -^. 7*— 1 27. #-i=3-i. 28 2^ + 3 _ «^ + 3 • x-2 2x-r 30. (3a;-l)2-7i«. ^, ^„ 3a;^-4 3^- «^+2=2^-q^r 32. (5.T + 2)2 -(2cc- 3)^ = 0, 33. l-(2x-iy = 0. 34. 10v' = bv. 35. Qx-m=^x\ 36. 1-1-221 37. 2a! + | = 2iK^ 38. ^x' = 42-6x. 39. ■+"m- >40. „=j-.. 41. a-a' = 0. 42. 3 _ 6 8 x—4 x — b x — S' 43. x-2 x + 2 10a;-8 x + 2 ' x-2 4-x' 44. , l + 2a;2 1— CC = -v— . 5 + a; 0. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 241 45 2 l-xx+l -g 8 + a; 4-2a; _8 5(a;-2) a; + 2 a;^-4' * 8-ic + 4 + 2a;~3' -c 7 — 3£c , 7-£c 2x + S ^^ ^ Sx-S ^ , dx-6 40. -j — 0-4-0 — TTJ^^'^i rt- 50. ox— ii = 2x-{- — o — 4— 3a; 2a; + 2 Ax— 2 x—S 2 47 1 , y _ y + 3 51 6 2 ^^- g + 4(^) = 2^- ^^- ^^-^=14- 2^4 Hmf. — Consider (x — 1) the unknown and this is a quadratic in 166. The discriminant. In the use of the formula of § 165, x= k-i 5 it was observed that the character of tUe 2A roots depended upon the part under the radical sign. This quantity, B'^—AAC, for that reason, is called the discriminant of the quadratic* Observing the formula, we see that (i) When B'—JfAG is a perfect square, the roots are real-, rational, and unequal. (2) When B'—Jf^AC is equal to zero, the roots are equal. {3) When B-—j^AC is positive but not a perfect square the roots are real and conjugate surds. (4) When B^ — J^AG is negative the roots are conjugate com- plex numbers^ * It is assumed in principles (1), (2), (3), and (4) that the co- efficients ^, J5, and Cof the general equation Ax'^-{-Bx^rC=^ are all real numbers. The principles may or may not be true when one or more of the coefficients are imaginary. 16 242 ALGEBRA From these observations we can determine the nature of the roots of any quadratic equation without solving it. Example 1. Determine the nature of the roots of a?^—5a^ + 6=0. Here-B= — 5, A=l and C=Q ; hence, B'— 4AC=1, a perfect square; hence, the roots are real, rational, and unequal. Example 2. Determine the nature of the roots of x^-\-x-i-3=0. Here JB^— 4AC= — 11, hence the roots are conjugate complex numbers. Without solving tell the nature of the roots of Examples 1 to 20 in Exercise 76. 167. The relation of the roots to the coefficients. By dividing both members of the general quadratic, Ax^-}'JSx+ 0=0, by A, B C we have x^ + -ix-\--j=0, which is of the form x}+px + q = 0, where ^ and q may have any finite values, integral or fractional. Solving 9?H7>a3 + 5' = by the formula Ave have _ —p ± y'jf — 4:q X 2 • Calling one root r^ and the other r^, we have _—p + l/2f—4q and r.=^I^PJZ^, Adding these we have r^ + r^=—p. Multiplying, we have .^^^^ Tzi^+vVzlil X r -i^-ri^'-4( y1 p'-(p'-4:q) 4 QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER ^43 From these relations we can state the following principles : When an equation is throicn into the form x^-\- px-\-q = ; (1) The sum of the roots is the coefficient of x with its sign changed ; {2) The product of the roots is tlie term free from x* From these relations we may form equations with given roots. For example, form an equation whose roots are 5 and 3. Since the coefficient of a? is — (5 + 3) and the term free from x is 3 • 5 , the required equation is x^ — 8.r + 1 5 = . EXERCISE 76. Write the quadratic equations whose roots are : 1. 6, -5. ^ _3^ 6. 2, -f. 2. -2, 7. * -^' 7. -4, -6. 3. \. 5. i, -|. 8. -f, -4. 9. i, 1. 10. -i -i 11. By the use of principles 1 and 2, § 167, show what the signs of the roots of ic^ — 11.^ + 24 = must be. 12. What are the signs of the roots of £c' + 5a;— 24==0? Of a;'^-7i^-18 = 0? 13. For what value of a will the equation aic^ + 3.7;— 5 = have equal roots ? 14. For what values of m will the equation 2a;^ + mic + 32 = have equal roots ? 15. For what values of c will 3a;'^— 2£c + c=0 have real roots ? For what value of c will the roots be equal ? * The term of an equation free from the unknown is called the absolute term. -2h±y4b' + 16x -8 -2b±2]/b'+4x -8 24:4 ALGEBRA 168. Solving a quadrate formula. In § 119 we solved some formulae which were linear with respect to certain general nmn- bers involved. By the methods of the preceding sections, formulae which are of the second degree with respect to a certain general number may now be solved for that number. Example 1. Solve x=2a{2a—h) for a. Removing the sign of grouping, x=4a^—2ah. Arranging in type form, — 4a^ + 2a6-f a?=0. Here A=-4, B=2b, C=x. Hence, by § 165, a= Example 2. Solve 1=-^ for f . Multiplying by c^ Ic^—af^. Hence, —at^=—lc^. Dividing by —a, ^^=77' Hence, f=±l/— , or ±c|/Z, or ±^y^W, ^ a f^ a a^ EXEBCISE 77. 1. ax'^—c=Q. Solve for x. . 2. aW-a'=^. Solve for h. 3. ah = a'^—A. Solve for a. 4. 2ic^ — 3a' = bax. Solve for a;, and for a. 5. x'^^-2a^ = Mx. Solve for aj, and for a. 6. a3^ + 2«£c=^- + 2«A. Solve for 5, and for x. 7. £c' + ra; + s = 0. Solve for a;. 8. mx^-\-nx^t=^. Solve for aj. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 245 9. a?+-=aH— Solve for a. ' a X The following formulse express important laws of physics. 10. 5=^/7^. Solve for ^. 14. i^=-^. Solve for f?. 11. /='^1 Solve for t. 16. E=\mv'. Solve for v. 12. F=^ — . Solve for v. 16. r=h-^. Solve for d. r d- 13. F=^. Solve for v. 17. s^vt +\gt\ Solve for t. 18. ?i^=-T2--T. Solve for ,9, /, and n. 169. Problems which lead to quadratic equations. Some problems can be solved by the use of quadratic equa- tions. It has been seen in § 166 that the solutions, or roots, of a quadratic equation may be whole numbers or fractions, positive or negative, rational numbers or surds, real numbers or imaginary numbers, depending upon the relation among the coefficients. A problem may by its nature require for its solution a certain kind of number. For example, a certain problem might require for its solution real numbers. Now, if, by solving the equation, imaginary solutions are obtained, they must be discarded. Plence, so7ne of the solutions of ayi equation may satisfy the equation and yet not satisfy all of the requirements of the problem. It is necessary, therefore, in solving problems by the use of equations, to examine the solutions to see if all of them satisfy the requirements of the problems. Example 1. Sixty-four times the number of students in a class exceeds 3 times the square of the number by 21. Find the number of students. 246 ALGEBRA Let x= number of students. Then Ux=3x^ + 21. Solving, we obtain x=21, or x=^. Evidently the problem requires for answer a ivhole number. Hence, the solution x=:^ must be discarded ; and the required number of students is 21. Example 2. A man walks 25 miles at a uniform rate. If he had walked f of a mile per hour faster, the journey would have taken 2 hours less. Find the rate of his walking. Let x= number of miles traveled per hour. Then — = number of hours required for the journey. 25 ^nd ——5— number of hours the journey would have re- x-j- ^ quired had he walked | mile per hour faster. 25 25 Hence, ^=;F4:^ + 2. . Solving, we get x=2l, or — 3^. The rate must be an arithmetical number. Hence, the solution a?=— 3| must be discarded ; and the required rate is 2^ miles per hour. Example 3. Find the real number whose square increased by 32 equals 8 times the number. Let x= the number. Then, £cH32=8iC. Solving, we get a?=4 + 4l/^ and 4— 4|/^. These expressions are not real. Hence the problem is impossible. QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 247 EXERCISE 78. By the use of quadratic equations solve the following prob- lems. 1. Find two arithmetical numbers whose difference is 7 and product 330. 2. Divide 31 into two parts the sum of the squares of which is 541. 3. Find two arithmetical numbers whose sum is 50 and product 336. 4. One of two numbers exceeds 25 by as much as 25 exceeds the other, and their product is 561. What are the numbers ? 5. Find two consecutive integers whose product is 5852. 6. Find two consecutive whole numbers the sum of whose squares is 313. 7. The difference between two numbers is 10, and the sum of their squares is 212. Find the numbers. 8. The denominator of a certain fraction is 5 more than the numerator. If the fraction be added to the fraction inverted, the sum will be 2J|. Find the fraction. 9. If 64 be divided by a certain number, and the same num- ber be divided by 2, the second quotient will exceed the first quotient by 4. What is the number ? 10. A pupil was to divide 12 by a certain number, but by mistake he subtracted the number from 12. His result was 5 too great. Find the number. 11. One-half a number plus the square of the number is 68. What is the number ? 12. The side of one square exceeds that of another by 3 inches, and its area exceeds twice the area of the other by 17 square inches. Find the lengths of their sides, 248 ALGEBRA 13. A rectangular field is 96 feet longer than it is wide, and it contains 298,000 square feet. What are its dimensions ? 14. One side of a rectangle is 4 inches longer than the other, and its diagonal is 20 inches. How long are the sides ? 15. A lawn 25 feet wide and 40 feet long has a brick walk of uniform width around it. The area of the walk is 750 square feet. Find its width. 16. There are two square lots ; the side of one is 42 feet longer than the side of the other. The two together contain 2146 square yards. What are their dimensions ? 17. A floor can be paved with 200 square tiles of a certain size ; if each tile were one inch shorter each way, it would re- quire 288 tiles. Find the size of each tile. 18. The printed portion of the page of a book is 4 inches wide and 6 inches long. How wide must the margin be in order that the whole page shall contain 48 square inches ? 19. In the center of a rectangular room is a rug 9 feet by 12 feet ; around this is a border of uniform width. The area of the floor is 208 square feet. What is the width of the border ? 20. The length of a rectangle is 6 inches greater than its width ; and if its width be doubled and its length diminished by 3 inches, the area will be increased by 36 square inches. What are the dimensions ? 21. The perimeter of a rectangular field is 184 rods, and the field contains 12 A. What are its dimensions ? 22. Two men start at the same time from the intersection of two roads, one driving south at the rate of 3 miles an hour, and the other west at the rate of 4 miles an hour. In how many hours will they be 25 miles apart ? 23. Two trains are 100 miles apart on perpendicular roads, and are running toward the same crossing. One train runs 10 QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 249 miles an hour faster than the other. At what rates must they run if they both reach the crossing in 2 hours ? 24. There are two numbers whose difference is 4. If 240 be divided by each of them, the difference between tlie quo- tients will be 10. What are the numbers ? 25. There are two arithmetical numbers whose sum is 33. If 36 be divided by the smaller number, and the larger number be divided by 4, the sum of the quotients will be 10. Find the numbers. 26. A man bought a certain number of books for $7.50. If he had paid 25 cents apiece more for them, he would have ob- tained 1 fewer for the same money. How many did he buy ? 27. A man finds that by increasing his speed 1 mile an hour it takes 6 hours less to walk 36 miles. How fast does he walk ? 28. A grocer paid $2.70 for eggs. Pie found that if he had paid 3 cents less for a dozen, he would have received 3 dozen more for the same sum. Find the price per dozen and the number of dozen. 29. A merchant paid 12160 for some carriages, all of the same price. By selling all but 2 of them at a profit of $36 each, he received the amount he paid for all of them. How many did he buy ? 30. A man sold a lot for $1125, thereby gaining i as many per cent as the lot cost him dollars. What did the lot cost ? 31. In a certain number of two digits the tens' digit exceeds the ones' digit by 2, and when the number is divided by the sum of its digits, the quotient exceeds twice the ones' digit by 3. Find the number. 32. A tree was broken over by a storm so tliat the top touched the ground 50 feet from the foot of the stump. The 250 ALGEBRA stump was | of the height of the tree. What was the height of the tree ? 33. A tree which stood at the edge of the bank of a stream fell with its top in the water. The tree was 60 feet high, the \^ bank on which it stood was 15 feet above the water level, and the body of the tree passed under the surface of the water at a point 20 feet from the bank. What portion of the tree was under water? 34. The hypotenuse of a right triangle is 4 inches longer than one leg and two inches longer than the other. Find the sides of the triangle. -y 36. One pipe can fill a cistern in 6 minutes less time than is 'required for another pipe to fill it. The two together can fill it in 10|^ minutes. Find the time required for each pipe alone to fill the cistern. ^ '^"^ . \ ' ^- "^ 36. One of two pipes can fill a tank in 28 minutes, and the time required for the other pipe to fill it is 19| minutes longer than is required for the two pipes together to fill it. Find the time required for the two pipes together to fill the tank. -^ > 37. It would take B four days longer to do a piece of work than it would take A to do it. The two together could do it in 6|^ days. In how many days could A alone do the work ? 38. With John's help Henry could remove the snow from a piece of side- walk in 6^ minutes less time than would be re- quired for Henry to do it alone. John could do it alone in 21 minutes. In how many minutes could Henry do it alone. 39. A could do a piece of work in 2 days less time than would be required for B to do it. A works 4 days, then leaves; and B takes his place and finishes the work in 5 days more. In what time could each one do it alone. 40. The sum of two numbers is 24, and the quotient of the QUADRATIC EQUATIONS— ONE UNKNOWN NUMBER 251 less divided by the greater is f| of the quotient of the greater divided by the less. Find the numbers. 41. Two trains on the same road start at the same time from stations 225 miles apart. One takes | minutes longer than the other to run a mile, and they meet in 3 hours. Find the speed of each train ? 42. A teamster having 12 miles to drive increased his speed one mile an hour after he had traveled two miles. He thus finished the distance in half an hour less time than it would have taken had he not increased his speed. How long did it take to drive the 12 miles ? 7^0 Suggestion. — Let his first rate be x miles an hour. 43. A merchant bought a certain number of mirrors for 1 36, and, after breaking one, sold the rest for 50 cents apiece more than they cost him, thus making $2.50 by the transaction. How many did he buy ? . 44. The rate at which a man can row in still water is twice the rate of the current of a river. He rows 6 miles down the river and back in 4 times as many hours as he could row miles per hour in still water. Find the rate of the current and the rate of his rowing in still water. 45. The number of square inches in the area of a square ex- ry ceeds the number of inches in its perimeter by 32. What is its area ? 46. The side of a square is the same as the diameter of a circle. The area of the square exceeds that of the circle by ap- proximately 3.4336 square inches. Find the diameter of the circle. 47. The base of a triangle exceeds its altitude bj^ 2 inches, and its area is 112 square inches. Find its altitude. 48. An engraving whose length was twice its width was so i- U 252 ALGEBRA mounted on Bristol board as to have a margin 3 inches wide, and equal in area to the engraving, lacking 36 square inches. Find the width of the engraving. 49. A certain number consists of two digits whose sum is 12 ; and the product of the two digits plus 16 is equal to the number expressed by the digits in reverse order. What is the number ? 50. The glass of a mirror in the shape of a rectangle whose length Avas twice its breadth cost $1.25 a square foot. The frame, measured on the inside, cost 10.75 a linear foot. If the glass cost $22 more than the frame, what were the dimensions of the glass ? 61. A certain farm is a rectangle, whose length is twice its breadth; If it were 20 rods longer and 24 rods wider, its area would be doubled. How many acres does the farm con- tain ? 52. A square lot has a gravel- walk around it. The side of the lot lacks one yard of being six times the breadth of the gravel-walk, and the number of square yards in the walk exceeds the number of yards in the perimeter of the lot by 340. Find the area of the lot and width of the walk. CHAPTER XVI. HIGHER EQUATIONS. EQUATIONS INVOLVING SURDS.— ONE UNKNOWN NUMBER. 1 70. It is not within the limits of this work to consider general equations of degree higher than the second^ but some special equations of higher degree may he thrown into a quadratic form^ i. e., we may consider tlie unknown as being a power of some single unknown, or as an expression involving an unknown. Thus, x^ + 3a?^ + 8=0 is a quadratic in iir^, for if we let J(^=y, an unknown, we have y'^ + Sy + 8=0, a quadratic in y. {x' + 3x—2y + 2{x^ + 3x—2) — 3=0 is a quadratic in x' + 3x—2. Letting y=x'^ + 2x—2, we have y^ + 2y—o=0, a quadratic in y. Such equations may be solved as quadratics. Example 1. Solve x^-13x'2 + 36=0. This equation has the form of a quadratic in x^. Solving for x^ Now solving Solving Hence, Example 2. Solve _13±T 169- Ml x- 2 13±5 - 2 = 9 or 4. x'^d, * x=±3. x^=4, x=±2. x=S, -3, 2, or -2. x^ x + 1 , X+1+" x^ -^ 253 254 ALGEBRA Here the second fraction is the first fraction inverted. Putting y for -— -^, the equation becomes Multiplying hj y, y''—3y + 2=0. Factoring, (y—2){y-l)=0. Hence, y=2 or 1. Therefore the given equation is equivalent to the equations -— -7=2 and — r^=l. x+1 x+1 x^ Solving ^+1""^' ^=^ ± y^- Solving ^^=1, .=1^. Example 3. Solve 2x^ + 2x + l ^^ x^ + x' This may be written 2{x'^ + x) + l= .^ . X -\- X Multiplying by x"^ + x, 2{x^ -{-xY-\-{x^ + x) — \Q=Q. This is a quadratic in x^ + x. Solving for x^-\-x, x^ + ^.= -l±l/l + 80 4 =2 or -f. Hence, the given equation is equivalent to the two equations x^ + x=2 and x^-{-i Solving x^ + x=2, x=l or —2. Solving x'^ + x=—^,x= 2~^ * Note. — In every equation which we have solved in this book, the number of roots, or solutions, is equal to the degree of the equation. In the Theory of Equations it is proved that this is true in general. A linear equation has one root, a quadratic tico roots, a cubic three roots, etc. The student should see that in every equation he gets a number of roots equal to the degree of the equation. HIGHER EQUATIONS. EQUATIONS INVOLVING SURDS 255 171. Equations of higher degree than the second can some- times be solved by factoring. Example 1. Solve xHl=0. (1) Factoring, {x+l){x^—x+l)=0. Hence (1) is equivalent to a:? + 1=0 and x^—x-\-l=0. Solving a? + 1=0, x= — l. Solving x^-x + l=0, x=^±^]/'^. Example 2. Solve x^-U=0. (1) Factoring, (x + 2){x- 2) {x^ + 2x+4)(x'-2x-{-4)=0. Hence, (1) is equivalent to i»+2=0, a?— 2=0, x^ + 2x + 4:=0, and x^—2x + 4=0. Solving each of these, x=—2, 2, — 1 ± j/— 3, or 1 ± ]/— 3. Examples. Solve ir' + 5a?=6a?^— 12. (1) Adding -6.^2 + 12, ar'-6icH5if + 12=0. Factoring by the remainder theorem, {x-3){x + l)(x-4)=0. Hence, (1) is equivalent to a?— 3=0, a7+l=0 and a?— 4=0. The solutions of these equations are 3, —1, and 4, respectively. Solve the following : 1. i«*-16-0. 2. x*-2x' = l^, 3. 2a;*-5a;^-12 = 0. 4. Q^ + 2x'=x + 2, 5. ar*-a;'-4a; + 4 = 0. 6. a;^-2a;^-5a; + 6 = 0. 11. 1--X .X EXERCISE ! 79. 7. 1 .x'^-l x^'^ 6 " = 1. 8. 0.^ + 1=3. 9. (a;^-iy + 24= = ll(a;''- -!)• 10. {x^^tcy- -2(a;^ + 2a;): -3. -11( n ^ X , 1+30 = 0. 256 ALGEBRA 1^ l4s + =^'-?'=B^ 17. x' = \. 18. a;*-81 = 0. 19. i««-l = 0. j4 2x + l «' _„ 20. .T^ + 1 = 0. ^*' x' ' 2a! + l ^• 15. .^ + 1-^4,. 16..' + . 1-/^^. 21. £c^-256-0. 22. i««-16 = 0. 23. ;«'' + l = 0. 24. (Saj'^ + Ga;)^ = 1 -(.?.''^ + 3a^ + 2). 25. {x^-x + 4.) ^ + {x'-x) = 2. 26. (. + ?)' + (' 4)=- 27. a3H-a^ + -4- X 1 = 4. Suggestion, x' «-|.=("S'- 28. a;* + 3ic^-2ic^-3a;+l = 0. Suggestion. Divide by x^ and arrange as in Ex. 27. 29. x'-%x^^^^x^-Zx-\-l = ^. 172. Equations which involve surds. To solve equations which are *>ra^/o?ia^ with respect to the unknown number, it is necessary to free the unknown number from the radical sign. To do this both members of the equation must be raised to the same power. (Ax. 5.) The following rule may be used in general : (1). Transform the equation into an equivalent one in v^hich one member consists of a single surd. This is called isolating the surd. (2). Raise both members to such a power as will free this iso- lated surd of the radical sign. (3). If surds still remain., repeat the operation. HIGHER EQUATIONS.— EQUATIONS INVOLVING SURDS 257 173. Introductionof new solutions. In general^ icheri both members of an equation are raised to a higher power ^ new solutions are introduced. For, let any equation be represented by A = B. Then, squaring both members, A' = B\ or A'-B' = 0. Factoring, {A-B){A + B)=0, This is equivalent to A — B = ^ and A^^B = 0. Of these two equations A—B=0 alone is equivalent to the given equation. Hence, all of the solutions of /l+B = are introduced by squaring. It is necessary, therefore, in solving equations which involve surds, to examiyie all solutions and discard those which do not satisfy the equation in its original foi'm. Example 1. Solve |/ 2^—3 = 5. Squaring, 2x—S=25. Whence 2x=28. And ir=14. The solution 14 satisfies the original equation, and therefore was not introduced. Example 2. Solve ^ + y'x-\-7=x. Isolating the surd, \/x + 7 = a? — 5. Squaring, x+7=x'^ — 10a:j + 25. Or x'—llx-^l^^O. Factoring, {x—^){x—2)=0. Therefore, ic=9 or 2. Now 9 satisfies the given equation, but 2 does not. Solution 2 was introduced by squaring. 17 258 ALGEBRA Example 3 . Solv^e y 2ic -f 8 + 2 + 2 ]/a? + 5 = . Isolating second surd, i/2x + 8-{-2= — 2yx-{-5. Squaring, 2x + 8 + 4: + 4^/2x + S=4x + 20. Therefore, 2\/2x-\-8=x + 4.. Squaring again, 8aj+32=ic'' + 8^7+16. Or, 0^2^16. Whence, 07=4 or —4. Now neither 4 nor —4 will satisfy the given equation. Both solutions were introduced by squaring. Hence the equation has no solutions ; it is an impossible equation. Example 4. Solve 3ic'— 4ic+]/3a;'— 4a;— 6 = 18. Equations such as this, where the unknown enters alike in the part free from radicals and under the radical, may be solved as a quadratic. By adding —6, we have 3a?^— 4a?— 6 + ]/3a?''— 4ic— 6=12, a qua- dratic in y2,x^ — 4:X—%. l±l/l + 48 Hence, ]/3x^— 4a7— 6 = -^ = 3 or —4. The solutions of |/3x^— 4£C— 6=3 are 3 and — |. And |/3ic^— 4ic— 6=— 4 will be found to be an impossible equa- tion. Hence the required solutions are 3 and — |. Example 5. Show that 1 has three cube roots, and find their values. We are to find what numbers cubed will equal one. If we let X stand for the cube root of 1, then a^=l. Adding -1, a^-l=0. Factoring, {x—l){x^ + x-\-l)=0. Therefore, a;— 1=0, 3(^ + x+l=0. The solution of a?— 1=0 is 1. The solutions of a;Ha?+l=0 are — | + ^i/— 3 and — |— ij/— 3. HIGHER EQUATIONS.— EQUATIONS INVOLVING SURDS 259 Hence, there are three cube roots of 1, one a real number and the other two complex expressions ; viz. : 1, — 2 + ^1/— 3, —\ ^]/ — 3. EXERCISE 80. Solve the following : 1. i/aj+l==2. 4. 4i/ic + 5 = 3]/3i«4-4. 2. 8-|/a;-l = 6. 5. 5i/ic + 2-3i/4ic-3 = 0. 3. 6 = 10-2i/5a;-l. 6. iri3 + «=2. 7. f/3^M=T6-if2^M=3^T20 = 0. 8. i/15 + 3a; + 2 = i/23-a-. 17. yx + 2-l = \/x-^. 9. 2i/9T^=3i/^T24-10. 18. ^=-f2x-^. 10. Vx' + ^x-r^=x + l. 1^' V^=^=J-^' 20. i/aj+i/32 + iK = 16. 11. cc— 3 = i/2i«"'— ic + lO — 5. 12. i/5iK + 4-i /12ic + 21. 21. i/£c + 4=-2 + |/a;. 13. \/x=yx + 'lb — l. 22. \/x+Q = vl2 + x, 23. ^/x-S2 = lQ-i/x. 14. i/«.-12 = i/-«-2. 24 1/^^3-1/^+12::=-^ 15. 1/^-7 + 1/^+7 = 0. 25. a!+i/^T5 = 5. 16. 2j/^+2 = i/^. 26. 1/2^^ + 1/2^+9 = 8. Ql 27. i/3a; + i/3a; + 13 28. i/aj-i/a;-3-^- l/3a; + 13 2 29. i/a;^-3a; + 5-i/a;^-5ic-2 = l. 30. ■~T~ = \/x+i/x-l, 31. i/cc + i/ic-6 = l/aj •^-•-- - -"• ^^•^^'/^-"-l/-^:=r6- 260 ALGEBRA 32. x'-Sx + 1 + 2yx'-^x=4. 39. y^^:f^j^^y^:f^ = Q^ 40 ^—= — =2. 34. i/i« + 2-5--i/ic-3. ' yl + x-yl-x Vx-\ _ X 41. :^— =__!—._ 8 _^ 00. ■ ,— I -, — riTT' 1 — |/£c 1 + 1 /cc 1— .« y'x + 1 iG — 1 y'x+\ — \/x—\ 36. ■\/x+yl+x=^/Yjf^: ' yx + l + y'x—1 Vl + x 2 43. ^/x+Vl — i/x + x^^l. 37. i/£c— 1 i«— l/a;— 8 44. i/l—x+i/2—x = yl—4x 38- |/^^^^a^-r*/^H^"3V. 45. i/a^^ + 3ic + l-l-2a;^-6a;. 47. 46. yx + 2 + i/x—S = y'2x+ll. 20 _ ._ 12 = -l/«.= v/15+a.. 48. -^^==.= ,/^^ + s + x. 1/15 49. i/xTlO + i/x^^=yQx. Solve for a? : 60. i/ic + a'^— i/ic— a^ = l/26. 52. y^x+ya—x=y'a-\-b. _ 2a3 61. i/x-\-\/x + a=—=^- 53. i/aj— a = ^>— i/^. 54. Find the four fourth roots of 16. 66. Fhid the six sixth roots of 1. CHAPTER XVII. SYSTEMS INVOLVING QUADRATIC AND HIGHER EQUATIONS. 174. As in the case of linear equations, so in general, solv- ing a system of quadratic equations requires, first, the elimina- tion ofallhutone of the unknown numbers ; and second, so?y*?i^ the resulting equation for that unknown number. When a system involves quadratic or higher equations, the method of elimination depends upon the forms of the equa- tions. 175. One equation linear and one quadratic. When one equation is linear and one is quadratic, either of the unknown numbers may be eliminated by substitution ; and the elimination leads to a quadratic equation. Since a quadratic equation in one unknown number can alicays be solved, then a system consisting of a linear and a quadratic can always be solved. Example 1. Solve | ^.-^I?7. g| Solving (1) for a?, ir=6 + 22/. (3) Substituting this value of x in (2), 42/' + 241/ + 36 4-2/'= 17. (4) Solving (4) , 2/ = - 1 or - »/. ^ v Substituting these values of y in (3), \X.""*v when 2/=— 1» ic=6— 2=4; ^ when 2/=-¥, a^=6-^8-=-f- ^^ ^ Hence, there are two solutions : v-^ "*\ a?=4, 2/=-l; ^=-1, 2/=--/. Let the student check the result. 261 262 ALGEBRA EXERCISE 81. Solve: 4. 5. 7. 10. 11. y=-l/12; {y=V7; \y=.-y7- \y\\V'^\ J x=-h _ j x=2, j x=-2, i y=-v7; 1 2/=o; 1 2/=^; etc. QUADRATIC AND HIGHER EQUATIONS 2Y5 Now proceeding as in Example 1, we get the curve shown in Fig. 8. This curve is called an ellipse. Astronomers have found that all of the planets move in ellipses. Example 3. Draw the graph of a?H i/^=9. By proceeding as in Example 2, we get the circle shown 11 Fig. 9. Fig. 9. 183. Solutions of systems. The solutions or roots of a sys- tem are the coordinates of the points where the graphs of the equations cross^ or intersect. Example 1. Interpret the solutions of the system a? + ; 4. 276 ALGEBRA The solutions are ^=l±?i^, ^^ 8-1/29 ^^^ 5 5 4-21 29 ,._ 8 + l/29 ^ 5 ' y- 5^' The graphs of these equations are the circle and the straight line PQ in Fig. 9. The points whose coordinates are their two solutions are the points P and Q^ the points where the graphs of the equations intersect. Observe that the straight line intersects the circle twice. This is as tnany times as the system has solutions. Example 2. Interpret the solutions of the system ix^ + y'=9, x4-l/32/=6. The solutions are, x=§, y=§i S\ and x=^, y=^i/3. These solutions are the same, i. e., the system has a pair of equal solutions The graphs of these equations are the circle and the straight line AB, Fig. 9. It is seen that the line AB does not cross the circle at all^ but touches it at one point -K, whose coordinates (I, |l/3), are the solutions of the system. Example 3. Interpret the solutions of the system ( icH 2/^=9, I x—2y=S. The solutions are, ^^8 + 2i/^19^ ^^-16 + l/-19. ^^^^ 8— 21/-19 -16-1/ -19 -D 4-1 1 4.- x= ^ y= -• . Both solutions are imagmary. The graphs of these equations are the circle and the straight line MN, Fig. 9. It is seen that the circle and the line MN do not intersect or touch at any point. In general, ivhen the solu- tions of a system are imaginary expressions., the graphs of the equations neither cross nor touch. QUADRATIC AND HIGHER EQUATIONS 277 Example 4. Interpret the solutions of the system This system has four solutions : a?=T?^l/l3, 2/=Al/l3; x=-^\\/n, I/-TJ y= _-^6j|/i3 ; and ic= — t«^i/13 The graphs of these equations are the two ellipses in Fig. 10. The four solutions are the co- ordinates of the four points, A, B, C, X), where the curves meet. Example 5, Interpret ^ the solutions of the sys- J 4xH9?/^=36, jl/lo; X — y^ •Vt/13, tern X^ + ' This system has four solutions : x=i\/-15, y=l\/lO x=iV--15, y=-ii/lO x=-ii/-15, y=iVlO _ and a?=— f]/ — 15, y=—^i/l{). These solutions are imaginary. The graphs of these equations are the ellipse (1) and the circle in Fig. 10. It is seen that the circle lies entirely within the ellipse, and does not meet it at all. This is to be expected, since the abscissas of the points where they should meet are imaginary. (1) c=0. (2) Eliminating y from this system, observe that we have the general quadratic equation ax'^ + bx + c = 0. Since all solutions of the system are solutions of the quadratic ax'^ + bx + c=0, The graphic solution of the system j ^^ f^'^., 278 ALGEBRA it follows that the abscissas of the intersections of the graphs of equations (1) and (2) will be the solutions of the quadratic. Now since the equation y=x^ will be the same for all systems formed from the quadratic, to solve any quadratic we need merely to find where the straight line ay + bx + c = cuts the fixed parabola. Example 6. Consider the equation x^—x—2=0. This is equivalent to the system I y-x-2=0. The graphs are (a) and (&), Fig. 11. The abscissas of their points of intersections are —1, and 2, the solutions of the quadratic. Example 7. To solve the qua- dratic, ic^— 6a?+9=0, we find the intersection of the line y—Gx + 9=0 with this same parabola. The straight line touches the parabola at the point x=3 ; hence, the solutions are 3 and 3. It is thus seen that to solve any quadratic we have simply to find the graph of ay + bx + c=0, the parabola remaining fixed for all quadratics. In using this method make on coordinate paper, ruled to centimeters and millimeters, a perfect graph of y=x'^ for your fixed parabola. Now find two points on the graph otay + bx-\-c=0. Connecting these by your ruler find the abscissas of the points where the ruler crosses tlie parabola, and these abscissas will be the solutions of the given quadratic. -1— Y : 1 . 1 qi ' L J . I r- i - 1 I : T f . it _ i ^2 t 4 2_ Y I._Z . T t^ - ^ _^K I 2 t 71 V t 'V Zt 2 V 2 (tjZ z )• Fig. 11. QUADRATIC AND HIGHER EQUATIONS The solution of the system } p ^'^2 + ^^ _|_ ^^ 279 Here again the elimi- nation of y gives the quadrati c, ax^ + ^aj + c = 0. Hence, the abscissas of the intersections of the graph of 2/=0, which is the x — axis^ with the parabola^ y — ax^ -^hx-\- c, will be the solutions of the qua- dratic equation. Example 8. Solve the equation a;'-^ ^ 4x' + 3 = . The graph of y=x^— 4a? + 3 is the parabola of Fig. 12 which is cut by^ the x—axis, or y=0^ in points x=l and x=3 which are solutions of Y *. .g the quadratic. Note. — The method of the preceding system is preferable to this one, since by the method of this system a new parabola must be found for each equation. It is evi(ient that the graphic solution of equations may be employed with equations of degree higher than the second. Example 9 . To sol ve x'' — 3x + 2 ^ . To solve this, we may solve the system ( 2/=^-3^ + 2, (1) t 2/=0. (2) 280 ALGEBRA Some solutions of (1) are as follows I:* > y ^ ik x= — 3, i=-16; x=l, y=0\ x=—2, y=0; x=0, y=2\ j x=S, I y=2o X 20. From these we get the graph in Fig. 13. The x—axis cuts this graph at the point— 2 and touches it at the point 1. Hence, the three solutions of the equa- tion are —2, 1, and 1. Y Note. — In this and pre- Fig. 13. ceding chapters it has been seen that an equation in two unknowns may represent some kind of line, straight or curved, and that the solution of a system of such equations may be obtained by carefully plotting the curves and measuring the distances of theii- points of intersection from the two axes. Since the accuracy of tlie solutions depends upon the accuracy with which the curves are con- structed and the accuracy with which the coordinates of the points of intersection are measured, the graphic method is useful chiefly in dis- cussing the nature of the solutions rather than in determining the actual solutions. That an algebraic equation may represent a curve was first discov- ered by Descartes in 1637. The subject of Analytical Geometry is founded upon his discovery and is a discussion of curves by use of the equations which they represent. QUADRATIC AND HIGHER EQUATIONS 281 EXERCISE 87. Draw the graphs of the equations : 1. x' + i/' = 16. 4. a;'^ = 8y + l. 7. x'-f = ^. 2. Wx'+f = 16. 5. 2/^ = 8iK + l. 8. x'-\-^x + f = S, 3. 4ic2 + 25/ = 100. 6. x' = ii/ + 4. 9. x'-^xy-2y' = 0. Draw the graphs of the following systems, and interpret their solutions : 10. |"+'^r^' 11. |»^ + .'/=25, ,2^ |-^+.V'=4. ^'^' |i«-22/ = 4. ■^*'* |a;^ + 36/-36. (4a^^ + V = 36, .y (2/^-2^-1, 14, 16, f a;'+a;y-6y2 = 0, ^g ( £c^ + 16y^ = 16, lQx' + i/ = U. By the use of the graph find the solutions of Tl^ ic^ + 4« + 4 = 0. 22. a;2-2a5-4 = 0. 26.^ x'-2x-l = 0. X "2©» £c^-2a;-8 = 0. 23. x' = lQ. 26. a^='-5i« + 3 = 0. 21. a!'^-2a; + 4 = 0. ^V + £c-l = 0. 27. a;''-2a;=0. 28. What is the geometric interpretation of an imaginary solution of a system ? 29. What is the geometric interpretation of equal solutions of a system ? 30. How does the graph show that in general a system which is composed of a quadratic and a linear equation has two solutions ? 31. How does the graph show that in general a system composed of two quadratic equations has four solutions ? 282 ALGEBRA 184. Problems solved by systems involving quadratics. Some problems may be solved by solving systems involving quad- ratic equations. Example 1. Divide 9 into two parts the sum of the squares of which shall be 45. Let the parts be represented by x and y. Then from the conditions of the problem we get x^y=^, (1) and a^' + 2/'=45. (2) Equations (1) and (2) form a system whose solutions are These two solutions of the system give only one solution to the problem. The required parts are 3 and 6. Example 2. In running a mile a drive wheel of a locomotive makes 264 revolutions fewer than a wheel of the tender ; but if the wheel of the tender were 2 feet greater in circumference, the drive wheel would make only 176 fewer revolutions in a mile. Find the circumferences of the wheels. Let x= number of feet in the circumference of the drive wheel, and 2/= number of feet in the circumference of the other wheel. Then in running a mile the drive wheel makes revolu- tions, and the other wheel makes — - — revolutions. Hence, from the first condition, ^-5|«=264. (1) If the wheel of the tender were 2 feet greater in circumference, it would make — — ^ revolutions in a mile. Hence, from the second condition, 2/ + 3 X ' QUADRATIC AND HIGHER EQUATIONS 283 Equations (1) and (2) form a system whose two solutions are x=20, y=10 ; x=-7^, y=-'^2. From the nature of the problem the solutions must be positive. Hence, the second solution of the system must be discarded. Therefore, the circumference of the drive wheel is 20 feet, and the circumference of the other wheel is 10 feet. EXERCISE 88. 1. Find two numbers such that their sum shall be 13, and the sum of their squares 97. 2. Find two numbers such that the sum of their squares shall be 146, and the difference between their squares 96. 3. Divide 8 into two parts such that the sum of their cubes shall be 152. 4. Find two numbers such that their difference shall be 4, and the difference of their cubes 208. 5. Find two numbers such that the square of the greater shall exceed the square of the less by 64, and such that the square of the less shall exceed twice the greater by 16. 6. The sum of two numbers multiplied by their product is 160 ; and their difference multiplied by their product is 96. Find the numbers. 7. The sum of two numbers is 58;. and the sum of their square roots is 10. Find the numbers. 8. The sum of two numbers equals the difference of their squares; and their product exceeds twice their sum by 8. Find the numbers. 9. The difference of the fourth powers of two numbers is 255 ; and the sum of their squares 17. Find the numbers. 10. If a number of two digits be multiplied by its ones' digit, the product will be 124. If the digits be interchanged 284 ALGEBRA and the resulting number multiplied by its ones' digit, the product will be 156. Find the number. 11. Find the number of two digits which equals the square of the ones' digit, and which also equals 4 times the sum of its digits. 12. If the suni of the squares of two numbers be divided by the first number, the quotient Avill be 11 and the remainder 6. The first number exceeds the second by 6. Find the numbers. 13. The area of a rectangle is 36 square inches. If its length be increased by 3 inches and its width by 2 inches, its area will be doubled. Find its dimensions. 14. A rectangle is 20 inches long and 16 inches wide. How much must be added to its width and how much must be taken from its length, in order that its area may be increased by 22 square inches and its perimeter by 2 inches? 15. The hypotenuse of a right triangle is 10. If one leg be increased by 3 and the other leg by 4, the hypotenuse will become 15. Find the sides of the triangle. 16. A rectangular lot which contains 880 square yards is 2.2 times as long as it is wide. How mucl\ will it cost to build a fence around it at 25 cents a linear foot ? 17. A guy-rope to a derrick is attached to a stake 30 feet from the foot of the derrick. If the rope were 16| feet longer, it would reach to a stake 53 i feet from the foot of the derrick. Find the height of the derrick and the length of the rope. 18. A flower garden contains 1000 square feet, and is sur- rounded by a path 5 feet wide. The area of the path is 750 square feet. What are the dimensions of the garden? 19. If the numerator of a fraction be increased by 1 and the deuominatov be diminished by 1, the resulting fraction will be QUADRATIC AND HIGHER EQUATIONS 285 equal to the given fraction inverted ; and 3 times the numer- ator of the given fraction exceeds 2 times the denominator by 3. Find the fraction. 20. A laborer received $32.50 wages. If he had worked 5 days longer, and had received 50 cents a day less, he would have received $41.25. How long did he work, and what were his wages a day ? 21. A certain number of men do a piece of work in a certain number of days. If there were 2 fewer men, it would take 4 days longer to do the work ; and if there were twice as many men, it would take 4 days less to do the work. Find the num- ber of men and the number of days required for them to do the work. 22. A grocer bought apples and potatoes for 154. He sold the apples for $36.80 and the potatoes for $18.70. He gained as many per cent on the apples as he lost on the potatoes. How much did he pay for each ? 23. A certain sum of money placed at simple interest for one year amounted to $265. If the principal had been $50 more and the i*ate 1% less, the interest would have been the same. Find the principal and the rate. 24. Two men can do a piece of work in 4| days. It would take one four days longer to do the work than it would take the other. How long would it take each of them to do the work ? EXERCISES FOR REVIEW (V). 1. What is the difference between a surd and a rational ex- ession f %, What determines the order of a surd ? pression f 286 ALGEBRA 3. When is a surd in the simplest form? By what prin- ciple may a surd be reduced to the simplest form ? 4. Simplify y'x'ip ; 2 1 1 65* {a —lif\ v^x^ — ^x^y + ^xy"^ ; y_l_ /a + 6 (x—yy y a—Q 5. What kind of surds may be added or subtracted ? 6. Simplify i/Sa-2VSla' + Syl92ab\ 7. Add i/a'x\y^z)\ i/9aV(y + 2;), and 3i/16(y + s)^ 8. Add tVv "72, -ii i, and 6i/21i. 9. By what principle do you change the order of a surd ? 10. Change i/5, y 11, i/13, to surds of the same order. 2d 3/27^ 11. Write as an e7itire surd o Sx 12. Which is the greater, i 5 or v TO"? 13. Write as one surd|/^ y 257 14. By what principle are surds multiplied ? Illustrate. 15. Find the product of i/Sx, y 27x\ and y9x\ 16. Find the product of Si/i + S]/ 5 and Qy2 + 7yW. 17. What is the corijugate of 2 + |/5 ? 18. Give a method for dividing by a surd. Illustrate. 15 + 3i 3 19. Simplify 15-2|/3 20. Find the value of (2i/a=^6)^; {bx\ 96a;«y; (v 48icy)l 21. Find the value of y^/^lxY^i i/vV + 6a;+9; l/^lx'yxXx-yY. QUADRATIC AND HIGHER EQUATIONS 287 22. Simplify: (''> 7I/3-5V2- ^^ Vf^x- 31/5 + 51 g (/) (I'/^^M?)/ ^ '' V/5-V3 ^) 2)/n5-3i/63 + 5v"28. W ^/s-!v^^) ^(f7)3i/f+i/TV + 4i/^V W l/2Xi/3xi/iXl/i. ^ 23. What is an imaginary number ? Illustrate. 24. What is the typical form of an imaginary number ? 25. Reduce ]/— 4ic* to the typical form. 26. What are the values of the successive powers of i/— 1? How is ^/^-i usually represented ? 27. Simplify: (a) V- -36a^- -l/-49a^ + l/- 81«^ ip) v~- ^•1/- -4 i/-l(>. (^) (l/' -2 + 1 /-5)(i/-2-i/ -5). (^0 (1/ -12- -l/-15)-i/-3. 28. , Write the conjugate of Z-\-y ■ -5. 29. What is a complex number f Illustrate. 30. Square i/^^S-j/'^^. 31. Divide 3 + |/^=5 by 4-i/^=^. 1/^^+3 32. Simplify ^-^_^^-=^' 33. How many roots has a quadratic equation in one unknown ? What kind of numbers may they be ? 34. Give an example of a 2yf(re quadratic equation in a. 288 ALGEBRA 35. How do you solve any pure quadratic ? How is the solution of any equation checked? 36. Solve and check : {a) 4£c^-3 = 0. {h) 3«2 = 9a2-54. (c) ^{t^\)-t{t-\)=^t. 37. What is a complete quadratic equation? Give two general methods of solving such an equation. 38. Solve by factoring : {a) 2a;^ + 9a;-5. {h) 6y^ + 5y-6 = 0. (c) 1-\%x = ^x\ 39. Solve by completing the square. (Always check your solutions.) (a) i«^-4£c = 32. {h) 2ic^ + 9;K-5 = 0. (c) SGaj'^-aea. + S^O. 40. Solve by the quadratic formula : {a) 8a;^ + 2i« = 3. (c) 3ic2 + 35-22a; = 0. {h) x^-\x=\. (d) 4x' + 17x=U. Solve the following for the general number : 41 J_ = _L+^ ' 42 ^zL^.?:z8=_?^+l \\ • a-\ a-2^a-'f ^- x+3^ x-3 x^-9^'2' ■ y-\ 2y ■ 44. In changing a, /*r«c^/o/2a/ equation to an integral equation, by what expression, in general, must the members of the equation be multiplied ? 45. How can you tell the nature of the roots of a quadratic without solving the equation? What is meant by the dis- criminant of a quadratic ? 46. Solve ax^-^hx^c = ^ and give the relations that must exist among the coefficients for the different kinds of roots. 47. What must be the value of m if the roots of ^tx^—lx-^m = are equal ? QUADRATIC AND HIGHER EQUATIONS 289 48. What relations exist between the roots and coefficients of x''+px + q = ? 49. Construct an equation whose roots are 3 and — ^. One whose roots are l + j/f and 1 — i |. 50. Explain the method of stating and solvmg a problem by means of an equation. 51. Can you solve an equation of a?i(/ degree ? What special kinds of equations of higher degree can you now solve ? 52. Solve the following : (a) x*^Q = bx\ 53. How many solutions has an equation of the 4th degree in one unknown ? One of the 5th degree ? One of the nth degree ? 54. What is an irrational equation in a; ? 55. Give the general method of solving an irrational equation. 56. When is a solution said to be introduced in a derived equation ? 57. Show that, in general, when both members of an equation are squared, 7ieia solutions are introduced? 58. Solve and check : (a) '6\/x=i/xTS + VxTQ. {b) 1/7 -a; 4-1 = 1/205 + l-l/l+a; i/l-a; ^''^ i+i/rT^~i/rF^* 59. How many cube roots has a number? How nmny fourth 290 ALGEBRA roots? How many nth roots? Find the cube roots of 1. Of 8. Of 27. 60. How many solutions, in general, has a system of one linear and one quadratic equation ? A system of two quadratic equations ? How is this illustrated by the graphs ? 61. What is a defective system? Illustrate. 62. What represent the solutions in the graphs of a system of two quadratic equations ? 63. Solve the system { f +/;t^lr'' Cr 64. The area of a circle is equal to -^, where G represents the number of units in the circumference, and r the number of Cr units in the radius. From ^=_-, and (7=2;:r find ^ in terms of r and r. Sr 65. From F=-q-j ^==4-^^, and r=\I>^ find V in terms of r o and TT ; also in terms of U and r. CHAPTER XVIII. INEaUALITIES. 185. Definitions. If a— ^ is positive, a is said to be greater than h. If a— ^ is negative, a is said to be less than b. The symbol for ''Hs greater than^^ is >, and the symbol for "is less than^^ is <. Thus, a>6 is read, "a is greater than 6." And a<&isread, " a is less than 6." From the above definitions it follows that when a — h is positioe^ ay>b, and when a—b is negative, a<^b. Thus, since 6—4=2, therefore 6>4; since 5— 8=— 3, therefore 5<8. The statement that two expressions are not equal, i. 6., that one of the expressions is greater or less than the otlier, is called an inequality. The expression at the left of the inequality sign is called the first member of the inequality, and the expres- sion at the right of the inequality gign is called the second member. Two inequalities which have the same sign of inequality are called inequalities of the same species. Two inequalities which have opposite signs of inequality are called inequalities of opposite species. Thus, 15>12 and 7>— 2 are of the same species ; and 6>— 4 and 9<17 are of opposite species. In this chapter, the letters used in the members of an inequality represent 07ily positive real nimibers. A negative number will be denoted by a negative sign. 291 292 ALGEBRA 186. Principles of inequalities. The following are some useful principles of inequalities. (1) If equal numbers be added to, or subtracted from, the members of an inequality, the result will be an inequality of the same species ; that is, . if fl>6, then a-\-c^b^c, and a—d^b—d. For, since a> b, then a — b\& positive. Hence, (a + c) — (6 + c), which equals a — b, is also positive. Therefore, a + c>^» + c. § 185. In like manner, a—d^b—d. Evidently the proof would have been similar, had the given inequality been a7, therefore 10 + 6>7 + 6, and 10-4>7-4. {2) If the members of an inequcdity be tnidtiplied, or divided, by equal positive numbers, the result ivill be an inequality of the same species ; that is, if a>6, then ac^bc, and->-. c c For, since a>^, therefore a — ^ is positive. Hence, c(a — b), or ac — bc, \s positive when c is positive. Therefore, acybc. § 185. In like manner, ->-, when c is positive. Thus, if o + 2>4,then multiplying both members by 3, we have i» + 6> 12. And if ocf -f 2xy Qx, then dividing both members by a?, we have aj + 2>6. (3) If the corresponding members of tivo inequalities of the same species be added, the residt will be an inequality of the same species ; that is, if fl> b, and c> (/, then a + c> 6 -h (/. INEQUALITIES 293 For, since a>^, and c>r/, therefore a — h and c—d are both positive. Hence, their sum, a— b + c—d, is positive ; i. e., (a-\-c) — (b + d) is positive. Therefore, a + c>i + (^. §185. Thus, if 2a?— 2/>10, and x+y^4^ by adding them we have 3x>U. (4) If the members of an inequality be subtracted from the members of an equation^ the result loill be an inequality of oppo- site species ; that is, if a>6/ and c = d, then c—a9, therefore 20-15<20-9. • (5) If the members of an inequality be subtracted from^ or divided by the members of another inequality of the same species., the residt is not necessarily an inequality of the same species ; that is, if fl>6 and c>(/, then a—c may or may not >6— (/ and - may or may not >;^. C Q Thus, 8>4 and 7>2, but 8-7<4-2, and |<|. Also, 8>4 and Q>2, but 8-6=4-2 and |<|. {6) If the members of an inequality be multiplied^ or divided., by equal negative numbers., the residt will be an inequality of opposite species ; that is, if fl>6, then acJ, therefore a—b is jyositive. Hence, ac—bc is negative when c is negative. Therefore, ac<,bc. § 185. Similarly, -<-• Thus, since 9>6, therefore 9(-2)<6(-2), i.e., -18<-12. (7) If the signs of all terms of an inequality be reversed., then the symbol of inequality must also be reversed This follows directly from (6) when c equals —1. Thus, if 3ic-22/ + 6>a-46, then 2y-^x-Q<4.b-a. Note. — These are but a few principles of inequalities. There are many others which could be established by similar processes. 187. Identical and conditional inequalities. Inequalities which are true for all values of the general numbers involved are called identical inequalities. Thus, a + 10>a is an identical inequality. Inequalities which are }iot true for all values of the general numbers involved are called conditional inequalities. Thus, 6a? + 2>0 is true only under the condition that x is greater than —^. Note. — It is to be observed that identical and conditional inequalities are analogous respectively to identical and conditional equations. 188. Proofs of some identical inequalities. Some identical inequalities may be established by use of the pJ-inciples in § 186. Example 1. Show that a'^ + b^^2ab, if a and b are unequal. Since a and b are unequal, therefore, {a—byy>0, or a'-2ab-i-b'>0. Adding 2a6, a' + b'>2ab. §186, (1) INEQUALITIES 296 Example 2. If a and h are unequal, and a + 6>0, show that We have a^ + h''>2db. Ex.1. Subtracting a6, a^-ab + W^ah. §186, (1) Multiplying by a + 6, d'-\-lf^a'h^ah\ §186, (2) Example 3. The sum of any positive number n and the quo- tient - is greater than 2. We have a' + h'''>2db. Ex.1. Now let d^ be n and let If be -. n Then, substituting in the above inequality, we have - » + i>2- 189. Solving conditional inequalities. A conditional in- equality is said to be solved when the possible values of the unknown numbers which will satisfy it are found. The range of these values is discovered by means of principles such as those established in § 186. Example 1, Find the possible values of x in 37-2^ ^ 3x-8 ^ 3— + a'> ^-9. Multiplying by 12, 148-8^+ 12a;>9x-21-108, § 186, (2). or 148 + 4;r> 9^-132. Subtracting 148 + 9aj, -5ic>— 280. §186,(1). Dividing by -5, xQx+16. Subtracting 6ic + 16, x''-6x-16>0. §186,(1). Factoring, (x-8){x + 2)>0. Hence, the factors a?— 8 and x + 2 must both be positive or both be negative, 296 ALGEBRA To make both positive, x must be greater than 8. To make both negative, x must be less than —2. Hence, a^>8 or <— 2. Example 3. Find what values of x will satisfy the system 10a?>3x + 49, (1) a7+5<|+55. (2) From (1), x>7. From (2), x<75. Hence the system is satisfied by any value of x between 7 and 75. Example 4. Find what values of x and y will satisfy the system \ 3j! + i/>10, \x^y=Q, Subtracting (2) from (1), 2x>2. Whence, ic>l. Multiplying (2) by 3, 3a: +3?/= 24. Subtracting (3) from (1), -2i/> -14. Dividing by —2, 2/<7. Hence, any solution of (2) in which x>l and y2i/aJ. 6. a* + b*:^a'b\ Suggestion : Let a=x^ and h=y'\ (1) (2) §186, (1). §186, (2). (3) §186, (1). §186, (6). INEQUALITIES 297 Find what values of x will satisfy the following: 7. 2aj-3>7. 11. a;2-2a;>3. 8. ^-hx<:^x-l\. 12. _l_3- 13. 6 _ 3 8 a?— ic— 4 tc — 3 10. -1- <_L 14. r^^a<^_\ Find the values of x that will satisfy the following systems : .. (3ii;-5<2a;+l, r3a;-16 5 ^^- l3a^ + 4>x + 8. ^g I— ^<3' 17 (3(0^-2X2(0^-3), r4^ ^ a; ^ l2a3-7>5.^+5. !^l>^=r2 + 3, .„ j^(^ + 3)>ic(i«-5) + 16, 1 2a; , p^ 4x ^°- |3a;(a; + 2)<3a;(a;-l). L.'« + 3'^'^^a; + 7' Find the values of a? and y that will satisfy the following systems : 21 ji«+2/=10, ric + 2 24 22. ^ 3a; + 2/>14, 3" +4y>2, y + 11 ^+1^-, 11 2 |a; + 2y=13. (7a; + 4y = l, , __ _ 23. Il^ + f^^i' 25. |?^ + *y>l. CHAPTER XIX, RATIO AND PROPORTION. 190. The Ratio of one number to another number of the same kind is the quotient obtained by dividing the first num- ber by the second. It follows from the definition that the ratio of two numbers is an abstract numher indicating the num- ber of times one contains the other^ or the part one is of the other. Thus, the ratio of 6 to 2 is 6-^2, or 8. The ratio of $8 to $4 is 2. The ratio of 5 ft. to 8 ft. is 5-i-8, or |. The ratio of a to Z> is a-=-6, • a OV y It follows that an indicated ratio is a fraction. Hence, a ratio may be expressed by any of the signs used to express a quotient or a fraction. Thus, the ratio of 3 to 4 may be written 3^4, 3/4, |, or 3:4. In the ratio between two numbers the dividend, or numera- tor, is called the antecedent, and the divisor, or denominator, is called the consequent. The two together are called the terms of the ratio. The ratio of a to b is sometimes called the direct ratio of a to b. And the ratio of J to « is called the inverse ratio of a to b. Thus, the direct ratio of 3 to 7 is f . The inverse ratio of 3 to 7 is |. The inverse ratio of one number to another is the direct ratio inverted. 298 RATIO AND PROPORTION 299 191. It is clear from the definition of a ratio that all of the laws established for fractions must apply to ratios. Ratios may be reduced to higher or lower terms. They may be added, subtracted, multiplied, divided, raised to powers, and have roots extracted. Ratios may be compared by reducing them to a common denominator, and comparing the resulting numerators. Ratios are compounded by taking their product. Thus, the ratio compounded of | and f is -^%. 192. Commensurable and incommensurable numbers. Two numbers whose ratio can be exactly expressed by two whole numbers are called commensurable numbers, i. e., two numbers are commensurable when there can be found some third number of the same kind, called their common measure^ that is contained an integral number of times in each. Two numbers whose ratio can not be exactly expressed by two whole numbers, i. 6., two numbers that have no common measure^ are called incommensurable numbers. Two fractions are commensurable if their numerators and denominators are commensurable. For the ratio of two such fractions can be expressed as the ratio of two whole numbers. Thus, the ratio of | to A=l-A-|-V-=!|. If the ratio of two numbers is a surd, the numbers are incom- mensurable. Thus, the ratio of v 2 to V^=y-^ = ^—^- The ratio can not be expressed by two whole numbers, hence \ 2 and i/5 are m- commensurdble. The ratio between two incommensurable numbers is called an incommensurable ratio. 193. Proportion. A proportion is an equation each of whose members consists of a ratio. Four numbers, a^ b, c, d. 300 ALGEBRA are said to be in proportion when the ratio oi ato b equals the ratio of c to d. The most used forms in which the proportion may be writ- ten are (1 c T = -vand a : b=c : d, sometimes written a : b::c : d. This proportion is read " a divided by b equals c divided by f?," or " a is to ^ as c is to t/." The terms of the ratios in a proportion are called the terms of the proportion. The first and fourth terms are called the extremes, and the second and third terms are called the means of the proportion. Thus, in 2:x=Qx'^:Sixf, 2 and So(^ are the extremes, and x and 6x^ are the means. a c In the proportion T=-p d is called the fourth proportional to ay by and c. 194. In the following sections are given the most im- portant principles in proportion. The student should master these principles. • • . 195. In any true proportion the product of the extremes equals the product of the means. Suppose a : b=c : d. a c Written in fractional form, ^=^. Multiplying by bdy ad^bc, which proves the principle. Thus, in 2: 4=6: 12, 2x12=4x6. 196. If the product of two iiumbers equals the product of two other numberSy then these four numbers form a proportion in ichich the extremes are the factors of either product and the means are the factors of the other product. ad= be. a c b-d' c a d b i> d RATIO AND PROPORTION 301 Suppose that Dividing by bd, or dividing by ac, -=7:^ or 7, = -j etc. ; which proves the principle. Thus, since 2x10=4x5, therefore 2: 5=4: 10, or 5: 10=2; 4. It follows from this principle that to test the correctness of a proportion it is sufficient to shoio that the product of its extremes equals the product of its means. Thus, 2 : x=6a?^ : Zj? is a correct or true proportion, for 2 • Zd^=x ■ Qx\ 197. Remark. It should be noted that by the principle of § 196 we get from the equation ad=bc the eight proportions : 1. a : b = c : d; 3. d : b = c : a; 5. b : a = d : c ; 2. a : c=b \ d\ ^. d : c = b : a; Q. b : d=a : c; 7. c : a = d : b; 8. c : d=a : b. It should also be noted that sii>ceany one gi\esad=bc, from any one of them all the others follow. Observe that (2) was obtained from (1) by interchanging the means, and that (3) was obtained from (1) by interchanging the extremes. The old mathematical terms for such changes are, respectively, mean alternation and extreme alternation. Observe also that (5) was obtained from (1) by interchanging the terms of each ratio. This process is called inversion. In (4) we observe that it is obtained from (1) by inter- changing the means and the extremes. Illustrate each of the 8 proportions given above with numbers. 198. The products of the corresponding terms of two or r>iore proportions form a proportion. 302 ALGEBRA Suppose that a c b-7( mp and x_io y~lj' Multiplying, am X cpw buy d q V or amx cpw bny cJqv Hence, amx^ buy, cpw, and dqi^ are in proportion. Axiom 3. 199. Like powers or like roots of the terms of a proportion are in proportion. Suppose that a : b = c : d. or a c 1>-71' Then, it)- -($)-. Axiom 5. or \ / \ / b"" d"' Again, n-vi Axiom 6. n — n — \/a 1 c or „ _ yb yd Thus, since 2 : 3=4 : 6, therefore 2=^ : ^^=^'' : 6^ or 8 : 27=64 : 216. And since 4 : 25=16 : 100, therefore |/4 : |/25=i/T6 : i 100, ©r 2 : 5=4 : 10. 200. The terms of a proportioii are in proportion by addi- tion ; i. e., the sum of the first two terms is to the first (or second) as the sum of the last tico terms is to the third (or fourth). RATIO AND PROPORTION 303 § 195. Suppose that a c b-d' Then, ad— he. Adding bd^ ad-^hd=bc-\^bd. or d(a + b)=b(c-i-d). Hence, a-\-h c + d b d ■ In like manner, a-{-h c^d § 196. § 196. a c Thus, since 3 : 5=12 : 20, therefore (3 + 5) : 5= (12 + 20) : 20, or 8 : 5=32 : 20. The old mathematical term for this process is composition. 201. The tenuis of a 2^^02)ortio)i are in pro2^ortion by svh- tr action ; i. e., the difference of the first tico terms is to the first {or second) as the diff'erence of the last tvm terms is to the third {or fourth). a c Suppose that t— ;/• Then, ad=bc. § 195. Subtracting bd^ ad—bd=bc — bd, or d{a — b)= b(c — d) . Hence, ^^JlI^. § 196. T vi a—b c—d In like manner, = a c Thus, since 10 : 3=30 : 9, therefore (10-3) : 3=(30-9) : 9, or 7 : 3=21 : 9. The old mathematical term for this process is division. 202. The terms of a proportion are in proportion by addl- tion and subtraction ; i. e., the sum of the first two terms is to their differenae as the sum of the last tico terms is to their dif- ference. 304 ALGEBRA Suppose that a c Then, h d ' And Dividine' a—h c—d b d ' a + b_c-{-d a—o c — d' §200. §201. Axiom 4. This and the preceding sections will enable us to obtain an equivalent proportion from a given proportion. (X c Example. Suppose T,~d' ^^^ By §199, p=^. (2) By § 201, ^JL^. (3) By § 199, ^ -^ (4) 203. Continued proportion. When, in a series of numbers, the first is to the second as the second is to the thirds and so on, the numbers are said to be in continued proportion. Thus, a, 5, c, d, e, are in coyitinued proportion, if T— 7=^— J' In the continued proportion ^=-, which is called a mean pro- portion, b is called a mean proportional between a and c, and c is called the third proportional to a and b. Thus, 4: 10=10: 25 is a mean proportion. 10 is the mean propor- tional between 4 and 25. And 25 is the third proportional to 4 and 10. 204. The tnean proportional between two mimhers equals the square root of their product. RATIO AND PROPORTION 305 Let b be the mean proportional between a and c. mu ah Then, -t = - * . he And }f = ac. §195. Hence, h=\/ao. a m _/>_ X b 71 <1 y a = 7*. Thus, in the proportion 9: 18=18: 36, we have 18 = 1/9 x 36. 205. In a series of equal ratios the sum of the antecedents is to the sum of the consequents as any antecedent is to its con- sequent. Suppose that Let Then —=^r.~ = r.- = r. etc. Axiom 7. n ' q ' y Clearing of fractions, a=rb^ m=rn^ p = rq^ x — ry^ etc. Adding, a + m-fp + a;+ • • • • =rb^rn^rq^ry^r • • • • Dividing by b^rn-\-q-^'y^r • • • • » a-Vm-\p^-x■^- am r^"—. , =r=7; =— =etc. o-\r n-^-q^y^ o n 2_4 8 16 2 + 4 + 8 + 16 30 ihus, 3~6~12~24'-3 + 6 + 12 + 24 ^^ 45' 206. The representation of a ratio by a single letter., as in § 205, is often a useful process. In order to test the accuracy of a proportion, we may proceed as in § 196, or as in the example in § 202. But often it is more convenient to rep- resent the equal ratios by a single letter, and then show that the proportion reduces to an identity, 20 306 ALGEBRA Example. If -r = :^, show that / .. , ~ 3. - Let r=r. Then ^=r. Hence, a=br^ and c=dr. Substituting, }AV+£^vV^+£ VdV + d' y^d'r^ + d' Factoring and reducing, b\/?Tl b V'r^ + l dVr^^Ti~d^^r^ or d~d' ^^ identity. Hence, the proportion is true. 207. Since a proportion is an equation, it may be solved for any number in it. Example 1. a : b=c : d. Find the value of d. a c Writing in fractional form, x=^' Multiplying by 6d, ad=bc. be Dividing by a, ^~a' Example 2. Find the fourth proportional to 3, 2ic^ and 21a?. Let a represent the fourth proportional. Then. ^3=-- Multiplying by 2ax^, Za=42o(?. Dividing by 3, a=14a?^. Example 3. Find the third proportional to a' and a6'. Let X represent the third proportional. Then, • 4=2^. ' db^ X Multiplying by db^x^ a^x=a^b*. Dividing by a^, x=b^. RATIO AND PROPORTION 307 EXERCISE 90. 1. What is the inverse ratio of 6 to 11? 2. Reduce the ratio of 10 to 18 to lowest terms. Find the simplest expression for the ratio of : 3. bx to Ihx^. 4. - to -T. 5. 7 rr, to a h' ' \x—yy {x—yf 6. Which ratio is the greater, -J^ or || ? 7. Write in descending order of magnitude |, i|, f ^. 8. Write the ratio co^npounded of the ratios J, f^^ |. 9. Write the ratio compounded of the ratios -, 'A, tt-t^. (a\l>V 12 10. Write the ratio compounded of the ratios ^^. — --, —ry 11. What must be the value of a, if the ratio -— -^ equals the ratio | ? 12. What must be added to each term of I to make it equal to I? 13. Two numbers are in the ratio of 3 to 4, and their sum is 35. Mn4 the numbers. 14. Two numbers are in the ratio of 7 to 5, and their differ- ence is 6. Find the numbers. 15. Two numbers are in the ratio of 2 to 9, and the difference of their squares is to the square of their difference as 11 is to 7. Find the numbers. 308 ALGEBRA Find i\\Q fourth proportional to : 16. 3, 4, and 36. 18. p>, v, pv. 20. «, h, c. 17. X, ^x\ bx\ 19. 1, 7, 10. 21. x, y, z. Find the third proportional to : 22. 1, 2. 24. p, pq. 26. a, Z>. 23. X, '6x\ 25. 10, 1. 27. ^, y. Find the mean p)roportional between : 28. 2, 8. 30. x\ ^x\ 32. a, h. 29. 15, 135. 31. 6, 9. 33. x, y. 34. In the proportion T^p'^ 17772' solve for h. 35. Show that | and 5 are com.mensnraMe numbers. 36. Show that ^ and 2J are commensurable numbers 37. Show that 7/2 and \^\^ are commensurable. 38. Show that f^'I and f/3 are incommensurable. 39. Show that ]/15 and 7/5 are incommensurable. 40. Show that 7 : 12 = 21 : 36 is a true proportion. \i a : b=^c : (7, show that : 41. a^ : b''=-ac : M 42. a& : bd'^ = a^c : ^V?. 43. (5a + Z>) : ^>=(5c + (^?) : ^?. 44. a : (a + c) = (a + ^>) : {a^b^c^d). 45. c : d=-f : -. 46. 2a : 5c = 2J : 5r?. 47. a : b=\a~c : 1/^. 48. {a^b) : (c + f?) = 7/aMT' : v"eTd\ 49. 7/c«^-^»=^ : 7/c^-' : yc'^d\ RATIO AND PROPORTION 309 50. {a-b) : {c-d)=^d^^' : ^&^^\ 51. {a'c + ac') : {b'd+bd') = {a + cf : {b^-d)\ 62. The sides of a triangular field are as 4 : 5 : 6, and the distance around it is 1,200 yards. What are the sides ? 53. What must be added to 2, 5, and 12, in order that the fourth proportional to the sums may be 24. 54. Express the ratio of 3^ to 1^ by the ratio of two whole numbers. 55. The difference between two numbers is 4, and their product is to the sum of their squares as 6 is to 13. Find the numbers. 56. A tree casts a shadow 80 feet long, when a rod 4 feet high casts a shadow 5 feet long. How high is the tree ? CHAPTER XX. VARIATION. ALGEBRAIC EXPRESSION OP LAW. 208. Variables and constants. In many problems in mathe- matics, numbers are involved %chose values are changing. Such numbers are called variable numbers, or variables. For distinction, numbers whose values do not change during the investigation of a certain problem are called constants. For example, one's age is a variable magnitude, expressed by a 'number which is always increasing. The weight of a stone lying on the ground is a constant magnitude, expressed by a r>umber of pounds which does not change. Two or more variables may be so related that a change in the value of one will cause corresponding changes in the values of the others. This relation between the variables is expressed by means of an equation. Thus, in case of a running train, the distance, time, and speed are varnables, and their relation is expressed by the equation vt=d, where v, ^, and d represent the speed, time, and distance, respectively. If the variables, x and y, are connected* by the equation y=x^^ then when x assumes the successive values 1, 2, 3, 4, 5, 6, 7, 8, 9, etc., y assumes the corresponding values 1, 4, 9, 16, 25, 36, 49, 64, 81, etc. If one variable depends for its values upon the values of another, the first is called a dependent variable, and the second is called an independent variable. 810 VARIATION. ALGEBRAIC EXPRESSION OF LAW 311 Thus, in the example above, if x is an independent variable, y is a dependent variable, for as x changes in value y also changes, and the value which y assumes depends upon the value which X assumes. 209. Direct variation. If two variables are so related that during all of their changes, their ratio remains constant^ each is said to vary as the other, or to vary directly as the other. As the first increases, the second also increases ; and as the first decreases, the second also decreases. If X varies directly as y, their relation is expressed by the equation ~=k, or x=ky, where ^ is a constant. The symbol oc is called the siyn of variation, and is read " varies as." Thus a cc b is read " a varies as 5," and is equiv- alent to the equation t=^, or a=kb- Example 1. xcc y^ and when y=4t, x=20. Express the rela- tion between the variables. Let x=hy. This equation must be satisfied by 2/=4 and x=2Q. Substituting, 20=4/b. Hence, A?=5. Therefore, x=5y is the required expression. 210. Inverse variation. One variable is said to vary in- versely as another when, during all of their changes, their j^roduct remains constant. As one increases, the other must decrease. If X varies inversely as y, their relation is expressed by the equation xy=k, where ^ is a constant. Example 1. Given that x varies inversely as y, and when a?=2, 2/=4. Find the relation between them. 312 ALGEBRA Let xy—k. This equation must be satisfied by ic=2, 2/=4. Substituting, 2-4=A;. Hence, fc=8. Therefore, xy=8 is the required equation: 211. Joint variation. One variable is said to vary as two others jointly when the first varies directly as the product of the other tico. If X varies as y and z jointly^ their relation is expressed by the equation x=kyz. Example 1. Given that a varies as h and c jointly, and when a=36, 6=3 and c=2. Express the relation between the variables. Let a=kbc. This equation must be satisfied by a=36, &=3, c=2. Substituting, 36=:fc-3-2. Hence, k=6. Therefore, a=66c is the required equation. 212. One variable is said to vary directly as a second and inversely as a third, when it varies as the quotient of the second divided by the third. If t varies directly as d and inversely as y, the relation be- tween them will be expressed by the equation ^=Z;-, where ^ is a constant. Example 1. Given that t varies directly as d and inversely as r, and when f=20, d=8, and v=2. Find d when t=\i) and v=^. Let t=k -■ V This equation is satisfied by f =20, cZ=8, u=2. Substituting, 20=|A?. Hence, k=5. VARIATION. ALGEBRAIC EXPRESSION OF LAW 313 Therefore, the equation becomes When ^=10 and i;=3, this equation becomes 10= sf. Hence, d=6. EXERCISE 91. 1. fcocy, and when 2/==5, x=lb. Write the equation be- tween tliem. 2. tcoc 2/, and when 2/ = 2, x=VI. Find x when y = l. 3. X varies inversely as y, and when 2/ = 3, a; = 8. Write the equation between them. 4. X varies inversely as y, and when i/^lO, £c=5. Find a; when 2/ ==2. 5. X varies jointly as y and ^, and when 2/ = 5, and^=l, £c=40. Write the equation between them. 6. X varies jointly as y and z^ and when y = ^ and 2 = 5, 05=150. Find z when a; = 120 and 2/ = 6. 7. X varies directly as y and inversely as ^, and when y=l^ and 2=2, 93 = 24. Find y when a:; = 14 and 2; = 2. 8. «^cc2/^ and when 2/^4, £c = 16. Find the equation be- tween X and 2/. 9. The velocity of a body let fall toward the ground varies as the time during which it has fallen from rest, and the velocity at the end of 3 seconds is 96 feet per second. Write the equation between the velocity and time. 10. The area of a triangle varies jointly as the altitude and base. When the altitude is 4 inches and the base 9 inches, the area is 18 sq. inches. What will be the area when the base is 6 inches and the altitude 10 inches ? 3l4 ALGEBRA 11. The area of a circle varies as the square of its radius. When the radius is 6 feet the area is 113.0976 square feet. What is the area when the radius is 8 feet ? 12. The vohime of a gas varies inversely as the pressure upon it. When the pressure is 8 lbs., the volume is 8 cubic inches. What is the volume when the pressure is 4 lbs.? 13. The distance a body falls from rest varies as the square of the time. In 2 seconds it falls 64 feet. How far will it fall in 4 seconds ? How far will it fall during the fourth second ? 14. Tlie number of oscillations made by the pendulum of a clock in a given time varies inversely as the square root of its length. A pendulum 39.1 inches long makes one oscillation in a second. How long must a pendulum be to make 4 oscil- lations in one second ? 15. In playing see-saw the two persons must sit on the board at distances from its point of support which vary inversely as the weights of the persons. Where must the support be placed under a board 16 feet long in order that two girls Aveighing respectively 65 lbs. and 80 lbs. may play see-saw? THE ALGEBRAIC EXPRESSION OP LAW. 213. Law expressed by the equation. An equation contain- ing two or more variables expresses a law according to which the variables change their values. Since there is a relation between the variables, they can not each assume any values at random. There is a fixed laio according to which they must change. This laio is expressed hy the equation. Thus, Newton's law of bodies falling from rest is expressed by the equation s=^t'\ where a is a constant, and s and t variables, s representing the number of feet through which the body will fall in t seconds. VARIATION. ALGEBRAIC EXPRESSION OF LAW 315 It evidently expresses the law that the distance through ivhich a body will fall from rest varies directly as the square of the time. Again, Boyle's law of gases is expressed by the equation VP=k, where Ar is a constant, and V the volume of a gas when under the pressure P. The equation evidently states that the volume of a gas varies inversely as the pressure upon it. When a law is known, the equation which expresses it can be written. Example 1. The distance through which an object will move with uniform motion varies as the velocity and time jointly, and it is found by trial that the object with a velocity of 20 feet per second moves 60 feet in 3 seconds. Write the equation which expresses the law of motion. If s= distance, v= velocity, t= time, we have s=kvt. Since s=60 when v=20 and ^=3, 60=A;-20-3. Hence, k=l. Therefore, s=vt expresses the law. EXERCISE 92. Write the equations which express tlie following physical laws. 1. The force with which moving bodies having the same velocity strike a stationary body varies directly as the masses of the moving bodies. 2. The attraction between two bodies varies directly as the product of their masses, and inversely as the sqmire of the distance between them. 316 ALGEBRA 3. If an object is revolved at the end of a string, the tension exerted upon the string by the object varies directly as the product of the mass of the object and the square of its velo- city, and inversely as the length of the string. 4. The time of vibration of a pendulum varies directly as the square root of its length. 5. The downward pressure upon the bottom of a vessel con- taining a liquid of given density varies jointly as the depth of the liquid and the area of the bottom of the vessel. , Y , \ / \ / \ / ^ / y / \ / \ / A. V. y V Fig. 14. 6. The amount of heat re- - ceived by a body from % radiat- ing source varies inversely as the square of the distance of - the body from the source of heat. 7. The resistance offered by a wire to a current of electri- - city varies directly as the X' length of the wire, and inverse- ly as its cross-sectional area. 214. As another illustration - of the expression of law by an equation, let us consider the graph of an equation with two unknown numbers. The graph of the equation y=x^ is shown in Fig. 14. Here x and y represent the coordinates of any point P on the graph. If we consider x and y as variables, by changing their values the point P can be made to move. As the absolute value of VARIATION. ALGEBRAIC EXPRESSION OF LAW 317 X increases, the point moves away from the line FP, and as the absolute value of y increases the point moves away from the line XX'. By making x and y assume as values all possible solutions of the equation, the point Pcan be made to take up in succession all positions on the curve. Hence, the graph may he considered as the path of a point moving subject to a law. And the equation between the coordinates expresses the law governing the m,otion of the point. The point must so move that its coordinates continually satisfy the equation. This principle is of great use in mathematical sciences such as astronomy and physics. Example 1. A body is ||^trn horizontally from a high point, at a velocity of 16 ft. per second. Find the path of its fall to the ground. Let the distance which it moves horizontally in t seconds be represented by a?, and the distance which it falls vertically in the same time be represented by y. Then, x=im. By Newton's law of falling bodies, y=lW^. Eliminating t between these two equations, x^=16^. Measuring posi- tive ordinates downward, and positive abscissas to the right, the graph of this equation, which is the path of the moving object, is found to be the portion of the parabola shown in Fig. 15. oJ 1 I + ^^ \ \^ N \ + \. 1 V Fig. 15. 318 ALGEBRA Example 2. A body is thrown from the ground at an angle of 45°, and with a force such that its horizontal velocity is 16 ft. per second. Find the point at which it strikes the ground. Let X represent the distance the object will move horizontally in t seconds, and y the vertical distance it will move in the same time. Then x=lQt. The vertical distance would be the same as the horizontal dis- tance if the force of gravity did not make the body tend to fall the distance 16^^ Hence, y=16t-iet\ Eliminating f, x^=lGx—lQy. The graph of this equa- tion will be found to be - the portion of a parabola, - shown in Fig. 16. Hence, this curve is the path of the object's flight. ^ Now the distance from f ^ ■ — .... > y N v / \ ^ — — — — ^ ^ the point of ascent to the Fig. 16. required point is the dis- tance AB of the figure. At the point 5, 2/=0- Hence, the distance is the value of x ob- tained by making 2/=0 in the equation x'=16^-16i/. When 2/=0, x^=z\^x. Hence, a?=0 or 16, Therefore, the body strikes the ground 16 feet from the point of ascent. 215. The limit of a variable. If the value of a variable changes according to some law, such that the difference be- tween the variable and a constant may become and remain less than any assigned small value, which may be taken as small as one wishes, the constant is called the limit of the variable. VARIATION. ALGEBRAIC EXPRESSION OF LAW 319 Suppose a point to move from X toward F, moving | the distance X Y XY the first second, arriving at A ; then ^ the remaining dis- tance AY the next second, arriving at B ; then }^ the remaining distance 5F the third second, arriving at O ; and so on indefi- nitely. In an infinite number of seconds, the point may be brought as near as we please to F, since we may continue the process of halving the distance from the point to Fas long as we choose. Evidently the distance from X to the moving point gradually increases^ and approaches as near as we please to the distance XY. Also the distance from the point to F gradually decreases and can be made less than any fixed small distance, which can be made as short as we please. Hence, the variable distance through which the point has moved from X approaches the whole distance XY as a limit ; and their difference^ the distance from the point to F, approaches zero as a limit. The sign, = , placed between a variable and a constant, indi- cates that the variable approaches the constant as its limit. Thus, a? ^« is read " ic approaches a as its limit " and means that, as X changes according to some law, the difference be- tween X and a can become and remain less than any small value whatever that may be assigned. 216. It follows from the definition of the limit of a vari- able, § 215, that if the limit of x is a, the limit of x — a is zero., i. e., the difference between a variable and its limit is another variable whose Vimit is zero. That is, If x=a, then x—a = x', where x'=0. Conversely, if x—a = x\ or jc=a + jr', vjhen x' is a variable whose limit is zero and a is a constant^ the limit of x is a. That is. If x=a + x', where jr'=0, then x=a. 320 ALGEBRA 217. Finite, infinitesimal and infinite num'bers. A variable whose limit is zero is an infinitesimal. Thus, if the value of a variable can be made to decrease indefi- nitely, and become and remain less than any assigned value, which may be taken as small as we please, it is an infinitesimal. A number, however small, which does not approach zero as a limit is not an infinitesimal. If the value of a variable can be made to increase indefi- mtely according to some law, and become greater than any assigned (or fixed) number, which may be taken as great as we please, it is called an infinite number, or infinity. An infi- nite number is represented by the symbol go . An infinite number does not approach a limit, hut increases without limit. The statement that x increases without limit is expressed by a?= go . Any number which is neither an infinitesimal nor an infinite number is called a finite number. All^ice^or definite numbers and all numbers heretofore con- sidered ?iYe finite numbers. 218. If.) in a fraction inhose numerator is any finite num,- ber^ except 0^ the denominator increases without Ibnit^ then the fraction approaches as a limit / that is, if jif=oo , then ~=0. X For, evidently as x increases, the fraction - decreases in value, and, by making x sufficiently great, - can be made less X than any assigned value, which may be taken as small as we please. Thus, y\, T^o, iwo, TO loo, TTTo^offo-) ©tc. , are a series of fractions in which each fraction is one-tenth the preceding one, and each denominator is ten times as large as the preceding one. VARIATION. ALGEBRAIC EXPRESSION OF LAW 321 If this series of fractions is carried out indefinitely, a fraction will be reached whose value is less than any assigned value, however small. 219. Properties of zero. Zero may be defined as the differ- ence between two equal numbers ; that is, 0=n—n. Thus defined, it leads to some interesting results. (1) Let a be any finite number. Then a(i = a{n—ri) =^an—an = 0. Hence, the product obtained by inidtiplying any finite number by zero^ equals zero. (2) If a be any finite immber, then ^ _n — n a a n n ~^ a a = 0. Hence, the quotient of zero divided by any finite number^ equals zero. (3) The symbol ^ represents a number which multiplied by gives 0. But, according to (1), any finite immber multiplied by gives 0. Hence, ^ represents any finite number ichatever. The fraction ^ is called an indeterminate fraction.* * In like manner it can be shown that the symbol § is indeterminate. 322 ALGEBRA (4) The fraction - where a is any finite number^ represents a number which multiplied by gives a. But, according to (1), no finite number multiplied by gives a. Therefore, tt has no finite value lohatever. If a is finite and a? is a variable whose limit is zero, then by taking £c smaller and smaller, - can be made to exceed any fixed number which may be chosen as great as we please. Hence, the limit of - when cc = 0, cannot be found; that is, it exceeds all definite fixed numbers^ or is infinite. This is expressed by a. Mut this is not an equation in the ordinary sense. (5) Since ^ is indeterminate, and ^y is not finite, therefore, it clearly is not allowable to divide by 0. See foot-note to § 24. Let the student show the fallacy in the following reasoning. Let a=h. Multiplying by a, a^=ab. Subtracting 6^, a^—b'^=ab—¥. Factoring, {a-\-h){a—b)=h{a—h). Therefore, 2b{a—b)=b{a—b). Divide by 6(a-6), 2=1. 220. Indeterminate fractions. For certain values of the variable, a fraction will sometimes take the indeterminate . form -^ VARIATION. ALGEBRAIC EXPRESSION OF LAW 323 Thus, when a = 2, _i) = m ^^ indeterminate form. This fraction, however, is equal to ^^ _,^ Now, as long as a is 7iot equal to 2, we may divide both terms by a— 2, hence -^^ ^^^ - = a + 2, when a is not equal to 2. If a=2, by the definition of a limit, a cannot become 2, although it differs from it ever so little ; hence we may divide by a— 2. Now, as a approaches 2, evidently a + 2 approaches (f 4 2 + 2 or 4. Hence we say ^=4 as a=2. Hence, •^ a —2 a'^—4. 4 is called the value of -. when a=2. a —2 The value of such a fraction, for any particular value of its variable, is defined as the limit which the fraction approaches when the variable approaches this particular value as its limit. Example. Find the value of — r-— - when x=a. oc — a When£c=a we have =-, the indeterminate form. x—a For any value of x other than a, we have x—a Now, as X approaches a this expression evidently approaches n^^ O'i a^ + aa + a^ or 3a^, the value of -— — — when 0?= a. X ^~a CHAPTER XXI. FRACTIONAL AND NEGATIVE EXPONENTS. 221. The following five fundamental laws of exponents have been established in Chapter V, Chapter VI, and Chapter VII: (A) «"'a" = a'^+"; (C) («'")" = «'»"; (D) (aby = a''b''l (-) (!)"=? In all the work heretofore, the exponent has been a positive integer. Where the exponent is a positive integer, by defini- tion it indicates the number of times that the base is to be used as a factor. Thus, d^ means a- a- a. The proofs of these five laws were based upon the definition of an exponent. And therefore these laws were established only when the exponents were positive 22t2t. Fractional and negative exponents. We have seen that by extending the operations with numbers so as to make them general, new kinds of numbers have been conceived ; viz., fractional, negative, and imaginary numbers. Likewise in the attempt to extract a root of a power of a number by the law, a fractional exponent, a neAV kind of exponent, is sometimes 824 FRACTIONAL AND NEGATIVE EXPONENTS 325 obtained if this law be extended to hold for all j^ositive integral values of m and n. Thus, y'd^=a^^'^ or a^- If the law a'"^«" =«"»-" be extended to hold when n is greater than m, then a negative exponent will result. Thus, a^-^-a5=a=*"^=a"2. It thus appears that the extension of one process gives rise to fractional exponents, and the extension of another to negative exponents. It is clear that a fractional or negative exponent cannot he given the same meaning as an exponent which is a j^ositive integer. 3 For example, a^ cannot indicate that a is to be used as a factor I of a time. Such a statement is absurd. Also, a-* cannot in- dicate that a is to be used as a factor —4 times. We shall define operations with negative and fractional ex- ponents so that fundamental law A, § 221, shall be true also Avhen m and n are fractional or negative. We may then in- vestigate the meanings for such exponents as are required by this law. 223. Meaning of a positive fractional exponent. Since fractional exponents are to be so defined, that law A shall be true for such exponents, then if m and n be any positive integers. m , m . m • to n terms a".a".a" . . . to ^i factors =«" " " (A), m or. Hence, a*'~|/a'", by the definition of a root, = (f'ar. § 147. 326 ALGEBRA Therefore, a positive fractional exponeiit indicates a root of a power ^ or a poioer of a root^ of its base ; the numerator indi- cating the power ^ and the denominator indicating the root. Thus, a"^=^i^a^, or (i^a)^ a^=i^^ 16^= (1^16)3= 2^ =8. It follows that in any expression, any indicated roots may be changed to fractional powers, and vice versa. ^ g .___ 3/"~" -2 11 2 Thus, y x'^ + y xy + y y^ meiy be written x^^+x^^'y^ + y^. Note. — Keducing a fractional exponent to higher or lower terms does not change the value of the expression. TO For, a" = V^ =17 «^^ § 148. mx Hence, fractional exponents that may occur in using laio A tnay he reduced to a common denominator^ then added. 224. Meaning of a negative exponent. Since a negative ex- ponent is to be so defined that law A holds, if n be any positive integer or a fraction, then,, a" •«"" = «""" Law A 48. Dividing both members by w\ «""=—-. a Hence, a negative exponent indicates a power., or a power of a root, of its base that is to be used as a divisor. Thus, a-^=\ or in-a*; a^lr^ ^ — ova^-^lf; a~^=~^-^. « '6^ at pa' FRACTIONAL AND NEGATIVE EXPONENTS 327 225. Atiy factor may he changed from the dividend to the divisor^ or from the divisor to the dividend, by changing the sign of its exponent. This follows from the interpretation of a negative exponent. Example 1. 7y^z-.^ 7a' Example 2. ^i^=^ip or a-^h-x-y-^,ov ^,^-,^,y, , _6 1 1 11 Examples. 4 ^=— = — ^~— 17^—7:7^' 4I (^4)5 25 32 EXERCISE 93. Express as roots of integral powers ; 4 2 2 1. a'T. 4. b^. 7. x'^. i i_o 5 2. a2. 5. ^7 . 8. 83. 3. x^, 6. yi 9. 27i 10. it\i. Write with positive exponents : 11. a;2y-3. ^^ lar^b-^c ^^ 2x(x'^-y^)-^ 12 a-i^>2c-2 ' ^xy-'^z-'^' ' Sy(x-y)-'' ' *^^ ^ (a— ^)(a4-^)^ 14. 3x-5y22-3. ^_1 05-4^-1 2 _3 ly. -. — 5 — . ^, x^y '^ 2 xy-^z-^ 24. ^ 15 -r27/-is-2 ^ ^-1 16 5^' 21 (^^-^)"' 25 <^'"^'>"^ . 328 ALGEBRA Write as roots of positive integral powers. 26. a~t 2 29. ■■ 2 ■* 31. 1 i* 27. x~^. ^ 2/ « 28. A. C 30. 33. (^^-^)"f. 32. 2(a + 5)-t Find the value of : 34. di 37. 625"* 40. (-64)-*. 35. si 38. (»)-'■ 41. 125~*X3^ 36. 27" i 39. ©-'■ 42. 8*x9-t. 43 25"^ • 81""^ " & /1\| 226. Having defined operations \oith fractional and negative exponents as being required to satisfy the laic a'"-a" = a'"^", toe loill 71010 shoio that they icill satisfy the other latos of exponents^ also : viz : .^''' . '■' Law B^ ar-T-a"- a'""". Law (7, (a'"y = a'"". Lawi>, (aby = a"b". Law JS; (» -F I. When the exponents are fractional. (1) a-n-^. -a'» X>, § 221. =i;^^i/ F § 67. Hence, law D is satisfied. =K^ js;§22i. =^r^ § 68. y b'" m Hence, law J^ is satisfied. II. When the exponents are negative. (1) 6«-"*^a-"=a-'"a'\ § 225. =«-'»+". § 222. 330 ALGEBRA Hence, law B is satisfied. (2) (^a-)-^=(L}j-^ §225. . = (<:r)" § 225. = cr\ or a'-"'''-''' E, § 221. Hence, law G is satisfied. (3) W-"-=(i'^ §225: 1 1 1 §221. = a-''-b-\ § 225. Hence, law D is satisfied. (4) vv y«y' §225. ^* = ^* a" Hence, law E is satisfied. 225. 227. The consideration of the laws when the exponents are surds or imaginary numbers will not be taken up in this book. 228. In the following exercise we may make use of the five laws of exponents, which have now been interpreted for nega- tive and fractional exponents, as well as for integral exponents. FRACTIONAL AND NEGATIVE EXPONENTS 331 All radical signs should first he replaced by fractional expo- nents. Example 1. Simplify a-^- a*. 2 3 8 9 1 7 =a;^^. ' Example 2. Simplify a?~^-^a7"^". -2 _ 3 _3_/'_ 3\ 4,3 ■x To Example 3. Simplify (4a"*6~^)~^. 3' 4^ 8 3/-^ / I-. 2 _1 Example 4. Simplify y=I:-(^) -'^• _i 332 ALGEBRA 3,— 3, 3,— Example 5. Multiply yx^—Vxy + vy^ by ^/x+^/y. 3/— 3 3 — 2 1_ 1 a Vx^—Vxy+Vy^=x^—x^y''^+y^. 5./73.r/2 3 3 -J ^ 8 It 2 x^—x^y^ + y'^ x^ + y^ 2 1 12 _? 1 12 + a?^ 2/ x^y'^ + y X +y EXERCISE 94. Simplify : 1 9^-2.7.-3.^7 _7 3 1 1 2. a 1 .3 16. 6a 2^.3^(3a */> i). 3. a2.^-2.«2. ^^ ^^ 2 1 16. (« 2^ V^- 4. 5iK a-3ie2. _2 1 3 4 1 17. (4.^ 3y 2^)-^^. 6. a-2-^a-4. 18. {-^la-^x 4) 3. 7. x-i^x"^. 19. (32^-tyH^)"i 8. a2^-3_^f^-i.7.-i, 20. i/av'aae. 9. a-lJ-2c-3-^«3^;2e-4. 21. T^-P^-^1>^. _4 _7. 5 , 5 10, 711 ^—771 5. 22. i/a-2a;-i---va-i 23.-6. 1^- y'-^y-'- 23. i>v^-^-2. 12. 5tV^(^-i^.-A). 24, |/^3^^^2. 13. aHic~^--(aZ»"3c"i). 26. y x^i/^x^y'x^^. FRACTIONAL AND NEGATIVE EXPONENTS 333 o/j V^^ ya^b^ 30. Qi/a-^---2V cr^. 26. "i/—^-^ 12 Va^b Va^b^ „^ _i ^ 31. (0^2 _y2) 2.^/(^_^.y)3. 28. i^a/«V«^)^ 33 ^/^JaW 29. a"(-«-'^). • I ^-3 • \^,-i^ 4 2.2. 44 22 4 34. {x^^-x^y^^y^){x^—^y^-^y^). 35. (a^— a^ + a^-a~i)(a^ + a^). 36. {x^ ^x^y^ -\-y^){^ —y^). 37. (2aj-i-3cc-7)(5 + 4i«-i). 38. (i>a4 + T/a2i>^2 4_-^^4-)(^^^^2_^V^2). 39. {Vx'^ — \^'xy^\/y''--){Vx^ yy). / 4 6 \ / 2 \ 40. (-X^— 3^ + 9)(-x:r-+3]. \l/a;4 yx'^ JWx'^ J 41. (a + 2i^a2_3^>--)(2-4a~i-6a~4). 42. {x^+y^)^(x^ + yi), 44. (x'^—2 + x~^)~{x-^—x~'^). 45. (^>5 + 2^»-5 + 7)(5-3/>»-5+2^>5). 46. (i/^ + v'^+l)(£c~^+a;~i + l). 47. (8a-2-8a2+5a6_3^-6)^(5^2_3^^-2)^ 48. (a^ + Z>~*)2. 50. (x^ + xi + l + x~~^-{'X~^)^-. 49. (a;~i-2/^)2. 51. (ai + ic~^)(a^-a;~4). 334 ALGEBRA 62. (3£c~^-4y^)(3a;~* + 4y^). 57. {a'' -\-b^)-^{a^-\-h^), 53. {a-4:h)^{ya-'li/b). 68. {x-y)^{\/~x-vy). ^ ^ ^ ^ 59. 8 3(252-81M. 55. (a-2--9)--(a-"-3). ^ ^ 66. (a-4-l)--(a-i-l). 60. 27^(27^+3^). EXERCISES FOR REVIEW (VI). 1. Define ratio ; terms of a ratio. Illustrate each. 2. What are commensurable numbers? Incommensurable numbers ? Illustrate. 3. What is a j^?rop(9r^^o?^? Illustrate. Name the ^erws of a proportion. 4. Show that the product of the extremes in any proportion equals the product of the means. 6. What is a test of the correctness of a proportion ? 6. Is 2a V : 3aic^=6aic^ : Stc^" a true proportion? Give proof. 7. Give five fundamental principles in proportion, and illus- trate each. 8. Define mean proportional ; third p>roportional ; fourth proportional. 9. Find the mean proportional between ^x and ^x^ ; between 72aj^y and ^x^\f\ between 4(a + ^) and 9(a + ^)^; between a and X. 10. Find the third proportional to 1 and 8 ; tox* and l^x^a^ ; to -l^^y ^^^ 25y^; to — t~x ^'^^ {a-\-by. 11. Find the fourth proportional to 9, 7, and 3; to 1, x^^ and 2y ; to !«, 2aJ, and bx ; to x—y^ ^^~y^ and 1. 12. What is a continued proportion ? FRACTIONAL AND NEGATIVE EXPONENTS 335 13. Show that if -=-==-=-, then — , — =—, — X y z w* x-\-y z + w 14. Show that if t = ^i^ then -j= \ ^ b cV d ^c'-^d' 16. Define a variable. A constant. Illustrate each. 16. What is direct variation? 17. If £c oc 2/, and £c=6 when y=4, find y Avhen £c=2. 18. If icoc y, and a;=10, when y=l, write the equation con- taining X and y. 19. If icocy, and when i»=f, y=9, express the equation between x and y. 20. What is inverse variation ? 21. Given that x varies inversely as a^ also that when iK = 2, a =5. Find aj when a = 10. 22. Given that x varies directly as y and inversely as z. W^hen jc=3, y=5, and z=l. Find 2 when aj=3 and y=15. 23. What is meant by the limit of a variable ? 24. What is an infinitesimal? An infinite number? A finite number ? 25. What is the value of - ? Of^? Of^? Of—? iC QO 26. Express in symbols the five fundamental laws of ex- ponents. Prove each when m and n are positive integers. 27. Simplify ««•«"; a'^a'-, {ay-, {ct'b'f; (^^' 2, m 3 28. What is the meaning ofa^? Of «'» ? Show that a^ means ya^. ^ 29. Simplify (a^)^^' ; {x'Y ; (2aM)«. 30. Find the value of 4* ; 8^ ; 25^ ; (J^)i 336 ALGEBRA 31. What is the meaning oia-'^ ? Of «-" ? Show how this interpretation of a negative exponent is derived. ct ^ 2 1* (/ — ^ 32. Write with positive exponents -Trr2 ; q-^'^^zTg ; 2(a—b)-^. 33. Find the value of 4~^; 5-2^5-3; 4~^; 25"i;(-8)t- 34. Simplify (a-^y^ ; (a'^fi; (i/^^T^; (aJ^y-^)-^ ; 35. Write the value of (x'i-^-y-^y; (2a-^-Sb~^y; 36. What is an ineqiiality ? Distinguish between con- ditional inequalities and identical inequalities. To which class does a^-^b^^^ab belong? To which class does ^3>^| + 5 belong? 37. What is meant by the solution of an inequality ? 38. Solve the conditional inequality in Exercise 36. 39. Given | ^""12^=-^^' ^^"<^^ ^^^ ^"^^^''^ ^* ^ ^^^^ V- 40. A teacher, being asked the number of his pupils, replies that twice their number diminished by 7 is greater than 29, and three times their number diminished by 5 is less than twice their number increased by 15. Find the number of pupils. 41. Three times a certain number plus 16 is greater than twice that number plus 24, and | of the number plus 5 is less than 11. Find the* number. ia — 23 CHAPTER XXII. PERMUTATIONS AND COMBINATIONS. 229. In this chapter we shall discuss an important class of problems which the following examples will illustrate. Example 1. In how many ways can a program be arranged consisting of a solo, a debate and an oration ? By putting any one of the three numbers first and each of the remaining two numbers after it, we get the following six ar- rangements : (1) Solo, debate, oration ; (2) solo, oration, debate ; (3) debate, solo, oration ; (4) debate, oration, -solo ; (5) oration, solo, debate; (6) oration, debate, solo. Example 2. How many different numbers of two digits each can be formed from the digits 2, 3, 4, 5, using each digit but once in the same number ? We can take each with one of the others ; hence, we get the following twelve numbers: 23, 24, 25, 32, 34, 35, 42, 43, 45, 52, 53, 54. These examples illustrate the general problem of finding the number of arrangements of a certain number of things taken from a given number of things. 230. Permutations. All the possible arrangements that can be formed from the different groups of ..r things which can be taken froni n different things, are called the permutations of n things taken r at a time. Thus, the permutations of the three letters a, &, c, taken one at a time, are a, 6, c Taken tn^o at a time, they are a6, 6a, ac^ 22 337 338 ALGEBRA ca, he, cb. Taken three at a time, they are abc, acb, bac, bca, cab, cba. It is clear that two permutations are different unless they con- tain the same things arranged in the same order. Thus, ab and ba are different permutations. The number of permutations of n distinct things taken r at a time is represented by the symbol ^Z'^. Thus, 3P2 represents the number of permutations of 3 things taken two at a time. 4P2 represents the number of permutations of 4 things taken 2 at a time. jf^Pg represents the number of permutations of 10 things taken 6 at a time. 231. The value of ^Z',., From the illustrations in § 230 we have seen that ^P^ = S, ^P^—^^ 3^3=6. There is a law by which the number of permutations in any case may be writ- ten. The law may be derived by the use of the following principle : Jff" a thing can be done in a dijlfere7it tcai/s, and a second thing can he done in b different ways without interfering with the firsts there vnll he ab wags of doing the tico things. The truth of this principle is evident. To illustrate it, suppose that 5 boats are plying between two cities. Find the number of ways in which a person may go and return from one city to the other by a different boat. Evidently, in going, he has the choice of any one of the 5 boats. In returning he has a choice of any one of the 4 not used in going; hence, with any one of the five choices he has four others, or 5 x 4 in all. The number of ways that n things can be taken from n things one at a time is evidently n. Hence, nP.=n. (1) PERMUTATIONS AND COMBINATIONS 339 From 9\ things one thing can be taken in n different ways, and after tliis is done a second thing can be taken from the remaining n—1 things in 71 — 1 different ways. Hence, from the preceding principle, there are ?i(/i—l) ways of taking the two things. That is, ^P,=?i(n-1). (2) Thus, 5P2=5-4=20; i2P2=1211=132. After the first two tilings have been taken in any one of the n(n—l) ways, the third thing can be taken from the remain- ing ?i—2 things in. n—2 ways. Hence, there are n{?i—l)(n—2) ways of taking the three things. That is, ,,P,=n(n-l){n-2). (3) Thus, 4P3=4-3-2=24. ^,P^=10-9-8=720. In like manner, ,P,=n(n-l){n-2)(n-8)(n-4.); ^' (5) ,P,=n(n-l)(?i-2)(7i-S)(n~4:)(n-b) ; (6) and so on. ^5 4 ^ X / Kow from (1), (2), (3), (4), (5), and (6), Ave see that ^P, has 1 factor ; ^^P^ has 2 factors ; ,,P^ has 3 factors ; ^P^ has 4 fac- tors ; jjPg has 5 factors ; ^Pg has 6 factors ; and so on. And in general „P^ will have r factors, the last one being 7i—(r—l), or n — r-\-l. That is, ^ <>- „ = /7(/7-l)(/7-2) 4 3 21. (8) 340 ALGEBRA 232. Factorial-/?. In (8) of § 231, the product . ?i(7i-l){7i-2) 4-3-21 is called factorial—/?, and is usually designated by the symbol I )i, or n\ . Thus, |^ = 6-5-4-3-21=720; 5! = 5-4-3-21 = 120. Hence, from § 231, we have Example. In how many ways can 5 books be arranged on a shelf ? The number =,P,, or |_5=5 4-3-21=120. 233. When the things permuted are not all different. In many problems the things permuted are not all different. We sliall now determine the number of permutations in such cases. Suppose that a of the n things are alike, and suppose that we form the iV^ permutations of the ??- things taken ?i at a time. Now, if in any one of these permutations the a like things be replaced by a unlike things, different from all the rest, then by changing the order of these a new things only, we can form I a new permutations from the one permutation. This can be done in the case of each of the JV permutations. Ileuce, in all, ]}^\a new permutations can be found. Therefore, JSr\a = „P,, (all different), and N = '^=^. Similarly, it can be shown that, if a of the n things are alike, and h others alike, then PERMUTATIONS AND COMBINATIONS 341 And if a of the n things are alike, h others alike, and c others alike, then \n ^ \a \h \c ' and so on. Example 1. How many different numbers can be formed by- using all of the figures 2, 2, 2, 3, 3, 4, 5, 5, 5, 5 ? 110 The number = -• =12600. li It li Example 2. Find the number of permutations of the letters of the word Indiana. Here there are two e's, two n'fe, two a's, one d. Hence the number = j^ g ^^ =630. EXERCISE 96. Find the value of : 1- li- L^ li 12. ,p,. ' 2. 16. '• ^ ■ ''■ '^^- . , |2 13 14 14. 1.A- V' 3. |3|5. 8.^^- 15. „P,. 4. |4 |2. [6 |8 16- i«^«- 17 9. -n^- 17. ,,A. '• T 14 15 16 ''• 'J- [8_ ^°- Tir' 19- # 342 ALGEBRA Show that 21. n(n-l){n-2) \n-d \n-l 22. 1 --\n-2. y^ !y('^^y^^X n-l 23. \a • \a • (a + l)~a = \ a + l ■ \ a — l. 24. How many numbers of three digits can be formed by using the digits 1, 2, 3, 4, 5, using each digit but once in tlie same number ? 25. How many numbers of two digits can be formed bj^ using the digits 2, 4, 6, 8 ? 26. How -many permutations can be formed of the letters of the alphabet taken twji-at a time ? 27. In how many different ways can 5 boys stand in a row ? 28. If 8 steamers ply between Liverpool and New York, in how many ways can I go by steamer from New York to Liver- pool and return by a different steamer ? 29. Six ladies and six gentlemen are to be seated about a circular table. In how many different positions can they be seated so that there shall be a gentleman at the right of each lady? 30. In how many ways can a class of 15 pupils be seated in 15 seats? 31. In how many ways can 2 different prizes be awarded to 10 boys so that no one boy gets both prizes? 32. How many permutations can be made of the letters of the word A?ina ? The word Missouri ? 234. Combinations. The different groups of r things that can be taken from n things, when the arrangement is not jcmi- sidered, are called the combinations of the ?^ things taken r at a time. PERMUTATIONS AND COMBINATIONS 343 Thus, the combinations of the letters a, 6, c, d^ taken two at a time, are ah, ac, ad, be, bd, cd. Taken 3 at a time, they are abc, abd, acd, bed. It is clear that two different combinations can not contain the same things arranged in different orders. Thus, abc and acb are the same combination. 235. The number of combinations of n things taken r at a time is represented by the symbol „Cr. Thus, 4C3 represents the number of combinations of 4 things taken 3 at a time. 236. The value of „^^, The number of combinations of n things taken r at a time is easily found by establishing the relation between „ C^ and „P^. Suppose n different things combined r at a time. . Every combination of the r different things will have |r permutations, taken r at a time. Hence, the total number of permutations will be „CV Ir ; that is, Therefore, Thus, 237. If, in the value for ^^C\ found in § 236, we replace „P,. by its value found in § 231, we get _ n{n-1)(n-2) ■ (/i-r+/) nC/^ It;; • v-"-; It is sometimes useful to express the value of „ C^ in a dif- ferent form. nCr L=nP. ,Cr = C?a = 5-4-3 ~3-21 10. 344 ALGEBRA Multiplying the numerator and denominator of the fraction in (1) by \?i—r, we get 7i(n—l){n—2) (n—r-i-l)\?i—r r \n—r or ^C=r-^=— (2) r \n—7' n^ r If we replace rhjn—r in (2), we get \n In C = ^= =_!=___. (3) ^ ""'' \7i—r \7i—n + r \n—r \r From (2) and (3) it follows that nCr = nCn-r' (4) Thus, 50C48 = 50C2=^ = 1225. EXERCISE 96. Find the value of : 1. ^Cy 3. 80 ^n- 5. 12 Cg. 7. gOg. 9' 25620- 2. joCg. 4. 15612. 6. 8 63. 8. 206^3. 10. ,C',. 11. Find the number of combinations of 12 things taken 2 at a time ; taken 3 at a time ; taken 9 at a time. 12. How many selections of 3 books can I make from 5 books ? 13. How many combinations can be made from the letters a, b, c, d, e, taken three at a time ? 14. How many combinations can be made from the 26 let- ters of the alphabet, taken 2 at a time ? 15. How many . different products can be formed from the numbers 2, 4, 6, 8, if each product contains 3 unequal factors ? PERMUTATIONS AND COMBINATIONS 345 16. In how many ways can a committee of 3 be selected from 8 men ? 17. How many different committees can be formed from 10 Republicans and 6 Democrats, if there are 2 Republicans and 1 Democrat on each committee ? 18. There are 8 points in a plane, no three of which are in the same straight line. Find how many lines can be drawn, each connecting 2 points. 19. In how many ways can 3 red balls and 2 white balls be selected from 8 red balls and 5 white balls ? 20. In an algebra class of 25 students, 20 recite each day. In how many ways can these 20 be selected for one day ? 21^' A farmer has 7 Borses. In how many ways can he hitch a two-horse team ? 22. In an examination a teacher gives a pupil the choice of any 8 questions out of 10. In how many ways can the pupil choose his 8 questions. 9 ^ C. ^ 1° Ti n ^/ CHAPTER XXIII THE BINOMIAL THEOREM. 238. In § 63 it was shown that any positive integral powier, of a binomial can be written clown by some laws which taken collectively constitute the binomial theorem. Thus, by these laws, {a' + 2by={ay + 4{a'f{2b) + Q{a'y{2bY + 4{a'){2by + (26)* =a^ + 8a^b + 24a*b' + 32aW + H)b\ In the general case, it will be found that the laws in § 63 will give It will be recalled that no rigorous proof of this theorem has been given. The binomial theorem can be proved to hold true for all exponents, integral or fractional, positive or negative. We shall prove the theorem here for the case when the exponent is a positive integer. 239. Proof when the exponent is a positive integer. From the rule for obtaining the product of two expressions, which is based upon the distributive law, it follows that a term in the product of any finite number of expressions can be ob- tained by multiplying a term of any one of the expressions by a term from each of the other expressions. A repetition of this process in every possible way will give all of the terms in the product of the expressions. 346 THE BINOMIAL THEOREM 347 For example, consider the product {a + b){p + q){x + y). In finding the product ot a + b andp + g we multiply each term of a + bhy each term ot p + q. Each of these resulting terms is then multiplied by x and by y to obtain all of the terms in the product of the three binomials. This process evidently amounts to the use of the above rule. . Consider the expression (a + Z>)^ This may be written in the form {a + b)(a-\-b)(a-\-b) to ^z factors. If we select a tenn from each of the n factors in this product^ and multiply these terms together^ and do this in every pos- sible way^ we shall obtain all of the terms in the product. (1) Now the term a can be selected from all of the factors in just one way. Hence, the product of these a's, which is a'\ is iho, first term of the product. (2) The term b can be selected from one factor and the term a from each of the other n — 1 factors ; and this can be done in as many ways as b can be selected from the n factors, which is n or „ C^. Hence, a"~^6 can be selected in n ways, or ^ C^ ways ; that is, na'^~^b^ or „ C^a'^~'^b^ is the second ter^tn of the product. (3) The term b can be selected from each of two factors and the term a from each of the other n — 2 factors ; and this can be done in as many ways as two 5's can be selected from the n factors, which is „ C^. Hence, a'^-'^b^ can be selected in „ C^ ways ; that is, „ C^a'^'^b'^ is the third term of the product. (4) The term b can be selected from each of three factors and the term a from each of the other n—^ factors; and this can be done in as many ways as three ^'s can be selected from the n factors, which is „ C^. Hence, a'^'W can be selected in „ C3 ways ; that is, „ O^a/^'W is th-e fourth term of the product. 348 ALGEBRA (5) Let this process be continued. In general, the term h can be selected from each of r factors and the term a from each, of the other n—r factors ; and this can be done in as many ways as r i's can be selected from the n factors, which is „(7^. Hence, a'^-^'W can be selected in ^G^ ways; that is, ^C/i^'-'lf is the (r+l)^A term, of the product. (6) Finally, the term h can be selected from all of the fac- tors in just one way. Hence, If can be selected in just one way ; that is, 5" is the last term of the product. Hence, (aH-6)"=fl'* + „(?ia'^-^6-h„(?2a"-='6'+„(?3a""'6'+ +Xa"~'*6"+ +6"- (1) The expaiision in (1) expresses in symbols the binomial theorem. If, in this identity, „ Ci, „ C^^ „ G^ etc., are replaced by their values, the identity becomes ^ . . ^^ n{n~1){n-2)-^ ■ ■ ■ („-.+ / )^_^^^ _^^„ ^^^ It is seen that (2) conforms to the laws of § 63. When n is a positive integer it is easily shown that there are always /? + 1 terms in the expansion. When r^ 71, the coefficient of the (r+l)th term (t. 6., the (n + l)th term) is 1 ; but when r=?i + l, the coefficient of the (r + l)th term becomes n{n—l){n—'^) {n—n — 1 + 1) n + 1 which is 0, since the last factor in the numerator is 0. Hence, a term does not exist in which r is greater than n ; that is, there are only n-\-l terms. Example 1. Expand (a^ + i/^)^ Here n=5. And fi^=^, 5^=10, 503=10, ^C,=^. THE EilNOMIAL THEOREM 349 Hence, {x' + y'f={x'r + Mx^W) + ,C,{aff{yy 4- .C^ix^Yifr + \c\{x^){yy+(yr =ic" + 5x^y^ + lOx^y* + lOxV + 5x^y^+ 2/^^ Example 2. Expand (2— ar*/. Here, ti=4. And ^0^=4, ^0^=6, fi^^^. Therefore, {2-x^)'=i2y + ^{2f{-x^)+^C^{2fi-0(^Y + ,C3(2)(-^/ + (-^r =lQ-32x^ + 2Ax^ - %x^ + x^\ Examples. Expand (a — 2& + c)^ Grouping terms, this becomes [(a— 26) + c]^ Hence, [{a-2h) + cY={a-2hf + fi^{a-2hfc-\-fi^{a-2hy + (^ =(a-26)» + 3(a-26)2c + 3(a-26)cHc^ Expanding each term, we have [(a-26) + c]^=a3-6a2& + 12a&2-86^ + 3a='c-12a6c + 126='c + 3ac^- 66cHc^ 240. The general term. The (r+l)th, or general term, in(l)of §239is„(?^a"-'-6'-,or-^ ^- ^— — ^ ' — By substituting the values of a, ^, ?i, r, in this expression, for the (/•+l)th term, we may write down at once any desired term of the expansion of any power of any binomial. Example 1. Find the 8th term in the expansion of {2x—iy^. Here, a=2x, 6= — 1, w=10, r=7. Hence, the 8th term =^^C,{2x)\-iy=^^C.,{2xf{-iy =i|^-8x»(-l) = -960a^. EXEBCISE 97. Expand : 1. {x-Vyy. 3. (l + 2aj2)*. 6. (l-3a=')». 2. (2ic-3y)«. 4. {^a'-Vhy. 6. {x'-ay.. 350 ALGEBRA 7. (i+2xy. ^^ ^^_^i^e. . 14. (^^-wy. 8. (l-ixy, ^^ (a-2+5-2)^ 15. ra-i + dy, 9. {x-^+xy. /,j 2by ^/ 5 10. (2a;-2 + l)^ -^^V V^^"^/ * 16.' {x'-y'^)\ 17. Find the third term in the expansion of (l + 2a^y. 18. Find the sixth term in the expansion of (cc^ — 2y)^*'. _i 2 19. Find the eighth term in tlie expansion of (x '^—a^y. 20. Find the fifth term in the expansion of (-^ — *^ 21. Find the fourth term in the expansion of (|«^^ + |c~2)i^ -2. 2. 22. Find the seventh term in the expansion of {x '^+x'^y^. f V^ ^*^ 23. Find tlie sixteenth term in the expansion of { 1 X 24. Find the twelftli term in the expansion of (2— la?^)'*. By grouping terms, express as binomials and expand : 26. {\-^x-xy, 27. {2-a-\-hy. 29. {ci~h-\-c-dy. 26. (£6^ + 2/' + ^')'. 28. {X^x^-x^-xy. 30. {2a-h^Zey. 24 J. Binomial theorem,— exponent negative or fractional. When the exponent -is a negative number or a fraction the ex- pansion of a power of a binomial as in § 239 gives an inde- finitely large number of terms. This follows from the fact that, for such an exponent, the coeflBcient, n{n-V){n-2) • • '\n-r^-\) of the general term can never become zero. It can be shown, however, that the use of the binomial theo- rem in such cases is allowable, pravided that the absolute value of a is greater than that of h. It is advisable to exclude the proofs from this book. The student may assume that the theorem holds in the exercises that follow. THE BINOMIAL THEOREM 351 Example 1. Expand {2—x^)-^ to 5 terms. We have i2-x')-^={2)-'+i-3)(2)-'{-x') + (_3)(--4)r-5)(-6)^^^_,^_^y^ — 1 I 3 '«»2 1 3 /y.4 I 5 ^6 r 1 5 /y.8 I — ¥ ' T'S^*^ ^TS"^ ^H'S'^ ^TS'^'^ ^ _3 Example 2.- Expand (a + 2a?)* to 5 terms. ^ We have {a + 2x)^={a)^ -\-{f){a)~^{2x)+^i^i:^{a)~^{2xy + ci)(z =a*+|a ^a?— fa ^a^^ + ^a ^o?-*^— jVgCt ^ a?*+ 1 2- O , ^ Example 3. Expand (i—x) ^ to 4 terms. We have (l.-a?r^==(l)"^ + (-i)(ir"2(-a^) + ti)H)(l)-|(_ic) 4-tMJ.Kd)(l)-i(-;rf • (iK=i^t:4)(«)-l-(2^)3^(J0G_^ . . . . ' + =l+4a?+|a?'.+ A^ + EXERCISE 98. Expand to four terms : 1 /I — oA-2 2 4 ^ ^ * 6. (1-x'y. 10. (ic + 2r's-. 3. K-.r^ '' ^^^'^^^' . 11. (8 + .)i 4. (2x.-3y)-^. 8- (20.^-1)"^. 6. (1 + a;^)-^ 9. (i + a;)"^ 12. (2-ic2)-J. In its expansion find : 13. Tlie 6th term of (2-a;)-3. 352 ALGEBRA 14. The 10th term of (a'^c^)^. 15. The 5th term of {ei-e'^)-^. 16. The 8th term of {l-2x^yi. 242 . Extraction of roots by use of the binomial theorem. The binomial theorem may be used to extract roots of arithmetical numbers. The process is best shown by an example. Example 1. Find to 4 places of decimals the value of y'W. (i)H)H)(25)-V-(_3).+ . . 2- 3 9 81 80 6400 1024000 =2-.0B75-.0014-.0001- =1.9610 approximately. A similar process may be used in any case. Hence the fol- lowing rule : ^Separate the number into two parts^ the first of ichich is the nearest possible perfect power of which the required root can be found ; then expand the resulting binomial by the binomial theorem^ and combine the values of the terms thus obtained. Thus, v65 = v'64 + l = (43 + l)^; i>623=|/625-2=(54-2)*. It is evident that the first few l^erms of the expansion will give a close approximation to the value of the root, if the suc- cessive terms decrease in value rapidly ; i.e.^ if the second term of the binomial is much smaller than the first. By § 241, no correct approximation of the root can be found by . THE BINOMIAL THEOREM 353 this method unless the given number is expressed as a bino- mial in which the second term is less than the first. Note. — A shorter method of finding the approximate value of any root of a number is by the use of logarithms, discussed in Chap- ter XXVI. EXERCISE 99. Find to four places of decimals the value of : 1. 1^15. 3. v2T9. 5. vMT. 7. 1/3120. 9. t>730. 2. 1/240. 4. yd. 6. y^M. 8. i/26. 10. 1/21. 23 CHAPTER XXiy. PROGRESSIONS. 243. Series. A succession of terms, in which each term after the first may be obtained from one or more of the preceding terms by some fixed law ; ^.e., obtained in the same way for all terms, is called a series. Thus, 2 + 4 + 6 + 8 + 10 + 12+ is a series. Each term after the first may be obtained by adding 2 to the preceding term. Also, l + x-]-x^ + oc^ + x*+ ' ■ ' • is a series. Each term after the first may be obtained by multiplying the preceding term by x. A finite series is one that has a finite number of terms. An infinite series is one that has an infinite number of terms. A series is called convergent either when the sum of all of the terms equals a fixed finite number ; or when the sum of the first n terms approaches a certain fixed number as a limit, when n is indefinitely increased. A series is called divergent when the sum of the first n terms can be made greater than any assigned number which may be as great as we please, by taking n sufficiently great. A finite series is ahoays convergent. In the series 2 — 2 + 2 — 2 + - • • •, the sum of the first n terms is either 2 or according as 7i is odd or even. Such a series is called an oscillating series. 244. We shall discuss in this chapter three special forms of the simpler series, known as arithmetical progressions, geometri- cal progressions, and harmonical progressions. 354 PROGRESSIONS 355 ARITHMETICAL PROGRESSIONS. 245. An arithmetical progression is a series of terms in which the difference between any term and the preceding term is the same for all terms of the series. This difference is called the common difference, and may be either a positive or a negative number, integral or fractional. The name arithmeti- cal progression is usually abbreviated to A. P. Thus, the series 1 + 3 + 5 + 7 + 9 + 11+ • • • • • is an A. P. in which the common difference is 2. And the series 12 + 8 + 4 + 0-1-8-12—14— • • • is an A. P. in which the common difference is —4. 246. The nth term of an A. P. Let a stand for the first term of _an A. p., / for the nth term, and d for the common difference. Then, by the definition of an A.P., the second term = a-\-d, the third term = a-\r'2id^ the fourth term = a + 3c?, the fifth term — a + 4id^ etc. It is evident from these expressions that the coefficient of d in the expression for any term is less by one than the number of the term. Hence, the nth term = a-{'(n — l)d; that is, /=a + (n—l)(f. Formula A. This equation, or formula, will enable us to find the value of any one of the four numbers, I, a, n, d, if the values of the other three are known. 356 ALGEBRA Example 1. Find the 20th term of the series 4 + 7 + 10 + 13 + • • • . Herea=4, d=3, n=20. Substituting in formula A, we have Z=4 + (20-l)3=61. Example 2. The 4th and 15th terms of an A. P., are 9 and 31, respectively. What is the series ? Here the 4th term is a + 3d=9, (1) and the 15th term is a + 14S=i-25-| 2(-9) + (25-l)4 1=975. Example 2. The first term of an A. P. is 2, and the sum of 20 terms is 135. Find the common difference. Here, a=2, n=20, iS=135, d is unknown. Froin formula C we have 135=10(4 + 19d)., Solving this for d, we get d=^. Example 3. How many terms of the series 3 + 1 — 1 — 3—5— • • • must be taken to make —140? Here, a=3, d=—2, S= — l'iO, n is unknown. From formula C we have -U0=^n{Q + {7i-l)(-2)}. Simplifying this gives the equation n^-4n-U0=0. The solutions are 14 and —10. The negative solution has no meaning in this problem, hence 14 terms must be taken. 358 ALGEBRA Example 4. In a certain A. P. the common difference is 1^, and the 12th term is 12^. Find the sum of the first 7 terms. We first find a from formula A, then find S from formula C. From formula A, 12^=a + lll|. Hence, a=— 4. Then from formula O, Example 5. Find the sum of all the numbers between 100 and 500 which are multiples of 6. The numbers between 100 and 500 which are multiples of 6 are 617, 6-18, 619, • • • • 6-83. Hence, the sum is 6-17 + 618 + 6194- +6-83, or 6(17+18 + 19+ • • • • • +83). The sum of the A. P. enclosed in parentheses is obtained by formulas A and C. From formula A, 83=17+ (n-l)l, whence 7i=67. From formula O, >9=-y (34 + 66)=3350. Hence, the required sum is 6-3350, or 20100. Example 6. Show that the sum of r + 2*-^ + 3'+ n^ =inin + l)i2n + l). Note. — While this series is not an A. P. it illustrates an application of it. Since (x-\-lf—ocr^=3x^ + Sx + l is true for all values of .t, we may- give to £c a succession of values 1, 2, 3, etc., and by adding the n identities, obtain the given series. That is, from (ic + 1)*— a?^=3a?H3a?+l, we have when x=l, 2'-l=»=3r + 31 + l, when x=2, S^-2^=3-2^ + S2 + l, when a?=3, 4^— 3^=3-32 + 3-3 + l, when x=n, (ii + l)^— n''=3-nH3-n + l. PROGRESSIONS 359 Now adding columns, and observing that the second term of each left hand member cancels the first term of the member above it, we have (?i+l)^-P=3(P + 22 + 3H • • • • w'') + 3(l + 2 + 3+ • • • • n) + (1 + 1 + 1+ • • • • to n terms) =3(P + 2'^ + 3^+ • • • • n')+pi(n+l) + n. _ Solving for 12 + 2^ + 32+ • • • • »^^ we obtain |n(n + l)(2n + l). (Let the pupil show how this solution was obtained.) 248. Arithmetical means. If thr^e numbers form an A. P., the middle term is called the arithmetical mean of the other two terms. Thus, in the series 3 + 8 + 13, 8 is the arithmetical mean of 3 and 13. If ic is the arithmetical mean of a and b, then, by the defini- tion of an arithmetical progression, a + b whence, x= ^ - Hence, the arithmeiical tnean betioeen two numbers equals half their sum. 249. In an A. P. all the terms between any two terms are called the arithmetical means of those two terms. Thus, since 2 + 4 + 6 + 8 + 10 is an A. P., 4, 6, and 8 are all arithmetical means of 2 and 10. Any number of arithmetical means may be inserted between any two given numbers. Example 1. Insert 7 arithmetical means between 10 and 30. Since there are to be 7 arithmetical means, 30 must be the 9th term of an A. P. of which 10 is the first term. Hence, we have a=10, /=30, n=9. 360 ALGEBRA Substituting in formula A, we have 30=10 + 8d. Solving, d=2l. Hence, the required series is EXERCISE 100. 1. Find the 30th term -in 1 + 6 + 11 + 16+ 2. Find the 16th term in -8-5—2 + 1 + 4+ 3. Find the 23rd term in — |— 1- + ^ + |+ 4. Find the 54th term in 11 + 17 + 23+ ••• •. 5. The 6th term of an A. P. is 17, and the 15tli term is 44 Pind the common difference. 6. The 3rd term of an A. P. is 0, and the 9th term is 22. Find the common difference. 7. The fifth term of an A. P. is 21, and the 8th term is 33. What is the 12th term ? The 20th ? 8. The 2nd term of an A.P. is 7, and the 11th term is 20^. What is the 7th term? The 15th term? 9. Which term of the series 1 + 4 + 7 + 10+ • • • is 46 ? 10. Which term of the series 10 + 6^+3-1- is -25? 11. In an A. P. whose common difference is 6, the 11th term is 72. What is the first term ? 12. Find the sum of the first twenty terms of the series 42 + 39 + 36 13. Find the sum of the first thirteen terms of the series 8 + 12 + 16+ 14. Find the sum of the first fifty odd numbers. PROGRESSIONS 361 15. Find the sum of the first fifty even numbers. 16. Find the sum of the first ten terms of a series whose first term is —6 and tenth term 25i. 17. How many terms of the series 15 + 12+9+ • • • • niust be taken to make 45 ? 18. In an A. P. whose first term is 8, the sum of tlie first 9 terms is 324. What is the 9th term ? 19. Find the sum of all odd numbers of two digits. 20. Find the sum of all even numbers between 100 and 300. 21. Find the sum of all numbers between 50 and 250 which are divisible by 4. 22. Show that the sum of the first n odd numbers is 71"^. 23. Show that r+2^ + 3^+ n'= ||(^ + 1) V Suggestion. See Example 6, §247. Remember that (ic+1)*— £c*= 4a73+6a?2+4£t?+l. 24. Insert four arithmetical means betweeji 5 and 25. 25. Insert six arithmetical means between —10 and \. 26. Find three numbers which are in A. P., such that their sum is 18, and such that the product of the first and last is greater than the second by 14. 27. A man had a cistern dug 12 feet deep. The first foot cost $1, the second 11.25, the third $1.50, and so on. What did the digging cost ? 28. A body falls toward the earth at the rate of a feet the first second, 3a feet the second second, 5a feet the third sec- ond, and so on. How far will it fall in t seconds ? 29. A man pays $50 of a debt the first year, $75 the second year, $100 the third year, and so on. In this way he pays the whole debt of $1100. How many years does it require ? 362 ALGEBRA GEOMETRICAL PROGRESSIONS. 250. A geometrical progression is a series of terms in Avhicli the ratio of any term to the preceding term is the same for all terms of the series. This ratio is called the common ratio, and may be either positive or negative. The name geometrical progression is usually abbreviated to G. P. Thus, the series 1 + 2 + 4 + 8 + 16+ • • • ■ is a G. P. in which the common ratio is 2. The series 2— 1 + 1 — 1 + 1—3^4- • • • • is a G. P. in which the common ratio is — |. In either series, if any term be multiplied by the common ratio, the product will be the next term. 251. The nth term of a G. P. Let a stand for the first term of a G. P. ; I for the nth term ; and r for the common ratio. Then, by definition of a G. P., the second term = ar, • the^ third term — ar'^^ the fourth term = ar^^QiG. It is evident from these expressions that the coefficient of a in any term is r, with an exponent less by one than the num- ber of the term. Hence, the nth t€rm = ar''-^ \ that is, l=:ar"-^. Formula yl. This equation, or formula, will enable us to find the value of any one of the four numbers, Z, a, r, n, if the values of the other three are known. Note. — Since n is an exponent, its value cannot in general be found, except by inspection, without the use of logarithms. See Chapter XXVI. In all of the problems in this chapter n can be found by inspection. PROGRESSIONS 363 Example 1 . Find the nioth term of the series 2 + 6 + 18+ Here a=2, r=3, n=9. Hence, Z=2-3«=13,122. Example 2. The tenth term of a G. P. is ^-f^, and the first term is 1. Find the common ratio. Here a=l, n=10, 1=^\y- Hence, -^-g =lr^j and r=}. Example 3. The fifth and eighth terms of a G. P. are ^f and — Ill, respectively. Write the series. Here the fifth term =ar*=^, (1) and the eighth term =ar^=— |ff. (2) Dividing (2) by (1), r'=-^; whence *'=— I- Replacing r by -| in (1), aff =ff ; whence a =3. Therefore, the series is 3 0i 4 8 I 1 6 33 I 64 128 i ^'^5^^fc 252. Sum of n terms of a G-. P. Let the sum of n terms of a G. P. be represented by /S. Then, jS=a + ar'\-ar'' + ar'-{- +ar"-2 + ar»-\ = a(l-^r + r' + r'+ • • • +r"-2+-^n-i^ =a(l:Z^\ §76. 1-r ^ Hence, 5=-^^—^. Formula^. 1— r But, by formula ^, l=ar''-^. Hence, from formula A and formula B, we get A a—r/ 1-r' The five numbers, a, r, ?i, /, /^, of a G. P. we call the elements 364 ALGEBRA If any three of the five elements of a G. P. be known, then the other two may be found by the use of formulas A, B^ and C. Example 1. Find the sum of the first six terms of the series .4M-64-9 + -V-+ We have a=4, r=|, /i=6. Hence, ^^ 4{l-(fr} ^ Example 2. The first term of a G. P. is 3, the sixth term is 9375, and the sum of the first six terms is 11,718. What is the common ratio ? By formula (7, • •11,718 = 3=^^; 1— r ' whence r=5. 253. Sum of an infinite number of terms of a G. P. From § 252 we have \—r This may be written in the form S=- 1—r \—r Now, if r he less than 1, r'* will be less still, and by increas- ing n sufficiently, r" can be made less than any assigned value which we may take as small as we please. Thus, if r=i, t^^tV, ^''=^4, ^'^fK, '^'=-toW, ^"=1™, etc. Now, at the same time that r^ becomes less than any as- signed value, ^ will also become less than any assigned value, however small. Hence, if n be taken sufficiently great, 5 will approach indefinitely near in value to = . 1—r 1—r 1—r PROGRESSIONS '6^)0 Consequently, in a G. P. where the common ratio is less than 1, by taking 71 sufficiently great, the sum of n terms, can be made to differ from :^—-- by a number less than any assigned value, which may be taken as small as we please. Hence, the sum of an infinite number of terms of a G. P, ichose common ratio is less than 1 is defined as the limit ^* That is, when n is infinite and r is less than 1, 1-r S=A-' Formula J9. 1 — r Example 1. Find the sum of the infinite series of terms 9 + 6 + 4+ • • •. Here r=|; hence, the formula >S=T-— may be- applied. We 9 have ^=3— — g=27. This formula can be used to find the value of a repeating decimal. Example 2. Find the value of .3333 This is a G. P. Avhose first term is y\, and common ratio ^V- Hence, ^=-^=1=,]. Example 3. Find the value of .12232323 We have. 12232323 =TVo+To¥oT7+ro¥o^+To^¥oiroo + - Hence, .12232323- •Too T^-J-9 00— 9 900' 254. Geometrical means. If three numbers form a G. P., the middle term is called the geometrical mean of the other two terms. 366 ALGEBRA Thus, in the series 4 + 12 + 36, 12 is the geometrical mean of 4 and 36. If X is the geometrical mean of a and ^, then by the defini- tion of a geometrical progression, we have h X Solving, x=i/ab. Hence, the geometrical mean of two numhers equals the square root of their product. 255. In a G. P. all of the terms between any two terms are called the geometrical means of those two terms. Thus, since 1 + 3 + 9 + 27 + 81 is a G.P., 3, 9, and 27, are geomet- rical means of 1 and 81. Any number of geometrical means may be inserted between any two given numbers. Example 1. Insert three geometrical means between ^ and 32. Since there are to be three geometrical means, | must be the first term, and 32 the fifth term, of a G. P. Hence, by formula A we have ir*=32; whence r=4. Therefore, the required series is 1 + ^ + 2 + 8 + 32. EXERCISE 101. 1. Find the ninth term of the series 1 + 6 + 36+ 2. Findthetenth term of the series Jg—i + 1—4+ 3. Find the eighth term of the series 2 + 3 + 4i+ PROGRESSIONS 367 4. Find the twelfth term of the series 1 2 + -^4~ * ' * ' • X X 5. The second term of a G. P. is i, and tlie eighth term is 2^g. Find the tenth term. 6. Tlie first term of a G. P. is 3, and the tliird term is 6. Find the sixth term. 7. The second term of a G. P. is 3, and the fifth term is ^-j-. Find the fourth term. 8. The common ratio of a G. P. is 3, and the seventh term is 81. Find the first term. 9. The third term of a G. P. is i, and the eighth term is 128. Write the first eight terms. 10. Insert two geometrical means between 125 and —8. 11. Insert three geometrical means between 1 and 4. 12. Insert five geometrical means between 2 and ||-. 13. Find the geometrical mean of 6 and 96. Find the sum of : 14. Five terms of i + i + |+ 15. Twelve terms of 2—4 + 8- 16. Ten terms of If + 22+5/3 17. Six terms of 64-32 + 16- ..••.. 18. Ten terms of -f +i.-5_|_ ...... 19. In a G. P. whose first term is 1, and common ratio 4, how many terms must be added to make 21 ? 20. In a G. P. whose first term is 1, and common ratio — i, how many terms must be added to make |^ ? Find the sum of an infinite number of terms of : ,21. 15 + 5 + 1+ ...... 22. -3-^i-^V+ • 368 ALGEBRA 23. I. + 1 + -J+ , 24. i + i + i+ • . 25. 5-3 + 1- 26. In a G. P. the common ratio is i. What must be the first term in order that the sum of an infinite number of terms may be 80 ? 27. In a G. P. the first term is 5. What must be the com- mon ratio in order that the sum of an infinite number of terms may be 4y-p=o, or A=a, B=b, C=c,- P=p. (2) One, or both, series infinite. In the convergent infinite series a + bx-\-cx- + dx^ + ex*+fx^^gx^+ * ' * '? which is written in ascending, positive, integral powers of x, any term whose value is not zero may be made greater than the sum of all terms that follow it by making x sufficiently small. Let us choose the term clx^. Now let k be greater than any coefficient following d. Then kx\l + x + x^ -}- x^ + • • • • ) is greater than ex* + fx^ + gx^ -{- Or, since l + x + x' + x'^ • • • • =Y37-, X X kx*YZ^ is greater than ex*+fx^+gx^+ * 1 kx But dx^ is greater than kx*YZZ~'> i^ i _ is less than d. X X X X Now, since d is not zero, .. _ can be made less than d, for by taking x sufficiently small, the numerator may be made less than any assigned value, while the denominator will approach 1. * When some of the terms are negative and the negative signs before the terms are changed to positive signs, a new series greater than the oUl arises; lience if the principle is proved for all terms positive, it is evidently true when part of the terms are negative. 378 ALGEBRA Hence, dx" can be made greater than ea;*+/£c^ + <7£«^-f In like manner, ayiy term whose value is not zero in the above series can be made greater than the sum of all the fol- lowing terms. If, now, A^Bx^Cx^^Dx"-^ =a-\-hx-^cx^-^dx^\ be true independently of the value of a?, we have {A-a)^{B-h)x-^{C-c)x^-V{D-d)x'^-\- • • • • • =0. Now, A— a must be either zero or not zero; and if ^— a were not zero, by taking x sufficiently smalL^— « could be made greater than the sum of all terms that follow. But this can not be possible, since the sum of the whole series is zero. Therefore, ^— a=0, qy A=a. Removing A— a, it can be sho^n in like manner that B=h\ thence removing B—h^ it can be shown that C=c ; and so on. 263. Expansion of fractions into series. A fraction may be expanded into a series by the use of the principle of undetermined coefficients. Example 1. Expand ^ ■ in ascending powers of x. Let us assume =A + Bx+Cx' + Dx^- 1 + x where A, B, C, etc., are independent of x. Multiplying by 1+a?, l + x'=A + {A + B)x+{B+C)x' + {C^-D)3(^ + Comparing coefficients, A=l. A + B=0\ whence ^= — 1. ji5+C=l; whence C=2. O+i)=0; whence i)= -2. UNDETERMINED COEFFICIENTS 379 Hence, the required series is =l-x + 2x^-2^ + 1 + x To determine what power of x shall occur in the first term of the expansion, we arrange both numerator and denominator in ascending powers of a?, and perform the first step of the division. Example 2. Expand » ~o' q 4 i^ ascending powers of x. ZX — itr — oX Dividing, the first term becomes ^xr^. Hence, we assume ^~f^ ^ =^3(r^ + A + Bx-\-Cx'+Dx^+ ...... aX — X^ — oX Multiplying by 2iP— x"^— 3a?*, Comparing coefficients, 2A=0; whence A=0. 2^—1 = 0; whence B=l. 2C-A-|=-4; whence C=K^ + |-4)=i. 2D-B-^A=Q; whence D=^{B + ^A)=l. Hence, the required series is 3_4a^ -1^1 + 3^+^^2+3^+ . 2£c-ar*-3a?* EXERCISE 103. Expand to five terms in ascending powers ofx: r=^' l-x^' l + SiC + a;^ „ l-\-X K £C + ic' g ^X — X^-^-X" "^^ T+^* • 2 + a;' ^' x'-x'' $80 ALGEBRA 10. 1;Z^. 12. f-^/-f.. 14. ^^+^ ,, l-2a;^ + a^ ,o 1-2^' ik ^x'^x' 11. -5 — T-i — -^- 13. .^ — ZT-S- 15. ar*— 4a;*— ic'^' ' cc + a;^ — 2£c^* * 1 + a;— aj^* 264. Expansion of surds into series, and extraction of roots. The principle of undetermined coefficients may be used to expand surds into series. The expansion is true only for those values of x which make the series convergent. Example 1. Expand yi + x in ascending powers of x. Assume V'l + x=A + Bx + Cx^ + D,Tcr^ + Ex* + Fx^+ Squaring, by the rule for squaring a polynomial, l+x=A' + 2ABx+{B' + 2AC)x^ + {2AD-h2BC)2(^ + {C' + 2AE+2BD)x*-{-{2AF+2BE+2CD)x^+ Comparing coefficients, A^=l; whence A =1. 2AB=1; whence B=A^=^. 2AC+B'=0', whence C=^^=-h 2A ^ 2AD + 2BC=0; whence D=-^^=^t^, 2AE+2BD+C'=0- whence E=-^'^^±^'=-^j^. ZA . Hence, i/l + a?= 1 + \x—\^ + -^^x^'—^^^x" + Note.— if, in the above example, we use ^=-1, different values for B, C, etc., also will be obtained, and the resulting series will be the other square root. The same method may be used to find the square root of a polynomial which is a perfect square. In this case the series is finite. UNDETERMINED COEFFICIENTS 381 Example 2. Find the square root of 4:—4:X + Qo(f—8a^ + 6x* Evidently the highest and lowest terms of this expression must be the squares of the highest and lowest terms of the root. Hence, the highest term must contain x^' Accordingly, we as- sume V4—4x+Qx'—8x'-{-6x*—4x^+x''=A + Bx+Cx^+Dj(^. Squaring, 4-4x+9x''—8x' + Qx*—4:X^ + x^=A^ + 2ABx+{B^ + 2AC)x^ + {2BC+ 2AD)x' + {C' + 2BD)x' + 2CDx^ + D'x^, Comparing coefficients, 2A5=-4; J52 + 2AC=9; 2BC+2AD=-8', C' + 2BD=6; 2CD=:-4:; D'=l. Each of these equations is satisfied when A =2, B= — 1, C=2 D=-l. Hence, y4—'ix+9x'—8x^ + Qx'—4x^ + x^=2—x+2x^—x\ EXERCISE 104. Expand to five terms in ascending powers of x : 1. y'l-x. 4. i/l-^x-x\ 7. i/l-2a;. 2. i/l + 9a;. 5. yl-x + x\ 8. i/l+2x-x\ 3. i/l-4a;. 6. i/^ + x". 9. i/8 + a^. 382 ALGEBRA Find the square root of : 10. l + 2x+Sx' + 2x' + x\ 11. l + Ax + lOx'-i-Ux' + dx'. 12. lQx*-'60x-nx' + 2^x' + 2b. 13. U-4x'-2ix-^Ux'-22x'-j 17x' + 4x^. 14. l-4«-32a^ + 64a«-64a^ + 12a^ + 48a*. 265. Reversion of series. The series i/=A + I^x+ Cx^-i-Dx^+ • • • • is said to be reverted wlien x is expressed in tlie form of a series of terms written in ascending powers of y. A series may be reverted by use of the principle of unde- termined coefficients. Example 1. Revert the series y=x+x^ + x^ + x*+ • • • • . Assume x= Ay + By^-{-Cy^+ Dy^+ Substituting, x=A{x+0(y'-^x^ + x*+ • • ')+B{x^ + 2x^ + Sx^+ . . . .) + C{x^ + 3x'+ ' ■ .) + D{x'+ • • •)• or, x=Ax+{A + B)x^ + {A + 2B+C)x^ + {A + 3B + SC+D)x*+ • • • •. Comparing coefficients, A=l. A +1?=0; whence B=—A=—l. A + 2B+C=0; whence C=-A-25=l. A + 3B+3O+i)=0; whence D=-A-^B-SC=-1. Hence, x=y—y'^-\-y^—y*+ • • • • . EXERCISE 105. Revert to four terms : 1. y=x—x'^-{-x^—x*-i- • 2, y=x + 2x' + ^x' + 4:X*-{- UNDETERMINED COEFFICIENTS 383 ^y»2 /y»3 ,-y,4 3. y = . + |+|-+|+ . 5. y=x—2x'^-^4x^—Sx*- rv* ^7*2 ™3 ^y,4 -J rO tl/ *o *o 7. y=ic + £c^ + aj''^ + a;^+ • 8. y = a.-g- + ^-y- 266. Partial fractions. Two or more fractions whose sum is a given fraction are called partial fractions. The process of separating a fraction into partial fractions is the opposite of addition of fractions. A fraction may be sep- arated into partial fractions by use of the principle of undeter- mined coefficients. 267. We consider first those cases in which the numerator of the given fraction is of lower degree than the denominator. (1) When each factor of the given denominator is of the first degree^ and no two factors are equal. Since the denominator of the given fraction must be the common denominator of all of the partial fractions, assume that the given fraction equals the sum of all the fractions whose denominators are the factors of tlie given denominator, and whose numerators are expressions independent of the general number involved. Example 1. 1 ft '¥•_!_ Q Separate ^ ^^ \^ . into partial fractions. 4x^-9 Since 4ic2-9=(2ic + 3) (2aj -3), assume 18ir + 3 A B 4^_9-2iC + 3'''2a?-3' 334: ALGEBRA Multiplying by 4a^—9, ■[8x + S=A(2x-S) + B{2x + S), or i8x+S=(2A + 2B)x + 3B-SA. Comparing coeflScients, - 2A + 2B =18, (1) and SB-SA=3. (2) Solving the system (1), (2), we get J.=4, B=5. Hence 1?^±-^=_A_ , _A_. ^®"^®' 4x^-9 2ic + 3 + 2x-3 Since the assumed equation is to be true for all finite values of the general number, we may usually shorten the work by assigning particular values to the general number that will make certain undetermined coefficients vanish. This is shown in the following example. Example 2. Separate -5 — ^—- into partial fractions. Since x^—25x=x{x—5){x + 5), assume x'-75 _A jB_ C x^—25x~ X a? + 5 X— 5* Multiplying by x^—25x, x^ — 75 = A(x + 5)(a?— 5) + Bx{x — 5) + Cx{x + 5) . Now to make the terms containing B and C equal 0, let x=0. Then — 75=— 25A; whence A=3. To make the terms in A and C equal 0^ let x= — 5. Then -50=50B; whence 5=-l. To make the terms in A and B equal 0, let x=5. Then -50=50C ; when C=-l. Therefore, x^-75 ^3__1___1_ ir— 25a? x x + 5 x—5 (2) Wheyi one or more factors of the given denominator are of higher degree than the flrst^ and no two factors are equal. UNDETERMINED COEFFICIENTS 385 In this case evidently the denominators of the partial frac- tions will be the factors of the given denominator as before ; but since the assumed fraction must be general enough to in- clude all fractions with the given denominator, the assumed numerator whose denominator is of the nth. degree in the gen- eral number involved must be the most general expression of the degree n—1 in that general number. Thus, a partial fraction whose denominator is x^ + 2 must be assumed in the form — ,. : if the denominator is £c* + 3, the numerator iC'^ + S ' must be assumed of the form Ax^ -{- Bx^ ^ Cx-\-D\ and so on. Example 1. Separate ' .^ into partial fractions. Since qc^-\-1 = {x-\-^){x'^—x-\-\), assume 5x^ + 1 A . Bx+C + x^ + i ~x 4- i x^—x + 1' Multiplying by o?"^ + 1 , ^x'' + l = A{x''-x^l) + {x + \){Bx-^C), or ^x' + l = {A + B)x'' + {B-A + C)x + A + C. Comparing coefficients, A + B=5; B-A + C=0\ A + C=l. Solving this system in A, B, and C, we get A=2, B=S, C=-l. nence, • ^^i-^^i + ^^_x+l' (3) When ttco 0?' inore /actors of the gw en denominator are the same^ i. e., ichen certain factors occur to a poicer. In this case a repeated factor in the denominator may have as many partial fractions corresponding to it as the number of times the factor is repeated, i. e., as the power of the factor. Their denominator will be this factor, raised, respectively, to \hQ first power, second power, and so on, up to a power equal to 25 386 ALGEBRA the number of times the factor is repeated. This is evident S 4 2 since any such fractions as _— ^ + — -^2 + T^^rjya ^^^ ^^^ ., , . , Sx'-2x + l be united into — ^ ^^-^ — • {x-iy So(^ + Sx^ 18x 8 Example 1. Separate ^^+io^ " ^^*^ partial fractions. Since ic* 4- 43?* = orXa? + 4) , assume 8.x^ + 8x''-18a;-8 _ A BCD x* + 4:3(^ ~x + 4: X x^-af' Multiplying by a?* + 40?^, 83(^ + 8x'-18x-S=Ax^ + Bx\x + 4:) + Cx{x+4:) + D{x+4), or 8o(f + 8x'-lSx-S={A + B)x' + {4:B+C)x' + {4:C+D)x + 4:D. Comparing coefficients, A + B=8,4:B+C=8- 40+D=-18; 4D=-8. Solving this system in A, J5, C, and D, we get A=5, B=3, =0-4, D=-2. Hence, x'-\-4.oe' ~x + 4:'^x x' x^' ' - 268. When the degree of the numerator is equal to, or greater than that of the denominator, the fraction must first be reduced to a mixed expression, then the fractional part separated into partial fractions by the methods of § 267. 9a?^ + 9x^ 6 Example 1. Separate .. 2 , k^_2 ^^^^ partial fractions. Reduced to a mixed expression, by division, :3X — 2+, Since 3icH5.T— 2=(a?+2)(3a?— 1), assume 16^-10 A B 'dx'-\-^x—2~x+2^2>x—l' By the method of §267, we get A=6, B=—2. „ 9£c' + 9i»2-6 006 2 UNDETERMINED COEFFICIENTS 387 EXERCISE 106. Separate into partial fractions : \.A, 2£c^-cc-l • a3^(£cH-l) 2 + 3a; - a;''-4a; + 3 5 + 38a; .. 2a;^- 1 3a;-12 ^- 6^q^5^=^* 27-8af* 13a;-21 ^g 2a3^ + a;^-3a; + 4 _ ^- (ir-l)(a;-2)(a^ + 3)' ' {x' -^-l^ix-X) ^x'-\-\hx 17 6^!z:4^i. ^- \x'^-^:x'-x-\ ' ^X^-l)' 7 -- + ^- + 1^ . 18. Si-SI- '• (a;+l)(i« + 2)(a:+3) a, +a; +i ■ 1 19 -^ + 2«' + 5 8. ^j^- ^*'' (a?-l)(a;+l) „ Sa^'-ia 20 2a;^-g«^-^ ^- (2x-3)»- (a:» + 2)(x' + 2) 10. . r^ ,.. • 21- "'"' 23. 24. 8 + 12a;-2a;^-14a;=^-10a;*-2ar^ ■ x\x+2y CHAPTER XXVI, LOGARITHMS. 269. Exponential equations. In all the equations which we have discussed, the unknown numbers have appeared as bases, with known coefficients and exponents. There are problems which lead to equations in which the unknown numbers ap- pear as exponents. Such equations are called exponential equations. Thus, 2^=16 is an exponential equation, the unknown number appearing as an exponent. The solution of this equation is a? =4. 4^=8 is an exponential equation. Its solution is ir=|, because 4^=V?=8. 270. The solutions of some exponential equations can be found easily by inspection. But in general this is impossible. Thus, to solve 3^=243 is to find the power to which 3 must be raised to give 243. This is seen by trial to be 5. Hence, x=5. But 2^=12 cannot be solved by inspection, x is more than 3, because 2-^=8; and x is less than 4, because 2*=16. Hence, x=3+a fraction. In fact, x here is what is called an incommen- surable number whose exact value cannot be found. The general exponential equation can best be solved by the aid of a set of numbers called logarithms. 271. Logarithms. T\\q logarithm of a number i'^ the expo nent which indicates the power to which a given base must be 388 LOGARITHMS 389 raised to produce that number. That is, if a*=w, then x is the logarithm of n to the base a ; and is written Thus, since 2=^=8, log28=3. Since 3*=81, log381=4. j-v r '- --£■ ^^'=' - ^-? Since 5*= 625, log5625=4. '>*-^ ^- -^ ^ ^' Since 4-2=i:,^-~, log4T'^=~2. Since 9-^=~|=.^V,log9^V=-|. ^x^.^;^- ~%, / It follows that the exponential equation a^ = n and the logarithmic equation aj=log,,?i are equivalent. EXERCISE 107. Express the followijig relations in terms of logarithms 1. 2^=^32. ^ 32-sr3. 7^=343. 5. 5« = 15625. 2. 3* = 81. '^ 4. 10^ = 10000. 6. 4-'=gL-. 7. 5-2 =J-^. 8. 10 2_ 1 2T- ^'^ -^^ — TO 0- ^ -^f^ Vo I 60 ^Express the following relations by means of exponents : 9. log39-2. "3'i;^ 11. log,16 = 2. 13. log,^V=-2. Z'^:^-^ 10. log2l6=4. 12. logs4=f. 14. logio.001 = -3. 15. log,oo.001 = -f. Find the values of the following logarithms: 16. logaS. 20. log,64. 24. log6216. 17. log,327. 21. log, .5. 25. logsyi^. 18. logiolOO. 22. log2.25. 26. log^ol. 19. logio-OOOl. 23. logger 27. log.l. 390 ALGEBRA To the base 4 what numbers have the following logarithms ? 28. 1. U 30. f 32. -i. 29. 3. 31. -2. 33. 5. 34. -4. 272. Fuiidamental principles of logarithms. Fi'om the definition of a logarithm it follows that any posi- tive number, except 1, may be used as the base of the logarithm of any arithmetical number. The following fundamental principles apply to logarithms to any base. (A) The logarithm of 1 to any base is ; that is, This principle follows from «"=1. {£) The logarithm of the base itself is 1 ; that is, %„a = l. This follows from a^ = a. ( C) The logarithm of a product equals the sum of the loga- rithms of Its factors ; that is, ( loa^mn = loq.m + loa^n. \^ To pjrove this, let a^=m^ and ay = n. \ Then m)i = a''-ay=a^+y. Hence, lQgam?i=a; + y = log„m + log„?i. In like manner, this principle can be proved to hold for any number of factors. {!>) The logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor; that is, l^9a{z] =logam—log^n. To prove this, let «^=m, and a«'=n. Then -.=a^-^a?'=Q-c-y^ LOGARITHMS 391 Hence, log„ (^J =a;-y =log„m-log„w. (^) The logarithm of a poicer of a number equals the loga- rithm of the number, multiplied by the exponent of the power; that is, To prove this, let aJ^^n. — Then 7i^ ={a''Y= ap*. Hence, \og^{n^=px=p log^^w. {F) The logarithm of a root of a number equals the logarithm of the nwnber^ divided by the index of the root; that is, To prove this, let Then l/^=l/a^=a''. - Hence, , log„i/ 7i=~ -^ , or - log„;i. Note. —Principle {F) might be considered a special case of {E) , yn being written as the power n^. By the use of the above principles we shall be able to replace the operations of multiplication and division by those of addi- tion and subtraction, and the operations of involution and evolution by those of multiplication and division. Example 1. Express loga"^ in terms of logaO?, loga?/, log„2;, and lo^a^v. \og~~= =\<^axy^-^o^aZ\/'^ By (D). Zy tV / =l0gaa? + l0ga2/'-l0ga2;-l0gal/w By (0). =logaa; + 2 loga2/-loga2;-i log„?<^. By {E) and {F), 392 ALGEBRA Example 2. Express 2 logaic— | logaV + h ^^SaZ as a single log- arithm. 2 logaX-f loga2/ + i log„2;=log„ar^-log„2/^ + loga2^^ By (^). =log„^ By(C)and(i)). Examples. If logio2=.3010, logio3=.4771, find logio|/6T logio|/6=^ Iogio6=|(logio2 + logio8) =^(.3010 + . 4771) = .3890. Example 4. If logio2=.3010, logio3=.4771, logio5=.6990, 3y find log,o^^. 1/800 Factoring, 360=23-3='-5; 800=2^-5^ I/' 360 3 . ' ^^^i«^*^^=logioi/360-logioT/800 =ilog,o2^-3=^-5-ilogio2^-5' = K3 logio2 + 2 logio3 + logioS) - I(5logio2 + 2logio5) =K.9030 + .9542 + .6990)-i(1.5050 + 1.3980) =.1263. EXERCISE 108. Express the following logarithms in terms of log„£c, log.y, log„2j, log„^ .• ''^- ^ 6. log^-!^. 2. log^a^y. /_ ^_ 1/^1/^ K 1 1/^1/ y a; LOGARITHMS 893 B.log.g. 10.1ogA-* ^^ ^ 9 loff (^^V. Ill -" «5~* 2a;y'' 13. l„g„/^+logy_^. 14. log„3^^^ Express as single logarithms : 15. log„a5+log„y + log„2-log„?^. 16. log„aj-log„y + log„^-log„^. 17. 2 log^a; + 2 log„y-2 log„2-2 log„«^. 18. i log„a;-i log„y. 19. }log„^() + |lbg„2. 20. 3 1og„a;-ilog„(y + ^). 21. 3log„g)+2 1og.(| If logio2 = .3010, logio3 = . 4771, logio5 = . 6990, logio7 = .8451, find to the base 10 the logarithms of the following numbers : 22. 18. 29. 70. l/98 3d. ■ ., 23. 15. 30. 210. y 144 24. 20. 31. 1000. . _ 3 ^ • 25. 50. 32. 6|. 37. ^;^l'g . 26. 24. 33. i/rs. 1^5 1 7 27. 21. 34. yW^ ^ (5^3 28. If. 35. 1^180. ''°- (5|)^- 273. Common logarithms. The logarithm of a number to the base 10 is called a common logarithm. Thus, logio6, \og^f,124:, logjo t\j logjo-lOOO, are common- loga- rithms. 394: ALGEBRA The logarithms of all arithmetical numbers to any one base constitute a system of logarithms. The common logarithms of all arithmetical numbers constitute what is called the common system of logarithms.* The common system of logarithms is used in practical calculations. This system is superior to other systems for practical use because its base 10 is also the base of our decimal system of numbers. The rest of the discussion in this chapter will be confined to commo?i logarithms ; and in the following sections the base will be understood to be 10, and will not be written. 274. Characteristic and mantissa. From the definition of a logarithm we have : since 10" = 1, log l'=0 ; since 10^ = 10, log 10 = 1; since 10=^ = 100, log 100 = 2; since 10=^=1000, log 1000 = 3; etc.; and since 10-^ = .l, log .1 = — 1 ; since 10-^ = .01, log .01 = -2; since 10-=' = .001, log .001 = -3 ; etc. It Is evident from the above that the logarithm of a positive integral power of 10 is a positive integer, and that the * The system of common logarithms was introduced in the seven- teenth century by Henry Briggs. Accordingly, logarithms to the base 10 are also called Briggs* logarithms. Another important system is the Napierian system, named after Napier, a contemporary of Briggs. Tlie base. of the Napierian system is the sum of the infinite series . , 1 1111 fI~'~J2'*"p'^|4'^J5 • "^^'^ s""^ ^^ this series is approximately 2.7182818, and is represented by the letter e. Napier himself, however, did not use the base e in his system. While the common system is used in practical calculations, the Napierian system is used in theoretical investigations. LOGARITHMS 395 logarithm of a negative integral power of 10 is a negative integer. Moreover, since 65 is greater than 10^ and less than 10^ log 65 will be greater than 1 and less than 2. Hence, log 65 = l + rt decimal. Also, since 382 is greater than 10^ and less than 10^, log 382 will be greater than two and less than 3. Hence, log Z%'± = 1-\r a decimal Evidently, the logarithm of any positive number, except a positive or negative integral power of 10, will consist of an integral ami a decimal part. Thus, log 825=2.9165, to four decimal places.* The integral part of a logarithm is called its characteristic, and the decimal part is called its mantissa. 275. Determination of tlie characteristic A number having one figure in its integral part lies between 10" and 10\ Hence, its logarithm lies between and 1 ; i. e., it equals + « decimal. A number having two figures in its integral part lies between 10^ and 101 Hence, its logarithm lies between 1 and 2 ; i. 6., it equals 1 + a decimal. Similarly, if a number has three figures in its integral part, its logarithm lies between 2 and 3; i. e., it equals 24-« decimal^ and so on. Therefore, if a number is greater than i, the characteristic of its logarithm is positive, and is less by 1 than the number of figures in the integral ]Kirt of the number. Thus, in log 6841.27 the characteristic is 3. In log 362.781 the characteristic is 2. . Again, a number less than 1, and having no zero imme- * It should be remembered that the decimal part is an incommen- surable number, and may carried to any degree of accuracy required. 396 ALGEBRA diately following the decimal point, lies between lO"" and 10-^ Hence, its logarithm lies between and— 1 ; i. e., it equals — 1 + « decimal A number less than 1, and having one zero immediately following the decimal point, lies between 10"^ and lO-l Hence its logarithm lies between — 1 and — 2 ; ^. e., it equals — 2 + «<^ecma^. Similarly, if a number less than 1 has two zeros immediately following the decimal point, its logarithm will lie between —2 and — 3 ; i. e., it equals — 3-f-a decimal ; and so on. Therefore, if a number is less than i, the characteristic of its logarithm is negative^ and is greater by 1 than the number of zeros immediately following the decimal point. Thus, in log .0683 the characteristic is —2. In log. 00.08 the characteristic is —4. In log .3974 the characteristic is —1. In writing the logarithm of a number the mantissa is always positive, and if the characteristic is negative, the minus sign is written above the characteristic to signify that it applies to the characteristic alone. Tl^s, log .00357=3.5527. This means -3 + .5527. 276. IVie common logarithms of 7iumbers which do not differ^ except in the position of the decimal pointy have the same mantissa. This follows from the fact that changing the position of the decimal point in a number is equivalent to multiplying or dividing the number by some integral power of 10. Thus, 3.261 X 10=32.61; 3.261 x 102=326.1 ; 3.261--102=. 03261. But when a number is multiplied or divided by a power of 10, an integer is added to, or subtracted from, its logarithm; hence the mantissa is not changed. LOGARITHMS " 397 For example, it has been found that log 2680=3.4281; lience, log 268 =2.4281; log 26.8=1.4281; log 2.68 = . 4281; log .268=1.4281. 277. Tables of Mantissas. The common logarithms of sets of consecutive integers have been computed and tabulated. At the end of this chapter is a table which contains the mantissas of the logarithms of all integers from 1 to 1000. Since the mantissa of the logarithm of a number depends only upon the sequence of figures, and not upon the position of the decimal point, only the mantissas of the logarithms of in- tegers need be tabulated. Since the characteristics may be found by the rules of § 276, they are left out of the table. Logarithms may be computed to any number of decimal places, the number depending upon the degree of accuracy re- quired in their use. In the table in this book the manttesas are computed to four decimal places. In this table the first two figures of each number are found in the column headed JV, and the third figure in the horizontal line at the top of the table. The mantissas, with decimal points omitted, are found in the columns headed 0, 1, 2, 3, etc. In finding the logarithm of a number, find its characteristic by § 275, and look in the tal)le for the mantissa. In looking- for the mantissa of a number containing less than three figures, annex ciphers until it has three figures. Thus, to find the mantissa log 38, look for the mantissa of 380. To find the mantissa of log 3, look for the mantissa of log 300. 398 ALGEBRA 278. Use of the table ; to find the logarithm of a given number. {a) When the given number contains not more than three figures, the mantissa of its logarithm is obtained directly from the table. Example 1. Find log 32.7. By § 275 its characteristic is 1. By § 276 the required mantissa is the mantissa of log 327. Look for 32 in the column headed N. Looking along the hori- zontal line of numbers opposite 32, to the column headed 7, we find .5145, the required mantissa. Hence, log 32.7=1.5145. Example 2. Find log .91. By § 275 the characteristic is —1. The required mantissa is the mantissa of log 910, This is opposite 91 in the column headed 0, and is seen to be .9590. Hence, log .91=1.9590. {h) When the given number contains more than three figures, use is made of the principle that when the difference of two numbers is small compared with either of them, the difference of the numbers is approximately proportional to the difference of their logarithms. Example 3. Find log 2.8465. Shift the decimal point until it follows the third figure. The required mantissa is that of log 284,65. Now 284,65 is greater than 284 by .65. The mantissa of log 284=, 4533. , The mantissa of log 285 =,4548. Subtracting, .4548— ,4533=. 0015. Hence, an increase of 1 in 284 causes an increase of ,0015 in the corresponding mantissa. Therefore, an increase of .65 will cause an increase of .65 X -0015, or approximately .0010, in the mantissa. LOGARITHMS Adding .0010 to the mantissa of log 284 gives .4543. Attaching the characteristic, we have log 2.8465=0.4543. Example 4. Find log .008214. The characteristic is —3. The required mantissa is the mantissa of log 821.4. Mantissa of log 821^.9143. Mantissa of log 822 =.9 149. .9149-.9143=.0006. .4X.0006 = .0002. .9143 + . 0002=. 9145. Hence, log .008214=3.9145. S99 EXERCISE 109. Find the logarithms of: 1. 215. 11. 100. 21. .06843. 2. 673. 12. 900. 22. 4268.4. 3. 940. 13. 1. 23. 1.096. 4. 717. 14. 2. 24. .00012. 5. 4,62. 15. 1684. 25. 99.99. 6. 19.9. 16. 34.27. 26. 2031.7. 7. 830. 17. 100.5. 27. .0083326 8. 16. 18. 926.81. 28. ,50416. 9. 29. 19. .8632. 29. 68593. 10. 8.5. 20. .00315. 30. .074803. 279. To find a number whose logarithm is given. («) When the given mantissa is found in the table^ the sequence of figures of the required number may be obtained by reversing the process (a) of § 278. The position of the decimal point is determined by reversing the rules in § 275. 400 ALGEBRA Example 1. Find the number whose logarithm is 1.9325. Looking in the table, we find the mantissa .9325 opposite 85 and in the column headed 6. Hence, .9325= the mantissa of log 856. Since the characteristic is 1, the number must contain two figures to the left of the decimal point. Hence, 1.9325=log 85.6. Example 2. Find the number whose logarithm is 2.5289. From the table we have .5289=mantissa of log 338. Since the characteristic is —2, the number must be less than 1, and must contain one zero immediately to the right of the decimal point. Hence, 2.5289=log .0338. (b) 'When the given mantissa is not found in the table^ the required number is obtained by reversing the process of (i), § 278. Example 3. Find the number whose logarithm is 3.6496. The mantissa .6496 is not in the table, but the mantissas of the table between which it lies in value are .6493 and .6503. Now .6496 is greater than .6493 by .0003. Also .6493= mantissa of log 446, and .6 503= mantissa of log 447. Subtracting, .6503-. 6493=. 0010. Hence, an increase of .0010 in the mantissa .6493 causes an in- crease of 1 in the corresponding number. Therefore, an increase of .0003 will cause an increase of .0003-^.0010, or .3, in the number. Adding .3 to 446 gives 446.3. Therefore, .6496=mantissa of log 446.3. Since the characteristic is 3, we have 3.6496=log4463. LOGARITHMS 401 Example 4. Find the number whose logarithm is ^3.8684. This mantissa is not in the table, but the next less mantissa is .8681, and the next greater is .8686. Now .8681=mantissa of log 738, and .8686= mantissa of log 739. .8686-. 8681 = . 0005, and .8684-. 8681 = . 0003. .0003-^. 0005 = . 6. ■ 738 + . 6=738.6. Hence, .8684= mantissa of log 738.6. Locating the decimal point, we get 3.8684=log .007386. EXERCISE 110. Find the numbers whose logarithms are : 1. 1.7582. 6. 6.6884. 11. 4.0096. 16. 5.0220. 2. 3.8615. 7. 2.4786. 12. 1.4703. 17. 3.0392. 3. .1847. 8. 3.6021. 13. 2.9765. 18. 4.4756. 4. T.4609. 9. 3.7251. 14. 2.8460. 19. .8735. 5. 2.6804. 10. 2.8976."^ 15. 1.4072. 20. 1.5734. 280. Cologarithms. The cologarithm of ic is the logarithm of — From this definition it follows that colog x=/og(-J =log l-/og x=—log x; that is, the cologarithm of a number may he obtained by changing the sign of its logarithm. Since the mantissa of a logarithm is alwaj^'s written positive, to change the sign of the logarithm would give a negative man- tissa. In order to avoid a negative mantissa in the cologarithm, 26 402 ALGEBRA usually in place of —log £c, its equivalent, (10— log cc) — 10, is used. Evidently, therefore, the cologarithm of a number may be found by subtracting the logarithm of the number from 10, and indicating the addition of —10 to the remainder. Example 1. Find colog 485. log 485=2.6857. Hence, colog 485=(10-2.6857)-10=7.3143-10. If log X lies in absolute value between 10 and 20, then in order to make the mantissa positive we use for colog x the form (20-log£c)-20. In general, if convenient, we may use the cologarithm in the form {a— log x)— a, where a is any number that will make the mantissa positive. Example 2. Find colog 267000000000. ' log 267000000000=11.4265. Hence, colog 267000000000=(20-11.4265)~20=8.5735-20. If the characteristic of the logarithm is negative, then the — 10, or —20, will disappear from the value of the cologarithm. Example 3. Find colog .00814. log .00814=3.9106. Hence, colog .00814= (10-3.9106) -10, = (10 + 3-.9106)-10, =12.0894-10, =2.0894. It has been shown that to obtain the logarithm of a quotient, we subtract the logarithm of the divisor from the logarithm of the dividend. Since colog x= —log «, instead of subtracting the logarithm of the divisor ice may add its cologarithm. Example 4. Find log 1 6827 ^^^ 8iT6 Hence, LOGARITHMS 6827 81.6' log 6827 + colog 81.6. log 6827= 3.8342. colog81.6= 8.0883-10. log ^=11.9225-10, = 1.9225. 403 EXERCISE 111. Find the cologarithm of : 1. 72.8. 6. 68.27. 11. .00321. 2. 691. 7. 1375. 12. .6847. 3. 4.56. 8. 261.3. 13. .0315. 4. 326.7. 9. 18329. 14. .0000623. 5. 12.34. 10. 43165. 15. .0004721. 16. .0005638. 281. Computation by logarithms. By the use of logarithms long computations can be avoided. Multiplication^ division, involution, and evolution, may be replaced by addition, sub- traction, midtiplication, and division, respectively. In using logarithms with negative characteristics it is some- times convenient to add some number to the logarithms, in order to make the characteristics positive, then to indicate the subtraction of the number. Thus, 3 .6271=^7.6271-10. Example 1. Find |^ ."00327. We first find log i^. 00327. log 1^.00327 =i log .00327 § 272, F. =i (3.5145) =^ (2.5145-5) =^5029-1 =1.5029. 40^ ALGEBRA ]Sl4>w 1.5029=log .318555. 621.3X.03247 71.8 Hence, x7.00327=. 318555. Example 2. Find the value of Taking the logarithm, ^^^621.3X.03247_^^^ 621.3 + log.03247 + colog 71.8. 71.8 log 621.3=2.7933, log. 03247 =2.5115=8.5115-10, colog 71.8= 8.1439-10 , Sum=19.4487-20 ^ =1.4487. But 1.4487=log .281. rru — ^e^^^ 621.3X03^47 ^q^ „^r^l„^^ir^n^■a^■^ xj.x\:/iCJLvi.c, " 71 8 .«^*, «^|^a.v^^x*xxo*wv.u.j . Example 3. Find the sixth power of .428. log .428«=6 log .428 =6(1.6314) = 6(9.6314-10) = 57.7884-60 • =3.7884. But 3.7884=log .00614. Therefore, .428«=.00614, approximately. Example 4. Find the value of 2.476 X (-1.724). The sign of the product must be determined by the laws of signs. By logarithms we obtain merely the absolute value of the product. . We have log 2. 476 =.3938 log 1.742=. 2410 .6348 Now .6348=log 4.313. Therefore, 2.476 X (-1.742) = -4.313, approximately. LOGARITHMS 405 > 5/- ^ ^ ^^H^f-J-y Example 5. Find the value of ^ / • Q27^ X 32 . 6 X |/ 54^«-j:„ , ,or / .Q27^X32.6Xt/ 54Kc^. or ' •' Eepresent this expression by x. Then log a?=i(4 log .027 + log 32.6 + 1 log 542 + 1^ colog 4.12 + 1 colog 7.14+^ colog 6.28) 4 log .027=7.7256=3.7256-10 log 32.6=1.5132 i log 542= .6835 i colog 4.12=^29. 3851-30) = 9. 7950-10 i colog 7.14=K49.1463-50) = 9.8292-10 ^ colog 6. 28=i(59.2020-60) = 9. 8670-10 loga?= 5) 35.4135-40 7.0827-8 = 1.0827. But r.0827z =log .12097. Therefore, x= =.12097, approximately. EXERCISE 112. Find by logarithm s the value of : 1. 71.6X.327. 8. .1965--18.97. 2. 1.068 X. 0039. 9. 6.765-f-(-.01286) 3. 4. 5. 6. 681.7X4.235. 3.1416X1728. •1.414x(--0632). (-4617)X(-.03269). 10. 11. 12. 5334X.02374 -47.43X3.246* 2.476x73.81 .524x6184* 1657 7. 7.631--6214. 1.025X326.81* 13. 17.86^ 16. 219.4^ 17. V 9.268". 14. .06814'". 16. (- -.2596)*. 18. i>6278.i: 406 ALGEBRA 19. i> -20035. 21. .684*. ^^- ('^A)*- 20. .068^ 22. (^)«. 24. 3.69i-^. 25. (.6827X.114)^ 26. y 27* X -028^X16.75^ y i/629Xi>87I0Xt5/12C3* 5 / 27. 98.7i/ .068X1/28.59 ' ^ i^6Xi>206:4Xi/:009l* 282. The solutions of exponential equations. The solution of an exponential equation may be obtained by the aid of logarithms. Example 1. Solve 52^-11(5^) +24=0. Factoring, (5^-8)(5^-3)=0. ■ Hence, 5^=8, (1) or 5^=3. (2) From (1), taking the logarithm of both members, X log 5=log 8. Therefore * a,-l2g_§_:i9031_ ±nererore, ^~log 5~.6990--^''^^^^- From (2), a;log5=log3. Therefore r-^-^^-^i^- 882^ ineretore, x-^^^ .--^gg^-.6825. Example 2. Solve 5*^-^=4^-1. Taking the logarithms of both members, (3^-l)log5=(.r-l)log4. Removing parentheses, Sx log 5— log 5=x log 4— log 4, or ar(3 log 5 -log 4)= log 5— log 4. Hence, ^^ Jog 5 -log 4 3 log 5-log 4 _ .6990-. 6021 ■"2.0970 -.6021 =0648. LOGARITHMS 407 EXERCISE 113. Solve : 1. 3^=81. 6. 23^=25. 9. 4^+1 = 8-2^+2. 2. 5'--25 = 0. 6. 4^-1 = 3^+1. 10. '\/'2^^=i/¥=^. 3. 2-^ = 64. 7. 2^-1 = .32^-5. 11. 1/3^1=^ =3»-^. 4. 3^=15. 8. 52^—12^+1=0. 12. 2'««'^=8. 13. 5'«'^^ = 625. 14. 176.82 = 2.36. 15. 32-^-4(3-)-12 = 0. 16. 2*^-3(22-^) =4. 17. 2(4^)-4^-6-0. EXERCISES FOR REVIEW (VIII). 1. Find the value of a in a+2 4-ffl_7 a — 1 2a 3' 2. Find the value of t in 16 l/«^-l + 6=; 3. Solvea!(£c''-4) + (a3-2)=0. . . 4. Solve for x and y ^^^ + y^-*- 5. Solve (a;-l)(£c + 2)(aj^-6ic + 9)=0. 6. On how many nights may a different guard of 5 men be taken from a body of 26 men ? 1\2« 7. What term in (x-] — j does not contain x ? 8. What is 'dn undetermined coefficient ? 408 ALGEBRA 9. What is the theorem of undetermined coefficients ? Upon what principles did the proof in this book depend ? 10. What use can be made of the theorem of undetermined coefficients ? Make use of the method of undetermined coefficients in the following : 11. Expand .. _o to five terms in ascending powers of x. 12. Expand ^_ _ i mto a series. \—x 13. Expand .. . , ^ into a series. 14. Expand j/l—Sx to five terms in ascending powers of x, in two Avays. 15. What is reversion of series ? Revert to four terms y=x—Sx^-i-bx^—7x*+ • • • . 16. What are partial fractions ? 17. In what way can a given fraction be separated into partial fractions ? 18. In what form must the partial fractions be assumed when each factor of the -given denominator is of the first degree, and two or more of the factors are equal ? Illustrate. 19. In what form must numerators be assumed when a factor of the given denominator is of higher degree than the first, and no two factors are equal ? Why is this ? 20. Separate into partial fractions : (n\ ^ + ^^-^' r^^ 2^ + 2 ^""^ {i-x){i-^xy' (^) ^(^17* LOGARITHMS 409 21. What is an exponential equation ? Illustrate. 22. Solve 2^ = 128; 4- = G4; 9'^ = 27. 23. Give the integers between which the solutions of the following equations lie : 2^ = 3; 7-^ = 100; 5^ = 600. 24. How can the solutions of exponential equations be obtained when they can not be seen by inspection? 25. What is a logarithm 'i 26. To what exponential equation is logio65 = i:c equivalent? 27. Find the value of 10^^49 ; log g 216; loggyV^ ^^^a^^ ^ log„a^; log J; log^a. 28. What is a system of logarithms ? 29. What are common logarithms ? 30. Give in terms of log^aj and log„y, {a) log^icy ; ih) iog„(^^) ; (c) logx; (^0 log^r X. 31. Prove the following identities : («) log,.(m-f-/i) —\og^m—\o^^n ; {h) log6mP=^logftm; (c) logioeXlog,10 = l. 32. Give the equivalents of the following and show how you get them : {a) log„a; {h) logj; (c) lcg„(l-~a) ; {d) log,o.001; (6) ilog^S. 33. What is the value of \ log39-2 log^S + log^a? 34. If a number is not an exact power of 10, what kind of number is its common logarithm ? 36. Define characteristic ; mantissa. 36. How do you obtain the characteristics of the common logarithms of numbers? 37. Give the characteristics of the following logarithms : log 628.75; log 1.864; log .00031 ; log .681; log 6931.7. 410 ALGEBRA 38. Does a change in the position of the decimal point in a number affect the value of the mantissa of its logarithm ? What determines the value of the mantissa ? Why is this ? 39. Explain how to find the mantissa by using the table of this book, when the given number has (a) less than three digits; (b) three digits; (c) more than three digits. Upon what principle does this last depend ? Is the result absolutely correct ? 40. Find the logarithms of 61 ; 372 ; 4 ; 3180 ; 96.4 ; 132.67 ; 4166.8; 1.726; .065; .0002; .68532. 41. Explain how to find the number corresponding to a given logarithm, {a) when the given mantissa is given in the table, (b) when the given mantissa is not given in the table. 42. Find the numbers whose logarithms are 2.3385 ; 1.8998 ; 3.7528 ; 3.8594 ; .6300 ; T.4835. 43. Define cologarithms. When would you use the colog- arithm of a number? Are cologarithms necessary ? 44. Find colog 6.73. 45. What is the advantage in using logarithms in long computations ? "46. By use of logarithms find 32.61 X7.26--403. 47. By use of logarithms find i^671.4. 48. By use of logarithms solve («) 3*"— 13-3^ + 36 = 0. {b) 30^+145x^52^-23.^ 49. Transform |/ ^-^ into a form adapted to computation by tables whea a, h, and c are definite numbers. / TABLE OF MANTISSAS 411 N. I 2 3 •4 5 6 7 8 9 10 0000 0043 0086 0138 0170 0212 0253 0294 03^ 0374 ii 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 12 0793 0828 0864 0899 0934 0969 1004 1038 1073 1106 13 1189 1173 1206 1339 1371 1303 1335 1367 1399 1430 1 14 1461 1493 1523 1553 1584 1614 1644L 1673 1703 1733 1 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 3014 i 16. 2041 2068 2095 3133 8148 2175 2201 2227 3353 3379 17 2304 3330 235o 3380 2405 2430 2455 2480 2504 8539 18 2553 2577 2601 3635 3648 2672 2695 2718 2T4^ 2765 19 2788 2810 2833 2856 2878 2900 2923 2945 2967 2989 2iU .3010 3033 3054 3075 3096 3118 3139 3160 3181 3201 21 3223 3>43 3363 3384 3304 3324 3345 3365 3385 3404. 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 24 3802 3830 3838 3856 3874 3892 3909 3927 3945 3962 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 26 4150 4166 4183 4300 4216 4233 4249 4265 4281 4298 27 4314 4330 4346 4363 4378 4393 4409 4425 4440 4456 28 4472 4487 4503 4518 4533 4548 4564 4579 4594 4609 29 4624 4639 4654 4669 4683 4698 4713 4728 4742 4757 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 31 4914 4928 4943 4955 4969 4983 4997 5011 5024 5038 32 5051 5065 5079 5093 5105 5119 5132 5145 5159 5172 iT 5185 5198 5311 5334 5237 5350 5263 5276 5289 5302 34 5315 5338 5340 5353 5366 5378 5391 5403 5416 5428 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 36 5563 5575 5587 5599 5611 5623 5635 5647 5658 5670 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5786 -sa- 5798 5809 5933 5831 5933 5833 5944 5843 5955 5855 5966 5866 5977 5877 5988 5888 5999 5899 6010 39 5911 40 6021 6031 6043 6053 6064 6075 6085 6096 6107 6117 41 6128 6138 6149 6160 6170 6180 6191 6201 6212 6333 42 6233 6343 6353 6363 6274 6284 6294 6304 0314 6335 43 6335 6345 6355 6365 6375 6385 6395 6405 6415 6425 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 6533 45 6532 6543 6551 6561 6571 6580 6590 6599 6609 6618 46 6628 6637 6646 6656 6665 6675 6684 6693 6702 6713 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6808 48 6812 6831 6830 6839 6848 6857 6866 6875 6884 6893 49 6902 6911 6930 6938 6937 6946 6955 6964 6972 6981 50 6990 6998 7007 7016 7024 7033 7042 7050 7059 7067 51 7076 7084 7093. 7101 7110 7118 7136 7135 7143 7153 52 7160 7168^ 7ir7 7185 7193 7202 7310 7218 7226 7335 53 7243 7351 7359 7267 7275 7284 7393 7300 7308 7316 54 7324 7332 7340 7348 7356 7364 7373 7380 7388 7396 'ii\ 41^ ALGEBRA N. 55 1 2 3 4- 5 6 7 8 9 7404 7412 7419 7427 7435 7443 7451 7459 7466 7474 56 7482 7490 7497 7505 7513 7520 7528 7536 7543 7551 57 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 E8 7G;J4 7642 7649 7657 7664 7672 7679 7686 7694 7701 59 7709 7716 7723 7731 7738 7745 7752 77G0 7767 7774 69 7783 7789 7796 7803 7810 7818 7825 7832 7839 7846 61 785a 7860 7868 7875 7882 7889 7896 7903 7910 7917 62 7924 7931 7938 7945 7952 7959 7966 79^3 7980 7987 63 7993 8000 8007 8014 8021 8028 STJ^ 8041 8048 8055 64 8062 8069 8075 8082 8089 8096 8102 8109 81^6 8122 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 66 8195 8202 8209 8215 8222 8228 8235 8241 8248 8254 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 70 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 71 8513 8519 8525 8531 8537 8543 8549 855.';' 8561 8567 72 8573 8579 8585 8591 8597 8603 8609 8615 8621 8627 73 8633 8639 8645 8651 8657 8663 8669 8675 8681 8686 74 8692 8698 8704 8710 8716 87^2 8727 8733 8739 8745 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 76 8808 8814 8820 8825 8831 8837 8842 8848 8854 8859 77 8865 8871 8876 8882 8887 8893 8899 8904 8910 8915 78 8921 8927 8932 8938 8943 8949 8954 8960 8965 8971 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9C25 80 9031 9036 9042 9047 9053 9058 9063 9069 9074 9079 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 83 9191 9196 9201 9206 9212 9217 9222 9227 9232 9238 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 87 9395 9400 9405 9410 9415 9420 9425 9430 9435 9440 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 9489 89 9494 9499 9504 9509 9513 9518 9523 9528 9533 9538 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 91 959a 9595 9600 9605 9609 9614 9619 9624 9628 9633 '92 9643 9647 9652 9657 9661 9066 9671 9675 9680 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 95 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 97 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 98 9912 9917 9921 9926 9930 9934 9939 9943 9948 9952 99 995C 9961 9965 9969 9974 9978 9983 9987 9991 9996 «'n APPENDIX A. SQUARE AND CUBE ROOTS BY FORMULA. (SUPPLEMENTARY TO CHAPTER VII). 1. In Chapter VII the processes of extracting the square and cube roots of expressions by inspection were discussed. We shall here show how to extract the square root or the cube root of a polynomial by use of a formula. This is a gen- eral process which may be used in case the method by inspec- tion fails. We may then show how to extract the square root or cube root of any arithmetical number by use of a formula. 2. Square roots of polynomials. The square root of any pol- ynoniial will he a polyno7mal. If the root has only two terms, it will be of the form a-Vb. In that case the rule for obtaining the root comes from the identity The work is usually arranged as follows : a2 + 2aZ» + d' | a + ^, root j^ la^h I 2ab-^h- 2ab-\-b'= 2ab + b'' When arranged in descending powers of a, the first term of the root, a, is the square root of the first term of the polynomial, a^. Subtracting a"^ from the polynomial, we get the remainder 2ab+b'^. Evidently the second term of the root, &, may be ob- 1 2 APPENDIX tained by dividing the first term of this remainder, 2a&, by 2a, OT twice the part of the root already found. The divisor 2a is called the trial divisor. If to the trial divisor we add &, the new term of the root, we get 2a + b ; and if this in turn be multiplied by 6, the product, 2ab + b^, will be the above remainder. Hence, 2a + 6 is the whole or complete divisor. And a + ?> is the entire square root. The above process will determine the square root of any polynomial whose square root is a binomial. Example 1. Find the square root of 9a?^— 12xi/ + 4i/2. Here, as in all cases, the given polynomial should first be ar- ranged in descending or ascending powers of some letter. This expression is already written in descending powers of x. The work will be as follows : a^ + 2ab + b^= 9x'-12xy-\-4y^ a'= 9x' a + b 3x—2y, the root 2a + b=6x-2y 2ab + b'= -12xy + 4y' — 12xy + 4y^ Since 9x^—12xy + 4y^ is of the/or?7i a"^ + 2ab + b\ therefore a^=9x\ Hence, a=Sx, the first term of the root. Subtracting 9x^ we have for remainder, —12xy + Ay'^. Now 2a, the trial divisor, becomes 6ic. Dividing —12xy, the first term of the remainder, by 6x, we obtain —2y, the value of b of the formula. Adding this to 6x, the complete divisor be- comes 6a;— 2^/. Multiplying this by —2y, we have —12xy + 4:yK Subtracting this from the remainder —12xy + 4:y^ leaves no remainder. Hence, the entire root is 3x—2y. If the root contains three terms, we have by grouping terms, the identity APPENDIX 3 Hence, when the root has three terms^ the first two may he found as above ; then hy grouping terms these two tnay he used as one, and the third term ohtained hy a repetition of the p>ro- cess used to ohtain the second term,. In like manner, when a root contains four terms the first three may he used as one, and the fourth ohtained by another repetition of the process used to ohtain the second. This process may he extetided to any number of terms. Note. — The student should take care first properly to arrange the order of the terms in any case, either in ascending or descending powers of some letter. Each remainder should also he arranged like the original expression. Example 2. Extract the square root of 16x* + 1 — 3a?+24a^ + Sa?^. First arrange in descending powers of x. The work is as follows : a +b +c 4:X^ + SX — ^ a'= 16x* + 2^x^ + 5x'- 16x' -3a? + i 2a + b=8x^ + Sx 2ab + ¥= 24x' + 5a?2- -3x + i 2{a + b)+c=Sx^ + 6x- 2cla + b) + c^= -\ -40^=^- -4x2- -3x + i -3x + i Since a^=\^x^, the first term is 4x^ Hence, the trial divisor, 2a, is 8x^. Dividing 24ar* by this gives 3x, the second* term of the root. Adding Sx to the trial divisor gives the complete divisor, 8x^ + 3x. Multiplying the complete divisor by Sx gives 24:X^ + 9x^. Subtracting this product from the first remainder leaves — 4a?^— 3a?4-5, the second remainder. Now using the root found, 4^x^ + 3x as one term, the new trial divisor becomes 8x^ + Qx. Dividing — 4x^ first term of the remainder, by 8x^, the first term of the trial divisor, gives — |, the third term of the root. 4, APPENDIX Then the complete divisor becomes Sx^ + Qx—^. Multiplying this by — ^, and subtracting, the third remainder becomes zero. Hence, the entire root is 4a?^ + 3j?— ^. Note. — Since every expression has two square roots, differing only i!i sign, a second expression for the root in any case may be obtained by changing tlie signs of the terms in the root found. Thus, in Ex- ample 2, another root is ^—dx—ix^. EXERCISE 1. Extract the square roots of the following : 1. Ax' + 4x + l. 5. 8«-4a^ + a* + 4. 2. ic^ + 25y- — lOicy 6. l + %x—x'' + Sx' — 2x' + x\ 3. ^x' + x'-2x' + 4:-4x. 7. 49aj« + 42£c«-19£c*- 12.^^ + 4. 4. 4a* + 49-3a'-70a + 20«l 8. x*-2x'i/-2xf + SxY + y\ 9. x'-Qx^>/ + UxY-12xi/' + 4:i/\ 10. y* + 4y + 4y^ + 2y + 4 + i,. 12. x^-^Sx + 12-^ + i, 11 o . «' I o A I A4 , 2^' I ^' 19 2 lla^^3«^^9a* Find to three terms the approximate square roots of : 14. 1 + x. 15. 1-2CC. 16. a' + l. 17. m^ + 3. 3. Square roots of arithmetical numbers. A^ii/ arithmetical number is in nature a polynomial. Thus, 5263=5000 + 200 + 60 + 3 .= 5-1000 + 2-100 + 6-10 + 3 = 5-103 + 2-102 + 6-10 + 3. APPENDIX 5 Hence, the square root of an arithmetical number will be ob- tained in practically the same manner as the square root of a polynomial. The squares of the numbers 1, 2, 3, • • * * 9, 10, are 1, 4, 9, • • • • 81, 100, respectively. Hence, the square root of an integer of one or tivo figures is a number of one figure. The squares of the number 10, 11, • • • • 99, 100, are 100, 121, • • • • 9801, 10000, respectively. Hence, the square root of an integer of three or /our figures is a number of two figures. Likewise, the square root of an integer otfive or six figures is a number of three figures. And so on. Therefore, ^/* the figures of an integer are marked off from right to left in groups of two^ the number of figures in the square root will he equal to the number of groups^ any one figure remaining on the left being counted as a group. Thus, marked off into groups, 21904 becomes 2'19'04'. Since there are three groups, the square root of 21904 will contain three figures. If the square root of a number be a number of two figures, we may denote the tens of the root by a and the ones by b. Then the root \fill be represented by a-^b. Hence, the given number will be represented by a^-^^ab + b"^. The use of the identity \/a' + 'lab^b^ = a-\-b to extract square roots of arithmetical numbers is best shown by examples. Example 1. Find the square root of 3969. Pointing off, we have 39'69. Since there are two groups^ the root must be a number of two figures. Since 60^ is less than 3969, and 70^ is greater than 3969, the root must lie between 60 and 70; i.e., the tens' figure of the root is 6, the square root of Q APPENDIX the greatest square in the left-hand group of the given number. Therefore, in the above identity, a denotes 6 tens, or 60. The work may then be performed as follows : I a + b 39'69' | 60 + 3=63 , root a''= 36 00 2a=120 2a + 6= 123 2ab+b'= 369 369 Since a=60, a^=SeOO. Subtracting 3600, the remainder is 369. The trial divisor^ 2a, becomes 120. Dividing 369 by 120 gives approximately 3. Hence, &is proba- bly 3. Therefore, the complete divisor, 2a 4- 6, becomes 123. Multiplying 123 by 3 gives 369, Subtracting this from the first remainder leaves zero. Hence, 60 + 3, or 63, is the required root. If the square root be a number of three figures, we have, by grouping terms, (a + b + cy={a + by + 2(a^b)c + c\ Hence, when the root is a member of three figures., the first two may he found as above ; then their sum may he used as one, and the third term obtained by a repetition of the process used to obtain the second. In like maimer., when a root contains four figures., the sum of the first three may be used as one, and the fourth obtained hy another repetition of the process used to obtain the second. The process may be extended to any number of figures. APPENDIX 7 Example 2. Find the square root of 203401. Pointing off we have 20'34'01'. Hence, there are three figures in the root. The work is as follows : 20'34'01' a"^ 16 00 00 a-\-h-\-c 400 + 50 + 1=451, root 2a = 800 2a + 6=850 2ah + h''= 43401 42500 2(a + 6)=900 2(a + 6) + c=901 2c{a + h) + c''= 901 901 The largest square in 20 is 16. Hence, a^= 160000. Therefore, a = 400. Subtracting 160000 leaves 43401 for remainder. The first tiHal divisor, 2a, is 800. Dividing 43401 by 800 gives approximately 50. Hence, the complete divisor, 2a + b, is 850. * Multiplying this by 50, we get 42500. . Subtracting this from 43401 leaves 901 for second remainder. Now adding 400 and 50 gives 450, a + 6. The second trial divisor, 2{a + b), is 900. Dividing 901 by 900 gives approximately 1, the third figure of the root. Hence, the complete divisor, 2{a + b) + c, is 901. Multiplying this by 1, we get 901. Subtracting this from the second remainder leaves 0. Hence, the required root is 451. When the square root of a number has decimal places^ the number itself will have tioice as many. Thus, (.23)2=. 0529. 8 APPENDIX Therefore, to mark off a number which contains a decimal^ begin at the decimal point and mark to the left and to the rights putting two figures in each group. Thus, 723.618 will become 7'23'.61'80'. Example 3. Find the square root of 50.9796. a+ 6+ c 7.00 + .10 + .04=7. 14, root 50'.97'96 a^= 49.00 00 2a= 14. 00 2a + 6=:14.10 2ah + h''= 1.97 96 1.4100 2(a + 6) = 14.20 2(a + 6) + c=14.24 2c(a + 6) + c'= .5696 .5696 An approximation to the value of the square root of a number which is not a perfect square may be obtained to any degree of accuracy desired. Note. — For convenience in writing, each new trial divisor, which is the sum of the parts or the root already found, may be denoted, respec- tively, by a, a', a", etc., and each new term denoted by h, h', h", etc., as in the following example. Example 4. Find to three decimal places the square root of 2. Since we want three decimal places it is convenient to annex 6 ciphers. The work is as follows: 2.'00'00'00' 1.00 00 00 1 + .4 + . 01 + . 004=1. 414 2a=2. 2a + 6=2.4 2ah^h''= 1.00 .96 2a'=2.8 2a'+6'=2.81 2a'h' + h'^= .04 .0281 2a"=2.S2 2a" + &"=2.824 2a"h" + b"'= .0119 .011296 APPENDIX EXERCISE II. Find the square root o^ : I. 7396. 2. 1849. 3. 26244. 4. 41209. 5. 12.25. 6. 146.41. 7. 125.44. 8. 4.6225. 9. .026244. 10. 2611.21. Find to three decimal places the square root of : II. 5. 12. 3. 13. .6. 14. 105. 15. 371. 16. .75. 17. 11.8. 4. Cube roots of polynomials. If the cube root of a polynomial has only tico terms, it will take the form, a-\-b, and the rule for extracting the cube root will come from, the identity __ The work is usually arranged as follows : d^ + ?,ci^h-{-?>ab'^ + Jf \a + h, root a^ ■ "■ 3a2 + 3a6 + 62 ^a^h + ?>ah^-^lf: 'Sa'b + 3ab' + ¥ 'Sa'b + Sab' + h^ When arranged in descending powers of a, the first term of the root, a, is the cube root of the ^rs^ term of the polynomial, a'\ Subtracting a^ from the polynomial leaves the remainder Sa''b + 3ab'' + b\ The second term of the root, 6, may be obtained by dividing the first term of this remainder, 3a^6, by 3a^, or three times the square of the part of the root already found. This divisor, 3a^ is the trial divisor. If to the trial divisor we add 3a6, or three times the product of the new term of the root by the old part of the root, and &^ or the square of the neiv term of the root^ we obtain 3a^ + 3a6 + b^ ; and if this in turn be multiplied by 6, the product, Sa^b + Sab^ + b^, will be the above remainder. 10 APPENDIX Hence, Sa^-\-^ab+b^ is the complete divisor. If the complete divisor be multiplied by b and the product subtracted from the first remainder, the second remainder becomes zero. The above process will determine the cube root of any polynomial whose cube root is a binomial. Example 1. Find the cube root of 8x^—S6x*y^ + 54x^y*—27y^. This is arranged in descending powers of x. The work is as follows : I a + b 83C^-S6xY + UxY-27y^ \2x'-Sy\ root a»= 8x^ Sa'=12x' Sa' + Sab + b'=12x'-18xY + ^y' 3aH7+3ab'-\-b'= -36xY + 54^V— 272/^ -36;rY + 54a;V-27?/« Since a'=8a?*, then a=2x^ the first term of the root. Subtracting 8a^ leaves —36xY + 54x^y*—27y^. The trial divisor, 3a^, becomes 12x*. Dividing —SQx^y^, the first term of the remainder, by 12a:* gives — 32/^ the second term of the root. 3a6, three times the product of 2x^ and — 3?/^ is —18x^y\ The b^ of the formula, the square of the new term —Sy^, is 9?/*. Hence, the complete divisor becomes 12x*—18xY + 9y^. Multiplying this by —3y^ gives — 36icV + 54xV— 272/« ; and subtracting this from the first remainder leaves 0. Hence, the cube root is 2x^—3y^. When the root has three or more terms, by grouping the terms of the root already found, these terms may be used as one, and the next term found by a repetition of the process used to obtain the second. APPENDIX Example 2. Find the cube root of 11 I a+b+c x^-6x^+l5x*-20af + 15x^-6x-\-l \x'-2x+l a'= x^ -63(^ + 15x*-20x' + 15x'- -6x^ + 12x*- 8x^ -Qx + 1 a'=a + b=x^—2x Sa'''=3x*-12.x^ + 12x^ 3a''-\-t^a'c + c^=3x*—12x^ + 15x'—6x+l 3^*-12ie + 15x2- 3x'-12x^ + 15x'- -6x+l -607+1 EXERCISE III. Find the cube root of 1. x' + Qx'i/ + 12x}/-i-Sy^. 2. x'-Sx' + Sx'-l. 3. x' + 9xy + 27xy^27i/', 4. 27a'-10Sa'b' + U4:Ctb*-Q4b\ 5. l-6a + 12a^— 8a\ 6. x'-Sx' + Qx'—7x' + Qx^-Sx-^l. 7. l~9a + SSa'-QSa' + Q6a'-S6a' + Sa\ 8. S + x^^9x' + Sx' + Ux^ + 12x + lW. 9. a' + Ua + -,-112 + —— -12a\ 10. 4ic=^-9ic^-6iB-l+ic«-6£c^ + 9a;*. 11. a;« + 3aj* - 54a; + 28ic'-9£c'- 6x^-27. a;' ^y ,^y'' y^ ^^' 27""6" • "T'S"- 19 APPENDIX 5. Cube roots of arithmetical numbers. The cubes of the numbers 1, 2, 3 • • • • • 9, 10, are 1, 8, 27 729, 1000, respectively. Hence, the cube root of an in- teger of one, two or three figures is a number of one figure. The cubes of the numbers 10, 11 99, 100, are 1000, 1331 970290, 1000000, respectively. Hence, the cube root of an integer of four^ five or six figures is a number of two figures. Likewise, the cube root of an integer of seven, eight or nine figures is a number of three figures. And so on. Therefore, if the figures of an integer are marked off from right to left in groups of three^ the number of figures in the cube root icill be equal to the number of groups^ one or two figures remaining o?i the left being counted as a group. Thus, marked off into groups, 2515456 becomes 2'515'456'. Since there are three groups, the cube root of 2515456 will contain three figures, i.e., ones, tens and hundreds. If the cube root of a number be a number of two figures, we may denote the tens of the root by a and the o?ies by ^, and hence, represent the root by a + b. Hence, the given number is represented by a^ + 3a^b + ^ab'' + b\ Therefore, the cube root of the number may be obtained by use of the identity ^a' + Sa'b + ^ab' + b' = a + b. The process is best shown by examples. Example 1. Find the cube root of 39304. Pointing off, we have 39'304'. Hence, the root will be a num- ber of two figures. Since 30^ is less than 39304, and 40^ is greater than 39304, the root lies between 30 and 40; i. e., the tens' figure of the root is 3, the cube root of the greatest cube in the left-hand group of the given number. Therefore, in the above identity a denotes 3 tens, or 30. APPENDIX The work is then performed as follows: 39'304' a'= 27 000 13 30 + 4=34, root 3a2=2700 Sa' + 3ab + b' =307 Q Sa''b + 3ab^ + b^= 12304 12304 Since a=30, a^=27000. Subtracting 27000 leaves 12304. The trial divisor^ 3a^ becomes 2700. Dividing 12304 by 2700 gives approximately 4. Hence, b is probably 4. The complete divisor, 3a^ + 3a6 + 6^, becomes 3076. Multiplying 3076 by 4 gives 12304. Subtracting this from the first remainder leaves 0. Hence, 30 + 4, or 34, is the required root. When the root is a 72 umber of three figures the first tv^o may he found as above ; then their sum may be used as one number^ and the third obtained by a repetition of the process used to obtain the second. And so on. Example 2. Find the cube root of 2515456. Denoting the parts of the root already found by a, and a' respectively and the new parts by b and b\ we have the following work : 3a' =30000 3a2 + 3a& + ?>-^=39900 3d'b + ?>ab''-\-¥= 2'515'456' 1 000 000 11515456 1197000 ! a + ?) +c 100 + 30 + 6 = 136, root 3a ■-'=50700 3a'2 + 3a'6' + 6''''= 53076 3a'26' + 3a'6'=' + 6'2= 318456 318456 There will be three figures in the root. 14: APPENDIX The first number is found to be 100. Then the first trial divisor is 30000. Dividing 1515456 by 30000 gives approximately 50. But it will be found that the resulting complete divisor^ when multiplied by 50, will give a number greater than 1515456. Hence, 50 is too large. It will be seen by trial also that 40 is too large. Therefore we use 30. Then the complete divisor becomes 39900. Multiplying 39900 by 30 gives 1197000. Subtracting gives a second remainder 318456. Now letting a' be 130, the trial divisor, 3a'^ becomes 50700. Dividing 318456 by 50700 gives approximately 6. The second complete divisor, 3a'^ + 3a'6' + 6'^, becomes 53076. Multiplying 53076 by 6 gives 318456. Subtracting this from first remainder leaves 0. Hence the cube root is 136. For reasons similar to those given in § 3, to mark of a miniber which contains a decimal^ begin at the decimal point and mark to the left and to the rights putting three figures in each group. If necessary, ciphers may be annexed to the right of the decimal point. An approximate value of the cube root of a number which is not a perfect cube may be obtained to any required degree of accuracy. Example 3. Find the cube root of 1860.867. 1'860'.867' [ 10. + 2. + .3:^12.3, root a^= 1 000 . : 3a^=300. 3a2 + 3a6 + 6'=364. 3a26 + 3a62 4-6'= 860.867 728. 3a'^=432. 3a'2 + 3a'6' + 6''=442.89 132.867 132.867 APPENDIX 15 EXERCISE IV. Find the cube root of : I. 103823. 2. 2744. 3. 15625. 5. 571787. 6. 340068392. 7. 523606616. 9. 328.509. 10. 41.063625. Find to two decimal places the cube root of : II. 2. 12. 10. 13. 106. 14. 3.7. 4. 148877. 8. 59.319. 15. .15 16 APPENDIX B. HIGHEST COMMON FACTORS AND LOWEST COM- MON MULTIPLES BY DIVISION. (SUPPLEMENTARY TO CHAPTER X). 1. In Chapter X we showed how, by factoring the expres- sions, to find the highest common factor^ {11. C. F.) or the lowest common multiple (X. C. M.), of two or more expressions. The highest common factor of two expressions may be ob- tained also by a division process, known as the Euclidian Pro- cess, which we shall here discuss. It may be used in those cases where the factors are not readily found. • 2. H. C. F. by division. The process involves the following principle : If A = BQ^ R, ichere A, B, Q, and B are integral expres- sions in the same general number, then the H. C. F. of A and B is the same as the IL C. F. of B and R. This principle is established by showing that every common factor of A and 5 is a common factor of B and i2, and every common factor of B and J2 is a common factor of A and B. Since A=J5^+i2, it follows from the Distributive Law that every common factor of B and jR is a factor of A, and hence is a common factor of A and B. Again, R—A—BQ, and therefore it follows as above that every common factor of A and jB is a factor of R, and hence is a com- mon factor of B and R. This proves the principle. Now suppose that A and B are two integral expressions whose highest common factor is required, the degree of JB being not higher than that of A. APPENDIX 17 Divide A by B, and call the quotient ^i, and the remainder R^. Divide B by R^, and call the quotient Q2, and the remainder R^. Divide Ri by R2, and call the quotient ^3, and tlie remainder R^; and so on. NoAV since the remainder must be of lower degree in the letter of arrangement than the divisor, by this process the successive remainders must diminish in degree. Hence, finally a point will be reached when either the remainder is 0, or does not contain the letter of arrangement. Jf the remainder is 0, then the last climsor is the required highest common factor. For, since the dividend equals the product of the divisor and quotient plus the remainder, we have from the above divisions J3=R,Q, + R,, i?„_2 — i?n-l Qri + ^n-> where i?,^ is the last remainder. From these equations it follows by the preceding principle that the H, C, F. of B and B^ is the same as the 11. G. F. of A and i?; the H. C. F. of B^ and B^ is the H. C. F. of B and i?„ and hence of A and B ; the II. C. F. of B., and B^ is the II C. F. of 7?i and B^, and hence of A and B ; and so on. That is, the II C. F. of A and B is the II C. F. of any two consecutive remainders in this succession of divisions. But if A\=0, the H. C. F. of B,,_^ and i?„ is B,,_, itself. Therefore, the II. C. F. of A and B is B„_i the last divisor. If the last remainder is not 0, but merely free of the letter of ;|^g APPENDIX arrangement, then there is no common factor containing the letter of arrangement. For, if 7?„ is merely free of the letter of arrangement, then J?„_i and i?„ can have no common factor which contains the letter of arrangement. Therefore, A and B can have no com- mon factor which contains the letter of arrangement. *3. It is clear that the above process consists simply of re- placing the two given polynomials by two new polynomials which have the same H. C. F., then replacing these by two other polynomials which have' the same IT. C. .F., and so on. Hence, it is allowable at any stage of the work to multiply either the dividend or divisor by any expression u^hich has not a factor belonging to the other. Likewise, it is allovxible to remove from either the dividend or divisor a factor 10 J lich is not common to both. A factor common to both dividend and divisor may be rernoved, provided that it is introduced into the H. C. F. as finally found. Example 1. Find the H. C. F. of Gx^ + llir + S and 2x' + 7x + Q. 6a^2 + lla;+ 3 | 2a?^ + 7a?+6 2x^ + 70.' + 6 | 2jg + 3, H. C. F. 6x"'^ + 21x-4-18 3 2x' + ^x x + 2 — 5 )-10.r-15 4x + 6 2x+ 3 4a^ + 6 In the first division we obtain the remainder —10a?— 15. From this remainder we remove the factor —5, since —5 is not a factor of the divisor 2x'^ + 7x+Q. Now the H. C. F. of the given expressions is the H. C F. of 2ic' + 7x + 6 and 2x + 3. Dividing 2x'^ + 7x-\-Q by 2a? + 3 leaves no remainder. Hence, their H. C. F. is 2a? + 3 itself. Example 2. Find the H. C. F. of 2x' + 3(^-7x'-20x and ^af-12x^ + 5x. APPENDIX 19 Removing the common factor x the work is as follows : 2ie + a?2-7a?-20 2 40?^- o? |4a?=^- 7 -12a?+5 -12a? + 5 4a?2-12a? + 5 4x=^-10a? — 2a? + 5 — 2x+5 |2a?-5 |2x-l 4ar*+ 2a-''-14a?-40 4:.x^-12x'+ 6x 14ar^— 19a?— 40 2 28a?2-38a?-80 28x*2-84a? + 35 23) 46a?-115 2a?- 5 H. C. F=x{2x-5), ov2x^-5x. Here, in order to avoid fractions, we multiply 2x^ + x'^—7x—2Q by 2 before dividing. We then seek the H. C. F. of 4a?2— 12a? + 5 and the remainder 14a?=^-19a?-40. Again, to avoid fractions, we multiply 14a?^— 19a?— 40 by 2. The second division gives the remainder 46a?— 115. From this we remove the factor 23, since 23 is not a factor of 4a?^— 12a?+5. We then seek the H. C. F. of 20-- 5 and 4a;=^ — 12a?+ 5. Dividing leaves no remainder. Hence, their H. C. F. is 2a?— 5 itself. Now to get the H. C. F. of the original expressions we must multiply 2a;— 5 by the factor x which was removed in the beginning. Example 3. Find the H. C. F. of ^oc'y-^xy' + y^ and 3a?^ — Sx^y + xy"^ — y^. Since y is a factor of the first expression, but not of the second, it may be removed. 20 APPENDIX We now seek the H. C. F. of 4:X^—5xy+y^ and 3ar^ — 3x^y + xy^ — y^. The work is as follows : Sa^ — Sx^y + xy^ — y^ ___^ 4 12a?' — 1 2x^y + 4xy^ — 4?/^ 12a^-15x'y + Sxy^ 3x^y-\-xy^—4:y'^ 4 12a?^2/+ 4£C2/^— 16?/^ 19^/'' ) 19xy'-19y' x-y 4xl—5xy + y^ 4iX^—4txy 4:X^—5xy + y^ 3x 4:X^—5xy + y^ 32/ -y, H. C. F. 4x-y — xy + y' — xy + y"^ 4. The H. C. P. of three or more expressions. The 11. C. F. of three or more expressions may be obtained by finding the H. C. F. of any two of the expressions, then finding the 11. C. F. of this IT. C. F. and the third expression ; and so on. Example 1. Find the H. C. F. of a' + a'-Ua-24:, a^-Sa' — 6a + 8, anda^ + 4aHa— 6. Thei?. C.F.old' + a'-Ua-24:Siuda^-3a'-Qa + 8isa''-2a-8. The H. C. F. of a'-2a-8 and d' + 4a' + a- Hence, a + 2 is the H. C. F. required. ■6 is a + 2. EXEBCISE FindtheH. C. F. of: 1. a^-Sx-2, x^-x^-4:. 2. aj'^ + ic-e, ar'-2a;^-£c + 2. 3. ar'' + 7ic^ + 5a;-l, Zx^ -^ hx' + x-1. 4. 2a2-5a + 2, 4a^ + 12«2-a-3. 5. 4aj^-12a;-27, eaj'^-Slaj + lS. APPENDIX 21 6. 2a^ + 9a'-6a-6, Sa' + 10a'-2Sa + 10. 7. Sx' + x-2, 4x' + 2x'-x + l. 8. 2a;'-7a'-2 + 7a, ba' + a'-Za'-A.a + A. 9. 2m'' — 3m' — 8?72 — 3, 3m* — 7m^ — Sm'- m— 6. 10. 9a=^^-22a5=^-85*, 3«=' + 13a'^> + 12a^»l 11. 2x''-^x''y-2xy\ 2x^ + lxhj-\-Zxy\ 12. a^-^alf-2h\ 2a'-^a'b-ab' + Qb\ 13. a;y-6ajy + 6ecy-3a;y + 2.^/, 6a!V-15xV + 21a;V-12a;y 14. 2lab- 17 ab'-bab'-\-ab\ baly'-SAab'-7ab. 15. ic* + 4£c^ + 4:x\ x'y + 5a;\y + ^x-y. 16. a;^ + 3a;' + 4, £c=' + 2£c'-4a; + 8. 17. ic=' + 3a; + 2, jc*-6a;=^-8aj-3. 18. x'-x^^\,x'^x^-^\. 19. x''-x'-%x^^\2, x''^4x'—Zx^- 20. a?-\, 2a'-a-l, Sa'-a-2. 21. 2m' + 3m -5, 3m'- m- 2, 2m' )n — S. 22. x' + x-Q,x' + Sx'-Qx-^, x'-2x'-x + 2. 23. 3«' + 5«' + a— 1, a' + 3a — 1— 3«^ «' — l + 7a' + 5a. 24. a'-9a-10, a'-30-7«, lO + a'-lla. 26. 2y' + Qy' + 4y\ 3?/=^ + 9y' + 9y + 6, 3^=* + 8y' + 5y + 2. 5. L. C. M. by use of H. C F. The L. C. M. of tico exjyressions may be obtained by cUviduiy their product by their H. C. F. This may be proved as follows : Let A and B be the expressions whose lowest common midtiple is required, and let ^be their highest common factor. Then A and B may be written A = Hq,, and B=IIq^^ in which q^ and q^ have no common factor. (Why ?) 22 APPENDIX Hence, AJB = Hq^Hq^ = IT{Hq^q.^. Now the L. C. M. must contain all factors common to A and i?, all other factors of A^ and all other factors of B. Hence, the L. C. M. equals Hq^q^. Therefore, from the above equation, AB==H(Hq,q,), we see that the product of A and B equals the product of their 11. C. F. and their L. C. M Therefore, their L. C. M. ma (J he obtained hy dimding their product by their H. C. F. 6. L. C. M. of tliree or more expressions. The Z CM. of three or more expressions may be obtained by first finding the Z.C.M. of any two of the expressions, then finding the If. CM of that result and a third expression ; and so on. EXERCISE VI. Find the Z. C M of : 1. 2a'-5a4-3, 2«=^— 7a + 5. 2. ^a' + Ua+lO, Qa' + ba-U. 3. a^-x'-Ux + 2i, x'-2x'-bx-}-e. 4. x'-12x+lQ, x'-4x'-x' + 20x—20. 5. 12a^ + 13a^ + 6a + l, lQa' + lQa' + 7a^l, 6. 4:rv'-12n' + bn, 2n* + n^-7n'—207i. 7. x'-Sx+2, x*-Qx' + Sx-Z. 8. 2x' + 6x-^, 2x'-{-x'-6x + 2. 9. 2a' + Sa-b,Sa'-a-2,2a'-Va-S. 10. a^+x-Q,x^-2x^-x + 2,x' + Sx'-Qx-S. 11. 2x' + Sx-2, 2£c'' + 15a;-8, iB^+10i« + 16. 12. 2a*+W + 4a\ 3«^+9a^+9a+6, Sa' + Sa' + ba+2. APPENDIX C. FUNDAMENTAL PROPERTIES OF NUMBERS. 1. In the following sections we shall establish some fun- damental laws of numbers which are at the foundation of arithmetic as well as of algebra. These laAvs were assumed in Chapter II. They apply to any numbers. We shall now establish these laws for commensurable mimhers. By the aid of the theory of limits they can be shown to hold for incom- rnensurahle numbers as well. We shall first establish the laws for integers. 2. Law of order in addition. Numbers integers. • Numbers to be added may be arranged in any order ; that is, The proof of this law is based upon the meaning of an integer. {a) When the numbers are positive integers. By definition, positive integers are merely arithmetical numbers, each repre- senting the number of units in a given magnitude. Let us consider two magnitudes of the same kind, one con- taining a units and the other b units. Then, a-\^b means the number of units in the new magnitude formed by putting the second magnitude with the first. And b-\-a means the num- ber of units in the new magnitude formed by putting the first magnitude with the second. But evidently the total number of units in the two magnitudes is the same whether the second be put with the first or the first put with the second. Hence, a+ ^=5 + «. By similar reasoning, this law can be shown to be true for any number of positive integers. 24 APPENDIX (b) When the numbers are negative mtegers. It has been proved that the sum of any number of negative integers is obtained by finding the sum of their absolute values, as in (a) above, and attaching to this sum the negative sign. But by {a) the sum is the same for any arrangement of their absolute values. Hence, the law is true also for any number of 7iegative integers. (c) Wheii the numbers are integers^ some i^ositive and some negative It has been proved that the sum of two numbers, one positive and the other negative, is obtained by finding the difference of their absolute values and attaching to this the sign of the number having the greater absolute value. This difference would be the same, whatever the arrangement of the numbers. Hence, the law of order in addition liolds for two integers^ one positive and the other negative. By (a), (^), and (c), it follows that an integer, positive or negative, may be brought to any position without changing the value of the sum. Hence, the law is true for any number of integers^ some positive and some negative. Note.— This law is also called the commutative law of addition. 3. Law of grouping in addition. Numbers integers. Numbers to be added may be grouped in any manner ; that is, a-h6 + c = a-h(6 + c). Since a, 5, c, are integers, we have by § 2, a-\-b-^c=b + c + a = (J + c) + a Similar reasoning will apply in the case of any number of integers. Note.— This law is also called the associative law of addition. APPENDIX 25 4. Law of order in multiplication. Numbers integers. The product of two or more numhers is not changed by chang- ing the order in ichich the midtiplications are j)erfornied ; that is, abc = acb = bac, etc. («) When «, ^, c, are positive integers, place a objects in a group, and form h rows of c groups to the row. c columns h rows a a a a a a a a a a a a Since there are a objects in each group and b groups in each column, there must be ab objects in each column. And since there are c columns, the total number of objects must be abc. In like manner, there are ac objects in each row, and b rows. Hence, the total number of objects must be acb. Hence, abc and acb represent the same number of objects. Therefore, abc=acb. If now a=l, this becomes bc=cb. (b) If there be any number oi positive factors, by (a) any two consecutive factors may be interchanged (considering the product of all factors preceding these two as the lirst factor a in (a) above) ; and by repetition of the process of intercliang- ing two consecutive factors, all of the factors may be arranged in any order without changing the value of the product. (c) The law hokls if some of the factors are negative. For, the absolute value of the product is the same whatever the 26 APPENDIX signs of the factors, and the sign of the product has been shown to depend only upon the number of negative factors. Any change in the order of the factors could not, therefore, change the sign or absolute value of the product. Note.— This law is also called the commutative law of multiplica- tion. 5. Law of grouping in multiplication. Numbers integers. Factors may he grouped in any manner / that is, ahc=a{hc). When «, J, c, are integers, we have by § 4, abc=hca = {bc)a. = a{bc). Similar reasoning will apply in the case of any number of factors. Note.— This law is also called the associative law of multiplica- tion. 6. Law of distribution. Numbers integers. The product of tic o expressions equals the sum of the products obtained by multiplying each term of either exj^ression by the other / that is, (a + b + c)x—ax + 6 jr + ex. (a) When x is positive. Since ic=l + 1 + 1+ to £c terms, then (a + b + c)x={a + b + c)^{a^b + c) + {a + b + c)-\- to X terms (Def. of multiplication.) =<^4-a + «+ • • • • to a; terms -]rb-\-b + b+ ••••tocc terms + c + c + c-{- ' ' ■ ' to ic terms §2 = ax + bx + cx. APPENDIX 27 Similar reasoning will apply to any number of terms in the multiplicand. (b) When x is 7tegatwe. Since — £c= — 1 — 1 — 1— .... to ic terms, therefore (a + ^ + c) ( — a?) = — (a + ^ + c) — (a + ^» + c) — (a + 6 + c) • • • • to a; terms (Def. of multiplication). = —a — a—a— • • • to £c terms —h—b—b— • • • to £c terms — c—c—c— • • • to £c terms § 2. = a{—x)^-b{—x)^-c{—x). Similar reasoning will apply to any number of terms in the multiplicand. 7. In order to prove the preceding laws when fractions are involved, it is necessary first to establish the following prin- ciples. m (1) a—=am-^n; that is, to multiply any expression a by the fraction — multiply a by m, then divide the product by n. This follows immediately from the definition of multipli- cation. (2) To prove abc = acb lohen a is a fraction, and b and c are integers. If b and c are positive, form b rows of the fractions a with c fractions in each row. Then there are c columns. 28 b rows ^ APPENDIX c columns A a a a a a a a a a Now, by definition of multiplication, ab represents the sum of the rt's in one column. And since there are c columns, abc represent the sum of all of the «'s. Again, ac represents the sum of the «'s in one row. And since there are b rows, acb represents the sum of all of the a's. Therefore, abc = acb. If b or c, or both b and c, are negative, the proof folloAvs as in (c), § 4. (3) To prove a m_m a b 71 n b This is established as follows : a ni ~b n By (1). = (y-) " I, m-~-nnb —nnb (quotient X divisor = dividend.) a ^-^■bm ., m a - Also— T-no^ n b (m a\ m a n b bn By (2). (quotient X divisor = dividend.) By (2) and § 4. APPENDIX 29 = — a-^hbn n By(i) = (^.«)^6.6.« m (quotient X divisor = dividend.) = m,a (quotient X divisor = dividend.) = €1-^1. §4. _, a tn ^ m a . Hence, -r — 710= — -rno. ' n n Axiom 7. Dividing by 5, then by w, a m Tn m a -n'~b' If n=l, this becomes a a = m~^. 8. Fundamental laws when fractions are involved. (1) The Imo of order in multiplication. In (3) § 7 we have established the law of order in multiplication for two factors, where one or both are fractions. By the same method the law can be shown to hold for any number of factors. (2) The law of grouping in multiplication., ^fhen. some of the numbers are fractions, now easily follows. Thus, abc=bca By (1). = {bc)a =a{bc). The same reasoning would apply to any number of factors. (3) The laic of distributioii. We are now able to complete the proof of the law of distribution. 30 APPENDIX We have x(a+b+c)—aia-\-xb-{-xc, whatever the expressions represented by x, a, b, c, because the multiplier a + b + c is obtained by first taking unity to form a, then to form b, then to form c, and adding the three results ; and hence to obtain the product x must be used in the same manner. Now, since x(a + b + c)=xa + xb + xc, (a + b + c)x=ax-{^bx + cx, by the laAV of order. The same reasoning would apply to any number of terms. (4) The law of order in addition noio follows. It has been shown that integers or rational fractions can be reduced to equivalent fractions having a common denominator, and such that the resulting numerators and denominators are all integers. Suppose that when reduced the fractions are etc. X Then, f+|+f+ .... = " + ^ + f • • ■ • §50, Chapt. VI. But the value oia + b + c • • • • is not changed by changing the order of the terms. Hence the value of - + ^+-+ Jb Jb tJu is not changed by changing the order of its terms. (5) The laid of grouping in addition, when fractions are in- volved, can now be established by the same reasoning that was used in § 3. INDEX [Numbers refer to pages.] Abscissa, 193. Absolute term, 243. Addition, 1, 35, 46, 47, 146. of negative numbers, 35. of fractions, 146. elimination by, 183. Algebraic expressions, 5. numbers, 32. sum, 35. Antecedent, 298. Arithmetical numbers, 33. means, 359. progressions, 354, 355. Arrangement of expressions, Axioms, 21. 60. Cologarithms, 401. Combinations, 342. Commensurable numbers, 299. Common difference, 355. Completing the square, 235. Complex fractions, 155. Complex numbers, 223. Conditional equations, 20. Conjugate surds and imaginaries, 219, 226. Consequent, 298. Constants, 310. Continuation, symbol of, 56. Coordinates, 192. Continued proportion, 304. Base of power, 12. Binomial, 15. square of, 77. Binomial theorem, 81. extraction of roots by, 353. general term of, 349. proof of, 346. Brace, bracket, etc., 6. Character of roots, 341. Checks, 48. Clearing of fractions, 165. Coefficients, 13. numerical, 12. undetermined, 375. 31 Degree, of terms, 131. of an equation, 161. Difference, 1. Discriminant, 241. Distributive law, 18. Division, 4, 64, 66, 153. Divisor, dividend, 4. Elimination, 181, 182, 184, 186. Equations, 19. equivalent, 179. exponential, 388, 406. fractional, 165. graphic representation, 192, 373. graphic solution, 197,375,378,279. 32 INDEX inconsistent, 179. independent, 179. indeterminate, 178. ill quadratic forms, 253. integral, 163. irrational and radical, 161. linear, 164. quadratic, 230. simultaneous, 179. solutions of, 178. symmetrical, 267. systems of, 180. Equivalent equations, 179. Evolution, 84. Exponents, 12. laws of, 56, 64, 75, 324. Expressions, 5. literal, 13. Extremes and means, 300. Evaluation of an expression, 6. Factoring, 103. equations solved by, 233. Factorial -n, 340. Factors, 12. Formula, the, 168, 169.^ Formula for solving quadratics, 238. Fourth proportional, 300. Fractional equations, 165. exponents, 324. Fractions, 71. complex, 155. partial, 383. Geometrical progression, 354, 362. means, 365. Graphic, representation of and solution of equations, 192, 197, 273, 275, 278, 279. representation of imaginary numbers, 228. Harmon ical progression, 354, 370. Highest common factor, 131. Homogeneous equations, 264. expressions, 131. Identical equations, identities, 20. Imaginary numbers, 223. Incommensurable numbers, 299. Inconsistent equations, 179. Independent equations, 179. Indeterminate equations, 178. fractions, 321, 322. Inequalities, 291-295. Inserting signs of grouping, 53. Integral equations, 161. expressions, 130. Involution, 75. Irrational numbers, 86, 212. Known and unknown numbers, 24. Law of order, 16. of exponents, 56, 64. of grouping, 16. of signs, 41. expressed by an equation, 214. Letter of arrangement, 60. Like and unlike terms, 15. Linear equations, 163. Literal numbers, 9, 11. advantage of, 9, 11. Logarithms, 388. characteristic of, 395. computations by, 403. common, 393. INDEX 33 Napierian, 393. Lowest common multiple, 135. Mantissa, 395, 397. Mean i)ioportional, 304. Means, arithmetical, 359. geometrical, 365. liarmonical, 370. Means and extremes, 300. Minuend, 1. Monomials, 15. Multiples, L. C. M. 135. Multiplicand, multiplier, 3. Multiplication, 3, 39, 57, 58, 150. Negative exponents, 326. numbers, 30-38. Numbers, definite, 9. algebraic, 33. commensurable, 299. constants and variables, 310. finite and infinite, 319, 320. general, 8, 9. imaginary and complex, 223. known and unknown, 24. natural, 9. opposite, 33. rational and irrational, 163, 212. real, 223. Opposite numbers, 33. Ordinate, 192. Parentheses, 6. Partial fractions, 383. Permutations, 337. Polynomials, 15. square of, 79, Powers, 12. Prime factors, 104. Progressions, 354. arithmetical, 354, 355. geometrical, 354, 362. liarmonical, 354, 370. Problems, directions for solving. 173. Products, 4. special, 92. Proportion, principles, 299. Quadratic equations, 230. graphs of, 273, 275, 277. principles, 242. systems of, 261. Quality, signs of, 32. Quotients, 5. special, 96. Radicals, 83, 86, 212. Ratio, 298. Rational expressions, 6, 86. Rationalizing factor, 220. Real numbers, 223. Remainder theorem, 123. Removal of si^ns of grouping, 51. Roots, 83. of a polynomial, 90. of an equation, 21. character of roots, 241. sum and product of, 242. Series, 254, 378. convergent, 354. divergent, 354. finite, 354. fractions expanded into, 378. oscillating, 354. reversion of, 382. Signs of grouping, 6. Simultaneous equations, 179. 34 INDEX Solution of equations, 21, 178. graphic method, 275. by special devices, 270. Square of binomial, 77. of polynomial, 79. Square roots, 79. Subtraction, 2, 37, 49. Subtrahend, 2. Surds, 86, 212. Symbol of continuation, 56. Symmetrical equations, 267. Synthetic division, 127. Systems of equations, 180. consistent or determinate, 180. defective, 269. equivalent, 180. impossible, 180, 190. indeterminate, 190. solution of, 180. Theorem, trinomial, 81. of undetermined coefficients, 375. remainder, 123. Third proportional, 304. Transposition ,165. Trinomial, 15. Undetermined coefficient^, S!7b. theorem of, 376. j Unknown numbers, 24. Variables, 310. limit of, 318. Variation, 310. direct, 311. indirect, 311. joint, 312. Vinculum. 6. Terms, 14. Theorem, binomial, 81, 346. Zero exponent, 64. operations with, 321. ANSWERS Exercise 1. 8. 8. 11. 33. 14. i. 17. 44. 20. 13. ^. 15. 12. 0. 15. 38. 18. 16. 21. 24. 10. 2^ 13. 48. 16. 4. 19. 6. 22. 0. 23. 216. 24. 1. 25. 18. Exercise 2. 26. 25. 1. 5. 4. 27. 7. 230. 10.. ^V- 13. 23. 16. 2, r 2. 11. 5. 40. 8. 37. 11. 28. 14. h 17. 17.. 3. 270. 6. C6. 9. 24. 12. 49. 15. liV 18. 576. 19. 20tli; X taken as a factor 20 times. 20. 4; 5; 1. 21. 7; x\ 3. Exercise 4. 5. 35a6a?2^. 12. 14a'. 19. ll^-'^a?. 6. 30a&c. 13, ^\x^y. 20. 484. 7. 2xyz. 14. 'S2x2j^. 21. 693. 8. 14abcxy. 15. mah. , 22. 24.8. 9. 15a^bx^y. 16. 28irV- 23. x; x^y; y\ 10. iabcxyz. 17. lOax. 24. x2; a; 6. 11. 15a. 18. loab^ Exercise 5. 25. 7?i2; 3m; 21. 1. 3. 3. 2. 5. 3. 7. 2. 9. 2. 11. 2. 13. I. 15. 1. 2. 7. 4. 4. 6. 3. 8. 1. 10. 1. Exercise 6. 12. 3. 14. 5. 16. 2. 1. 8 boys. 3. 10 yrs. and 30 yrs. 5. 16, 48. 2. $3. 4. 45, 80. 6. 8, 15. 2 ANSWERS [Ex. 6-9 7. B, $250. A, $500. C, $800. 8. 4 cows; 12 hogs. 9. 112 yds. by 224 yds. 10. 8 hrs. H- H ^^^' Per hr. 12. 64. 16. 65,72 20. 18 mi. 24. 12, 24. 13. 9. 17. 32,48 21. 12 hrs. 25. $24,000. 14. 48 lbs. 18. 57,58 , 59. 22. If hrs. 26. 7. 15. 24, 39. , 19. 6 da. 23. 15, 20, : 25. 27. 5 yrs., 35 yrs. 28. 12 mi. 29. 3 lbs. 30. A, 25; B, 27. Exercise 7. 1. 3. 4. -27. 7. -3V 10. -9. 13. 14. 2. 7. 5. 2i. 8. -.24.-^ 11. -2. 14. 4. 3. -22. 6. 0. 9. 9. 12. — 5. 15. 14. 16. -3. 18. 6. 19.-39. 23. -200 lbs; -50 lbs. 24. 25 ft ; +5; +25; -5; +15. 25. $210; - -$160. Exercise 8. 1. -14. 7. 540. 13. 81. 19. 24. 25. -3. 31. f. 2. -14. 8. -280 14. -64. 20. 108. 26. -xV 32. -54.-^ 3. 14. 9. ri 15. -1. 21. 576. 27. -2.5. 33. -5i. 4. 14. 10. 1. 16. 216. 22. -162. 28. -12. 34. 8. 5. 30. 11. 0. 17. .432. 23. -9.- .29. -i. 35. -192. 6. -12. 12. 32. 18. 72. 24. 9. Exercise 9. 30. 18. 36. -28f. 1. 3a?. 3. -4c3. 5. IQax^. 7. 10F<^. 2. -4a26. 4. -3a5cd. 6. lA. 8. -B^. 9. 2lpq. 13. -23m27i2. 17. — 5ahc, 10. 86f.4C. 14. -X. 18. ^yz. 11. -5ofiy\ 15. -18^3. 19. 12a-9x. 12. -Slpqr. 16. 5a^l^c. 20. Qah-ldxy. 21.^ -2a5c2+9a26c+7a52c. 28. 3a2+262. 24. {2+a+h)xyz. 23. (3a+56-7c)a;2, 25. (-7c+2+Sa)y\ Ex. 10-13] ANSWERS 1. -dx-^y+5z. 2. 7a— c. 7. 2oc^+2x. 8. 4a2. 9. 2i)d^+6x^y—Qxy^—2y^. Exercise 10. 3. -3P+dQ+4R-5S. 5. 15p+6(/-S 4. dac—ixy. 6. Sab+Qbc. 10. 23Vi-|6^ic. 11. 5x^+i^x-2. 12. -9a?4+4a?34-2£c2+7a7+4. 9. 7x-10y. 10. -a75-ic3 + .x2. n. 4a26+2a62+268. 12. 2AB+5xy-'7PQ. Exercise 11. 1. 6a%\ 5. 136?/2. 2. — CoTi/. 6. 24x^y'^z. 3. aic2. 7. 2x-7y+10z. 4. -lOabc. 8. 4a+14a6— 7c. 13. cc2-ll£c+13. . 14. x''+x^+oc^+x*+2o(^+x+l 15. a*4-3;rS+?-7+3c2s5. 16. a;3-2a;2+^-l; a_2^24.^_l. _2x^+x-l. 17. _as-a862+4a253_2ic. 18. a.'3_^7^2^_^,^_l. 19. 9.125a-7.6a;2-6.25?n3-5??i2-l. 20. —x^+x^+x—2. 24. —x^+Qx^—Qx—4. 27. —0734.407—6. 22. 2aH5-l|a:2+10 25. a,'8_2a;2+8a7-2. 28. 36. 23. 7a3+2a2-3a- -7. 2S. a;3-4a7+6. 29. -9. 30. 3. 32. -3. 34. 16. 36. 16. 31. -18. 33. 8. 35. 26. Exercise 12. 37. 8. 1. 3a- 6. _ 2. 507+5?/. 7. -2b. 3. 52+4^( 4. 4a-2b. 5. 3aj3-7£c2+l. 8. -2a?3- -4x^+x-4:. 6. -3(K8-a;2?/+6a;2/2+22/8. 9. 7x'+Sxy-5y^-2y». 10. a-b. 12. 5-2/. 14. -4a;+3a. 16. 2. 11. 15a-26. 13. e+ic. 15. 1. 17. x: 1. x^-oc^+oc^-ix^-x+l), 2. ax—{by+cz—dw). Exercise 13. 3. a-(-25+3o-4d). 4. -(10e-5/+9r). ANSWERS [Ex. 13-16 5. (2+a)c(^+{b-S)x^+(6-c)x. 6. 7+{5-2a)x+{l+Qb)x^+{d-4a)x^. 7. (a-'d)x*+ (2+a)c(^+ {l-b)x^+ {d-c)x-l, 8. _(5_2)7/-(l-a)?/+5. 9. -(-p+q-r)y-{dq-2p)y^-sy*. 10. -(6x+2)y^-{d-x^)y^-(dx^-5)y. 11. 10+(2+b+c)x-(a+2)x'^. 12. (a+Q)x^-5x. 13. 5ii;2-3a7-l. 14. {l+a)x^+ (a.-b)x^+ (b+c+l)x+4:. 15. (a-3)ic-^+(a2+a-3)a'2+2aa;. 17. (a-b)x'2+ib-c)x+c+d. 16. .Tio+(2-a+&)ic5+?>+c-d. 18. (b+2)x^-{c+S)x+5-a, 19. pa:;3_p^2_(gjfr7')a;_^_g. Exercise 14. 6. 4a^b^c^. 7. -^^5^4. 8. -P^Q^. 12 1. -«7;,i2. 2. 60a^66. 3. 210a;V- 5. -3a2364c7c?5. y. —pn^'n 10. 14.21875a;62,3. 48pV'- 2i^22^11. 2(a+b)5. -15a3(6+c)*. 11. 11. 12. 13. 14. 16. x'^\ 17. Qa\ 19. (a3)io. Exercise 15. 5 . da^y^zw — Sxf^j/zHv + Soc^yzw^. 6. 10p2g2,._15pg2^2+20p2gr2. 7. ^35-^53. 8. -80ir8+12ar5-8^. 10. 1 5a363c3ic22/2_ Qa^b'2c^x'^y'^+da^bcx^^+ dal^cx^y^+dabc^x^y^z. 11 . — 5a%22/4 + a^x'^y^— -ja^xy* + ^'^x^yK -14?/. 16. -6ir-12?/. 1. a^b^-2a^b^+a^b\ 2. ic'^— a^+.r^— a;4. 3. -6a6624.5a5j>3_2a364. 4. Ax'^y^+Qx^y^-Wx'Y- 9. -Ix^y^z^-lx^y^z^ 15. -26c. 1. a2+2a5+52. 17. 2ic3+3ic^?/-5ir2/2. 20. 24ic2-30iC2/. Exercise 16. 2. a2-62. 18. 2ab^-2a^b. 19. 7a?4-42/*. 3. 6a;24-iC2/-22/2. Ex. 16J ANSWERS 5 5. 307*4-072-4. 7. 20p2g2-14pV-6p2?-2. 8. .t2-h5o;4-6. 10. 16a2-49. 12. 12x^y^ -Sc-^d^. 9. 4a;2+4a;-35. 11. ic2-l00. 13. ia^_|f£c2+^. 14. ^3ga2_|a5-^\62. 17. 0.94a2-5.55a6-3.64&2. 15. ^od»-^x^y+^\xy^-j%y^. 18. a^x^-b^if-bcyz+acxz. 16. 5.625372+ 15.375^^-1 1.257/2. 19. p^q^+2pq^r-p^qr+q^r^-pqr\ 20. ait;4-5a;-f-ca;+a?/+b?/+e2/+a-2;+62;+c2;. 21. 2a7*+irS-8.T2+23;^7_i2. 22. a;5+£c4+ic3_a;2-a;-l. 23. 4a77-2a76+7£)75— 7a74+7a;3-7j;2+3o?-5. . 24. x8+;;c4^_i, 27. a4+a252+54. W/^^^ 25.^-a^2+<^23^_^^ ^ 28. ic^-l. "— «^~ ' 2a^2«®-«'^'^-14«i^*+19«I'^^-^^ 23. ^2^2_4^8C_ 4^3(7+ 16^502. 30. — 2p2c2+3gr2(^2_2gc?re-r2e2+pgcd+3jpcre. 31. -2a;8+3a;c6+2aj6-6a2a;2-3«ic2+4a£c4+2a2x4. • 32. a4-4a36+6a262-_4a53+54, 33. £c«4-aa7^— 4a2£t^— 46t3a'3+4a4o72+3a5o7. 34. a262_ 52c24_26c2fZ + 2b2cd-c^d^-2bc(P- bM^. 35. ai4-2aio6"3+2a666-a2&9." 41. |a3-ia2&+i.|a52- ^^^63. 36. a;"+i-f-a?«?/+a7y«+2/«+^- 40 a?* llx^a^ a* 37. a^2«_2^2n, ■ 30 900 30' 38. a6<^+a*^'^62«+a2o53._,.55c, 43. a73-a72+|£c-|. 39. x^"+^-x^''+^y^+x^y»-^-y^. U. Sx^+8x^-\^x^+%^x^-\^x+3. 40. a;» + 2_£pn+l4.2it:n_^^«_l + ^H-2^ 45. ^9^;;p4_4 3£p2_|„^_ 46. 2.8x'*-7.36;r3+5.7a;2+4.16a'-10.24. 47. (r2«— 07*2/*— ^"Z/'+Z/*^'- ^- ^^^4- 1007^4-35072+5007 +24. 48. 07»+'" + 07'«2/" + 07"2/'" + 2/"+'". 55. 1 07^ + ^0727/+ ^077/2 +^',^3, 49. x^+2x^—x—2. 56. 07*"— 2/*". ^ 50. o?8-l. 57. a6— &6. 51. 6a3+a25-lla62_663. 68. 4a6. 52. x^-1. 69. 072-907+6. 53. ai2_5i2, 60. 2x^-2xy-2y^. 61. 6£c8+12a;2-14a;-4. 62. 5a'^-oa^b-2a^b^-ab»+b*+a^b'^+a*b^-ab\ 5 ANSWERS [Ex. 17-19 • Exercise 17. 1. aK 6. -Aa^bc^. 11 -is^t. 16. -2st. 2. a^l^. 7. ip\ 12. 5a7. 17. 9r+2. 3. -2a863. 8. -12ia4. 13. a^. 18. ^fz. 4. -^xy^. 9. 7a262.. 14. -V^»r '. 19. |s». 5. 465c. 10. -|??i%p3, 15. £c<-?/«-i. Exercise 18. 20. Vs2M. 1. x-^l. 5. 10a4+964. 9. 2x'^y^+ix^y—%x^y^. 2. xy-l+4x\ 6. a2-2a&+3b2. 10. 2x-|2/+if^2/2. 3. -x^+5-dx^. 7. —a+b+c. 11. a— 6+c. 4. -3x^+2x^ 8. |j;7-2^'V. • 12. a2+a6+62. 13. —la^—'^-ixy'^+'^-ix^y. 15. 5a7«-2a;»+i. le, a«_a2«. 14. 13£c4-20.8a?3+39. 17. a7«+2/". 18. l+£f2+ic4. 1. 3a+l. 4. 2. 5.2^+1. 5. 3. y+1. 6. 10. s(^-\-xhj+xy'^+y^. 11. x^—xy+y^. 14. a^-a262_^54. 19. 15. 1207-1. 20. 16. x+1. 21. 17. x^—2xy+y^. 22. 18. .r4+£c2+i. 23. 29. —d-dx—x"^. 30. a2_,_2a6+a+4?>2_25+l 31. a;3_4^2+ii£t;_24. 32. x—y+z. 33. £C3_,_^2y_|_a72/24-2/8. 34. a4+3a3+9a2+27a+81. 35. 2a2+9a-5. 36. x^+ocf^y+x^y^+y^. Exercise 19. 5a+l. 3a2-5a-2. x^+2xy+y^. 7. c-4. 8. a-5. 9. a2+a5+62. 12. x^—x^y+x'^y^—xy^+y*. 13. a4+a252_^54. it'4+a73-|-.x2+a;H-l, 24. a;*— 072+!. 2ic4+aic2_3a2, 25. ^y-L 2a2-5a+7. 26. o^— 2a:2rt+2a?a2_a3. £C^+£c2//2+i/4. 27. l-3£c+2a;2-a^. a;2— 6^7+9. 28. a;2+2ir2/+%^- 37. a2+5a+6. 38. x'^—xy—xz+y^+z'^—yz. 39. it'2+2a??/+2/2— 0^2;— ?/2;+;2;2. 40. ^a2--Ja6+|62. 41. ?i+^+a^62+4a253+i664. lb 4 42. |£C2— ^.T2/+2/2_ 43. ^x^-lx^y+^xy^. Ex. 19-21] ANSWERS 44. a2'»+2a'"6'"— 62'n. 46. 2a;«-r3a;»-i— 5ic»-2. 1. a" 2. a28. 3. -aiofois. 4. a?6y/8. 5. x^hf^z-^. 6. a6a?3o?/i8. 7. Oojio. 9. 343aV- 10. ^2x^^y^. 11. -32a20o;^. 12. aj2o?/i2. 45. a^« — £c2a^a_j_^^2_f_9^. 29. a^-20.r2+100. x^y^+iaxy+ia^. 30. a,'8— 1007*4-25. in^-Qm+9. 31. x^-2x^+x*. i;j4_6„j,2_|_9. 32. 4a7*+4r8H-ic2. m^n^—imny+iy^. 33. 25ir*— 100:^+076, 36. 4a;2-126a?7/4-962^2. 37. 4a262+16a62c+1662c2. ir2o_|_2icio+l. 42. £c2«_^2a;»+l. ic24_2£pi2+l. 43. a;2»-2a7»+l. 45. 4a;2»+i2a;«a»+9a2'«. 8 ANSWERS [Ex. 21-23 46. a2n+6_3rt«+8|ia+8+^2<.+6. a* _2a^a^ ofi 47. 4a:2'»+2+4?iar2»+J+n2aj2». it'i'> , .T^ , a?* 49. a2»62n-2_2a2n-l&2»-l4.a2«-252». 53. 4a2+4G(b+?>2+4ac+26c+c2. a2 2ac cf °"- b^ bd d^' 54. 9a2+6a6-6ac+6^-26c+ca- 55. 16-16a-86+4a2+4a&+62. 56. 16-246+ 9&2_24c+186c+9c2. 67. if-i^+c^+ab-^-^+'^. Exercise 22. 1. l4-2a;+3ic2+2a^+a;*. 2. a8+2a6+5a*+4a2+4. 3. Ax^+12x^+2oa^+2ix^+iexf^. 4. a;W+2a;8+3^+2a^+2cc4+2a?3+a;2+2a7+l. 5. icc»-12x^+25x^-2^x^+16. 6. a2+4624-9c24-i6d24.25e2-4a&+6ac-8ac?+10ae-126c+16&d-206e- 24c- -6. 32. a^+2a?3-2a;2_3a;+2. 26. x^—2xy+y^+10x- ■10^/+21. 33. x^- Aa^x^+Sa*. 27. p2_,_2pg+^2_6p_| 6g-160. 34. a;2n_ -^nyn_2\y^\ 28. x*+2x^+ix^+'6x- ■18. 35. a;2'»+24-a7n+i2/»-i-62/2''-2. Exercise 29. 1. a;+3. 7. 5ab-i. 13. l+6m2n8. 2. £c-3. 8. l+7aa;. 14. 13a?t/3-12a*. 3. a2+5. 9. l+4a:4. 15. 5a +1562. 4. 63-4. 10. t^-1. 16. a»+6». 5. 2;z;+l. 11. 8a?2+92/2. 17. a?»+i-a»-i. 6. 3a-l. 12. lOa-lla^. 18. a+6+c. 19. a+b+x+y. 20. {a+b)^-{x-y)\ 10. m8+m2n,+mn2+w8. Exercise 30. 11. l-2/+2/2-2/8+2/*-2^. 12 ANSWERS [Ex. 30-33 14. a^-a^+a^-a+l. 17. 8a^-12a^h+18a%^-27b^. 15. a^+x'+x^+x+l. 18. 27a3-45a2Z>2+75a64- 12566. 16. a2+62. 19. a^-cc^y^+x^y^—x^y^+y^. 22. a^+a^2/+^2/'^+^2/'+^^2/*+^2/^+2/^- 23. x'^—a^y+x^y^—x^y^—x^y^—x^y^+xy^—y'^. 24. a^2/6+a;22/4a8+a^2a6_,_a». 25. 25a-862. 26. 64p6 + 32p5g8^_16p4g,6_,_8p3^_|.4p2gl2_f_2pgl5_^gl8^ 27. a^»— a;2"2/"+a7''2/*«— ^3». 31. a^"— a2»+a»— 1. Exercise 31. 4. ar4(a^+5). 5. x^y^(x^+y*). 6. 5a2(a-26). 1. a;(aj2— 2^+1). 2. ir2/(a;+3a?2!/— 5?/). 3. a?(a;2-3). 10. Sx'^{2x^+Sy^+xy). 11. 12a*5'»(262-3a2). 12. 4x2(ic2+l)i 13. 2a8(4a2+2a5+62). 18. ^^(l+a). 1. 5(a;-4)2. 2. 4(a-b)2. 3. a(.T8+3)2. 4. Mxy+h)^. 5. 5a(2a-3)2. 1. {x+Q){x+7). 2. (x+8){x-Q). 3. (a?-5)(a;-4). 4. {x-7)ix+A). 6. (07+8) (a?+9). 6. (a;-|-10)(aT-5). 7. (a;-14)(a?+4). 7. 7a;2(4£c2+?/2). 8. 9x^(2x^-1). 9. a*(a2-6a6+262). 14. a262c2(a253 + 52c4 + (j3c2). 15. 6a782/(4a722/2_2+7a?32/). 16. 9m?i(3m2+4mn+9?i2). 17. 14a2?/2(4?/2-a2/+2a2). 19. a»(a7«— 1). 20. 5a''-'^y''-'^{a+2y) Exercise 32. 6. -l(3a;-l)2. 7. -l(a2-4)2. 8. h^{5a-x^y)^' 9. Sxy{x-ly)^. 10. 8(6a2+56)'^. 16. 3a(a-6)6. Exercise 33. 8. (£c-4)(a?-8). 9. (a+15)(a-12). 10. (m-16)(m+15). 11. (^-21)(f4-20). 12. (a+l)(a+i). 13. (c+10)(c-7). 14. (6+14) (6+6). 11. 7ix+y)^ 12. 5a(a-6)8. 13. 2xy(x^+y)». 14. -5aa?(a-36)8. 15. 2(2a-6)*. 15. (c+21)(c-4). 16. (d-ll)(d+5). 17. (2+r)(l+r). 18. (6+s)(4-s). 19. (3+^) (5-2/). 20. (2+a)(l-a). 21. (2*^-2) (22+1). Ex. 33-34] ANSWERS 13 30. 49. 50. 51, 52. 53. 54. 55. 56. 57. 58. 59. 60. 1. 2. 3. 4. 5. 16. 17. 18. 23. 23. 24. {x''-2)(x^-5). (x^-8)ix^+2). ix'2+18){x'^-10). (x^+l){x^+2). (x^-S)(x^~2). (0^+13) (a^-3). (£C«-16)(a^+14). (ax+ll)(ax+2). (a+136)(85-a). (xy—Sah) (xy—2ab). {xy^+9ab) {xy^-2ah), (xy-5c){xy+2c). (x^y^+7a)(x^y^+2a). {x^y^-M^h) (ic3?/3-4a86). (am2+5c2)(am^+6c2). (l_2a)(l-a). (l+9.T)(l-3a;). (a+6-f-2)(a+6+5). {x-y-\Q){x-y+^). {{x+yy+W} {{x+y)^+^}. 31. (b7/-10)(6?/+3). 40. (a66-7)(a66+5). 32. (ab+10)(«6+20). 41. {xy^+n){xy^-4:). 33. {xy-8){xy-2(i). 42. (p5Jg4_l)(p2g4_2). 34. (mri+10)(mn— 6). 43. {x+2y){x+y). 35. (abcH-15)(a6c-2). 44. {x-^ij){x-2y). 36. (j72/2;-10)(a7^2f-9). 45. (a7+72/)(ic+107/). 37. (x2j/2-l)(a;22/2_2). 46. (a+136)(a-36). 38. (a^.v8+3)(aV+ll).47. (a-206) (a+26). 39. (aj42/*-14) (07*2/4+9). 48. (5+3a)(6-a). 61. {a+h—x—y){a+h-2x-2y). 62. 2(a;-12)(iC+7). 63. 3(£C+3)(a;-2). 64. 5(a?+4)(a;+5). 65. 2(a;-25)(a;+8). 66. a{x+1[){x—2). 67. a2(a;+7)(£C-5). 68. x{x+la){x—Qa). 69. 307(07— 24^) (aj+ 42/). 70. 0722/2(0^+72/) (07+22/). 71. 2o7(o72/+26)(o72/+5). 72. 2(10-'a;)(ll+a;). (2ir+l)(oj+2). (307+1) (0^+2). (2o7+5)(o;+2). (2o;-l)(cc+3). (2o;-3)(o;-l). (7a3-l)(2a3+l). (3a2ic24.3)(2a2.:c2+3) (a6c+4)(5a&c— 1). 3a (207+3) (07+1). 22/(2oj+7)(9oj-l). Exercise 34. 6. (5oj+l)(o;-2). 7. (3o;+2)(4a?-l). 8. (6o;-5)(o;-l). 9. {2t-l){t+\).- 10. (o;+2)(8o;-l). 11. (4c-3)(3c-4f. 12. (ft-16)(56+l). 13. (a+5)(7a+l). 14. (2/+3)(102/-9). 15. (r2-2)(2r2+3). 19. (ar2y'-3)(2a722/+5). 20. (2a263_5)(3(i253_4), 21. 3(a?+2)(4o;-l). 25. {x+y){2x+y). 28. {x-iy){^x+y). 26. (4o;-2/)(3.T-2/). 29. (2«i-3a)(6m+a). 26(a-9)(3a+4). 27. {2x+y){x-2 30. (20^2/— 5)(i>^+3). 14 ANSWERS [Ex. 34-35 31. (a;2+ 42/2) (3£c2- 2^2). 32. Sa(x-2y){9x—y). 33. 2('Sah-2c)(ab+c). 34. {ax+b)(x-l). 35. (y+2a)(2y+h). 36. {2z-h)(z-a). 37. (aa?+l)(a?+5). 38. 2cj(a;+a)(a?— &). 1. 2. 3. 4. 13. 14. 15. 16. 17. 18. 19. 20. 21. 24. 25. 26. 27. (.17+6) (07-6). (a;+10)(a;-10). (a;4-7)(a;-7). (x+8){x-8). {10y-7b){10y+7b). (9a-85)(9a+85). (25a;- 155) (250?+ 155). (3aa7-l)(3aa7+l). {ixy+3){ixy-d). (2ay-b)(2ay+5). (5a7a— 4)(5a?a+4). (10a^-95)(10£n/+95). (4072;— 5c) (4a72;+5c) . (13071/2;- 5a5c) (13o7i/2;+5a5c). (la-lb) (ia+ib). {lxy-ia)i^xy+^a). {i\ab-i^x) {^\ab+^%x). (f-fa^2/)(l+l^^). (I_i^0(j5c)(l+-ijpa5c). Exercise 35. 5. (o;+12)(o;-12). 6. (a;-14)(o;+14). 7. (07+a)(07— a). 8. (x+n)(x—n). 9. (2a;+a)(2o7-a). 10. (3o7-a)(3ir+a). 11. (2o7-3a)(2o;+3a), 12. (4o7— 5a)(4o7+5a). 31 /-E — ^^\ (^ -I. ^^\ ' \4:b yfUb'^yl' 32. l^-l\l^+l\ ''- {TWb-''y)(mb-'y)' 34 i^-W^+^\ \9y 2xJh)y^2xJ' 35. / ^_ 10xyz \L^ -JOxyz y c /\ c 36. (a+5-l)(a+5+l). 37. {a+b—c—d)(a+b+c+d). 38. (x+y—a—b){x+y+a+b). 39. (o7— 2/— a+5)(07— 2/+a— 5). 40. (2o;4-22/-3a-35)(2o;+22/+3a+35). 42 l <^+^ _ c+d \( a+b j^ c+d\ ' \a-b c-dl\a-b^c-df' 43 i 5(x+y) ijx-y) ) ( 5{x+y) 4(x-y) ) ' i 7(a+5) 9(a-5) J | 7(a+5) "^9(a-5) f ' Ex. 36-38J ANSWERS Exercise 86. 1. (a-b-x){a-b+x). 7. (a+5&-l)(a+55+l). 2. {{+x-y){l + x+y), 8. (a-h-2)(a-b+2). 3. (x-y-l)(x-y+l). 9. (a-4b-2c)(a-ib+2c). 4. (« + 3b-3c)(a+36+2c). 10. {l-a+dx)(l-\-aSx). 5. (l_aT-42/)(l+a;+%). 11. (3£c-2a+3c)(3a?+2a-3c). 6. (x+2a-y)(x+2a+y). 12. (a+6-c4-d)(a+6+c-d). 13. {x-2y-a-3b){x-2y+a+Sb). 14. (a-56— 2?w-f-3n) (a-56+2m-3n). 15. {x-Qa-2y-b) (x—Qa+2y+b). 16. (32/-a?)(2/+3a?). 19. (lla-46)(-a-26). • 17. {Sx-4y)(x-2y). 20. (-Sy) (2£c). , 18. (-2x-y){ix~9y). 21. (9a+36)(lla-6). Exercise 37. 1. ix^+x+l){x^-x+l). 2. (l-aj+2ic2)(l+a;+2a;2). 3. (x^-x^+l){x^-x+l){x'^+x+l). 4. (a2+2a6+362)(a2_2a5+362). 5. (a2+3a+3)(a2-3a+3). 6. (2^2+207?/+ 5?/2)(2a;2-2.T2/+ 5^2), 7. (2a2+562+4aZ>)(2a2+562-4a6). 8. (x^-nxy-dy^){x^^'6xy-dy^). 9. (6a2-4a?/-5^2)(6a2+4ay-52/2). 10. (7a4-3a2?>+262) (7a*+3a26+252). 11. (237*- 6a;22/3+ 52/6) (2a^+6a;2?/3+52/«). • 12. {8x^y^-2xy+l){8x^y^-\-2xy+l). 13. (a*-4a2a72/2_3a.22^)(a4+4a2iC7/2-3ic2^4). 14. (2ni^ii*—5ab»mn^-6a%^) (2m^u'^+r)ab^mn^-5a^b^). 15. (.T6-cT8i/r3+4)(a7«+a;Vl34-4). 16. (oa^-2ab»c+2b*c^) (5a^+2ab^e+2b*c^), Exercise 38. 1. (x+y){x^—xy+y^). 2. {x+y)(a^—oc^y+x^y^—xy^+y*). 3. (x+y)(x^-a^y+x*if-x^y^+x'^y*-xy^+y^). 15 l^ ANSWERS [Ex. 38-39 5. {x+y) {x^o-x^y+x^y^-x'^y^+afiy*-x^y^4x*y^—x^y''+x^y^-xy^+y^oy 6. (dx+a) {9x^-'6xa+a^). 9. 5(2a+36)(4a2-6a6+962). 7. {4x+^y) {Ux^-12xy+9y^). 10. 2(Q+x)(S6-6x+x^). 8. (5a+6&)(35fx2-30ab+3662). 11. 3(3a;+l)(9a;2~3a?-hl). 12. (b+i)(b*-b^+h^-b+l). 13. (i+x){i—x+x'^—x^+a^—x^+x^). 14. (2+!c)(16-8ic+4a;2-2ic3+a?*). 15. (l+5a)(l-5a+25a2-125a3+625a*). 16. {ah+l){a^t^-ah+l). # 17. (007+2) (a4a?4-2a3ic3+4a2x2-8aa;+16). 18. {a+5xy)(a^—5axy+25x^y^), 21. 0a+5c)(^a4-Ja36c+^a252c2-ia&3^34.54c4j. 22. (4a+i&)(256a4-16a36-}-a262-3-Va63+^^e&*). 23. (2x+j^y) (Q'ix^-lQa^y+ix*y^-x^y^+ix^y*-j\xy^+-i^y^). 24. (a7+2/'^)(a72— ic?/2_,_2/4), 25. (3a;2+22/)(9a?4-6a;22/+42/2). 26. (x+l){x^-x+l)(x^-x^+l). 27. (a+6)(a*-a36+a262-a53+b4)(ai'>_a555+5io), 28. (1+a) (l-a+a2-a3+a4) (I_a6+ai0-ai5+a20). Exercise 39. 1. (x-y)(x2+xy+y^). 2. {a-h){a^+ab+l^). 3- (a;-^)(a74+a?82/+a?2^2+a,^_^2/4). 4. (ic— y) (x^+afiy+x^y^+y^x^+x^y^+xy^+y^). 6. (x-y) {x^^+x^y+x»y'^+x^y^+x^y*+x^>y^+x^y6+o(fiy'i+x^?f+xy^+y^^). 7. (2a-y)(Aa^+2ay+y^). 10. {2d^-b)(ia'^+2a%+b^). 8. (3a;-4a)(9x2+12a7a + 16a2). n. (i_|;^)(i+|^+^a;2). 9. (5-7a)(25+35a+49a2). 12. {x-l)(x*+x^+x^+x+l). Ex. 39-40] ANSWERS I7 13. (2-y)(lQ+8y+4y^+2y^+y*). 14. (3-a)(81+27a+9a2+3a8+a4). 15. {x—2a)(x'^+2x^a+4x'^a^+8xa^+lQa'^), 16. (x^l) {x^+x^+x^+oc^+x^+x+l). 17. (l-a)(l+a+a2+a8+a4+a5+a6). 18. {2-xy) {Q4:+d2xy+16x^y^+8x^y^+4x^y^+2xY^+xf^y^). 19. 3(aT-l)(9a?2+3aj+l). 20. 3(6-a)(36+6a+a2). 21. ^(x-y)(x2+xi/+7j'^). «« / a\ / . a^a x^a^ xa^ a\ 22. (^-g|(a^+-^-+^+^+_). 23. 2(^x-y) (^i^x^+la^y+ix'y^+lxy^+y^). 24. 2a?(fa;-2/)(fa?2+3a.^+2/'^). 25. a8(^_(|r2/)(a;2+a?a?/+a22/2). 29. (a-b-c+d) {(a-by+ (a-b) (c-d) + (c-d)^} . 30. (x-2y-2a-b) {{x-22j)^+ {x-2y}^(2a+b) + {x-2y)^{2a+by^i- (x-2y){2a+b)^+{2a+by}. Exercise 40. 1. (a2+62+a6i/2)(a2+62-a6i/2). 2. (a2+62)(a2+52+a&i/3)(a2+62_a&i/3). 3. (a4+64+a262|/2) (a4+54_a262|/2). 4. (a2+62) (a8_a652_^aV-a266+68). 5. (a4+54+a262|/3)(a4+54_a252|/3) (a2+62+(^;^y^2) (a2+62-a5i/2). 6. (a^.+l)(a2+l+a\/d)(a^+l-ai/3). 7. (ic2+l)(ir;i2-;t;io+x8-ar6+£c4-a;2+l). 8. (x^+l+x\/2)(x^+l—x\/2). 9. (a;24-l)(a;8-aj«+a3*-aj2-|-i); 10. (l+a:*+a'2|/2)(l+aj*-ic2|/2). 11. {4x*+l+2x^\/2)(4x*+l-2x^l/2), 18 ANSWERS [Ex. 40-41 12. (l+9ic8+3ir*i/2)(l+9a38-3a^]/2). 13. (4^2+92/2) (ix^+gy+mxyy'S) (4a;^+9y^-Qxy\/3). Exercise 41. 1. (a2+62)(a+5)(a-6). 2. (a+6) (a-6) (a2+a6+52) (a2_a?>+52). 3. (a-f-6) (a-&) (a2+?)2) (a2+52+ct?,^/2) (a2+52_ot5-j/2). 4. (ci+b) (a-b) (a'^-a^+a^b^-ah^+b^) (a^+a^b+a^b^+ab^+b*), 5. (a+b) (a-b) (a^+b^) (a^+ab+b'^) {a^-ab+b'^) (a^+b^+ab\/d) (a2+52_c(5-^/3), 6. (a+b) (a-b) (a^+a^b+a'^b^+a^b^+a^^+ab^+b^) (a^-a^b+a'^b^-a^^+a^^-ab^+b^). 7. (a-^b^)(a^+b^). 9. {a^-'l^){a^+b^){a^+b^). 8. (a3-64)(a3+64). 10. (a5-62) (a5+62.). , 11. (a?2+l)(a;+l)(£c-l). yi2. (a;+l)(a?-l)(aj2_^a;+l)(a-2-a;+l). 13. (l+x)(l-x){i+x2){l+x'^+x\/2)(l+x'2-xi/2). 14. (l+a?)(l— a?)(l+a;+a?2-|-£c3_i_^)(l_a;_^^.2_£t.3_^^)^ ^15. (1-0?) (i+x) (l+a-2j (l+a?+a-2) ( 1 -xi-x^) (l+x^+xi/d) (l+x^-x\/d). 16. (2+a^2) (2-0^2). 17. (2+a;2)(2-a;2)(2+2iC+£c2)(2-2aj+a?2). 18. (9-073) (9+0^). 19^ (10_a8)(10+a8). 20. (2a-l)(2a+l)(4a2+l)(4a2-2]/'2a+l)(4a2+2i/2a + l). 21. {Sa^+2x»){Sa^-2a^)(9a^+Axfi). 22. (H-2a2) (l-2a2) (l+2a+2a2) (l_2a+2a2) (l+4a*+2a2|/2) (l+4a*-2a2]/2). 23. (K2ic)(l-2aj)(i-a;+4a?2)(iH-a;H-4a?2). -^24. {xy-ab){x'Y+abxy+a^b»){xy+ab)(x'^y^-abxy+a^b^). 25. (2+3a6)(2-3a5)(4+9a262). 26. (2-a) (2+a) (16+8a+4a2+2tt3+a4) (16-8a+4a2-2a3+a*). Ex. 41-43] ANSWERS. 19 29. (Factors of x^'+a'^) (factors of aj^—a"). / 30. (£c»-3— 2/»+2)(cc"-8+2/»+2). 31. {(a+6)2+c2}(a-f-&+c)(a+6-c). 32. {(j;_2/)2+ (a-6)2; {x-y+a-h) (x-y-a+b). 33. {(a+b)2+c2}(a+6+c)(a+5-c). 34. [(2x—yy^+ (a+4b)2} (2x—y+a+ih) (2x-y-a-4b). Exercise 42. 1. (a+b)(c-d). 5. (a2+i)(5_|_c). 9. (x-l)(y-l). 2. (a7-7/)(a-5). 6. (a+b)(x^+x+l). 10. (a-2)(6-3). 3. (a;+2)(a+6). 7. (2+d)(a-6+c). 11. (£C+b)(£c+a). 4. (x-y)(a+4). 8. (a;-l)(«-6). 12. (a?-2)(a?+a). 13. {x—a){x+a+c). 16. (a+6)(a7+l)(a7— 1). 14. (x-a)(l+x-a). 17. (a7+l)(a:-l)(a;+3)(ic2+l). 15. (a7+2/)(« + l)(«^-«+l). 18. (x-y)(x+y-'i). 19. (a-6)(£tf-l)(a.^+(r3+ic2+.T+l). 20. (a+1) (a-1) (6+1) (5-1) (a2+l) (b^+1). 21. (07— l)(acc+5). ' ' 24. (mn+l) (pq+2). 22. (2o^+2/)(^+l)(^^-^+l). 25. {2a-b) (2x-y). 23. 2fe2(a+6)(a-b). 26. (6+(i)(6-d)(a-c)(a2+ac+c2). 27. {a-l)(a+l)(x-l)(x^+x^+x^+x+l). 28. (aH^5)(£C+l)(ic+2). 29. {x-y)(x+2y\ 30. {m—q)\m+q) (n+p) (n'^—np+p^) (n—p) (n'^^+np+p^). 31. (a+&) (a-J) (x+y) (x-y) (x^+y^) {x^+y^+xy\/2) {x^+y^-xy\/2). 32. (a-b){a+b)(a^+b^){x-y)(x2+xy+y% ^ 33. a(a+6)(a;+l)(a;-l). 34. {x+y—z—w){x—y—z+w). 35. (a:+2/+'2^)(^+2/-^)(^-2/+^)(^-^+2/)- 36. {a-b)ia+b+cy 39. (ic2_2/24-i)(aj2+2/2). 37. (a+6+2c)(a+6+3c). 40. (x-y+z)(x^+xy-{-y^). 38. (a-5+c)(a-5-c). 41. {3-x~x^)(l+x). 42. (a+6)(aa;+62/+c). 20 ANSWERS [Ex. 43 1. 2. 3. 4. 13. 14. 15. 20. 21. 22. 23. 24. 28. (2a;+l)(ir-l). (Sx+2){x+l). (2a;-l)(aJ-2). (4x-l){x-h2). (x-i){2x-l)(x+e). (x-l)inx^+x+2). {a+2) (a^+a+l). Exercise 43. 5. •(2a;-3) (x-d). 9. ^(x+2) (x^-x-1). 6. (3a;+5)(a;+3). 10. (x-l){x+S){x+5). 7. (Sx+2)(x-n). 11. (x+l)(x^-2). 8. (x-l){x-2){x-d). 12. (a7_l)(£c4-l)(a?2+4). 16. (a+3)(a2-a-l). 17. (a+l)(a+4)(a-5). 18. (a+4)(a+3)(a3+l). 19. (a+1) (a-1) (a2-ai/3+l) (a^+a\/2+l). (2x-y) (x-y). {^x-y)(x+y). ix+y)(x-y)(x+2y). (x-dy){x+Sy)(x~2y). (a+db){a-h){a+2h). a(x^-Sa^) (x^+Sa% - (^-irnr- 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. ah(2x-l){x+1), (2a+35)(2a+13b). (ar-7)(ar+4). a{x—a)(x—5a). 35. {y-z){x-z). 36. (x+'S) (x^+6). .37. (2a-2b-y-z) (2a-2b+y+z). 38. (3£C-52/)(3a;-42/). 39. (a2-a54-62)(a2+a6+52). 40. {x^—ixy—y^){x^-{-4:xy-y^). 41. (a^3_6)(a?3+7). 42. {x^-5y^) (x^+ly"^). 44. £c(it?-3)(l-ir)(l+a!?). 45. (by+l+xy){l-xy). 46. (x-y-z) (x-y+z) (x+y-z) (x+y+z). (x+y)^(x-y). x(y-d){y+10). (c—b)(a+d). (x+d){x-2)(x^-x+Q). (8f+9)(9f-5). (a+1) (a-1) (5+1) (5-1). (2z-M){z+U). -a2(3a-£c)2. x{x^+c)(b+ax). (a2_b2_c2)2. 57. (ic-4a4) (a;-a252). 58. (£C— a— 5)(ic+2a). 59. (2x''-dxy-\-Sy^) (2x2+dxy+Sy^). 60. (a— c)(a+c)(5+c). 61. (x-S) {x^+2x+2). 62. (a+3)(a2-2a-l). 63. {x-l){x-Z)(x+b). 64. (a;-l)(a;-3)(ar-7). 65. (a;-l)(a;-2)(aj-3). 66.(a;+2)(ic+3)(a;-3). Ex. 43-45] ANSWERS 21 67. (y-5x-3z)(y-5x+dz). 68. (3a;»+52/")(2a7''— 32/"). 69. (3a-36-3)(a-6-4). 70. (7a + 76+3c)(a+6-2c). 71. (x+3)(x-i)(x^-5). 72. (a-1) (a+1) (a-h). 73. (a?-a)(a;-3). 74. aa7(a?— 2a) (a?— a). 75. {x-2)i2x^+x-7). 76. (5a;-7)(7ar-5). 77. x»(x-7){x+S). 78. (a7-3b-2)(ar-3[>4-2). 79. (.x-3_5)(a;24.7).^^ 80. (x-l){x-i-2)(x+S). 81. (a;-3)(a;2+7). 82. (x+l)(x^+24x-16). 83. (2a;+22/-2;)(5a7+52/4-6«). 84. (8a+8?)+9c)(3a+36-4c). 1. 2a262. 2. 3a363c2. 3. 15a^2/3. 4 8xyz^. 5. a%. 6. tzK 7. icV- 8. 7a53.^2. 9. Ixyz^ 10. dabd^. 11. xy^z. 12. 2a^W. 13. d6xyz. 14. 35a3&. 15. 6ma2. 16. 5a2a?. 17. 7a2Z>2. 18. 7CC2/2. 1. 6a268. 2. 30a36*. 3. 8a762c8. 85. (a;+2-2a+y)(a;+2+2a -y\ Exercise 44. 19. ar32/5. 37. 2a;-3. 20. 2(a+6)2. 38. ax{x—a). 21. 7{x-yy. 39. a2_^5. 22. 4(a+£c)2. 40. a(a+2a;). 23. x-1. 41. 1. 24. 5(a?+l). 42. 7m2+5m+5. 25. VS(x--i)^x+l). 43. a+h. 26. a;+2. 44. 1-a;. 27. a:-l. 45. a-&. 28. 2a;+l. 46.* m-2. 29. 4. 47. a?— 2y. 30. x-1. 48. a2+a5i/2+52. 31. a;-l. 49. (a+l)2. 32. a-&. 50. l+ir+a;2. 33. a-b. 51. a—x. 34. ic. 52. a;-5. 35. a;-3. 53. 2{x-\-y). 36. a;+l. 54. a-1. Exercise 45. 4. lQ2x^y^. 7. 100a2m2n«. 5. 42ic8?/822. 8. 60^6^8. 6. Sia'b^c^. 9. 144aW66. ^2 ANSWERS [Ex. 45-46 10. 108x^y\ 12. 3(a-6)8(a+?>)2. 14. oc^(l+x)^l-x). 11. 2(x+y)^ 13. 2£c2(a7-l)(ii7+l). 15. 30 (a+6)2(c-d)8. 16. 235a8a;(a-b)(2a+5)3. 23. (2x-l)(x+2)iSx+l). 17. (a;+l)(£t?-l)2. 24. (2x^-x-\0)(2x^+x-d). 18. (a?+l)(a?-l)(a;-2). 25. 2a7(a7+l) (072+50:4-6). 19. x{l-x){l+x)(l+2x). 26. 2a;2(a?+l)(a;-l)(£c2+a:+l). 20. ia-b)(a-\-h){2a+h). 27. (ic+l)(a;-l)(aj+2) (i»2+i). 21. (aH-2)(a+3)(a+4). 28. (l-aT^Xl+aj+icO- 22. (a;+6)(a;-5)(a7-l). 29. 2x^{l+x)(l-x){x-2){x^+l). 30. 12a;2(a7+7)(a;_2)2. 31. (a-&)(a+6)(a2+&2)(c[24.of5+52). 32. l-ic8. 37. 07(1-0?) (l+x)(l+a;2)(10-a;). 33. 076-1. 38. (07-l)(a;+l)(jr+2)(a;+3). 34. 6o;2(£c+3)(o;-l)(o;2+l). 39. 3o74(o7-l)(ic2+ir+l). 35. (a4-64)2. 40. (a+b)^a-hf. 36. (3a-2)2(3a+2)(9a2+6a+4). 41. (a;-3)(.T-12)(o;2-2). 42. (a+l)(a4-2)(a-l)2(a2-2a+3). 43. (6a3+a2-5a-^)(3a2+a-2). 44. (6o78-7o72z/-2a;y2)(3^2_,_^2/-4^2), , 45. (07+1) (07-1) (x+2) (x-2){x-r^) {x-S). 46. (07+2) (207-1) (307+1). 47. (2a+l)2(2a;-l)2. 49. (dx-^)(2x+7)(ix-n). 48. (a-b)(x-y)K 50. (4a;+l)(2a;+7)(3a?-8). Exercise 46. 1 « '7x^ ab-2 ' b' 6- 2p- 2 2a2 5o^ ^' 362- 7- "7^- 3 528 __3266c 4 — . 9 -^^ 6. f . 10. ^|. » 07+1 Ex. 46-48] ANSWERS 23 16. 17. 18. 19. 20. 21. x+2 x^ ' x-y xy ' x+1 x^+x+1' x—d X + 4:' X' ■x+1 X^ — 0(^+X^—X + l' 3.r+ x^' 1 2 3. 10. 3a +86+ 2 X—1 + ix 6x2' x-V 1062 3a -26* _3_ 6£C2' 22. 23. 24. 27. x+2 3x+i' 1+x+x^ x+x^+a^+oc^' x'^+2xy+y ^ x'^+xy-\-2y^' 1+x^ 1-x^' l-g 2+a* x+1 x+2' Exercise 47. 28 4 — 3? 2 + 07' x^+Zxy+y^ ^^- x^-2xy-iy^' b+c—a 30 31. b—c+a m-2 m+9' c—d c+d' x+2 . x-d' 4. £c2+2a?+l. 7. 2a;+l + 2a:2_i 5. x+y+ x-y 6. x^-dx—2+ 8X + 4: 2v^ ^ x+y a'2-2.r+4+ x+2 11. x^+2x:^+'lx+8+ 32 a;-2* 12. x^—xy+y'^ hx 6a;2- 2y3 i»+2/' Exercise 48. 8a* 9a3 20a8a; 24a3ic3' 24a3.T3' .r2— ,r ic^+a? 1' 1' 6a?3 24a«a^* 1 a;2-l- 3. 4. 18&4 3c3 12a262c' xyz' 6£C+3 5a? +5 12a262c' I2a262c* £C2 ?/2 (ri/2;' a?^/^;' 707+14 (a;+l)(a;+2)(2a;+l)' (a7+l)(a;+2)(2ic+l)' (a?+l)(a;+2)(2a:+l) (2a;+l)(a;-5) (l-a?)(a;-1) x (x+l) ' {x+l){x-l){x-5y {x+l){x-l){x-5y {x+l){x-i){x-5)' (a+b)(a'^+b-2) a^-b^ 2(a;+l)a (a;+l)*' 2(a-b)(ag+b2) 2a;2 {a—b){a+b) x{x+l) {x+1)*^ {x+iy' 24 10. 11. 12. 13. 14. 15. 16. 1. 2. 3. 4. 5. 17. 18. 19. 26. 28. ANSWERS [Ex. 48-49 xir a;(a;2_j,y2) .r^faT+y) ^^___ y{x^-y^y y{x^-y^y y{x'-y'^)' 6a?(a?+?/) -3?/(a?+y) x^ x(y-b)(z-c) y(x-a)(z-c) z(x-a)(y-b) -2?/2 {x-a}{y-b){z-cy {x-a){y-h){z-cy {x-a){y-b){z-cy x^-9 x^-1 (x-l)(x-2)(x-dy (x-l){x-2)(x-'dy 2a^+ia 4ac+12c ■(a+2)(a-2)(a+3)' (tt+2)(a-2)(a+3)* b—c c—a {a—b){a—c){b—c)' {a—b){a—c){b—c)' (a—b)(a—c)(b—c)' n-b y'^-z,^ z'^-x'^ x^-y"^ {x-y){x-z){y-zy {x-y){x-z)\y-zy {x-y){x-z){y-zy Exercise 49. - 4.^2+37 + 1 7.-r-2 - 2a;2 • abc Ux-j-9 lOa^ * x+y+z xyz Ax x^-1' a+2h 10^2+20? x'^-\ ' 1 2+y' 7. 8. 9. 10. a^—x 3a;«-2a;2+8a;-2 2oc^—5x^+2x ' 4a2 a2-a;2* x+1 X 2x+x^ 1+x 11. 12. 13. 14. 15. 16. 20. 21. (a;+4)(2a;-l)(3a;+l)' a;2_i- 2&2 22. 4bcd + 6acd —Sabd— 2abc ASabcd y^ + xy^ + Sxh/ — x* x^^ • ^^• 2xy—2y x^—xy^' 27. 0. oc^—x'^+dx+l x^~l 24. 25. 30. 1-x —x^+x^—2x 1-£C2 • 5— a;— 5a?2 l-X'2 * 4ab cC^-b^' -4 {x+\){x~'d){x+by 2a?2— 6a?— 14 a;2-2a?-8 * a^ x-r (a—b){a—c) 4ax+4bx—4:ah 452- a;2 • Ex. 31. 35. 37. 38. 49-51] x^-l 32 ANSWERS 26 24-8a;2 34. 1. 5. 9. 10. 11. 12. 13. 14. 15. 16. 1. 2. 3. 4. 14a;2+130 £C4-26£C2+25" 4.T3+2a;2+4a;-5 ad x+y x+2y x^+xy+y^ x^+ocy+y^ x^—xy+y^' a+1 a+o x+y x^ h^ ' ^x'^-\-^xy b-2+3ab+9a2 68 3a^ bxy 1 3a -36* x^—xy+y^ x^—5x'^y^+4y^' 2bc^-2bd^+2a^-2bm a^c^-b^c'^-a'^d'i+b^d^ ' ^^ 2iic*-2a*+2a^xi-2a^ 39. 40. y ^ a^xz' 2y+S x-1' 17. 2x^-x^ x^-1 ' x(x+y) x-y Exercise 50. 3. 1. 7. 07-1, x^—x^a^—x^+a'^ a^+a%+ab^+b^ 1 x-1 18. a+b. 19. x+y. 20. x^—x^y+xy^. a 21. 22. 23. 24. 5. 6. 7. a—b 2(x-y) x+y x+y-S X x-y x+y Exercise 51. x'^+4xy+4y^ 2x'^+xy—'Sy^' ab^-a^ x+\ x+^' 8. 1. 41. 42. a2-62 25. 26. 27. 1 2+x 4. a«6. 8. a^-x'^+x-\. 2ofi—x'^+Ax-2 a;2-4 225a2c2-16a?* 25c2 x^ x+2' oc^—Ax a?^- 9 29. 2x. 30. 6. 31. 6a;+9. 32. a;2+a;-4. 33. 4a2. x—2 10. a;'^+9a;+14' (a-6)2(CT2+6g ) (rt-6)2+l • ^^' 6xy^ ' 12. x+x^ 1-x' 26 ANSWERS [Ex. 52-55 Exercise 52. U X x-\ x+V 1+237 l+a7+a;2* 13. a 6 a^+4a72+7a7+6 .T3+5x2+9a7+5" 7. 8. a2+62 a2-62- a2-a+l 10. CT-4 a— 5* ?l2 11. 3a72-ir2/-32/2. 2a- 1 14. 307+3* Exercise 53. 1. 1. 6. 8. 11. ^. 16. I, 21. 3. 2. -2. 7. -6. 12. 15^. 17. 16. 22. If. 3. A. 8. 0. 13. 8. 18. -3 23. If. 4. 2. 9. 1- 14. -|. 19. V. 24. 12. 5. 2. 10. i. 15. -37. Exercise 54. 20. f. 25. 2. 1. 6. 7. I 13. 4. - 19. 4. 25. 8. 2. 6. 8. h 14. -H. 20. 1. 26. 4. 3. -f. 9. 4. 15. 19. 21. |. 27. -2. 4. 5. 10. 6f. 16. f. 22. \. 28. -10. 5. -f. 11. 1. 17. 13. 23. ^\. 29. -S^. 6. 1. 12. 15. . 31. 3 18. -7. Exercise 32 55. 24. 13. 30. ^. 1 ^+cy ' 2a ' 6. 7. 1. 2a- 1 - 11. a+b—c ' 2. y-2. 3. 1. 8. 4-a* —a. 12. a(x+S)+h(x-S) x-1 4 c(a;-l)- 07- -a(a;+3) 9. 2ax-h c 13. 4t+l t+2' 5. 3a;. ,^ 10. 0. 14. -t. Ex. 55-56] ANSWERS 2Y 15. 1 19. ?f. 23. ^. r ' v^' c 16.^. 20. f?. 24. £^'. 17. ^. 21. ?^. 25. |. 18. i!:. 22. E±D:, 26 .:^ 27. f(F_32). 28. rr r+r' Exercise 56. 1. $10. 6. 16 and 20. 11. 96. 2. 12 and 38. 7. -24. 12. 26. 3. -7 and 8. 8. 3 and 25. 13. 17 and 18. 4.-3 and 12. 9. 25 and 36. 14. j%. 5. 24. 10. 27. 15. 9. jg 16 yrs.=: John's age. 21. 293^ mi. from station. ' 10 yrs.= James' age. .« .« , .. 22. g^Vmin. 17. 10 yrs. and 40 yrs. ^ 18. $125. ^^' ^^' 19. $72000. 24. 10 rods by 16 rods. 20. 45 dimes ; 6 quarters. 25. 6. 26. 1 J mi. per hr. 10 mi. 3^ mi. per hr. 27. -36. 31. $350. 28. A, $120; B, $160; C, $80. 32. 8%. 29. 11 and 12. 33. Silk $1.10; linen $.55. 30. -2. 34. 18 ft. by 20 ft. 35. In 8| hrs. ; 21 J mi. from starting point of first pedestrian. 36. 21 y\ min. after 4 o'clock. 41. 60. 37. 5y\ min. before 5 o'clock. 42. $3750 and $2500. 38. 15 min. 43. 12 lbs. iron ; 60 lbs. lead. 39. 8f da. 44. 25 oz. 40. 6 da. 45. 246. " 28. ANSWERS [Ex. 57-60 Exercise 57. 1. x=2, y=S. 9. a=5, x-Q. 17. x=7, y=10. 2. x=-l,y=^. 10.y=5,b=7. 16. x=^2, y=16. Z. x=5,y=L 11. a=rf|,5=-if. 19. x=l, y=i. 4. x=-2, y=-S. 12. 07=36, y=d6. 20. x=-d, y=ll. 5. x=-h y=h 13. a=^, b=-'^. 21. x=iO, y=lb, 6. a- -2, b=-d. 14. ir=fff, 2/=fA. 22. x-1, y=A. 7. a7=4, a=ii. 15. jp=2, g=5. 23. x=a—b, y=a—b. 8. a=3, 6=7. 16. .t=12, ^=11. 24. aj=:a, y=b. 25. a;=a, 2/=|- 1. a;=13, y=17. 6. a=35, 5=20. 11. a=12, 6=-13. 2. x=4, y=-5. 7. a=7, 6=9. 12. 6=6, 2^=18. Z. x=il,y=^j\\ 8. a;=15, a=8. 13. m=Yi^, ?i=ff. 4. a;=8, 2/=-2. 9. a=-l, 2/-3. 14. a;=7, ?/=-2. 5. x=2, y=l. 10. aj=14, y=zU. 15. a?=1.8, 2/=1.4. 16. 07=6, 2/=12, 17. a=3, 6=-7. 18 x=^^^-tl^ V- 2n+m' ^ 2n+m Exercise 59. 1. a=17, 6=13. 7. a;=10, y=2A, 13. x=7, y=-2. 2. a=-3, £c=-7. 8'. aj=l, 2/=-f. 14. a? =5, 2/= 7. 3. a7=9, 2/=2. -^. a;=H, 2,=,?,. 15. a?=2, 2^=3. 4. m= — 11, n=7. 10. x=l,y=—\. 16. a;=6, 2/=4. 5. £c=8, 2/=l. 11. a=5, 2/=-3. 17. a;=9, 2/=-3. 6. a;=-V/-, 2/=/t- 12. a7=2, 2/=8. 18. a7=12, 2/= -3. 19. 0?=^, 2/=tV 20, aj=12, 2/=3. Exercise 60. 1. a?=7, 2^=3. 4. a?=i, y=l. 7. a?=20, 2/=32. 2. a?=7, 2/=4. 5. a=4, 2/=3. 8. £c=-l, 2/=3. 3. a;=5, 2/=7. 6. a=-2, 6=-3. 9. a?=ll, y=-4. Ex . 60-64] ANSWERS ( 10. 11. x=5, y=-5. x=8, y=l. 14. 15. a?=20, y=12. a;=12, y=10. 18. x- ^ ^,y- ^ ^. a+b '^ a+b 12. a=—%\h=—'l. 16. x=67, y=10d. 19. x=a^+b^, y=ab. 13. 00 = 4:, 2/=f. 17. a=10, x=8. 20. x=a^, y=b. 21. x=2p, y=iq. 22. x Exercise 61. =18a-246, y=SQb-2ia. 1. x=u y=2. 4. a=3, h=4. 7. x=2, y=7. 2. x=-2, 2/-3. 5. x=S, y-2. 8. a=l b=h 3. x=h y=h 6. a;=:5, 2/=2. 9.x=^^^,x=-^, 10. x=l, 2/=l. 11. 07=10, y=5. 29 Exercise 62. 1. x=l, 2/=3, z=^. 7. 07=4, 2/=5, 5r=6. 2. ic:=6, 2/=l, z=2. 8. a;=20, 2/=10, ;2;=30. 3. x=l, y=h z=h 9- a?=l, 2/-7, 2;= -4. 4. 0?=^, y=l z=i. 10. p=2, 3=3, r=4. 5. .T=-5, 2/=8, 2;=-9. 11. x=l, y=2, z=S, w=:4. 6. x^-l, 2/=8, ;s=l. 12. p--=2, q=-l, r=-3, s=5. Exercise 64. 1. 21 and 8. 9. Father 35 ; son 10. 2. 15 and -6. 10. 36 and 27. 3. 3i ft. and 8i ft. 11. 22 and 16. 4. 96 and 24. 12. 26 and 8. 5. Orange 4 cents ; peach 1 cent. 13. |. 6. A's age 36 ; B's age 21. 14. j\, 7. 4 cows ; 24 hogs. 15. 18 and 40. 8. 5 dimes ; 20 nickels. 16. 48 and 8. 17. 10 persons ; $16. 18. A, $125 ; B, $250 ; C, $200 ; D, $450. 19. 54. 20. Carriage, $175 ; harness, $25. 21. 12 doz. at 12 cents per doz. ; 8 doz. at 4 for 5 cents. 22. $575, $1250, $1825. 30 ANSWERS [Ex. 64-65 23. 6 men ; $2. 24. $750 ; 6%. 25. 8.5 in. first year ; 10 in. second year. 26. A*s, 2| mi. per hr. ; B's, 3 mi. per hr. 27. 24 mi. 28. 45 mi. per hr. ; 55 mi. per hr. 29. A's, 6 yds. per sec. ; B's, 4 yds. per sec. 30. 8 hrs. 31. 2 mi. per hr. 32. Current 2| mi. per hr. ; crew 4 mi. per hr. 33. 90 ft. per sec, and 60 ft. per sec. 34. 360 revolutions per min., and 540 revolutions per min. .35. A, 30 days ; B, 24 days. 36. A, 68f days ; B, 160 days ; C, 240 days. 37. Man $2.75 per day ; boy $1.25 per day. 38. A, $23,040 ; B, $7,680. 39. 35 hogs ; 60 days. 40. Alt., 9 in. ; base, 12 in. 41. Velocity in still air 350 yds. per sec. ; velocity of wind 6.5 yds. per sec. 42. One in 16 min. ; other in 20 min. 43. $1.25 and $1.65. 44. 6yx from first ; d^^ from second. 45. 4 from first ; 3 from second. 46. 42 lbs. tin ; 14 lbs. zinc. 47. 8j% oz. first ; 4|^ oz. second ; 2}f oz. third. 48. 432. 49. 24, 60, 120. 50. Length 30 rods ; width 20 rods ; area 600 sq. rods. 51. 384. Exercise 65. 1- 2l/2. 3. _2^;3; 5. 6l/£ 2. 51^5. 4. 3i>?. 6. 6i/3. Ex. 65-67] ANSWERS 7. 5i/4. 8. 42i/2. 9. soil's. 10. 21^102. 18. 2ah-\/2a-h. 28 19. a2v^2. 20. iy^. 21. ii/6. 3°- ^V9c.,2. {a—x)\/ a—x. (^+2/)l/4a. 11. 91^5. 12. Sxyi/Sy. 13. ia^b^i/dab. 22. ii/is". 23. ^^i^ii: 24. i^/^. 25. .-^al^^- 31. 32. 33. 3 / — r ■1 14. 2axy\/2xy^. 2^>/^-4^C. 15. -5ir22/4y^5a.2. 16. 2x»y^y. 17. a'^y^i/Zay'^. 2/^ 26. ^V'a^bcK 34. 27. 4^^'2'- 35. -i |l^l+2a;+2a^. Exercise 66. 1. 5i/2. 3. 1/5. 6. 4V3. 2. 21/3: 4. 5l/6. 6. -4/5. 7. Ii'/e: 14. fl/2-|/3 8. (3b2+2a6-a2)^^ 15. (3-2a+4a7)i^^. 9. (3a-2a25+262)|/26r 16. VV3. 10. 51/ 3a2. 17. -I1/5: 11. {x+2y=2)\/y. 18. -/^v^i: 12. (2a+3&)i/a6c-a6|/c. 19. 2|/5H-6-S ii;^4-: 13. {x'^-xy)y'x^-^yH/xy\ 20. 3v 2+11/2: Exercise 67. 1. 1^08, v^a*r 4. v^343, ^/144, 1^64. 2. y^^ k'^, I^^ 5. i/a8>6i«, y'a^bis, j/ a^b*. 3. F 5^ |^2«; J^P5; 6. 1/5. 31 32 ANSWERS [Ex. 67- 7. i/48. - ^i- 19. 51^25920. 8. 1/ iol 20. i/a^-b^. 9. 1/192. - i/i- 21. (a;-?/) i/a?+2/- ^ 22. a+b. 10. y'567. 16. 21^3. 23. -1. 11. l/f. 17. 61^500. 24. 21/2; 12. i/|. 25. 4. 13. \/h' 18. 1^55296. 26. 5. 27. 6+2i/2+2i/3+2l/6! 28. ai/2a6. 31. c2i^3a*62. 32. ^]/a6. aa'2 / 1 29. xy\/^' OK ^ ^ / ,0 J ^^- (a+6)2l/"^-^-- 30. a86ci/3c. 33- SaW^b^' Exercise 68. 36. ^i^2a2^2. 1. fl/3. 3. Uj/U-l/2i). 5. 4|/3. 2. ii/l5. 4. 3v/2. 6. 5i/3-5i/2. 7. 3+1/6-1/15-1/10. 8. 3+i/T4-v^6-fi/31. 9. 1/15-4. 13. l+|ir— |i/a;'^+3£C. 14. l+il/2-il/6"-Ji/3: 15- A-Ai/iH-tVi/S-tWS. 16. |l/3+tl/'2-il/7-il/42. 17. i+ll/6-^l/l5. 10. b-i/b2-a2. 11. l+2a;2+2a7i/l+a;2, 12. ^+ii/;i^z^ 13 (x+ l/xy+ -\/xz) (x+y—z—2\/xy y x^+y^+z^—2xy—2xz—2yz 19 n—b+j/ac— }/hc )(a+ h—c—2\/ob) 1. 21/2. 3. a7i/4a72/^. a2+6'2+c2-2a6-2ac-2bc Exercise 69. 3. 125a?V8^. 4. 64a666. 5. a«. 6. a^b^y'a^. Ex. 69-71] ANSWERS 7. 64a66?/5. 5 __ ^ 3/- •^ 12. 1/4. 16. 2x2 1/2. 8. 243ari'2/24. . V '*. ^ V 9. 256(a2-62)2. 13. i>5^. 1^' l^"'-^. 10- V^S. 14. v^3H^2. sa 18. |>2a;2. 19. 7.T3aV2a. 11. 1^2. 15. 2. „p- •^ 20. \/a. Exercise 70. '• '^^ '' ^Y^ 13. (a+6)^— 2. 10i/-l. 8. 7i/2l/-l. 3. 25,/=T. 9. aV:^ ''• (^-3^2/)>/-l. 4 ii/ZT 10- 4a?3?/|/_i. 15. 7|/-1. 5. ii/-i. ^^- 7;^i/-i- 16. 121/-1. 6. fi/3T. 12. 9(a;-2/)2|/ZT. 17. 2y^, 18. (2i/2+|/10+i/7)|/-:^. 22.' 16+3i/=T. 19. 5a2|/ZT. 23. 2a+26. 20. (2x^y-h6x^y^)i/~. 24. 2|/::6. 21. Sx+y}/'^. 25. 2a;-5?/i/^. Exercise 71. 1- -^^- 4. -24|/'^. 7. a363. 2. -12. 5. 420. 8. -2x^y2, 3. -70. 6. lOSOj/^. 9. 5. 10. 6+i/6+2i/^-3i/^. 11- 1- 12. 0. 13. -10i/l0-6i/5+5i/6+3i/3. ^^* ^~"' 19. -7|/^. 24. i]/IT. 20. -i/H^. 25. ^. 21. 2. 26. |, 15. y^-x^. 16. -5v^^. 17. -fi/'^. 22. |. 27. fic8 18. -5i/^. 23. 2. 28. -|v/^. 29. *-fl/^. 30. i+iv/^s+ii/iTs-ii/e. 34 ANSWERS 1 81. f- -fv/-5. 32. f +1]/ -3. 33. i- tW-5' 34. . -^■^/To-il/15-il/i4-il/21. 35. x^- -y+2xi/-y x^+y. ^ Exercise 72. 1. ±2. 11. ±2i/-l. 20. ±5. 2. ±5. 3. ±13. 4. ±7. 12. 13. 21. 22. ±il/30. ±ii/39: 5. ±|l/30. 14. ±iV2. 23. ±ll/55. 6. ±il/30. 15. ±Ai/-^09. 24. 0. 7. ±25. 16. ±l/-3. 25. ±1. 8. ±|. 17. ±3. 26. ±V/-14. 9. ±>/7. 18. ±7. 27. ±f. 10. ±2. 19. ±1. Exercise 73. 28. ±2. 1. 3, 4. 14. -hh 27. 3,^. 2. -5,2. 15. 11,11. 28. -2,5. 3. -1, -7. 16. -h -h 29. 11, 12. 4. -6,9. 17. 11,-7. 30. -10, f.. 5. 2, i. 6. -3, i. 7. -2,1. 8. 1, -h 18. 19. 20. 21. 5,6. -2,5. 3, -V-. \S -2. 31. 32. 33. 3,-|. 4,9. 9. h -f. 10. 2, -i. 22. 23. 2, -V-. 34. 35. — 1, 6. -1, -9. 11. -2, i. 24. 4,^. 36. 28, -20. 12. -2, 1. 25. i,i. 37. h -v- 13. 4, 7. 26. 6, -V-. 38. 5,-f. [Ex. 71-73 Ex. 74-75] ANSWERS Exercise 74. 1. 2, 4. 17. 1, -f. 33. 7, f. 2. -6,2. 18. tV±tVv"141. 34. ±5. 3. -4, -10. 19. -i±i|/-71. 35. 4, -W-. 4. -3,2. 20. tV±tVi/13. 36. -i±il/5. 5. 6, 5. 21. _4±|/-5. 37. -|±^|/185. 6. -7, 2. 22. -3±i/-2. 38. i±ii/37; 7. 9, -1. 23. i±T/3. 39. 2±ii/3. 8. 11, -2. ^^24. -3±4i/-l. 40. 3, -|. 9. -3, 18. 25. 14, -8. 41. 4, 11. 10. -2, -10. 26. 3, 5. 42. 3, -V. 11. h -1. 27. -5, -13. 43. -3, 5. 12. 3, -h 28. ±4. 44. 0, 4. 13. h f. 29. 4,-3. 45. |f±xVl/273. 14. -i, 3. 30. ±1. 46. ±4i/2. 15. f. |. 31. 5,7. 47. 2±ii/3: 16. -i 7. 32. 7, -V-. Exercise 75. 48. W±^Vl/209. 1. -2,5. 16. f±il/-ll. 31. 6, -1. 2. -3, -7. 17. tV±tV1/337. 32. 4, -|. 3. -2, f. 18. f±il/57. 33. 0, 1. 4. i, -2. 19. if, -1- 34. 0, h "^5. -l,f. 20. -f±il/l3. 35. l±3l/^l. . 6. 7, |. 21. -i±il/l7. 36. 7, -V-. 7. 1, -f. ^22. l±3i/-l. 37- 1, -h 8. f , |. 23. -i±i|/129. 38. 3, -V-. 9- I |. 24. 2±i/5. 39. 1, 1. 10. 8, h 25. f±il/-7. 40. -i±il/5. 11. -|±il/-23. 26. -l±2|/2. 41. 0, 1. 12. i±i^-34. 27. 3, -i. 42. 7, V-. 13. A±3W-1'^9. 28. -i±il/-3. 43. 0, -5. 14. 4±2i/3 29. 3, -V-. 44. h -2. 15. 2, f. 30. l|±iV|/l33. 45. 3, |. 35 36 ANSWERS [Ex. 75-78 46. 3, |. 48. 3, -5. 50. 4, -1. 52. 4, -|. 47. -¥, h 49. 4, - 4. 51. 2, -h Exercise 76. 53. 5, -|. 1. ic2_^_3o^o. 4. 2r)x^+30x+Q=0. 7. x''+10x+24:=0. 2. x^-5x-U=0. 5. Gx'^+x-2=0. 8. 12a?2+25a?+12r=0. 3. 16ir2_s.^+i::^0. 6. 6ir2-7a?-10rr0. 9. 4a?2_7a;4-3z=0. 10. 8^2+6j;c4.1-0. 11. Positive. 12. One positive and one negative in each. 13. -A ±|/^. ±a. 14. ± 16. 15. (a) c not greater than Exercise 77. (^)c=i. 3. i&± 11/62+16. 4. a7=:3a or —^a ; a=^a7or —3a;. 5. x=a or 2a ; a=x or 4^x. 6. ?>=:£c or — iT— 2a ; x=b or —5— 2a. 7. — ir±i-i/r2— 4.9. ^ 1_ ■2m'^2?^T 8- -9^±9^l/^'-4mf. ). X, 10. 11. 12. 13. 14. 15. ID mna 16. ± 17. ■v±\/v^+2c i8.5=±V|?^;'=±i|/|';-4|/f. Exercise 78. 1. 15 and 22. 3. 42 and 8. 2. 10 and 21. 4. 17 and 33. Ex. 78-79] ANSWERS 37 5. 76 and 77 ; or -77 and -76. 8. /^. 6. 12 and 13 ; or -13 and -12. 9. 16 or -8. 7. 4 and 14 ; or —14 and —4. 10. Either 3 or 4. 11. 8 or - V-. 12. 2 in. and 5 in.; or 4 in. and 7 in. 13. 500 ft. by 596 ft. 18. 1 inch. 14. 12 in. and 16 in. 19. 2 ft. 15. 5 ft. 20. 6 in. by 12 in. 16. 25 yds. and 39 yds. 21. 32 rods by 60 rods. 17. 36 sq. in. 22. 5 hrs. 23. 30 mi. per hr. and 40 mi. per hr. 24. 8 and 12; or -12 and -8. 32. 100 ft. 25. 9 and 24. 33. 35 feet. 26. 6. 34. 10 in., 8 in., 6 in. 27. 2 mi. per hr. 35. 24 min. and 18 min. 36. 15f min. 37. 12 days. 38. 15 min. 39. A, 8 days ; B, 10 days. 40. 10 and 14, or -60 and 84. 41. 45 mi. per hr.; and 30 mi. per hr. 45. 64 sq. in. 48. 12 inches. 46. 4 inches. 49. 84. 47. 14 inches. 50. 4 ft. by 8 ft. 51. 20 A. 52. 529 sq. yards ; 4 yards. Exercise 79. 1. 2, -2, 2i/^, -2i/~. 7. i/S, -1/^, 1/2, -1/2. 2. 1/5, -V/'S, 1/^, -l/^. 8- 1» -1. l/2, -V2. . 9. 2, -2, 3, -3. 3. 2, -2, il/-6, -il/-6. ^0 ^^ _^; ^^^ _^ 4. 1,-1,-2. jj -3±i/l0, -f±ii/29. ^' ^' ^' -^' 12 -^±y^ -i±i/5 6. 1, 3, -2. • 2 ' 2" • 28. 15doz.; 18 cents. 29. 12. 30. $500. 31. 42. 40. 4.1 42. 2ihrs. *i, 43. 12. 44. 1. 38 ANSWERS [Ex. 79-80 13 ^±V~ 1±V~ 17. 1, ZllAl^. 14. 1 ± 1/2, 1 ± 1/2. ^«- 3' -^' ^^ ^' -^^' _ 15. 1, -1, l/-3, -l/-3. 19- 1» -1> 2 ' T) . 16. 1, -2, zl±V^. 20. l/^±/-^ , -V2±i/-2 21. ±2, ±2i/~ l/2±l/^, -l/2±i/^. 22. ± i/2, ± |/^, 1 ± l/^, -1 ± l/^. 23. 24. ±l/-l, '^'%^ '-1 ~ -l/3±i/- 2 -1^ -|±iT^34+2i/- -15; - |±^l/34- •21/ -n 1. 25. l±l/-7 l±3i/- 2 ' 2 EI. 26. 2, 4, n! r±v^l7 2 • 28. -l±l/^ - - -i± 2 VI 27. 1, 1, — 3±l/5 3 • 29. 1, 1, 1± ^ zl. Exercise 80. 1. 3. 13. 144. 25. None. 37. 9. 2. 3. 4. 5. 6. 7. 5. 1. 4. 7. -5. 4, -1. 14. 16. 15. 9. 16. -|. 17. 7. 18. -12. 19. 4. 26. 27. 28. 29. 30. 31. 8. 12. 4. -1. I 38. 39. 40. 41. 42. 43 «'-^* ^' 2a2 • 13. 1- 16. None. 0. 8. -2. 20. 49. 32. None. 44. 2. 9. 10. 11. -8, 40. -4. 3,2. 21. 0. 22. None. 23. 81. 33. 34. 35. -h 7. i. 45. 46. 47. 0,-3. 7. 1. 12. 5, A. 24. 4. 36. i. 48. 1. Ex. 80-83J ANSWERS 39 49. 6,2. 51. None. 53. i^+^f , .0. 2^^ 52. OiV^. 54. 3, -2, ±2^ZT. 55. 1, -1, -'Y^ , i±J^- Exercise 81. 1. a;-l, y=:2 ; 0?=^^ 2/=- !• H- ^=^, 2/=2 ; a?=2, ^=3. 2. .r=3, y=—2 ; a7=— 3, 2/=2. 12. x—4:, y=l ; aj=2, 2/=:3. 3. x=l, y=i ; .T=4, y=^\. 13. 07=7, ?/=4 ; a;=— 4, y=—7. 4. ic=— 5, 2/=2 ; a7=-10, 2/=-8. 14. a?=l, i/=5 ; ir=-3, 2/=-3. 5. x=5, y=7 ; a7=-ff, 2/=-ff. 15. x=^, y=l ; ic=|, 2/=-|. 6. ir=2, 2/=l ; x=-l, y--=-2. 16. ir--=i, 2/=| ; x=i, y=l 7. £c=2, y=^ ; a^^-H, 2/=-2. 17. a;=ll, y=-8 ; a;=8, 7/=-ll. 8. x=^, y=2 ; £P=2, y=i. 18. a=3, 6=1 ; a=l, 6=3. 9. a;=-4, y=-3 ; ic=f, y=—\^-. 19. ^=4, 2i7=12 ; t=—^^, w=-\K 10. a?=6, ?/=l ; a:=l, ?/=6. 20. m =— 8,?i=— 3; m=-V, n=-V^. 21. A=5, B=z4:; A=z4, ©=5. Exercise 82. 1. 07=5," t/=3 ; 07=— 5, y='S ; a;=5, ?/=— 3 ; x=—5, y——S. 2. 07=6, ?/=l ; a?=-6, y=l ; .t=6, y=-l ; a7=-6, ?/=-l. 3. x=5, y=2 ; a7=— 5, 7/=2 ; a;=5, y——2 ; a;=— 5, y=—2. 4. a7=ii/2, 2/=il/2"; x=-U/2, y=iV2; x=n'^, y=r-ii/2; x=-^y2,y=-^l/2. 5. 07=2, 2/=3 ; a;=2, 2^=-3 ; £C=-2, ?/=3 ; 07=-2, y=-S. 6. 07=1, 2/=5 ; x=l, y=-o ; 07=-|, 2/=r) ; 07=— |, ?/=-5. 7. 07=6, y=i ; 0^=6, 2/= — | ; a-=— 6, 2/=| ; o?=— 6, o;=— f. 8. 07=i, y=i ; 07=i, 2/=-i ; x=-i, y=i ; 0?=-^ y=-h 9. £c=6, 2/=9 ; o;=6, y-—d ; 07=-6, jf=9 ; 07=-6, y=-9. 10. o;=4, y=2 ; a;=4, 2/=— 2 ; a;=— 4, y=2 ; 07=— 4, y=—2. 11. a?=4, 2/=3 ; a;=-4, ?/=3 ; a?=4, y=-3 ; a;=-4, 2/=-3. Exercise 83. 1. x=2, y=l', x=^2, y=-l; a;=||/2, y=iy2; a;=-fi/2; 2/=-|l/2l 40 ANSWERS [Ex. 83-85 2. x=0, y=yl9 ; x=0, y=-yid ; x=S, y--^ ; a?=:-3, y=2. 3. x=i/% y--0 ; x=-y5, y=0 ; a7=5, 2/=-3 ; x=-5, y=S. 4. a;=3, t/=4; ic=-3, 2/=-4; x=lV% 2/=|t/3; a7rr-||/3, ^/zzr— fi/i". 5. ic=2, 2/=4 ; x--'Z, y--^ ; a;=i/2, y=^\/2; .t~-|/ 2, 2/=-3i/2. 6. a;=il, 2/=2;.rr=-l, 2/=-2 ; a7=|/3; 2/=0 ; a;=:-|/3, ?/=0. 7. a?— 1, 2/=5 ; a7=: — 1, 2/=— 5 ; £c=14, 2/=— 8 ; 0?= — 14, 2/=8. 8. x—2, y=5 ; a;=— 2, y=—5 ; ar=4|/3, ?/= — 1/3 ; ic=— 4i/3, y=V'd. 9. ic=2, 2/==5 ; a;=:— 2, y——5. (Defective system). 10. 07=3, 2/=5 ; a?=: — 3, y=—5 ; x=S, y=—5 ; x=—S, y=5. 11. (j?=:4, y=5 ; x=—4:, y=—^ ; x=Si/'S, y=l/'S ; 0?=— 3y 3, ?/=: — ]/3. 12. 07=5, y=l ; a7=-5, 2/=-l ; x=iy -10, 2/=-il/ -10 ; Exercise 84. 1. x=~l, y=2 ; 07=2, 2/=-l. 2. a7=-2, ?/=5 ; £r=:5, y=-2. 3. 07=6, ?/=2 ; 07=— 2, ?/= — 6. 4. 07=5, 2^=-l ; 07=-5, y=l ; 07=-l, ?/=5 ; 07=1, ?/=-5. 5. 07=5, 2/=— 3 ; 07=3, 2/=— 5. 6. 07=6, y=z5 ; 07=5, y=G: 7 o^-l£l±i ^- 1^5-1 . ^_ t/5-1 l/5+l . \-Vl -l-j/s ". l-T/5 ^_ l-|/5 ^- 2 ' ^- 2 ' ^ 2 ' ^ 2— 8. 07=5, ^=4 ; 07=4, 2/=5 ; 07=— 5+i/— 14, y=— 5— l/— 14; 07=— 5— |/— 14, 2/=— 5+l/— 14. 9. 07=7, ^=-1 ; 07=-7, 2/=l ; a7=l, 2/=-7 ; 07=-l, 2/=7. 10. 07=6, y-^ ; 07=-9, 2/=-6. 11. a;=5, 2/=-7 ; o;=-7, 2/=5. 12. a;=4, 2/=5 ; o;=5, y^L Exercise 85. 1. a;=3, 2/=2 ; o?=2, 2/=3. 2. o;=0, 2/=-3 ; aj=3, 2/=0. 3. a?=5, 2/=-2 ; o;=2, 2/=-5. Ex. 85-86] ANSWERS 41 4. x=l, y=2 ; x=2, y=l ; a;=:f +il/-55, 2/=^-il/-55 ; 5. irrz— 2, 2^=:8 ; x=8, y-—2. 6. 37=4, y=l ; a7=-l, ^=-4 ; a;=f+il/-79, 2/=-|+il/-79 ; 7. a;— 5, 2/=3 ; a;=5, 1/=— 3 ; 0?=:— 5, y=d ; a;=:— 5, y=~d. 8. 07=3, 2/=^3 ; x=3, y=d ; a?=— 3, 2/=— 3 ; x=—d, y=—S. 9. 37=4, y=2 ; a7rr-2, 2/=-4. 10. x=2, y--S ; a7=3, 2/=2. 11. x=S, y=l ; a;=l, y=S ; a7=2+5|/^, 2/=2-5i/^ a;=:2-5v/-l', 12. x=S,y=2;x--2,'y=-S. Exercise 86. 1. x=25, y=9 ; x=9, ?/=25 ; ir=-81+8v ^97, 2/=-81-8]/^^97'; .T=-81-8l/^7, 2/:rr-81+8l/^I^. 2. x=l, y=4: ; a7=4, ^=1 ; x=-S-i ^, y=-S+y^; 3. x=4:, y='d ; 07=3, y=4: ; a?=— 3, 2/=— 4 ; a7=— 4, y=—d. 4. .T=i, 2/==i; 07=-^,?/=-^. 5. x=2, 2/=7 ; ^7=7, y=2 ; a7=f , 2/=! . ^^,-7^ 2/=|. 6. 37=2, 2/=l ; 37=1, 2^=2 ; 0;= — 1, y=—2 ; a7=:— 2, y=—l ; V-i-yn. ^- -V-^-yn -yz-i+yii ^- 2 ' 2^ 2 • 7. 07=8, 2/=2 ; a;=2, y=8. 8. a;=3, y=l ; 07=-3, y= — l ; 07=1, y=S ; a7=-l, 2/=-3. 9. x=-i, 2/^796. 10. a;=27, 2/=8 ; x=-8, y=-27, 11. ^=6, y=n ; a.=3, 2/^6 ; a.=Z:lW57, ^..ziW:^. ^_ -19-3v57 .._ -19+3v^57 42 ANSWERS [Ex. 86-89 12. x=l, y=—5 ; a?=5, y=-l ; x=l, y=4: ; x=-i, y=—l, 13. Impossible system. 14. x=W, y=Q. 15. x—4:, y=6 ; x=5, y=4: ; ^_ -9-l/T6i ^^ -9+1/161 . .^._ -9+ 1/161 .,_ -9-i/167 . 2'^ 2' 2'^ 2 16. a;=0, ^=0; x^i, y-\ ; a?-!, 2/=-2 ; x=—\, y= — h Exercise 88. 1. 4 and 9. 2. 5 and 11 ; 5 and — 11 ; —5 and 11 ; or —5 and —11. 3. 3 and 5. 4. 6 and 2 ; or -2 and -6. 5. 10 and 6 ; 10 and -6 ; -8 and 0. 6. 2 and 8. 7. 49 and 9. 8. 5 and 6 ; -2 and -1 ; or 2|/^ and -2]/^. 9. 1 and 4 ; 1 and —4 ; —1 and 4 ; or —1 and —4. 10. 62. 11. 36. 12. 10 and 4. 13. 4 in. and 9 in.; or 6 in. and 6 in. 14. 2 in. to width and 1 in. from length ; or 3 in. to width and 2 in. from length. 15. 6 and 8. 16. $96. 17. Derrick 40 ft.; guy-rope 50 ft. 18. 40 feet and 25 ft. 19. f . 20. $3.25 ; 10 days. 21. 6 men ; 8 days. 22. $32 for apples ; $22 for potatoes. 23. $250 ; 6%. 24. 8 days, and 12 days. Exercise 89. 7. a?>5. 9. a!<6. 11. a?>3or <~1. 8. x^\K 10. ic>ll. 12. a;>5or <-3. 13. X between Y- and 7. 14. X between a and -, when a is not 1. a 15. X between 2 and 6. 18. Impossible system. 16. X between —14 and —3. 19. x between 4 and 12. 17. x5 and ^<5 that satisfy equation. 22. Values of a7>3 and y<^5 that satisfy equation. 23. Values of a;— H that satisfy equation. Ex. 89-92] ANSWERS 43 24. Values of a;^ — |f and y^rii t^** satisfy equation. 25. Values of ic^-— ^ and y^:^r\y that satisfy equation. 1. -V-. 2. 3. f. 1 3i- 4. h a* 5. ._-_, 6. T^. 7. 11, li f. 8. /.. 9. |. 10. 3 2(a-5)' 11. 4. 52. 320 yds ; 4* 54. |. 55. 1. x=^y. 2. 42. 9. V=:32t. 12. 16 cu. in. 13. 256 ft., 113 ft. Exercise 90. 12. 4. 24. pq\ 13. 15 and 20. 25. tV. 14. 15 and 21. 26. ^ 15. 4 and 18. a ' 16. 48. 27. X ' 17, 150^. 28. 4. 18. i'2. 29. 45. 19. TO. 30. 3iu*. 20. he a' 31. 3v^6. 21. y_z 32. Vab. X 33. Vxy. 22. 4. 34. 2ad 23. 9a;6. "3c"' is; 480 yds. 53. 3 or 4. 55. 8 and 12 ; or -12 and -8. 56. 64 ft. Exercise 91. 3. xy=24:. 5. x=8yz. 7. 7. 4. 25. 6. 2. 8. x^=4y^. 10. 30sq.in. 11. 201.0624 sq. ft. 14. 2.44 approximately. 15. 7^^^ ft. from end of heavier. Exercise 92. ' 1. F—km, where -F= force, ?7i=mass, and ifcrra constant. 2. a— ^2 , wherea=attraction, d=distance between tliem, m=rmass of one, m'=:mass of other, and k=?L constant. 3. t= — j — , where f=tension, m^mass^ v=velocity, ?=length, and fc— a constant. 44 ANSWERS [Ex. 92-94 4. t=1c\/ 1, where t=time, l=length, k=£L constant. 5. p—khh, where p=pressure, 7i=depth, &=:area of bottom, ^-—a constant. 6. H8^=k, where H=hent, s— distance, A;=a constant. 7. R=k-, where 72= resistance, Z=: length, a = cross-sectional area. a and k=R constant. Exercise 93. "• ^- 12 ^ 20. A- _ "■ ac2- «+b 28. 8i/a. A 3— 13. -o^- " — ^ 1 5. ^gi, 14. M. '^- SMi^)- ^l/^ V. ^x2. ~ 24. -^• 16. |«V. a;")/^ «.- 52 31 y if . 17 '^^?/'^ 25. 1- • ,V- 9. y 27. 1 10. |/(|7. IS- 56^- ^^; "^4 32. ^>(^:^2 33 V^^+b . 34. 27. 36. i. l/a^' 35. 4. 37. ^k- 38. |. 39. V. 40. ^i^. 41. ^j%^, 42. /^. 43. |. 44. 24. Exercise 94. 1. 2a;2. 6. a^, 2. 1. 10. i/m\ 3. a. '• ^ -• 12. ab. 4. 4^. 61/c & ^ 15b 5. -5-^. 9. -^ 14. -rz::. l/a8 V a46** l/aii Ex. 94-95] ANSWERS 45 15. ?5. a le. a46i2. 17. ^^^j^^ 22. 23. 24. 25. a2aj. xyVx^^yK 30. 31. 32. 3 X+y V^-y' — s. 18. la'y'x. X 26. 27. 28. 1 33. 34. 35. a7 a;3+a;^2/^+2/«. ' a2-l. , ^ifiy^ 20. a. xyi/xy^' 21. a. 29. -1. 36. x-y. j 37. 8a^2_i8ar-i- 38. a2-62. 39. x+y. 47- -1507. 44. 45. xi- + 1965+36-115-5-6&-10. i 40. ^+27. 46. x-^ ^+2a;-* +3+2a!^+a;*. 41. 2a-20a^+18a~*. 47. a4- -1+a-* 42. a;^— £c32/^+2/^ 48. aJ+2aW ^6-1. 43. a;"^+a;~^+a7' -^+1. 49. a^i —2a? V+y- 50. X- i+2a;-^+3aj~^+4a;"^+5+4a^+ ^x' ^+2a;*+a?. 51. a^-a?~i 52. 9a;-' -162/5. 56. 57. a-3+a-2+a-i+l. 1 53. -\/a+2i/h. 58. x^+x^y^+x^y^+y^. \ 54. ar-2-3ar-i+9. 59. h ' 55. a-»+3. 60. 18. V Exercise 95. V ■^v,,^^^^ 1. 24. 6. 336. 11. 30. 16. 240. ^ 1 2. 720. 7. 60. 12. 180. 17. 12. 3. 720. 8. TliuU 13. 24. 18. 56. 4. 48. 9. A. 14. 1814400. 19. 56. 5. 7. 10. 8|. 15. 380. 20. 16380. 40 ANSWERS [Ex. 95-1 24. do. 25. 16. ANSWERS [E 26. 650. 28. 56. 30. !l£: 27. 120. 29. 103^800. 32. 6 ; 10080. Exercise 96. 31. 90. 6. 56. 11. 66i,2^.0 i ;22a. 16. 56. 7. 9. 12. 10. 17. 270. 8. 1140. 13. 10. 18. 28. ' 9. 53130. 14. 325. i 19. 560. 10. 2. 15. 4. 20. 53130. 21. 21. • 22. 45. Exercise' 97. 1. 4. 2. 210. 3. 82160. 4. 455. 5. 220. 1. oc^+5x^y+10x^^+10ai^+5xy^-]-y^, 2. 64ic«-576a76y-i-2160a-V~4320a^+4860a;V-2916a;^+T292/«. "' H-8a;2+24.r*+32if«4-16a^. •: l28ai*-h448ai2&-l-672ai062+560a823»»+280a6&*+84a4664.i4a266+57, 1 ~ 15a«+ 90a«-270a»+405ai2~ 343«i5. x'^-6aTlOa24- 15a^a*_20ic«a«4- 15a?*a8 -6a'2«io-+- a^^. 1 -f 16.T+112a?2+448a^+1120a^+1792a--5+1792cc«+1024a77+256a^. ar-*+4ar-2+6+4r2-fa?*. 32^10 ^_ 80ar-«-+-S0;r-«+ 40ar-4+ lOar-2-i- 1 . a«- 6a65^+15a4b- 20a«6^+ 15a262- 6aZ)^+fe8. ,. 5a^h 5am 20a2?)8 40a6* . 326* "^- 32^24*^ T"+ 27~~+~8r"^"2ir' • r f-9a W+ 36a ^b^+8ia 64-12('. '^^ " ^'^ ■ ~ ' ' - ♦-36a"'^&^+9«''V^+68. '. 18. -8064a;iV- 19- -36a?-ia '. Ex. 97-98J ANSWERS ^^ 20 ^*^^?^ oo ^•'>''>04 81 ' "* — — isT"" 21. ^%%ayc-^. 24 _,^^.22 22- 210a;i 25. l+dx-5o^+Safi-afi. 27. 16-82a+24a2-8a8+a4+3262-48a62+24a262-4a862-f.'M64-24a6* +6a264+86«-4a&6+58. >?8. l+3a!+6a;2+4ic3-6a^-2ic6+3a*-.'r9. 29. «3-53+c8-#-3a26+3a52+3a2c-3c2d+3cd2+362c-3a2d!-3t2d+3ae2 +3ad2_36c2-35d2_6a6c+6a&d-6acdH-66cd. 30. 1 6a4- 32a864- 24a262-8a68+ 64+ 96a8c2- 1 44a26c2+ TZab^c^- 12b»c^ +216a2c*-216a6c*+5462c4+216ac«-1086c6+81c8. Exercise 98. 1. l+2a7+3a;2+4a^+ 2. i-ix^+ix^-j\afi+ -' 3. a-8+4a-i0a;2+l0a-i2a^+20«-Wa^+ , "*• 32a:^ ^64a^^ 128.x' '^25()a,-8'^ 5. l-^x^+Gx*-10j(fi+ •- • . 6. l-lx^-lx*-^\afi- 7. a*+fa"^62_^^^-f54_^^.5^^-l56. 8. ^+-J_+ 3,5 l/2£P 4i/2£cs 32i/2ic6 1281/ 2a-7 9. v^-iv^a;+iv^2if2-j|v 2arV 1.0. a;~*-|a;~^+||a;~~'^*_ma;~'oV • 11. i/3+ — 7^- -+ -(-)' ' ' • ^ '^2]/3 241/3 1441/ 3^ ^ 12. i%>---37:~^ — H7:- 5--- ^ ~ 3i/4 18|/'4 3241/4 13. ^V?^'- . 15. 5e-5. 14. 5ligf||T«~'^'ci 16. — s^Vj-^i^". 48 ANSWERS [Ex. 99-102 1. 2. 3. 4. 5. 21. 26. 1. 2. 10. 11. 12. 13. 18. 19. 1.96794-.. 2.9925+. 146. 37. 20i. 329. 3. 7500. 10, 6, 2... 1679616. -16384. Exercise 09. 3. 6.0276+. 5. 4.0193+. 7. 4.9984+. 4. 2.0800+. 6. 1.9947+. 8. 5.0990+. 3.0006+ . 10. 2.7589+. Exercise 100. 6. 3^. 7. 49 ; 81. 8. 14i ; 26i. 9. 16th. 10. 11th. 24. 9, 13, 17, 21. 27, $28.50. 11. 12. 12. 270. 13. 416. 14. 2500. 16. 2550. 16. 97i. 17. 5 or 6. 18. 64. 19. 2475. 20. 19800. 25.' -8h -7, -H, -4, -2i, -1. 28. at^ Exercise 101. 5- jAi- 3. 34U. ^' -^- 6. ±12i/2. 9. T25+5V+i-l-i+2+8+32+128. 29. 8. 7. -8^. 8. i. -50, 20. 1/2, 2, 2i/2. 24. Iflll^. 20. 4. 3. 21. 22i. 26. 60. 28. |+|+^«^ + 3-V5+ 36-12+4-1+ A%' 34. IH. I "35. I 14. Hi 15. -2730. 16. 971.2+, 17. 42. 24. 1. 25. 3i. 27. -j\. or ^+l+^%+jh+ ' ' 31. A 30. |.' 37. 56^«^ ft. 38. 4 years. 39. $1.21i. 40. 149||ft. Exercise 102. 1. A. 2. ;,V 3. ^%. 4. ^V 6. -i'^-HKA + tV+3\ + iV+-. 6. 0+4+2+|+l + t+ • Ex. 102-105] ANSWERS 49 7. i. 8. 3. 9. -2, -6, 6, 2. 10. hh ^% 1%, ^» if. 11- 80+41/899, 80-4V/399. Exercise 103. 1. l+x+x^+x^+oc^+ ' ' ' . 2. l+2a:+2a;2+2ir8+2ii;4+ , 3. 2x—Bx^+dx^-dx^+Sx^- 4. l+x'^+x^+oc^+x^+ 5. x+x^+oc^+oc^+x^+ . . . . , 6. i+^x-^x^+j%x^-^%x^+ 7. l-x+Sx^-5x^+7x*- 8. 2x+5x^+Ux»+S'7x*+97x^+ • • • . 9. x-2+2x-^+2+2x+2x'^+ 10. ocr-^—Qc—^—x-'^—l—x—x^— 11. x-^-h4x-2+17x-^+7S+S09x+ ■ • • • . 12. x^-x^-2x^+7x^-Sx^- 13. x-^—l+x—dx^+5x^- 14. |aj-2-fa^i+i/-ifa;+|fic2_ • • . . . 15. 2xi^-2x^+5x*-7x^+12x^- Exercise 104. 1. l-ix-ix^-j^s^-j^^x^- 2. l+|a7—V-i»2+W^--rlF^+ • • • • 3. l-2£C-2ic2-4a^-10ii^- •••..•. 4. l+^x-^x'^+j\x^-j%%x^+ 5. l-j^x+^x2+j%x^+j^^x^- 6. 2+ix^—,\x^+^\-^a^-j^^^^x^+ . 7. 1—x—ix^-ix^-^x*- • • • • . 8. l+x~x^+x^—^x^+ 9 2i/34.j/L8_l/2^ . 1^3 9__5i/2 ,2 , . . . 10. l+x+x-\ 11. l+2a;+3a;2. 12. 4x^+3x-5. 13. 4.-dx+x^-2x\ 14. l-2a+4a2-8a8. Exercise 105. 1. x=y+y^+y^+y^+ . 2. x=y-2y^+5y^-Uy*+ 3. x=y-iy2+ly»-^\y^+ 50 ANSWERS [Ex. 105-107 4. a;r=2/-32/2+13^3-672/4+ 5. x=y+2y^+4y^+8y'^+ 6. 07=3?/+ V2/2+ W/2/^+-¥tV2/'+ . 7. x=y—y^+2y^—ny^+ 8. x=y+ly^+j%y^+^\\7f+ Exercise 106. 1 1 _ 1 12 1 ?_ • 1-x T+x' (a?-l)3 x-l' 2 -J_+_A_ 13. l--i- 3 8 _ 1 5__ 14 3__2 8 3(a?-2) 6(07+1) 2{x-iy ' x x+l (x+l)'^' 4 7 8 _ 1 __ 1 + 3^ • 3a;-2^2ii;+3' 2^-3 9+6^'+4i^'2- 5 _J_ + _:L__1_. 16 ^ 1 I -^'-1 ' a?-l a?-2 £c+3 ' ic-1 a'4-1 (ic^— ar+l)* A. ^5 35 10_ 17 A-1+I-_J_+ 3 4(2a?+l)^12(2a;-l) 3(a;+l)' ar a;2^a?3 a;-l^(x-l;2' ^' x+l a?+2 a;+3' 2{x-^+x+l) 2(x^-x+iy 8 _J x-2 19 _J 2 2 3(07+1) 3(.T2-a;-+-l)' ' 07-l a?+l 072+0;+ 1* 2 12 1 20 2a?-7 _ 2x'^+2x-4: ^- 207-3"^ (207-3)2 {2x-d)^' ' 3(x2+2) 3(0^3+2) ' ID 1 I 2 , 1_ 21 16 16 4 o;-!"^ (07-1)2 a;-2* ' {x+2y^ (x+2)^ x+2' 11 _i __i i_ 22 3 - 4 2 • 4(07-1) 4(07+1) 2(^24.1)- • (i_a;)3 (i_a;)2^i_aj- 23 1.1, 1 1 07+1 0;— 1 a;2+a7+l 072— 0?+l' 24. l-i + — 1 _L_ 078 07^(a;+2)3 o;+2' Exercise 107. 1. log232=5. 4. Iogiol0000=4. 7. logg^V^-^ 2. log381=4. 5. log5l5625=6. 8. logiojU = -'^' 3. log7343=3. 6. log4^\ = -3. 9. 32^9. Ex. 107-109J ANSWERS 10. 2*= 16. 16. 3. 22. -2. 28. 4. 11. 42=16. 17. 18. 19. 3. 2. -4. 23. 2. 24. 3. 25. -3. 29. 64. 12. 8^=4. ^0. 3) 13. 7-2-^V 31. !».. 14. 10-3=.001. 20. 3. 26. 0. 32. h 15. 100~^=.001. 21. -1. 34. 27. 0. 33. 1034. 51 Exercise 108. 1. logaOC+\ogay+\ogaZ+\ogaU^'- 3. log„2+ ]ogaX + 2 logaV + S loga 2. 2 \oga00+2 logaV. 4. 2 logaX+logay-^ logaZ. 5. i logaX+^ \0gay-l0gaZ-i logatV. 6. i \0gaX+t loga?/— i logaZ—^ logaW. 7. logaX-logay-logaZ-logaW. 8. l0gaa;+2 logay-^OgaZ-2 ]ogaW. 9. 2(l0gaa7 + l0ga2/-l0ga;^-l0gaU'). H- i \ogaX+^ ]ogay-i logaZ. 10. I ]0gaa?-f logay. 12. i I0ga^+l0ga2/ + i log„a?. 13. j logaX-i logay +i logaZ-i \ogaW. 14. l0ga2 + logaa;+2 loga?/-loga3— loga^:-^ logaW. 15. log„^^. 21. >og»| 30. 31. 2.3222. 3. ^«- ^^^«^- 22. 1.2552. 32. -.8239. 23. 1.1761. n. iog«,^:. 24. 1.3010. 33. .6276. 18. logj/--. 1/2/ 25. 26. 1.6990. 1.3801. 34. 35. .7385. .7517. 19. )logal/ ff^ ' ^loga "^ ^^. 27. .3680. 36. .2762. 28. .2219. 37. -.3460. Vy+z 29. 1.8451. 38. -.0537. Exercise ! 109. 1. 2.3324. 4. 2.8555. 7. 2.9191. 10. .9294. 2. 2.8280. 5. .6646. 8. 1.2041. 11. 2. 3. 2.9731. * 6. 1.2989. 9. 1.4624. 12. 2.9542. 52 ANSWERS [Ex. 109-113 13. 0. 17. 2.0021. 21. 2.8353. 25. 1.99996. 14. .3010. 18. 2.9670. 22. 3.6302. 26. 3.3079. 15. 3.2263. 19. 1.9361. 23. .0398. 27. 3.9208. 16. 1.5349. 20. 29. 3^4983. 4.8363. 24. 4.0792. 30. 2.8739. 28. r.7025. Exercise 110. 1. 57.3. 6. 488000. 11. 10223.8. 16. 105195.12. 2. 7270. 7. 301. 12. 29.53. 17. 1094.5. 3. 1.53. 8. 4000. 13. .09474. 18. .0002989. 4. .289. 9. .00531. 14. 701.5. 19. 7.4733. 5. .0427. 10. 790. 15. .2554. 20. 37.445. Exercise 111. 1. 8.1379- -10. 5. 8.9087- -10. 9. 5.7368-10. 13. 1.5017. 2. 7.1605- -10. 6. 8.1658- -10. 10. 5.3649-10. 14. 4.2055. 3. 9.3410- -10. 7. 6.8617- -10. 11. 2.4935. 15. 3.3260. 4. 7.4859- -10. 8. 7.5829- -10. 12. .1645. 16. 3.2489. Exercise 112. 1. 23.41. 4. 5428.6. 7. .001228. 10. .8224. 2. .004165 5. -.08936. 8. 103.58. 11. .0564. B. 150.9. 9. -526.05. 12. 4.947. 13. 1817500. 14. .000000000002, approx. 15. 14.81. 18. 4.295. 19. -4.1166. 16. .004541. 17. 1.561, approx. 20. .001204. 22. .04327. 24. 5.1811. 21. .387. 23. 3025.7. 25. .00000285. 26. 14.53. 27. 85.6, approx. Exercise 113. 1. 4. 4. 2.4651. 7. 2.165. ^ 10. 13. 2. 2. 5. 1.5481. 8. 3.3852. 11. 5. 3. 3. 6. 8.6336. 9. 2.998. 12. 1000. 13. 10000. 14. 2.209. 15. 1.631. 16. 1. 17. i. ANSWERS TO REVIEW EXERCISES Exercises for Review (I). 5. 16|. 21. 45 ft. and 15 ft. 27. -$500. 8. 8, 16, 32. 22. 105 and 21. 28. +50« and -15°. 11. (a) 13f ; (&) 230f. 25. +75 and -15. 30. -3. 12. (a) 46; (5)9. 26. -15. 60 lbs. 32. 32. 34. 64 ; -27 ; ^V ; tV- 36. -4 ; 4 ; -27 ; |. Exercises for Review (II). 1. -y^-xhj. . 5. lla;2-3a;+2. 2. 2ab-3a;-2c+5a;2. 7. a^+2a'^y +2x^+0^-^1/^, 3. 9a;2-92/+2a. 8. 2«8-6a2+a-5. 4. (3?/2+2a4-3cy)a?. 11. 2a-4&-2c. 13. (7-a)a.^+(7-a5).T8+(5-3c2)a;. 14. - {h-^+c)o(fi- (a+26-2)a;2- (4a-b)a;. 15. a2c^+{a-^)x^+{h-2)x+a^+h-^2. 16. a?; a"; x^K 17. -3a8aV- 19. 2x'5— 4a^+6.T3-10a;2. 21. _3a*+5a46-5a362+4a268-2a6*+266. 22. 40a2_35a5. 23. a^ ; a; a^ . ^8, 25. Quotient-a4+2a364-3a262^-4ab3-2a66+55*— 467 . Remainder=6a5S-6a68— 466+469. 29. |a;4^. 30. a^ ; a""* ; b'^'m^" ; S^. 31. £c5//5 ; £ci22/i8 ; 64a96a>. 32. -8a^2/i%i2 . 81a2464ci2ci8 . -^^. 33. a2+2a6-t-62; Ax^-\2x^j^+^if ; ^a*+2a26+962 ; 0^-4^2+4. 34. rt2_|_52_,_c2_|_2a5+2cic+26c; Ax'^-\-^y'^-\-z^—\2ocy+ixz—Qyz. 53 64 ANSWERS [Ex. for Rev. II-III 36. ±2a^; Sa% ; -2a^b^ ; {a—hY\ imaginary. 38. Ax'^-a or a-4x2 ; 9+5n3 or -9-5n3 ; 2x'^-^y\ 39. a2-b2; l6a^_9?/4; ^-86076. 40. x'^+{a+h)x+ah; a^+2a;2-24; 4a2624.20a6+21 ; i^c'^x'^+^^cx-^Z. 41. ic2_f.a:^^2/2 ; £t7^+£c2^2_|_|^4 . a:^_a:22/2_|_2/4 ; a^—a^+a^—1. 42. When ?i is even. 43. When n is odd. Exercises for Review (III). 3. (a) 2x{Zx^-2xhj+'f); (b) a^b(da-2h+l); (c) 2x'^(x^-3); (d) ab{a-b){a^+b^)', (e) x-y«-Hx^+y); (/) a-b-ia^+b'). 4. (a) {xy^-l)(xy^+l)', (6) ia;(a?-62/) (a!+6?/); (c) (a-6+2c3)(a-5-2c3); (d) (a+26+c) (a+26-c); (e) (ir+2)(ic-2) (0^2+4). (^) (l_£c2+22/2)(l+a;2-22/2). 5. (a) (a:4-2)(ir-2)(x2-2£c+4)(a;2+2a;+4); (6) (a2+9)(a4-9a2+81); (c) (l+ar2^2+a^)(l-a;2|/2+a^); . (d) (x-l-y)(x^-2x+l+xy—y+y^). 6. (a) (a;-6)(a;+5); (6) (Sx-2)(x+l); (c) (3+4a:)(l-2ir); (d) (5a?2-2/)(a^2+22/); (e) (2a:+32/)2; (/) (5-a^)(3+a;); (r/) (a4+a2+l)(a4-a2+l); (/i) (6-2a) (&+4a). 7. (a) (a+6)(a+l)(a-l); (6) (£c-a)(a;+l) (ct2-£c+1); (c) (a?— 3+a— b)(a'— 3— a+6); (d) (x+l)(a+b)(a+c). 8. (rt) (a;-l)(a;+2)(£t?-3); (6) (.t+2) (2a'-l)(a;+4); (c) (a-3)(a-3)(a-5); (d) (6+3) (d^.,, 4). 9. (a) 2aa;(2aa;+3a2^-2/); (6) 5a3a^(a+a;)2 ; (c) (2a:-y)2; (d) (a;i/+16)2; (e) (2a;2-1.5mn)2; (/) (2a7- 152/2) (207+151/2); (l-14ir2/=^)(l + 14cc2/2); (gr) (x^+y2)(x+y); (h) {x+2)(x-y); (i) (a-b){a^+b^); (j) {dn-4m)(2-7m'^); (k) {x-y+2)(x-y-2); (I) (a-b+5)(a-5-5); (m) (d+3c+l)(d+3c-l); (n) (5.T+72/2f)2 ; (o) (iri/3+3)(a:2/3+9); (1)) (a72/-llz)(aj?/-13^); (g) (x-b)(x-24)', (r) (itV+i2)(a-22,3._io); (s) (ir2+i)(a;+l); {t) (a-b)(a^+b^); (u) {a+b+2ab){a+b-2ab); (v) {x^-\-y^){a+b){,a-b); {w) (2a;- 3) (07+4); {x) (a+b)(c-d). Ex. FOR Rev. III-IV] ANSWERS 55 10. (a) (3a8+l)2;(6) (4a+56)(3a-26); (c) (a+5)(a-b)(a+25)(a-2?>); (d) {2x^-4xy+dy^)(2x^+ixy+2y^); (e) (8a+96)(2a-6); (/) (a-i-h+c-d)(a+h-c+d); {g) (x-+,l)^; (h) (a+h){a-h-l)', (i) {x^+xy+y^)(x^-xy+y^); (j) (Sx-2y)(2x-Sy); (k) {7x-Sy)ix+4ty)', (l) (4a-7)(4a-5); (m) (a2'»-?>'»)(a2'»-hi!>"); (n) (fl+6)(a2-6a+36;; (o) (x^+2xy+y^+l)(x+y+l)(x+y-l); (p) (a2+a+l)(a2-a+l); (g) 3a;2^(a;2+(CT/+2/2)(a;2-a;2/+i/2); (r) a(a-l-a5-5); (s) (l+rt)(l-b); (f) (a+5+c)(a+6-c) (c+«— b)(c— a+6); (w) {2a+b){dc—2d)', (v) {ax+hy){bx+ay); (M^) (a+cc)(a-a;)(b2+a:2); (a?) (3a-l)(2a+36). 11. (a) aj-2; (6) a?— 5 ; (c) 2ic-l ; (d) £t'+3. 12. (a) 6a;2(i»-3)2; (5) 10a26(a-5)(a+l); (c) a?2(a;-3)2(a;+2); (d) {dx+5){2x-l)(x^+^). iPi /^\ 4ic2-19a;+24 . /m 18-6^_ . /^x 22a?+42 art-6x'3+7a;2+6a;-8 ' ' ' a?2-2ic-48 ' ' ' a?2_,_i0ic+21* 26-2a 18. (a)x+7; {b)dx+Q; (c) 3a?+30. 19. rJr-::,. 20. € 21. a. - 1- x2^yr «"• a;2 • ""' 2ic2_i' no / XA /^,^^ , . a^+h ,,.a^—Qax+x^ . . x^z+x—y^z—z ^.. 23. (a) ; (6) 1 ; (c) ^:p^, ; (d) ^,^_^^, ; (e) ;^3|^— ; (/) 1. Exercises for Review (IV). 5. (a) ^', (6)4, (c)i. ^ , , 1+x 1 ... 2ci . a+2d , 2at—a 6. (a) a=-2^, a;^^^^^ ; (5) a=^^-^, i^-^ST^ ^=—2" ' ... . i * J * J a d . d (c) t=pr^, i)=-^, r=-^, ^=- ; d=vt, v=-^, t=-. 7-2 8 — 9 ^^ ^- "• ^- bx ^' 2 ' ,n f—M- r,-f^ n-J^ 11 7«-:ST „-V^ r-— 12. 1. 18. (a)x=S,y=-4; {b)x=hy=^. 19. (r=— 1, 2/:=a+6;a=^, 6=r^. 56 ANSWERS Ex. for Rev. IV-V] 20. x=z4, y='S. 21. x=a-\-h, y—a—b. Exercises for Review (V). 6. (l-6a2+126)x?^3a. 7. (aic2?/+aaj22;+3a2a;+ 12?/+ 12;2;) 1/^+7. 8. 61/ 7. 13. ^/ 635000. 8/ 6,—— 6/-- 10. 1/125, 1/121, l7l3. 15. 3j;i/243£rii. 11. i^8a^ 16. les+isiy'Io+ssi^m 12. v^fO. 17. 2-1/5" 19. f{+|fi/3. 20. Z2aWi/ab'', 50a^i^3aj; 2yy'3x^y. 21. i/3a;22/3; i7a;+3 ; dx^\/x-y. 22. (a) 5+1/6 ; (6) 15+4i/l5; (c) 111/? ; (fi) x'ul/loT 27 (e) i/a?s^io ; (/) Aa^ ; (gr) 37-2+^ ; (h)i/2. 2b'. 2ic3|/Zi. 27. (a) 8ai/^; (&) -24^^. ; (c) 3 ; (d) 2-i/5. 28. 3-1/^. 30. -8+21/15. 31. M-^Si/35+Ai/^+At/^. 32. -^1/3+1^:^-1+11/31/-^. 36. (a) ±il/3; (b) ±3; (c) ± l/3. 38. (a) i or -5 ; (b) | or -| ; (c) i or -|. 39. (a) 8 or -4 ; (5) i or -5 ; (c) a or |. 40. (a) i or -f ; (b) i±il/l3 ; (c) 5 or | ; (cl) f or -5. 41. Cor 3. 42. 13 or -J/. 43. l±v^^^^ . 47. ^. 6 49. 7a;2-19ar-6=0 : x'^-2x-ji=0. 52. (a) 1/3, -1/3, 1/2, -1/2 ; (5) -1, -1, ^±|i/IT; (c) 1, 1, -3±2i/2. Ex. FOR Rev. V-VII] ANSWERS 57 58. (a) i+ll/19; (6) |; (c) 0. 59. Cube roots cf 1: 1, -i±il/^;cube roots of 8: 2, -lij/^ ; cube roots of 27: 3, -f ±fl/^. 63. x-1, y=^2 ; x=2, y-\ ; £»7=-3+i/"^, 2/^-3-]/^ ; .T=-3-i/^, ^=-3+1/ -2^ 64. A-TTi'^ 65. F=|7rr3; Fi^^ttDS. Exercises for Review (VI). 6. No. 22. 3. 9. 4£n2 ; i2cc^y2 . 6(a+6)2 ; |/^. 37. «i5 ; «* . ^,15 ; a'^n . «^. 10. 64; 256«4; 8?/3 ; (a+6)5. ' ' ' ' ^^ ' ^^'* 11. t ; 2.T22/ ; GOZ>a; ; a^+2/. ^9. a^ ; 0-2 ; 64aV?^- 17. |. 30. 8; 4; 5; ^ 18. .^=10^. „„ b2 . 2a253 ^ 2 19. 60^=2/. • a2 ' 31/ ' (a-6)2' 21. 1. 33. i; 5;3V;i; -32. 3 — 4 — ?/^ ^^ 34. .aio ; i/a2 ; Va^ ; ^/-, ; g^^' 35. £c"3+2a;~^2/-i+?/-2 ; 4a-2-12a-i?>~^+95-i ; x-'^+^x'^y^ ^^x'^y^ +y ; a;"^-2/~^ ; a/^+h^ ; a^+a^b^+b^. 38. a greater than 2. 39. a? greater than V, 2/ greater than ^-f. 40. 19. 41. Between 8 and 15. Exercises for Review (VII). 3. n{n-\) ; 60; 5040; ?i(n-l)(n-2) (?i-r+l). 4. 16. 5. 120; 720. 6. 6 ; 24 ; 720 ; n{n-\) (?i-2) 3-2 1. 7. 19958400. 8. 50400. 9. 35;190;l; ^^V^) ; ^^^^-^>-V^^^-^^+^^ 1. 10. 4950. 11. 25. 12. 66. 13. 103740. 14. 581400. 19. l-2a^+|i»«-t?a?9+/Ta?i2_^\a7i5+,t^a7i8. 58 ANSWERS [Ex. for Rev. VII-VllI 20. l+^x+-g\x-2+-sUj^+ • 3*- '^''h 21. „Cr-ia»-'+i6'"-i. ' 36. 4+8+12+16+20. 22 — £c20 38. ?^. 23. --'|Ja'2o. 39. a-i;ri=— 4. 24. l+JJ.r-5.r3+3a^-£C«. 41. f||Mf- 25. 2.828 ; 2.924 ; 2.024. 42. 11|. 28. 39. 43. l+3i+6+8i+ll + J3i+16. 29. —V. ^' ^' 30. '^"^ ; 50. 45. 4i/3. 31. 11. . 46. 7 ±41/3. 33. 345. 47. 3+6+18+54+162. Exercises for Review (VIII). 1. 3or-t. 2. ±1/5. 3.-1,-1,2. 4 a+V2b-a'^ a-y2h-^\ ^_a- V2h-a'^ 7/=r«+l/26— a2 ^ (i;^=6 5. 1, -2, 3, 3. 6. 65780. 7. (?i+l)th. 11. l+6a7+18x2+54.T3+162.T*+ • • • ■ . 12. l+3a?+-4ic2+7ic3+lla?4+ 13. l-2a;+a?2+a?3-2a74+ 14. l-|ic-V-£c2— V/^-W/^^- • 15. a;=2/+37/2+13^3+672/*+ on (f,\ 1 I " 1 . /i)\ q 1 I " , • ^^ 2(1-.^) "^2(1+0;) (1+07)2'.^'^' 2(a+l)'^2(rt-l) ' (o\ 3 2 1. ,,'22 1 ^^ a;+l 2.'r+l"^2a;-l' ^ ^ a; a;+l'^(a?+l)2 ' (p\ ^-1 5 _ 4 . , -. 3 1_ 3 5a?-3 '"^ a? a:2 a;+i (a?+l)2' ^-^ ^ ir2 a; 2(a?-l)"^2(.T2+a;+i)* 22. 7 ; 3 ; f. 26. 10^=65. 23. 1 and 2 ; 2 and 3 ; 3 and 4. 27. 2 ; 3 ; -4 ; 2 ; a? ; ; 1. 30. (a) logaOJ+loga?/; (5) logaa;-loga^ ; (c) nlogaO?; (cZ) i- logaO?. 32. (a) 1 ; (5) ; (c) -1 ; (d) -3 ; (e) 1. 33. 1. 34. An integer plus a fraction. ANSWERS TO EXERCISES IN APPENDIX Exercise I. 1. ^x+1. 2. x^-5y. 6. 7. 1+x—x'^+x^. 1x^+8x^-2. 11. «+62+|. 3. x^-x+2. 8. x^-xy+y^. 12. X+4:--^. 4. 2a2+5a-7 9. x^-Sxy+2y^. X 13. .^-"-f ^ 2 2x 5. a2-2a-2. 10. y^+^y+l- 14. l+4^x-ix^+ • • . 16. l4-ia2. -ia4+ •••••. 15. 1-x—^x^- - • • • 17. m^+l,m-^-^^m-^+ • • • • . Exercise II. 1. 86. 5. 3.5. 9. .162. 13. .774 + . 2, 43. 6. 12.1. 10. 51.1. 14. 10.246+. 3. 162. 7. 11.2. 11. 2.236+ 15. 19.261 + . 4. 203. ,8. 2.15. 12. 1.732+ 17. 3.435+. Exercise III. 16. .866+. 1. x+2y. 6. X-2-X+1. 10. ar2-2;r-l. 2. 0^2-1.^ 3. a;2+32/2. 7. 8. l-3a+2a2. 2+i!t;+£t'2. 11. a;2-2a;-3. 4. 3a -462. 5. l-2a. 9. a-4+2. a Exercise IV. ''■ M 1. 47. 4. 53. 7. 806. 10. 3.45. 13. 4.73+ 2. 14. 5. 83. 8. 3.9. 11 1.25+. 14. 1.54+, 3. 25. 6. 698 9. 6.9. 12. 59 2.15+. 15. .53+. 30 ANSWERS [Ex. IN App Exercise V. 1. a;-2. 7. x+1. 13. x^y—xy. 19. x^+x^- 2. x-2. 8. a-2. 14. 1ab-ab^. 20. a-1. 3. x+1. 9. m^-2m-S. 15. £c3+2a;2. 21. m-1. 4. 2a-l. 10. da+4b. 16. None. 22. x-2. 5. 2x-9. 11. 2x^+xy. 17. x+1. 23. a+1. 6. a2+4a-5. 12. a2_a6-2b2. 18. None. 24. a -10. 25. y+2. Exercise VI. 6. 4n5-15n3-33w2+20n. 7. x^-2x^-Qx^+20x'^-ldx+e. 8. 2a?4+7a^-2a;2-13.r+6. 9. 12a4+44a8+21a2-47a-30. 10. x^+4x^-Sx^-29x^+2x+24. 11. 2x^+2Sx^+Q0x^+28x-'S2. 12. 18a8+84a7+162a6+186a5+132a4+54a8+12a2. 1. 4a8-20a2+31a-15. 2. 24a8+50a2_31a-70. 3. ;r6-17a?3+12£c2+52.T-48. 4. x^-'i7x^+lQx^+Q0x-80. 5. 48a4+64a3+37a2+10a+l UNIVERSITY OF CALIFORNIA LIBRARY BERKELEY Return to desk from which borrowed. This book is DUE on the last date stamped below. ^1 ,i^=,^ i90ct'5fLU .^'^ ^. &^^ ,tpa4l953LU 16 lu .Vbft^'f rec- .^V 2^^^^ a^ Ur'i'^ ^T REC'D L^ EC D LD APR& 1963 oto ,^7 1963 LD 21-100m-9,'47(A5702sl6)476 r UNIVERSITY OF CALIFORNIA LIBRARY ^ '' ' '^ ' i^M: ym-