# Full text of "Intermediate Algebra"

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TEXT FLY WITHIN THE BOOK ONLY 00 68071 > OSMANIA UMVERSITY LIBRARY Call No. tt^^ Accession No. ig, . Author WkU- , to! \VMVW, L. Title This book should be returned on or before the date last marked below. INTERMEDIATE ALGEBRA FOR COLLEGES BY WILLIAM L. HART PROFESSOR OF MATHEMATICS UNIVERSITY OF MINNESOTA D. G. HEATH AND COMPANY Copyright 1948, by D. C. HEATH AND COMPANY No part of the material covered by this copyright may be reproduced in any form without written permission of the publisher. PRINTED IN THE UNITED STATES OF AMERICA (408) PREFA Cf HIS BOOK offers a collegiate substitute for third semester high school algebra. The text was designed for a college student who will study it either (1) as a preliminary to taking college algebra, or (2) as terminal work hi algebra which is intended as a prerequisite for ele- mentary courses in various fields of natural or social science, or in business administration. A suitable selection of content from the book would provide a satisfactory algebraic foundation for a first course in trigonometry or in the mathematics of investment. In the case of a student of the assumed preparation, the text provides sufficient material for a substantial course utilizing from 40 to 60 class hours. The plan of the text was based on the assumption that the typical student involved is of a mature age but studied his elementary algebra so long ago that practically all fundamentals must be taught as if they were relatively new material for him. Hence, the early chapters of the book present a mature but frankly elementary treat- ment of the foundations of algebraic technique with a generous amount of discussion and problem material. Also, appropriate refresher work on arithmetic is provided incidentally hi the algebraic problems and explicitly in an early optional chapter devoted to computation. The tempo of the discussion in the text is gradually increased until, in the later chapters, distinctly collegiate speed is attained so that the student will find it easy to make the transition into a substantial second course devoted to college algebra. The text makes no attempt to present material which custom dictates as primarily within the sphere of college algebra, although such material frequently may enter the most substantial courses in third semester algebra at the secondary level. However, in the interest of efficiency and math- ematical simplicity, the terminology and general viewpoint of the text is distinctly collegiate. Emphasis is laid on the logical sequence of topics, accuracy of definitions, and the completeness of proofs. vi PREFACE SPECIAL FEATURES Adult nature of the presentation. The discussion in the text is couched at a level suitable to the maturity of college students. Hence, the available space and assumed class time are utilized mainly to explain and illustrate the mathematical principles involved and only the necessary minimum attention is devoted to artificial mo- tivation of the type which might properly be expanded for younger students. Terminology. Particular emphasis is given to the language con- cerning variables, functions, equations, and the most elementary aspects of analytic geometry because of the importance of this vo- cabulary in fields of application which the students will enter in college. The technical vocabulary of the algebraic content is limited by excluding terms which are of small or doubtful utility. Illustrative material. Extensive use is made of illustrative ex- amples to introduce new theory, to recall previous knowledge, and to furnish models for the student's solutions of problems. Emphasis on development of skill in computation. The viewpoint is adopted that the student needs refresher training in the operations of arithmetic, as well as* new mature appreciation of various features of computation. Hence, work with fractions and decimals is intro- duced quickly, and substantial early sections are devoted to a dis- cussion of approximate computation. Also, the exercises and applica- tions continue to demand computing skill throughout the text, and the chapter on logarithmic computation is made very complete. Supplementary content. A small amount of material not essential in the typical course is segregated into obviously independent sec- tions labeled with a black star, *. Also, the teacher will understand that several of the later chapters are optional and that their omission in whole or in part will not interfere with the continuity of other chapters. The book was planned to eliminate the necessity for fre- quent omissions by the teacher. The extent and grading of the exercises. The problem material is so abundant that, in many exercises, either the odd-numbered or the even-numbered examples alone will be found sufficient for the student's outside assignments, and the balance may be reserved for work in the classroom. In each exercise, the problems are arranged PREFACE vii approximately in order of increasing difficulty. Examples stated in words are emphasized at the appropriate places to avoid the develop- ment of an inarticulate form of algebraic skill. However, the text does not spend valuable time in specialized training to develop problem solving skills devoted to artificial or unimportant types of problems. Answers. The answers to odd-numbered problems are provided in the text, and answers for even-numbered problems are furnished free in a separate pamphlet at the instructor's request. Flexibility. The grading of the exercises, various features in the arrangement of theoretical discussions, and the location of certain chapters are designed to aid the teacher in adapting the text to the specific needs of his class. Composition and appearance. The absence of excessively small type, the generous spacing on the pages, and the special care taken in the arrangement of the content into pages create a favorable setting for the use of the book by both the teacher and the student. University of Minnesota WILLIAM L. HART CONTENTS CHAPTER 1. THE FUNDAMENTAL OPERATIONS 1 2. INTRODUCTION TO FRACTIONS AND EXPONENTS 22 Review of Chapters 1 and 2 46 3. DECIMALS AND ELEMENTS OF COMPUTATION 48 4. LINEAR EQUATIONS IN ONE UNKNOWN 58 5. SPECIAL PRODUCTS AND FACTORING 82 6. ADVANCED TOPICS IN FRACTIONS 104 Review of Chapters 4, 5> and 6 118 ' 7. RECTANGULAR COORDINATES AND GRAPHS 119 8. SYSTEMS OF LINEAR EQUATIONS 131 9. EXPONENTS AND RADICALS 142 10. ELEMENTS OF QUADRATIC EQUATIONS 170 11. ADVANCED TOPICS IN QUADRATIC EQUATIONS 186 r I2. THE BINOMIAL THEOREM 204 ^. RATIO, PROPORTION, AND VARIATION 210 14. PROGRESSIONS* ~~" "~~~ 221 15. LOGARITHMS ._. 240 1 6. SYSTEMS INVOLVING QUADRATICS 262 APPENDIX 277 TABLES I POWERS AND ROOTS 283 II FOUR-PLACE LOGARITHMS OF NUMBERS 284 in COMPOUND AMOUNT: (1-K*)* 287 IV PRESENT VALUE OF $1 DUE AFTER k PERIODS: (1 -M) ~* 88 ANSWERS TO EXERCISES 289 INDEX 315 CHAPTER THE FUNDAMENTAL OPERATIONS 1 . Explicit and literal numbers In algebra, not only do we employ explicit numbers like 2, 5, 0, etc., but, as a characteristic feature of the subject, we also use letters or other symbols to represent numbers with variable or undesignated values. For contrast with explicit numbers, we agree that number symbols such as a, b, x, and y will be called literal numbers. In this book, as a rule, any single letter introduced without a qualifying description will represent a number. 2. Signed Numbers * The numbers used in the elementary stages of algebra are called real numbers. They are classed as positive, negative, or zero, 0, which is considered neither positive nor negative. The word real is used with reference to these numbers in order to permit contrast with a type of number called imaginary, which will be introduced at a later stage. ILLUSTRATION 1. 17, f , and are real numbers. In arithmetic, the numbers employed consist of zero, the integers or whole numbers 1, 2, 3, , and other unsigned numbers which we express by means of fractions or the decimal notation. These num- bers, except for zero, will hereafter be called positive numbers. When we choose, we shall think of each positive number as having a plus sign, +-, attached at the left. ILLUSTRATION 2. The positive number 7 may be written -f 7 for emphasis. * For a logical foundation for algebra, see pages 1-78 in College Algebra, by H. B. FINE; GINN AND COMPANY, publishers. 2 THE FUNDAMENTAL OPERATIONS In a later section, we shall formally define the negative numbers, which will be described as the "negatives" of the positive numbers. ILLUSTRATION 3. Corresponding to + 6 we shall introduce the negative number 6. Positive and negative numbers may be contrasted concretely hi assigning values to quantities which are known to be of one or other of two opposite types. In such a case, we conveniently think of any positive number P and the corresponding negative number P as being opposiies. With this hi mind, we frequently refer to the signs "+" and " "as being opposite signs. ILLUSTRATION 4. In bookkeeping, if a gain of $5000 is assigned the value + $6000, a loss of $3000 could be given the value - $3000. We shall desire the results of operations which we shall define for signed numbers to correspond with our intuitions when the numbers are interpreted concretely. ILLUSTRATION 5. Let t indicate an increase of temperature when t is positive and a decrease when t is negative. Let time be considered positive in the future and negative in the past. Then, the following concrete state- ments should correspond to the indicated addition or multiplication. A decrease of 20 and then a rise of 8 creates a decrease of 12; or, (- 20) + 8 = - 12. A decrease of 10 per hour in temperature for the next 3 hours mil create a decrease of 30 ; or, (+ 3) X (- 10) = - 30. // the temperature has decreased 10 per hour for the preceding 4 hours, the temperature 4 hours ago was 40 higher than now; or, (- 4) X (- 10) = + 40. EXERCISE 1 Under the specified condition, what meaning would be appropriate for the indicated quantity with opposite signf 1. For 5 miles, if + 5 miles means 5 miles north. 2. For + $10, if - $15 means $15 lost. 3. For + 8, if 3 means a fall of 3 in temperature. THE FUNDAMENTAL OPERATIONS 3 4. For - 14 latitude, if + 7 latitude means 7 north latitude. 5. For - 170' altitude, if -f 20' altitude means 20' above sea level. 6. For -f 30 longitude, if - 20 longitude means 20 west of Greenwich. 7. For - 5', if + 5' means 5' to the right. Introduce your own agreements about signed values and express each of the following facts by adding or multiplying signed numbers. 8. A gain of $3000 followed by a loss of $9000 creates a loss of $6000. 9. A fall of 40 in temperature followed by a rise of 23 creates a fall of 17. 10. Thirty-five steps backward and then 15 steps forward bring a person to a point 20 steps backward. 11. If you have been walking forward at a rate of 25 steps per minute, then 6 minutes ago you were 150 steps back from your present position. 12. If the water level of a river is rising 4 inches per hour, then (a) the level will be 24 inches higher at the end of 6 hours; (6) the level was 36 niches lower 9 hours ago. 3. Extension of the number system At this point, let us start with the understanding that we have at our disposal only the positive numbers and zero. Then, we shall extend this number system to include negative numbers, properly defined, and shall introduce the operations of algebra for the whole new number system. Hereafter, when we refer to any number, or use a literal number without limiting its value, we shall mean that it is any number of the final number system we plan to develop. 4. Algebraic operations The fundamental operations of algebra are addition, subtraction, multiplication, and division. Whenever these operations are intro- duced, the results in applying them will be the same as in arithmetic when only positive numbers and zero are involved. 5. Multiplication The result of multiplying two or more numbers is called their product and each of the given numbers is called a factor of their product. To indicate multiplication, we use a cross X or a high dot between the numbers, or, in the case of literal numbers, merely write them side by side without any algebraic sign between them. 4 THE FUNDAMENTAL OPERATIONS We separate the factors by parentheses if the dot or cross is omitted between factors which are explicit numbers, or when a factor not at the left end of a product has a plus or a minus sign attached. ILLUSTRATION 1. 6-3 = 6X3 = 6(3) = 18. We read 6X3, 6-3, or 6(3) as "six times three." ILLUSTRATION 2. 4ab means 4 X a X b and is read "four a, b" If a = 2 and b = 5, then 4ab = 4(2) (5) = 40. If N is any number, we agree that (+ 1) X # = N; N X = 0. (1) 6. Negative numbers Let 1 be a new number symbol, called "minus 1," to which we immediately assign the following property: (-Ijx (-1) = +1. (1) By our standard agreement about multiplication by + 1, (4-1) x (-1) = -1. (2) If P is any positive number, we introduce P as a new number symbol, called "minus P," to represent ( 1) X P. That is, - P = (- 1) x P. (3) We call Pa negative number. Our number system now consists of the positive numbers, zero, and the negative numbers. ILLUSTRATION 1. Corresponding to + 6, we have the negative number 6, defined as ( 1) X 6. In concrete applications, corresponding to each positive number, we may think of multiplication by 1 as having the property of producing a number of opposite type, called negative. 7. Absolute value The absolute value of a positive number or zero is defined as the number itself. The absolute value of a negative number is the given number with its sign changed from minus to plus. The absolute value of a number N is frequently represented by the symbol | N |. ILLUSTRATION 1. The absolute value of + 5 is + 5. The absolute value of - 5 is also + 5. We read | - 3 | as "the absolute value of - 3." We have | - 3 | = 3 and = 0. THE FUNDAMENTAL OPERATIONS 5 8. Inserting signs before numbers If we insert a plus or a minus sign before (to the left of) a number, this is understood to be equivalent to multiplying it by + 1 or 1, respectively. ILLUSTEATION 1. -f 5 = (+ 1) X 5 = 5. - 16 = (- 1) X 16. + a = (+ 1) X a = a. - a = (- 1) X a. Hereafter, we shall act as if each explicit number or literal number expression has a sign attached, at the left. If no sign is visible, it can be assumed to be a plus sign because, for every number N, we have N = H- N. 9. Properties of multiplication We agree that the following postulates * are satisfied. I. Multiplication is commutative, or the product of two numbers is the same in whatever order they are multiplied. ILLUSTRATION 1. 7 X 3 = 3 X 7 = 21. ab = ba. ILLUSTRATION 2. (- 1) X (+ 1) = (+ 1) X (- 1) = - 1. II. Multiplication is associative, or the product of three or more num- bers is the same in whatever order they are grouped in multiplying. ILLUSTRATION 3. 5X7X6 = 5X(7X6) = 7X(5X6)= 210. abc = a(bc) = b(ac) = (a6)c. We read this " o, b, c, equals a times b, c, equals b times a, c, etc." ILLUSTRATION 4. The .product of three or more numbers is the same in whatever order they are multiplied: abc --= a(bc) (bc)a = bca = (ac)b = acb, etc. 10. Computation of products To compute a product of two numbers, find the product of their absolute values, and then I. give the result a plus sign if the numbers have like signs; II. give the result a minus sign if the numbers have unlike signs. * A postulate is a property which is specified to be true as a part of the defi- nition of the process. 6 THE FUNDAMENTAL OPERATIONS The preceding facts about a product are arrived at naturally in any example by recalling the multiplication properties of 1. State- ments I and II are called the laws of signs for multiplication. ILLUSTRATION 1. ( 5)( 7) = 4- 35 because (- 1) X 5 X (- 1) X 7 - (- 1)(- 1)(5)(7) = (+ 1)(36). (- 4) X (+ 7) - (- 1) X 4 X 7 - - 28. In a product, the result is positive if an even number (2, 4, 6, ) of factors are negative, and the product is negative if an odd number (1, 3, 5, ) of factors are negative. ILLUSTRATION 2. - 3(- 2)(- 5) (4- 6)(- 5) * - 30. ILLUSTRATION 3. (- 1)(- 1)(- 1)(- 1) - [(- 1)(- !)][(- 1)(- 1)] - (+ !)(+ 1) - 1. 11. Division To divide a by 6, where b is not zero, means to find the number x such that a = bx. We call a the dividend, b the divisor, and x the quotient. We denote the quotient by a -5- 6, or r> or a/6. The fraction a/6 is read "a divided by 6," or "a over 6." In a/6, we call a the numerator and 6 the denominator; also, a and 6 are sometimes called the terms of the fraction. The fraction a/6, or a -s- 6, is frequently referred to as the ratio of a to 6. ILLUSTRATION 1. 36 * 9 = 4 because 4 X 9 = 36. The absolute value and sign of any quotient are a consequence of the absolute value and sign of a corresponding product. To compute a quotient of two numbers, first find the quotient of their absolute values, and then apply the laws of signs as stated for products. 40 ILLUSTRATION 2. XTo = ~ 4 because 10 X ( 4) = 40. i" *" 9 __ 40 =4-5 because 5 X (- 8) - 40. o Note 1. Division is referred to as the inverse of multiplication. Thus, if 7 is first multiplied by 5 and if the result, 35, is then divided by 5, we obtain 7 unchanged. Or, division by 5 undoes the effect of multiplication by 5. Equally well, multiplication is the inverse of division. i THE FUNDAMENTAL OPERATIONS 7 EXERCISE 2 Read each product or quotient and give its value. 1. 7 X 8. 2. (- 3) X (- 5). 3, (- 2) X (6). 4. 8 X (- 3). 5. (- 9) X (- 4). 6. (- 3) X 5. 7. 4 X 0. 8. (+ 5) X (- 3). 9. (- 2) X (+ 4) 10. (+ 4)(+ 6). 11. (- 7)(- 8). 12. X (- 3). 13. (- 1)(- 5). 14. - (- 4). 15. - (+ 8). 16. + (- 7). 17. + (+ 4). 18. - (+ 1). 19. - (- 3). 20. - (- 1). 21. + (- 9). 22. (- 7)(- 4)(6). 23. (- 2)(- 7)(- 3)(4). 24. 5(- 2)(7)(- 3)(- 4). 26. (- 5)(- 3)(- 4)(- 2). 26. - 6(- 4)(- 3). 27. - 3(5) (2) (- 4). 28. - (- 7)(- 4). 29. - 4(- 5)(6)(- 3). + 16 . 31 ~ 16 39 -11. -+8 31 * 8 3 -3 12 -42 -36 -28 +39 ^-i8' 36 'T7' 38. (- 1)(- 1)(- 1)(- 1)(- 1). 39. (- 1)(- 1)(- 1)(- 5). State the absolute value of each number. 40. 16. 41. - 52. 42. - 33. 43. - }. 44. -f 14.2. 45. Find the product of 3, 5, and 4. 46. Find the product of 5.3, 4, and + 2. 47. Find the product of 1.8, 2, and 4. Read each symbol and specify its value. 48. | 7 |. 49. | + 4 |. 50. | - 6 |. 61. | - 31 . 52. - 1.7 . 53. Compute 4o6c ifa= 3, 6= 4 and, c = 2. 54. Compute 3xyz if x = 2, y 10, and 2 = 5. 65. Compute 2abxy if a = 3, b 4, x 3, and y =* 5. 56. Compute 5hkwz if h 3, k = 2, w = 5, and z 2. 8 THE FUNDAMENTAL OPERATIONS 12. Addition The result of adding two or more numbers is called their sum. Usually, to indicate the sum, we take each of the numbers with its attached sign, supplying a plus sign where none is written, and then write these signed numbers in a line. Each number, with its sign, is called a term of the sum. We usually omit any plus sign at the left end of a sum. ILLUSTRATION 1. The sum of 15 and 17 is written 15 + 17; the sum is 32, as in arithmetic. We can state that a plies sign between two numbers indicates that they are to be added, because we could write a sum by inserting a plus sign before each term and then writing the numbers in a line. However, this might introduce unnecessary plus signs. The most useful statement is that, when numbers are written in a line, con- nected by their signs, plus or minus, this indicates that the numbers are to be added. ILLUSTRATION 2. The sum of 17 and 12 is represented by 17 12. Later, we will justify saying that this equals 5, the value of the expres- sion in arithmetic. By using a needless plus sign we could have written 17 + (- 12) for the sum. We specify that the number 1 has the new property that the sum of 1 and + 1 is zero. That is, -1 + 1 = 0. (1) Also, for any number N, we agree that N + = N. The operation of addition satisfies the following postulates. I. Addition is commutative, or the sum of two numbers is the same in whatever order they are added. ILLUSTRATION 3. 5 + 3 = 3 + 5 = 8. a + 6 = 6 + o. ILLUSTRATION 4. 0= 1 + 1 = + 1 1. II. Addition is associative, or the sum of three or more numbers is the same in whatever order they are grouped in adding. ILLUSTRATION 5. 3 + 5 + 7 = 3 + (5 + 7) - 5 + (3 + 7) = 15. THE FUNDAMENTAL OPERATIONS ILLUSTRATION 6. = c -+ (a + &) = c + a -{- b, etc. Thus, the sum of three or more numbers is the same in whatever order they are added. Addition and multiplication satisfy the following postulate. III. Multiplication is distributive with respect to addition, or * a(6 + c) = ab + ac. ILLUSTRATION 7. 8 X (5 + 7) = (8 X 5) + (8 X 7) = 40 -f 56 = 96. % 1 3. Introduction of the negative of a number The negative of a number N is defined as the result of multiplying N by 1, so that the negative of N is N. The negative of a positive number is the corresponding negative number. The negative of a negative number is the corresponding positive number. ILLUSTRATION 1. The negative of + 5 is 5. The negative of 5 is + 5 because - (- 5) = (- 1) X (- 5) = + 5. Thus, we notice that, if~one number is the negative of another, then the second number is the negative of the first. We observe that the sum of any number and its negative is 0. ILLUSTRATION 2. By Postulate III of Section 12, - 5 + 5 = [(- 1) X 5] + [(+ 1) X 5] = 5 X (- 1 + 1) = 5 X = 0. - a + a = [(- 1) X o] + [(+ 1) X a] = a X (- 1 + 1) = a X = 0. 14. Subtraction To subtract b from a will mean" to find the number x which when added to 6 will yield a. This also is the definition used for subtrac- tion in arithmetic. Hence, the result of subtracting a positive num- ber from one which is no larger f will be the same in algebra as in arithmetic. - We read a(b + c) as "a times the quantity b + c." t This is the only case of subtraction which occurs in arithmetic because negative numbers are not used in that field. Thus, (15 25) has no meaning until negative numbers are introduced. ?0 THE FUNDAMENTAL OPERATIONS % ILLUSTRATION 1. The result of subtracting 5 from 17 is 12 because 12 + 5 - 17. If x is the result of subtracting b from a, then by definition a = 6 + x. (1) On recalling arithmetic, we would immediately like to write x = a b, which means the sum of a and b. To prove that x = a b, we add b to a, as given in equation 1 : _&=-6 + a = -&+(& + z) = (-6 + 6) + z = z. (2) T, In (2), we proved that the result of subtracting b from a is obtained by adding b to a. Thus, to subtract a number, add its negative. (3) We define the difference of two numbers a and b as the result of subtracting the second number from the first. If x is this difference, we proved in (2) that x = a - b. (4) Thus, we can say that the minus sign in (4) indicates subtraction, just as in arithmetic. However, it is equally important to realize that (a b) means the sum of its two terms a and b. ILLUSTRATION 2. The difference of 17 and 5 is (17 5). The difference (17 5) represents the sum of 17 and 5. Also, (17 5) represents the result of subtracting 5 from 17, which is 12. Note 1. In a difference a 6, the number b which is subtracted is called the subtrahend, and the number a, from which b is subtracted, is called the minuend. These names will not be mentioned very often. 1 5. Computation of a sum The computation of a sum of two signed numbers or, as a special case, the subtraction of one number from another, always will lead to the use of one of the following rules. I. To add two numbers with like signs, add their absolute values and attach their common sign. II. To add two numbers with unlike signs, subtract the smaller ab- solute value from the larger and prefix the sign of the number having the larger absolute value. THE FUNDAMENTAL OPERATIONS 11 ILLUSTRATION 1. 7 + 15 = 22, just as in arithmetic. EXAMPLE 1. Add 5 and 17. SOLUTION. The sum is - 5 - 17 = - (5 + 17) = - 22 by Rule I. To verify this, we use Postulate III of Section 12: - 5 - 17 = [(- 1) X 5] + [(- 1) X 17] = (- 1) X (54-17) - - 22. ILLUSTRATION 2. The sum of 6 and 6 is zero or, as in arithmetic, the result of subtracting 6 from 6 is zero: 6 6 = 0. ILLUSTRATION 3. The sum of 20 and 8, which is the same as the result of subtracting 8 from 20, is 20 - 8 = 12. ILLUSTRATION 4. The sum of 20 and 8 is, by Rule II, - 20 + 8 = - (20 - 8) - - 12. To verify this, we recall that 20 = 12 8. Hence, -20 + 8= -12-8 + 8= -12 + 0= -12. Essentially, + 8 cancels 8 of 20 and leaves 12. EXAMPLE 2. Subtract 15 from 6. FIRST SOLUTION. Change the sign of 15 and add: result'^ - 6 + 15 = 9. SECOND SOLUTION. Write the difference of 6 and 15: result = - 6 - (- 15) = - 6 + 15 = 9. CHECK. Add - 15 and 9: - 15 + 9 = - 6. 16. Algebraic sums An expression like c - 3 - 5a + 76 (1) is referred to as a sum, or sometimes as an algebraic sum to emphasize that minus signs appear. Expression 1 is the sum of the terms c, 3, 5a and 76. In connection with any term whose sign is minus, we could describe the effect of the term in the language of subtraction instead of addition, but frequently this is not desirable. To compute a sum of explicit numbers, first eliminate parentheses by performing any operations indicated by the signs. Then, some- time^, it is desirable to add all positive and all negative numbers separately before combining them. 12 THE FUNDAMENTAL OPERATIONS ILLUSTKATION 1. Since + ( 12) = 12 and ( 7) = + 7, - 16 - (- 7) + (- 12) + 14 = - 16 + 7 - 12 + 14 = - 28 + 21 = - 7. Or, we could compute mentally from left to right: -16 + 7w-9; -12 is -21; + 14 is - 7. Note 1. It is undesirable to use unnecessary plus signs. Thus, in place of + (- 3) + (+ 8) we write -3 + 8. EXAMPLE 1. Compute (5z 3o6) if x 3, a = 4, and 6 = 7. SOLUTION. 5x - 3ab = [5(- 3)] - [3(- 4) (7)] = - 15 + 84 = 69. Comment. Notice the convenience of brackets to show that the multi- plications above should be done before computing the sum. 1 7. Summary concerning the number zero We have mentioned that the operation a -*- 6 is not defined when b = 0. That is, division by zero is not allowed. However, no ex- ception has arisen in multiplying by zero, adding or subtracting 0, or in dividing zero by some other number. Thus, if N is any number, AT + = AT; N-Q = N; # X = 0. If N is not zero, then TT = because = X N. r ILLUSTRATION 1. 6 + = 0. 3X0 = 0. = = 0. 5 Note 1. Contradictions arise when an attempt is made to define division by zero. Thus, if we were to define 5/0 to be the number x which, when multiplied by zero, will yield 5, then we would obtain 5 = Q-x, or 5= 0, which is contradictory. EXERCISE 3 Compute each sum mentally. 1. 20 + 7. 2. - 28 - 9. 3. - 13 - 5. 4. 36-4. 6. 28 - 15. 6. 12 - 17. 7. - 16. 8. 6 - 25. 9. - 13 - 19. 10. -35 + 6. 11. 7 - 42. 12. - 25. 13. - 6 + 6. 14. -18 + 3. 15. -13 + 0. 16. 16.8 - 15.2. 17. - 13.7 - 4.5. 18. - 5 + 3.4. THE FUNDAMENTAL OPERATIONS 13 What is the negative of the number? 19. 7. 20. - 4. 21. - 13. 22. 0. Without writing the numbers in a line, (a) add the two numbers; (b) sub- tract the lower one from the upper one. 23. 45 24. 36 26. - 13 26. - 12 27. - 53 16 - 19 - 17 + 18 4- 17 28. - 13 -24 29. 17 30. - 15 31. - -4.3 7.6 32. -.87 1.35 Find the value of the expression in simple form. 33. + (- 5). 34. - (- 8). 35. - (+ 7). 36. + (- 6). Compute each sum. 37. - (- 3) + 7. 38. 4- (4- 4) - (- 8). 39. - 6 + (- 4). 40. - (- 4) 4- (- 9). 41. 16 - (- 6). 42. - 15 - (4- 12). 43. 12 - 7 - 5 4- 3. 44. - 16 - 3 + 5 - 7 4- 6. 46. - 10 4- 17 - 8 4- 14. 46. 43 - 25 - 6 4- 8 4- 12. 47. - 16 4- 14 4- 36 4- 8. 48. 3 - 16 4- 17 - 8 - 9. 49. - 3 4- (- 5) - (- 16) - (- 4). 60. 5 - 7 4- (- 3) - (4- 16) - 3.7. Find (a) the sum and (b) the difference of the two numbers. 51. 16 and 12. 62. 15 and - 3. 63. - 33 arid - 7. 64. - 14 and - 5. 66. 6 and 38. 66. 15 and 67. Compute the expression when the literal numbers have the given values. 67. 16 4- Bab] when a = 4 and b = 7. 68. 2a 3cd; when a = 2, c = 3, and d 5. 69. 26 7 4- 4oc; when a = 5, b = 7, and c = 3. 60. x 2yz 3; when x = 4, y = 2, and z 7. 61. 2a 4- 56 - 16.3; when a = 5 and 6 = - 6. Find (a) the sum of the numbers; (b) their difference; (c) their product; (d) the quotient of the first divided by the second. 62. - 60 and - 15. 63. and - 14. 64. - 12 and 4. 66. 6 and - 3. 66. - 52 and - 13. 67. and 23. 14 THE FUNDAMENTAL OPERATIONS 1 8. Real number scale On the horizontal line in Figure 1, we select a point 0, called the origin, and we decide to let this point represent the number 0. We select a unit of length for measuring distances on the line. Then, if P is a positive number, we let it be represented by the point on i i i i ixj i i i i i i i i i i i ix i i i 7 6 -543-2-1 1 2 3 4 56 Fig. 1 the line which is at P units of distance from to the right. The negative number P is represented by the point which is at P units of distance from to the left. The points at whole units of distance from represent the positive and negative integers. The other points on the line represent the numbers which are not integers. Thus, all real numbers are identified with points on the real number scale hi Figure 1. If M is any real number, it can be thought of as a measure of the directed distance from to M on the scale, where OM is considered positive when the direction from to M is to the right and negative when the direction from to M is to the left. 19. The less than and greater than relationships All real numbers can be represented in order from left to right on the scale in Figure 1. We then say that one number * M is less than a second number N, or that N is greater than M, in case M is to the left of N on the number scale. We use the inequality signs < and > to represent less than and greater than, respectively. This definition includes as a special case the similar notion used in arithmetic for positive numbers. The present definition applies to all real numbers, positive, negative, or zero. If a 7* 6,f then either a < b or a > b. ILLUSTRATION 1. In each of the following inequalities, we verify the result by placing the numbers on the scale in Figure 1. Thus, 7 < 3 because 7 is situated to the left of 3 in Figure 1. We read this inequality as " 7 is less than 3." 4<6; 0<8; - 4 < 0; - 6 < 5. To say that P > is equivalent to saying that P is positive, be- cause the numbers to the right of in Figure 1 are positive. To say that M < is equivalent to saying that M is a negative number. * Until otherwise indicated, the word number will refer to any real number. t We read 7* as "not equal." THE FUNDAMENTAL OPERATIONS 15 20. Numerical inequality We say that one number 6 is numerically less than a second number c in case the absolute value of bis less than the absolute value of c. To distinguish this relation from ordinary inequality, we sometimes place the word algebraically before greater than or less than when they are used in the ordinary sense. , ILLUSTRATION 1. 5 is numerically less than 9 because | 5 1 < 1 9 1. It is also true that 5 is algebraically less than 9, because 5 < 9. ILLUSTRATION 2. We see that 3 is numerically less than 7 because | - 3 | = 3 and | - 7 | = 7, and 3 < 7. On the other hand, - 7 < - 3. Thus, 3 is algebraically greater than 7 but numerically less than 7. In Illustration 2, we observe a special case of the fact that, if one negative number b is algebraically less than a second negative num- ber c, then b is numerically greater than c. EXERCISE 4 Construct a real number scale 10 inches long, and mark the locations of the positive and negative integers from 10 to + 10, inclusive. Then, read each inequality and verify it by marking the two numbers on your scale. 1. 5 < 9. 2. < 7. 3. - 3 <0. 4. - 3 < 8. 5. - 5 < 2. 6. - 7 < - 4. 7. 8 > - 9. 8. 5 > - 3. 9. - 3 > - 10. Mark the numbers in each probkm on your scale and decide which sign, < or > , should be placed between the numbers. 10. 7 and 9. 11. - 2 and 5. 12. and 8. 13. - 6 and 0. 14. - 3 and 3. 15. - 2 and - 7. 16. - 9 and 10. 17. 7 and 5. 18. 8 and - 3. 19. - 6 and - 3. 20. - 7 and - 9. 21. 8 and - 10. Read the inequalities and verify that they are correct. 22. |-7| <|8|and - 7 < 8. 23. | 4| < | 9| and 4 < 9. 24. | - 3 1 > 1 1 but - 3 < 0. 26. | - 8 | > 3 but - 8 < 3. 26. | -5| <|"-7|but -5> -7. 27. ] - 6 i < ] - 9 1 but - 6 > - 9. 76 THE FUNDAMENTAL OPERATIONS Stale which number (o) is algebraically greater than the other; (6) is nu- merically greater than the other. 28. - 8; 6. 29. - 5; - 3. 30. 7; 4. 31. 0; - 3. 32. 5; 0. 33. - 2; 7. 34. - 3; - 7. 36. 2; - 6. 36. - 8; - 3. 21 . Signs of grouping Parentheses, ( ), brackets, [ ], braces, { }, and the vinculum, , are symbols of grouping used to indicate terms whose sum is to be treated as a single number expression. Any general remark about parentheses which follows will be understood to apply as well to any other symbol of grouping. Parentheses are useful for enclosing and separating algebraic ex- pressions written side by side as an indication that they are to be multiplied. ILLUSTRATION 1. 3( 5) means 3 times 5, or -f 15. In reading algebraic expressions, the student may use the words "the quantity 1 ' as he comes to the left marker for any symbol of grouping enclosing more than one term. ILLUSTRATION 2. (3 5o)(2 + 6a) is read "the quantity 3 5a times the quantity 2 + 6a." If a = 4, (3 - 5o)(2 -f 60) - (3 - 20) (2 + 24) = (- 17) (26) = - 442. Parentheses can be employed to avoid ambiguity hi regard to the order of application of the fundamental operations. ILLUSTRATION 3. Doubt arises as to the meaning of (9 6 -f- 3). Does it mean (9 6) -s- 3, which equals 1, or does it mean 9 (6 -5- 3), which is (9 2) or 7? The two possible meanings were written without ambiguity. ILLUSTRATION 4. If a = 2, b = 3, and c = 5, then 5-3o& + 2oc = 5- [3(2)(- 3)] + [2(2) (5)] = 5 - (- 18) + 20 = 5 + 18 + 20 = 43. A factor multiplying a sum within parentheses should be used to multiply each term of the sum. ILLUSTRATION 5. - 3(2a) = (- 3) (2) (a) = - 6a. THE FUNDAMENTAL OPERATIONS 17 ILLUSTRATION 6. - 4(a - 26 + 7) - - 4a - 4(- 2fe) - 4(7) (1) - - 4a + 86 - 28. (2) The student should write (2) without writing the right-hand side of (1). ILLUSTRATION 7. - ( 5 + 3z y) = (' 1)(- 5 + 3x y) = 5 - Zx + y. The presence of a plus sign or a minus sign before any algebraic expression indicates that it is to be multiplied by + 1 or 1, re- spectively. Multiplication by + 1 would leave a sum unchanged, and multiplication by 1 would change the signs of all its terms. These remarks justify the following rules. I. To remove or to insert parentheses preceded by a plus sign, rewrite the included terms unchanged. II. To remove or to insert parentheses preceded by a minus sign, re- write the included terms with their signs changed. ILLUSTRATION 8. + (3r 7 -f %) = &c 7 -f - (- 2 + 5x - 7y) = (- 1)(- 2 + 5x - 7y) = 2 - 5x + 7y. ILLUSTRATION 9. In the sum on the left-hand side below, we enclose all terms after the first in parentheses preceded by a minus sign. 5a - 2c + 3d - 8 = 5o - (2c - M + 8). The signs of the terms enclosed were changed in order that, if the parentheses were removed, the original terms would be obtained. In performing an operation which removes parentheses, if only explicit numbers are involved, it is best to compute each sum within parentheses before they are removed. ILLUSTRATION 10. - 3(2 - 5 + 7) = - 3(4) = - 12. EXERCISE 5 f Compute the expression. 1. (- 2)(- 5). 2. 7(- 3). 3. (- 6)(4). 4. (0)(- 3). 5. - (- 5). 6. + (- 4). 7. 2(- 5 + 9). 8. - 3(5 - 14). 9. - 4(- 5 - 6). 10. (- 8 + 6)(- 5 -f 12). 11. (7 - 3) (16 - 5 - 2). J8 THE FUNDAMENTAL OPERATIONS 12. Compute (36 4oc 7) if 6 2, a 3, and c = 5. 13. Compute (- 6 + 3oc - 6c) if a - 3, b - 4, and c = - 2. 14. Compute (3 - 5a)(- 2 + 60) when a = 4. Compute (c 2d)(4d 3c) w#A <Ae pfoen vaJwes for c and d. 16. c = 4; d = 3. 16. c = - 2; d = 5. 17. c = - 2; d = - 3. Rewrite, by performing any indicated multiplication and removing paren- theses. Evaluate, if no letters are present. 18. - (17 - 8 - 3). 19. - (2 - 5 + 16). 20. + (3 - 6 + 15). 21. - (2a - 56 + c). 22. + (- 3 + 7a - b). 23. + (31 - 5a + y). 24. - (- 3 - 5z + 4y). 26. - (80 - 36 - c). 26. - (2x - 5y - 9). 27. 3(5a). 28. - 2(4c). 29. - 3(- 5o). 30. 5(- 3x). 31. 4(2o - 3). 32. - 2(x - y + 8). 33. - 5(3 - a - 6c). 34. 2(- 5 + 7a - 46c). 36. 3(6 - 4a + 56). 36. - 4(- 5 + 66). Rewrite, enclosing the three terms at the right in parentheses preceded by a minus sign. 37. _ 5 + 7 a _ 46. 38. - 60 -f 46 - c. 39. 6 - 3x - 4y. 40. 2a - 3 + 56 - c. 41. 16 - 4a - 6 -f 3c. 42. - 13 + 5 - c -f 06. 43. 2ac -f- 3 - 5a + 4c. 44. Compute (5 - 17) -5- 3; 5 - (17 -4- 3). 22. Similar terms Two terms such as 5a6c-and 7o6c, where the literal parts are the same, are called similar terms or like terms. In a term such as 5a6c, the explicit number which is a factor is called the numerical coefficient of the term, or, for short, the coefficient. ILLUSTRATION 1. The numerical coefficient of 5o6c is 5; of 7o6c is 7; of abc is 1. We never write the coefficient when it is 1. Thus, we would never write Io6c for abc. A sum of similar terms can be collected (added) into a single term, by use of the distributive property of multiplication. To collect a sum of similar terms, add their numerical coefficients and multiply the result by the common literal part. THE FUNDAMENTAL OPERATIONS 19 ILLUSTRATION 2. 5ab + lab - ab(5 -f- 7) 12o6. If we think of "06" as a concrete object, the result here is obvious: 5 of the objects plus 7 of them equals 12 of them. ILLUSTRATION 3. 9o6 + 4o6 = ab( 9 + 4) = o6( 5) = 606. In finding a sum of algebraic expressions, we collect similar terms. A direction to collect terms means to collect similar terms. EXAMPLE 1. Find the sum and the difference of 3a 5y - 8 and 3y 2a 6. FIRST SOLUTION. 1. The sum is (3a 5y 8) + (3y 2o 6) = 3a - 2a - 5y + 3y - 8 - 6 = a - 2y - 14. The student should learn to omit the intermediate details hi such a solution. 2. The difference is (3a 5y 8) (3y 2a 6) = 3a - 5y - 8 - 3y -f 2o -f 6 = 3a + 2a - 5y - 3y - 8 -f- 6 = 60 - Sy - 2. SECOND SOLUTION. To find the sum, arrange the given expressions with like terms in separate columns, and add. To find the difference, change the signs in (3y 2a 6), or take its negative, and add similarly. A 1 1 i wv vi/ <-* jiii *^O """ *^2/ *"" o ^Lrta: < . ^oa: < i ^<z ot/ + o r 3a _ 5y .. 8 r \ - 2a + 3y- 6 AflW< \ = a 2y 14. Difference = 5a Sy 2. In finding the difference, we could have changed the signs of ( 2o + 3y 6) mentally without rewriting at the right. Thus, from the columns for the sum: subtracting 2a from 3a gives 5a; etc. EXAMPLE 2. Perform indicated multiplications, remove parentheses, and collect terms: 3(4a - 3xy - 56 - 2) - 2(- 5 + 3o -f- 56 - fay). (1) SOLUTION. The sum equals 12o - 9xy - 156 - 6 + 10 - 60 - 106 + 12*y = 6a -f 3xy - 256 + 4. (2) Comment. We can obtain a relatively certain check on the work by sub- stituting values for the letters in (1) and (2). The results should be equal. In checking by this method, avoid substituting 1 for any letter. CHECK. Substitute a = 2, b = 4, x = 3, and y = 4: In (1): 3(8 - 36 - 20 - 2) - 2(- 5 + 6 + 20 - 72) = - 48. In (2): 12 -f 36 - 100 + 4 = 52 - 100 - - 48, This checks. 20 THE FUNDAMENTAL OPERATIONS 23. Nests of grouping signs If a symbol of grouping encloses one or more other symbols of grouping, remove them by removing the innermost symbol first, and so on until the last one is removed. Usually, we enclose parentheses within brackets, and brackets within braces. . ILLUSTRATION 1. [3t/ (2x 5 4- 2)] = - [3^ - 2x -f 5 - z] = - 3y + 2x - 5 + z. ILLUSTRATION 2. - [16 - (2 - 7) - (- 3 + 5)] = - [16 - (- 5) - (2)] = - [16 + 5 - 2] = - 19. EXERCISE 6 Collect similar terms. 1. 3o + 80. 2. - 56 + 76. 3. - 13z - 5x. 4. 19o6 -f 5o6. 6. - 2cd + 8cd. 6. 5xy 5xy. 7. 3x - a + 2z - 5a. 8. - 3a + 46 -f 7a - 96. 9. - 2o - 3c + 5a - 8c. 10. 6z - 4y + 12z + 3y. 11. lie - 5cd + 8c - 13crf. 12. Qh - 5kw - llh + 7kw. (a) Add the two expressions; (6) subtract the lower one from the upper one. After each operation, check by substituting convenient values for the letters in the given expressions and the final result. 13. - 4a + 76 - 3 14. 3a - 56 + 5 2a - 96 + 7 - 9a - 46 - 7 16. 3x - 5a6 - c 16. - 2a - 46c - 2d 6a6 - 3c 5a - 66c -f d 17. 3m - 5k - h 18. - 7r -f 3* - 6 - 6m + 4fc - 5h 8r - 5s -f 9 Find J/ie sum of the three expressions separated by semicolons. First arrange the expressions with like terms in separate columns. 19. 3x - 2y - 5; - fix + 7y - 8; - 12z - 8y -h 13. 20. - 3 + o - 6c; 26c - 3a + 5; 4a - 56c - 9. 21. 2oc - 5xz/ - 86; 4cy 3ac -h 56; 76 - 6xy - 4ac. 22. 7ad - 56 - 4a; 6a - 3ad + 56; 3o -h 5arf - 96. THE FUNDAMENTAL OPERATIONS 21 Remove parentheses and collect similar terms. 23. 2(a - 36) - 5(6 - 2a) + 3(- a - 36). 24. 5(x - 3y + 5) - 2(* + 2? - 4) - 3(- 4z - y + 6). 26. 6(2a - h - k) - 3 (a - 4A + 5Jfc) + 2(4A - 3a - &). By w$e of parentheses, write an expression for the quantity described; then remove parentheses and collect terms. 26. Subtract 3a - 26 and 2a + 56 from - a 56. HINT. The result equals (- a - 56) - (3a - 26) - (2a + 56). 27. Subtract 2a - 3y - 5 from 5a + 7y - 8. 28. Subtract 3a 7h and 5A 6a from the sum of 2o 3h and 5a -f 29. Multiply 3o - 5y - 3 by 2, and - 5o -f 7y - 5 by - 3, and then add the results. Remove all signs of grouping and collect terms. 30. - [4o - (2a + 3) J 31. - [2* + (3 - 40 J 32. - (a - 2) - [2o - (a - 3)]. 33. a + [6 + (a - 6)]. 34. 9 - - (6 - 2)]. 36. 2r + [> - ( + 4r)J 36. - [> - (y + 3)]. 37. 2i/ - 5 + [3 - 2(y - 2) J 38. 3o - [3a - 4(5 - a)]. 39. 3 - [2x - 2[> - (2x - 5)]). 40. - {a - [a - (2a - 7)]}. 41. - {26 - [6 - (36 - 4)]}. 42. - 3(2z - 3y) + 2[2 - (5x - y)] - (* + 7y - 8). 43. 2(3/i - A; - 5) - 3[A - 2(* - 3)] - 4(A - 2k -h 2). CHAPTER 2 INTRODUCTION TO FRACTIONS AND EXPONENTS 24. Fractions in lowest terms The basic properties of fractions as met in arithmetic are primarily consequences of the definition of division. These properties extend immediately to fractions as met in algebra, where the only essential new feature is the introdmction of negative numbers hi the fractions. We shall recall and use the properties of fractions without rehearsing the sequence of definitions and proofs which, logically, would be necessary hi building a foundation for the use of fractions hi algebra. FUNDAMENTAL PRINCIPLE. The value of a fraction is not altered if both numerator and denominator are multiplied, or divided, by the same number, not zero. ILLUSTRATION 1. 5 5X3 15 a ac 7 36 7X3 21 36 -M2 3 b~ be 84, 84-5-12 7 ILLUSTRATION 2. On multiplying numerator and denominator by 1 below, we obtain - 3 (- 1)(- 3) 3 -4 (- 1)(- 4) 4 We say that a fraction is in lowest terms if its numerator and de- nominator have no common factor except + 1 and 1. To reduce a fraction to lowest terms, divide numerator and denomina- tor by all their common factors. ILLUSTRATION 3. = = =- (Divide out ac) 7acy 7y INTRODUCTION TO FRACTIONS AND EXPONENTS 23 210 0X3X7X2 7X2 14 ILLUSTRATION 4. 135 3X3X3X3 3X3 9 In the preceding line we divided out the factor 5 and one 3 from numerator and denominator. 25. Change in sign for a fraction If the numerator or denominator of a fraction is multiplied by 1, the sign before the fraction must be changed. These actions are equivalent to multiplying the fraction by two factors 1, whose product is H- 1. This keeps the value of the fraction unaltered. T - 0-3 (- l)(o - 3) 3-Q ILLUSTRATION 1. = = - ~ = = 26. Multiplication and division of fractions To multiply one fraction by another, multiply the numerators for a new numerator and multiply the denominators for a new denominator. 3 ^ 6 18 a c ac ILLUSTRATION 1. = X = = ^=- i* j ** cr 5 7 oo o a oa To divide one fraction by another, invert the divisor and multiply the dividend by this inverted divisor. T o 4 . 3 4 7 28 ILLUSTRATION 2. F^JT^K'Q 7c' .) i O o ID a a _._ c _b _ a d _ ad b ' d c be be d It is frequently useful to recall that any number can be expressed as a fraction whose denominator is 1. By use of this fact we verify the following results. To multiply a fraction by a number, multiply the numerator by the number. To divide a fraction by a number, multiply the denominator by the number. 7 _/5\ 75 35 7. 571 7 ILLUSTRATION^ 7( g ) = rg =* T 5 "" 4 = 4 = 5*4 " 20* 1 24 INTRODUCTION TO FRACTIONS AND EXPONENTS T A , . 6 5 . 6 5 7 35 ILLUSTRATION 4. 5 = = T ^ = 7 = -- 7 1 / 1 o o /a\ c a ac ILLUSTRATION 5. c (r) = T ' I = T" \o/ 16 o f? _i. ?^_ ? I b ~ b ' I 6 c be EXERCISE 7 Express the result as a fraction in lowest terms without a minus sign in numerator or denominator. 1 18 10 30' 00 J.K 2 3 . 72 21 A 66 4 '7f 6 '35' fi 21 OfJ K7 7 . 8 9 63 - 10 42 - 6 ' 14* 120 38 9 ' 81 10 ' 63 11 ~ 5 >fl O Id ~ ^. 15 78 - 11. 3 . l2 ' - 7 13 * - 2 14 ' 35 15 ' - 26 16 M n^ 10 3% 19 27a . 16 ' 16c* OQ / >* / i / lo. ~ 6?/ 18 ' 6a6 he *>- 21 4a - 2 ^^ 5ad! oo . . ^ 1 * Q^ da 22> 66 Z3% - 3d u. 3 4 VL*L ^ 5 7 *{! 9R 15 21 26. y- T - 27 5.?. 4 '' 2 rf 28. H- 2i O 29 n -^- 29. y . 5 30 15 J- 5 - 31. ^^l 4 38 57 3 2 .5(|). 33. 21 (?). 34. g). 36. eg). 36. 5(4). 37. (6). 38. 1 + 2. 39. | + 6. 40. y + 15. 41. y - 6. 42. || -*- 2a. 43. y * d. .,325 .. 3o 56 2c ... x 2c 3d ' l_ 8 12a 48. - 49. 60. - 5c 15 INTRODUCTION TO FRACTIONS AND EXPONENTS 25 15 14 3A 4w 51. 1- 62 - fo' M k 53 -T ^ yet- 54. TT-- 2w 65. 6 -*- 1- 66. 5 * ~ 67. 5 + 4' 68. 5d -T- 3d 2 7 1U c 69 4- * 61. - < c AO 5w? 62. 7= 15w> 7 9 d 8 -Ml -H)H)H)H)- -(-SX-SX-8- 27. Positive integral exponents We write a 2 to abbreviate a -a, and a 8 for a -a' a. We call a 2 the square of a and a 3 the cube of a. ILLUSTRATION 1. 5 2 = 5-5 = 25. 5 3 = 5* 5' 5 = 125. If m is any positive integer,* we define a m by the equation 9 a m = a-a-a- -a. (m factors a) (1) We call a m the mth power of the base a and call m the exponent of this power. The exponent tells how many times the base is used as a factor. ILLUSTRATION 2. 3 4 = 3-3-3-3 = 81. (- 4) 2 = (- 4)(- 4)'= + 16. (_ 4)3 = (- 4)(- 4)(- 4) = - 64. (?)' = . f.f = |. We notice that, by the laws of signs, an even power of a negative number is positive and an odd power is negative. By definition, b 1 = b. Hence, when the exponent is 1 we usually omit it. Thus, 5 means 5 1 and y means y l . 28. Index lows The following laws for the use of exponents are called index laws. At present we will illustrate them, and verify their truth in special cases. The proofs of the laws will be given later. * Until otherwise specified, any literal number used in an exponent will represent a positive integer. 26 INTRODUCTION TO FRACTIONS AND EXPONENTS I. In multiplying two powers of the same base, add the exponents a m <i n = c m+n . ILLUSTRATION 1. a 3 o 2 = a 3+2 = a 6 , because a 3 o 2 = (aa-o)(a-o) = o0'O-a-a = a 5 . szV = x^x 4 = z 1+2+4 = a: 7 . II. 7n obtaining a power of a power, multiply the exponents: (<z m ) n = fl 17 " 1 . ILLUSTBATION 2. (a 3 ) 2 = a (3X2) = a 6 , because (a*) 2 = a 3 -a 3 = (a-a-a)(a-a-a) = a-a-a-a-a-a = a 6 . III. To obtain a power of a product, raise each factor of the produc to the specified power and multiply: (abc) n = a n 6 n c n . ILLUSTRATION 3. (o6) 3 = o 3 ^ 3 , because (oft) 3 = ab'ab'ab = (a-0'o)(6-6-6) IV. To obtain a power of a fraction, raise the numerator and denom inator to the specified power and divide: fa\ n _ a? W " &"' T A fa\* X 4 u /^\ 4 X X X X X 4 ILLUSTRATION 4. (-) =-7 because (-1 = ------- ~i* \yl y* \yi y y y y y 4 T c / 2 \ 4 24 16 /T ILLUSTRATION 5. (^) = ^ = o7* (Law IV (-!)'-[<-(!)]'= -"(1)'- -I- -fi- If we recall that an odd power of a negative number is negative, we ma; abbreviate the preceding line by omitting the first two steps. ILLUSTRATION 6. (2o 2 6) 3 = 2 3 (a 2 ) 3 6 3 = 8a 6 6 3 . ILLUSTRATION 7. <IAWS \ W / (A') 4 A" A 12 ,, INTRODUCTION TO FRACTIONS AND EXPONENTS 27 EXERCISE 8 Compute by the definition of an exponent. 1. 2 4 . 2. 5 3 . 3. 10 2 . 4. 10 s . 5. 10 4 . 6. 10 5 . 7. (- I) 2 . 8. (- I) 3 . 9. (- I) 4 . 10. (- 5)*. 11. (- 3) 3 . 12. (- 2) 6 . 13. (- 5) 3 . 14. 6 2 . 15. (- 3)*. 16. 10*. 17. (- 10) 3 . 18. () 4 . 19. (f) 3 . 20. ($). 21. (- |) 3 . 22. to) 4 . 23. - 2 4 . 24. - 3 3 . 25. - 6 s . 26. 2(3 4 ). 27. 3(- 5 2 ). 28. 6(10*). 29. 5(4 3 ). 30. - 2(- 5). 31. For what values of the positive integer n will ( l) n be positive and when will it be negative? 32. Compute 3a 2 6 if a = - 2 and b = 4. 33. Compute 5xy* if x = 3 and y 2. 34. Compute - 4W if h = - 3 and k = 2. 35. If x is positive or negative, what is the nature of x 2 , positive or nega- tive? If x is negative, what is the nature of x 3 and of x 4 ? Perform the indicated operation by use of the laws of exponents. 36. a b a 4 . 37. 3V. 38. xx 3 . 39. y*y 7 . 40. 10H0 6 . 41. xx*x*. 4$ T yy*y 7 . 43. 6*6*6. 44. 3 3 3 B . 45. a h a k . AA /.i/2fi vv. XX. 47. (a*) 4 . 48. (2 2 ) 3 . 49. (xyY. 50. (cd) 3 . 61. (3x) 2 . 62. (c 3 ) 6 . 63. (A 3 ) 2 . 64. (ox 2 ) 3 . 56. (36) 4 . /c\ 3 ' w 67 - 4 - 58. (?) 3 - \a/ . 60. (?) 4 . 61. (f). 62. (- f) 4 . 83. (- J) 64. I-}*- /2/i\ 4 66. I ; ) J% 4 / O*C \ 66. ( 67. (- S 68. (xV) 4 - 69. (3C 2 ) 3 . 70. (2o 2 6 4 ) 4 . 71. (3xu;) 4 . 72. (- 2A 2 ) 3 . 73. (- 3x 2 ) 4 . 74. (- 5xt/ 2 ) 3 . 75. (- (O/7\4 /^ 2 rr\' / 3 \* / S)- 77 -(^)- 78 -(-^)- ro -(- 80. Find the value of x 3 - 3x 2 -h 4x - 7 when (a) x = 3; (6) x * - 2. 28 INTRODUCTION TO FRACTIONS AND EXPONENTS 29. Integral rational terms At present, in any sum, the typical term will be either an explicit number or the product of an explicit number and powers of literal numbers, where the exponents are positive integers. The explicit number is called the numerical coefficient, or for short the coefficient of the term. A term of this variety, or a sum of such terms, is said to* be integral and rational * in the literal numbers. An algebraic sum is called a monomial f if there is just one term, a binomial if there are just two terms, and a trinomial if there are just three terms. Any sum with more than one term also is called a polynomial. ILLUSTRATION 1. 3z + 7ab is a binomial. ILLUSTRATION 2. In the trinomial 8 + x 3a6 2 , the terms are 8, x, and 3a& 2 , whose numerical coefficients are 8, 1, and 3, respectively. ILLUSTRATION 3. 5z 2 x 7 is an integral rational polynomial in x. To multiply two integral rational terms, multiply their numerical coefficients and simplify the product of the literal parts by use of the law of exponents for multiplication. ILLUSTRATION 4. 6a6 2 (3a 2 6 4 ) = (6)(3)(aa 2 6 2 6 4 ) = 18a 3 6 6 . To multiply a polynomial by a single term, multiply each term of the polynomial by the single term and form the sum of the results. ILLUSTRATION 5. 5(3z 2 - 2x - 5) = 15z 2 - lOz - 25. EXERCISE 9 Perform the indicated multiplications and simplify the results by use of the law of exponents for multiplication. 2. 32/(22/ 6 ). 3. ab(3a). 4. 5. - 5(32 7 ). 6. - 2a 2 (3a 9 ). 7. cd^cP). 8. x(- * The word integral refers to the fact that the exponents are -integers. The force of the word rational will be pointed out later. t This name need not be used very often because the simple word term is just as desirable. INTRODUCTION TO FRACTIONS AND EXPONENTS 29 9. - 4fl(4). 10. 3c(- c 3 ). 11. - 2z(+ xy*). 12. - **(- 2z 2 ). 13. - ax*(- 2o 2 x). 14. 5oy(- 2xy 2 ). 15. 2a6(- 4a 8 6 2 ). 16. - 8m 3 (- 2m). 17. - 4r z h(- 6rA 4 ). 18. - Gc 2 ^- 3cd). 19. 3(4z - y). 20. - S(5x - 2z 2 ). 21. - 5(3 - 4a). 22. 4(2a - 56). 23. - 3(- 5x - 4y). 24. x(2x 2 - 3x). 25. 2x(- 3z - 5z 3 ). 26. o(a 2 - a 3 ). 27. 2* 2 (3 - z 4 ). 28. - 4a(l - 2a6). 29. - 3fc(fc - M). 30. ah\a - 31. - 5w(2 - 3u> 4- 4i^). 32. - 33. 3a6 2 (2a 2 6 3 )(a6 4 ). 34. - 35. - 4m 3 n(- 3m)(2mn 3 ). 36. 4?/0 2 ( 37. 2aj3a- 38. 2a^- 3a 2 6 3 . 39. - Multiply the polynomial by the term which is beneath it. 40. 5z 3 - 3x 2 - 2x - 5 41. 6 - 3x 42. a 2 - 2a6 - 6 2 43. 3 - Gab - 5a 2 6 2 - a 3 6 3 - 2ab - Sab 44. 15(2z 2 - J + f). 46. 8(| - iy + Jif). 46. 12(f - f* + 2 2 )- 47. 16(Ja 2 - fa - J). 48. (12 - 6a + 24o 2 ). 49. J(- 16 -f 8x - 60. 30. Multiplication of polynomials To form the product of two polynomials, multiply one of them by each term of the other and collect similar terms. ILLUSTRATION 1. (2x 3y)(x 2 xy) = 2x(x 2 xy) 3y(x 2 xy) = 2s 3 - ILLUSTRATION 2. (x + 5) 2 = (x -f 5) (a: + 5) = x(x + 5) + 5(* + 5) = x 2 + 5x -f 5x + 25 = a? -f 10x + 25. Before multiplying, if many terms are involved, arrange the poly- nomials in ascending (or descending) powers of one letter. 30 INTRODUCTION TO FRACTIONS AND EXPONENTS ILLUSTRATION 3. To multiply (x 2 + 3x* - x - 2) (2s + 3) : 3x + x 2 - x-2 (Multiply) _ 2x + 3 6s 4 + 2x - 2x 2 - 4x (Multiplying by 2x) Ox 3 + 3x 2 - 3x - 6 (Multiplying by 3) (Add) 6x* + llx 3 + a* - 7x - 6 = product. EXAMPLE 1. Multiply SOLUTION. x 2 + - 2x*y - -f -f x* y 2 + 2xy by y 2 + 2 CHECK. Place x = 2 and y = 3. = 4 - 12 - 9 = - 17. = 4 + 12 + 9 = 25. Hence, the product should equal 25- (- 17) = -425 when x = 2 and y = 3. = 16 - 144 - 216 - 81 = - 425. Comment. The preceding numerical check does not absolutely verify the result, but almost any error would cause the check to fail. In checking, if 1 were substituted for a letter, we could not detect an error in any of its exponents because every power of 1 is 1. Hence, in numerical checks, avoid substituting 1 for any letter. EXERCISE 10 Multiply and collect similar terms. Check by substituting values for the letters, when directed by the instructor. 1. (x - 3)(x + 4). 3. (x - 5)(2x + 7). 5. (5a - 7)(4a - 3). 7. (2h - ZK)(2h + 3k). 9. (a + 36)(2a - 56). 11. (3r - 5s) (3r + 5). 13. (06 - 3)(2a6 -I- 5). 15. (cd - x)(cd +'x). 17. (a -f 3)*. 18. (2x - t/) 2 . 21. (3a - 26) 2 . 22. (2x + 4/0 2 . 26. (y - 2)(y -f- 5). 2. (x + 9)(x + 10). 4. (x - 3) (7 - x). 6. (3y - 2) (2 - By). 8. (3w - 4r)(2w + 3r). 10. (x + y)(x - y). 12. (a 2 - 2)(2a 2 - 5). 14. (3 - 5x')(2 + 3x 2 ). 16. (ay - 2z)(2ay + 3). 19. (h - 4fc) 2 . 20. (4 -f 23. (ax - 6) 2 . 24. (5 - x 2 y) 2 . 26. (c 2 - 2a 2 )(c 2 + 3a 2 ). INTRODUCTION TO FRACTIONS AND EXPONENTS 3? 27. (a s - 6 2 )(3a 3 + 46 2 ). 29. (y - 2)(i/ 2 + 2y + 4). 31. (x - 3)(x 2 + 2x - 5). 33. (2 + x)(3 - 4x - x 2 ). 35. (1 - 2x)(2x - x 2 + 5). 37. (5 - 6) (4 - 26 - 6 2 ). 39. 40. 41. 42. 43. (5 -h 3x 2 - 44. (x 2 -+ 3x + 45. (5z - 2y + 3) 2 . 47. (a + 6)(a 2 - db -f fe 2 ). 49. (a; + 3)(* - l)(2a? - 5). 61. (2a - 3)(3a + 5)(a - 2). 28. (x 2 - 30. (4a 2 - 2a + l)(2a + 1). 32. (a - 4)(3a 2 - 2a + 1). 34. (c + 2)(2c - 5 - 3c 2 ). 36. (h - 4)(2A 2 - 1 + A). 38. (2 - y)(y + 5 - - 2x 2 + 5x - 7)(2x - 1). 3). 3). 3 + x). a: 2 + 3 - 46. (x + 48. (x - 2y)(x* + 2xy + 50. (3x - y)(2x + y)(2x - y). 62. (x - a*)(x 2 + a*x + a 2 *). 63. (x n - 64. (x 4 + T/ 4 - x?y + X 2 y 2 )(x 2 - 2xi/ - 31 . Exponents in division By the definition of a power, a 6 a'CL'd'd' = a 3 a-a-a a-a-a a-a T" 1 a a-a-a-a-a a-a = a 2 ; (Divide out a-a-a) = -= (Divide out a a a) The preceding results illustrate the following index law. In dividing one power of a specified base by another power of the 6oae, subtract the exponents: (ifm>n); = (ifm<n). (1) 32 INTRODUCTION TO FRACTIONS AND EXPONENTS In addition to formulas 1, we note that a n = 1- (2) a 5 x* I 1 h? ILLUSTRATION 1. . = 1. -^ = -^-^ -: 77 a 6 x 10 x ia ~ 4 x 6 h* In dividing one integral rational term by another, we use formulas 1 and 2 to reduce the fraction to lowest terms. ILLUSTRATION 2. In the following simplification, we can think of dividing numerator and denominator by 5ox B . Or, we can think of dividing numerator and denominator by 5 and, also, of applying formulas 1 to the powers of a and of x, separately: - 15ax 6 3 a 3 x 6 Sa 3 " 1 3a 2 lOax 9 2 a x 9 2x 9 ~ 5 2x 4 16a 3 x 2 16 a 3 x 2 8a 2 ILLUSTRATIONS. ILLUSTRATION 4. / 1\ a w 6 = { ^) -r -= = \ 2/ a 4 w> 2 2a 3 where the intermediate details should be performed mentally. - 6ay 6 a 3 y* 2 ILLUSTRATION 5. f = ^= - ~ = 15 a d 32. Division by a single term To divide a polynomial by a single term, divide each term of the polynomial by the single term and combine the results. 6x 2 - 9x 3x 4 . 6x 2 9x ILLUSTRATION 1. = x 8 - 2x -f 3. EXAMPLE 1. Perform the division: (4o 2 6 4 - 8a 2 6 - 26 2 ) ^ (- 2O6 8 ). SOLUTION. Write the quotient as a fraction: 4a 2 6 4 8a 2 6 - 2a6 3 - 206 s INTRODUCTION TO FRACTIONS AND EXPONENTS 33 EXERCISE 11 Perform the division, expressing the result as a fraction or sum of fractions in lowest terms by use of the law of exponents for division. Express each final fraction without a minus sign in numerator or denominator. y "a 2 x 8 h 9 x 2 6 - ;r 7 - 5r 8 a o -<t Sw 3 .. 15a6 J.X* J.4. 4t/ 5a 91 /t2 Q/r^ IK ^rltt 1fi OX 96 15 ' 3a 16 ' 12X 3 ' 1ft . . 9A /i*4/)/2 nrfi oq * 4/ . Qd 276C 3 /y3 J)2 ^ ft/ ^* " c ^* flt xy 6 a 6 o ^^.^ jbOo 6 ^%^^ "" ^ A O>7 . OQ 3x 5a 7 ^21.5 ^^ 26. / on 36m *'*' To 18mn 32. ~ 9A3t * 35. (8x 3 ?/) -* (- 24X 3 ?/). 36. (- 49rV) * (- Trs 4 ). 37. (- 80) + (24a6 2 ). 38. (4ac) -s- (72a 3 c 2 ). 39. (48xV) ^- (- 8x2/ 2 ). 40- (- 36a6 2 ) -i- (- 4a6). . 6a H- 206 . 5/i - 25A; . 4a + 166 41. - -- 42. 5 4 7x 2 - 5x ._ 3a 3 - 2a 4 -- 45. - x a 2 - 15o 4 _ 3a2 60. (8a 3 - 6a 2 - 4a) + (~ 2). 61. (6x - 3x 2 - 9) 4- (- 3). 62. (x 4 - 3x 3 - 5x 2 ) H- x 2 . 63. (y* - 5y - y 3 ) -5- (- y). 54. (_ 36 + 126 - 96 2 ) ^ 3. 66. (x 4 - 3x 2 + 5x) - 66. (32a 2 6 4 + 48aV) ^- (16a 2 ). 67. (21a 2 6 2 - 34 INTRODUCTION TO FRACTIONS AND EXPONENTS _ - 02? - Sax 4- 5o RA 7x* - 4x* - 3x + 2 DO. ox 2x - 7a6 2 c 3 + 6c* c . 61 * M - 276" co a&d 3 - a 2 6d - a 3 62 ' - -- 63 ' 33. Fundamental equation of division In the following long division, we use the customary terminology of arithmetic: 15 (Quotient) (Divisor) 17 259 (Dividend) - 89 85 4 (Remainder) 259 = (17 X 15) + 4; = 15^. (2) In Section 11, when we met the notation a -5- 6, we called its complete value the quotient of a divided by 6. In (1), the complete value of the quotient is 15^. Hence, if there were danger of lack of clearness, in (2) we would call 15 the partial quotient. Frequently, the word quotient refers to a partial quotient. When appropriate, we shall use the qualifying word partial to prevent ambiguity. After any step in a division process similar to that met in (1), the remainder and partial quotient satisfy the equation dividend ,. . , remainder /ON . -7T-7 - = quotient H 37-7 - (3) divisor H divisor v ' Or, dividend = (quotient) (divisor) + remainder. (4) Equation 4 is frequently called the fundamental equation of division. The first equation in (2) is an illustration of (4). 34. Division by a polynomial The long division process for polynomials is similar to long division in arithmetic. We say that the division is exact if the final remainder is zero. INTRODUCTION TO FRACTIONS AND EXPONENTS 35 EXAMPLE 1. Divide: (x 2 + 3x - 40) -5- (x - 5). SOLUTION, (x 2 -f- x) x; this is the first term of the quotient. Then, x(x 5) = x 2 5x; we subtract this from the dividend, obtaining 8x 40. 2. (8x -T- x) = 8; this is the second term of the quotient. Then, S(x 5) = 8x 40. We subtract this, obtaining zero. x + 8 (Quotient) (Divisor) x - 5 | a 2 + 3x - 40 (Dividend) (Subtract) x* - 5x 8x -40 (Subtract) Sx - 40 (Remainder) CHECK. Since the division is exact, we obtain a check by verifying that (quotient) (divisor) = dividend. We find (x -f 8)(x - 5) = x* + 3x - 40, which checks. EXAMPLE 2. Divide: (4x 3 - 9x - 8x 2 + 7) -r (2x - 3). SOLUTION. 1. We first arrange the dividend in descending powers of a?, obtaining . 4x 3 - &c 2 - 9z + 7. 2. Since (4x 8 -s- 2x) = 2z 2 , this is the first term of the quotient; etc., as follows. 2x 2 - x 6 (Quotient) _ (Divisor) 2x - 3 j 4r* - Sx 2 - 9x -f 7 (Dividend) 2x*(2x - 3) -> (Subtract) 4s 3 - 6x 2 - 2x 2 ) + 2x] = - x. - 2x 2 - 9x x(2x - 3) - (Subtract) - 2x 2 + 3x [(- 12x) -r 2x] = - 6. - 12x + 7 - 6(2x - 3) - (Subtract) - 12x + 18 11 (Remainder) Conclusion. From equation 3, Section 33. 4x* - 8x 2 - 9x + 7 _ _ _ _ 11 2x-3 "^ X 2x-3 CHECK. Substitute x = 3 in the dividend, divisor, and quotient: dividend = 16; divisor = 3; quotient = 9; remainder = 11. Refer to equation 4, Section 33: does 16 = 3(9) - 11 - 27 - 11 - 16? Since this equality is satisfied, we conclude .that the solution is correct. 36 INTRODUCTION TO FRACTIONS AND EXPONENTS SUMMABT. To divide one polynomial by another: 1. Arrange each of them in either ascending or descending powers of some common letter. 2. Divide the first term of the dividend by the first term of the divisor and write the result as the first term of the quotient. 3. Multiply the whole divisor by the first term of the quotient and subtract the product from the dividend. 4. Consider the remainder obtained in Step 3 as a new dividend and repeat Steps 2 and 3; etc. Note 1. The numerical check in Example 2 does not constitute an ab- solute verification of the solution. To verify the result, we should (without substituting a special value for x) multiply the divisor by the quotient, add the remainder, and notice if we thus obtain the dividend. EXERCISE 12 Divide and summarize as in the solution of Example 2, Section 34. // the division is exact, check by multiplying the divisor by the quotient. If the division is not exact, check by substituting convenient values for the literal numbers. 1. (x* 4- 7z 4- 12) ^ (x + 3). 2. (</ 2 4- 15y + 36) + (y + 12). 3. (c 2 - lOc 4- 21) -h (c - 7). 4. (d 2 - 12d + 35) -J- (d - 5). 6. (s 2 4- 6s - 27) ^ (s 4- 9). 6. (2z 2 - 13x + 15) + (x - 5). 7. (y 2 4- Qy - 40) 4- (y + 10). 8. (54 + 3z - x 2 ) -f- (6 + x). 9. (4c 2 - 15) -h (2c + 3). 10. (6a 2 - ab - 6 2 ) ^ (3a + 6). 11. (x 4 + 3x 2 - 4) ^ (x 2 - 1). 12. (h* + 3h* - 10) -h (ft 2 -f 5). 13. (2z 2 -f 7z + 8) -i- (z -f 2). 14. (3a 2 - 7) + (a - 5). 15. (2a 2 - ab - 66 2 ) -*- (2o + 6). 16. (6z 2 + zy - 2t/ 2 ) ^ (3z + 2y 17. (4z 6 + Sz 3 - 6) -i- (4z 3 - 3). 18. (2a - a 8 - 15) -!- (2a 3 + 5). 19. (5x + 3x 2 - 3) -f- (x - 2). 20. (a + 6a 2 + 3) -f- (2a - 1). 21. (x 3 + 3x 2 - 2x - 6) -* (x 4- 3). 22. (60 - 2a 2 4- 3a 3 - 26) -* (a - 2). 23. (4x 3 - 9* - 8x 2 4- 7) + (2x - 3). 24. (36y - 9 - 19y 2 - 15^) -5- (3y 2 4- 5y - 4). INTRODUCTION TO FRACTIONS AND EXPONENTS 37 26. (Sy* - 1% 2 - 6 + lly) * (W - 3y + 2). 26. (2^ - 5y 2 - 12 + llr/) * (2y - 3). 27. (x 4 - 4z 3 + 3z 2 - 4x + 15) -J- (z - 3). 28. (&c 2 - 3 - 5z 3 ) -^ (7* - 2 + 5z 2 ). 29. (2z 3 + Wy + 12^ + 17zt/ 2 ) * (2x + 3t/). 30. (a 3 - 3a 2 6 - 6 3 + Soft 2 ) 4- (a 2 - 2a6 + 6 2 ). 31. (a: 3 + 27) (* + 3). 32. (a 3 + ft 3 ) -s- (o + 6). 33. (x 3 - 2/ 3 ) --(- y). 34. (IGx 4 - y 4 ) * (2x - y). 8^-27 - 43x 2 - 9x - 5 - 7 > 3*' + 22 40. (6a 3A + ISo 2 * - 4a A - 15) -5- (2a* + 3). 35. Fractions with a common denominator In a fraction, the bar should be thought of as a vinculum, a symbol of grouping, which encloses the numerator. We use this fact in adding fractions. 2 _ a ILLUSTRATION 1. In the fraction -- = > the bar encloses 3 a and 5 the fraction equals (3 a) * 5. SUMMARY. To express a sum of fractions with a common denominator as a single fraction: 1. Form the sum of the numerators, where each is enclosed in paren- theses and is given the sign of its fraction. 2. Divide by the common denominator. 8 3,9 8-3 + 9 14 ILLUSTRATION 2. = -= + = = - = - = -=- 5 o 5 o o T o^ 5 - x , 3 - 2x ILLUSTRATION 3. 7= --- r= -- =q lla lla lla 6 - (5 - x) + (3 - 2x) = 6-5 + s + 3-2s = 4 - x lla lla lla ' 38 INTRODUCTION JO FRACTIONS AND EXPONENTS 36. Alteration of a denominator To change a fraction to an equal one having an added factor in the denominator, we must multiply both numerator and denominator by this factor, in order to leave the value of the fraction unaltered. ILLUSTRATION 1. To change $ to 14ths, we multiply numerator and de- nominator by 2 because ^ = 2: 3 = 3 X2 6 7 7X2 14' ILLUSTRATION 2. To change the following fraction to one where the de- nominator is 6o 3 6, we multiply numerator and denominator by 2a 2 6, because -i- 3a = 5 - x 2a?b(5 - x) 10a 2 6 - 2a?bx 3a 2a 2 6(3o) 6a 3 6 37. Prime integers An integer is said to be prime if it has no factors except itself, and -f- 1 and 1, which are factors of any algebraic expression. Two factors are considered essentially the same if they differ only in sign, and then their product can be expressed as a power of either one of them. To factor an integer will mean to express it as a product of powers of distinct prime factors. ILLUSTRATION 1. 2, 3, 5, 7, 11, etc., are prime numbers. ILLUSTRATION 2. When factored, 200 = 2-2-2-5-5 = 25 2 . 38. Lowest common multiple At present, when we refer to a monomial, or single term, we shall mean an integral rational term whose numerical coefficient is an integer. The lowest common multiple, LCM, of two or more monomials is defined as the 'term with smallest positive coefficient, and smallest exponents for the literal numbers, which has the given term as a factor. As a special case, the LCM of two or more integers is the smallest positive integer having each given integer as a factor. ILLUSTRATION 1. The LCM of 3, 5, and 10 is 30. ILLUSTRATION 2. The LCM of 30& 8 and 5a 2 6 is INTRODUCTION TO FRACTIONS AND EXPONENTS 39 SUMMARY. To find the LCM of two or more terms: 1. Express each term with its coefficient factored. 2. Form the product of all letters in the terms and all the different prime factors of the coefficients, giving to each letter or integral factor the highest exponent it possesses in the given terms. EXAMPLE 1. Find the LCM of 20a 2 6 s and 70o 4 6. SOLUTION 1. In factored form, 20a 2 6 8 = 2 2 -5a 2 6 3 ; 70a 4 6 = 2-5-7a 4 6. 2. Hence, LCM - 2 2 -5-7a 4 6 3 = Note 1. In brief, the LCM of two or more terms equals the product of the LCM of their coefficients and the highest powers of the Utters seen in the terms. EXERCISE 13 Express the sum of fractions as a single fraction in lowest terms. ,3,79 ft 2 18 ,6 L i+i-r 2 -5-y + 5* 2,6 __a 4 c _d _ 7 3 *3+3 3* 888' .3 5,7-6 A 11 5 , 5. ---- -- o. ---- a a a z z z _ 3 2a - 56 5 6 - 3a 7 '7 -- 7 -- 8 '3~~T~* ^ 40 - 3 5 2y - 5 n . _ --- _ . _. 12. - - - . * 3 -a; 14 a a a - 206 3 Write the missing numerator or denominator to create equality. <i ^ 1ft 5 _ 10 3 _ 6 17. - - 18. - - 19. 40 INTRODUCTION TO FRACTIONS AND EXPONENTS A. - __ 24 -^- = __ 26 ~ 36x 7 Express the fraction as an equal one having the specified denominator. 2 5 26. ; denom., 21. 27. 5 ; denom., 32. 7 o 28. ; denom., 40. 29. |; denom., 35. 5 30. *j\ denom., be. 31. =-; denom., Qy. 32. -= -', denom., lOxy 3 . 33. ,5-5,; denom., 18xV- 5x?/ 2 3x 2 ?/ 4 Q 1 34. STTL; denom., 20ft 2 fc 4 . 35. ; denom., 20a 5 6 3 . Find the LCM o/ /ie given terms. As a partial check, verify that each term is a factor of the LCM. 36. 5; 4; 10. 37. 16; 24; 48. 38. 12; 54; 30. 39. 15; 12; 75. 40. 200; 36; 28. 41. 300; 27; 21. 42. e^; 9x 2 t/ 2 ; 15xy*. 43. 8a 2 6 6 ; 4a 3 6 4 ; 6a6. 44. 2o6; 14a 2 6 3 ; 66 4 . 45. 6a 2 x; 4a'x 2 ; 9ax 4 . 46. 3zV; 12xi/ 3 ; 20x. 47. 5W; 10^; 16^. 48. Change , ^, and ^ to the denominator 42, and then add the fractions. 39. Addition of fractions * To combine a sum of fractions into a single fraction, the given fractions must be changed to new forms having a common denomina- tor. We define the lowest common denominator! LCD, of a set of fractions as the LCM of their denominators. ILLUSTRATION 1. The LCD of J, f , and is the LCM of 6, 5, and 8 or 3-5'8 or 120. Hence, to add the fractions, we. change them to new fractions whose denominator is 120: 1 4_ ? 4- 1 = 2Q . 3-24 7-15 = 20 + 72 + 105 197 6 + 5 + 8 6-20 + 5-24" l "8.15 120 120* * To simplify the present chapter, we will deal only with the case where each denominator is an integral rational term. INTRODUCTION TO FRACTIONS AND EXPONENTS 41 SUMMARY. To express a sum of fractions as a single fraction: 1. Find the LCD of the given fractions. 2. For each fraction, divide the LCD by the denominator and then multiply numerator and denominator by the resulting quotient, to change to an equal fraction having the LCD. 3. Form the sum of the new numerators, where each one is enclosed in parentheses and is given the sign of its fraction. 4. Place the result of Step 3 over the LCD, remove parentheses in the numerator, and reduce the fraction to lowest terms. ILLUSTRATION 2. In the following sum the LCD is 20; we think of 2 as 2/1. We observe that ^ = 5 and f = 2. Hence, 2-20 5(3s) 2(7 - 2x) o _ _ 4 10 20 20 20 40 - 15x - 2(7 - 2x) 40 - 15s - 14 + 4x 26 - llx 20 20 20 3w 7 EXAMPLE 1. Express as a single fraction: Sax* SOLUTION. 1. The LCD is 15a 3 z 2 . We have 15a 3 z 2 Ida's* r , . = 3z; -r r = 5a 2 . 5a 3 x 3ax 2 2. Hence, we multiply by 3x and 5o 2 in the given fractions to change to the LCD: _ 7(5a 2 ) = 9xy - 35a 2 3aa: 2 (5a 2 ) EXERCISE 14 Combine into a single fraction in lowest terms. 1. 5 8" -! 2. f- ^fr 3. 5 6 1 3* 4. 3 5 9 10 5. 13 16 3 8* 6. 1_ 4 7. 2 7 5 "*"2l' 8. 1 6 3 5* 9. 3~ * 10. 2c 5 d 2* ,11. 3 k 4* 12. ^H 2 ^ ~5 14. i - - 1 15. ? _ 2 + 83 7 ^ 21 42 INTRODUCTION TO FRACTIONS AND EXPONENTS 1A 12 4 , 17 3 74 lg 0,1,4 16. y-5 + 3. 17 ' To-30 + 5' 18 ' " 2 + 6 + 15 - 3 3 3o-6 7 a-3 . 64-2 2a; - 3 , 3 - 3a a + 7 2 x - 3 g - 7 ~ 2 ~ 25 10 18 v\ b c <u h w - Ta ~ 6^' ^ S "" 25* . S ~ 33k 4 OB A- 1. 34 -I- A. 36' 12y 2y Gx Sxy 36 - - 37 _ 36> 37 ' 39 JL - ^ " 1 40 3a 2 5a6 _ 62^ 4 10 9 54 2X JIt a: - 1 x - 3 2o~3 5o + 3 47 - y-2 Ki - 2 - 4* 4a - 5 3a + 2 51. g -- 1 - -2J 52. ? 2 + 2-x 4 +. /NTRODUCT/ON TO FRACT/ONS AND EXPONENTS 43 40. Mixed expressions A sum of an integral rational part and of one or more fractions is referred to as a mixed expression. If a mixed expression occurs as a factor in a product or as the numerator or denominator of a fraction, it is usually desirable to combine the mixed expression into a single fraction before performing other operations. ILLUSTRATION 1. We refer to a number such as 5| as a mixed number. 23 17 23(17) 391 T o ILLUSTRATION 2. ILLUSTRATION 3. 3z 2 -h - is a mixed expression. ILLUSTRATION 4. (2 + |j(3 | 15 - a = (6 + a) (15 ~ a) 90 + 9a - a 2 15 15 '41 . Complex fractions * A simple fraction is one without any fraction in its numerator or denominator. A complex fraction is one having one or more fractions in the numerator and denominator. SUMMARY. To reduce a complex fraction to a simpty fraction: 1. Express the numerator and denominator as simple fractions. 2. Form the quotient of the simple fractions and reduce the result to lowest terms. - , 3 5+3 8 I _X^ __ A -t- T t 5 5 5 8 3 12 ILLUSTRATION 1. - = g ^ g - - = 333 Sometimes it may be convenient to reduce a complex fraction to a simple fraction by the single operation of multiplying both numerator and denominator of the complex fraction by the LCD of all simple fractions involved. * In this chapter, the problems will avoid questions of factoring which will be met in Chapter 5. 44 INTRODUCTION TO FRACTIONS AND EXPONENTS ILLUSTRATION 2. To reduce the given complex fraction of Illustration 1 to a simple fraction, we observe that the LCD of 3/5 and 4/3 is 15. Hence, we multiply both numerator and denominator by 15, observing that 15$) - 9 and 15(j) = 20: 1 + 5 15 + 9 24 12 4 30-20 10 5 , , 2 15o6 3a + yr T Q 56 56 ILLUSTRATION 3. = = 6o6-7 36 15o6 + 2 36 45a6 56 606 - 7 30a6 - 35 where we divided out 6 from numerator and denominator. 26-5 26-5 36 36 26 - 5 1 26-5 ILLUSTRATION 4. 4-6 4-6 36 4-6 126 - 36* 1 DEFINITION I. The reciprocal of a number H is = ILLUSTRATION 5. The reciprocal of 3 is J. r ILLUSTRATION 6. The reciprocal ofvis? = Y'o =:: o' The reciprocal of T is a 1 a a Thus, the reciprocal of a fraction is the fraction inverted. EXERCISE 15 Express the fraction or product as a simple fraction in lowest terms. 1. (4 + f)(5). 2. (3 - f)(4f). 3. (2f)(3f). 4. (6 -f |) (6 - f). 6. (f - 36) (f - 46). INTRODUCTION TO FRACTIONS AND EXPONENTS 45 A 7 15 ~* 6 - T+T 7 - T+T + -IS 7 2a 2 6 10. 12. f-r-^- 13. g^-4- 14. * i + 5 "* 1 - f "'5 5c~f J-2c' 2_ 3 4_ 3 5_ 3 16. 2 16. - r- 17. T- 5 . . e , o . , o - + 4 6 +: 4 + ^ X* Fj tif^ \Jb \J l/Vx J. , 2 _ _ + 1ft 2x y 19 2 TIT' a: ^ 01 _ 2 3 5x 5 x" h 4y 3 + 3z 3 5 5 3 a 2 26 w\ 2 y 2 3 23 ' 2 1 a 4a6 32/ 5 ' 2x 5 3 w; 2 1 2 2xw 9A 3 27 . . 28. . 29 . 2 + ? ' + ^ 3 2a 56 30. !i. 31. 1- 32. . - 2 a: -f 3y Find the reciprocal of the expression, and express the result as a simple fraction in lowest terms. 33. 75. 34. 17. 36. . 36. f 37. - . 38. 10. 39. 12? . 40. 2J. 41. - 5J. 42. a. 44. . 45. 46. By writing a fraction, show that to divide a number N by a number H means the same as to multiply N by the reciprocal of H. 46 INTRODUCTION TO FRACTIONS AND EXPONENTS EXERCISE 16 Review of Chapters 1 and 2 Compute each expression, leaving any fraction in lowest terms. 1. (- 3)(- 4)(- 5). 2. - (- 2)(- 5). 3. - (- 3)(- 4)(0). . (-7X-3) (- 2) (5) ( * -14 " (-3X-6) ~ (-2X-4) 7. - 7 + 19 - 16. 8. 3 - {- 4) - 7. 9. - 2 -f (- 3) + 6. Find the absolute value of the expression. 10. - 8. 11. (- 3)(- 2). 12. | - 14 |. 13. | 17 |. (a) Add the two numbers; (6) subtract the lower one from the upper number. 14. 17 16. - 23 16. - 15 17. 17 - 5 - 9 _ 29 25 18. Read the expressions 4 < 2 and 17 > 0, and verify their truth by marking the numbers on a real number scale. Insert the proper sign, < or > , between the numbers. 19. 11 and 19. 20. - 15 and - 27. 21. and - 6. 22. Which one of the numbers 15 and 7 is less than the other? Which one is numerically less than the other? Perform any indicated operation, removing any parentheses, reducing any final fraction to lowest terms, and employing the laws of exponents to simplify expressions. 23. - (3a - 26 - c). 24. 2(3 - 5a - c). 26. - a 2 (3a 8 6 - aft*). 26. 5(- 3*). 27. 3*y(2* - Wy). 28. - 2*y(- 3* - 5j/). 29. 3(2a - h + A) - 2(3a + 4h - 3k). 30. - x*(3xy* - 2xy + 3) + 4i/(2xV - 3x* + 5). 31. 3o - [2a - 3(5 - a)]. 32. - (6a - 6) - 2[3 - (2o - 46)]. - 15 4 14 3 6 37. 17. 38. | -4- 6. 39. 7 + |- 40. ~ 41. - 2AW(3iUPX- 4A 2 fc). 42. - 4/*V(2% 2 - 3/fy - 5A). INTRODUCTION TO FRACTIONS AND EXPONENTS 47 43. (Si 8 !/) 4 . 44. (- 2oV). 46. (- 46. (-2)'. 47. (I)'- .(-?)'. .(*)' ' ' -(I)- 66. (2* + 3) (2s - 7). 66. (3x* - 7x)(2x 57. (2z - 2/)(3z - %). 68. (2a - 36) 2 . 69. (3x* - 5x - 2r> + 3)(4z + 5). 64. (6o 3 - 19o 2 + 21a - 9) ^- (2o - 3). 66. (18 + 4x 3 - 26a; - 2x 2 ) ^- (2x - 5). A7 4y + 8 4x - 7s* 67 ' Express as a single fraction in lowest terms. 2 7 . 10 - ft 2h-3 4A-7 _ -4- / II. ' ti . i 33^3 5 10 -_2. 72 j! _ 4. A. 3 ' " 12 " 2xy 3y 2 4x 2 2 - 3x 5 - By -. 2a 3 3a - 6 iO ^ * rfc I4 Q * AV ^ y 5 .!__:>. 35 76. -J. 76. | |- 77. -J | 5 a 56 78. ^-~- 79. 9 J_ _ _. a x y 5 -L^. _ 3 3 1 81. Find the reciprocal of -=; of (a 5). o 6 CHAPTER 3 DECIMALS AND ELEMENTS OF COMPUTATION 42. Decimal notation The decimal notation * for writing numbers is called a place sys- tem, for which the base is 10. In this notation, each number is an abbreviation for a sum involving units and powers of ten, written by use of the Hindu-Arabic digits or figures (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). ILLUSTRATION 1. 3456 = 3(1000) + 4(100) + 5(10) + 6 = 3(10 3 ) + 4(10 2 ) + 5(10) + 6. We have 3456 as the sum of 6 units, or ones, 5 tens, 4 hundreds, and 3 thousands. ILLUSTRATION 2. 23.572- = 2(10) + 3 + ~ + ^ + 2 10 ' 100 ' 1000 10 ' 10 2 ' 10 3 Counting to the left of the decimal point in a number, the places are named the units' place or ones' place; tens' place; hundreds' place; thousands' place; ten thousands' place; etc. Counting to the right -of the decimal point, we have the tenths' place; hundredths' place; thousandths' place; etc. The name of any place is the value of a unit located there. These values should be remembered in terms of powers of ten. ILLUSTRATION 3. 1000 = 10 3 ; .001 = rrr = -; .01 = T- ILLUSTRATION 4. We read .5073 as tc point, 5, oh, 7, 3," or as "5073 ten-thousandths" * In this chapter, unless otherwise specified, any number referred to will be positive. DECIMALS AND ELEMENTS OF COMPUTATION 49 In this chapter we will think of each number in its decimal form. The part to the right of the decimal point is called the decimal part of the number. We consider each number as having a decimal part even when the decimal part is zero and the number is then an integer. The decimal places of a number are those places to the right of the decimal point in which digits are actually written in the number. ILLUSTRATION 5. 23.507 has three decimal places and its decimal part is .507. In any number, observing its digits from left to right, we may visualize an endless sequence of zeros at the right of the last digit not zero, if there is such a last digit. A number of this character is called a terminating decimal. An endless decimal is one in which, however far we proceed to the right, we never reach a digit beyond which all digits are zeros. Hereafter, unless otherwise stated, any number mentioned will be a terminating decimal. ILLUSTRATION 6. 35.675, or 35.67500 , is a terminating decimal. The familiar number TT = 3.14159 is an endless but not a repeating decimal. The fraction J is equal to the endless repeating decimal .333 If a unit in any place in the decimal notation is multiplied by 10, we obtain a unit in the next place to the left. If we divide a unit in any place by 10, we obtain a unit in the next place to the right. The preceding remarks justify the following convenient rules. I. To multiply a number by 10, move the decimal point one place to the right in the number. II. To divide a number by 10, move the decimal point one place to the left in the number. ILLUSTRATION 7. 10(315.67) = 3156.7. ~~ = 31.567. To divide by 1000, which is 10(10) (10), we move the decimal point three places to the left: 21.327 * 1000 = .021327. 43. Addition of decimals In finding a sum of decimals, after they have been arranged with the decimal points in a column, it is desirable to add once going up- ward and then downward to check in each column. 50 DECIMALS AND ELEMENTS OF COMPUTATION ILLUSTRATION 1. To find the sum of 31.457, 2.6, 3.15, and 101.41, we annex zeros to extend each number to the 3d decimal place, and then add. 3.150 EXAMPLE 1. Add: 1.573 + 3.671 - 1.157 - 4.321 + 10.319. Sum 138 ' 617 31.457 101.410 SOLUTION. The sum of the positive terms is 15.563; of the negative terms is - 5.478. (15 563 1 5 478 f sum - 10.085. EXERCISE 17 Write the number in decimal form. 1. 3 and 25 hundredths. 2. Point, oh, 1, 5, 3, 9, oh, 4. 3. 10*. 4. 10'. 6. ~ 6. i- 7. i- Write each number as a sum involving powers of 10, with one term corre- sponding to each digit (not zero) of the number. 8. 567. 9. 3149. 10. 16,342. 11. .319. 12. 27.0457. Write the number in decimal notation equal to the sum. 13. 5(100) + 3(10) +6 + A + A 14. 2(1000) -f 4(10) + 3 + 1 + JL + -J- 15. 5(10*) + 7(10') + 3(10) + 5 + + Compute the indicated sum. 16. 2.057 + 3.11 + 4.985 + 3.05 + 1.5 + 2.177 + .459. 17. 3.193 + .098 -f 1.567 -f 2.457 + 3.167 + 2.13 -f- 1.5072. 18. 21.675 - 14.521. ' 19. .0938 - .0257. 20. 1.721 - 2.468. (a) Add the numbers. (6) Subtract the lower number from the upper one. 21. 5.26 22. .357 23. - 43.8468 24. .02438 1.38 .2983 - 59.923 - .5729 25. Compute 2.156 - 3.149 + 4.183 + 2.147 - 4.159. DECIMALS AND ELEMENTS OF COMPUTATION 51 26. Compute 13.083 + 2.148 - 41.397 - 2.453 + 12.938. Perform the multiplication or division mentally. 27. 10(.532). 28.1000(1.0219). 29. 10*(32.653). 30. 100(.00415). .0317 13.257 57.38 .0498 oo ' wo IS" ISSo* 44. Multiplication of decimals Any number whose decimal part is not zero can be written as a fraction whose numerator is an integer and denominator is a power of 10. The exponent of 10 hi the denominator can be taken equal to the number of decimal places in the given number. ILLUSTRATION 1. 1.21(.205) = _ 121 205 (121) (205) 24,805 10 2 ' 10 s 10 2+ * 10 6 We conclude that the five decimal places in the result are a consequence of the law of exponents for multiplication, because the factors had two and three decimal places. This result is a special case of the following rule. SUMMARY. To multiply two decimals: 1. Find the digits of the product by multiplying the factors with their decimal points disregarded (or even removed). 2. Add the numbers of decimal places in the factors to find the number of decimal places in the product, and insert the decimal point. ILLUSTRATION 2. To multiply .0238 X 112.75, we find the digits of the product, at the right, and then point off (2 -f 4) or 6 decimal places to obtain 2.683450. Notice that the final zero had to be written and counted in fixing the decimal point. In stating the final result, we could then omit the zero. 11275 (X)238 90200 33825 22550 2683450 The digits in the product of two decimals depend only on the digits of the factors. If the decimal points are moved hi the factors, this only alters the position of the decimal point in the result. ILLUSTRATION 3. In Illustration 2, .0238(112.75) = 2.683450. Hence, 2.38(1127.5) - 2683.450. 52 DECIMALS AND ELEMENTS OF COMPUTATION EXERCISE 18 Perform the following multiplications. 1. 3.51(1.4). 2. .46(.107). 3. 13(.461). 4. .0142(3.6). 5. 21.38(.024). 6. 156.1(1.38). 7. 398.4(.0342). 8. .00175(.2147). 9. .0346(.00157). 10. 85.2(1.356). 11. 9.137(.2346). 12. 74.308(.00259). 45. Significant digits In any number N, let us read its digits from left to right. Then, by definition, the significant digits or figures of N are its digits, in sequence, starting with the first one not zero and ending with the last one definitely specified. Notice that this definition does not involve any reference to the position of the decimal point in N. Usually we do not mention final zeros at the right in referring to the significant digits of N, except when it is an approximate value. ILLUSTRATION 1. The significant digits of 410.58 or of .0041058 are (4, 1, 0, 5, 8). 46. Approximate values If T is the true value and A is an approximate value of a quantity, we agree to call A T the error of A. ILLUSTRATION 1. If T = 35.62, and if A = 35.60 is an approximation to T, then the error of A is 35.60 - 35.62, or - .02. The significant digits in an approximate value A should indicate the maximum possible error of A. This error is understood to be at most one half of a unit in the last significant place in A, or, which is the same, not more than 5 units in the next place to the right. ILLUSTRATION 2. If a surveyor measures a distance as 256.8 yards, he should mean that the error is at most .05 yard and that the true result lies between 256.75 and 256.85, since the error (plus or minus) might be =fc .05. In referring to the significant digits of an approximate value A, it is essential to mention all final zeros designated in A. ILLUSTRATION 3. To state that a measured weight is 35.60 pounds should mean that the true weight differs from 35.60 pounds by at most .005 pound. To state that the weight is 35.6 pounds should mean that the true weight DECIMALS AND ELEMENTS OF COMPUTATION 53 differs from this by at most .05 pound. Thus, there is a great distinction between 35.6 and 35.60 as approximate values although there is no difference between 35.6 and 35.60 as abstract numbers. 47. Rounding off a number In referring to a place in a number, we shall mean any place where a significant digit stands. In referring to a decimal place, the word decimal will be explicitly used. To round off N to k figures, or to write a fc-place approximation for N, means to write an approximate value with k significant digits so that the error of this value is not more than one half of a unit in the kth place, or 5 units in the first neglected place. This condition on the approximate value of N leads us to the following method. SUMMARY. To round off a number N to k places, drop off the part of N beyond the kth place and then proceed as follows: 1. // the omitted part of N is less than 5 units in the (k -f- l)th place, leave the digit in the kth place unchanged. 2. // the omitted part of N is more than 5 units in the (k -f l)th place, increase the digit in the kth place by 1. 3.* // the omitted part of N is exactly equal to 5 units in the (k + l)th place, increase the digit in the kth place or leave it unchanged, with the object of making the final choice an even digit. ILLUSTRATION 1. The seven-place approximation to TT is 3.141593. On rounding off to five places (or four decimal pldles) we obtain 3.1416. We changed 5 to 6 because .000093 > .00005. On rounding off TT to three places we obtain 3.14. ILLUSTRATION 2. In rounding off 315.475 to five figures, with equal justification we could specify either 315.47 or 315.48 as the result. In accordance with Item 3 of the Summary, we choose 315.48. 48. A notation for large numbers For abbreviation, or to indicate how many digits in a large number are significant, it is sometimes convenient to write a number AT as the product of an integral power of 10 and a number equal to or greater than 1 but less than 10, with as many significant digits as are justified by the data. * Item 3 could be replaced by various similar and equally justified agreements. 54 DECIMALS AND ELEMENTS OF COMPUTATION ILLUSTRATION 1. If 5,630,000 is an approximate value, its appearance fails to show how many zeros are significant. If just five digits are significant we write 5.6300C10 6 ), and, if just three are significant, 5.63C10 6 ). 49. Accuracy of computation By illustrations, we can verify that the following rules do not underestimate the accuracy of computation with approximate values. On the other hand, we must admit that the rules sometimes over- estimate the accuracy. However, we shall assume that a result ob- tained by these rules will have a negligible error in the last significant place which is specified. I. In adding approximate values, round off the result in the first place where the last significant digit of any given value is found. II. In multiplying or dividing approximate values, round off the result to the smallest number of significant figures found in any given value. ILLUSTRATION 1. Let a = 35.64, 6 = 342.72, and c = .03147 be approxi- mate values. Then, a + b -\- c is not reliable beyond the second decimal place because both a and b are subject to an unknown error which may be as large as 5 units in the third decimal place. Hence, we write a + b + c = 378.39147 = 378.39, approximately. ILLUSTRATION 2. If x = 31.27 and y = .021 are approximate values, then,*by Rule II, we take xy = .66, because y has only two significant digits: xy = 31.27(.0ll) = .65667 = .66, approximately. ILLUSTRATION 3. If a surveyor measures a rectangular field as 385.6' by 432.4', it would give an unjustified appearance of accuracy to write that the area is (385.6) (432.4) = 166,733.44 square feet. For, an error of .05 foot in either dimension would cause an error of about 20 square feet in the area. By Rule II, a reasonably justified result would be that the area is 166,700 square feet, to the nearest 100 square feet, or 1.667(10*) square feet. In problems where approximate values enter, or where approxi- mate results are obtained from exact data, the results should be rounded off so as to avoid giving a false appearance of accuracy. No hard and fast rules for such rounding off should be adopted, and the final decision as to the accuracy of a result should be made only after a careful examination of the details of the solution. DECIMALS AND ELEMENTS OF COMPUTAT/ON 55 EXERCISE 19 Round off, first to five and then to three significant digits. 1. 15.32573. 2. .00132146. 3. .3148638. 4. 5.62653. 6. 195.635. 6. .00128558. ' 7. .0345645. 8. 392.462. Tell between what two values the exact length of a line lies if its measured length in feet is as follows. 9. 567. 10. 567.0. 11. 567.4. 12. 35.18. Assuming that the numbers represent approximate data,* find their sum and product and state the results without false accuracy. 13. 31.52 and .0186. 14. .023424 and 1.14. 15. .0047(11.3987126). HINT for Problem 15. In proceeding to multiply or divide approximate values, there is no advantage in keeping many significant digits in one value when other values have relatively few significant digits. A conservative rule is that, before multiplying or dividing, we may round off any given value to two more significant digits than appear in the least accurate of the given values.^ Write the number in ordinary decimal form. 16. 100(3.856). 17. 27.38(10 2 ). 18. 1.935(10<). 19. 2.056(10). Write as the product of a power of 10 and a number between 1 and 10. 20. 38.075. 21. 675.38. 22. 153,720,000. 23. 45,726. Given that the number is an approximate value, write it as the product of an integral power of 10 and a number between 1 and 10, assuming, first, that there are five significant digits and, second, that there are three significant digits. 24. 9,330,000. 25. 453,120. 26. 23,523,416. 27. 72,200,000. In the following problems, state each result without false accuracy. 28. The measured dimensions of a rectangular field are 469.2 feet and 57.3 feet. Find the perimeter (sum of lengths of sides) and area of the field. 29. The measured dimensions of a rectangular box are 20.4 feet, 16.5 feet, and 7.8 feet. Find the volume of the box in cubic feet. 30. Given that one cubic foot contains approximately 7.5 gallons, how many gallons are contained by 2.576 cubic feet? * In this book, unless otherwise stated, the numerical data in any problem should be assumed to be exact. Results obtained from exact data may some- times be rounded off. t See Note 3 in the Appendix for a convenient abridged method for multiplyic two numbers with many significant digits. 56 DECIMALS AND ELEMENTS OF COMPUTATION 50. Division of decimals When one decimal is divided by another, the division process some- times gives a zero remainder if the work is carried to a sufficient number of decimal places. Usually, however, we may expect a remainder not zero however far we continue the division. We can always arrange the details so that the actual work amounts to division by an integer. This arrangement is useful in locating the decimal point in the result. ILLUSTRATION 1. To compute 372.173 -5- 1.25 with accuracy to two decimal places, we first indicate the division as a fraction, and then multiply its numerator and denominator by 100, to obtain an integer as the divisor: 372.173 = 372.173(100) 37,217.3 1.25 1.25(100) ~ 125 (1) 1.25, 297.738 (Quotient} 372.17*300 (Dividend) 250 122 112 67 75 At the right, we insert the original decimal points in dividend and di- visor and mark with "A" the new lo- cations of the decimal points observed in the final fraction in (1). The mul- tiplication by 100 in (1) is equivalent to the action of moving the decimal point two places to the right in both dividend and divisor. In the process of division, the integral part of the quotient ends when we begin using digits of the dividend at the right of the new decimal point , " A ." If we place each digit of the quotient directly above the last digit being used at that stage from the dividend, the decimal point in the quotient occurs exactly above the altered decimal point, A , in the dividend. We find the quotient to three decimal then round it off to 297.74, which we say is correct to To check, we compute 1.25(297.74) = 372.175. 92 3 87 5 4 80 3 75 1 050 1J300 50 places, 297.738, and two decimal places. We do not obtain exactly 372.173 because 297.74 is not the exact quotient. Note 1. In any division, estimate the result before dividing, to check the location of the decimal point in the quotient. Thus, in Illustration 1, a convenient estimate would be 375 -5- 1.25, or 300. Then we are sure that the actual answer should have three places to the left of the decimal point. DECIMALS AND ELEMENTS OF COMPUTATION 57 Note 2. In dividing approximate values, obtain the quotient to one more figure than is specified as reliable by Rule II, Section 49. Then, round off the quotient in the preceding place. Any terminating decimal can be expressed as a fraction, which then may be reduced to lowest terms. Conversely, we can express any common fraction as a decimal by carrying out the division indicated by the fraction, to obtain either a terminating or an endless decimal. ILLUSTRATION 2. On carrying out the division, we obtain J = .875, exactly. 3125 = 25 1000 8 ILLUSTRATION 3. 3.125 = ILLUSTRATION 4. To express ii as a decimal we divide, at the right. After reaching .458 in the quotient, we meet 8 each time as the only digit in the remainder. Hence, 3 will be obtained endlessly in the quotient. The result is the endless repeating decimal .4583, where the dot above 3 means that the digit repeats endlessly. 24 .45833 11.00000 96 140 120 200 192 80 72 80 etc. EXERCISE 20 (a) Write each expression as a fraction, and then alter it to make the divisor an integer. (6) Divide until the remainder is zero. 1. 3.562 -T- 2.6. 2. 2.849 -f- .74. 3. 140.14 -^ 2.45. Obtain the result of the division correct to three decimal places. 4. 381.32 -5- 58. 5. .083172 -5- .316. 6. .5734 *- 12.8. Assume that the numbers in the following problems are approximate values. Carry out the division and state the result without false accuracy, according to Rule II, Section 49. 7. 573.2 * 3.83. 8. 19.438 -s- 2.21. 9. .09734 -f- 3.265. 10. 98.3 -s- 21.473. 11. .003972 4- .0139. 12. .01793 ^ .5634. Change the decimal to a fraction in lowest terms. 13. 2.75. 14. .0125. 16. 2.375. 16. .0175. 17. .0325. Obtain the decimal equal to the given fraction. If the decimal repeats endlessly, carry the division far enough to justify this conclusion. 18. f . 19. f . 20. A. 21. A- 22. ft. 23. CHAPTER 4 LINEAR EQUATIONS IN ONE UNKNOWN 51 . Terminology about equations An equation is a statement that two number expressions are equal. The two expressions are called the sides or members of the equation. An equation in which the members are equal for all permissible values of the letters involved is called an identical equation, or, for short, an identity. An equation whose members are not equal for all permissible values oi the letters is called a conditional equation. ILLUSTRATION 1. In the following equation, by carrying out the multipli- cation on the left-hand side, we verify that the product is the same as the right-hand side. This is true regardless of the values of a and b. Hence the equation is an identity. (a + 26) (a + 36) = a 2 + 5a6 + 66 2 . ILLUSTRATION 2. The equation x 2 = is a conditional equation be- cause the two members are equal only when x =-2. The word "equation" by itself will be used in referring to both identities and conditional equations, except where such usage would cause confusion. Usually, however, the word "equation" refers to a conditional equation. At times, to emphasize that some equation is an identity, we shall use " s " instead of " = " between the members. A. conditional equation may be thought of as presenting a ques- tion: the equation asks for the values which certain letters should . , values are requested, are called unknowns. Some of the letters in an equation may represent known numbers. ILLUSTRATION 3. x* + 3x - 5 = is an equation in the unknown x. LINEAR EQUATIONS IN ONE UNKNOWN 59 52. Solution of an equation An equation is said to be satisfied by a set of values of the unknowns if the equation becomes an identity when these values are substituted for the unknowns. A solution of an equation is a set of values of the unknowns which satisfies the equation. A solution of an equation in a single unknown is also called a root of the equation. To solve an equation in a single unknown means to find all the solutions of the equation. ILLUSTRATION 1. 4 is a root of the equation 2x 3 = 5, because when x = 4 the equation becomes [2(4) 3] = 5, which is true. 53. Equivalent equations Two equations are said to be equivalent if they have the same solutions. EXAMPLE 1. Solve: 3 - 3x = - 7 - 5x. (1) SOLUTION. 1. Add 5x to both members: 1 - 3x -V 5x = - 7 - 5x -V 5x, or 3 + 2x = - 7. (2) 2. Subtract 3 from both members, or, which is the same, add 3 to both sides: - 3 + 3 -f- 2x = - 7 - 3, or 2x = - 10. (3) 3. On dividing both sides of (3) by 2 we obtain x = - 5, (4) and conclude that this is the only solution- of (1). CHECK. Substitute x = 5 in (1). Left-hand side: 3 - 3(- 5) = 3 + 15 = 18. Right-hand side: - 7 - 5(- 5) = - 7 + 25 = 18, which checks. Comment. Each equation obtained from (1) was equivalent to it and this would justify us in saying that (1) has just one root, x = 5, without any necessity for the check. The equivalence of (1), (2), (3), and (4) is a consequence of the following familiar facts: if equal numbers are added to or subtracted from equal numbers the results are equal; if equal numbers are multiplied or divided by equal numbers the results are equal. 60 LINEAR EQUATIONS IN ONE UNKNOWN ILLUSTRATION 1. Any value of x satisfying (1) will also satisfy (2), be- cause we pass from (1) to (2) by adding equals, 5x and 5x, to the equal sides of (1). And, any value of x satisfying (2) will satisfy (1) because we can pass from (2) back to (1) by the inverse operation of subtracting 5x from both sides of (2). Hence, (1) and (2) are satisfied by the same values of x, and thus are equivalent. In solving an equation in a single unknown, by use of the following operations we pass from the original equation to progressively simpler equivalent equations, which finally yield the desired roots. SUMMARY. Operations on an equation yielding equivalent equations. 1. Addition of the same number to both members. 2. Subtraction of the same number from both members. 3. Multiplication (or division) of both members by the same num- ber, provided that it is not zero and does not involve the unknowns. Note 1. We observe that Operation 2 is a special case of Operation 1 because subtraction of a number N is equivalent to addition of N. The division part of Operation 3 is a special case of the multiplication part, because division by a number N is equivalent to multiplication by 1/N. Convenient mechanical processes, and corresponding terminology, grow out of Operations 1, 2, and 3. A term appearing on both sides of an equation can be canceled, by subtracting the term from both sides. ILLUSTRATION 2. Given: x -j- 3 = f -f 3. Subtract 3 from both sides: x = f. A term can be transposed from one member to the other, with the sign of the term changed, by subtracting it from both members. ILLUSTRATION 3. Given: x + 5 = 7. Subtract 5 from both sides, or transpose 5: x 7 5, or x = 2. ILLUSTRATION 4. Given: x 4 = 9. Transpose 4: x 9 + 4, or x = 13. The signs of all terms on both sides may be changed, by multiplying both sides by 1. LINEAR EQUATIONS IN ONE UNKNOWN 61 ILLUSTRATION 5. Given: 3x 6 = 5 ox. Change all signs (multiply both sides by 1) : 6 = 5 H- ax. 54. Degree of a term The degree of an integral rational term in a certain literal number is the exponent of the power of that number which is a factor of the term. If the term does not involve the number, the degree of the term is said to be zero. The degree of a term in two or more letters together is the sum of their degrees in the term. The degree of a polynomial is defined as the degree of its term of highest degree. ILLUSTRATION 1. With x as the literal number involved, the degree of Sx 3 is 3. The degree of 2x is 1 because x x l . The degree of (5x 3 3z 2 -f 2x 7) is 3 ILLUSTRATION 2. The degree of 3#V in x is 2, in y is 3, and in x and y together is (3 -f- 2) or 5. ILLUSTRATION 3. A polynomial of the first degree in x is called a linear polynomial in x and is of the form ax + b, where a and b do not involve x. 55. Linear equations An integral rational equation is one in which each member is an integral rational polynomial in the unknowns. A linear equation, or an equation of the first degree, is an integral rational equation in which the terms involving the unknowns are of the first degree. ILLUSTRATION 1. The equation 3x 7 = 5 is a linear equation in x. SUMMARY. To solve a linear equation in one unknown: 1. If fractions appear, place parentheses around each numerator and clear of fractions by multiplying both members by the LCD of the fractions; then, remove parentheses and combine terms. 2. Transpose all terms involving the unknown to one member and all other terms to- the other member. Combine terms in the unknown, exhibiting it as a factor. 3. Divide both sides by the coefficient of the unknown. 4. To check the solution, substitute the result in the original equation. 62 LINEAR EQUATIONS IN ONE UNKNOWN r. *oi x x 3 3 + x n EXAMPLE 1. Solve: --- ~ = r -- 2. SOLUTION. 1. The LCD is 30. Hence, multiply Ijoth sides by 30, ob- serving that 30 -p = 10(z-4); 30-- = 15(s - 3); 30 -_ = 3(3 + 10(x - 4) - 15(x - 3) = 3(3 + x) - 60; 10* - 40 - 15* + 45 = 9 + 3x - 60; - 5x + 5 = 3x - 51. / 2. Subtract 3z and 5 from both sides: - 5x - 3x = - 51 - 5; - 8x = - 56. 3. Divide both sides by 8: x = 7. CHECK. Substitute x 7 in the original equation. r * 1 j ; 7 ~ 4 7 - 3 3 4 1 o 1 Left-hand side: ^ --- j; = ___ == 1_2= 1. o A o 34-7 10 Right-hand side: ^. -- 2 = -r 2=1 2= 1. This checks. In the case of a linear equation in a. single unknown x, if the un- known remains in the equation after Step 2 of the standard method of solution is applied, the equation is then of the form ex = 6, where c j* 0, and b and c do not involve x. On dividing both sides of ex b by c we obtain x = b/c. That is, a linear equation in a single un- known has just one root. Note 1. In directions for solving an equation, in this book, a specification to add, subtract, multiply, or divide will mean to perform the operation on both sides of the preceding equation. EXERCISE 21 Solve the equation for the literal number in it. 1. 5x - 3 = x + 7. 2. 3z - 6 = 18 - x. 3. 5 - 2y = 2 - 3y. 4. 3 - 3x = - 7 - 5x. 5. 2 + 5 - 4(2 - ). 6. 2y - 4 = 1 - 4y. 7. 5 - 5y = 5 - 4y. B. 2(4 - x} = 8 - 3x. 9. Sy + 3 - 5y - 4. 10. 7 - x = 2(1 - LINEAR EQUATIONS IN ONE UNKNOWN 63 11. 2(7 + a) - 1 - 7x. 13. 9 - 7h = 14A - 12. 16. 52 - 11 + 32 = 2 - 3. 17. 5x -- 1 - 4* -f 2. 19. 4* + * - 3* - }. 21. to - f - 3s + ft. , 3 h 25 -4- _ 25 . - - 4 - 3 27. _ _ _ 10 2 6 2 31. __ 3 " T "" 5* - 7 _ 4 + y _ 35. 5 - 25 5x - 3 3 -a: _ 5 a; - 2 ~6~ ~ 6 " ~2~ 6 3x - 2 a; - 5 T- + 43. .26 - z = .98 - 3. 46. .26a: - .2 = .53* - .38. 47. 2.5* - 3.7 = 13.5 - 1.8x. 49. .19* - .358 = .032 - .07*. 12. llh - 5 - 8fc - 4. 14. 7k + 12 = 2k - 7. 16. 11 - 6w> - 34w - 9. 18. 42 - i = 32 + f . 22. 5* + Y = 17* - 5* 3* 3* -------. * 15 3 ~ 5 5 28. - = - - 10 3 2 12 3 5 13 TO 32. 3 - 2x 9 x - 3 37 ^ 2a; + 7 TO + ~~5~~ - 5 8 13 x - 2 - 3 2s - 1 16 44. 3x - .55 = .33 - 46. 2.3z - 2.4 = 1.6 - 1.7*. 48. .21* - .46 = .79 + .96*. 60. 4.088 + .03* - 3* - .07. 64 LINEAR EQUATIONS IN ONE UNKNOWN 61. 2.035 - Mix = - .212* - .215. 62. 3(5z + 2) - 12 = 25(2z + 1) - 3. 63. 2y -I - 10(y + 1) = 2(2 - 3t/) - 1. 64. 82 - 2(32 - 1) = 7z - 3(z - 1). 66. 2x - 6(z + 1) = 1 - 3(3* - 1). 66. S(w + 2) - 5(2w - 1) = 6(w - 2) + 3. 68. 4r - 9 = 7(2 - r) - 6(r + 1). Solve for P, or A, correct to two decimal places, by first clearing of fractions. 69. 300 = P[l + f (.07)]. 60. 250 = A[l - J(.05)]. 61. 500 = A[l - A(.06) J 62. 750 = P[l + tt(.07) J 56. Simple Factoring of polynomials If all terms of a polynomial contain a common factor, we may express the polynomial as a product of this factor and a second factor. The second factor is the sum obtained by dividing each term of the polynomial by the common factor of the terms. ILLUSTRATION 1. 3x + ax -f bx = z(3 + a + 6). + 2xy + 4zy 3 = xy(3y + 2 At present, in solving equations in an unknown x, we will be in- terested in factoring polynomials only where a: is a common factor of the terms. If we express such a polynomial as the product of x and another factor, we then refer to this factor as the coefficient of x. ILLUSTRATION 2. On factoring, we obtain 2x + 4ox + %cx = 2x(l -f 2a + c). (1) In (1), we say that the coefficient of a: is 2(1 -f- 2a -f c). 57. Constants and variables In a given problem, a constant is a number symbol whose value does not change during the discussion involved. A variable is a number symbol which may take on different values. Any explicit number, such as 7, automatically is a constant wherever it appears. LINEAR EQUATIONS IN ONE UNKNOWN 65 ILLUSTRATION 1. The volume V of a sphere is given by the formula V = Jur 3 where r is the radius. In considering all possible spheres, r and V are variables but TT is a constant, approximately 3.1416. 58. Literal equations Sometimes an equation in an unknown x may involve other literal numbers besides x. In such a case, during the process of solution for x, we assume that the other literal numbers are constants. The summary of Section 55 still specifies our method of solution. EXAMPLE 1. Solve for x: 3bx 2 = 2cx + SOLUTION. 1. Transpose 2 and 2cx: 36z - 2cx = 2 + a. (1) 2. Factor on the left in (1) : x(3b - 2c) = 2 + a. (2) 3. Divide both sides of (2) by (36 2c), the coefficient of x: _ 2 + a X ~ W^2c' 2x 3 x EXAMPLE 2. Solve for x: ^r -- = ao a 2a SOLUTION. 1. The LCD is 2ab. Hence, multiply both sides by 2ab, noticing that 2a6(|) = 4*; 2o6(|) - 66; 2^) - te. We obtain 4z 66 = bx. 2. Transpose terms, and solve for x : fiA 4x - bx = 66; x(4 - 6) = 66; x = ^y 37 5 EXAMPLE 3. Solve for x: s -- =- = 7- (4) 4 SOLUTION. The LCD is 12z. Multiply both sides by 12x: 18 - 28 = - 15*; - 10 = - ISz; x = . (5) CHECK. Substitute x = in (4) : 3 797 5 ____ _ , 5 Right-hand side: T- This checks. 66 LINEAR EQUATIONS IN ONE UNKNOWN Comment. In this chapter, the unknown will occur in a denominator only under the most simple conditions. In solving (4), an incorrect choice of the LCD, containing an unnecessary factor, might have prevented the equations in (5) from being linear in x. 59. Formulas Frequently, the data in a problem come to us as the values of a set of variables, which we represent as literal numbers. Sometimes we are able to write a mathematical expression for one of the variables in terms of the others. The resulting equation is referred to then as a formula for the first variable. An algebraic formula is one involv- ing only the operations of algebra. ILLUSTRATION 1. The Fahrenheit reading, F, and centigrade reading, (7, in degrees for a given temperature are related by the equation 5F = 9(7 -f 160. On solving for F, we obtain a formula for F in terms of C: F = f C + 32. (1) To find F corresponding to 36 centigrade, substitute C = 36 in (1) : F = f (36) + 32 = 64.8 + 32 = 96.8. EXERCISE 22 Solve for x, or y, or z, whichever appears. Reduce any final fraction to lowest terms. 1. bx = 3 H- c. 2. 16z - 4 = h. 3. ex 5a = 3h. 4. ay by = 5. 6. 3x = ax 4- 26. 6. 2z = bz a. 7. 2ay 5c = 3by -f 4a. 8. 7x d = Sax + 8. 9. ax - Sax = 5c - bx. 10. 2dx - 3 = d?x + b. 11. = a. 12. 3x = ~ 13. = d. b be - ax c A ... 26 3x n .,_ a?x , 14. -T j = 0. 15. --- = 0. 16. -r- = 2a 3 . b d c a 3 17 _ _ _ 18 _ = o 19 - - JL f 3T7" mo U. AO rt i \J AU - 2c 6 "' *" 4 ~ BC a x x a 2 24.--- = -1 o 3 2c c 2 1% LINEAR EQUATIONS IN ONE UNKNOWN 67 26. 3a(2x - 36) - 5(cz - 3) = 26. 26. 4c(ax - 6c) - 2a(bx + a 2 ) = 0. 97 4 2 1 5 29 3 V/ - - = - Ho* ' - == - - * 3z 15 3z 24 4x 29 - = + . 10* 12 ^ 15z 662 15az 31. In the Fahrenheit-centigrade equation, 5F = 9C + 160, solve for C in terms of F. Then, use the resulting formula to find the centigrade tem- perature correct to tenths of a degree corresponding to the following Fahren- heit temperatures: (a) 32; (6) 212; (c) 80; (d) 50. 32. Let an object be shot vertically upward from the surface of the earth with an initial velocity of v feet per second, and let us neglect air resistance and other disturbing features. Then, it is proved in physics that s = vt %gP, where s feet is the height of the object above the surface at the end of t seconds and g = 32, approximately, (a) Find s if v 100 and t = 6. (6) Solve for v to obtain a formula for v in terms of s and t. (c) From (6), compute the velocity with which the object must be shot to attain a height of 1000 feet in 5 seconds. 33. Solve / = ma for a. 34. Solve 8 = k -{- vt for v. 36. Solve I = a + (n l)d for a; for ; for d. I M 36. Solve I = for M; for t\ for J. 37. Solve S - - r- for a. r 1 38. Solve S = - r for I: for a. r 1 Each of the following problems states a rule for computing values of a certain variable in terms of others. State the rule by means of a formula. 39. The average, A, of three numbers M, N, and P is one third of their sum. 40. The cost of electricity for a house in a certain city is 7jf per kilowatt- hour for the first 10 kilowatt-hours, 5^ per kilowatt-hour for the next 20 kilo- watt-hours, and 2%i per kilowatt-hour for all over 30 kilowatt-hours. Write an expression for the total cost, C, of (a) 60 kilowatt-hours; (6) n kilowatt- hours where n > 30. 41. On any order for more than 200 units of a certain manufactured product, the cost is 15^ per unit for 200 units and 12^f per unit for the re- main(}er of the order. Write a formula for the cost, C, in dollars, of n units if n > 200. 63 LINEAR EQUATIONS IN ONE UNKNOWN 60. Algebraic translation In applying equations in the solution of problems stated in words, we translate word descriptions into algebraic expressions. ILLUSTRATION 1. If x is the length of one side of a rectangle and if the other dimension is 3 units less than twice as long, then the other dimension is (2x 3) ; the perimeter (sum of lengths of sides) is 2z + 2(2z - 3), or 6z - 6, and the area is x(2x 3). ILLUSTRATION 2. If x, y, and z are, respectively, the units', tens', and hundreds' digits of a positive integer with three digits, the value of the integer is x + Wy + lOOz. SUMMARY. To solve an applied problem by use of equations: 1. Introduce one or more letters to represent the unknowns and give a description of each one in words. 2. Translate the given facts into one or more equations involving the unknowns, and solve for their values. 3. Check the solution by substituting the results in the written state- ment of the problem. EXAMPLE 1. $350 is to be divided between Jones and Smith so that Jones will receive $25 more than Smith. How much does each receive? SOLUTION. 1. Let x be the number of dollars which Smith receives. Then, Jones receives (x + 25) dollars. 2. The sum of the amounts received is $350, or x + (x + 25) = 350. .. (1) On solving (1), we obtain x = 162.50. Hence, Smith receives $162.50 and Jones receives $162.50 + $25 or $187.50. These results check. EXAMPLE 2. Find two consecutive even integers such that the square of the larger is 44 greater than the square of the smaller integer. SOLUTION. 1. Let x represent the smaller integer. Then, the larger integer is x + 2. Their squares are x* and (x -f 2) 2 . 2. From the data, (x + 2) 2 - x 2 = 44. (2) Expanding: x* + 4z + 4 x* = 44; 4z = 40; x = 10. 3. The integers are 10 and 12. LINEAR EQUATIONS IN ONE UNKNOWN 69 CHECK. 10 2 - 100; 12 2 = 144; 144 - 100 = 44. EXAMPLE 3. How long will it take Jones and Smith, working together, to plow a field which Jones can plow alone in 5 days and Smith, alone, in 8 days? SOLUTION. 1. Let x days be the time required by Jones and Smith, working together. 2. In 1 day, Jones can plow J and Smith J of the field. Hence, in x X 3/ days Jones can plow ^ and Smith can plow ^ of the field. 5 o 3. Since the whole field is plowed in x days, the sum of the fractional parts plowed by the men in x days is 1 : x = 3^ days. EXERCISE 23 Solve by use of an equation in just one unknown. 1. A line 68 inches long is divided into two parts where one is 3 inches longer than the other. Find their lengths. 2. A rope 36 inches Sng is cut into two pieces such that one part is 4 inches less than twj 'as long as the other part. Find the lengths of the parts. ;he 3. Find the dinu J P| ( jns of a rectangle whose perimeter is 55 feet, if the altitude is f of the base. 4. One dimension of a rectangle is the other. Find the dimensions of the rectangle if its perimeter becomes 130 feet when each dimension is increased by 5 feet. 5. What number should be subtracted from the numerator of fj to cause the fraction to equal f ? 6. Find two consecutive positive integers whose squares differ by 27. 7. Find three consecutive integers whose sum is 48. 8. Find the angles of a triangle where one angle is three times a second angle and six times the third angle. 9. A rectangle and a triangle have equal bases. The altitude of the rectangle is 25 feet and of the triangle is 20 feet. The combined area of the triangle and the rectangle is 280 square feet. Find the length of the base. 10. J?ind two consecutive positive odd integers whose squares differ by 32,* 70 LINEAR EQUATIONS IN ONE UNKNOWN 11. The length of a rectangular lot is three times its width. If the length is decreased by 20 feet and the width is increased by 10 feet, the area is increased by 200 square feet. Find the original dimensions. 12. A triangle and a rectangle have the same base. The altitude of the rectangle is 4 feet longer, and the altitude of the triangle is 5 feet shorter than the base. The area of the rectangle is 90 square feet greater than twice the area of the triangle. Find the length of the base. 13. A sum of money amounting to $13.55 consists of nickels, dimes, and quarters. There are three times as many dimes as nickels and three less quarters than dimes. How many of each coin are there? 14. If (59 3z) is divided by the integer x, the quotient is 5 and the remainder is 3. Find the value of x. HINT. Recall: dividend = (quotient) (divisor) + remainder. 16. A peddler sold 7 bushels more than f of his load of apples and then had 9 bushels less than f of the load remaining. Find his original load. 16. In 1 hour, Jones can plow J of a field and Roberts -fa of it. If they work together, how long will it take them to plow the field? 17. A room can be painted in 21 hours by Smith and in 14 hours by Johnson. How long will it take them to paint the room working together? 18. How long will it take two mechanical d f .chdiggers to excavate a ditch which the first machine, alone, could con lete in 8 days and the second, alone, in 11 days? Jo* * 19. How long will it take workers A and B, together, to complete a job which could be done by A alone in 7 days, and by B alone in 9 days? 20. How long will it take to fill a reservoir with intake pipes A, B, and C open, if the reservoir could be filled through A alone in 6 days, B alone in 8 days, and C alone in 5 days? 2JL. If 1000 articles of a given type can be turned out by a first machine in 9 hours, by a second in 6 hours, and by a third in 12 hours, how long will it take to turn out the articles if all machines work together? 22. An integer between 10 and 100 ends in 5. By writing an equation, find the integer if it is 5 times the sum of its digits. 61. Percentage The words per cent are abbreviated by the symbol % and mean hundredths. That is, if r is the value of h%, then h % = 156 = ' LINEAR EQUATIONS IN ONE UNKNOWN 7? fi 4 7*i ILLUSTRATION 1. 6% = = .06. 4J% = = .0475. From equation 1, we obtain h = lOOr; hence, to change a number to per cent form, we multiply r by 100 and add the % symbol. ILLUSTRATION 2. If r = .0175, then lOOr = 1.75 and .0175 = 1.75%. 18 9 ILLUSTRATION 3. cnn = xrv\ = - 0225 = 2 - 25 %- oLMJ The description of a ratio M/N in per cent form is the background for the following terminology. If M is described by the relation M = Nr, where r is the ratio of M to N, we sometimes say that M is expressed as a percentage of N, with r as the rate and AT as the base for the percentage: M = Nr, or percentage = (base) (rate) ; (2) M . percentage /0 , r = -TT* or rate = * r - (3) W base v ' ILLUSTRATION 4. To express 375 as a percentage of 500, we compute the rate r = fj$ = .75. Hence, 375 = .75(500), or 375 is 75% of 500. EXAMPLE 1. Find the number of residents of a city where 13% of the population, or 962 people, had influenza. SOLUTION. Let P be the number of residents: .UP - 962; P - , = 7400. .lo lo Note 1. In the formation of a mixture of different ingredients, we shall assume that there is no change in volume. Actually, a slight gain or loss of volume might occur, for instance, as a result of chemical action. In the typical mixture problem, where one special ingredient is involved, the equation for solving the problem frequently can be obtained by writing, in algebraic form, the statement that ( the sum of the amounts of the] _ ( amount of the ingredient } , . \ ingredient in the parts / \ in the final mixture, j If the price of a mixture is the fundamental feature, the equation may be obtainable by using equation 4 with the costs of tHe parts thought of in place of the ingredients. 72 LINEAR EQUATIONS IN ONE UNKNOWN EXAMPLE 2. How many gallons of a mixture containing 80% alcohol should be added to 5 gallons of a 20% solution to give a 30% solution? SOLUTION. 1. Let x be the number of gal. added. In x gal., 80% pure alcohol, there are .80x gal. of alcohol. 2. In 5 gal., 20% pure, there are .20(5) gal. of alcohol. 3. In (5 4- x) gal., 30% pure, there are .30(z + 5) gal. of alcohol. 4. The alcohol in the final mixture of (5 -f- x) gal. is the sum of the alcohol in the x gal. and in the 5 gal. Or, .30(z + 5) = .80* + .20(5); x = 1. EXAMPLE 3. What percentage of a 20% solution of hydrochloric acid should be drawn off and replaced by water to give a 15% solution? SOLUTION. 1. Think of the solution as consisting of 100 units of volume; then the solution contains 20 units of acid. 2. Let x% be the rate for the percentage which should be drawn off. Then, from the 100 units we should draw off x% of 100, or x units. 3. In x units there are .20# units of acid. There remain (20 .20z) units in the final solution of 100 units, after water is added. Therefore, .15 = 2 ; 15 = 20 - .20z; x = 25. Or, we should draw off 25% of the original solution. EXERCISE 24 Change to decimal form. 1. 5%. 2. 4J%. 3. 3|%. 4. 45%. 5. 126.3%. 6. Change to per cent form. 7. .07. 8. .0925. 9. .025. 10. .0575. 11. 1.35. Compute each quantity. 12. 6% of $300. 13. 3^% of 256. 14. 110% of 1250. Express the first number as a percentage of the second. 15. 75, of 200. 16. 400, of 640. 17. 350, of 200. 18. The average price of copper per pound in the United States was approximately $.138 in 1926, $.081 in 1931, and $.215 in late 1947. Find the per cent of change in the price from 1926 to 1931; from 1931 to 1947. Solve each problem by using an equation in just one unknown. 19. If 385 is 85% of x, find x. 20. If 268 is 24% less than y, find y. LINEAR EQUATIONS IN ONE UNKNOWN 73 21. After selling 85% of a stock of dresses, a merchant finds that he has 84 dresses left. What was his original stock? 22. A merchant buys 100 dozen shirts at $13.20 per dozen. He sells 90 dozen at a markup of 30% over the purchase price. At what price per shirt could he afford to sell the remaining 10 dozen to clear his stock if he desires his total receipts from the shirts to be 25% greater than the cost? 23. $3000 of Smith's income is not taxed by the state where he lives. All of his income over $3000 is taxed 2% and all over $8000 is taxed 3% in addition. If he pays a total tax of $800, what is his income? 24. Under the taxes of Problem 23, with an additional surtax of 5% on all income over $20,000, Johnson's tax is $1400. Find his income. 25. A merchant has some coffee worth 70^f per pound and some worth 50^. How many pounds of each are used in forming 100 pounds of a mixture worth 65^ per pound? 26. How many gallons of a mixture of water and alcohol containing 60% alcohol should be added to 9 gallons of a 20% solution to give a 30% solution? 27. How many gallons of a solution of glycerine and water containing 55% glycerine should be added to 15 gallons of a 20% solution to give a 40% solution? 28. How many ounces of pure silver must be added to 150 ounces, 45% pure, to give a mixture containing 60% silver? 29. A feed merchant wishes to form 200 bushels of a mixture of wheat at $1.25 per bushel and wheat at $.80 per bushel, so that the mixture will be worth $1.00 per bushel. How much of each kind should he use? 30. How many pounds of cream containing 35% butterfat should be added to 800 pounds of milk containing 3% butterfat to give milk con- taining 3.5% butterfat? 31. An automobile radiator holds 8 gallons of a solution containing 40% glycerine. How much of the solution should be drawn off and replaced by water to give a solution with 25% glycerine? 32. What percentage of a 30% solution of sulphuric acid should be drawn off and replaced by water to give a 20% solution? 33. What percentage of a 40% solution of alcohol and water should be replaced by pure alcohol, to give a 75% solution? 34. What percentage of a mixture of sand, gravel, and cement, con- taining 30% cement, should be replaced by pure cement in order to give a mixture containing 40% cement? 74 LINEAR EQUATIONS IN ONE UNKNOWN 62. Lever problems A lever consists of a rigid rod with one point of support called the fulcrum. A familiar instance of a lever is a teeterboard. If a weight w is attached to a lever at a certain point, the distance h of w from the fulcrum is called the lever arm of w, and the product hw is called the moment of w about the fulcrum. The following statement is demonstrated in physics. LEVER PRINCIPLE. // two or more weights are placed along a lever in such a way that the lever is in a position of equilibrium, then, if each weight is multiplied by its lever arm, the sum of these products for all weights on one side of the fulcrum equals the sum of the products for all weights on the other side. In other words, the sum of the moments of the weights about the fulcrum is the same on both sides. Note 1. In all lever problems in this book, it will be assumed that the weight of the lever is negligible for the purpose in view. ILLUSTRATION 1. In Figure 2, the sum of the moments for the weights at the left is (5-80 -h 4-250) or 1400, and for those at the right is (4- 100 + 5-200), or 1400. Hence, this lever is balanced. MIAMI Fig. 2 > x- EXAMPLE 1. Two girls, weighing 75 pounds and 90 pounds, respectively, sit at the ends of a teeterboard 15 feet long. Where should the fulcrum be placed to balance the board? SOLUTION. 1. Let x feet be the distance from the fulcrum to the lighter girl. Then, the lever arm for the other girl is (15 x) feet. 2. Hence, from Figure 3, 75* = (15 - z)90; x = 83^ feet. EXERCISE 25 1. A weight of 300 pounds is placed on a lever 20 feet from the fulcrum. Where should a weight of 275 pounds be placed to balance the lever? 2. A weight of 60 pounds is placed on a lever 8 feet from the fulcrum, How heavy a weight should be placed 12 feet from the fulcrum on the other side to give equilibrium? LINEAR EQUATIONS IN ONE UNKNOWN 75 3. A teeterboard is balanced when one girl weighing 80 pounds sits 4 feet from the fulcrum, another girl weighing 100 pounds sits 7 feet from the fulcrum on the other side, and a third girl sits 6 feet from the fulcrum. How much does the third girl weigh? 4. Jones and Smith together weigh 340 pounds. Find their weights if they balance a lever when Jones sits 5 feet from its fulcrum on one side and Smith sits 6 feet from the fulcrum on the other side. 5. A 40-pound weight is placed 6 feet from the fulcrum on a lever, and a 60-pound weight 8 feet from the fulcrum on the other side. Where should a 30-pound weight be placed to give equilibrium? 6. How heavy a weight can a man lift with a lever 9 feet long if the fulcrum is 2 feet from the end under the weight and if the man exerts a force of 140 pounds on the other end? 7. How many pounds of force must a man exert on one end of an 8-foot lever to lift a 300-pound rock on the other end if the fulcrum is 1J feet from the rock? 63. Uniform motion When we say that a body is moving in a path at constant speed, we mean that the body passes over equal distances in any two equal intervals of time. Such motion is referred to as uniform motion in the path. The velocity * or speed or rate of the body in its path is defined as the distance traveled in one unit of time. If v is the veloc- ity, and d is the distance traveled in t units of time, d = vt. (1) j Since v = - , the velocity is referred to as the rate of change of the t distance with respect to the time. In stating a velocity, the units for the measurement of time and of distance must always be mentioned. ILLUSTRATION 1. If an airplane flies 1250 miles in 5 hours at uniform speed, the speed is d 1250 _ n ., , v = - = > or 250 miles per hour. t o The speed of the airplane per minute is 2 - or 4J miles. * In physics, velocity is denned as a vector quantity, possessing both magnitude and direction. In this text, wherever the word velocity is used, it will refer to the magnitude (positive) of the velocity vector. 76 LINEAR EQUATIONS IN ONE UNKNOWN Note 1. All motion considered in this book will be uniform motion. If the velocity of a moving body is variable, a discussion of the motion must bring in more advanced notions met in physics and calculus. EXAMPLE 1. A messenger, traveling at a speed of 65 miles per hour, pur- sues a truck which has a start of 2 hours and overtakes the truck in 3 hours. Find the speed of the truck. SOLUTION. Let x miles per hour be the truck's speed. Then, 3. (65) = (3 + 2)z; 195 = 5z; x = 39 miles per hour. The equation d = rt applies in the discussion of any variable quantity d which changes uniformly at a specified rate r with respect to change in the time t. Thus, we may refer to a rate of increase or a rate of decrease under various conditions. EXAMPLE 2. A motorboat went 70 miles in 4 hours when traveling at full speed upstream on a river whose current flows at the rate of 6 miles per hour. How fast can the boat travel in still water? SOLUTION. 1. Let x miles per hour be the speed of the boat in still water. In travel upstream, the rate of the current reduces the speed of the boat to (x 6) miles per hour. 2. From d = vt, with t = 4, v = x 6, and d = 70, 70 = 4(s - 6). (2) On solving (2) we obtain 4x = 94; x 23.5. Hence, the boat can travel 23.5 miles per hour in still water. 64. Radius of action of an airplane Suppose that an airplane flies with a groundspeed * of Gi miles f per hour in a particular direction from a base B and then back along this path at a groundspeed of (? 2 miles per hour, when the engines are working at full power, and while the wind maintains a constant direction and speed. G\ and <? 2 in general would be different because of the effects of the wind. Suppose that the gasoline tanks of the airplane permit it to operate at full power for only T hours after leaving B. We refer to T as the available fuel hours. Then, it is of interest to consider the maximum length of time, h hours, during which the airplane may fly out if it is to return to B by the end of * Speed with respect to the ground as contrasted to the airspeed, or speed with respect to the air, which itself may be in motion because of a wind. t The subscript 1 on GI is just a tag. We read "Gi" as "G sub 1 " or "G, 1," LINEAR EQUATIONS IN ONE UNKNOWN 77 T hours. The distance R which the airplane flies out from B in h hours is called the radius of action of the airplane in the specified direction, with the given wind. It can be proved * that EXAMPLE 1. How many fuel hours must be available in order to have 900 miles as the radius of action in a direction where the groundspeeds out and back are, respectively, 300 and 200 miles per hour? SOLUTION. Substitute R = 900, Gi = 300, and G z = 200 in the second equation in (1) : _ 7X300X200) 900 " 100 + 200 ' r Hence, T = = 7J hr. EXERCISE 26 1. At what rate does an automobile travel if it goes 450 miles in 9 hours? 2. Jones and Smith travel toward each other from points 500 miles apart, Jones at the rate of 60 miles per hour and Smith at 50 miles per hour. When will they meet if they start at the same instant? 3. Two men start at 7 A.M. from the same place, in opposite directions, at speeds of 36 miles and 48 miles per hour, respectively. When will they be 600 miles apart? 4. At 6 A.M., a motorcycle messenger starts from a city at a speed of 45 miles per hour to meet a regiment which is 120 miles away and is ap- proaching at a speed of 5 miles per hour. When will the messenger meet the regiment? 6. One man can run 400 meters in 54 seconds and a second man can run the distance in 60 seconds. How long will it take the faster man to gain a lead of 12 meters on the slower man if they start together in a 400-meter race? 6. An airplane leaves the deck of a battleship and travels south at the rate of 230 miles per hour. The battleship travels south at the rate of 20 miles per hour. If the wireless set on the airplane has a range of 800 miles, when will the airplane pass out of wireless communication with the ship? * See page 51, in WILLIAM L. HART'S College Algebra, 3d Edition, D. C. HEATH AND COMPANY. 78 LINEAR EQUATIONS IN ONE UNKNOWN 7. How many seconds will it take for a man to travel y miles if he travels x miles in t seconds? 8. In an 800-meter race between two men, the winner's time is 2 minutes, and his lead is 40 meters. How many seconds would it take the loser to run 800 meters? 9. An airplane flew 850 miles in 2J hours against a head wind blowing 30 miles per hour. How fast could the plane fly in still air? 10. Two men start together in a race around a 300-yard oval track, one man at a speed of 9 yards per second and the other man at 7J yards per second. When will the faster man be exactly one lap ahead? 11. When will Johnson be twice as wealthy as Smith if each has $4000 now and if their estates are increasing at the annual rates of $400 for Smith and $1200 for Johnson? 12. Jones can run around a 400-meter track in 65 seconds. How long does Smith take to run the 400 meters if he meets Jones in 35 seconds after they start together in a race around the track in opposite directions? In each problem, (a) find the radius of action for a flight by an airplane in a direction where the groundspeeds have the indicated values; (b) find the number of hours flown on the maximum outward trip. 13. Sixteen fuel hours available; groundspeed out is 240 miles and back is 210 miles per hour. 14. Twelve fuel hours available; groundspeed out is 190 miles and back is 225 miles per hour. 16. Fourteen fuel hours available; groundspeed out is 175 miles and back is 200 miles per hour. // an airplane is to have the specified radius of action in a direction corre- sponding to the given groundspeeds, find the number of fuel hours which must be available. 16. Radius of action is 1350 miles; groundspeed out is 200 miles and back is 225 miles per hour. 17. Radius of action is 750 miles; groundspeed out is 195 miles and back is 175 miles per hour. 18. How many fuel hours must be available to permit an airplane flight out from a field for 5J hours in a direction such that the groundspeed out is 200 miles and back is 185 miles per hour? 19. At how many minutes after 2 P.M. will the minute hand of a clock overtake the hour hand? 20. After 10 P.M., when will the hands of a clock first form a straight line? LINEAR EQUATIONS IN ONE UNKNOWN 79 65. Interest Interest is income received from invested capital. The capital originally invested is called the principal. At any time a'fter the investment of the principal, the sum of the principal and the interest due is called the amount. The rate of interest is the ratio of the interest earned in one year to the principal. If r is the rate and P is the principal, then _ interest per year _ T - J interest for one year = Pr. (2) Thus, the interest is a percentage of the principal, with r as the rate. ILLUSTRATION 1. If $1000 earns $36.60 interest in one year, then, from equation 1, r =^^ = .0366, or r = 3.66%. 66. Simple interest If interest is computed on the original investment during the whole life of a transaction, the interest earned is called simple interest. Suppose that P is invested at simple interest for t years at the rate r. Let / be the interest and F be the final amount at the end of the t years. Then, the interest for one year is Pr and, by defini- tion, the simple interest for t years is t(Pr) or Prt', that is, / = Prt. (1) Since amount equals principal plus interest, F = P + /. (2) From (1), P + / - P + Prt = P(l + rt). Hence, from (2), F = P(l 4- rt). (3) In equations 1 and 3, t represents the time expressed in years. If the time is described in months, we express it in years assuming a year to contain 12 equal months. If the time is given in days, there are two varieties of interest used, ordinary and exact simple interest In computing ordinary interest we assume a year to contain 360 days, and, for exact interest, we assume a year to contain 365 days. 80 LINEAR EQUATIONS IN ONE UNKNOWN To find the amount F when P, r, and t are given, first find the interest from 7 = Prt and then compute P + / to find F. Note l'. Unless otherwise specified, the word "interest" in this book will refer to simple interest. ILLUSTRATION 1. If $5000 is invested for 59 days at 5%, the ordinary interest due is 5000(.05)^y = $40.97; the final amount due is 5000 + 40.97 = $5040.97. EXAMPLE 1. If $1000 accumulates to $1250 when invested at simple interest for 3 years, find the interest rate. SOLUTION. 1. We have P = $1000; F = $1250; I = 1250 - 1000 = $250. 2. From 7 = Prt with t = 3, OKA 250 = 1000(r)(3); 250 = 3000r; r = = .08J = In F = P(l + rt), the principal P is frequently called the present value of the amount F because, if P is invested today at the rate r, the accumulated amount at the end of t years will be F. EXAMPLE 2. Find the present value of $1100 which is due at the end of 21 years, if money can be invested at 4%. SOLUTION. 1. We have F = $1100, r = .04, and t = 2. Hence, from (3), 1100 = P[l + f(.04)]; 1100 = P(l + .10); 1.1P = 1100; P = ~ = $1000. JL .L CHECK. 7 = 1000(.04)(f) = $100; F = 1000 + 100 = $1100. EXERCISE 27 Find the ordinary interest and the final amount. 1. On $5000 at 6% for 216 days. 2. On $8000 at 4J% for 93 days. Find the exact interest and the final amount. 3. On $3000 at 4% for 146 days. 4. On $2500 at 51% for 27 days. 6. Find the amount due at the end of 8 months if $150 is invested at 9%. With the given data, solve F = P(l + rt) for P, to the nearest cent. 6. F = $1000; r = .03; t - f 7. F = $3000; r = .05; t = LINEAR EQUATIONS IN ONE UNKNOWN B1 8. At what rate will $750 be the interest for 5 years on $6000? 9. Find the invested principal if it earns $375 interest in 3 months when the interest rate is 10. Find the principal if it earns $150 interest in J year at 8%. 11. (a) Find the principal which will amount to $1300 by the end of 6 years when invested at 5%. (6) Verify the result by computing interest on it for 6 years. 12. Find the present value of $1888 which is due at the end of 4 years, if the interest rate is 4%. 13. Jones agreed to pay Smith $6000 at the end of 5 years. What should Jones pay immediately to cancel his debt if Smith agrees that he can invest money at 4%? 14. Roberts buys a bill of goods from a merchant who asks $2000 at the end of 2 months. If Roberts y wishes to pay immediately, what should the seller be willing to accept if he is able to invest his money at 8%? 15. A debtor owes $1100 due at the end of 2 years and he requests the privilege of paying an equivalent smaller sum immediately. At what simple interest rate would the creditor prefer to compute the present pay- ment, at 5% or at 6%, and how much would he gain by the best choice? 16. How long will it take a given principal to double itself if invested at 5% simple interest? 17. A man invests $7000, one part of it at 5% and the balance at 4%. If the total annual interest is $320, how much is invested at each rate? CHAPTER 5 SPECIAL PRODUCTS AND FACTORING 67. Square root If R 2 A, we call R a square root of A. ILLUSTRATION 1. 4 is a square root of 16 because 4 2 = 16. Every positive number A has two square roots, one positive and one negative, with equal absolute values. The positive square root is denoted by H- VZ, or simple VA, and the negative square root b VZ. We call VZ a radical and A its radicand. Unless other- wise stated, the square root of A will mean its positive square root. By the definition of a square root, If x is positive or zero, ~ - x. (2) ILLUSTRATION 2. V9 = 3 because 3 2 = 9. The two square roots of 9 are V9 and V, or V9, or 3. We read " rfc" as "plus or minus." ILLUSTRATION 3. VJ J because (|) 2 = 4- 68. Perfect squares In the square of an integral rational term, each exponent will be an even integer because, in squaring, each original exponent is multiplied by 2. ILLUSTRATION 1. (3xV) 2 = 3 2 (x 2 ) 2 (i/ s ) 2 An integer is said to be a perfect square if it is the square of an integer. The student should learn the most common perfect square integers, with the aid of Table I, page 283. SPECIAL PRODUCTS AND FACTORING S3 ILLUSTRATION 2. The perfect square integers are 1, 4, 9, 16, 25, 36, etc. Their square roots are, respectively, 1, 2, 3, 4, 5, 6, etc. Thus, \/25 = 5 because 5 2 = 25. An integral rational term is said to be a perfect square if it is the square of some other term of the same variety. Hence, in a perfect square, each exponent is an even integer. ILLUSTRATION 3. 25a 2 6 4 is a perfect square because 25a 2 6 4 = (Safe 2 ) 2 . I. To find the square root of a perfect square monomial: 1. Rewrite the literal part with each exponent divided by 2. 2. Multiply by the square root of the numerical coefficient of the given term. ILLUSTRATION 4. Vl6x*y* = VT6Vx*y* = 4#y, because II. To find the square root of a fraction, find the square root of the numerator and of the denominator and divide: / - v^4 2 ILLUSTRATION 5. \/T^ = 7= = ^* V25 5 IWQa 6 \/100a 6 10o 3 ILLUSTRATION 6. Note 1 . For the present, we shall consider \/A only where A is a perfect square monomial, or where A is a fraction whose numerator and denominator are perfect square monomials. All literal numbers in A will be supposed positive or zero. EXERCISE 28 Find the two square roots of the number and check by squaring the results. 1. 25. 2. 49. 3. 121. 4. 64. 6. fc. 6. Find each square root and check by squaring the result. Inspect Table I if necessary. 7. V9. 8. VIOO. 9. \/81. 10. \/l44. 84 SPECIAL PRODUCTS AND FACTORING 11. Vl96. tf. Vf. 13. \/if. 14. 15. V. 16. Vg. 17. VS- 18. 19. Vz 4 . 20. V. 21. Vo 2 ". 22. 23. Vtf*. 24. V^ 5 . 25. Viol 26. 27. \/4^. 28. \/16^. 29. \/49?. 30. 31. \64a. 32. fo 2 . 33. \49w*c 4 . 34. 39. tv 40. t/=?. 41. \(1- each quantity. 51. (\/37) 2 . 52. (\/142^) 2 . 63. (V^) 2 . 64. (>/659)2. 55. A negative number can have neither a positive nor a negative square 'root. Why is this true? 69. Products of binomials When desired, the product of two binomials may be found by longhand methods. ILLUSTRATION 1. (3s - 5y)(2x - 7y) = 3x(2x - ly) - 5y(2x 7y) = 6x 2 - 21xy - Wxy + 35t/ 2 = 6x 2 - Zlxy + 35y 2 . ILLUSTRATION 2. (x -f y)(x y) = x(x y) + y(x y) x z xy + xy y z = x 2 y 2 . ILLUSTRATION 3. (a + 6) 2 = (a -f fe)(a + 6) = a(o + 6) + 6(a + 6) = a 2 + 2ab + ft 2 . 70. Special products The student should be able to dispense with the longhand methods of the preceding section and should form the product of two bi- nomials mentally. Products of the following types occur frequently. The student should verify each right-hand member. SPECIAL PRODUCTS AND FACTORING 85 I. a(x + y) = ax -f II. (*4-y)(*-if) = x 2 - III. (a + b)* = a 2 + 2a& + IV. (a - &) 2 = a 2 - 2ab + V. (x + a)(x + 6) = x 2 + (ax -f 6x) -f ab. VI. (ax + b)(cx + d) = acx 2 -f (aa*x + bcx) + 6<f. It proves convenient to memorize Types II, III, and IV as formulas and also in words. ILLUSTRATION 1. Type II states that the product of the sum and the differ- ence of two numbers is the difference of their squares. ILLUSTRATION 2. Type III states that the square of the sum of two numbers equals the square of the first, plus twice the product of the numbers, plus the square of the second number. ILLUSTRATION 3. (c - 2d)(c + 2d) = c 2 - (2d) 2 = c 2 - 4#. (Type II) ILLUSTRATION 4. From Type III with a = 3x and 6 =* 2y, = 9x 2 The right members of Types V and VI should not be committed to memory. However, the nature of the right members should be memorized, with (ax 4- bx) in Type V and (adx + bcx) in Type VI remembered as the sum of the cross products. ILLUSTRATION 5. To obtain (2# 5) (3x + 7) : f 2* -5 ' j^XT I ( Product = 6z 2 x 35. 3z -F7 Sum of the cross products: I4x 15x = x. The diagram and auxiliary computation of the sum of the cross products should be omitted and replaced by mental computation as in the next il- lustration. ILLUSTRATION 6. (2x - 7fc)(3z -f 2h) = 6x 2 - I7hx - because the sum of cross products is 2lhx + 4hx, or I7hx. ILLUSTRATION 7. (x 2 - 3y*) 2 - (z 2 ) 2 - 2(a; 2 )(3^) + W) 2 (Type IV) 86 SPECIAL PRODUCTS AND FACTORING ILLUSTRATION 8. (x 2 - 2y)(x* + 2y)(x* + 4t/ 2 ) (Type II) /)](z< + 4s/ 2 ) = (x* - 4i/ 2 )(z< -f 4s/ 2 ) - (4i/ 2 ) 2 - x 8 - ILLUSTRATION 9. (- 3x - 4)(- 3s + 4) = - (4 + 3z)(4 - 3x) = - (16 - 9z 2 ) = - 16 + 9z 2 . EXERCISE 29 Expand and collect terms, performing as miich of the work as possible mentally. 1. 5(3a - 4t>). 2. 3c(2 - 6c). 3. ab(4x - ax). 4. - 5x(2y - 3z). 6. (c - d)(c + d). . 6. (h - 2k)(h + 2k}. 7. (a + y)\ 8. (c - 3x)(c + 3x). 9. (4 - y)(4 + y). 10. (5 - 2y)(5 + 2y). 11. (3 + 2r)(3 - 2r). 12. (3x - 4) (3 + 42). 13. (a 2 - 36) (a 2 -f 36). 14. (a6 - 2) (06 + 2). 16. (a - 2) (a - 4). 16. (c + 3). 17. (x + 5) 2 . 18. (y - 4) 2 . 19. (2a - 5) 2 . 20. (3x - 2) 2 . 21. (2z - 22. (3x - 4y) 2 . 23. (2a + 6) 2 . 24. (x - 26. (3 -h *)(2 4- x). 26. (2x + 5y)(- 2x - 5y). 27. (x - 5)(* -f 9). 28. (x + 13)(* - 4). 29. (a + 26) (a + 36). 30. (w - 2z}(w -f &). 31. (2x + 3)(3x + 4). 32. (3x - 5)(2x - 3). 33. (4y - 3x)(2y - x). 34. (2^/ + w)(y + 5w). 36. (2y - 3)(3y -h 5). 36. (y - 3)(2y + 7). 37. (3u> -f 5)(7w ~ 2). 38. (3 - 4z)(2 + 5x). 39. (4x - 3y)(2x + 3y). 40. (3u - 5w)(2w + 3io). 41. (6 - 5x)(~ 2 -h x). 42. (- 3 - 2x)(2 - x). 43. (3* - 4)(- x + 5). 44. (- y - 3)(fy + 4). 45. (x 2 -h 2) 2 . 46. (4 + 36 2 ) 2 . 47. (2xy - 48. (4a^ - 2/) 2 . 49. (3 + 46x) 2 . 60. (x - 2wxr 2 ) 2 . SPECIAL PRODUCTS AND FACTORING 87 61. (x - J) 2 . 62. O/ + i) 2 . 63. (J - 2*)'. 64. (we 2 - 2a)(wx* 4- 2o). 66. (cd - 3^)(cd + 3x). 66. (x + .l)(z + .5). 67. (a; - .2)(x -f .5). 68. (.3 + z)(.2 - *). 69. (3 - .2z)(2 -f .5x). 60. (a - 6)(o + b). 61. (Jo - #>)(Ja + J6). 62. (& + fe)(fz - /). 63. (.4* - .3)(.2x - .5). 64. (x - y)(x 4- 2/X* 2 + ^ 2 ). 66. (2 - a; 2 )(2 + z')(4 66. (w - 3)(w 4- 3)(w> 2 + 9). 67. - 7x(2ax - 3x 2 - 68. - 3yz(2y* - 3yz + 2 2 ). 69. (- 3 - 4)(- 2 -f fa). 70. (- x - t/) 2 . 71. (- 2s - 3y) 2 . 72. (- 3 - 73. [2(* - ?/)] 2 . 74. [3(o + 6)] 2 . 76. [5(2c 76. (3o: 2 - 8)(x* + 2). 77. (4z 2 - 3)(3x 2 + 2). 78. (2x 2 - 3i/ 2 )(x 2 -f 4?/ 2 ). 79. (2a 2 + 56 2 )(a 2 - 36 s ). 80. (z 3 + 3)(3x 3 - 4). 81. (3a - 26 s ) (7a 3 + 66 s ). 82. (2u 4 - 3v 2 )(3w 4 + 2v 2 ). 83. (4x - 3^)(3x + 2y*). 84. (2x 2 - 5i/ 2 )(2z 2 -f 5y 2 ). 86. (3w 2 - 7v 2 )(3w 2 -f 2V 2 ). 86. (2a6c 2 - 71. Grouping in multiplication The method of the following illustrations is particularly useful in applications of Types II, III, and IV of Section 70. ILLUSTRATION 1. (c -f 2d - lla)(c -f 2d -f Ho) = [(c + 2d) - llo][(c + 2d) -f- llo] (Type II) = (c + 2d) 2 - (Ho) 2 = c 2 + 4cd + 4cP - 121o 2 . ILLUSTRATION 2. (2x + y - 3z) 2 = [(2x + y) - 3] 2 (Type IV) = (2x + t/) 2 - 2(32) (2* -f y) + (3s) 2 = 4s 2 . + 4x + t/ 2 - 12x2 - 6z + 9s 2 . ILLUSTRATION 3. (a + 6 + c + 2) = [(a + 6) + (c + 2)? (Type III) - (a + &) 2 + 2(o + 6)(c 4- 2) + (c + 2) 2 = a 2 4- 2o6 4- 1 2 4- 2oc 4- 4o 4- 26c 4- 46 4- c 2 4- 4c 4- 4. 88 SPECIAL PRODUCTS AND FACTORING ILLUSTRATION 4. (2x - 3 + 2y)(2x + 3 - 2y) = [2* - (3 - 23,)] O + (3 - 2y)] - (3 - 2y) 2 = 4x* - 9 4- I2y - EXERCISE 30 Expand and collect terms by use of preliminary grouping. 1. [(* 4- y) 4- 2] 2 . 2. [(o - 6) + 5J. 3. [3 - (2x - y)J. 4. (2 + a 4- w)\ 6. (3s + y + 5) 2 . 6. (* - 2y - 3) 2 . 7. (4a - b - c) 2 . 8. (- 2 + a + 6) 2 . 9. [2x - 3 (a - fc 2 )] 2 . 10. (2x - 3* 2 + 3?/) 2 . 11. [(ar + y) - 3][> + y) + 3J 12. [(c + 2x) - 2][(c + 2x) + 2], 13. [4 - (2o + 6)] [4 + (2o + 6)1 14. (a + w -f 4) (a + w - 4). 15. (a + 6 - x)(a + b + x). 16. (3* + y - 2) (3* + y + 2). 17. (3a - y + 4)(3o - y - 4). 18. (a - fc 2 + )(a -f t 2 4- 2). 19. (z 2 - y + 2)(a; 2 + y - ). 20. [(a 4. 6) + (c - 3)] 2 . ' 21. [2s + y + o - 3] 2 . 22. (a 4- c 4- 6 - 5) 2 . 23. (2x - z + y - 2) 2 . 24. (o 4- b 4- c 4- d)(a + 6 - c - d). 26. (2x + y-z + 3)(2x + 2 26. (a + 3y 5)(a 4- 3y - 27. (c - 2d - a - x)(c 2d + a + x). 28. Expand (x -\- y + z) 2 and state the result in words. Use the formula of Problem 28 to expand each square. 29. (2a 4- 36 H- 4c) 2 . 30. (3a - 26 4- 3c) 2 . 31. (w - to 4- 3a) 2 . 32. (42 s - 3*i/ 2 - 5X 8 ) 2 . 72. Terminology about factoring In our discussion of factoring, unless otherwise stated, the coeffi- cients will be integers in any polynomial referred to. Such an expres- sion will be called prime if it has no integral rational factors except itself, or its negative, or 1. No simple rule can be stated for determin- ing whether or not an expression is prime. SPECIAL PRODUCTS AND FACTORING 89 ILLUSTRATION (. We shall say that (x y) is prime although x - y = (Vx -f Vy)(Vx - Vy), because these facltors are not integral and rational. Other prime expressions are (x + y), (x* --f- y 2 ), (z 2 + xy + ?/ 2 ), and (x* xy + To factor a polynomial will mean to express it as a product of positive integral powers of distinct prime factors. ILLUSTRATION 2. To factor 4c 4 46 2 z 2 , we write After an expression has been factored, the factors should always be verified toy multiplying them to obtain the given expression. 73. Factoring by inspection Each ty/pe formula of Section 70 becomes a formula for factoring when read! from right to left. I. ax + ay + az = a(x -f y -f z). iLLUS-niiATiON 1. by 2 + 3y + % 2 = t/(6?/ + 3 + Sy). ILLUSTRATION 2. If a factor 2x z y 3 is removed from the term the remaining factor can be verified by division: Hence, 14xV = 2xV(7x?/ 2 ). ILLUSTRATION 3. In the following factoring, we remove the common factor hxy 2 from each term : 3 - II. T/ie difference of two squares equals the product of the sum and the difference of their square roots: ILLUSTRATION 4. # 2 9 = (x 3)(z -f 3). ILLUSTRATION 5. To factor 25s 2 Oy 4 , we observe that 25a? = (6x) 2 and Oy 4 = (32/ 2 ) 2 . Hence, H- ILLUSTRATION 6. a 4 IGt/ 4 = (a 2 - 4y 2 )(a 2 4- - (a - 90 SPECIAL PRODUCTS AND FACTORING 74. Perfect square trinomials * j An integral rational polynomial with three tei;ms is called a trinomial. The square of any binomial is a perfect s}quare trinomial. A trinomial of this type can be recognized and factored by the formulas of Types III and IV of Section 70. Perfect square trinomials: III. a 1 + 2ab + fc* = (a + &); IV. a 1 - 2a& + b* = (a - &). In a perfect square trinomial, we notice that 1. two terms are perfect squares, and 2. the third term is plus (or minus) twice the product of the square roots of the other terms. To verify that a trinomial is a perfect square, take the square roots of the terms which are perfect squares, compute the third tlerm which should be present, and check by inspection. \ ILLUSTRATION 1. To factor 4z 2 2Qxy -f 25y 2 , we observe perfect squares 4z 2 and 25y 2 , whose square roots are 2x and 5y. Hence the third term should be 2(2x)(5y) = 2Qxy, which checks, and gives - 2Qxy + 25y* = (2x - ILLUSTRATION 2. 162 4 + 242^ + 9w 2 = (4Z 2 -f 3w)*. ' EXERCISE 31 Factor by use of Types I and II. // fractions occur, leave the factors in the form which arises most naturally by standard methods. Check by multiplying the factors. \ bx. 2. 2cx + 4<fc. 3. 6zy 2 -f 2ax. 4. bx -4- x -f- c 2 ^. 5. 2cy 6. 3o& + 2a - 5a 2 . 7. - ac + 3bx + ex 8. - 5y* - 3y -f ay 2 . 9. - 4o< -f < 2 - cf 3 . 10. 46V -f 6te 2 + 86cx 2 . 11. 3aV - 2o?/ 2 + ay * See Note 5 in the Appendix for an explanation of the process for finding the square root of a number expressed in decimal notation, by means of pure arith- metic. This process is intimately related to the formula of Type III. SPECIAL PRODUCTS AND FACTORING 91 12. 6aV - 3<w 2 + 40^. 13. 2t0 4 z - 6w*c 2 + 5w*r*. 14. x 2 - a 2 . 15. w> 2 - s 2 . 16. y* - 25. 17. 64 - x*y\ 18. 36 - 2 s . 19. 4x* - y\ 20. 9x 2 - 25s 2 . 21. 36# - 121. 22. 9s* - 1. 23. 4a 2 - 9ft 2 . 24. 1 - 25x 2 . 25. 256a* - 1. 26. 9s 2 - J. 27. J - w*. 28. 9a 2 6*u> - 16s*u>. 29. 25W? 2 - c 2 ^. 30. 49u 2 - 16v^. 31. 36a 2 6 s - 64x 8 . 32. 9 2 - 144a 2 6 2 . 33. ax* - ay*. 34. Which trinomials are not perfect squares? 36. x 2 + 3x + 4. 36. a 2 + a + 1. 37. 4z 2 + 6z + 9. 38. 9x 4 - 6x 2 39. 3z 2 + fay + 4y*. 40. 4z 2 - Insert any missing term to complete a perfect square. Then factor by use of Types III and IV. 41. x 2 -f 2bx + b*. 42. rf 2 + 2<fy + y 2 . 43. a 2 - 2a + 1. 44. x 2 -f ( ) + 16. 46. u? - ( ) -h 36. 46. 4z 2 - 20xz + 25 8 . 47. x 2 -h 81 - l&c. 48. x 2 + x + J. 49. 49x 2 + 14ox + a 2 . 60. 1 + z 2 z 2 - 2x2. 51. 64 - 16a6 + a 2 6 2 . 62. 9a 2 + ( ) + 256. 53. 4x 2 + ( ) + 9 2 . 64. 166 2 - ( ) + 49x 2 . 66. 4c 2 (P - ( ) + 25a 2 . 56. 9x 2 - ( ) + hW. 67. - 30xy + 9x 2 + 25y 2 . 68. 24ax + 9x 2 + 16a 2 . 69. 4x 4 - 28x 2 + 49. 60. 25 - 30x 2 + 9x 4 . 61. 4a 4 - 12a*b* + 96 4 . 62. Factor. 63. 49 2 - 4ft 2 . 64. 75o 2 - 3a 2 6 2 . 65. . 25s 2 - 30^2 + 9u 2 . 67. x 2 -f 25^ - 68. 9X 4 + 49s/ 4 - 42xV- 60. 4a 2 x - 4ax -f x. 92 SPECIAL PRODUCTS AND FACTORING 70. x* - 9y*> 71. 98u 4 - 50. Y2. 9oz 2 - lay*. 73. 25x 2 - 1006 4 . 74. 3a 2 x 2 - 5ay. 76. 16x 4 - 76. 25wV - TOu 2 ^ 2 + 49t^. 77. 18u 2 - 60wt> 78. 2w 6 - 12t*V + 18y 8 . 79. 147x 2 - First factor and then compute. Check by expanding the original expression. 80. 23 2 - 17 2 . 81. 52 2 - 48 2 . 82. 27 2 - 23 2 . 83. 104 2 - 96 2 . 84. 45 2 - 55 2 . 85. 37 2 - 33 2 . 75. Factoring trinomials by a trial and error method We recall the formulas of Types V and VI of Section 70. V. x* + (a + b)x + ab = (x + a)(x + b). VI. flex* -f (ad + bc)x + bd = (ax + b)(cx + d). Certain trinomials of the form * gx* -f hx + k can be factored by a trial and error method suggested by the preceding formulas. \ EXAMPLE 1. Factor: x 2 2x 8. SOLUTION. 1. We wish to find a and 6 so that (x + a)(x + 6) = x* + (a + 6)a? + ab = x 2 - 2x - 8. 2. Hence, ab = 8; thus a and 6 have opposite signs and are factors of 8. Since the sum of the cross products is 2x, we guess that a = 4 and 6 = 2. This is correct because (x - 4)(* + 2) = x* - 2x - 8. EXAMPLE 2. Factor: 15x 2 + 2z 8. SOLUTION. 1. We wish to find a, 6, c, and d so that (ax + 6) (ex -f d) = oca; 2 + (ad + bc)x + bd = 15z 2 + 2z - 8. Hence, oc = 15, 6d = 8, and the sum of the cross products is 2x. 2. First trial. Since oc = 15, choose a = 15 and c = 1 ; since bd = 8, choose 6 = 2 and d = 4. This selection is wrong because (15z + 2)(z - 4) = 15s 2 - 5&c - 8. 3. Second trial. Choose a 3, c 5, 6 = 2, and d = 4. This selection is correct because (3x - 2) (fir + 4) = 15x 2 -f- 2z - 8. * If g, h, and k were chosen at random, without a common factor, the trinomial would probably be prime. Later, we shall discuss a condition which g, h, and A; satisfy when and only when the trinomial is not prime. SPECIAL PRODUCTS AND FACTORING 93 If one prime factor is merely the negative of another, we do not consider them as distinct prime factors; we combine their powers into a single power of one of them. ILLUSTRATION 1. In ( x 2}(x -f- 2) = x* 4x 4, we notice that ( x 2) = (x + 2). Hence, we write - x z - 4x - 4 = - (x + 2)(z + 2) = - (x + 2) 2 . Note 1. The preceding factoring methods apply to polynomials in which the coefficients are any real numbers, not merely integers as in the illus- trations. The nature of the coefficients which we agree to allow in a poly- nomial and its factors affects our definition of a prime expression but not our general factoring procedure. EXAMPLE 3. Factor: 6z 4 x* 15. SOLUTION. By trial and error, 6s 4 - z 2 - 15 = (3z 2 - 5)(2z 2 + 3). EXERCISE 32 Factor by trial and error methods. 1. x* + Sx + 15. 2. z 2 -f lOz + 21. 3. a 2 - 8a + 12. 4. y* - 1y + 12. 6. x 2 - Sx + 15. 6. 2 2 - 52 - 6. 7. < 2 + 4* - 21. 8. w* - 5w - 24. 9. z 2 - 3x - 18. 10. a 2 + 6a - 16. 11. w* + 2w - 48. 12. 4 - 3y - ^ 2 . 13. 15 - 2w - w> 2 . 14. 8 - 7a - a 2 . 16. 24 + 2w - w 2 . 16. 6 2 + 36 - 28. 17. 32 - 4y - y\ 18. 27 + Qw - w? 2 . 19. 54 - 3& - fc 2 . 20. 36 + 5h - K. 21. x 2 - 6z - 72. 22. '2o: 2 + 7z + 3. 23. 5a 2 + 12a + 7. 24. 3o 2 + 8a 4- 5. 25. 10z 2 - llz + 3. 26. 3a 2 - lOa + 7. 27. 8x 4 - 10z 3 + 3z 2 . 28. 2x* - x* - 29. 3 + 2 2 - 5. 30. 94 SPECIAL PRODUCTS AND FACTORING 31. 3x* + x 9 - 10. 32. 15j/ 2 4- 4y - 4. 33. 8u* 6I0 8 9. 34. 5 4- 3x 4- 2z 2 . 35. 150 4 - a 2 - 28. 36. 8 4- 2y 2 - 15^. 37. 7 - 19z - 6z 2 . 38. - 12fc 2 - 8fc 4- 15. 39. 27z 2 4- 3x + 2. 40. 5a 2 4- J2a& 4- 7b*. 41. 3s 2 4- 5xy 4- 2y 2 . 42. 3x 2 4- 7ax 6a 2 . 43. Sw 2 4- 14u 15z 2 . 44. ISwr 4 4- 9t^ 20. 45. 5w 2 28wu> 4- 12^. 46. 45fc 2 Sxy 4t/ 2 . Factor by the appropriate method. 47. 6a 2 - 13o6 4- S6 2 . 48. 4a; 2 - 7xy + 3t/ 2 . 49. lOOa 2 - x*. 60. 49 2 - 46 2 . 61. 52. Tc 2 + 19cd - 6tf. 63. 64a 2 - 48ac + 9c 2 . 64. - 2z 2 + 15 4- x. 65. - 6z 2 + 20 - 7*. 66. 9 4- 250 2 io 2 30tw. 57. 2x? 58. 8a 2 c - 18c. 59. W + 60. 3a 4- 13a6 4- 10a6 2 . 61. 25x 2 - 1006 4 . 62. 75ccP + 30c 2 rf + 3cX 63. 2r - llhr 4- ISAV. 64. .Sic 4 - .16d 4 . 65. ^ - 16y*. 66. Sx 4 - 16x 2 H- 3. 67. 31x - 5x 2 - 3. 68. xV 4- 9xy 52. 69. 3s 4 7z 2 20. 70. 6s - 1 9s*. 71. 72. 9s 2 * - 4. 73. 74. - 4x 2 4- 12s - 9. 76. - 9a 2 4- 30a6 - 2S6 2 . 76. 3s 4 - 17x 2 -f 10. 77. 2X 4 + x 2 - 15. 78. 3s 4 - 5xV - 2y*. 79. 3O 4 76. Factoring by use of grouping The following methods make frequent use of the fact that an ex- pression enclosed within parentheses should be treated as a single number expression. - SPECIAL PRODUCTS AND FACTORING . 95 ILLUSTRATION 1. 5(x a) 3(x a) *= 2(x a). ILLUSTRATION 2. To factor the following expression, we observe the common factor (a 6), and remove it from each term: 2c(o - b) 4- d(a - b) - (a - 6)(2c + d). ILLUSTRATION 3. After grouping, we observe a common binomial factor, and then complete the factoring: bx + by 4- 2&c 4- 2hy = (bx + by) + (2kc + 2%) (* 4- y) = (6 + 2A)(* + y). ILLUSTRATION 4. The second term below was altered by changing signs (or, multiplying by 1) both within and without the parentheses in order to exhibit the same binomial factor as the first term: a - 6) + 40(6 - 2a) - 3x(2a - b) - 4y(2a - b) 4t/)(2a - 6). ILLUSTRATION 5. In order to factor below, we group two terms within parentheses preceded by a minus sign, and hence change the signs of the terms, in order to exhibit the same factor as observed in the other terms: xz kx 4- kw wz (xz wz) (kx kw) z(x w) k(x w) = (z k)(x w). ILLUSTRATION 6. 6 - 3s 2 - 8x + 4z 3 = (6 - Sx) - (3z 2 - = 2(3 - 40) - x*(3 - 4c) = (3 - 40) (2 - &) ILLUSTRATION 7. We factor below as the difference of two squares: (c - 2z) 2 - (6 - a) 2 = [(c - 2x) - (b - a)][(c - 2) + (6 - a)] = (c - 2x - 6 + a)(c - 2x + 6 - a). ILLUSTRATION 8. a 2 - c 2 + 6 2 d 2 - 2db - 2cd = (a 2 - 2a6 -|- 6 s ) - (c 2 + 2cd + <P) = (a - 6) 2 - (c -f d) 2 = [(a - 6) - (c + d)][(a - 6) + (c + = (a 6 c d)(o 6 + c H EXERCISE 33 Factor. - 5(* 4- 2|/). 2. 4(3A + *) - 9(3A 4- 3. c(x 4- y) 4- <*(* 4- J/). 4- 5a(c - 3d) ~ 26(c - 3d). 6. 2A(m - 2) - 3fc(m - 2). 6. 2c(x 4- 40) 96 SPECIAL PRODUCTS AND FACTORING 7. - 5c(r + ) + 2d(r + s). 8. - 2x(a + h) - 3y(a + h), 9. 3A(w ) (> *) 10. 2x(h - 2fc) + 3% - 6ky. 11. 3a(w - 2k) + 26u> - 46fc. 12. fee + fa/ -h 2hx + 2%. 13. 3ac + 3bc + ad + bd. 14. 2a + 2ay + bx + by. 15. cr cs + 3dr 3ds. 16. 4&c 46A + bcx 56c. 17. 2cx + cy 2dx %. 18. box + 26z lOod 4bd. 19. 4Ax 4bh 8cx + 8bc. 20. 36tt> 360 4aw -f 4az. 21. (x 3 - 2a; 2 ) - (x - 2). 22. (ax 3 + fee 2 ) - 4(ax + b). 23. x* + 2x* + x + 2. 24. ax 2 + 6z 2 -h ad 2 + b&. 26. x 8 - 3x 2 + x - 3. 26. 2x 2 - 4x + 1 - 8z 3 . 27. a 3 - 3a 2 - 3 H- a. 28. 2 + 4x - lOx 4 29. 3X 3 - 2x 2 + 6x - 4. 30. 4 - &c 2 - 5x + 31. 2(r s) x(s r). 32. a(x y) + 6(?/ 33. x 2 - (s H- 3) 2 . 34. (w - I) 2 - 16& 2 . 35. (2z + wY - y*. 36. (4a - fe) 2 - (2x - ?/) 2 . 37. (c - 3d) 2 - (2x + y) 2 . 38. (4x - 3y) 2 - 25. 39. z 2 + 20 H- 1 - 9* 2 . 40. 4i^ + 20w + 25 - 81 2 . 41. 2/ 2 + 2y + 2 2 - 4x 2 . 42. 9w> 2 - 4a 2 - 4ab 6 2 . 43. 4a 2 - 92 2 - 62 - 1. 44. 16i/ 2 - a 2 -f 2ab - fe 2 . 46. 9x 2 2/ 2 + 2y z 2 . 46. t^ 2 4x 2 y* 4xy. 47. 16a 2 - 1 - 9x 2 H- 6x. 48. a 2 c - a 2 d 49. bx 4 -by 4 + ex* - cy 4 . 60. a 2 - ft 2 - a + 61. 2* - w 2 + t*> - 2 2 . 62. ch -f 6dfc - 53. r 2 + 6r< + 9/ 2 - a 2 - 2a6 - fe 2 . 64. 4z 2 + 4x^ -f 2/ 2 - 9a 2 - I2at - 4< 2 . 66. c 2 + 4c 4- 4 - 9eP - 6dh - ft 2 . 66. 16x 2 - 24xj/ H- 9j/ 2 - 9a 2 - 12a - 4. 67. 9x 2 - Qxy + 2/ 2 - 25a 2 + 10a6 - 6. SPECIAL PRODUCTS AND FACTORING 97 68. 4z 2 - 4xy -f y* - 9s 2 + 6w? - w 2 . 59. 6 2 - 9z 2 + 2ab + a*. 60. 4# - 1610 2 - 4cd + c 2 . 61. 4o 2 + 96 2 - 4z 2 - y 2 - 4zy - 12o6. 62. a 2 - 96 s - d 2 - 2o - 1 - 66d 63. 16s 4 - Sly 4 + 4z 2 - 9y\ '64 cV - 81c 2 + 324 - 4s 4 . 77. Cube of a binomial We verify that (x + y)* =(x + y) 2 (x + y) - (z 2 + 2^ 4- 2/ 2 )(z 4- y) = x 3 -t- 2x 2 !/ + xt/ 2 + x?y + 2^!/ 2 + y 3 . On collecting terms we obtain (1) and, similarly, we could verify (2) : (x + y) 8 = y* + 3x*y + 3xy* + y 8 ; (1) (x - y) 8 = x 8 - 3x a # + 3xy a - y 8 . (2) The student should memorize these formulas. ILLUSTRATION 1. From formula 1, with x = 2a and y = 6, (2a + 6) 3 = (2a) 3 + 3(2a) 2 (6) + 3(2a)(6 2 ) + 6 s 12a 2 6 ILLUSTRATION 2. From formula 2, (4 - xY = 64 - 78. Sum and difference of two cubes By long division we could verify that a - a 2 Hence, we have the following formulas, useful for factoring when read from left to right, and useful in multiplication when read from right to left. 98 SPECIAL PRODUCTS AND FACTORING ILLUSTRATION 1. By use of (1), read from right to left, with 6 3, (a - 3)(a 2 4- 3a + 9) - a 3 - 3* - a 8 - 27. ILLUSTRATION 2. From formula 2 with a = 3s and b 27s 8 + 8s/ 3 = (3*)' = (3* 4- 4- ILLUSTRATION 3. 1 - 64s 8 - I 8 - (4z) 8 = (1 - 4z)(l + 4z 4- 16z 2 ). EXAMPLE 1. Factor: j/ 6 - 19?/ 8 - 216. SOLUTION. tf - lty - 216 = (y 3 - 27) (^ + 8) 4). EXERCISE 34 Divide by long division and check by use of Section 78. a 3 -A 8 o + 276 3 . 8s 3 O. *' x + y ~ a-h ~ a 4- 36 * 2x - 3y Multiply by inspection. 5. (c 4- w?)(c 2 - cw 4- w*). 6. (u v)(w 2 4- uv 4- t> 2 ). 7. (3a - c)(9o 2 4- 3ac 4- c 2 ). 8. (1 - w)(l 4- w 9. (1 3x)(l 4- 3x 4- 9x 2 ). 10. (2 3u)(4 4- 11. (6 2x)(6 2 4- 2bx 4- 4x 2 ). 12. (4# 4- l)(16y 2 4y 4- 1). Factor. 13. d 3 - ^. 14. A 3 4- 2*. 15. y 3 - 27. 16. u 3 4- 1. 17. 1 - v 9 . 18. 8 - x 8 . 19. s 3 4- 1000. 20. 64 - t^. 21. 1 - 27s 3 . 22. 125 4- 8^. 23. s 3 - SwA 24. 8 - 27x 3 . 25. 216s 3 - yV. 26. x 8 - 64^. 27. 343a 3 - Sz 3 * 3 . Expand each cube by use of the formulas of Section 77. 28. (c 4- d)*. 29. (h - A;) 3 . 30. (2 4- y) 3 . 31. (u 4- 3) 3 . 32. (5 - y) 8 . . 33. (2x 4- w) 8 . 34. (y - 3x) 8 . 35. (4x 4- y) s . 36. (a - ft 2 ) 8 . 37. (a 2 - 2x) 8 . 38. (x 2 - i/ 2 ) 8 . 39. (c - 2ft 2 ) 8 . 40. (o - 2s 8 ) 8 , 41. (2c* - 3s) 8 . 42. (.1 - 2x) 8 . SPECIAL PRODUCTS AND FACTORING 99 Factor. 43. y* + 7z* - 8. 44. 276 + 26fc - 1. 46. 8z - 19xV - 27y. 46. 64a - 16a*6 + 6 6 . 47. a - 3o 2 + 3a - 1. 48. s 3 + 62? + 12s + 8. 49. w - 9w*x + 27w*c 2 - 27s 8 . 60. 125u s - 75u 2 -f 51. (c - d) 3 - a s . - 62. (h - x)* - (y - *79. Trinomials equal to differences of squares An expression of the form z 4 -f kx 2 y 2 + y 4 can be written as the dif- ference of two squares if the expression becomes a perfect square after the addition of a perfect square multiple of x 2 y 2 . EXAMPLE 1. Factor: 64o* - 64a 2 6 2 4- 256*. SOLUTION. 1. A perfect square involving 640 4 aqd 25ft 4 is (8a 2 -f 56 2 ) 2 = 640* + 80a 2 6 2 -f 256 4 . 2. Hence, 64a 4 64a 2 6 2 + 256 4 becomes a perfect square if we add 144a 2 6 2 . Therefore, we add 144o 2 6 2 and, to compensate for this, also subtract 144a 2 6 2 : 64a* - 64o 2 fe 2 + 256 4 = (640 4 - 64O 2 ?) 2 -f- 256 4 -h 144a 2 6 2 ) - 144a 2 6 2 = (64a 4 + 80a 2 6 2 + 2S6 4 ) - 144a 2 6 2 = (8a 2 + S6 2 ) 2 - 144a 2 6 2 = (8a 2 + 5ft 2 - 12a6)(8a 2 H- S6 2 + 12a&). EXAMPLE 2. Factor: . Qx 4 - I6x*y* + 4y*. (1) SOLUTION. 1. The perfect squares involving 9s 4 and 4y* are = 9s 4 db In order to obtain + 12a:V from (1), we would have to add 2&rV> but this is not a perfect square. To obtain 12#V we must add 4afy 2 , which is a perfect square. 2. Add, and also subtract, 4zy in (1) : Ox 4 - 16zy + 4^ = (9s 4 - 16*y + 40* 2xy). *EXERCISE 35 Factor by reducing to a difference of two squares. 1. a* + 2 + L 2. ^ - 3 2 + 1. 3. Qo 4 + 2o l -f 1. 4. ftc 4 + lla; 2 + 4. 5. s 4 + A 2 ^ + h*. 6. 9s 4 - 10*' -h 1. 700 SPECIAL PRODUCTS AND FACTORING 7. 4t0* + Sa 2 ^ -h 9a 4 . 8. a< - 9oV + Ify/ 4 . 9. 25a 4 - 5aW -f 46*. 10. 4# + 4d?h* + 25A 4 . 11. s* + 4. 12. w^ + 4s 4 . 13. s 4 -h 64A 4 . 14. 625Z 4 + 4w*. 16. 81* 4 + 64s 4 . 16. x* - 12aW + 160 4 . 17. Oa 4 - 16a 2 c 2 + 4e*. 18. 19aV + 4x* -h 490 4 . 19. 25O 4 + 9^ - 34ay. 20. 4a^ - 24 + 25. 21. *80. Perfect powers An integral rational term is said to be a perfect nth power if it is the nth power of an integral rational term. ILLUSTRATION 1. IGa 4 ^ 8 is a perfect 4th power because 16o 4 6 8 = (2O6 2 ) 4 . ILLUSTRATION 2. 8o 6 6 6 is a perfect cube because 8o 6 6 6 = (2a 2 6 2 ) 3 . The original exponents have 3 as a factor. In a perfect nth power, each exponent has n as a factor because in raising a term to the nth power we multiply each of the original ex- ponents by n. ILLUSTRATION 3. (2 3 a 2 6 4 ) n = 2 3n a 2n 6 4n . *81 . Special cases of sum or difference of perfect powers I. // n is even, commence factoring (a n b n ) by recognizing it as the difference of two squares. ILLUSTRATION 1. x 6 - t/ 6 = (x 3 ) 2 - (y 3 ) 2 = (a: 3 - ^)(x 3 + y*) = (x - t/)(x 2 + xy + y*)(x + y)(x* - xy + y*). We could have commenced by factoring (x 8 t/ 6 ) as the difference of two cubes, but this would have been an inefficient method. ILLUSTRATION 2. To factor 16o 4 6 4 81, we observe that each term is a perfect square. Hence, 16a 4 & 4 - 81 = (4a 2 6 2 - 9)(4a 2 6 2 + 9) = (2afe - 3)(2afe + 3)(4a 2 6 2 + 9),' where the final factor is a prime sum of perfect squares. II. // n is odd and has 3 as a factor, we can commence factoring (a n 6 n ) by recognizing it as the sum or difference of two cubes. SPECIAL PRODUCTS AND FACTORING 707 ILLUSTRATION 3. x 9 + y 9 = (z 3 ) 3 + (y 3 ) 3 = (x 3 + 2/*)(z 6 x 3 !/ 3 + |/*) (Using Section 78) *EXERCISE 36 Express the perfect power as the 3d or 4th power of some other term, which- ever is the case. 1. 8a 3 6 3 . 2. 27ay. 3. 16a 4 6 4 . 4. 81xy. 5. 125xy. 6. 64xy. 7. 256W 8 !; 12 . 8. 625ay. Factor each expression which is not prime. 9. a 4 - x 4 . 10. y 4 - 81. 11. 16 - w 4 . 12. Six 4 - y 4 . 13. x 8 - y 8 . 14. x 4 -f y 4 . 15. 81 - 16x 4 . 16. y - x*. 17. w - 1. 18. a 6 - 64. 19. z - 64/. 20. a 6 + 64. 21. x 6 + 1. 22. 729 - a 6 . 23. 729 + x 6 . 24. 125 - <z fl . 25. 256 - a 8 . 26. h 9 - k*. 27. a 9 + 6 9 . 28. a 8 + 6 s . 29. Six 8 - y 4 . 30. 16x 4 - Sly 8 . 31. 625 - 16X 8 . 32. x 8 - w 8 * 8 . 33. a 6 - 646. 34. 64 + xy . 35. 8a 3 - 27x 6 . 36. x u - y u . *82. Properties of factors of a" db b n We have verified special cases of the following results, where n represents a positive integer. Any special case of the results can be checked * by long division. I. For every value of n, (a n b n ) has (a b) as a factor; in other words, (a n b n ) is exactly divisible by (a b). ILLUSTRATION 1. a 3 b 3 (a 6) (a 2 + ah -f a* - b 4 = (a - 6)(a 3 + o 2 6 + a 4 - 16 = a 4 - 2 4 = (a - 2)(a 3 + 2a 2 + 4a + 8). II. // n is even, (a n b n ) has (a -{-b) as a factor. ILLUSTRATION 2. a 2 6 2 = (o 6) (a + b). 4 _ 54 = ( a + 5)( s _ 2& 4. a &2 _ * A convenient method for giving a general proof of the results is met in a more advanced section of algebra. J02 SPECIAL PRODUCTS AND FACTORING III. Ifnis odd, (a n + b n ) has (a -f 6) as a factor. ILLUSTRATION 3. a 8 -f 6 s = (a + 6) (a 2 ab -f a 7 + & 7 - (a + 6)(a - a 6 6 + o 4 ^ - a 8 6 + a'6 4 - IV. If n is even, (a n 4- 6 n ) does not have either (0 6) or (a + 6) as a factor. ILLUSTRATION 4. (a 2 + fe 2 ) and (a* + o 4 ) are prime, (a 6 H- 6*) is not prime but it does not have either (o + 6) or (a 6) as a factor: 6 + 6 ( a 2 + &)(a4 _ 2&2 _|_ J|) f where each factor is prime. Special cases of the following general properties were exhibited by the second factors in Illustrations 1, 2, and 3. A. When (a* b n ) is divided by (a 6), all coefficients in the quo- tient are -f 1. B. When (a n + b n ) or (a n b n ) is divided by (a + 6), the coefficients in the quotient are alternately -f 1 and 1. Factors obtained by reference to (I), (II), and (III) are not always prime. Also, as seen in Illustration 4 and Section 81, an expression of the type a n 4- & n , with n even, may be factorable although (IV) is true. In finding the prime factors of a n b n , first use the methods of Section 81 if possible, before employing (I), (II), and (III). ILLUSTRATION 5. z 6 + 64 = (x 2 ) 8 + 4 3 = (z 2 + 4)(x* - 4s 2 + 16). ILLUSTRATION 6. x 9 - y 9 = (z 8 ) 8 - (y 8 ) 8 = (z 8 - ^(x 6 + xV -f = (x - ILLUSTRATION 7. x 4 y 4 = (x 2 ) 2 (y 2 ) 2 = (x 2 # 2 )(x 2 + y 2 ) - (* - y)(* + y)(* 2 + 2/ 2 ). (1) By use of (I), x 4 - ^ - (x - 2/)(x 8 + x*y + xy 2 + y 8 ). (2) Equation 1 shows that the second factor in (2) is not prime; this factor could be factored J>y grouping: = &(x -f y) + j/ 2 (x + y) Thus, we finally arrive at the factors obtained in (1) but by a much less desirable process. SPECIAL PRODUCTS AND FACTORING f03 *EXERCISE 37 Find the quotient by long division, and the remainder if the division is inexact. x 4 + 16 . . - s + y s + 2y z + 2 eocft reswft without using long division t by use of properties A and B o/ Section 82, and cAeefc 6y multiplication. ~ 1 u * - 16 13. ( - y>) + (a + y ). 14. (i6x* - a 4 ) -*- (2a? + a). 15. (a 8 - 8) * (a - 2). 16. (243* 6 - 1) * (3x - 1). 8 J?. 19 8 " y 19. - sr - ^ oo s" " 26 ' a;* - 2a 166* Factor eacA expression which is not prime. 28. a 5 - c 6 . 29. a 4 - w 4 . 30. u 7 - v 7 . 31. u* + t^ 6 32. 32 + x 6 . 33. 1 - ^. 34. a* - 256s/ 8 . 35. u - . 36. t> 6 - 32u 8 . 37. 32a 6 - 1. 38. x n + y l \ 39. 128 + 40. x 6 - 243j/*. 41. a 8 - 27z. 42. 16x 4 + 43. 4x* + 1. 44. lutx 4 + 810 4 . 45. w 3 * + y 46. 32x w + y*. 47. x 16 + y". 48. u ffi - 49. 512 - x 9 . 50. z + 512a'. 51. u CHAPTER 6 ADVANCED TOPICS IN FRACTIONS 83. Reduction of fractions to lowest terms Whenever we make a reference to factoring in a fraction, it will be assumed that the numerator and denominator are integral rational polynomials with integral coefficients. In the final result of any operation on fractions, we agree to leave any expression in a factored form if it arises naturally. SUMMARY. To reduce a fraction to lowest terms: 1. Factor the numerator and denominator. 2. Divide both numerator and denominator by all their common factors. ILLUSTRATION 1. In the following fraction, we divide both numerator and denominator by 3z 4y and indicate this by cancellation. 3x 2 + 2xy - Sy z &xr=^Sy)(x + 2y) x + 2y ILLUSTRATION 2. In reducing the following fraction to lowest terms, we first notice that one factor in the numerator is merely the negative of a fac- tor of the denominator. x 2 -9 (x-3)(x 12 + 2x - 2x 2 2(3 - z)(2 + x) _. x - 3) In the preceding line, we obtained (x 3) in the denominator by multiplying (3 x) by 1, and hence it was necessary to change the sign before the fraction to keep its value unaltered. ADVANCED TOPICS IN FRACTIONS 105 EXERCISE 38 Reduce to lowest terms. 1 , wy a*b(x - 2y) 6c -f 6d irv n * r ni A/ r\ \ * 3 3c + 3d c 2 d(o 4- 36) c 4at/ 26y cz(z + y) cd*(a + 36) * 2ac 6s z 2 - y 2 ax - ex A ~ a 2 -c 2 1ft ^ ^. t1 4o 2 - 96 2 10. : 11. ^ ^* 12. cx + cy ' 2ax 36x * 4ax 26x m 2 -m-42 a 2 + 2a - 15 13 ' m 2 - 3m - 28 14 - a 2 + a - _ w f 13a; - 10 -A 3x 2 - 7ax 4- 4a 2 lo. 17. 4 3a: 2 4" 2ac 8a 2 - 12a 2 -A a 2 - 4aa; ' 2a 2 - 9ax - i i t a IW ; ; 4#2/ 4" ty x* 4" bxy -\- ax + ay . U rt , 2a 26 22. Reduce to lowest terms with as few minus signs as possible remaining in the numerator and denominator. -3 OK -2z-2y Ofl 2a + 2d 27. a o 4- 6 ax 5a y) 2 5 - z a; 2 - Ox 4- 9 * 3v - 3w 18 - 2z 2 (y - 2z) 2 h 3x - 9 . 9 - 15z 4- 4a; 2 94* * - 15 2a: 2 15ca; + 20dx 9c - 27 10x 2 + 29x - 21 ~ 5 8a + 276 s 39. 26x 2 - 26 3 - x 3 4- 6^ x* - ?0<$ ADVANCED TOPICS IN WACT/ONS 84. Lowest common multiple of polynomials The LCM of two or more integral rational polynomials is defined as the polynomial of lowest degree in all the literal numbers, with smallest integral coefficients, which has each given polynomial as a factor. Two results for a LCM which differ only in sign will be considered essentially identical because usually the sign of a LCM is of no importance. To find a LCM, first factor the polynomials. \ ILLUSTRATION 1. The LCM of 2(3 - z)(3 + a), 4(z - 3)(s - 1), and 3(z - 3)' is 4-3(x - 3) 2 (z + 3)(z - 1). We did not consider (3 - x) and (x - 3) as distinct factors because 3 x (x 3). The LCD of two or more fractions is the LCM of their denomina- tors. We shall deal with the notion of a LCM only where it is a LCD. Note 1. The highest common factor (HCF) of two or more integral rational expressions is the expression of highest degree, with largest integral coefficients, which is a factor of each of the given expressions. Thus, the HCF of 6sV and 4xy* is 2xy 8 . We shall not find it essential to use the HCF terminology. 85. Addition of fractions with polynomial denominators SUMMARY. To express a sum of fractions as a single fraction: 1. Find the LCD; that is, factor each denominator and form the product of all different prime factors, giving to each factor the highest exponent with which it appears in any denominator. 2. For each fraction, divide the LCD by the denominator and then multiply both numerator and denominator by the resulting quotient, to express the fraction as an equal one having the LCD. 3. Combine the new numerators just obtained, with each numerator placed in parentheses preceded by the sign of its fraction, and divide by the LCD. Note 1. To check the addition of fractions, substitute explicit values for the literal numbers in the given sum and the final result. EXAMPLE 1. Express as a single fraction: 4x -9 x* + x-6 ^ ADVANCED TOP/CS IN FRACT/ONS 107 SOLUTION. 1. Factor the denominators: * 2 - 9 - (* - 3)(x + 3); z 2 + x - 6 - (x + 3)(* - 2). Hence, LCD (a - 3) (a; -f 3) (a? - 2). 2. In the 1st fraction, LCD * (x 2 - 9) = x - 2. 3. In the 2d fraction, LCD -*- (x 2 + x - 6) = * - 3. 4. We multiply numerator and denominator by x 2 in the 1st fraction, and by x 3 in the 2d fraction: x* - 9 z 2 + x - 6 * - 2) 3z(a; - 3) (* - 3)(* + 3)(x - 2) (as - 3)(x + 3)(* - 2) 4sQc - 2) - 3a(:c - 3) = x* + x (* - 3)(* -f 3) (a? - 2) (a; - 3)(x + 3)(a; - 2) CHECK. When x - 4, we obtain: 16 -12 16 6 10 (2) (3) In (1) : 16-9 16 + 4-6 7 7 7 For the result in (3) : j-. ow , , ow , ^r -TT which checks. (4 - 3)(4 + 3)(4 - 2) 7 Comment. With practice, the student should be able to omit details such as those on the right in (2). ILLUSTRATION 1. In the following addition of fractions, we change signs in the second denominator in order to exhibit the identical nature of two factors in the denominators: 5 7 5757 X g\ 1 I A J f* .. 3(3c-2d) 6c - 4d 3(3c - 2d) 2(3c - (5-2) - (7-3) _ 11 3-2(3c-2c*) 6(3c-2d)' EXERCISE 39 Change the fraction to an equal one with the specified denominator. 1. 3x/(x 2); new denominator, (x + $)(x 2). 2. 2y/(y 4); new denominator, (y - 4)(3y 1). 3. 3z/(2z 3) ; new denominator, 4x 2 - 9. f 4. 2/(a -f 2) ; new denominator, 2o 2 -f 6. 5. (3 a)/(2 a); new denominator, 2a 4. 708 ADVANCED TOPICS IN FRACTIONS Combine into a single fraction in lowest terms. Where letters are involved, check by substitution when directed by the instructor. 6 _Z. _ 4-- 7 5? _ 8 ? 3a 4- 2 10 30 + 5* v 2a 46* 3 x - 5 ' 12 32 " nf -i\ ^p - / iv * XU. _ . 3(a - 6) 5(a - 6) 7x 4- 00 29. 11. s-- - v--- 12. 3x 3y 5x 5y a 6 a 4- 6 3 2 - 3s 15. s - FJ + ^j - 16. 2c - 6d ' 3d - c 2a - 46 66 - 3a 4 2 - 1 ^ 6x - 3 3x - 2 2x 20. 6x 4- 6 9a 2 d 2 6a a s a K 21. * 2x 23. 3a-+l. 24. !- - 3 OK 4 5 10 26. _ , ~ + - - -; 26. -f 2y ' a: 2 - y 2 4c - z 2 ' 3a; 2 - 48 07 2fl ~ n i 3a - 4n a - 4 2 - llo 2a - 2n "*" 6n - 6a* 2a - 4 "*" 2 - a ' 4- 2x - 1 2x4-1 x 4- 4 x 2 4- x - 12 x 2 4- 4x - 60 x - 6 " 3n - 3 ~~ n 2 4- 3n - 4* 32> o 2 - 16 ~ a 2 4- 8a 4- 16 33 2c ~ 3 I 4 34 g + 5x2 I 3 ***** no 10 I o > i 1 ^ i ? *" ^ ^ .2 i^ - 18 ' 3C 2 - lie 4- 6 & - ^ ' 2x - 2y 5 - 27 4x 2 - 12x 4- 9 x* + 8 x 2 - 2* 4- 4 37 3s* 5s 2 - 3 2s* - 3 s 4- 3 " t " :c 4 -4""2x 4 4-x 2 -6 2^4-3^-2 x 6 - 4 ADVANCED TOPICS IN FRACTIONS J09 39 3x-2 2s + 5 2z 2 - x - 3 3s 2 + 6* + 3 ~ r x - 2 3s + 5 , 2a; - + x - 6 2z 2 - x - 6 4* 2 - 9 7 3 - < ** 49 ** & _o _r ro i 8s 2 - 18 ' 2x* - toe + 9 2z 2 - 3z - 9 a + 36 a - 26 6a 2 - 06 - fe 2 ' 3o 2 + 7a6 + 26 2 2a 2 r - o 2r - a ___ r 2 - 6ar + 9a 2 r 2 - 9a 2 q 86. Factoring in multiplication or division of fractions To multiply or divide fractions involving polynomials, factor the numerators and denominators and divide out all common factors from the numerator and denominator of the final result. 7x - 15 2x 2 - 19z + 42 ILLUSTRATION 1. ^ _ ^ _ u -- ^ _ 12 (2x - 3) (a; + 5) (2s -'7) (a - 6) = (x + 6)(s - 6) (2x - 7)(x + 2) ' 4(2z - 3) 4(x + 2) ' where we divided both numerator and denominator by (2x 3)(2z 7). xy* - y 8 T n x* + x z y xy* w 3 x* xy 2t/ 2 ILLUSTRATION 2. -5 ^ 75 = -f-j ^ -5 ** x* 2xy + y z x* + x*y x 2 xy = y 2 (a? - y) (a; - ' (^ + y) (x - y) 2 a: 2 (x - where we divided both numerator and denominator by (x y)(x T - 4 . t ON 2a; - 4 . x -2 ILLUSTRATION 3. -T - =- - (x 2) = -^ - = -. -- : z 2 5 v a; 2 5 1 = 2(g - 2) 1 2 a; 2 - 5 'x - 2 x 2 - 5* where we divided out (re 2). Whenever a mixed expression is involved in the numerator or de- nominator of a fraction, or as a factor in a product, it is advisable to change the mixed expression to a single fraction as the first step hi simplification. 710 ADVANCED 7OP/CS IN FRACTIONS At EXAMPLE 1. Reduce to a simple fraction: u SOLUTION. Express the numerator and denominator of the complex fraction as simple fractions and divide: 25 u 2 _ 9 9 9u-25 3u - 5 9 3u - 5 5) 9 (3t< - 5) 3 Comment. A somewhat shorter solution is obtained if, as the first step, we multiply both numerator and denominator by the LCD, 9, of the fractions involved in them. ILLUSTRATION 4. (1 -f ~ J * (2 =-) 2 Zx 2 1 2 + 3s . 4 - 9s* *"^ *""' 2 (2 -f- 3z)(2 - 3z) 2 - 3s EXERCISE 40 Perform the indicated operation and reduce to a simple fraction in lowest terms. Check by substituting values for the letters, where directed by the in- structor. 3a - 36 a + 26 2c-4da6-3a a-&' 6-3 *6c-26c 4 tf- hx-hy cw-bw, - ab-ac 3x-Zy ch - ex K "" ^2 IA\ A 5. -= - r- (a; 2 lo). o. -= - r- . . = ^ - r a; 2 4a; 5w aw 5k ok - 1 . 6y - 2 2s - 2y . (a? - - 16 ' / 2 * A / K o *N IA 9 9. (5x 3x 2 ) 4- - r- =- 10. = - =- ? -- - v ' x + 3 3s 3y 6 2 - 9 a* + 06 4a* - 96 14. 3x-l ADVANCED TOPICS IN FRACTIONS _ ox-f 6x 17. 20. 23. 26. a IU 29 . 9x 2 - 1 o 2 a 2 XO* ." ' " 4x + 5 ill. -4-3 3x 18. LI-. ?_ -* z ^ 2o a 2 ._! 19. ^ j- 6 a 2 l-i 1 erf 1 I-* OK W . 5y i T * 24_ fl ' +6 r.~* a+1"" r a 1 i - ? u 2x 2 -f 5x - 12 x a AI 2 2o Q7 IMMiMMWMW Oft xa + 2x + 1 A ^5x 2 4- x 6 X CL /, , 4 M 4 4a * * \ L , 5 r/if\ /Olf A'l'X '* ju / y ^ * C L -I- 2 32. (l - \ o 2 1- 33. 2s* + 5z - 12 35. nx + 34. on at; x ex ac). y 36. *~ 2? on- 1 37. _i-- _ 112 ADVANCED TOPICS IN FRACTIONS 41 / a 4 - 816 4 __ a + 36 \ _._ a 2 + 606 + U 2 c - 3o6c + 96^ ' a 2 - 6a6 + 96 2 / * a 3 + 276 8 42 c 4 ~ 2c 3 d + 4c 2 <P . (c*d - c* c*'+ 8d \ ac + 6W - 2ad - 36c T \c 2 - 4cP ' a 8 - 276 3 / 44 3o H- 46 5oc H- 5o t~ y """ ~~~* a 16o 2 __ z o 2 ^ a 47. 4- 7^ 2 Find <Ae reciprocal of the expression and reduce the result to a simple fraction in lowest terms. /3a:+l g . 3 \ l2^^2 - 5 + JTT/' Reduce to a simple fraction in lowest terms. a* x 4 + 4x 2 + 8 6* x* - 4 51. _L_ -.. 62. * . 4 a 4 x + 2 .. A-2 2_V_? L_\ V a + 3/U + 2 3 -a/ 64. ? 66. A 2a x r-7 4o x 1 + 2a 66. _ T _. 67. 4o - 1 4o - a- 1 ADVANCED TOPICS IN FRACTIONS 773 87. Equations involving fractions To solve an equation involving fractions, we proceed as follows. 1. Factor all denominators and form the LCD in factored form. 2. Enclose each numerator in parentheses and multiply both sides of the equation by the LCD to clear the equation of fractions. 3. Remove parentheses and solve. G, i i 2x 2x + 28 , EXAMPLE 1. Solve: 3 ~ 2^3 = (1) SOLUTION. 1. The LCD is (2z + 3)(2z - 3), or (4z 2 - 9), 2. Multiply both sides by the LCD: 2x(2x - 3) - 2x(2x + 3) = 2x + 28, (2) Or because (2x + 3)(2z - 3) = 2x(2x + 3); etc. 3. Expand in (2) and collect terms: 4x* - Qx - 4x 2 - Gx = 2x + 28; - 28 = 14x; x = - 2. The student should check by substituting x 2 in (1). 88. Operations leading to extraneous roots A. // both members of an equation are divided by an expression involving the unknowns, the new equation may have fewer roots than the original equation. ILLUSTRATION 1. By substitution, we verify that x 1 and x = 2 are roots of x z 3x H- 2 = 0. On dividing both sides by (x 2) we obtain x* - 3x + 2 A (x - A - A - ^ = 0. or - - ^^ - - = 0, or x 1 = 0. . x 2 x 2 The final equation has just one root, x = 1. The root x 2 was lost by the division. In solving algebraic equations, we usually avoid operations of Type A in order that roots may not be lost.* B. // both members of an equation are multiplied by an expression involving the unknowns, the new equation thus obtained may have more solutions than the original equation. * See Note 4 in the Appendix for a "proof" that 2 = 1, in which the fallacy involves an operation of Type A which conceals a division by zero. 114 ADVANCED TOPICS IN FRACTIONS ILLUSTRATION 2. The equation x 3 * has just one root, x 3. If both sides of x 3 are multiplied by (x -f 2) we obtain (x + 2)(x - 3) - 0, or x* - x - 6 = 0. By substitution, we verify that this equation has two roots, x = 3 and x 2. The root 2 was introduced by the multiplication. A value of the unknown, such as x * 2 in Illustratioij 2, which satisfies a derived equation but does not satisfy the original equation, is called an extraneous root. Whenever an operation of Type B is employed, test att values obtained to reject extraneous roots, if any. EXAMPLE 1. Solve: -5 = ; = H 17 = 0. S)*Z MMB I 1l** ^" I O* ^M I v A *C/ X C/ |^ JL SOLUTION. The LCD is x 2 1; multiply both sides by x 2 1: s - 1 + 2(x - 1) = 0; 3z - 3; qr a: - 1. TEST. Since x 1 makes z 2 1 = in the denominators of the given equation, 1 cannot be accepted as a root because division by zero is not ad- missible. Hence, 1 is an extraneous root and therefore the given equation has no root. EXAMPLE 2. On a river whose current flows at the rate of 3 miles per hour, a motorboat takes as long to travel 12 miles downstream as to travel 8 miles upstream. At what rate could the boat travel hi still water? SOLUTION. 1. Let x miles per hour be the rate of the boat in still water. Then the rate of the boat in miles per hour going upstream is (x 3) and downstream is (x + 3). 2. From the standard equation d = vt of uniform motion, we obtain t d/v. Hence, the time in hours for traveling g 8 miles upstream is =; X u 12 12 miles downstream is r-x 3. Hence, j- = --. (1) 4. Multiply both, sides of (1) by (x - 3)(s + 3): S(x + 3) = 12(x -3); 8x + 24 = I2x - 36; 4x . 60; x = 15. (2) Thus, the boat travels 15 miles per hour in still water. ADVANCED TOPICS IN FRACTIONS 115 EXERCISE 41 Each equation will reduce to a linear equation if cleared of fractions properly. This reduction may be prevented and extraneous roots may be introduced i] unnecessary factors are employed in the LCD. Solve each equation and check. * 61 . ^ x 3 .75 *" x - 2 ~ 2 '* x~=3 " 2 A 7 3 A K I 7 1 4. !SS 4. 6. 7: If* <7* J9 9 mLf ml/ mm& 4mr 2* - 2 w ~ + 1 3< + 4 .a 10. 2 5 12. * + 2 2x + 2 a; x - 3 14 2h + 1 3x - 3 3* - 6 2 _ 1 14< *-3 m ** + 20- - * x 2 - 1 "I - 1 t* + t-2 -* + 14 17. j^j - 1 - g^' W. 2 ^ 6* 2 -h6 3 ~ 3:C 4.2. 2^ . O A^r2 .. /r _ 9 IT 9 " j \JJb ^^ J(/ ^^ mi IJvU ^^ mm \ - - 3 "" x 02 4x 2 + 3 = 1 } .3 + * 8x 3 + 1 2x + 1 * J ' ft 2* + 3 3 - * 2 * - 5 * 2 - 6* + 5* 3 3* 2 + 3* 4 - 9* 2 4* + 3 7 - 2* 2 26. ' * - 2 2* - 3 2x 2 - 7x + 6 -7 - 14 w 6 5 _ 5-4to ^ _3 2 2s-l 27 : ^ TT T"3 ii ?r^* * ^ i_ yi ^. c **" X + 4 JC 176 ADVANCED TOPICS IN FRACTIONS 29. In a certain fraction, the denominator exceeds the numerator by 5. If the numerator and denominator are both increased by 3, the fraction equals f . Find the original fraction. 30. On six quizzes in mathematics, a student has obtained 70% as his average score. How many scores of 86% each must be obtained to bring his average score up to 80%? 31. In one hour, Jones can plow J of a field. If Jones and Smith both work, they can plow the field in 2 hours and 24 minutes. How many hours would it take Smith alone to plow the field? 32. When the wind velocity is 40 miles per hour, it takes a certain airplane as long to travel 320 miles against the wind as 480 miles with it. How fast can the airplane travel in still air? 33. When the wind velocity is 20 miles per hour, it takes a certain air- plane 90% as long to travel with the wind to any destination as it would to return to the starting place against the wind. How fast can the airplane travel in still air? 34. A fuel tank has one intake pipe which fills it hi 8 hours. A second intake pipe is installed and it is found that, when both are in use, they fill the tank hi 2 hours. How long would it take the second pipe alone to fill the tank? 35. On a river whose current flows at the rate of 3 miles per hour, a motorboat takes as long to travel 12 miles downstream as to travel 8 miles upstream. At what rate could the boat travel in still water? 36. Two rivers flow at the rates of 3 miles and 4 miles per hour, re- spectively. It takes a man as long to row 13 miles downstream on the slower river as to row 15 miles downstream on the faster river. At what rate can he row in still water? 89. Solution of literal equations involving factoring In solving a linear equation in a single unknown x, when other literal numbers occur in the equation, it may be necessary to factor either in clearing of fractions or in simplifying the final result. EXAMPLE 1. Solve for x: b(b + x) = o 2 ax. SOLUTION. 1. Expand: fc 2 + bx = o 2 ox. 2. Add ox; subtract 6 2 : o& -f 6x = o 2 6 2 . 3. Factor: x(o + 6) = (o - 6) (a + 6). 4. Divide by (a + 6) : x = a b. Comment. To check, substitute x = a 6 in the original equation. ADVANCED TOPICS IN FRACTIONS 117 EXAMPLE 2. Solve for *: ? + L+* = !L. 6 2 a 2 a 6 a + 6 SOLUTION. The LCD is (6 a) (6 + a). We first rewrite the equation to change the sign in the denominator (a 6) and then clear of fractions by multiplying by the LCD : aw + 6* w -\- b w + a 6 2 a 2 6 a a + 6 Multiply by (6 2 - a 2 ) : 010 + 6 2 - (w + 6) (6 + a) - (w + a)(6 - a). (1) The student should expand in (1) and solve, to obtain w = o. EXERCISE 42 otoe for x or # or is, whichever appears. 1. ex 3o = 2/i. 2. 7x a = 3ox 5. 3. 3o* - bz = 9a 2 - 6 2 . 4. 36x - 96 2 = 2ax - 4o 2 . 6. 4a3 a 2 = 4z 1. 6. mnx a anx m. 7. x 2 - 3n 2 = (3n - z) 2 . 8. 6(6 - x) = a 2 + ax. 9. a6x a 2 = 6 2 o6x. 10. bx bd ad ax. 11. 6(6* - a) = a*x + 6 2 . 12. hz - A 2 = kz - k*. 13. 2bx + 6a 2 = 3ac + 4a6. 14. ax - ab - a 2 = b(x - 26). 15. acx + adz + d 2 = c 2 6cx 6ax. x(a - 46) - 6 2 x + 6 _ x + a b ' a 2 -6 2 ~^a- b*~*a + b 17 c = d 18 '* * 4A 19. - d 2x - c x + 26 x - 2a x 26 2x + h 4x 2 - W 2x - h * 2oc + x a - b x a 2 x + 2a 21. 26 2 + 06 - a 2 6 + a 26 s + 4a6 + 2a 2 22. Solve C = r- for 6. 23. Solve S - - ^ for r. 6 a r 1 24. Solve s = r for 6. c b 118 ADVANCED TOPICS IN FRACTIONS EXERCISE 43 Review of Chapters 4, 5, and 6 Perform the indicated operation and collect terms. 1. (3x - 5j/)(3x + 5y). 2. (4x 2 - 3yz)(4x* + 82/2). 3. (2* + 3) 2 . 4. (x - 2*)*. 5. V - 3w>) 2 . 6. (2a + 56) 2 . 7. (a - 4)(a 2 + 4a 4- 16). 8. (2x - 3s)(4x 2 + 6x -f 92 s ). Factor. 9. j/ 2 - 25s*. 10. 4z* - 9W. 11. * 12. M 4 - Slj/ 4 * 4 . 13. o* - 276. 14. 8w + 15. % 2 + 122/2 2 + 4z*. 16. 2/ 2 + y - 12. 17. 8 + 42 - 21. 18. 6x 2 + x - 15. 19. 2 - 12x 2 + 5x. 20. 4A 2 - 28/wo + 49t^. 21. S* 2 - 30; -h 45u>*. 22. 06 + 26c + 3ad H- 6cd. 23. 2a 8 -h 4a 2 - 2o - 4. 24. (m + w?) 2 -f 4m + 4w; -f- 4. 25. x* - a 2 - 6a6 - 96 2 . 26. x 2 + 4x -h 4 - 9a 2 . 3 4- 5 1 2 -- j- . j^ Reduce to a simple fraction in lowest terms. l-~ . 5 Q * T 29 - O j * _ _ (j -~ . $ -f- _ x y 9 ox 3x - 1 2x + 3 M 2a - 6 , 56 3y-2 2y-5 3c - 6 36 or x and 3x 2 + x . 12 4 H- 2x ' 3+lx 3 . 1 + 4x 2 2 ~- " 7 - 2x 4x 2 - 16x + 7 a - 6 a 6 . A 3X 2 - 2x -h 9 3 + 2x CHAPTER 7 RECTANGULAR COORDINATES AND GRAPHS II I I I I I I 4 --3 - - 2 90. Rectangular coordinates On each of the perpendicular axes OX and OF in Figure 4, we lay off a scale with as the zero point on both scales. In the plane of OX and Y we shall measure vertical Y distances in terms of the unit on OF and horizontal distances in terms of the unit on OX. We agree that horizontal distances will be consid- ered positive if measured to the right and negative if to the left; vertical dis- tances will be considered positive if measured upward and negative if downward. Let P be any point in the plane. The horizontal coordinate, or the abscissa of P, is the perpendicular distance, x, from OY to P; this di- ~ 1 -6-5-4-3-2-1 1 + III 2 3 -4.. 123456 I I I I I. I V IV Fig. 4 rected distance is positive if P is to the right of OF and negative if P is to the left of OF. The vertical coordinate, or the ordinate of P, is the perpendicular distance, y, from OX to P; this directed distance is positive if P is above OX and negative if P is below OX. Each of the lines OX and OF is called a coordinate axis, and the abscissa and ordinate of P together are called the rectangular co- ordinates of P. The point at which the axes intersect is called the origin of the coordinate system. When the axes are labeled OX and OF as in Figure 4, we sometimes refer to the abscissa as the z-coordinate and to the ordinate as the y-coordinate. 720 RECTANGULAR COORDINATES AND GRAPHS Notice that there is no necessity for using the same unit of length for the scales on OX and OF. ILLUSTRATION 1. In Figure 4, the coordinates of P are x == 5J and y = 2. The coordinates of a point are usually written together within pa- rentheses with the abscissa first. Thus, we say that P is the point (5J, 2). In Figure 4, R is the point ( 3, 4). Note 1. The coordinate axes divide the plane into four parts called quadrants, which we number I, II, III, and IV, counterclockwise. To plot a point, whose coordinates are given, means to locate the point and to mark it with a dot or a cross. EXAMPLE 1. Plot the point ( 3, 4). FIRST SOLUTION. At 3 on OX, erect a perpendicular to OX. Go up 4 vertical units on this perpendicular to reach the point R in quadrant II which is ( 3, 4). SECOND SOLUTION. At + 4 on OF, erect a perpendicular to OF. Go to the left 3 units on this perpendicular to reach ( 3, 4). Note 2. The word line in this book will refer to a straight line unless otherwise specified. EXERCISE 44 Plot the following points on a coordinate system on cross-section paper. 1. (3, 4). 2. (3, 0). 3. (1, - 2). 4. (- 3, - 5). 6. (0, - 2). 6. (- 5, 0). 7. (0, 7). 8. (- 3, 4). 9. (- 2, - 3). 10. (- 3, 5). 11. (4, - 4). 12. (- 2, 1). 13. Three corners of a rectangle are (3, 4), ( 5, 4), and (3, 1). Find the coordinates of the 4th corner and the area of the rectangle. Find the area of a triangk with the given vertices. 14. (4, 3); (4, 7); (- 2, 3). 15. (0, - 4); (3, - 4); (3, 2). 16. (- 2, 1); (3, 1); (5, 5). 17. (0, 0); (5, 3); (5, 7). 18. A square, with its sides parallel to the coordinate axes, has one corner at ( 3, 2) and lies above and to the left of ( 3, 2). If the units of length on the axes are the same and if each side of the square is 4 units long, find the coordinates of the other corners. RECTANGULAR COORDINATES AND GRAPHS In which quadrant does a point lie under the specified condition? 19. Both coordinates are negative. 20. The abscissa is negative and the ordinate is positive. 21. The abscissa is positive and the ordinate is negative. 22. A line is parallel to OX and passes through the point where y = 3 on OF. What is true about the ordinates of points on the given line? 23. A line is perpendicular to OX at the point where x 2. What can be stated about the abscissas of points on the given line? 24. How far apart are the lines on which the abscissas of all points are 3 and 4, respectively? 25. How far apart are the lines on which the ordinates of all points are 7 and 3, respectively? 91 . The function concept We recall that, in a given problem, a constant is a number symbol whose value is not subject to change during the course of the dis- cussion, and a variable is a number symbol which may take on different values. When desirable, we may think of a constant as a variable which can assume only one value. If a first variable, x, and a second variable, y, are so related that, whenever a value is assigned to x, a corresponding value (or corre- sponding values) of y can be determined, we say that y is a function of x. Then x is called the independent variable and the second variable, y, which is a function of x, is called the dependent variable. To say that y is a function of x means that the value of y depends on the value of x. ILLUSTRATION 1. In the formula A = Trr 2 for the area of a circle, if r is a variable then A is a variable and A is a function of r. Any formula in a variable x represents a function of x; the values of the function can be computed from its formula. ILLUSTRATION 2. (3z 2 H- 7x -f- 5) is a function of x. If x = 2, the value of the function is (12 +14 + 5) or 31. Note 1. If just one value of y corresponds to each value of x, we say that y is a single-valued function of a;; if just two values of y correspond to each value of x, then y is a two-valued function of x; etc. 722 RECTANGULAR COORDINATES AND GRAPHS 92. Graph of a (unction Let y represent any function of x. Then, each pair of corresponding values of x and y can be taken as the coordinates of a point in an (x, y) coordinate system. This leads us to adopt the following terminology. DEFINITION I. The graph of a function, y, of x is the set of all points (or the locus of points) whose coordinates form pairs of corresponding values of x and y. To graph a function will mean to draw its graph. In graphing a function, we usually plot the values of the independent variable on the horizontal axis of the coordinate system. A linear function of x is a polynomial of the first degree in x and hence has the form ax 4- 6, where a and 6 are constants. In Illustra- tion 1 below we meet a special case of the fact that the graph of a linear function of x is a straight 'line. This fact, whose proof we shall omit, is the basis for the name linear function of x. ILLUSTRATION 1. If x is the independent variable, in order to graph the function (J:c 3), we introduce y to represent it. That is, we let y \x 3. If x = 5, then y = f( 5) 3 = 6. Hence,, one point on the graph is (5, 6). Similarly, we let x 0, 2, etc., and compute the corresponding values of y given in the following table. We plot (- 5, - 6), ( 2, 4J), etc., in Figure 5 and join them by a straight line, which is the graph of the function. From the graph, we read that the value of the function is zero (the graph crosses the z-axis) when x = 5. The func- tion equals 2 when x = 1, approximately. X = - 5 -2 3 6 y = -6 -4i -3 -li f o Fig. 5 If y is a linear function of x, we need only two pairs of values of x and y to obtain the graph, because a straight line is definitely located if we know two points on it. However, in graphing any linear func- tion, we shall compute three values of the function hi order to check the arithmetic involved. If the corresponding three points do not lie on a line, an error is indicated. RECTANGULAR COORDINATES AND GRAPHS 123 Note 1. In graphing, do not choose the position of the origin or the scales on the coordinate axes until after a reasonably complete table of values has been prepared. Then, make the appropriate selections of origin and scales so that as large a graph as possible may be placed on the available paper. If a function of x is defined by a formula, in general its graph is a smooth curve* To graph such a function, we introduce some letter, such as y, to represent the function, compute a table of cor- responding values of x and y, and draw a smooth curve through the corresponding points on a coordinate system. ILLUSTRATION 2. To graph x 2 4x -f 6, we let y represent the function, y = x* - 4x + 6, compute the following table of values, and plot the points. The graph, in Figure 6, is a curve called a parabola. x = - 1 1 2 3 5 y = 11 3 2 ' 3 11 Fig. 6 93. Functions not defined by formulas Functions not defined by formulas arise frequently. Sometimes the only information concerning a function consists of a table of corresponding values of the function and the independent variable, where the table may be obtainable by experimental means or obser- vation. In drawing the graph of such a function, sketch a smooth curve through the points obtained from the given values, unless otherwise directed. Instead of drawing a smooth curve through the points, it is sometimes desirable to connect them by segments of straight lines and thus to obtain a broken-line graph. Note 1 . The intersection of the'coordinate axes may be selected to repre- sent any convenient value, not necessarily zero, on either scale. ILLUSTRATION 1. The second row of the following table gives the general wholesale price index number of the United States Department of Labor for the critical depression months from June, 1930, to June, 1931. A value like 86.8 means 86.8% of the average level in 1926. To graph the index * Or, m some cases, two or more disconnected smooth curves. 124 RECTANGULAR COORDINATES AND GRAPHS number as a function of the time, in Figure 7, we choose coordinate axes, with time plotted horizontally and index number vertically. We take 1 month to be the unit of tune. We let the intersection (origin) of the axes represent June, 1930, on the axis of abscissas and 60 on the vertical axis, and assign units on the axes to suit the size of the figure. Then, for December, 1930, we plot the point (6, 78.4), etc. We join the plotted points by a reasonably smooth curve, which is the graph of the function. From the graph, extended as a guess to July, 1930, we estimate that the index number then was 68.6. JUNK '30 Jui/r AUG. SEPT. OCT. Nov. DEC. JAN. '31 FEB. MAR. APR. MAY JUNE 86.8 84.0 84.0 84.2 82.6 80.4 78.4 77.0 75.5 74.5 73.3 71.3 70.0 f=0is June, 1930 1 is July, 1930 8 9 10 11 12 13 . 7 EXERCISE 45 The letter x represents the independent variable in all problems where it appears. Clearly indicate the scale on each coordinate axis employed. 1. Graph the function (2x -\- 3). From the graph, (a) read the values of the function when x = 2J and x = 3J; (6) read the values of x corre- sponding to which the values of the function are 2, 0, and 3. Graph the function of x and, from the graph, read the value of x for which the function equals zero. 2. 3x + 5. 3. 3 - 4z. 4. - 2 - 5x. 5. 6. - 2x. 10. 7. 7. -2 11. -4. 3*. 8. 4 - 2x. 12. 0. 9. - 3 - 2x. 13. x. HINT for Problem 10. Any constant can be considered as a function of any variable x, with just one value for the function. The graph is horizontal. RECTANGULAR COORD/NATES AND GRAPHS 125 14. Graph the function of y defined by (3y 4), with the y-axis horizontal and with z used as a label for the function. 15. Graph (x* 6z + 7) by computing its values for the following values of x: 1, 0, 2, 3, 4, 6, and 7. From the graph, (a) read the values of the function when x = 5 and x 1; (6) read the values of x for which the func- tion equals or 10. 16. Graph ( x 2 4x + 6) by computing its values for the following values of x: 6, 5, 4, 3, 2, 1, 0, 1, and 2. From the graph, read the values of x for which the function (a) equals 0; (6) equals 3. 17. The table gives the total mileage of hard-surfaced roads forming parts of state highway systems in the United States at the ends of various years. Graph the mileage as a function of the time. YEAR 1926 1929 1931 1934 1939 1940 1942 MILEAGE 54,000 75,000 96,000 110,000 120,000 122,000 130,000 18. The table gives the time it takes money to double itself if invested at certain rates of interest, compounded semiannually. Graph the time as a function of the rate. From the graph, find the time for money to double at 3i%. TIME, YEARS 46J 34f 28 23J 14 ill RATE 1% 4% 19. The velocity of sound in air depends on the temperature of the air. By use of the following data, graph the velocity as a function of the temperature. From the graph, read the velocity if the temperature is 35; 8.5; 120. VELOCITY, FT. PER SEC. 1030 1040 1060 1080 1110 1140 1170 TEMP. (FAHRENHEIT) -30 -20 20 Q 50 80 110 HINT. Let the origin represent 1000 feet on the vertical axis. 20. The table gives the number of divorces per 1000 marriages in various years in continental United States. Graph the number of divorces as a function of the time. YEAR 1890 1900 1916 1922 1930 1934 1937 1940 DIVORCES 62 81 108 131 170 157 173 169 726 RECTANGULAR COORDINATES AND GRAPHS 21. The weight of a cubic foot of dry air at an atmospheric pressure of 29.92 inches of mercury, under various temperatures, is given in the follow- ing table, where weight is in pounds, and temperature is hi degrees Fahren- heit. Graph the weight of air as a function of the temperature. TEMP. 12 32 52 82 112 152 192 212 WEIGHT .0864 .0842 .0807 .0776 .0733 .0694 .0646 .0609 .0591 22. The following table gives the "thinking distance" t, and the "braking distance" b involved when a motorist, traveling at s miles per hour, decides to stop his car. The value of t is the distance traveled by the car in } second, the interval which elapses between the instant an average driver sees danger and the instant he applies his brakes. The sum d t + b is the total distance the car will travel before stopping after danger is seen. On one coordinate system, draw graphs of t as a function of s and d! as a function of 8. 8 (mph) 20 30 40 50 60 70 t (feet) 22 33 44 55 66 77 b (feet) 21 46 82 128 185 251 94. Functional notation Sometimes we represent functions by symbols like /(#), H(x), K(s), etc. The letter in parentheses tells what the independent variable is. The letter to the left is merely a convenient name for the function. ILLUSTRATION 1. We read "f(x)" as " the /-function of x," or for short "/ of x." We may represent 3s 2 5 by f(x) and write f(x) = Zx 2 5; we read this "/ of x is 3x 2 5." H(y) would represent a function of y. For instance, we may let H(y) 7y* + 6. If F(x) is any function of x and a is any value of x, then F(d) represents the value of F(x) when x = a. ILLTTSTRATION 2. "F(a) " is read F of a." If F(x) - 3s 2 - 5 - x, F(3) - 3-3 2 - 5 - 3 - 19; - 3) - 3(- 3) 2 - 5 + 3 = 25; ;- 6 s ) = 3(- fc 2 ) 2 - 5 - (- 6 2 ) - 36 4 - 5 -f 6*; [F(~ 2)] 2 - (12 - 5 + 2) 2 = 81; 5F(2) - 5(12 - 5 - 2) - 25. RECTANGULAR COORDINATES AND GRAPHS 127 A variable z is said to be a function of two variables x and y in case a value of z can be determined corresponding to each pair of values of x and y. Similarly, we may speak of a function of three variables, or of any number of variables. The functional notation just introduced for functions of a single variable is extended to func- tions of more than one variable. ILLUSTRATION 3. F(x, y) would be read "F of x and y" and would repre- sent a function of the independent variables x and y. Thus, we may let 2. Then, F(2, 1) - 2 + 3 + 2 - 7. EXERCISE 46 Vf( x ) = 2z + 3, find the value of each symbol. 6. /(- f). // G(z) = 2z 3 2 , find the valve of the symbol or an expression for it. 1. 0(- 3). 8. 0(6). 9. 0(J). 10. 0(o). 11. 0(2c). 12. 0(3z). 13. If F(x) = *' - x + 3, find F(- 2); F(6); F(c 2 ); F(* - 2). 14. If 0(ii = , find 0(2); 30(1); [0(3)?; 15. If KM - , find K(2); 2K(4); [JC(3)? ; 16. If F(x, y) - ac + 2y, find F(2, - 3); F(- 1, 4); F(o, 6). 17. If F(x t y) - x 2 + 3xt/, find F(- 1, 2); F(- 3, - 2); F(c, 26). 18. If F(x) = x* - ar, find F: 19. If /(x) =x*-4x + 5, graph /(a?) by use of /(- 1), /(O), /(I), /(2), 20. If /(x) = ^ - 12x + 3, graph /(x) by use of /(- 4), /(- 3), /(- 2), /(- 1), /(O), /(I), /(2), /(3), and /(4). t 95. Functions defined by equations A solution of an equation in two variables x and y is & pair of cor- responding values of x and y which satisfy the equation. Usually, an equation in two variables has infinitely many solutions. 728 RECTANGULAR COORDINATES AND GRAPHS ILLUSTRATION 1. Consider 3z 5y = 15. If x = 3, then 9 5y = 15, or y = f . Hence, (x = 3, y = f ) is a solution of the given equation. If y 0, then 3x = 15 or x 5; hence (x = 5, y, = 0) is another solution. Thus, by substituting values for either variable and computing values of the other variable, we could find as many solutions as we might desire. In case x and y are related by an equation, then usually we may think of y as a function of x and, likewise, of as a function of y. This is true because, in general, for each value of either variable we can find corresponding values of the other variable by use of the equation. In particular, a linear equation in x and y defines either variable as a linear function of the other variable. ILLUSTRATION 2. From 3z 5y = 15, on solving for x we obtain x = 5 + fe/; on solving for y we obtain y - & - 3. Hence a; is a linear function of y and, equally well, y is a linear function of x. 96. Graphical representation of an equation The graph, or the locus, of an equation in two variables x and y is the locus of all points whose coordinates (x, y) form solutions of the equation. If we think of a: as an independent variable, the graph of the equation is identical with the graph of the function, y, of x, defined by the equation. In particular, if a, 6, and c are constants, the graph of the linear equation ax + by = c ISQ. straight line. For, the graph of this equation is the graph of the linear function of x, or of the linear function of y, defined by the equation. ILLUSTRATION 1. From Zx 5y 15, we obtain y = far 3. The graph of 3x 5y = 15 is the graph of the linear function fx 3; this graph is found in Figure 5, page 122. The abscissa of any point where a graph on an (#, y) coordinate system meets the a>axis is called an x-intercept of the graph. The ordinate of any. point where the graph meets the 2/-axis is called a y-intercept of the graph. To find the x-intercept (or intercepts) of the graph of an equation hi x and y, place y - in the equation and solve for x; to find the y intercept (or intercepts), place x = and solve for y. RECTANGULAR COORDINATES AND GRAPHS 129 SUMMARY. To graph a linear equation in x and y: 1. Place x and compute y, to find the y-intercept. 2. Place y = and compute x, to find the x-intercept. 3. Find any other solution of the equation and draw the line through the points determined on plotting the solutions obtained. ILLUSTRATION 2. To graph 3x % = 15, first let x and obtain 5y = 15, or y 3; hence, (0, 3) is a point on the graph. If y = then 3x = 15, or x 5; the x-intercept is 5, or (5, 0) is a point on the graph. The graph is shown in Figure 5, page 122. ILLUSTRATION 3. The graph of the equation x 5 = consists of all points (x, y) in the coordinate plane for which x = 5, and the value of y is of no importance because it does not occur hi the equation. Hence, the graph of x 5 = is the line perpendicular to the x-axis at the point where x = 5, as shown in Figure 8. -H 1 o - 44 44 x Fig. 8 97. Equation of a line An equation of a curve on an (x, y) coordinate plane is an equation in the variables x and y whose graph is the given curve. ILtwo equa- tions have the same graph, in general the equations differ only in nonessential features. Hence, although a given curve may have infinitely many different equations, we shall refer to any one of these as the equation of the curve. ILLUSTRATION 1. 3x + 2y = 7 is the equation of a certain straight line. This line also is the graph of 6x 4 4y = 14 because these two equations have the same solutions. Frequently we refer to a function of a variable x, or to an equation in x and y, by giving the function or equation the name of its graph. ILLUSTRATION 2. Thus, we may refer to the line 3x + 2y = 7, or to the parabola y = x 2 4x + 6 (see Figure 6, page 123). We shall assume without proof the fact that the equation of any straight line on an (x, y) coordinate plane is of the form ax 4 by = c where a, 6, and c are constants. The equation of a line is a linear 130 RECTANGULAR COORDINATES AND GRAPHS relation between x and y which is true when and only when the point (x, y) is on the line. F ILLUSTRATION 3. The equation of the vertical line 3 units to the left of the y-axis is x 3. ILLUSTRATION 4. Let P, with coordinates (x, y), be any point on the line through (0, 0) and (1, 2). Then, from similar right triangles hi Figure 9, V % - = or y x I this is the equation of the line. Graph each equation. 1. 3x + 2y - 6. 4. 2x + 7y - 0. 7. 4x - 5w - 20. EXERCISE 47 2. 3y - 4s - 12 0. 6. 3x - y - 9. Q O/M __ O* <6C 10. s-7. 14. y - 0. 11. y - 5. 15. 12. a; = - 3. 16. &c -f 9 = 0. 3. &c - 5y - 0. 6. 3x - 15 + 6 9. 5j/ 4- s = 10. 13. y - - 4. 17. 3y + 4 - 0. Give fte equation of the line satisfying the given condition. 18. The horizontal line (a) 6 units above OX"; (6) 4 units below OX. 19. The vertical line (a) 5 units to the right of OF; (6) 4 units to the left of OF. 20. The line on which the ordinate of each point is (a) the same as its abscissa; (6) the negative of its abscissa. Without graphing, find the coordinates of the points where the graph of the equation cuts the axes. 21. Zx + 5y = 15. 22. 2x - 5y = 10. 23. - 3x + 2y - 5 = 0. 24. Find an expression for the linear function of y defined by the equation 2x + 7y = 9. 25. Find an expression for the linear function of x defined by the equation 3a? - 5w - 11. CHAPTER 8 SYSTEMS OF UNEAR EQUATIONS 98. Graphical solution of a system of two equations A solution of a system of two equations in two unknowns, x and 2/, is a pair of corresponding values of x and y which satisfy both equations. If a system has a solution, the equations are called simultaneous. / x - y = 5, (1) 1 x + 2y - 2. (2) EXAMPLE 1. Solve graphically: SOLUTION. 1. In Figure 10, AB is the graph of (1) and CD is the graph of (2). AB consists of all points whose coordinates satisfy (1) and CD consists of all points whose coordinates satisfy (2). Hence, the point of intersection, E, of AB and CD is the only point whose coordinates satisfy both equations. 2. We observe that E has the coordinates (4, 1). Hence, (x = 4, y = 1) is the only solution of the system. These values check when substituted in (1) and (2). O B D Fi 9 . 10 SUMMARY. To solve a system of two equations in two unknowns graphically: 1. Draw the graphs of the equations on one coordinate system. 2. Measure the coordinates of any point of intersection of the graphs; these coordinates form a solution of the system. Usually a system of two linear equations in two unknowns has just one solution, as was the case in Example 1, but the following special cases may occur. 132 SYSTEMS OF LINEAR EQUATIONS A. // the graphs of the equations are parallel lines, the system has no solution and the equations are called inconsistent equations. B. // the graphs of the equations are the same line, each solution of either equation is also a solution of the other and hence the system has infinitely many solutions. In this case the equations are said to be dependent equations. Note 1. Usually a graphical solution gives only approximate results, because in obtaining them we estimate certain coordinates visually. EXERCISE 48 Solve graphically. If there is no solution, or if there are infinitely many, state this fact with the appropriate reason. , 2 , , \ y + 2x = - 3. \ 2y - x = - 5. \3y + 4z = 23. f 2y - Zx = 0, f 3x + 8 = 0, f 5y - 3 = 0, \ 4y + 3z = - 18. \ 6x + 7y = 5. \ Wy + 3z = 4. / 2y - 5x = 10, R f 2z - 3y = 0, f 3z + 5y = 2, \ 2y - 2a; = 3. \ to + 7y = 0. \ 2x - 3y = 5. , n , ' - y = 6. \ 2y - 4x = 5. \ 4x - 6 = 2a; + 2y = 7. \ 6* - % = 3. \ 10y - 2a; + 4 = 0. 16. (a) Graph x + 3y = 5. (6) Multiply both sides of the equation by 2 and graph the new equation, (c) By inference, state how two linear equations are related if they have the same graph. 99. Elimination by addition or subtraction T, 1 a i r j f 4* + 5y = 6, (1) EXAMPLE 1. Solve for a; and y: < , / 4. SOLUTION. 1. Multiply (1) by 3: 12x + 15^ = 18. (3) 2. Multiply (2) by 5: 10z + I5y = 20. (4) 3. Subtract, (3) - (4): 2x = - 2; x = - 1. (5) In obtaining equation 5, we have eliminated y by subtraction. 4. On substituting x = 1 in (2) we obtain 3y = 4 -f 2 or y = 2. 5. The solution of the system is (x = 1, y = 2). The student should check this solution by substitution in (1) and (2). SYSTEMS OF LINEAR EQUATIONS 133 SUMMABY. To solve a system of two linear equations by elimination by addition or subtraction: 1. In each equation, multiply both members, if necessary, by a properly chosen number to obtain two equations in which the coefficients of one unknown have the same absolute value. 2. Add, or subtract, corresponding sides of the two equations obtained in Step I so as to eliminate one unknown. 3. Solve the equation found in Step 2 for the unknown in it, and sub- stitute the result in one of the given equations to find the other un- known. If two linear equations in x and y are inconsistent or dependent, then, in eliminating one unknown, the other will also be eliminated. If the equations are dependent, an identity = results from this elimination. If the equations are inconsistent, a contradictory equation such as = 36 is obtained. We shall omit proving these facts but shall exhibit special cases of them. Note 1. Hereafter, to solve a system of equations will mean to solve algebraically, unless otherwise stated. If the given equations involve fractions, clear of fractions before applying the preceding method. EXAMPLE 2. Solve for x and y: \. SOLUTION. 1. Multiply (6) by 2: 6z + 4t/ = 12. (8) 2. Subtract, (7) - (8) : = 12. (9) Hence, the given equations are inconsistent because a contradictory statement, 12 = 0, results from the assumption that a pair of values of x and y exists which satisfies (6) and (7). COMMENT. In Figure 11, AB is the graph of (6) and CD is the graph of (7). It is observed that these lines are parallel and hence do not 3x + 2y = 6, + 4y = 24. D Y (6) (7) Fig. 11 intersect, which agrees with the preceding algebraic proof that (6) and (7) have no solution. 134 SYSTEMS OF LINEAR EQUATIONS 1. 4. EXERCISE 49 Solve by elimination by addition or subtraction and check. Zx - y = 7, 2x + 3w = 12. - 2* = 0, -f 2y = 0. 3s - 2y = 2, - 3z = 2. 2. 5. 8. -3 = 7: _ _ 322' 10. 10 /3z + 5y = 9, . 13 ' \102,-7*=-8. 11. 2x = 3y + 12, 8 = 0. 9, H- Jy - x 1 6. 9. 10, = 6z 4- 14. -I- 2x - - 3, -f 2s - - f 12. 2z x 2" 14. = - 3, 11* + 5y = - 15. 15. 7 3 - y = 8, 7x + 4y = 43. Proceed with the solution until you recognize that the equations are incon- sistent or dependent. Then check by graphing the equations. 2* - - 10. * - 7 = 19. < 1 00. Elimination by substitution EXAMPLE 1. Solve for x and y: SOLUTION. 1. Solve (2) for x: 4x - 6, = 4. 2. Substitute * - J(4 - 3y) in (1): 4(4 - - 6. - 2y - 3 = 0, 6. (1) (2) (3) (4) In obtaining equation 4, we have eliminated x by substituting for x from one given equation into the other. 3. Solve (4) for y: 4. Substitute^/ = 2 in (3): * - 1(4 - 6) - - 1. Hence, the solution of the system is (x = 1, y - 2). SYSTEMS OF LINEAR EQUATIONS 135 SUMMARY. To solve a system of two linear equations by elimination by substitution: 1. Solve one equation for one unknown in terms of the other and sub- stitute the result in the other equation. 2. Solve the equation obtained in Step 1 for the second unknown. 3. Substitute the value of the second unknown in any equation in- volving both unknowns and find the value of the first unknown. 1. EXERCISE 50 Solve by elimination by substitution and check. x = 3y - 1, ^ / 2x + y = - 3, - y - 15. - 3 = 4. fw = 2t; + 4, \ 100 - 2u - 1. I i L I f \ u w + 1, f 2w> - 0. 5. 8. - x = 0, 4.y = 0. = -3, 5x-2y = 21. 6. 9. -f 3x + 4 = 0, x + 2y = 0. y = 14. 3z + 5y = 0. 10-15. Solve Problems 1-6 of Exercise 49 by substitution. Clear of fractions if necessary and solve by any method. Do not restrict your choice to just one of the two available methods. <^Js m ^" tJ 1/ " ^\ ^^ I \JJb mm ^ m 9J u m **^ ^J * ^ ^*. I is i "*T^ JL^Xo r * ^*!^^t i^wv \J ^^ f ^ ^y I ^*~*r ij ^^ f * ^f J j 2x - 7y = 3. * \ 4y - 9x = 5. " \ 6r + 21s = - 7. - 4y = .5, == .O. - 1.5. ~.6a; = 3.45, ~~"~ "~ O I 22. 25. 04 Q ^7 2 I = i 26 27. - 2y 4- 3 - J = 0, ' -? = o. * H- Jy = A- 2x + 5y 29. *-2 2-f y ' 4 IZ + i ^2 3a? + y + 4 . 3 _ 10 "^5 - 1 - 1 _ 8 - y - te - 2 3x - 3 136 SYSTEMS OF LINEAR EQUATIONS 101. Systems involving literal coefficients If a system involves other letters than the unknowns, it is usually best to solve by finding each unknown in turn by elimination through addition or subtraction. T> i c i / j ( ax + by = e t (I) EXAMPLE I. Solve for x and y: < . ' , ;; \cx-\-dy = f. (2) SOLUTION. I. Multiply (I) by d: adx + My = de. (3) 2. Multiply (2) by b: bcx + bdy = of. (4) 3. Subtract, (3) (4) v x(ad be) = de bf. (5) 4. Suppose that ad be 7* and _ de bf divide by ad be in (5). ad be 5. By similar steps [multiplying (I) _ of ce by c and (2) by a] we find y. ad be 1 02. Systems linear in the reciprocals of the unknowns - + - = 17, (1) x y EXAMPLE 1. Solve: { - + - = 2. . (2) c y 10 ^ SOLUTION. 1. Multiply (2) by 5: + - = 10. (3) x y 2. Subtract, (3) - (1): - = - 7; 7 = - 7x; x = - 1. X 3. Substitute x = - 1 in (1) : - 3 + - = 17; y = ~ The solution of the system is (x 1, y = i). Comment. Equation 1 is said to be linear in l/x and l/y because, if we let u = l/x and v = l/y, equation 1 becomes 3u -j- 50 = 17. Similarly, equation 2 becomes 2u H- t> 2. In place of the preceding solution, we could first solve for u and v; then their reciprocals would give x and y. EXERCISE 51 Solve for the literal numbers without first clearing of fractions. 1. 3 - 5 - 17 - 17 ' x y 3. y 2 4 _ 4- - 5w ^ y 3 4- 5 u + 2i SYSTEMS OF LINEAR EQUATIONS 137 x y 6. (x y Solve for x and y,orforw and z. M-H, -4. 6. 10 J_ 9 -- h - = u v u ax - 2y = 2 + 6, = 2 - 26. 7. = a -j- 6, *' 1 2abx - aby = a? - 6 2 . 11. / ??* + ? = * 12. 8. 10. 2cz dy = c 2 + d 2 , -f y = 2c. 610 as 6 s = 0, aw; 4- bz bw = a 2 . ax + by = 3, 14. 2aw = 4a 2 + 6 2 , 16. \ w> - 22: = 2a - 6. = 2a_ 6 26 a 6 a* f 2aw + 62 = 06, oi/ = 3. *"" \ w - bz 3a6 + 26. aw + bz = a 2 + 6 s , 15. 6w> - a = a 2 + 6 2 . 17. _ 2a 1. 2m n , 2n + 3j 1 m x m 2n n + x 2m - 3y . 103. Solution of a system of three linear equations A system of three linear equations in three unknowns usually has one and only one solution. In special cases, however, such a sys- tem may have no solution, in which case the equations are called in- consistent, or infinitely many solutions, in which case the equations are called dependent. Such cases will not be considered in this book.* EXAMPLE 1. Solve for x, y, and z: 3z + y - z = 11, x + 3y - z = 13, x + y - 82 = 11. 2x - 2y = - 2. + 9t/ - 82 = 39. 2x -f 8y = 28. (1) (2) (3) (4) (5) (6) = 30; y = 3. SOLUTION. 1. Subtract, (1) (2): Multiply (2) by 3: Subtract, (5) - (3) : 2. Solve (4) and (6) for x and y: Subtract, (6) - (4) : Substitute y = 3 in (4): 2x - 6 = - 2; x = 2. 3. Substitute (x = 2, y = 3) in (1) : 6 + 3 - z = 11; z = - 2. The solution of the given system is (x = 2, ?/ = 3, z = 2). * For a more complete treatment, see College Algebra, Third Edition, by WILLIAM L. HART; D. C. HEATH AND COMPANY. 738 SYSTEMS OF LINEAR EQUATIONS SUMMARY. To solve a system of three linear equations in three un- knowns: 1. From one pair of the equations, eliminate one of the unknowns; eliminate this unknown from another pair of the original equations. 2. Solve the resulting equations for the two unknowns in them. 3. Substitute the values of the unknowns found in Step 2 in the simplest of the given equations and solve for the third unknown. irNote 1. To solve a system of four linear equations in four unknowns, first we would obtain three equations in three of the unknowns by eliminating the other unknown from three different pairs of the original equations. Then, we would solve the new system of three equations and, later, obtain the value of the fourth unknown. A similar but more complicated method would apply to systems in five or more unknowns. A more elegant method is presented in advanced college algebra. 1. EXERCISE 52 Solve. Do not commence by clearing of fractions. ' 3y - 5x = 1, [ 2x + y = 2, 2. \ 2y - 5z = 7, Qx + 2z = 1. 10. 3a? + - 1, 3. ' 12x - 3, x y 22 = 1, - 2 = 0. ' x - y + fa = 7, 6. | 2x + 3y + 62 * 0, ^ + 22/ + 92 = 3. ' 2s - y + 2 = 2, 8. < 12x -f y - 3 = 3, k 6x - y + 62 = 12. 7 - + = 3, = 2. 5. x y z 60 o l & m - H = 5. L* y z HINT for Problem 10. Let - x #> + 6c = 14, 4c - 6 = 2 + 3a, 14c - lOa - 96 = 10. 7. { 2A + 3J5 - 2C + 1 = 0, .A+ 3B + 2C= 1. 9. 3e = 3, - 6z - 62 + 3 = 0, , 2 - y - x - 2. 11. x y 2 5+1--; * M, - = v, and - ~ w in all equations. y z SYSTEMS OF UNEAR EQUATIONS 139 *12. 5x y = 6, 4y + * = 10, -f 62 - 3w = - 3. 4s - y = - 2. *13. x + 2y + w = 4, 2 - 3u = 6, - x 6 = - 2, + - = - 2. 104. Applications of systems of linear equations EXAMPLE 1. The sum of the digits of a certain two-digit number is 9. If the digits are reversed, the new number is 9 less than 3 times the original number. Find the given number. SOLUTION. 1. Let t be the tens' digit and u be the units' digit of the number. Then, the number is 10J -j- u. 2. If the digits are reversed, u becomes the tens' digit and t the units' digit. The new number is IQu -f t. 3. From the problem, \ 1n w ~ ' I lOtt t = 3(10< + u) - 9. We obtain t = 2 and u = 7. The original number is 27. EXAMPLE 2. Workmen A and B complete a job if A works 2 days and B works 3 days, or if both work 2f days. How long would it take each to do the job alone? SOLUTION. 1. Let x be the number of days it takes A to do the job alone, and y the number of days for B alone. 2. The fractional part of the work which A does in one day is l/x and, for B, is 1/y. Since they do the whole job, under each set of given data, 2,3 , , 12 1 , 12 1 t - + - = 1, and -= h -5 * x y o x 5 y On solving the system consisting of the preceding equations by the method of Section 102, we find x = 4 and y = 6. Any line on an (x, y) coordinate plane which is not parallel to the y-axis has an equation of the form y mx + 6, where m and 6 are constants. We can use this fact to obtain the equation of a line through two given points. ILLUSTRATION 1. To find the equation of the line through (4, 3) and (6, 9), we substitute each pair of coordinates for (x, y) in y = mx + 6: (when x 4 and y = 3) f 3 = 4m + 6, (1) (when x = 6 and y 9) \ 9 = 6m -f b. (2) The solution of [(1), (2)] is (m = 3, b = - 9). Hence, the equation of the desired line is y - 3x 9. It can be verified by graphing or by substitution that the given points lie on the graph of this equation. 140 SYSTEMS OF LINEAR EQUATIONS EXERCISE 53 Solve by introducing two or more unknowns. 1. One angle of a triangle is 30 and a second angle is four times the third angle. Find the unknown angles of the triangle. 2. The width of a rectangle exceeds its length by 5 feet and the per- imeter of the rectangle is 25 feet. Find the dimensions. 3. Two angles are complementary and one exceeds the other by 7. Find the angles. 4. A contractor has a daily payroll of $73 when he employs some men at $6 per day and the rest of his workers at $5 per day. If he should double the number receiving $5 and halve the number receiving $6 per day, his daily payroll would be $74. How many employees does he have? 5. How much of a 20% solution of alcohol and how much of a 50% solution should be mixed to give 8 gallons of a 30% solution? 6. How much milk containing 2% butterfat and how much contain- ing 6% butterfat should be mixed to form 100 gallons of milk containing 3% butterfat? 7. If each dimension of a rectangle were increased by 5 feet, the area would be increased by 95 square feet and one dimension would become twice the other. Find the original dimensions. 8. If a two-digit number is divided by its units' digit, the result is 16. If the digits of the given number are reversed, the new number is 18 less than the original one. Find this number. 9. A weight of 5 pounds is 6 feet from the fulcrum on the right-hand side of a lever. It is balanced if we place a first weight 4 feet from the fulcrum on the right and a second weight 7 feet from the fulcrum on the left, or if we place the first weight 8 feet to the right and the second weight 9 feet to the left of the fulcrum. Find the unknown weights. 10. If we seat a boy at 5 feet and a girl at 8 feet from the fulcrum on one side of a teeterboard, they balance a man weighing 160 pounds who is seated 6 feet from the fulcrum on the other side. Balance is maintained if the boy moves to 8J fe6t and the girl to 4 feet from the fulcrum o'n their side. Find their weights. 11. When we divide a certain two-digit number by its tens' digit, the result is 13. If we reverse digits in the number and then divide by the original number, the result is 31/13. Find the original number. 12. The sum of the reciprocals of two numbers is 1/2 and the difference of the reciprocals is 1/6, Find the numbers. SYSTEMS OF LINEAR EQUATIONS 13. How much silver and lead should be added to 100 pounds of a mix- ture containing 15% silver and 30% lead to obtain an alloy containing 25% silver and 50% lead? 14. How much chromium and nickel should be added to 100 pounds of an alloy containing 5% chromium and 40% nickel to give an alloy con- taining 15% chromium and 50% nickel? 15. A man divides $10,000 among three investments, at 3%, 4%, and 6% per annum, respectively. His annual income from the first two in- vestments is $80 less than his income from the third investment and his total income is $460 per year. Find the amount invested at each rate. 16. Workmen A and B complete a certain job if they work together for 6 days or if A alone works for 3 days and B alone works for 10 days. How long does it take each man to complete the job alone? 17. In a three-digit number which is 31 times the sum of the digits, the units' digit is one half the sum of the other digits. If the digits are reversed, the new number obtained is 99 greater than the original number. Find its digits. Find the equation of the line through the given points on an (x, y) coordinate system by solving a pair of equations for two unknowns, or by inspection. 18. (2, - 3); (4, 3). 19. (- 3, 1); (- 2, - 3). 20. (3, - 2); (- 3, - 12). 21. (- 4, 5); (8, - 4). 22. (- 3, 5); (- 3, - 2). 23. (4, - 2); (9, - 2). 24. An airplane, flying with the wind, took 2.5 hours for a 625-mile run and took 4 hours and 10 minutes to return against the same wind. Find the velocity of the wind and the speed of the airplane in calm air. 25. An army messenger will travel at a speed of 60 miles per hour on land and in a motorboat whose speed is 20 miles per hour in still water. In delivering a message he will go by land to a dock on a river and then on the river against a current of 4 miles per hour. If he reaches his des- tination in 4J hours and then returns to his starting point in 3J hours, how far did he travel by land and how far by water? CHAPTER 9 EXPONENTS AND RADICALS 105. Proofs of the index laws We have already employed the following results, called index laws, which govern the use of positive integral exponents. 1. Law of exponents for multiplication: a m a n = a m+n . Proof. 1. By definition, a m a- a a; (ra factors) a n = a a a a. (n factors) 2. Hence, a m a n = (a-o a)(a*a-a a) = a m+n . \_(m 4- n) factors a] II. Law for finding a power of a power: (a m ) n = a mn . Proof. 1. (a m ) n = a m -a m a m ; (n factors a m ) (By Law I) = a m+m+ ' * +m . (n terms m) 2. Since (m + m 4- + m) to n terms equals mn, (a m ) n = a mn . III. Laws of exponents for division: = a "(ffm>n); _ = _ ILLUSTRATION 1. -5 = a 6 . 15 " ~V d Or a Proof, for the case n > m. By the definition of a m and a n , a m (m factors) o" ** a -a- -a-fil'ji- $' (n factors) 1 1 _____ n m) factors a] a*a a a n-OT EXPONENTS AND RADICALS 143 IV. Law for finding a power of a product: (ab) n = a n b". Proof. (ab) n = ab-ab afc; (n factors ab) (n factors a, and b) = (a a - a) (b b b) ~ a n b n . Law IV extends to products of any number of factors. Thus, (abc) n = a n b n c n . ILLUSTRATION 2. (4a 2 6 4 ) 3 - 4 3 (a 2 ) 3 (6 4 ) 3 = Q4a*V*. (a\ n a n h) ^ M* T /a\ 4 a 4 /a 2 \ 2 (a 2 ) 2 a 4 ILLUSTRATION 3. = . - D r /^ n a a a ( , a\ Proo/. ^) ^6T"6 ; (n factors^ (n factors a) _ a-a - a __ a^ (n factors 6) 6 6 6 b n Note 1. The determination of powers of numbers is called involution. EXERCISE 54 Find each power by use of the definition of an exponent. 1. 2 6 . 2. (- 5) 4 . 3. (- 3)*. 4. (.I) 4 . 6. (j) 2 . 6. (- ). 7. What is the sign' of an odd power of a negative number? Perform the operations by use of the index laws. 8. a 6 a 8 . 9. zV. 10. 2 3 2. 11. (x 3 ) 5 . 12. (3z) 4 . 13. (2a 2 ) 6 . ' 14. (46) 8 . 15. (5afy) 4 . 16. ddW. 17. (- 2z J ). 18. (- fy 4 )'. ' 19. (- 2a) 4 . 20. (b*)\ 21. (a*) 2 . 22. (c*) 3 . 23. (d 2 )**. 24. (o6 2 ). 26. (cd)*. 26. (- .2a 2 6) 3 . 27. (- . - er- 144 EXPONENTS AND RADICALS - (' _ '. 62. (^'?. 63. \- yz 64. Prove that part of Law III which applies if m > n. 66. (a) Compute ( 2) 4 and 2 4 . (6) Under what condition on the positive integer n will ( 3) n 3 n ? 106. Imaginary numbers We have called R a square root of A if R 2 A. If A is positive, it has exactly two square roots, one positive and one negative, denoted by =t VA. ILLUSTRATION 1. The square roots of 4 are db Vi or =t 2. If a negative number P has R as a square root, then R 2 P. But, if R is either positive or negative then R z is positive and thus cannot equal P. Hence, P has no positive or negative square root. Therefore, in order that P may have square roots, we define the symbol V P as a new variety of number, called an imaginary num- ber, with the property that (V^P)* - - P and (- V^) 2 = - P. Thus, P has the two imaginary numbers V P as square roots. As an immediate extension of the preceding terminology, we agree that, if M is a real number, each of the expressions (M + V P) and (M V P) will be called an imaginary number. ILLUSTRATION 2. The square roots of the negative number 5 are the imaginary numbers =fc V 5. (7 + V 18) is an imaginary number. Unless otherwise stated, any literal number in this book will represent a real number. Imaginary numbers will not enter actively into our discussions until we meet them in the solution of equations in a later chapter. EXPONENTS AND RADICALS 145 Note 1. The somewhat unfortunate name imaginary number is inherited from a time when mathematicians actually considered such a number to be imaginary in the colloquial sense. In a similar fashion, our common negative numbers, at their first introduction into mathematics, were also called illusory or fictitious. The student will soon appreciate that imaginary numbers deserve consideration on an equal footing with real numbers. Imaginary numbers are indispensable not only in pure mathematics but also in important fields where mathematics is applied. Imaginary numbers will be studied in more detail in a later chapter. 107. Roots We call R a square root of A if R 2 = A and a cube root of A if R z = A . If n is any positive integer we say that R is an nth root of A if R" = A. (1) ILLUSTRATION 1. The only nth root of is 0. 2 is a 5th root of 32 because 2 5 = 32. - 3 is a cube root of - 27. The following facts are proved in college algebra when imaginary numbers are treated in a complete fashion. 1. Every number A, not zero, has just n distinct nth roots, some or all of which may be imaginary numbers. 2. If n is even, every positive number A has just two real nth roots , one positive and one negative, with equal absolute values. 3. Ifn is odd, every real number A has just one real nth root, which is positive when A is positive and negative when A is negative. 4. // n is even and A is negative, all nth roots of A are imaginary numbers. If A is positive, its positive nth root is called the principal nth root of A . If A is negative and n is odd, the negative nth root of A is called its principal nth root. ILLUSTRATION 2. The real 4th roots of 81 are =fc 3 and + 3 is the principal 4th root. The principal cube root of + 125 is + 5 and of 125 is 5. All 4th roots of 16 are imaginary numbers. ILLUSTRATION 3. The real cube root of 8 is 2. Also, by advanced methods it can be shown that 8 has the two imaginary cube roots ( 1 + V 3) and 146 EXPONENTS AND RADICALS 108. Radicals The radical A, which we read the nth root of A , is used to denote the principal nth root of A when it has a real nth root, and to denote any convenient * nth root of A if all nth roots are imaginary. In v^L, the positive integer n is called the index or order of the radical, and A is called its radicand. When n 2, we omit writing the index and use VA instead of \^A for the square root of A. I. A is positive if A is positive. II. ^A is negative if A is negative and n is odd. III. v~A is imaginary if A is negative and n is even. ILLUSTRATION 1. ^'sl = 3. ^32 = 2. v / 8 is imaginary. ILLUSTRATION 2. y^y = ^ because ^) = 3* = gj' By the definition of an nth root, (VA) n = A. (1) ILLUSTRATION 3. (V3)* = 3. v / (169) 7 = 169. (^ctf) 8 - 2cd*. (v'c 2 + cd + d 2 ) 4 = c 2 + cd + #. Note 1 . To avoid ambiguous signs and imaginary numbers in elementary problems, the following agreement will hold in this book unless otherwise specified. If the index of a radical is an even integer, all literal numbers in the radicand not used as exponents represent positive numbers, and are such that the radicand is positive. By the definition of A as a principal nth root, it follows that, under the agreement f of Note 1, = a. (2) ILLUSTRATION 4. ^5 = x. v'S 4 = 5. * This matter of convenience is discussed in more advanced treatments of imaginary numbers. If all nth roots of A are imaginary, it is not usual to call any particular one of them the principal nth root. f If a is negative and n is even, then a n is positive and the positive nth root of a n is a, or "^a* a. This case is ruled out by Note 1 and does not come under formula 2. For instance, if a is negative, Vo* = a. EXPONENTS AND RADICALS 747 EXERCISE 55 State the two square roots of each number. 1. 64. 2. 49. 3. 81. 4. 121. 5. J. 6. &. 7. .01. State the principal square root of each number. 8. 16. 9. 144. 10. 100. 11. &. 12. &. 13. State the principal cube root of each number. 14. 8. 16. - 27. 16. 27. 17. 125. 18. - 1. 19. - 216. 20. - State the two real fourth roots of each number. 21. 81. 22. 16. 23. 625. 24. 10,000. 26. A. 26. .0001. Find the specified power of the radical, or the indicated root. 27. Vtf. 31. V&. 35. (v^) 6 . 39. v^=~8. 43. ^64. 47. v^~T. 51. V400. 65. v'J. 59. VX)1. 60. 109. Rational and irrational numbers A real number which can be expressed as a fraction m/n, where the numerator and denominator are integers, is called a rational number. All integers are included among the rational numbers be- cause, if m is any integer, then m can be expressed as the fraction m/1. A real number which is not a rational number is called an irrational number. ILLUSTRATION 1. 7, 0, and f are rational numbers. Any terminating decimal fraction is a rational number. Thus, 3.017, or 3017/1000, is a rational number. IT and V% are irrational numbers. A proof of the irration- ality of V2 is given in the Appendix, Note 1. Any irrational number can be expressed as an endless but not a repeating decimal. Thus, v = 3.14159 and V% - 1.414 are endless but not repeating decimals. 148 EXPONENTS AND RADICALS If A is not the nth power of a rational number, and v^A is real, then VA is irrational and is called a surd of the nth order. ILLUSTRATION 2. V3 is a surd. v^64 is not a surd because v'd? 2. A surd of the second order is sometimes called a quadratic surd and one of the third order a cubic surd. The values of various quadratic and cubic surds are given to a limited number of decimal places hi Table I. 110. Rational and irrational expressions An algebraic expression is said to be rational in certain letters if it can be expressed as a fraction whose numerator and denominator are integral rational polynomials in the letters. An algebraic expres- sion which is not rational in the letters is said to be irrational in them. T . - . ,. . . , ILLUSTRATION 1. ~ - = - is rational in a and x. oa ILLUSTRATION 2. V3x -h y is not rational in x and y. ILLUSTRATION 3. The expression x\ /r 2 3x 2 \/5 is an integral rational polynomial in x. The presence of irrational explicit numbers, A/2 and V5, is of no concern. Hereafter, unless otherwise stated, in any integral rational poly- nomial we shall assume that the numerical coefficients are rational numbers. 111. Perfect powers of rational functions A rational expression is called a perfect nth power if it is the nth power of some rational expression. Also, we say that a rational number is a perfect nth power if it is the nth power of some rational number. If an integral rational term is a perfect nth power, the numerical coefficient separately is a perfect nth power. Also, each exponent in the term has n as a factor, because in obtaining the nth power of a term each exponent is multiplied by n. In verifying that a term is a perfect nth power, first factor the coefficient. ILLUSTRATION 1. 32y 15 is a perfect 5th power: 32y 15 = 2V 5 = (2I/ 3 ) 6 . X* ILLUSTRATION 2. ^r* is a perfect square: e ^ X 2 ^ 9o 6 \3o 3 / EXPONENTS AND RADICALS 149 112. Elementary properties of radicals The following Properties I and II have already been met in Sec- tion 108 as direct consequences of the definition of an nth root. II. Va n = a. (If a is positive when n is even) III. Vob = vWb. ILLUSTRATION 1. V&x? = VsVtf = 3z, because (3x) 2 = 9z 2 . ILLUSTRATION 2. \fab = ^fc^fb because (vWfc) 3 = (^a) 3 ^) 8 = 06. IV " 1V * T o 4/81 81 3 , /3\ 4 3 4 81 ILLUSTRATION 3. ^ = = -, because (- j = ^ = T , 4 u a ILLUSTRATION 4. = -r-> because -= = .' = _ 6 n / * V. // m/n is an integer, va m = a n . ILLUSTRATION 5. v^o^ = a 1 * 1 = a 4 , because (o 4 ) s * a u . i . The following proofs are complete if a and 6 are positive, be- cause then all principal roots involved are positive. The interested student may consider the possibility of negative values for a and 6. To complete the proofs, it should be demonstrated that in all cases the two sides of each formula in (III), (IV), and (V) are either both positive or both negative and hence are equal. Proof of (III). Raise (v) to the nth power: (V~ a ^b) n = (^a) n (^6) n = ab. Hence, by the definition of an nth root, ^a^fb is an nth root of ab, or n/- n/r n/ r Vav 6 = v ab. Proof of (V). By Law II, page 142, (a n ) n a"' n - a m . Hence, in m a" is an nth root of a m , or a" = Va"*. 150 EXPONENTS AND RADICALS We observe that Property II is a special case of Property V, with m = n. However, we dignify Property II by special attention be- cause, if we are able to express A as a perfect nth power, Property II gives VA by mere inspection. ILLUSTRATION 6. V&xfiy 9 - ^(2xV) 8 = 2zV . (Property II) Or, by Properties III and V, EXERCISE 56 Each radicand is a perfect power. Find the specified root. 1. VP. 2. 3d*. 3. Vol. 4. Va*. 5. 6. Vy. 7. . 8. **. 9. . 10. 11. <&. 12. v^io. 13. ^x 5 . 14. Vy*. 16. 16. V*2. 17. -^. 18. -^7?. 19. \/A. 20. 21. tff. 22. ^Jf. 23. ^3fc. 24. ^p. 25. 26. Vm*. .27. V/8E 5 . 28. ^^. 29. 30. V^. 31. tfrV. 32. \/9^. 33. 34. v/*. 36. ^- .001. 36. v 5 . 37. 38. ^SE 1 " 2 . 39. ^- 32 10 . 40. ^0625. 41. 42. ^I6a. 43. VSS 5 . 44. ^06. 46. 113. Fractional powers We have previously defined a? only when p is a positive integer. We shall now introduce other types of powers in such a way that all the types, together, will obey laws of the same forms as those for positive integral exponents. If fractional exponents are to obey the law of exponents for multi- plication, then, for example, akcfr o^ a, or (afy a; thus, a& must be a square root of a* Similarly, we should have - a 6 , or (a$) 2 a 6 , EXPONENTS AND RADICALS 151 so that a$ should be a square root of a 5 . Accordingly, if m and n are m any positive integers, we define a n to be the principal nth root of a m : m a" = -v/a 1 "; (1) i [when m = 1 in (1)] a n = v^a. ^ (2) Thus, we may use fractional exponents instead of radicals to denote principal roots. The defining equation 1 is consistent with Property V of page 149, which was proved for the case where m/n was an integer. ILLUSTRATION 1. 8* = v^ = 2. (- 8)* = v^"8 = - 2. ILLUSTRATION 2. x = v. 8* = \& = te = 4. ILLUSTRATION 3. (- 8)* = v / (- 8) 2 = vlJ4 = 4. m In (1), we defined a n to be the nth root of a m . It is very useful to m prove the theorem that, also, a n is the mth power of the nth root of a, or m a" = (v^)". (3) To prove (3) we must show that the right members of (1) and (3) are identical, or that . (4) ILLUSTRATION 4. To show that \/aJ = ('^a) 4 , raise the right member to the 3d power and use the laws for integral exponents : [(>^)4]3 = (^)i2 = [t^) 3 ] 4 = (a) 4 - a 4 . Hence, (^a) 4 is a cube root of a 4 . If we assume that a is positive, then this cube root is positive and hence is identical with the principal cube root represented by v^o 4 . In accordance with Note 1 on page 146, we agree not to deal with m the symbol a" if n is evenjand a < 0. With this case eliminated, we prove (4) by raising the right-hand side to the nth power, showing that we obtain a m , and thus demonstrating that (\^a) m is an nth root of a m : ~~ - a*. 152 EXPONENTS AND RADICALS ILLUSTRATION 5. From (3), since \/64 = 2, 64$ = (^64) 6 = 2 6 = 32. Notice the relative inconvenience of the following evaluation by use of (1) 64* = ^oT 6 = &(* = v^ = 2 6 = 32. if Note 1 . The difficulties^ met if a n is used when a is negative and n is eve are illustrated below where a contradiction " + 1 = 1 " results from reel less use of the symbol ( - 1 = V^I = (- l)i = (- 1)* = v'Pl)' 2 = V 114. Zero as an exponent If operations with a are to obey the law of exponents for multiplies tion, then we should define a so that a n aa n = a+ n = a n , or aa n = a n , or a = = 1. ' ' a n Hence, if a ^ 0, we define a by the equation a = 1. ILLUSTRATION 1. By definition, 5 = 1. (17z) = 1. 115. Negative exponents If a negative exponent is to obey the law of exponents for mul tiplication, then, for instance, we should have a 3 cr 3 = a 3 " 3 = a = 1 Hence, if p is any positive exponent of the types previously intro duced, we define or* by the equation a p a~ p = 1, or ILLUSTEATION 1. -- Br - I _ 1. In a fraction, any power which is a factor of one term (numerate or denominator) may be removed if the factor, with the sign of it exponent changed, is written as a factor of the other term. That is a _ ax~ n bx ~ b n f a a I a _^ ax"" Proof. 7 T == r ar* = r ' bx n b x n b b EXPONENTS AND RADICALS 153 We may use negative exponents in avoiding fractions. ILLUSTRATION 2. -rj- = 17o TJ = 17o6~ 2 . ILLUSTRATION 3. To express the following fraction wityi positive ex- ponents, we may use (2) mechanically: 3a" 2 6 3 _ 3c*b 3 _ Sc 3 ^ 3 fQ . IF 3 ^ 4 " = ~oJoJ = ~HT' ( ' In more detailed fashion we obtain (3) as follows: Q/7~ 27)3 /I \ / 1 \ O7)3 ^3 2J)3/^ Otv C/ / ^ JL i I / j -*- 1 OC/ O Ot/ O x *v Pw "r*-*')-'-^-?)--*-;?--*-- (4) The student should act as hi (3) but should also appreciate (4). EXERCISE 57 Find the value of the symbol by changing from a fractional exponent to a radical or from a negative to a positive exponent if necessary. 6. 4- 1 . 10. 81*. 16. (i)*. 20. 16-*. 25. (.36)*. '. 30. (- S)- 4 . 31. (- 2)~ 6 . 32. (- 5)- 3 . 33. (- 1)*. 34. (- 8)*. 35. (- 8)-*. 36. (- 27)-*. 37. (- 125)*. 38. (.0081)*. 39. (.0001)-*. Find the value of the symbol by use of formula 3, page 151. 40. 8*. 41. 25*. 42. 4*. 43. 36*. 44. 81*. 45. 125*. 46. ft)t. 47. (^)*. 48. (- 27)*. 49. (- 64)'*. Express with positive exponents. 1. 9*. 2. 25*. 3. 8*. 4. 27*. 6. 2~ J . 7. 35- 1 . 8. 16. 9. 4*. 11. 3-2. 12. 5~ 3 . 13. S- 4 . 14. 16*. 16. (*)*. 17. GM*. 18. (- 8)*. 19. 7. 21. 9~*. 22. 27-*. 23. ft)-*. 24. ()-' 26. (.09)*. 27. (- S)- 2 . 28. 725. 29. (- 5 60. a~ 3 . 51. b~*. 62. c-. KS f t r ^ / ii i/W ftls t^ 54. a~ 3 6 2 . 66. c 2 d~ 8 . 66. 62T 2 . 67. 3A~ 4 . 68. 2o6- 3 . 69. 4x~*y. 60. arV 5 ' 61. 2oy~ 5 . 62. Zax-*y-*. 63. 4- 1 aar 3 . 64. 5~ 2 ca;- 3 . 66 _. 3 miljm 67 a 3 69 * b~~* f? a~ 8 rf-2* ' 5 754 EXPONENTS AND RADICALS 6a~ 8 75.^- 76. ^r- 77. ^br 78. Write without a denominator by use of negative exponents. 80. i- 81. i- 82. - 83. - 4 - 84. ^ y* & x* y 4 a* 4a M 2a ftft &c* OO. -.i oft O v 90. /- rvo\K* 8L (L03)" 8 (1.04)" "* (1.05) 8 Rewrite, expressing each fractional power as a radical and each radical as a fractional power. 94. x*. 95. z*. 96. a*. 97. &i 98. 3ai 99. 5c*. 100. ax*. 101. fcci 102. v/^ 3 . 103. v/6. 104. (56)*. 105. (6c)*. 106. Vy>*. 107. (2^)*. 108. (4C 3 ) 9 , 109. (7a')i 110. (2a^)t. 111. <Q>. 112. ^fe 11 . 113. >/5?. 114. -C^HM. 116. Vo 2 - 36. 116. #(a 4- 6) 2 . 117. ^(c - 3d) 3 . 118. Va 2 -f 6 2 . 119. ^a 8 - fe 8 . 120. ^ST^ 3 . 121. vT^ll 122. Compute (- i)~ 8 ; (- J)-; (- .04)~. 116. Extension of the index laws We have defined a p if p is any rational number but we have proved the index laws only for positive integral exponents. A detailed discussion, which we shall omit (see Appendix, Note 2), shows that the formulas of Laws I to V of pages 142 and 143 apply if the ex- ponents are any rational numbers. Hereafter, unless otherwise specified, to simplify an expression involving exponents will mean to perform indicated operations as far as possible by use of Laws I to V, and to express the result without zero or negative exponents. Moreover, unless specifically requested, we shall not introduce radicals for powers having fractional exponents. In operations involving exponents, it is frequently convenient to express the numerical coefficients of terms as products of powers of prime factors. EXPONENTS AND RADICALS 755 ILLUSTRATION 1. (x*)$ - aH = x 4 . /21te\* /8-27afa\* = /23 8 x\* 2*3*3* \125arV V 125 / V 5 8 / " 5* "" 25 " 1 i 1 !"* 1 1 / 1 Y" 17 iV (-125) -Lrs/ EXAMPLE 1. Simplify: .-a ^f FIRST SOLUTION. Change to positive exponents: J_ J_ 1 q*y* 1/ 2 a* y 2 a V SECOND SOLUTION. Multiply both numerator and denominator by to eliminate negative exponents : (<rV*)(oV) (a- 2 + if*) 1 + ay t/ 2 + a 2 We may use the special products of Chapter 5 in multiplying or factoring polynomials involving negative or fractional exponents. ILLUSTRATION 2. (ar 2 y$)(x~* + y$) (Type II, page 85) ILLUSTRATION 3. (z* - 2?/- 1 ) 2 (Type IV, page 85) ILLUSTRATION 4. 2x- 2 + z" 1 - 6 = (2s;- 1 - 3)(xr l + 2). EXERCISE 58 Simplify and, if no letters are involved, evaluate. 1. a&e 2 . 2. a&c*. 3. aAr 4 . 4. a s a. 5. (a*) 4 . 6. (3 4 )t. 7. (2)1 8. (x*) 4 . 9. (4z 2 )*. 10. (3x~^. 11. (5a~ 2 ) 8 . 12. (cr 1 * 2 ) 8 . 13. (5)i 14. (*t)t. 16. (o*)*. 16. (ax- 1 ) 4 . 17. (a*)- 1 . 18. (a 2 )" 2 . 19. (&)-*. 20. (a")-. 21. (2X- 2 ) 8 . 22. (5a-) 2 . 23. (oV)-*. 24. ?56 EXPONENTS AND RADICALS 26. (3x-y) 2 . 26. (5arV 3 ) 2 . 27. (Gar 1 !/- 3 )- 2 . 28. 29. 32i 80. 271 31. 125*. 32. 216i 33. (4-V)t. 34. (a:- 4 !/ 2 )*. 35. (27a- 8 a*)i 36. (25ar 2 )-*. 2-5 a \l n 1 VT . Qfl - QQ q of. r* oo. TT: w rr 1 f ^ ;j~ 44. ^- 46. -^. 46. 47.^. 48.^. 49. ^S. 60. 5L -^f-s- 62. i~ 63. 67. PT-r - 68. 69. (27w)i 60. (32a 6 6 6 )*. 61. (125x 6 )^. 62. (216ar 6 )i Simplify to a single fraction in lowest terms without negative exponents. 63. a- 1 + b~ l . 64. 3a~ 2 + 6. 66. a^ - Zr 3 . 66. 5a~ l + &-' A7 a ~ lb Aft ^-^ AQ < !r V 2 7n S- 2 -4-2 67 ' a -l + ' ** /,-2 J.A* 69 ' x^J_,r-2* 70 - rt -l J_ A-l 4-1 _ a - 71. " + L- 72. T^ fl _.. 73. " " 74. a -2 _ 5-2 4-2 _ a -2 a - 8 _ 6 -- 3-2 + ^-2" 76. (Sat)- 1 . 76. (c + M)' 1 - 77. (cr 1 + b^}~ 1 . 78. (4a~ 8 - &)->. Expand and simplify, without eliminating negative exponents. 79. (ar 1 - j/- 1 )^- 1 + 2/" 1 ). 80. (3a - 6' 81. (4x - y)(4a; + y). 82. (jg - 83. (xi + y*)(a:* - y*). 84. (2xi - 3)(3x* + 1). 86. (5a* - 2)(3ai + 4). 86. (a* - 26*)(a* + 36*). 87. (<r + 6) 2 . 88. (or 1 + 3) 2 . 89. (a* + Z>*j 2 . 90. (2a* + 36ty. 91. (a' 1 + y 2 ) 2 . 92. (x 4 - 2?/- 1 ) 2 . 93. (a- 1 + 6) 3 . 94. (2 + ar 2 ) 3 . 96. (3 - y~ 1 ) 3 . 2 + 4). 97. (3 - 6arH2 EXPONENTS AND RADICALS . 157 98. (z 3 V*)i 99. (3a;*) n . 100. (3*a*6 n )*. 101. (4a~*6 n ) m . / a 6 * \* * / 4z 4 * \* /.125;r 9 \3 -- /. 102 - to 103 - 455= l04 - (* 106. (a - 6*) (a 2 + a6* + 6). 106. (c* + <*)(< - <*d + if Factor into two factors involving fractional or negative exponents. When possible, factor as the difference of two squares or as a perfect square. ILLUSTRATION 1. x y = Or, x - y = (a*) 3 - (yty = (x* - y*)(x* + zfyi + 107. a; 2 - jf. 108. a~ 2 - 6 4 . 109. Oar 2 - 110. ^ - x*. 111. 4x^ - jft. 112. 9a* - 113. 9x* - 25^. 114. 16a* - 49y*. 116. 4o - 96. 116. x* - Zxy- 1 + tr 2 . 117. 2 - Gzor 1 + 9x~ 2 . 118. 4a~ 2 - 4a~ 1 6~ 1 + b~\ 119. 9o~ 2 - tor*-* -f b~*. 120. 4ot - 20ai6V + 256*. 121. 122. a:- 2 - 4X- 1 - 5. 123. 124. - tyf . 125. Sa + 276. 126. e - By. 127. 216 - if Find the quotient by use of factoring. 128 . - r> 129 . uc!. 130 . - - 117. Simplification of a radicand Although it is possible to express a radical as a power with a frac- tional exponent, in some operations it is convenient to retain the radical form. This remark applies in particular to the following operation. SUMMARY. To remove factors from the radicand in a radical of order n: 1. Separate the radicand into factors of which as many as possible are perfect nth powers. 2. Find the nth root of each perfect nth power and express the final result by use of the property 'Vab = v^v^S. 158 EXPONENTS AND RADICALS ILLUSTRATION 1. A/147 = A/49-3 = A/49A/3 = 7A/3; A/147 = 7(1.732) = 12.124. (Table I) T o 3 /o , 5 3/3az 3 ILLUSTRATION 2. \/3a + -7 = \; - \ x 3 \ x Hereafter, unless otherwise specified, if a radicand involves frac- tions, reduce it to a single fraction. If a radical is of order n, simplify the radicand by removing from it any factor which is a perfect nth power. Also, in a radical of odd order, change the radicand to a form where all signs are " + " if possible. ILLUSTRATION 3. v / ^"2 = v / ^~Iv / 2 = - 1 v'- fl - 26 = v'- (a + 26) = v'^Tv'a + 26 = - v'a + 26. In a sum, two or more terms involving the same radical as a factor may be combined by factoring. ILLUSTRATION 4. 5V/3 + 26\/3 = (5 + 26)\/3. V20 -f 2\/45 = V5\/5 + 2V9A/5 = 2\/5 + 6^5 = 8>/5. (\/3 V7 + 3V^5) cannot be simplified in form. ILLUSTRATION 5. 7\/5 - 3V5 + 6\/5 = V5(7 - 3 + 6) = 10\/5. (Using V5 = 2.236 from Table I) = 10(2.236) = 22.36. EXERCISE 59 Simplify the radical and then compute by use of Table I. 1. A/18. 2. A/75. 3. V20. 4. A/24. 6. A/200. 6. A/500. 7. A/27. 8. A/108. 9. A/72. 10. 11. A/5. 12. A24. 13. A6. 14. A54. 16. v108. 16. AV3T3. 17. xVZTs. 13. A^~l6. 19. A^^54 20. Simplify by removing perfect powers from the radicand. 21. Vx\ 22. Vo*. 23. \Vy*. 24. \V5. 26. v^io. 2 6. 27. A/9o*. 28. A/8^. 29. v^. 30. 32. A20o*. 33. V2. 34. EXPONENTS AND RADICALS ?59 36. 4 39. 43. <> 47. v^ 61. * ^V12V 69. V9 + 92/ 2 . 62. 66. 36. V75?. 40. V- 27a 8 . 44. VsiaV 1 ' 48. VxJla 5 . 62. 37. 41. 46. 49. 63. 38. V375J 5 . 42. ^ - 128x. 46. V/320V 5 . 60. 64. 66 - V- |i- 60. V4 - 4o 2 . --S- 68. 63. 61. 64. + 5a 2 6. 66. 67. 68. 2d , d ' ** * ab ^ a 2 Simplify and then collect terms, exhibiting any common radical factor. 71. 5A/2 + 3A/2. 72. 3\/3 - a\/3. 73. 2\/l8 + V50. 74. Vl2 + V75. 76. Vl47 - VS. 76. ^24 -h 2V/8I. 77. aV2 - 56V2. 78. VS + V25a. 79. 80. ^Sa 4 + *VVa. 81. ^48^ - V48i/. 82. 118. Products and quotients of radicals The product or the quotient of two radicals of the same order can be expressed as a single radical by use of the following properties of radicals. Also, we recall that, by the definition of a root, (\^J) n A. ILLUSTRATION 1. (2V3)(5V6) = loV3\/6 = 10Vl8 - 30V2. ILLUSTRATION 2. (5^3) - 5(^3) 8 - 125(3) - 375. . ILLUSTRATION^ V S ^J/oS /^6 /a 1 G ^ - Vg' ^p ~ W " V^ " b\b' 760 EXPONENTS AND RADICALS ILLUSTRATION 4. EXAMPLE 1. Multiply (2\/3 + 3V2x)(3V3 - \/2z). SOLUTION. The product is 6(V3) 2 - 2\/3\/2z -f = 18 - 2Vfo -f 9\/6i - 3(2x) = 18 - Comment. The student may prefer an expanded solution: 2V3 + 3\/2z - V2s (multiply) 4- 18 + 7\/3\/2x - 3(2z) = 18 - Qx + 7\/6x. If we remove a positive factor P multiplying a radical \^A we must multiply by P n under the radical sign because ILLUSTRATION 5. 3\/6 = EXERCISE 60 Express by means of a single radical, and then compute by use of Table I if necessary. 2. V2\/3. 3. \/5vTo. 4. \/3Vl5. 6. vWl2. 6. (-^2) 8 . 7. (2^6). 8. (3>/5) 2 . 9. 3V3(2\/6). 10. 5V6(2V2l). 11. V30>/35. 12. (2V3V5) 2 . 13. (3^2) 3 . 14. 16. 3>? / 36(v / 45). 16. ^^l^ls. 17. v^ VT/s VT4 v'To \/QQ 18. -^- 19. -^ 20. ^?- 21. 4^- 22. Express as a single radical and then simplify. 23. -j^- 24. ^- 26. -^- 26. -^=- 27. - /rfc V4C ^ 7 ^^ a/^^r Multiply, simplify, and collect terms. 28. V3aVl5a. 29. ZVxVSx*. 30. EXPONENTS AND RADICALS 76? 31. zxfa*. ' 32. VfoV25xy. 33. 34. (3VS). 36. 5\/3a~'3. 36. 2v / 4^ 4 . 37. 38. (5i~+i;) 2 . 39. (6V?~+~1) 2 . 40. (- 41. (2 + 3V6)(3 - V5). 42. (2\/2 - 3)(5\/2 + 2). 43. (\/3 - \/2)(\/3 + V5). 44. (2\/3 - V5)(3V3 -f V5). 46. (3\/2 + V3)(V2 + 4V). 46. (Vg - \/5)(V6 + Vg). 47. (2V3 - V7)(2V3 + V?). 48. (5\/2 - 3\/3)(5V2 -f 3\/3). 49. (2\/3 + V5)(V2 + 4). 60. (3^2 + V3)(\/2 + V6). 61. (v^ + 5) 2 . 62. (3 + 5\/2)2. 63. (\/2 - 2\/3) 2 . 64. (V - 2Vi)(Vx + 2V2). 65. 56. (2#x - 3)(5v^ + 2). 67. 68. (Vx - 5V^)2. 69. (Va + 6v^) 2 . 60. (aVy - 61. \^^^- x- 62. v^- Replace the coefficient by an equivalent factor under the radical sign. 63. 3V2a. 64. sVfo. 65. aVbx. 66. 67. S^P. 68. 2^a. 69. 2V/36. 70. 119. Rationalization of denominators To rationalize a denominator in a radical of order n, after the radicand has been expressed as a simple fraction, multiply both nu- merator and denominator of the radicand by the simplest expression which will make the denominator a perfect nth power. In particular, if the radical is a square root, we make the denominator a perfect square; if a cube root, we make the denominator a perfect cube. T 1 /3 /S 7 ? VSI 4.583 A , K frr , , T , ILLUSTRATION 1. 1/7 = \~7T = ~~T~~ ~ 7 = -655. (Table I) Notice the inconvenience of the following attempt at computation of without rationalization of the denominator; a long division is required. 1<732 ' (Table T ILLUSTRATION 2. 762 EXPONENTS AND RADICALS ILLUSTRATION 3. 3 /64ar4 __ 3/64~ _ 8/43.32? _ "^4 3 -3a; 2 _ 4^33* V 9 "" \9x* ~ \9z 4 -3z 2 "* ^oO = 3a; 2 + <* a 2 6 oW EXERCISE 61 Compute by use of Table I a/ter rationalizing the denominator. 1. V}. 2. V|. 3. Vf . 4. V|. 5. Vf. 6. VJV. 7. ^}. 8. ^J. 9. ^S. 10. ^5. ^~dW 12. ^pj. 13. V^. 14. ^. 15. ^T03. 16. Vl)07. 17. \/^12. 18. ^- .128. Eliminate any negative exponents, rationalize the denominators, and col' lect terms involving a common radical factor. 19. \K- 24. 29. 34. . 26. i/T- 26. S^- 31 "V5S?" s=s- 36. 22. \ F?> 27. 32 32 - w * 37. 28. 33. 8/6 9o 4 tf-1 44. v/ 47. VF 5 . 48. VF. 5 41. 46. 42. 46. 49. 60. v^a~ 7 6~ 2 . c 3 ^ V^Tft' 61. 66. V J + ar'. 67. 60. ^^ 63. 68. v^| + or*. 69. v'a -f 61. 5VJ -h V45. 62. 10V| $4. ^ -h <^. 66. ^F 1 EXPONENTS AND RADICALS 163 120. Additional devices for rationalizing denominators The method of the following illustration is frequently equivalent to the procedure of the preceding section. ILLUSTRATION 1. The denominator below is multiplied by ^2 in order to make the new radicand a perfect cube: T V3 V3 \/5 Vl5 ILLUSTRATION 2. 7= = = = = = V5 \/5 V5 5 If a denominator has the form aV& c V5, we can rationalize it by multiplying by aVb + c Vd because (aVb - cVd)(aVb + cVd) - (a\/6) 2 - (c\/5) 2 = a'6 - 3V^ - \/3 3\/2 ~ V3 2V2 + \/3 ILLUSTRATION 3. p - 7= = 7= - 7= 7= - 2V2 - \/3 2\/2 - \/3 2>/2 + \/3 6(\/2) 2 + (3 - 2) V5 - (V3) 2 9 + V6 9 + 2.449 _ - (V3) 2 8 ~ 3 5 In finding the quotient of two radicals, it may be desirable to write the expression as a single radical before rationalizing. ILLUSTRATION 4. EXERCISE 62 Rationalize the denominator and, if no letters are involved, compute by use of Table I. Collect terms in any polynomial. 1 1 . JL y * 2 -L. 4B - *v 5 < 6 s 'vl' 1 3 * rn* V7 6. A. V3 fi 3 D. . vU , V5 7 ' Vl' 8.^- *' vf' ia^. 2V5 11 lg , i 13 2 ~ V3 14.; 2V3 + 5 15. 1 JLU* 3 + V3 V3-4 3 + 2V2 1ft. 3 17 . 1ft V^-2\/3 2>/2 + V3 V2 + V3 164 EXPONENTS AND RADICALS V2 + 3V3 V2 + V5 * y - 7=" 4v - - * . - - 2V3 - V5 2V3 + 3 V2 3V5 - V2 22. . 23. - 24. 2\/3 - 3\/7 2V6 + V7 2 V? + \/3 25. (3 - V2) -7- (2 + V2). 26. (\/3 + VTI) + (1 - VII). 27- : S 28. -|L 29.-,^=. 30. -^=. 31. 32. ~^> 33. -^:. 34. ~ 36. -^^. 37. l . 38. X K Ol O */ *-v r" - 40. 41. o/ ^ y . 42. = "v 16a 3 6 2 v27xy* 2\fc \^2a 43. -^p=. 44. ^g- 1 . 46. 8 - + ^ 3 + SVa; + 2 Vo + 6 - V2o 3 2 46. -= -= 7=' 47. =: p: >. = =.< V3-V2 + V5 V3-V6 + V5 V^-VJ 49. vS- 50. v fc - 61. hfW - m ^ m 1 21 . Changing from fractional exponents to radicals To change a product of powers involving fractional exponents to radical form, first change the fractional parts of the exponents to fractions with their lowest common denominator. ILLUSTRATION 1. 1 22. Radical operations performed by using fractional exponents In this section the results will be desired in radical form. I. To find a power or a root of a radical: 1. Express each radical operation by use of a fractional exponent applied to the radicand. 2. Simplify the indicated power of the radicand and express the result as a radical. EXPONENTS AND RADICALS 165 ILLUSTRATION 1. ILLUSTRATION 2. (2v / 5x) 2 = 4[(5z)*] 2 We recognize that, in simple cases, it may be unnecessary to in- troduce fractional exponents in an operation of Type I. Also, with experience, one observes simple rules such as "the wth root of the nth root of A is the mnth root of A." (1) ILLUSTRATION 3. v^ 3 = 3. A^Va = v'a. II. T'o ^/ki the product or the quotient of two radicals of different orders: 1. Express each radical as a fractional power of its radicand, and change the resulting fractional exponents to their LCD. 2. Rewrite in radical form and combine into a single radical. ILLUSTRATION 4. V^3\/2 = 3*2* = ILLUSTRATION 5. (3a6)t 27a 3 3a III. To reduce the order of a radical, when possible: 1. Change to fractional exponents in lowest possible terms with a common denominator. 2. Rewrite finally in radical form. ILLUSTRATION 6. ^625 = In reducing the order of a radical, it is convenient to commence by expressing the radicand as a power of some expression. ILLUSTRATION 7. \/16z 2 = v"(4z) 2 = 166 EXPONENTS AND RADICALS 1 23. Simplest radical form As far as problems in this text are concerned, we agree that an expression is in its simplest radical form if all possible operations of the following varieties have been performed, with any negative ex- ponents eliminated. SUMMARY. To reduce a radical expression to simplest form: 1. .Express any power or root of a radical, or product of radicals, as a single radical. 2. Reduce each radicand to a simple fraction in lowest terms. 3. Rationalize all denominators. 4. Remove from each radicand all factors which are perfect nth powers, where n is the order of the radical. 5. Reduce each radical to the lowest possible order. 6. Combine any terms with a common radical factor. It must not be inferred that the preceding operations need be performed in the specified order. To simplify a radical expression will mean to reduce it to simplest radical form. ILLUSTRATION 1. To simplify the following radical we rationalize the denominator, and finally notice that the order of the radical can be reduced. ef^~ = 6/q2.4c 2 = v \16c l \16-4c 12 2c 2 2c 2 2c 2 EXERCISE 63 Change to simplest radical form. 1. atb*. 2. x*y*. 3. 5ai 4. 2z*. 6. 6. a$bt. 7. a$b%. 8. xly$. 9. xty$. 10. Reduce to a radical of lower order. 11. \^. 12. ^wT 2 . 13. v^. 14. Vy\ 15. 16. v^. 17. 1??. 18. Vtf. 19. #9. 20. 21. ^27. 22. 4T25. 23. ^36. 24. ^49. 25. 26. ^8l. 27. J/W. 28. v/iF 2 . 29. V%a*. 30. EXPONENTS AND RADICALS Change to simplest radical form. 31. (V&. 32. VS. 33. V5. 34. 167 47. (2v / 3) 4 . 48. (v^ 2 ) 6 . 49. 36. (v'a)'. 37. (\/3) 4 . 38. (V2)'. i > 41. (V5). 42. (\/6) 4 . 43. 4 fAj/0^4 IBV* \ V ^t/y 61. V^. 52. V\/5. 63. 66. -V^a;. . 67. 61. 66. 69. v^ -5- Va. 72. V2 -5- -^2. 75. v^ 4- \/2. 78. ^ 81. >^S. 82. 39. (v"5) 4 . 44. 54. A/V1. 35. 40. 45. 50. 66. 62. 66. ^2^2. 70. v 73. 76. v 79. v 83. 59. 63. ^yv 67. V 84. 86. vx^. 90. 94. 87. 3 91. (v^x 2 ) 6 . 95. 88. v 92. 99. 100. 101. 46 4 60. 64. 68. 71. v^-r- 74. ^3 -i- 77. V6~ 80. %Vy -. ~ 85. 89. 93. 97. Vx~* + 102. 103. (v^) 7 . 104. (2v / 5) 4 . 107. Wjty 2 . 108. 111. (cv/4) 3 . 112. v 105. 109. (av^) 6 . 113. v 106. 110. (6V/3) 6 114. V3a m . 119. - + a 118. - 120. -j 122. 4 + 9y~ 2 . 123. Va + b - \/3a ;+ EXPONENTS AND RADICALS EXERCISE 64 Chapter Review Find the value of each symbol, using Table I if necessary. 16. 21. 26. 6~ 3 . 2. (- 2)~ 4 . 3. (- 15). 4. 125i 5. 27*. 4*. 7. 25*. 8. 9~*. 9. (- 8)-i 10. Vf Vf. 12. (VIM) 2 . 13. (V239) 3 . 14. Vg. 15. V^. y*- 17. V^. 18. V^^. 19. VV|. 20. ^Vlf , 2V27 22. 3VI * 23 V ^ 4% 4 ^^ O W 9K V6 Vl25a 27. ^~ ^ -. 5! 4O. j-j- -\/ JQ O "^ O * ., y - y - V3 + V2 2\/2 - Write without fractions by use of negative exponents. 06' Ovr OJL* ' a + 26 Express without radicals, or negative signs or zero in the exponents, and simplify by use of the laws of exponents. 33. &?. 34. <Vf. 35. V9. 36. Vi^. 37. 38. v^p. 39. v^. 40. ty&o}. 41. ^C^. 42. 43. &r 8 . 44. W~ 5 . 46. (16z)*. 46. (a6-)*. 47. K1 51. 4 62. (2o + 6- 2 )- 1 . 63. S^- 1 + 2y)~ l . 64. 5(a~ 2 + 36- 1 )' 2 . 65. State the principal 4th root of 256 and principal cube root of 27. Change to simplest radical form. 68. ^iaF 3 . 69. v'- 62. #&. 66. (V2a). 66. V96V. 60. 64. aW. 68. (2V3X) 3 . 72. -C^^. 76. V5V5. 67. ^32^. 61. 65. 73. 77. \/2lfyr\ 70. 74. v/49. 78. 63. \/J 4- Vj. 67. 71. 76. 79. EXPONENTS AND RADICALS m 84. 3\/ + 86. (V2- 89. 91. + &V8. Va . 6 - Va -b 93. 96. 96. V(a 82 - 86. 87. - . 5VI2 - a 88. (Vo + 90. 92. 276 3 -i- 36. 3 v 2 94. x -J- x^a 2 - 6 2 . -f- 97. V6- 1 - a~ 2 ^- (V a - 169 CHAPTER 10 ELEMENTS OF QUADRATIC EQUATIONS 1 24. Terminology and foundation (or imaginary numbers By definition, R is a square root of 1 in case # 2 = 1. But, if R is either positive or negative, R 2 is positive and hence cannot equal 1. Obviously R = is not a square root of 1. Hence, no real number is a square root of 1. Similarly, if P is positive, any square root R of the negative number P would satisfy the equation R z = P. But, R 1 is positive or zero for all real values of R and hence P has no real number R as a square root. Therefore, in order that negative numbers may have square roots, we proceed to define numbers of a new type, to be called imaginary numbers. Let the symbol V 1 be introduced as a new variety of number, called an imaginary number, with the property that * - 1 V - 1 = - 1. For convenience, we let i = V 1. Then, by definition i*i 1 or i 8 = 1. We agree f that the operations of addition, subtraction, and multiplication will be applied to combinations of i and real num- bers as if i were an ordinary real literal number, with i 2 =A1 Then, in particular, (- ,') = ,-=_!, (2) so that i, as well as -h i, is a square root of 1. Any positive integral power of i can be easily computed by recalling that i 2 1 and hence < -(*)(*) -(-1)(-1)-1. (3) * Refer to the introduction of negative numbers in Section 6 and observe the similarity of the present discussion, t This procedure can be arrived at logically by a more advanced discussion. ELEMENTS OF QUADRATIC EQUATIONS 171 ILLUSTRATION 1. t 3 = i(i 2 ) t( 1) = i. t 13 = W - ILLUSTRATION 2. ' (3 + 5i)(4 + t) = 12 + 23t 4- 5t 2 - 12 + 23i - 5 = 7 4- 23t. If P is any positive number, we verify that - - P; (- VP) = t^P = - P. Hence, the negative number P has the two square roots db t' Hereafter, we agree that the symbol V P or ( P)^ represents the particular square root t'VP. Then, P has the two square roots db V P = iVP. This agreement about the meaning of V P is equivalent to saying that we should proceed as follows in dealing with the square root of a negative number: (4) ILLUSTRATION 3. The square roots of 5 are = ILLUSTRATION 4. V- 4V- 9 = (iV4)(i\/9) = 6i 2 = - 6. ]Vofe /. The formula VaVfe = Vo6 was proved only for the case where Va and VJU are real. We can verify that the formula does not hold if a and 6 are negative. Thus, by the formula, , V^I\A=~9 = V(- 4)(- 9) = V36 = 6, which is wrong, because the correct result is 6 (in Illustration 4). If a and b are real numbers, we call (a + bi) a complex number, whose real part is a and imaginary part is bi. If b 7* 0, we call (a 4- W) an imaginary number. A pure imaginary number is one whose real part is zero; that is, (a -f bi) is a pure imaginary if a = and 6^0. Any real number a is thought of as a complex number in which the coefficient of the imaginary part is zero; that is, a - a -f Oi. In particular, means (0 4- Oi). 772 ELEMENTS OF QUADRATIC EQUATIONS ILLUSTRATION 5. (2 3i) is an imaginary number. The real number 6 can be thought of as (6 + Oi). Note 2. In this book, unless otherwise stated, all literal numbers repre- sent real numbers, except that hereafter i will always represent V 1. Any literal number in a radical of even order will be supposed positive, if this is possible and adds to our convenience. itNote 8. The student has seen that we introduce imaginary numbers in order to provide square roots for negative numbers. It might then be in- ferred, incorrectly, that still other varieties of numbers would have to be introduced to provide cube roots, fourth roots, etc., of positive and negative numbers and also roots of all orders of imaginary numbers. An extremely interesting theorem is that the real numbers and the imaginary numbers, as just introduced, provide all the numbers we need in order to have at our disposal roots of all orders of any one of these numbers. Explicitly, in more advanced algebra,* it is proved that, if k is any positive integer, then any complex number N has just k distinct kih roots, which are also complex numbers (including real and imaginary numbers as special cases). irNote 4- In the theory of electricity, it is customary to use j for V 1 because the letter i is reserved for a different purpose. EXERCISE 65 Express by use of the imaginary unit i and simplify the remaining radical. 1. V^~9. 6. V^ 11. \ 16. 17. v^^OO. 21. V-^36. 22. V-- 27. V^ 29. V- a 2 ** 2 . 30. V- Ifo*. 31. v 32. V 33. V- 34. V- 12t0. o 2 ^ 38.^ 36. V A / cW 89 - V- 26 ' 36. V- 4ay. 37. V-27F. 4L V- 5- 46. -63. 46. -f * See De Moivre's Theorem and related topics in any college algebra. State the two square roots of each number. 42. - 81. 43. - M. 44. - ELEMENTS OF QUADRATIC EQUATIONS 173 Perform the indicated operation and simplify by use of i 1 1 until i does not occur with an exponent greater than 1. 47. i 6 . 48. i 7 . 49. %. 60. t 8 . 61. t 18 . 62. i. 63. (3 - t)(3 + t). 64. (3i + 5) (4 - 3i). 66. (3 + 2i)(3 - ft). 66. (2i + 3) 2 . 67. (3i - 2)(5i + 7). 68. (4t + 3)(- ft + 5). 69. (5 - 2i) 2 . 60. (3i - 4)*. 61. (4t + 5) 2 . 62. (2i 2 - 3t -f 5) (ft - 3). 63. (i 3 - 2i 2 + 3)(3i 2 - 5). 64. (4t - 7)(ft -f 5i 2 ). 66. (ft + 4i - t 2 )(2 + 3t). 66. V^~2V^~8. 67. V^V- 75. 68. V- 27\^ : ~3. 69. V^2(3 - 5V^1). 70. V^~3(5 - V^27). 71. (5 - V^~8) 2 . 72. Substitute x =* 3 + 6i in (a; 2 - 60; + 34). 73. If /(*) = 3a: 2 + 2x - 7, find /(ft); /(- 3i); /(2 - 5i). 74. If /(x) = a* - a? - 3, find /(ft); /(I + i). Obtain an expression of the form a + bifor the fraction. 3i-4 HINT for Problem 75. Multiply numerator and denominator by 5 4i. 125. Equations of the second degree A quadratic equation, or an equation of the second degree, is an integral rational equation in which, after like terms are collected, the terms of highest degree in the variables are of the second degree. A quadratic equation in a variable x can be reduced to the standard form ax a + bx + c = 0, (1) where a, b, and c are constants and a 7* 0. A complete quadratic equation in x is one for which 6 7* 0, and a pure quadratic equation is one for which & = 0. ILLUSTRATION 1. 3z 2 5x + 7 = is a complete quadratic equation and 5x 2 7 = is a pure quadratic in x. 1 26. Pure quadratic equations To solve a pure quadratic equation hi x, solve the equation for x 2 and extract square roots. 174 ELEMENTS OF QUADRATIC EQUATIONS EXAMPLE 1. Solve: 7y* = 18 + 3y 2 . SOLUTION. 1. ly 1 - 3y* - 18; 4y 2 = 18. 2. Divide by 4: y* = f . 3. Hence, j/ must be a square root of j. On extracting square roots and using Table I, we obtain - 2.121. EXAMPLE 2. Solve: 2j/ 2 + 35 = - 5y*. SOLUTION. 7y 2 = 35; 2/ 2 5. Hence, y = =b V 5 = iV5. EXAMPLE 3. Solve for x: a*x* + ft 2 = o&c 2 + a 2 . SOLUTION. 1. o 2 ^ 2 a&c 2 = a 2 ft 2 . 2. Factor: (a 2 - ab)x* = a 2 - 6 2 . 3. Divide by (a 2 06) and reduce to lowest terms: 2 . a?-& = (a ~ b)(a + 6) X tf-ab a(a - 6) '* . + 6 or. a^ = ! a 4. Extract square roots and rationalize the denominator: K a a 1. If the coefficients in a quadratic equation are explicit numbers and if a radical occurs in any solution which is a real number, always com- pute the decimal value of the solution by use of Table I. If it is desired to check such a solution, substitute the radical form instead of the decimal value, unless otherwise directed by the instructor. The approximate decimal value, as a rule, could not lead to an absolute check. EXERCISE 66 Solve for x, or otherwise for the letter in the problem. 1. 5x 2 - 125. 2. 3x 2 = 12. 3. x 2 = - 9. 4. 4x 2 - - 9. 6. 9x 2 - - 25. 6. 2x 2 - 3. 7. 5x 2 - 7. 8. 3x 2 - 11. 9 A 7*2 == / in QT^ Jt 11 ^/vr2 = h 19 9hr% 1 ft t JU - G. XV* 7* / XX* OCftv " fii% JLm% &WU - 1Q. 13. 15 - 162 2 - 4. 14. 9 - 7* * 6. 15. Jx 2 - 1 Jx 2 . 16. 9x 2 + 49 - 0. 17. 7X 2 - 5 - 3s 2 . 18. Jx 2 - } * fcc 2 . ELEMENTS OF QUADRATIC EQUATIONS 175 19. l&r 2 + 64 - 0. 20. f z* + 4 = a*. 21. fcc - | - 22. 4oz 2 - c - rf. 23. 4a + 2cz 2 - 4d. 24. 9ac* - 46 25. 4z 2 - 25a = 4&r*. 26. 9az* - 46 + 9cz*. 27. 4x 2 + 25a = 25 + 4a 2 z*. 28. 2cx* + 4d - c* - 4<fc*. 29. Solve /r = rov 2 for v. 30. Solve S J#* for . 31. Solve A = ?rr 2 for r. 32. Solve A Jir 2 A for . or x. 07 ^ 2 + 2 _5 04^.5 x 49 33. -g- - g. 34. j - 35. j - - - 0. 8 4 2* + 3 2 as + 6 127. Solution of an equation by factoring We know that a product of two or more numbers equals zero when and only when at least one of the factors is zero* This fact is the basis for the f ollowing method, which applies to integral rational equa- tions of any degree. SUMMARY. To solve an equation in x by use of factoring: 1. Clear the equation of fractions if necessary by multiplying both sides by the LCD of all fractions involved. 2. Transpose all terms to one member and thus obtain zero as the other member. Factor the first member if possible. 3. Place each factor equal to zero and solve for x. EXAMPLE 1. Solve: 6 - 5z - 6x 2 - 0. SOLUTION. 1. Multiply both sides by 1, to obtain convenience in factoring with a positive coefficient for # 2 . 6* 2 + 5x - 6 = 0. 2. Factor: (3z - 2) (2* + 3) - 0. 3. The equation is satisfied if 3x - 2 or if 2x + 3 = 0. 4. If 3x 2 = 0, then 3x 2; x = is one solution. 5. If 2x 4- 3 = 0, then 2x = 3; x f is a second solution. * This fact holds for a product of complex numbers. 776 ELEMENTS OF QUADRATIC EQUATIONS EXAMPLE 2. Solve: 4x* + 20s -f 25 0. SOLUTION. 1. Factor: (2* + 5) 2 - 0; or (2z + 5)(2* + 5) - 0. 2. If 2z -f 5 = 0, then x = \ . Since each factor gives the same value for x, we agree to say that the equation has two equal roots. In solving an equation, if both sides are divided by an expression involving the unknowns, solutions may be lost. EXAMPLE 3. Solve: 5z 2 &t. SOLUTION. 1. Transpose 8z: 5z 2 - Sx = 0; x(5x - 8) = 0. 2. Hence, x or 5x 8 = 0; the solutions are and f . INCORRECT SOLUTION. Divide both sides of 5z 2 = &c by x: 5x = 8. Then, incorrectly, we obtain x = f as the only solution. In this incorrect solution, the root x = was lost on dividing by x. Some literal quadratic equations can be solved by factoring. EXAMPLE 4. Solve for x: 2a?x* + 3a6x - 26 2 = 0. SOLUTION. 1. Factor: (2ax ft) (ax + 26) = 0. 2. If 2ax - b = 0, then 2ax = 6; x = H-- 3. If ax -f 26 = 0, then ax = - 26; x = --- a 4. The equation has the two solutions 5- and --- ^ 2a a EXERCISE 67 Solve 6y factoring. 1. x 2 - 3z = 10. 2. y 2 - 5y = 14. 3. s 2 -f * = 12. 4. z 2 -f 3z - 28. 6. 21x = 14z 2 . 6. W - 144 = 0. 7. 3x 2 - 7x - 0. 8. 6s 2 = Ifo. 9. 5z 2 - 9z - 0. 10. x 2 + 8 - 6x. 11. 4x 2 - 25 = 0. 12. x 2 + 15 - 8x. 13. 2s 3 + 5z - 3. 14. 3z 2 - 2z - 5. 15. 8z 2 -f 3 = ELEMENTS OF QUADRATIC EQUATIONS 177 16. 16z 2 - 24x - 9. 17. 25y 2 - 20y - 4. 18. z 2 + 6* - - 9. 19. 4y* + 40 = - 1. 20. 3z 2 + 2 = - 7*. 21. 2z 2 + 7x - - 6. 22. lOz + 3 -f &c 2 - 0. 23. 12 - 5z 2 - 17z = 0. 24. 6z 2 - 19* + 15 = 0. 25. 16z 2 + 40z + 25 = 0. 26. 8 - 22z + 15z 2 = 0. 27. 15 - 7w - 4w 2 = 0. 28. 8z 2 + 2x - 15 = 0. 29. 7z 2 + 9z - 10 = 0. 30. 49x 2 + 28x = - 4. 31. 4 + 5x - 9x 2 = 0. 32. 8 - 2x - x 2 = 0. 33. 6 + 5x - 6x 2 = 0. Solve for x or w or z. 34. 3fo 2 + ex = 0. 35. 2or 2 - 3dx = 0. 36. x 2 + or - 6a 2 = 0. 37. x 2 + 56x + 66 2 = 0. 38. 2z 2 + bx - 36 2 = 0. 39. 3w; 2 - bw - 46 2 = 0. 40. 46 2 z 2 -h 4a6x + a 2 = 0. 41. 66 2 z 2 - 7bx - 3 = 0. 42. 2aV - abx - 36 2 = 0. 43. 9aV + 12a6x + 46 2 = 0. 3 , 7 5 46 8 2 + 3w? 1 " 4*" ' 8z 2 " - " 3i H h 3 ' 3^ + 1 1 6^ 9 4. A7 3 5 * # -f- 4 2 ' ** 1 2x + 5 3 2111 1 r 2 l li l 40 2x 1 * 2 - 2 12 22-2 s + 1 1 -4x x + 1 10u> 4w+l 4w>-7 2x4 -11 3x - 1 n - 2 2w - 1 w 2x + 8 * - 1 62. (x + 3)(2z - 5)(3z + 7) = 0. 63. 6z 8 -f ar 2 - 15* = 0. 64. (2x - 3)(3x -f 5) = 2x + 7. 65. (3z - l)(2x + 5) = 3z + 19. 128. Completing a square A binomial x* + pa: becomes a perfect square if we add the square of one half of the coefficient of x. That is, we complete a square if we add g)' or f : 178 ELEMENTS OF QUADRATIC EQUATIONS ILLUSTRATION 1. x 2 fa becomes a perfect square if we add the square ofi(6), or 3: x 2 - Ox + 9 - (x - 3) 2 . ILLUSTRATION 2. To make z 2 7x a perfect square, we add (J) 2 or a* - 7z + ^ - (x - I) 2 . SUMMARY. To sofoe o quadratic equation in x by computing a square: 1. Transpose all terms involving x to the left side and all other terms to the right member and collect terms. 2. Divide both members by the coefficient of x 2 . 3. Complete a square on the left by adding the square of one half of the absolute value of the coefficient of x to both sides. 4. Rewrite the left member as the square of a binomial. 5. Extract square roots t using the double sign on the right. EXAMPLE 1. Solve: a; 2 + 4x + 1 = 0. (1) SOLUTION. 1. Subtract 1: x 2 + 4x 1. (2) 2. Since 4 -*- 2 = 2, add 2? or 4, to complete a square on the left: * 2 + 4z + 4 = 4- 1; (3) (x + 2) 2 - 3. (4) 3. Extract square roots: x + 2 = V3 t or x - - 2 db V3. (5) Thus, the roots are irrational numbers. To compute approximate values for the roots to three decimal places, we obtain V3 from Table I : x * - 2 + 1.732 = - .268 x = - 2 - 1.732 = - 3.732. % EXAMPLE 2. Solve: 3x* - &e + 2 * 0. SOLUTION. 1. 3x* - &c ** - 2. 8 2 2. Divide by 3: s a 53 = ~- 3. Since * 2 J, add (|)* or ^ to complete a square. 8 tr Hence, /4\* 16 2 16 6 I _ I - .. . _ . ^ .. ^ . \3/ 9 39 9 4 * 10 ELEMENTS OF QUADRATIC EQUATIONS 179 4. Extract square roots: 4 From Table I, VIo = 3.162. Hence, 3.162 OOQ>y . 4-3.162 O . n SB 2.387, and x - = - .279. ., o o EXAMPLE 3. Solve by completing a square: x* -f 4s + 7 0. SOLUTION. 1. x 2 4- 4x 7. 2. Since (4 4- 2) * 2, we add 2* or 4 to both sides: & + 4* + 4 4 - 7; or (x + 2)' - - 3. 3. Hence, x + 2 - =t V^IJ - * fx/3; x - - 2 *V3. EXAMPLE 4. Solve for x: ox* -f bx + c * 0. SOLUTION. 1. Subtract c: ax* + fez c. 6 c 2. Divide by a: x 1 -f- - --- a a O AJJ/^V 6* 1 , & , /&\* &* C 3. Add (TT-) , or -:-=: rr* + -x + (^-) - ^-r --- \2a/ ' 4a 2 a \2a/ 4o a o c . ... / , 6 \ 8 6 - 4ac Simplify: (x + ^j -TT"' 4. Extract square roots: 6 2o _ , , . 6 fe Vfe 2 4oc fe =fc V6* 4oc 5. Subtract JT-: a; = =- =b ~ = s 2a 2a 2a 2a Note 1 . An equality A 2 = B 1 is satisfied if A 5, that is, if A=B arif A = -B. (7) Equally well, A 2 = B 2 if - A 5, that is, t/ - A - J? or t/ - A * - B. (8) If both sides of each equation hi (8) are multiplied by 1, we obtain equa- tions (7). Therefore, on extracting the square roots of both sides of A* = B 1 , we obtain all possible information by writing A = B, instead of writing A = B, where we read d= as " + or ." That is, it is necessary to use the double sign on just one side if the square roots of both sides of an equation are extracted in solving it. 780 ELEMENTS OF QUADRATIC EQUATIONS EXERCISE 68 Find what must be added to the expression to make it a perfect square, and then write this square. 1. x* - Sx. 2. x* + 10*. 3. * 2 - 2cx. 4. * a + 4dx. 6. * 2 - J*. 6. x 2 + f*. 7. * 2 + j*. 8. x* - J*. Sofoe fo/ completing a square. 9. a; 2 + Ox = 7. 10. a: 2 4- 10s -f 24 = 0. 11. * 2 + 4* = 21. 12. a; 2 + 9 = 6*. 13. * 2 + 4* + 4 = 0. 14. 2w^ + 3 = 8w;. 15. 2y* 4- 4y = 5. 16. x 2 + 13 = 6*. 17. x 2 + 5 = 4z. 18. 6x 2 - 2 = 4z. 19. 9x 2 + 1 = 12*. 20. 9s 2 + 6* = 1. 21. 4* 2 + 13 - 12*. 22. 4z 2 4- 4z - 3 = 0. 23. 3z 2 -h 8z = 1. 24. 16x 2 + 9 = 24*. 26. 9* 2 - 12* = 21. 26. 4* 2 - 12* = 5. Verify the statement by substitution. 27. * = (- 2 + V2) and * = (- 2 - \/2) satisfy * 2 + 4* + 2 = 0. 28. * = (2 3i) are solutions of * 2 4* + 13 = 0. Solve for x by completing a square. 29. * 2 - 2o* - 15a 2 . 30. * 2 + bx = 66 2 . 31. 2* 2 - 56* = 36 2 . 32. 6* 2 - 4* - c = 0. 33. 3* 2 -f 2o* - b = 0. 34. .3* 2 - .06* - .144 = 0. 36. o* 2 + 4* - c = 0. 36. 2* 2 + bx + c = 0. 37. Hx* + Kx + P - 0. 38. A* 2 + 2B* + C = 0. 1 29. The quadratic formula In Example 4 on page 179, the quadratic equation ax a 4- bx + c = (1) was solved by the method of completing a square; the solutions were found to be x = -n. We call (2) the quadratic formula. In (2), it is permissible for a, 6, and c to have any values, with a j* 0. ELEMENTS OF QUADRATIC EQUATIONS 181 SUMMARY. To solve a quadratic equation in x by use of the quadratic formula: 1. Clear the equation of fractions and reduce it to the standard form ax* + bx + c = 0. 2. List the values of the coefficients a, b, and c. 3. Substitute the values of a, b, and c in the formula. ILLUSTRATION 1. To solve 3z 2 62 2 = 0, we observe that a = 3, 6 = 6, and c = 2. Hence, from the quadratic formula, _ - (- 6) V(- 6) 2 - 4-3- (- 2) _ 6 2VT5 _ 3 3.873 X ~ 6 6 ~ 3 ; x = 3 + 3 ' 873 = 2.291, and x = 3 " 3 ' 873 = - .291. o o ILLUSTRATION 2. To solve 2z 2 4x + 5 = 0, we notice that = 2, fe = 4, and c = 5. Hence, from the quadratic formula, 4 db Vl6 - 40 4 V- 24 4 2t>/6 2 4 ~4~4~2 EXAMPLE 1. Solve for x: x* 3ex -f 5dx I5de = 0. SOLUTION. 1. Group terms in a:: z 2 4- x( 3e -f 5d) 15de = 0. 2. In the standard notation, a = l, 6= 3e + 5rf, and c = From the formula, - (- 3e 4 The radicand becomes 9e 2 + 60efe = 25(^4- 30de + 9e* /ox (3) Hence, from (3), - (- 3e + &0 (5d + 3e) o* o*? .1 i .. . i i ............... * x 2 ' 3e - 5d 4- 5d + 3e 3e - 5d - 5d x = 2 The solutions are 3e and R . x = - - ** " . In deriving the quadratic formula, we showed that an integral rational equation of the 2d degree in x has just two roots (which sometimes are identical). This result is a special case of a general theorem that, if the degree of the equation is n, the equation has just n roots (with repetitions of values possible among them). 782 ELEMENTS OF QUADRATIC EQUATIONS EXERCISE 69 Solve by use of the quadratic formula.* Check if directed by the instructor. 1. 6z a + x - 2 - 0. 2. 3y + 2y - 5 0. 3. 6y 2 - 7y - 3 =* 0. 4. 7j/ 2 - 8y_ = 12. 6. j/ 2 - 2y 4- 10 - 0. 6. z 2 + 13 - 4z. 7. 4* 2 + 9 - 12*. 8. 16z 2 - 25 - 0. 9. 4z 2 - &c + 1 - 0. 10. 9* 2 + &c + 1 - 0. 11. 36z 2 - 49 = 0. 12. By - 18 - Ify 8 . 13. 9z 2 + te - 1. 14. 4 + 4x - 6z 2 . 15. 2z 2 + 3 - 8z. 16. 2* 2 - 2s - 7. 17. 4z* 4- 3 - 2z. 18. 9s 2 + 6z = 7. 19. 9s 2 + 16 - 0. 20. 4z* + 13 - 40. 21. 3s 2 = 4z - 8. 22. 4s* + 5 - 82. 23. a* + .15 = .8z. 24. x* + 6 = 6z. 25. 4x 2 -f 53 = 4x. 26. 3x 2 -f 2x = 9. 27. 4x 2 + Sx = - 9. 28. 9z 2 - 27 - &c. 29. 4x 2 -f 13 = 12x. 30. 18z 2 + 33x = 40. 31. 25z 2 -f 4 - 20c. 32. 21x 2 + 19a; - 12. 33. 9y z + 23 = 3(ty. 34. 24y 2 + 2y = 15. 36. 4r 2 -h 29 = - 8x. 36. 16a; 2 + 34x = 15. Solve for x or y by use of the quadratic formula. 37. &c 2 - 5dx - 6tf - 0. 38. 2z 2 + hx - 16A 2 = 0. 39. ox 2 - dx + 3c - 0. 40. 2oc 2 -f 36x - c = 0. 41. 5&2/ 2 - 3Jty + 6 - 0. 42. 2/uc 2 -f 3x - 5A = 0. 43. y 2 + 2cy + dy -f 2crf * 0. 44. y* - 46y + Say - 12o& = 0. 45. 5fcc* - 6 + lOfcc - 3a: - 0. 46. 8% 2 4- 12y - 15 - 10% = 0. 47. 6% 2 - 4hy -f 10 - 15y - 0. 48. 2x 2 - 3hx + A* - x = 1. 49. 3z 2 -f 3&c 2 - 6x + 5/uc - 10 + 5x = 0. 50. Check the solutions ( 6 V6 J 4oc) -i- 2a by substitution in the equation ax* + 6x -f c = 0. 51. Solve for y in terms of a: a; 2 2y 2 2 + xy + a; + 5y = 0. 52. Solve for x in terms of y: 2y 2 + 15z 2 - 2 - x + 3y IZxy = 0. 63. Solve for a; hi terms of y in Problem 51. 64. Solve for y in terms of in Problem 52. * Table I is useful in detecting perfect square numbers. Thus, if we meet V1764, by reference to Table I we observe that 1764 * (42)*. ELEMENTS OF QUADRATIC EQUATIONS 183 130. Outline for solution of quadratic equations A pure quadratic equation should be solved by merely extracting square roots, as in Section 126. Any other quadratic equation should be solved by factoring if factors can be easily recognized. In all other cases, solve by use of the quadratic formula, unless otherwise specified. The method of completing the square is not recommended in any problem unless specifically requested; this method was in- troduced mainly as a means for deriving the quadratic formula. 1 31 . Applications of quadratic equations From geometry, we recall the Pythagorean theorem, which we associate with the triangle 5 in Figure 12. . R 9 . // a and b are the lengths of the perpendicular sides and c is the length of the hypotenuse of a right angled triangle, then a 2 -f fc 2 * c 2 . Applications of the preceding theorem frequently introduce quad- ratic equations. EXAMPLE 1. Find the length of a side of an equilat- eral triangle whose altitude is 3 feet shorter than a side. SOLUTION. 1. In Figure 13, let ABC represent the triangle. Let x feet be the length of a side of AABC. Then, the lengths of AD and DC are, re- spectively, \x and (x 3). 2. From the Pythagorean theorem, AC 2 = Iff + or (X ~ (1) 3. Simplify in (1) : x* = ^ -f z 2 - fa + 9; x 2 - 24* + 36 - 0. 4. Solve (2) by the quadratic formula: 24 (2) X rf 171 * 8 - 12*6^8; = 22.39 and 1.61. (Using Table I) The smallest root has no significance in the problem because (1.61 3) is negative. Hence, the side of the specified triangle is 22.39 feet long. 184 ELEMENTS OF QUADRATIC EQUATIONS EXAMPLE 2. An airplane flies 560 miles against a head wind of 40 miles per hour. The plane took 28 minutes longer for this flight than would have been the case in still air. How fast could the airplane travel in still air? INCOMPLETE SOLUTION. 1. Let x miles per hour be the speed of the air- plane in still air. In flying against the wind, the speed is (x 40) miles per hour. From the equation distance = (rate) (time), the flight times for a distance of 560 miles against the wind and in still air are, respectively, 560 , 560 and x _ 40 x 2. From the statement of the problem 560 560 28 " /JA " x - 40 x '60 The student should clear of fractions and solve for x. MISCELLANEOUS EXERCISE 70 Solve each equation by three methods, (a) by factoring, (b) by completing a square, and (c) by use of the quadratic formula. 1. 2* 2 + 5* = 3. 2. 3* 2 + 5 = 16*. 3. 12* 2 + 11* = 15. Solve for x or y or z by the most convenient method. 4. 2/ 2 - 33 = ty. 6. x* - 4x = 45. 6. 16</ 2 + 9 = 24p. 7. 14* 2 - x - 3. 8. 3 2 - 6 - 2*. 9. 49* 2 + 4 = 14*. 10. 4y* + 17 = I2y. 11. 1 + 25* 2 = 10*. 12. 6* 2 + 5* = 56. 13. 25* 2 - 20* = 1. 14. 16 - 5* 2 = 0. 16. 6* 2 = 7*. 16. 6* 2 + 5 = 0. 17. 9 - II* 2 - 0. 18. 4* 2 = 49*. 19. 5y 2 + 36 - 0. 20. 16y 2 + 1 = &y. 21. 20* 2 + 13* = 21. 2- * =3. . 3 x 5 + x 3* o OB 7 * - 2 * 2 - 4 X 2 1 5 ~* 3-2* - 4 "" 2*T~I ~ 6T= 2 - * 2* " l 28. \g& + a* 3S. 29. 4* 2 -f 26* + b - 1. 30. &* 2 - 2fcc + 2 - *. 31. 3* 2 4- hx + 3fc* -f W? 0. 32. a* 2 - 26* = 2* + 3. 33. c* 2 + 2hx = 5 + 4kx. ELEMENTS OF QUADRATIC EQUATIONS 185 34. Solve for x by completing a square: hx* + 2fcc m 0. 35. Solve for x by completing a square: dx* 3cx + h = 0. eocA problem by introducing only one unknown number. 36. Divide 45 into two parts whose product is 434. 37. The area of a rectangle is 221 square feet and one side is 4 feet longer than the other. Find the dimensions. 38. Find two consecutive integers whose product is 306. 39. Find a number which is ^^ less than its reciprocal. 40. Find the length of a side of a square where a diagonal is 6 feet longer than a side. 41. Find the length of a side of an equilateral triangle whose altitude is 2 feet shorter than a side. 42. After plowing a uniform border inside a rectangular field 50 rods long by 40 rods wide, a farmer finds that he has plowed 60% of the field. Find the width of the border. 43. The diameter of a circular field is 40 yards. What increase in the diameter will increase the area by 440 square yards. (Use IT = 3^.) 44. A circular field is surrounded by a cinder track whose width is 20 feet and area is J of the area of the field. Find the radius of the field. 46. An airplane flew 660 miles in the direction of a wind and then took 40 minutes longer than on the outward trip to fly back against the same wind. If the plane flies at the rate of 200 miles per hour in still air, how fast was the wind blowing? 46. Jones travels 4 miles per hour faster than Smith and covers 224 miles in one hour less time than Smith. How fast does each man travel? 47. A motorboat takes 2 hours to travel 8 miles downstream and 4 miles back on a river which flows at the rate of 2 miles per hour. Find the rate at which the motorboat would travel in still water. 48. If- A is the measured cross-section area of a chimney, its so-called effec- tive area E is the smallest root of the equation E* 2AE + A* .36A = 0. Solve for E in terms of A and, from the result, find E if A =20 square feet. 49. If an object is shot vertically from the surface of the earth with an initial velocity of v feet per second, and if air resistance and other dis- turbing factors are neglected, it is proved hi physics that $ = vt %g&, where s feet is the height of the object above the surface at the end of t seconds and g 32, approximately, (a) Solve for t in terms of s. (6) If v = 200 feet, use Part a to find when = 500 feet and feet. CHAPTER ADVANCED TOPICS IN QUADRATIC EQUATIONS 1 32. Graph of a quadratic function A quadratic function of x is a polynomial of the second degree in x and hence has the form ax* -f bx + c, where a, b, and c are constants and a ^ 0. EXAMPLE 1. Graph the function z 2 2x 3. SOLUTION. Let y = x z 2x 3. We select values for x and compute the corresponding values for y. In Figure v 14, we plot the points (- 3, 12), (- 2, 5), etc. In the table of values, we arrange the values of x in their natural order as they appear on the z-axis, because then the corresponding points on the graph are met hi their natural order as we draw the curve. The curve through the plotted points is the graph of the function and is called a 2? == -3 - 2 1 2 4 5 y = 12 5 -3 4 -3 5 12 Fig. 14 parabola. The point V at the rounded end is called the vertex of the parabola. Since V is the lowest point of the graph, the ordinate of V, or 4, is the smallest or minimum value of the function, and we call V the minimum point of the graph. The vertical line through V is called the axis of the parabola. The part of the curve to the right of this axis has exactly the same shape as the part to the left. That is, the parabola is symmetrical with respect to its axis. The equation of the axis of the parab- ola y = x 2 2x 3 shown in Figure 14 is x 1. ADVANCED TOPICS IN QUADRATIC EQUATIONS 187 If a parabola is concave downward (open downward), instead of concave upward as shown in Figure 14, then the vertex of the parab- ola is its highest point and is called the maximum point of the curve. At a more advanced stage, we meet proofs of the following facts: I. The graph of ax* + bx -f c is a parabola with its axis perpendicular to the x-axis; this parabola is concave upward if a is positive and concave downward if a is negative. II. The abscissa of the vertex of the parabola is x = 5-; when x has this value, the function has its minimum or its maximum value according as a is positive or negative. 2 ILLUSTRATION 1. In Figure 14, at V, x = 7:-^- = 1. A- \ SUMMARY. To form a table of values in graphing a quadratic func- tion f(x): 1. Find the coordinates of the vertex of the graph. 2. Choose pairs of values of x where, in each pair, the values are equi- distant from the vertex, one value on each side; the values off(x) cor- responding to each pair will be equal. ILLUSTRATION 2. In Example 1, the abscissa of V is x 1. Then, we selected z = 1 db 1, orz = 2 and z = 0; x I 3, or z = 4 and x = 2; etc. The corresponding pairs of values of y are equal. *EXAMPLE 2. Divide 50 into two parts such that their product will be a maximum. SOLUTION. 1. Let x be one part; (50 x) is the other part. 2. Let/Or) represent the product #(50 x), or f(x) = 50z x 2 . 3. The maximum of f(x) is attained at the vertex of the parabola which is the graph off(x), or when x = [50 * ( 2)] = 25. Hence, the product of the parts of 50 will be greatest when they are equal, each being 25. The corresponding largest product is 625. Note 1. After having formed a table of values for graphing a quadratic function of x, select the scales on the coordinate axes with care. Choose the unit for distance on the z-axis large enough to spread out the parabola in order to make it generously open. Choose the vertical unit independently of the previous choice of the z-unit in order to be able to plot all points from the table of values on the available part of the cross-section sheet. 188 ADVANCED TOPICS IN QUADRATIC EQUATIONS EXERCISE 71 For each function, (a) find the coordinates of the vertex of the graph and the equation of its axis; (b) graph the function, with values of x extending at least 4 units on each side of the vertex; (c) state the maximum or minimum value of each function. 1. x*. 2. 4z 2 . 3. - x\ 4. - 6z 2 . 5. * 2 + 5. 6. x* - 4. 7. a* + 6z + 5- 8. z 2 - 4x + 7. 9. - 2z 2 + 4* + 3. 10. - 3z 2 + I2x. 11. 2z 2 -f 8z + 3. 12. - 3s 2 + 6z - 5. 13. 4z 2 - 12*. 14. 2z 2 - 20z + 4. State whether the function has a maximum or a minimum value, and obtain this value without graphing by finding the coordinates of the vertex. 16. 4z 2 - IQx + 3. 16. - 3z 2 + 24z - 7. 17. - 6z 2 + 8. 18. If an object is shot vertically upward from the earth's surface with an initial velocity of 96 feet per second, (a) draw a graph of the distance s as a function of t\ (b) from the graph, find when the object commences to fall, the maximum height which it reaches, and when it hits the sur- face. (Recall the formula of Problem 49, page 185.) 19. If an object is shot vertically upward from the earth 's surface with an initial velocity of 80 feet per second, find when the object reaches its maximum elevation, without graphing. (See Problem 18.) 20. Graph the function z 3 I2x + 3 by use of the integral values of x from 4 to 4 inclusive. ifGraph each of the following functions, with enough computed points to obtain a graceful curve. 21. x 3 . 22. x 4 . 23. - x*. 24. - z 3 . 26. x s + 2x* - 4* + 3. 26. - 3x* - 4z 3 + 12* 2 + 6. if Solve each problem by introducing just one unknown x and then finding the maximum of a quadratic function of x, without graphing. 27. Divide 60 into two parts whose product is a maximum. 28. Find the dimensions of the rectangular field of largest area which can be inclosed with 600 feet of wire fence. 29. In forming a trough with a rectangular cross section and open top, a long sheet of tin is bent upward on each long side. If the sheet is 30 inches wide, find the dimensions of the cross section with the largest possible area. 30. Divide H into two parts whose product is a maximum. ADVANCED TOPICS IN QUADRATIC EQUATIONS 789 1 33. Graphical solution of an equation If x has a value for which the graph of f(x) meets the x-axis, then with this value of x we have/(x) = 0. Hence we are led to the follow- ing procedure. SUMMARY. To find approximate values of the real roots of an equa- tion in x graphically: 1. Simplify and transpose all terms to one member to obtain an equa- tion of the form f(x) = 0. 2. Graph the function f(x) and measure the abscissas of the points where the graph meets the x-axis; each of these abscissas satisfies the equation f(x) = 0. EXAMPLE 1. Solve z 2 2x 3 = graphically. SOLUTION. 1. Let y = x* 2x 3 and consider its graph in Figure 14, page 186. The graph crosses the x-axis at x = 3 and x = 1. 2. Since y = when x = 3 and when x = 1, these are values of x for which z 2 - 2x 3 = 0. That is, 3 and 1 are roots of the equation. If the roots of f(x) = are imaginary) this would be indicated by the fact that the graph of f(x) would not meet the x-axis. 1 34. Graphical solution of a quadratic equation In order to solve the equation ax 2 + bx + c = (1) graphically, we construct the graph of the quadratic function ajc 2 + bx + c. (2) The parabola, which is the graph of this function, I. cuts the x-axis in two points when and only when equation 1 has unequal real roots; II. touches the x-axis in just one pointy or is tangent to the x-axis, when and only when the roots are equal; III. does not meet the x-axis when and only when the roots are imaginary. Note 1. A parabola can be defined geometrically as the curve of inter- section when a right circular cone is cut by a plane which is parallel to a straight line on the cone through its apex. 790 ADVANCED TOP/CS IN QUADRATIC EQUATIONS ILLUSTRATION 1. In Figure 15, parabolas I, II, and III are, respectively, the graphs of the functions in the left members of the fol- lowing equations. (I) z 2 - 2x - 8 - 0; (II) z 2 - 2x + 1 = 0; (III) *'-2z + 5 = 0. From the graphs, we see that (I) has the roots x - 4 and x - - 2; (II) has equal roots, x = 1; (III) has imaginary roots. In graphing a specified quadratic func- tion, we have no license to simplify its form by multiplication or by division. But, before solving a quadratic equation graphically, we may (1) clear the equation of fractions; (2) divide out any common constant factor from all terms; (3) make the coefficient of z 2 positive. Operation 3 would cause the corresponding graph to open upward. Fig. 15 EXERCISE 72 Find the real roots of the equation graphically. 1. 2x - 5 = 0. 2. x* + 2x - 8 = 0. 4. 7. 10. + 4x + 7 - 13 - - 12x - 9. 5. x* - 2x + 3 = 0. 8. 4c - 2z 2 - 5. 11. 2z 2 = 4x + 3. 3. x 2 + 6# + 9 = 0. 0. 5. x* - 2x + 3 = 0. 6. f z = 2 - z 2 . 12. 3z 2 = 6x - 5. grrap/& o/ x 2 4- 4z -f 4 and then t in Problems 13 and 14, find specified results by use of the graph. 13. Find the values of x for which the value of the function is 1. 14. Solve re* + 4x -f 4 = 6 by inspection of the graph. 15. By use of a single graph, solve each of the following equations graph- ically: 2s 2 - 5x = 0; 2s 2 - 6x + 3 = 0; 2z 2 - 5x - 7 = 0. 1 35. Character of the roots Let r and s represent the roots of ax 2 -f- foe + c 0. Then, from the quadratic formula, ADVANCED TOPICS IN QUADRATIC EQUATIONS 197 - b + V6 2 - 4ac 2a ; 8 b - Vb 2 - 4oc 2a (1) We assume that a, 6, and c are real numbers and that a j^ 0. Then, the roots are imaginary when and only when b 2 4oc is negative; if one root is imaginary, the other is also. If ft 2 4oc 0, then r = s = 6/2a. Moreover, if r , on sub- tracting the expressions in (1) we obtain = r s = 2Vb 2 - 4oc 2a S a 0; V6 2 -4oc - 0. Hence, if r s then fc 2 4ac = 0. From the preceding remarks and Section 134, we see that the items in any row of the following summary hold simultaneously. THE ROOTS OF ax 2 + bx + c = real and unequal real and equal imaginary THE VALUE OP 6 2 - 4ac > 6 s - 4oc = 6 2 - 4ac < THE GRAPH OF ax z + bx + c cuts x-axis in two points is tangent to x-axis does not touch x-axis If a, 6, and c are rational numbers, the roots are rational when and only when V& 2 4oc is real and is a rational number. That is, the roots are rational numbers when and only when fe 2 4oc is a perfect square. We call 6 2 4oc the discriminant of the quadratic equation ax* + bx + c = 0, or of the quadratic function ax 2 -f bx 4- c, because, as soon as we know the value of 6 2 4oc, we can tell the general character of the roots of the equation without solving it, and the general nature of the graph of the function without graphing it. ILLUSTRATIONS OF THE USE OF THE DISCRIMINANT EQUATION DISCRIMINANT HENCE, THE ROOTS ARE 4z 2 - 3z + 5 = (-3)2 _ 4-4-5 = - 71 imaginary numbers 4z 2 - 4x + 1 = (- 4) 2 -4-4 = real; equal; rational % * 4 4z 2 - 3x - 5 (-3) 2 + 4-4-5 - 89 real; unequal; irrational a 2 - 2x - 3 = (- 2) 2 - 4(- 3) = 16 = 4 2 real; unequal; rational 192 ADVANCED TOPICS /N QUADRATIC EQUATIONS Before computing the discriminant in any equation, it should be simplified by clearing of fractions and combining terms. EXAMPLE 1. State what you can learn about the graph of the quadratic function 3a 2 -\- 5x 6 without graphing. SOLUTION. 1. The discriminant of the function is 25 72 = 47. 2. Hence, the graph would not touch the z-axis. Since the coefficient of z 2 is r- 3, the graph is concave downward and therefore must lie wholly below the a>axis. 136. Conjugate imaginarics If two imaginary numbers differ only in the signs of the coefficients of fyeir imaginary parts, then either of the given numbers is called the conjugate of the other. ILLUSTRATION 1. The conjugate of (3 + 5i) is (3 5i). The conjugate of (a + bi) is (a bi). When the roots of a quadratic equation are imaginary, these roots are conjugate imaginary numbers, because the imaginary parts come from V6 2 4oc in the quadratic formula. ILLUSTRATION 2. The roots of x* -f- 4z + 5 = are - 4 =t v 16 - 20 . . , . . . x = ~ = 2 i, conjugate imaginanes. EXERCISE 73 Compute the discriminant and tell the character of the roots, without solving. 1. y* - 1y + 10 = 0. 2. y* - 4i/ - 21 = 0. 3. z 2 + 2z - 2 - 0. 4. 3s 2 - 5z + 7 = 0. 6. 9z 2 + 12z + 4 = 0. 6. 4x* + 4z = 3. 7. 30 - % 2 = 2%. 8. 3z - 2 = 5z 2 . 9. 25 + 4z 2 = 20z. 10. 2x - 3 - fo 2 . 11. 5x 2 + 1 - 2z. 12. 25z 2 + 1 13. 8a? - 7 - 0. 14. 5z 2 - 3x = 0. 15. 1 - 2x - 16. 3 + 5z 2 = 0. 17. Qx = ftc 2 + 4. 18. z 2 -f .4a; 4- .3 = 0. Solve graphically; check the graph by computing the discriminant and thus determining the character of the roots. 19. x 2 - 4x = 6. 20. z 2 + 7 - 4x. 21. 4z 2 + 4z = 1. ADVANCED TOPICS IN QUADRATIC EQUATIONS 193 Compute the discriminant of the function and, without graphing, state all facts which you can learn about its graph. 22. 4x* - I2x + 9. 23. 2z 2 - 3z - 5. 24. 3** - 4x. 25. - 3z 2 + 5z - 7. 26. 4z 2 + 5x + 7. 27. - 3s 2 - 2x + 4. Specify the conjugate number for the imaginary number. 28. 3 + 7t. 29. - 2 - W. 30. - 2 + V^~9. 31. 6\^"T. 1 37. Sum and product of the roots By use of - 6 + V6 2 - 4oc , - 6 - Vfe 2 - 4oc r = - o - and s = - ~ - ' 2a 2a - 26 6 , . . we obtain r rs = 2a a' - b + Vb 2 - 4oc - 6 - Vb 2 - 4oc 2a ' 2a = - (b 2 - 4oc) ^ 4oc = c 4a 2 4a 2 a Hence, for the equation ox 2 -f 6x + c = 0, swm o/ the roots eqvals -- : r-fs= -- (1) c c product of the roots equals -: rs = (2) ILLUSTRATION 1. For 3z 2 5x + 7 = 0, we find r + s = and rs = J. 1 38. Factored form of a quadratic function THEOREM I. If r and s are the roots of ax 2 + bx + c = 0, ax* + frt + c = a(x - r)(x - 5). (1) Proof. 1. We can write ox 2 + bx + c = at x 2 4- -a? 4- - 1 \ o a/ 5 c 2. From Section 137, - = (r + s) and - = rs. Hence, (Z d 4- bx + c = a[x 2 (r + s)x 4- rs] = a(x r)(x s). 194 ADVANCED TOPICS IN QUADRATIC EQUATIONS ILLUSTRATION 1. A quadratic equation whose roots are 5 and 3 is (x + 3)(z - 5) - 0, or s 2 - 2x - 15 - 0. [a - 1 in (1)] ILLUSTRATION 2. A quadratic equation whose roots are J(2 =b 3t) is > - i(2 - 30] - 0. To eliminate fractions we use a = 4 = 2-2, and then group within paren- theses to exhibit the sum and difference of two quantities, as an aid in multiplying: 0; [(2z - 2) - 3t][(2s - 2) + 3t] - 0, or (2s - 2) - 9i - 0. Since i 2 = 1, we have 4z 2 - &c + 4 -h 9 - 0, or 4s - &c + 13 - 0. Under certain circumstances, we have seen how to solve the equa- tion ax 2 -f bx + c = by first factoring the function ax 2 + bx 4- c. Formula 1 permits us to reverse this process and to /actor the funo tion by first solving the equation (of course, not using factoring in the solution). EXAMPLE 1. Factor 6x 2 23a? -f- 20 by first solving an equation. SOLUTION. 1. Solve 6z 2 23z + 20 0, by the quadratic formula: 23 V49 23 db 7 * " 12 - l2- ; 5 A 4 a? jr ana a; A O 2. From formula 1 : 62* - 23* + 20 = 6(z - $)(* - j) - (2* - 5)(3* - 4). Formula 1 states that any quadratic function of x can be expressed as a product of factors which are linear in x. However, these factors involve rational, irrational, or imaginary coefficients depending on the nature of the roots r and s. In particular, from the facts about rational roots, on page 191, we draw the following conclusion: // a, b, and c are rational numbers, ax* 4- bx + c can be expressed as a product of real linear factors with rational coefficients when and only when the discriminant ft 2 4ac is a perfect square. ADVANCED TOPICS IN QUADRATIC EQUATIONS 795 EXERCISE 74 Find the sum and the product of the roots of each equation, in the unknown x, without solving for x. 1. x 2 + 5x - 3 = 0. 2. 2x - 5x* + 7. 3. 4z* = 3z - 6. 4. 7 &c - 2z 2 . 5. 7 - 3x = 4z*. 6. 5s 2 - 17 = 0. 7. 18 = 5x 2 . 8. 12z 2 + 3 = 0. 9. 5 - 9z* = 7s. 10. 2z 2 - 3 - 5z. 11. 5 = 7z 2 - 4x. 12. 4z 2 - 12s - 9. 13. ax* + dx = h. 14. cz 2 = 3z 6. 15. 4z* = ox + c. 16. 2z 2 + 3z + as + c - 0. 17. 5z* + <w 2 + 3x + d = 0. 18. 2z 2 + 3a; + 2a + d = 0. 10. & + ex* - 2x + ex - d - 5. Compute the indicated product. 20. 4(* - f)(x + f). 21. 6(x - J)(* + 22. (a; - 2 + 3i)(x - 2 - 3i). 23. (x + 1 - 2V2)(z + 1 + 2\/2). Form a quadratic equation with integral coefficients having the given numbers as roots. 24. 3; - 7. 26. - 2; - 3. 26. J; - f . 27. J; 2. 28. -f; -f. 29. 2; -. 30. - f; - f . 31. \/2. 32. }V3. 33. 3V2. 34. 2i. 35. t. 36. 1 db V2. 37. - 2 =t V5. 38. 3 2\/2. 39. J j}V3. 40. - J =fc JV2. 41. 3 5t. 42. - 2 3i. 43. 4 2i. 44. - Ji. 45. 2 2iV5. 46. i f tV3. 47. - Jt V2. Factor, after first solving a related quadratic equation by use of the quadratic formula. 48. 12z' + llz - 36. 49. 27z* + 2Lc - 40. 50. 27x 2 - fay - 16j/ 2 . 51. 24x 2 - 13* - 60. 62. 48x 2 + 50s - 75. 53. 27x 2 + 12z - 32. Without factoring or solving any equation, determine whether or not the ex- pression has real linear factors with rational coefficients. 64. 8z* + 7* - 2. 56. llz 2 + 12* - 5. 56. 6z* + 25xy + 26j/ J . if Factor, perhaps by use of imaginary or irrational numbers. 57. a? + Ox + 10. 58. 4s* - 12x + 7. 59. 2z* - 2x + 6. 796 ADVANCED TOPICS IN QUADRATIC EQUATIONS 1 39. Equations in quadratic form EXAMPLE 1. Solve: x* - 5z 2 + 6 = 0. (1) SOLUTION. 1. Factor: (z 2 - 3)(z 2 - 2) = 0. 2. If x* - 3 = 0, then x = \/3; if z 2 - 2 = 0, then x = V2. The given equation has four solutions, Vjji and =t V2. SECOND SOLUTION. 1. Let y x 2 ; then i/ 2 = z 4 and, from (1), 2/ 2 - % + 6 = 0. 2. Solve for y: (y - 3)(y - 2) - 0; hence, y = 3 and ?/ = 2. 3. If y = 2, then z 2 = 2 and z = A/2. 4. If y = 3, then x 2 = 3 and x = \/3. Comment. The given equation is said to be in the quadratic form in z 2 because we obtain a quadratic in y on substituting y x*. EXAMPLE 2. Solve: 2ar 4 - z~ 2 - 3 = 0. SOLUTION. 1. Let y = or 2 ; then 2/ 2 = ar 4 and 2# 2 # 3 0. 2. Solve for y: (2y - 3)ft/ + 1) = 0; hence, y = 1 and y = f . o ^1^9 1 3 Tf 11 - thpn r~ 2 = - = - r 2 *= - rr = -4- -Vfi ^ "~ 2 2' a: 2 2' 3* 3 v 4. If 2/ = - 1, then or 2 = - 1; = - 1; x* = - 1; x = t. 5. The solutions are =fc t and EXAMPLE 3. Solve: (x* + 3z) 2 - 3x 2 - ftc - 4 = 0. INCOMPLETE SOLUTION. 1. Group terms: (z 2 + 3z) 2 - 3(x 2 + 3x) - 4 = 0. 2. Let y = x* + 3z; then y 2 3y 4 = 0; hence, y = 4, and y = 1. We should then solve z 2 3x = 4 and x 2 + 3z = - 1. 1 . In solving an equation of the form x k = A where A; is a positive integer greater than 2, we agree for the present that we desire only real solutions unless otherwise specified. The real solutions, if any, of x k = A are the real A?th roots of A. Thus, x 4 = 8 has no real solutions while z 6 = 64 has the real solutions x = 'v/ci = 2. ADVANCED 7OP/CS IN QUADRATIC EQUATIONS 197 EXAMPLE 4. Obtain all roots by use of factoring: 8z 3 + 125 = 0. SOLUTION. 1. Factor: (2x + 5)(4z 2 - lOx + 25) = 0. 2. Hence, 2x + 5 = 0, or 4z 2 lOz + 25 = 0. 3. The solutions are x = - J and x = i(10 VlOO - 400) = EXAMPLE 5. Find the four 4th roots of 625. SOLUTION. 1. If z is any 4th root of 625, then x* 625. 2. Solve for x: x 4 - 625 = 0; (z 2 - 25)(z 2 + 25) = 0; z 2 = 25 or x* = - 25.' Hence, x 5 and x = 5i are the desired 4th roots of 625. Note 2. In this section the student has met further illustrations of the truth of the theorem that an integral rational equation of degree n in a single variable x has exactly n roots (we admit the possibility that some of the roots may be equal). Also, we have seen illustrations of the related fact that, if n is a positive integer, every number H has exactly n distinct nth roots, some or all of which may be imaginary. EXERCISE 75 Solve by the method of page 196, without first clearing of fractions when they occur. Results may be left in simplest radical form. 1. x* - 5z 2 + 4 = 0. 2. & - 10z 2 + 9 = 0. 3. x* - 8x* + 16 = 0. 4. 9z 4 + 4 = 13z 2 . 5. 4z 4 + 15z 2 = 4. 6. y 4 - ?/ = 2. 7. y 4 4- 7y z = 18. 8. x 4 - 9 = 0. 9. Sly 4 - 16 = 0. 10. x 6 - 8 = 7r*. 11. 27z + 1 = 28X 3 . 12. 8/ + 39y 3 = 5. 13. 4ar 4 - liar 2 -3 = 0. 14. 36X" 4 - 13ar 2 + 1=0. i 16. 250T 4 - 26x~ 2 + 1 = 0. 16. 2 + 17ar 2 - 9Z- 4 = 0. 17. 8z 6 + 35z 3 + 27 = 0. 18. 1 - 2ar 2 - 3or 4 = 0. 19. (x 2 - z) 2 - (&c 2 - &c) + 12 = 0. 20. (x 2 + 3z) 2 - 3z 2 - 9z - 4 = 0. 21. 2(2z 2 - x) 2 - 6z 2 + 3z - 9 = 0. 22. z 2 + 4z 2 - 17z 2 - 60 - 68z - 0. 798 ADVANCED TOPICS IN QUADRATIC EQUATIONS 26 25 ' x* + 3* 29. a; 4 + 2x* + a; 2 - (14z 2 + 14z) + 24 = 0. 30. 4x* - 4s 3 + * 2 + 4z 2 - 2x - 15 = 0. 31. z w - 30z 5 = 64. 32. 6x 6 + 7s = 20. 35 2 T HINT for Problem 33. Let y = (2 - x)/x 2 . 36. 2x* - llax 2 + 12a 2 = 0. 36. (2z 2 - 3az) 2 - 2o 2 a; 2 + 3a 3 x = 2a 4 . if Find all roots by first using factoring. 37. 27z* - 8 - 0. 38. z 8 + 8 = 0. 39. x 3 - 27 = 0. 40. 16z 4 - 81. 41. 81 - 625z 4 = 0. 42. Sy 3 - 125 - 0. 43. 125z + 27 = 0. if Find the three cube roots of each number. 44. - 27. 46. 64. 46. 1. 47. - 1. 48. 8. 49. J. 60. ^y. if Find the four 4th roots of each number. 61. 1. 62. 16. 63. 81. 64. 625. 66. 16. 66. 256. 67. Jf . 1 40. An operation sometimes leading to extraneous solutions Let M N represent any equation. On squaring both sides, we obtain M 2 = N 2 , which is satisfied if3f = ATorifM = N. Hence, the solutions of M 2 =* N* consist of all solutions of M = N together with those of M = N. ILLUSTRATION 1. x - 5 is the only root of x 3 2. (1) On squaring both sides, we obtain (x 3) a * 4. (2) On solving (2) for x we find x 3 = 2; hence, x = 5 or x 1. Therefore, (2) has the root x = 1 besides the root x = 5 of (1). If an operation on an equation in x produces a new equation which is satisfied by values of x which are not roots of the given equation, we have agreed to call such values extraneous roots. From ADVANCED TOPICS IN QUADRATIC EQUATIONS 199 the preceding discussion, we observe that, if both members of an equation are squared* extraneous roots may be introduced. ILLUSTRATION 2. In Illustration 1, z = 1 is an extraneous root. 141. Irrational equations An irrational equation is one hi which the variables occur under radical signs or hi expressions with fractional exponents. EXAMPLE 1. Solve for x in the folio whig equations (a) and (6). (a), 2* -2 = 4. (6) 2x - 2 = - 2 z* + 4. SOLUTION. 1. Square both sides: 2. 4x* - 8x .+ 4 = 2x* + 4. 3. 2z 2 - Sx = 0; 2z(z - 4) = 0. 4. x = and # = 4. TEST. Substitute x = in (a) : Does - 2 = V4? Or, does -2 = 2? No. Substitute x 4 in (a) : Does 8 - 2 = V36? Yes. x = is not, and = 4 is a root. SOLUTION. 1. Square both sides: 2. 4z 2 - Sx + 4 = 2x 2 + 4. 3. 2z 2 - &c = 0; 2z(z - 4) = 0. 4. x = and x = 4. TEST. Substitute x = in (6): Does - 2 = - VI? Yes. Substitute x *= 4 in (6) : Does 8 - 2 = - V36? Or, does 6 - 6? No. z = 4 is not, and x = is a root. Comment. We met the extraneous roots x = in solving (a) and x = 4 hi solving (6). The test of the values obtained in Step 4 in either solution was necessary in order to reject these extraneous roots. The necessity for the test is also shown by the fact that, although (a) and (b) are different equations, all distinction between them is lost after squaring. SUMMARY. To solve an equation involving radicals: 1. Transpose the most complicated radical to one member and all other terms to the other side. 2. // the most complicated radical is a square root, square both mem- bers; if a cube root, cube both members; etc. 3. Repeat Steps 1 and 2 with the effort to eliminate all radicals in- volving the unknowns. Then, solve the resulting equation. 4. Test each value obtained in Step 3 by substitution in the given equation to determine which values are roots. * Also true if both sides are raised to any positive integral power. 200 ADVANCED 7OP/CS IN QUADRATIC EQUATIONS Note 1. Recall that, if A is positive, VA, or A$, represents the positive square root of A and that VA represents only the principal nth root of A. m Also, in testing for extraneous roots, remember that we are using a to represent only the principal nth root of a m . EXAMPLE 2. Solve: (x - 2)* - v2* + 5 = 3. SOLUTION. 1. Vx - 2 = 3 -f V2z + 5. 2. Square: * - 2 = 9 + 6v2*45 + 2x + 5. 3. Simplify: - * - 16 = 6V2* 4 5. 4. Square: * 2 + 32* + 256 = 36(2* + 5) ; x*- 40* + 76 = 0; (x - 38)(* - 2) = 0. Possible roots of the given equation are x = 38 and x = 2. TEST. Substitute x = 2 and * = 38 in the original equation: * = 2: does V2 - 2 - vT+5 = 3, or does -3 = 3? No. * = 38: does V38 -2 - V76 + 5 = 3, or does 6-9 = 3? No. Hence, neither x = 2 nor * = 38 is a root. Therefore there are no solutions for the given equation. EXERCISE 76 Solve for x or y or z. 2. V3 -z = 5. 1. Va+2 = 3. 8. v'J+l = 1. 11. 3* = 5\/2. 10. (3 + z 14. 2V3 5s = 0. 13. (2* + 3)* = - 5. 3. V2 - 7z = - 4. 6. v"6* - 2 = 4. 9. (2 -f z)* = 4. 12. (2 - 2)* = - 2. 15. Vy = 6 - y. = 3-2*. 17. 3V* + 9 = 2*. 18. 4* 2 + zV3 = 0. - 2* -2 = 0. 20. v / * ? ^"24* -3 = 0. 21. 2i/ 2 - 3j/\/5 = 0. 4- 1 = 1 - Vi. 26. V2* 4 4 4- V^ = 1. ^2 - V2*4-3 = 2. 27. V? -2* - V3 - * = 1. 28. V3 -2* + V242* = 3. 29. V2 - 4* + 2\/l - 3* = 2. 30, g*4 2* = 2. 31. 2*45- = 2. ADVANCED 7OP/CS IN QUADRATIC EQUATIONS 201 32. V3 + x - (3 - x)* = x*. 33. 2Vx 2 + x - 2 - x = x 2 - 2. 34. Solve for x: V3x + a - 3Vx + Va = 0. 35. Solve for z: Vz - a + V2z + 3a = V5a. 36. V3-f 3x = 2V3x - 2 - \/3 - x. 37. Solve for x: Vx" + V3x + 46 = 2V2x -h b. 38. Solve v = V20s, (a) for s; (6) for 0. 39. Solve t = 7r\/-> (a) for Z; (6) for g. *40 Solve: 4x* + 7x - 2 = 0. SOLUTION. 1. Let y = xi; then y 2 = x$ and 4g/ 2 + 7y 2 = 0. 2. Solve for y: (4y - l)(y + 2) = 0; y = - 2 and y = J. 3. If y = 2, then xi = 2 or 'v'x = 2; hence, x = 8. 4. If y = J, then x^ = J or -Six - J; hence, x = (J) 3 = irSolve by reducing to a quadratic in some new variable. 41. 52 + 3Vz = 2. 42. 2x* + 9xi = 5. 43. 3X" 1 + 5 = 44. 3x + 7xi = 6. 45. 2x* = 6 + a. 46. 2x -1 + x~* = 6. 47. 3xi = 8xi - 4. 48. 4x^ = 7x* + 2. 49. 4x~ l + 3x~* = 1. 60. x 2 + 2x - Vz 2 + 2x - 6 = 12. 61. 2x 2 + 3V2x 2 + 3 = 7. if Find all real roots. 62. #* = 8. 53. x* = 32. 64. 2$ = 16. 65. x* = - 8. 66. y* = 4. 67. x* = 9. 68. & = - 243. 69. x* = - 27. 60. (2x + 1)* = 4. 61. (5 - 3x)* = 27. 62. (2 + 3x)* = 8. 63. 2x~ 3 + 15x-* -8 = 0. 64. 3X 3 + 26x* -9 = 0. *1 42. Miscellaneous problems about the roots EXAMPLE 1. If c is a constant and 2 is one root of the equation 3x 2 - 7x + c = 0, find the other root. SOLUTION. Let the roots be r and s, with r 2. Then, from page 193, r + 8 = : or, 2 + s - *; * - 202 ADVANCED TOPICS IN QUADRATIC EQUATIONS EXAMPLE 2. Find the constant h if one root of the following equation exceeds the other by 5: x 2 x Zh = 0. SOLUTION. 1. Given condition: r s = 5. (1) 2. Sum of the roots: r + s 1. (2) 3. Product of the roots: rs 2h. (3) 4. Solve (1) and (2) for r and s: r = 3; 8 = - 2. (4) 5. Substitute (4) in (3): - 6 = - 2A; h = 3. (5) EXAMPLE 3. Find the values of k for which the following equation in x has equal roots: kx* + 2s 2 - 3kx + fc = 0. SOLUTION. 1. Group the terms in standard form: (k + 2)z 2 - 3kx + fc = 0. Hence, the standard coefficients are a = k + 2, 6 = 3k, and c = &. 2. If the roots are equal, the discriminant 6 2 4oc is zero: discriminant = (- 3&) 2 - 4(& + 2)(fc) =0; or 5k* - Sk = 0. 3. Hence, k(5k 8) = 0; or k = and k = f . THEOREM I. // one root of ox 2 + bx + c = is 2fte negative of the other root, then 6 = 0. Proof. 1. If r and s are the roots, then r = s, or r + s = 0. 2. Hence, r + s = = 0, or 6 = 0. Therefore 6 = 0. ' a ' THEOREM II. 7/6 = 0, then one root of ax 2 + bx + c is the negative of the other root. Proof. Since 6 = 0, then =0 = r + s. Hence, r = s. . Note 1. Theorem II is the converse of Theorem I. Theorem II could have been abbreviated by adding the words and conversely at the end of Theorem I. Or, both theorems are included in the following statement: " One root of ax* + bx + c = is the negative of the other root when and only when 6 = 0." In this statement we justify the phrase only when by Theorem I, and the word when by Theorem II. EXAMPLE 4. Find the values of the constant h so that the equation + 9A 2 a? 3 -\- x will have one root the negative of the other. SOLUTION. 1. Write in standard form: hx* -f x(W 1) 3 0. 2. From Theorem II: W - 1 = 0; h - d= . ADVANCED TOPICS IN QUADRATIC EQUATIONS 203 *EXERCISE 77 By use of the discriminant, find the values of the constant k for which the equation will have equal roots for the unknown x. 1. 4x* - Zkx + 1 - 0. 2. 4z 2 + 5kx + 4 = 0. 3. 2fer 2 + 9 - \2x. 4 fcc 2 + 3fo + 5 = 0. 5. x 2 - fcr 2 - 6. 5s 2 - 2fcc - k = 0. 7. & 2 :t 2 - fcc - a; 2 - x = 3. 8. x 2 kx + a; k - 0. 9. fee + x 2 + Axe 2 - 2x = 4. Find <Ae values of the constant k for which the graph of the function of x will be tangent to the x-axis. 10. 5z 2 - 2kx + k. 11. 2kx* - 3fcc + 5. 12. s 2 - 3x - k - fcc. 13. 2z 2 + 2z - 3fc - 2A;x. 14. 2x 2 - 2Jb 2 - 5kx + 5. In all probkms, x is the unknown and all other letters are constants. 15. If one root is 3, find the other root: 2z 2 5x + d = 0. 16. If one root is 2, find the other root: 3x 2 -f dx + 5 = 0. 17. If one root is 5, find the other^oot: 2z 2 + bx 3 = 0. 18. If one root is J, find the other root: 3z 2 + 7x + h = 0. Find the value of the constant h under the given condition. 19. The sum of the roots is 5: 3hx 2 4x 5hx + 6 = 0. 20. The sum of tl|e roots is 7: * 5x 2 - hx* -f Mhx + 4 = 0. 21. The product of the roots is 9: 2x 2 3/u: 2 - 6z + 4/i = 0. 22. The product of the roots is - 6: 3/w; 2 + 5x + h - 1 = 0. 23. One root exceeds the other by 2: 2x 2 4A + 5x = 0. 24. One root exceeds the other by 3: 3z 2 5x + 3h 6 = 0. 26. One root is four times the other: 2x 2 + 20z + A 2 = 13. One root is the negative of the other; find h. 26. hx* - Ox + Zhx - 5 = 0. 27. 2to 2 - ihx - 5h?x + 6 = 0. n 28. x 2 + 12z - 3A 2 x + h = 0. 29. h*x* +^h*x + 5hx - 4 = 0. 30. hx* + Wx = 3 + x. 31. x 2 - 3A 2 z = A - 2x. 32. 3z 2 + 5h?x = 2 + x. 33. x 2 - hx - /i 2 a; + 2x = 0. 34. Prove that, if ox 2 + bx + c = has one root zero, then c = 0, and conversely. CHAPTER THE BINOMIAL THEOREM 143. Expansion of a positive integral power of a binomial By multiplication, we obtain the following results: (x + y) 1 = x + y\ (x + yY = & + 3x*y 4- (x + y) 4 = x* + 4x*y 4- 6z 2 2| 4- 4XI/ 3 4- (re 4- y) 6 = z 5 4- 5z 4 i/ + lO^y + lOz 2 */ 3 4-.5Z2/ 4 + y*. We see that, if n = 1, 2, 3, 4, or 5, the expansion of (x + y) n con- tains (n 4- 1) terms with the following properties: I. In any term the sum of the exponents of x and y i& n. II. The first term. is x n , and in each other term the exponent of x is I less than in the preceding term. T r III. The second term is nx n ~ l y, and in each succeeding term the ex- ponent of y is 1 more than in the preceding term. IV. // the coefficient of any term is multiplied by the exponent of x in that term and if the product is divided by the number of that term, the quotient obtained is the coefficient of the next term. * ILLUSTRATION 1. In (at^ y} 4 , the third term is 6zV* By Property IV, we obtain (6-2) -5- 3, Or 4, amhe'^pefficient of the fourth term. In (x + y) B , the fourth term is lQx*y*; m^rqjperty IV, we obtain (10-2) -s- 4, or 5, as the coefficient of the fifth term.* s V. The coefficients of terms equidistant from the ends are the same. ILLUSTRATION 2. The coefficient of the second term equals that of the next to the last term, etc. THE BINOMIAL THEOREM 205 We shall assume that Properties I to V are true if n is any positive integer, although we have merely verified their truth when n - 1, 2, 3, 4, and 5. The theorem which justifies this assumption is called the binomial theorem, which we shall accept without proof in this text. EXAMPLE 1. Expand (c -f w) 7 . SOLUTION. 1. By use of Properties I, II, and III, we obtain (c + w) 7 = c 7 + 7cfiw + c 5 2 -f cV + eW + c 2 ^ 6 + cufl + w 7 , where spaces are left for the unknown coefficients. 2. By Property IV, the coefficient of the third term is (7-6) 4- 2, or 21; that of the fourth term is (21 -5) -5- 3, or 35. 3. By Property V, we obtain the other coefficients; hence, (c + w>) 7 = c 7 + 7c*w + 21cV + 35cV -f (w\ 6 2a -r 1 SOLUTION. 1. (2a - |) = [(2a) + (- |) ]' 2. We use Properties I to V with x 2a and y = ^ and keep the terms o of the binomial within parentheses in finding the coefficients: _ 6 = (2a) J ' - I or 80 . - 160 ... 20 ATote 1. The following array of numbers is called Pascal's Triangk. The successive rows give the coefficients in the successive positive integral powers of x -f y. To form any row after the second, we first place 1 at the left; the 2d number is the sum of the 1st and 2d numbers in the pre- ceding row; the 3d number in the new row is the sum of the 2d and 3d numbers in the preceding row; etc. This triangle was known to Chinese mathematicians in the early fourteenth century, and it appeared in print in Europe for the first time in 1527. 1 1 1 121 1331 14641 1 5 10 10 5 206 THE BINOMIAL THEOREM The preceding diagram exhibits the fact that the largest coefficient hi any power of x + y is the coefficient of the central term oMerms. We observe that the signs are alternately plus and minus hi the expansion of a power of a binomial where one term bears a plus sign and the other term bears a minus sign. 144. The factorial symbol The symbol n\ is read "n factorial," and is an abbreviation for the product of all integers from I ton inclusive, where n is a positive integer. ILLUSTRATION 1. 5! = 1-2-3-4-5 = 120. 31 = 1-2-3 = 6. 1L = 1-2-3-4-5-6-7 1 J_ 10! 1-2-3-4-5-6-7-8-9-10 8-9-10 720* EXERCISE 78 Expand each power by use of Properties ItoV. 1. (a + 6) 6 . 2. (c - d) 6 . 3. (x - y) 9 . 4. (c + 3) 6 . 6. (2 + a) 4 . 6. (x - 2o) 7 . 7. (36 - y). 8. (2c + 3d) 8 . 9. (a + 6 2 ) 8 . 10. (c 8 - 3d} 4 . 11. (a 2 - 6 2 ). 12. (c - x 3 ) 6 . 13. (x - i) 6 . 14. (1 - a) 9 . 16. (Vz - v^). 16. (z* + a) 6 . 17. (- a + JT 2 ) 4 . 18. (2T 8 - x) 5 . 19. (z* - 2a~ 1 ) 4 . 20. + . 2 1 1.(?-36>V , \a / Find only the first three terms of the expansion. 23. (a + 12) 16 . 24. (c - 3) 25 . 26. (a 2 + 6 3 ) 20 . 26. (1 + 2a) 10 . 27. (1 - .I) 22 . 28. (1 + .2) 12 . 29. (1 - V2) 12 . 30. (1 - 3s 3 ) 18 . 31. (2x - a 2 ) 80 , 32. (z* + &)". 33. (a" 1 + 3) M . 34. (x - a" 2 ) 11 . 36. (* - y) n . 36. (a + z)*. 37. (z 2 - y) m . 38. (w; 2 .-h )*. Compute each factorial expression. 71 ot 39. 6! 40. 8! 41. 11! 42. ^ 43. ^fVi 3! 5! 4! 1 45. General term of the binomial expansion By use of Properties I to IV of Section 143, we obtain THE BINOMIAL THEOREM 207 , , v. n , i , n(n - 1) (z T y) % + nx n ~ l y H ~-jr ^ , n(n- (1) In (1) we read the dots " " as "and so forth." In the terms in y, y* t and y 8 , we observe special cases of the following facts, which we shall accept without proof. SUMMARY. Description of the term involving y r , in (x + y) n : A. The exponent of x is n r. B. The denominator is the product of all integers from I tor inclusive; that is, the denominator is rl C. The numerator of the coefficient has r factors, the first being n and each other being 1 less than the preceding factor. The last factor is n r + 1. f- When (A), (B), and (C) are combined, they state that . . t . r n(n 1) (n r 4* 1) . r /\ . the term involving y r is - p x n r y r . (2) By use of formula 2, we may write (x + y) n = *" + nx~*y + n(n x n ~*y* -f , n(n - 1) * (n - r + 1) vn _ r , ir H -- -? - AC" r y r . . . J_ l*n r! - ir i-y- (3) We refer to (3) as the binomial formula. By use of (2), we can write any term of (3) without writing the other terms. Hence, we refer to (2) as the general term of the expansion of (x + y) n . ILLUSTRATION 1. The term involving y 4 in the expansion of (x + y) 7 is 7.5.5.4 jj x*y* or 35zV. EXAMPLE 1. Find the 8th term of (3o* - 6) 11 . SOLUTION. The 8th term will involve the 7th power of the 2d term of the binomial. Hence, use (2) with r = 7, x - 3a*, and y - b: 8th term is '' <3*ty<- W - - 26,730a'&'. 208 THE BINOMIAL THEOREM Note 1. To derive a formula for the rth term in (3), we notice that this term will contain y*" 1 as a factor. Hence, we substitute (r 1) for r in (2) and find that *i ~*u t n(n 1) (n r 4 2) , tte rth term is ' ^. ixn-r+i^r-i. (4) We may call (4), as well as (2), the general term. Example 1 could have been solved by use of (4) with r = 8. EXAMPLE 2. Find the term involving z lz in the expansion of (v z 3 ) 7 . SOLUTION. Since 2" = (z 3 ) 4 , we use formula 2 with n = 7, r = 4, x v, and y z 3 : the term is T~o~q~l t;3 (~~ 2;3 ) 4 > or 35V 3 ;? 12 . EXAMPLE 3. Compute (1.01) 6 correct to 3 decimal places. SOLUTION. (1.01) 6 = (1 4 .Ol) 6 = I 6 + 6(1) 6 (.01) 4 15(1) 4 (.01) 2 4 = 14 6(.01) + 15(.01) 2 -f 20(.01) 3 4 = 1 -f .06 + .0015 + .000020 H (negligible terms) = 1.06152 = 1.062, approximately. EXERCISE 79 Find only the specified term. 1. Term involving y 6 in the expansion of (a 4 y) 9 . 2. Term involving x 6 in the expansion of (z + x) 10 . 3. Term involving y* in the expansion of (x y) 7 . 4. Term involving z 6 in the expansion of (x 4 3y) 8 . 5. 4th term of (a x) 9 . 6. 3d term of (w z) 11 . 7. 6th term of (a 2 4 x) 7 . 8. 10th term of (x 2 4 y 3 ) 10 . 9. 4th term of (x - 5y) 7 . 10. 5th term of (1 - .02) 7 . 11. 6th term of (1 4 .I) 8 . 12. 4th term of (Jx - 13. Term involving 2 s in the expansion of (x 2 2 ) 6 . 14. Term involving w 10 in the expansion of (w* 4 I/ 8 ) 8 . 15. Term involving y 8 in the expansion of (x y) n . THE BINOMIAL THEOREM 209 16. Term involving z 3 in the expansion of (o a 17. Term involving x% in the expansion of (y + 18. Middle term of (x* - y) w . 19. Middle term of (a + 3* 2 ) 8 . 20. Middle terms of (a* + 2/) 7 . 21. Middle terms of (2s* + j/*). 22. Term involving -j in the expansion of ( r ) 23. Term involving -g in the expansion of (sp -\ ) (y x\* o) ' Find the term or terms with the largest coefficient in the expansion of the power. 25. (a + x)*. 26. (c - w>) 10 . 27. (a 2 + 6) 9 . 28. (c - rf 3 )". Compute by use of the binomial theorem. In any problem involving decimals t use only enough terms to obtain the result correct to three decimal places. 29. (10 - a) 4 . 30. (100 - 2) 3 . 31. 99 4 . 32. 39 4 . 33. 51 3 . 34. (1.01) 7 . 35. (1.01) 12 . 36. (1.02). 37. (1.03) 7 . 38. (.99) 6 . 39. (.98) 6 . 40. (1.04) 10 . 41. C1.02) 11 . 42. (.52) 8 . 43. (.49) 9 . 44. 101 B . 45. 62*. CHAPTER 13 RATIO, PROPORTION, AND VARIATION 146. Ratio The ratio of one number a to a second number 6 is the quotient a/6. The ratio of a to 6 is sometimes written a:b. A ratio is a frac- tion, and any fraction can be described as a ratio: a:b = |- (1) The ratio of two concrete quantities has meaning only if they are of the same kind. Their ratio is the quotient of their measures in terms of the same unit. ILLUSTRATION 1. The ratio of 3 feet to 5 inches is ^. 147. Proportion A proportion is a statement that two ratios are equal. That is, a proportion is merely a statement that two fractions are equal. The proportion * a:b = c:d means that -r = 3- (1) o a The proportion a: 6 = c:d is read "a is to 6 as c is to d." We say that the four numbers a, 6, c, and d form a proportion. 'In a proportion a: 6 = c:d, the first and fourth numbers, a and d, are called the extremes, and the second and third, b and c, are called the means of the proportion. ILLUSTRATION 1. To solve the proportion a;: (25 x) 3:7, we first change it to fractional form, and then solve the resulting equation: x 3 Z * *'* J x = 75 3z; ICte 75; hence, x - 7.5. *"~ X f * RAT/0, PROPORTION, AND VARIATION 211 EXAMPLE 1. Divide 36 into two parts with the ratio 3:7. SOLUTION. 1. Let x and y be the parts; then x + y " 36. (2) x 3 2. Also, x\y 3:7, or - =* = Hence, 7x 3y. (3) 3. On solving the system [(2), (3)] we obtain (x - 10.8, y = 25.2). ATote Jf . If two triangles (or polygons of any number of sides) are similar, then (a) the ratio of any two sides of one triangle equals the ratio of the corre- sponding sides of the other triangle, and (6) the area of one triangle is to the area of the other as the square of any side of the first triangle is to the square of the corresponding side of the other triangle. EXAMPLE 2. The sides of a triangle are 12, 8, and 15 inches long. In a similar triangle, the longest side is 40 inches long. Find the other sides. SOLUTION. 1. Let x and y be the lengths in inches of the sides of the similar triangle corresponding to those sides which are 8 and 12 inches long in the first triangle. Then, y:12 = 40:15 or - - ; (4) r 4ft z:8 = 40:15 or | - ~ (5) o JLO 2. Solving (4) and (5) we find y = 32 feet and x - 21J feet. EXERCISE 80 Express each ratio as a fraction and simplify. 2. J:. 3. 5:7. 4. x:- 6. Find the ratio of the given quantities. 6. 75 pounds to 160 ounces. 7. 27 days to 156 hours. 8. 51 pints to 17 quarts. 9. 25 miles to 3175 yards. 10. 72 cubic feet to 1320 cubic inches. Change to fractional form and solve. 11. 3:(20 - 2x) = 5:2. 12. (2 - 3y):(4 + 5y) = 3:2. 13. x:(x - 25) - 6. 14. (2 - x):(3 + x) - (4 - *):(2 -f x). 15. 2*:(5 - 3z) * 2. 16. (4 + x):(3 + x} - - 2:5*. 17. A line 18 niches long is divided into two parts whose lengths have the ratio 5: 4. Find the lengths. 272 RATIO, PROPORTION, AND VARIATION Solve by introducing one or more unknowns. 18. Divide 45 into two parts whose ratio is 4: 11. 19. Divide 90 into two parts such that the ratio of one part decreased by 5 to the other part decreased by 10 is 1:4. 20. Find two numbers whose difference is 18 and whose ratio is 4:3. 21. The sides of a triangle are 12, 8, and 18 inches long. In a similar triangle, the shortest side is 40 inches long. Find the other sides. 22. The sides of a polygon are 10, 7, 4, and 8 inches long. If the long- est side is lengthened by 2 feet, by how much should the other sides bo lengthened to obtain a similar polygon? 23. A triangle whose base is 15 inches long has an area of 220 square niches. Find the area of a similar triangle whose base is 6J feet long. 24. The area of a quadrilateral is 49 square feet and its longest side is 12 feet long. Find the area of a similar quadrilateral whose longest side is 15 feet long. 25. The area of a triangle is 150 square feet and its shortest side is 12 feet long. Find the shortest side of a similar triangle whose area is 30 square feet. 26. A man 6 feet tall stands at the foot of a tower and casts a shadow 10 feet long. How high is the tower if its shadow is 69 feet long? 27. A man 5J feet tall stands 40 feet from a street light and casts a shadow 9J feet long. How high is the light? 28. Solve the proportion 3: a; = z:27 for #. 29. Solve the proportion, a:x = x:b for x. If a:x x: 6, then x is called a mean proportional between a and b. If a:x = x:b, then x* = ab or x = =t Vo&; or, if neither a nor b is zero, there are two mean proportionals between a and b. Find the mean proportionals between each of the following pairs of numbers. 30. 64 and 4. 31. - 4 and - J. 32. 2 and 8. 33. 25 and 25. 34. - 2 and 8. 36. - 3 and 27. 36. 2o 3 and 4o. 37. y* and x~ 4 . 38. zy and -- ,.-,,-12. _ * + 2z + 4 ' 4 y 2-t/ ' 2-2 *// a:b = c:x, then x is called the fourth proportional to a, b, and c. Find the fourth proportional to each set of numbers: 41. 2, - 5, and 14. 42. 5, 4, and 7. 43. 3, 6, and a'6. RAT/O, PROPORTION, AND VARIATION 213 *// a : 6 = 6 : x, then x is called the third proportional to a and b. Find the third proportional to each pair of numbers: 44. 18; 50. 45. 2J; . 46. 2xy, y. 47. 5m*n; 3m 3 . *// a:b = c:d and if no denominator involved is zero, prove the following properties of the proportion. 48. PROPERTY I. ad = 6c; or, in any proportion, the product of the means equals the product of the extremes. 49. PROPERTY II. - == -?; or, the means may be interchanged without de- C tt stroying the proportion. (The resulting proportion is said to be obtained from a:b = c:d by alternation.) 50. PROPERTY III. - = - (The resulting proportion is said to be ob- tained from a:b = c:d by inversion.) 51. PROPERTY IV. T = -7 (Said to be obtained by composi- tion. To prove, add 1 to both sides of a: b = c:d.) 52. PROPERTY V. r = -7 (Said to be obtained by division.) 148. Direct variation Let x and y be related variables. Then, we say that y is proportional to x, or y varies directly as x, or y is directly proportional to x y or y varies as x } in case there exists a constant k such that, for every value of x, there is a corresponding value of y given by y = kx. (1) We call k the constant of proportionality or the constant of variation. ILLUSTRATION 1. The circumference C of a circle varies directly as the radius r because C = 2irr. The constant of proportionality is 2ir. 11 From y = kx, we obtain k = - Hence, if y is proportional to x, X the ratio of corresponding values of y and z is a constant. Con- RATIO, PROPORTION, AND VARIATION versely, if the ratio of corresponding values of two variables y and x y is a constant, then y is proportional to x, because the equation - * k X leads to y = kx. ILLUSTRATION 2. If y is a function of x and if it is known that - = 4, X then y = 4x, and y is proportional to x. If y is proportional to x, then x is proportional to y. In other words, the proportionality relationship is a reciprocal property. This is true because, if y kx, then * - T y. (2) Hence, if y varies as x, with fc as the constant of proportionality, tnen x varies as y, with l/k as the constant of proportionality. If y varies directly as x, so that equation 1 is true, then the graph of the relationship is a straight line, because (1) is linear in x and y. We observe that, for any value of k, the graph of (1) passes through the origin, because (x 0, y = 0) is a solution of (1). ILLUSTRATION 3. If y is proportional to x, with 3 as the constant of proportionality, then y = 3x. The graph of this equation is given in Figure 16. 1 49. Inverse variation Fis. 16 We say that y is inversely proportional to x, or y varies inversely as x, in case there exists a constant k such that, for every value of x, there is a corresponding value of y given by (i) k y = * x From this equation, k - xy, or the product of corresponding values of x and y is a constant. RATIO, PROPORTION, AND VARIATION 215 ILLUSTRATION 1. The time t necessary for a train to go a given distance d varies inversely as the speed s of the train because t =* d/8. The constant of proportionality here is d. If y varies inversely as x, with k as the constant of proportionality, then likewise x varies inversely as y, because the equation A; xy, which comes from (1), leads to both of the equations y * K - and = - (2) x y ILLUSTRATION 2. If y varies in- versely as x, with 4 as the con- stant of proportionality, then y = 4/x or xy = 4. The graph of y as a function of x is the graph of the equation xy = 4. This graph has no points for which x = or y = because in such cases xy = 0, and hence xy j* 4. We make up the following table of val- ues by substituting the values of x in y = 4/z. The graph, in Figure 17, extends beyond all limits upward and downward, approaching the y-axis as shown. Similarly, as x grows numerically large without bound, either through positive or through negative values, y approaches zero and the curve approaches the z-axis. The curve in Figure 17 is an illustration of a hyperbola.* Fis. 17 X -8 4 - 2 1 - i * i 16 1 2 4 8 y -i j -2 4 - 16 4 2 1 i 1 50. Joint variation We say that z varies jointly as x and y, or z is directly proportional to x and y, or z is proportional to x and y, or z varies as x and y, in case z is proportional to the product xy, or z'= hxy, * Graphs of hyperbolas are considered in detail in Chapter 16. d) (2) 216 RATIO, PROPORTION, AND VARIATION where k is a constant of proportionality. Notice that the significance of the word and hi each statement hi (1) is that x and y are multiplied in (2). Any of the various types of variation may be combined. ILLUSTRATION 1. To say that z varies directly as x and y and inversely as uP means that z kxy/vP. ILLUSTRATION 2. If P = lOafy/s 3 , then P varies directly as # 2 and y, and inversely as z 3 . 1 51 . Applications of variation equations Suppose it is known that certain variables are related by a variation equation, with an unknown constant of proportionality, k. Then, if one set of corresponding values of the variables is given, we can find A; by substituting the values in the variation equation. EXAMPLE 1. If y is proportional to x and w 2 , and if y = 36 when x = 2 and w ~ 3, find y when x = 3 and w = 4. SOLUTION. 1. We are given that y - ku?x, where k is an unknown constant. 2. To find k, substitute (y = 36, x = 2, w = 3) in y = kw*x: 36 = &(3 2 )(2); 36 = 18fc or k = 2. (1) 3. From (1), y = 2w*x. (2) 4. Substitute (a? - 3, w - 4) in (2): y - 2- 16-3 = 96. Notice that the following steps were taken hi Example 1. 1. The variation statement was translated into an equation involving an unknown constant of proportionality. * 2. The unknown constant was found by substituting given data. 3. The value of the constant of proportionality was substituted in the equation of variation, and this equation was used to obtain the value of one variable by use of given values of the other variables. Useful information can be obtained by use of an equation of varia- tion on many occasions when the data are not sufficient to enable us to find the value of the constant of proportionality. EXAMPLE 2. The kinetic energy of a moving body is proportional to the square of its velocity. Find the ratio of fye kinetic energy of an automobile traveling at 50 miles per hour to the kinetic energy of the same automobile traveling at 20 miles per hour. RATIO, PROPORTION, AND VARIATION 217 SOLUTION. 1. Let E be the energy, and v the velocity in miles per hour. Then, E = kv 2 , Where A; is a constant of proportionality. (The data do not permit us to find the value of &.) 2. Let EI be the energy at 20 miles per hour, and E 2 the energy at 50 miles per hour. Then, #1 - *(20) 2 or E t - 400fc; (3) # 2 = &(50) 2 or # 2 = 2500&. (4) 3. From (3) and (4), E 1 _ 2500fe _ 26 ^ Ei ~ 400& ~ 4 " **' Thus, E 2 is 6J times as large as EI. Note 1. In applications of an equation of variation, the constant of pro- portionality will depend on the units in terms of which the variables in the problem are measured. Hence, if the constant is determined for one set of units, care must be exercised to employ the same units whenever this value of the constant is used. EXERCISE 81 Introduce Utters if necessary and express the relation by an equation involv- ing an unknoum constant of proportionality. 1. H varies directly as x and inversely as wP. 2. B is proportional to x 2 and inversely proportional to z. 3. Z is proportional to \ /r x and varies inversely as y*. 4. K is proportional to z and w 2 and inversely proportional to xy. 6. (x + 2) is inversely proportional to (y + 3). 6. The area of a triangle is proportional to its altitude. 7. The volume of a sphere is proportional to the cube of its radius. 8. The volume of a specified quantity of gas varies inversely as the pressure applied to it, if the temperature remains unchanged. 9. The weight of a body above the surface of the earth varies inversely as the square of the distance of the body from the earth's center. 10. The power available in a jet of water varies jointly as the weight of the water per cubic foot, the cube of the water's velocity, and the cross- section area of the jet. 11. The maximum horsepower of the boiler which can be served by a chimney of given cross-section area is proportional to the square root of the height of the chimney. 218 RATIO, PROPORTION, AND VARIATION For each formula, give a statement about the variable on the left side in the language of variation. All letters except the constant k represent variables. 12. y - 7w. 13. z 3x\ 14. z = 5xy*. 16. u = 4* 4 - <> rt 16. w = -T- 17. 10 r- 18. u = 19. w - y # y 2 z By employing all data, obtain an equation relating the variables. 20. P is directly proportional to x* and P 18 if x = 4. 21. R is inversely proportional to x and directly proportional to y, while R = 4 when x 3 and y 5. 22. U varies directly as 3 and y, and inversely as z 2 ; C7 = 15 when a; = 5, y = 2, and 2 = 3. 23. # varies jointly as x and y and inversely as Vz; H = 6 when z = 2, y 3, and 2 = 9. 24. If w is proportional to z and if w = 5 when x = 7, find w when * - - 6. 25. If y is inversely proportional to x and if y 5 when x = 20, find y when x 15. 26. If H is proportional to x and inversely proportional to y, and if H = 3 when 2 and y = 4, find H when y = 9 and x = 5. 27. The distance fallen by a body, starting from a position of rest in a vacuum near the earth's surface, is proportional to the square of the num- ber of seconds occupied in falling. If a body falls 256 feet in 4 seconds, how far will it fall in 7 seconds? 28. The kinetic energy E, of a mass of m pounds moving with a velocity v, is proportional to mv 1 . If E = 2500 foot-pounds when a body weighing 64 pounds is moving at a velocity of 50 feet per second, find the kinetic energy of a body weighing 30 pounds whose velocity is 2400 feet per minute. 29. If one body is sliding on another, the force of sliding friction is pro- portional to the normal pressure between the bodies (if this pressure is moderate). If the sliding friction between two cast-iron plates is 60 pounds when the normal pressure is 270 pounds, find the normal pressure when the sliding friction is 600 pounds. 30. The maximum safe load of a horizontal beam supported at its ends varies directly as its breadth and the square of its depth and inversely as the distance between the supports. If the maximum is 2400 pounds for a beam 4 inches wide and 10 inches deep, with supports 15 feet apart, find the maximum load for a beam of the same material which is 3 inches wide and 5 inches deep, with supports 25 feet apart. RAT/0, PROPORTION, AND VARIATION 219 31. How far apart may the supports be placed if a beam 5 inches wide and 8 inches deep, like those in Problem 30, supports 6000 pounds? 32. A beam like those in Problem 30 is 6 inches wide and the supports are 12 feet apart. How deep must the beam be to support 3500 pounds? 33. The approximate amount of steam per second which will flow through a hole varies jointly as the steam pressure and the area of a cross section of the hole. If 40 pounds of steam per second at a pressure of 200 pounds per square inch flows through a hole whose area is 14 square niches, (a) how much steam at a pressure of 260 pounds per square inch will flow through a hole whose area is 20 square inches; (6) what is the area of a hole which allows 30 pounds of steam to flow through it when the pressure is 300 pounds per square inch? 34. The electrical resistance of .a wire varies as its length and inversely as the square of its diameter. If a wire 350 feet long and 3 millimeters in di- ameter has a resistance of 1.08 ohms, find the length of a wire of the same material whose resistance is .81 ohm and diameter is 2 millimeters. 36. If y is proportional to x and if y = 16 when x = 4, graph the relation between x and y. Make a statement about the change in the value of y t (a) if x varies from any given value to a value three times as large; (6) if x increases by 25% from a given value. 36. Repeat (a) and (b) of Problem 35 in case y is inversely proportional to x and y = 16 when x = J. 37. The approximate velocity of a stream of water, necessary to move a round object, is proportional to the product of the square roots of the ob- ject's diameter and its specific gravity. If a velocity of 11.34 feet per second is needed to move a stone whose diameter is 1 foot and specific gravity is 4, how large a stone with specific gravity 3 can be moved by a stream whose velocity is 22.68 feet per second? 38. Read Example 2 in Section 151. Find the ratio of the kinetic energy of a skater whose speed is 20 miles per hour to his energy when his speed is 15 miles per hour. 39. The horsepower that can be safely transmitted by a solid circular steel shaft varies jointly as the cube of its diameter and the number of revolu- tions it makes per minute. If a shaft 1.5* in diameter rotating at 1520 revolu- tions per minute can' transmit 135 horsepower, find the speed at which the shaft could transmit 162 horsepower. 40. The illumination received from a source of light varies inversely as the square of the distance from the source, and directly as its candle power. At what distance from a 50 candle power light would the illumination be one half that received at 30 feet from a 40 candle power light? 220 RATIO, PROPORTION, AND VARIATION 41. Newton's Law of Gravitation states that the force with which each of two masses of m pounds and M pounds attracts the other varies directly as the product of the masses and inversely as the square of the distance between the masses. Find the ratio of the force of attraction when two masses are 8000 miles apart to the force when they are 2000 miles apart. 42. As a first approximation, it is found that the wind pressure on a surface at right angles to the direction of the wind varies jointly as the area of the surface and the square of the wind velocity. What wind velocity would be necessary to cause the pressure on 40 square feet of surface to be double the pressure exerted on 10 square feet by a wind velocity of 30 miles per hour? 43. The current in an electric circuit varies directly as the electromotive force and inversely as the resistance. In a certain circuit, the electromotive force is A volts, -the resistance is b ohms, and the current is c amperes. If the resistance is increased by 20 ; %, what per cent of increase must occur in the voltage to increase the current by 30%? Note 1. The statement x is to y is to z as r is to s is to t, or x, y, and z are proportional to r, s, and t is abbreviated by x: y:z = r:s:t, and means that there exists a number k 9* such that x =? AT, y = ks, and z kt. if Find x, y, and z under the given conditions. 44. x:y:z = 4: 2:5, and x + %y + z = 40. HINT, x 4k; y = 2&; z = 5k. Substitute in the given equation and find the value of k. 46. x:y:z = 5: 3:2, and x y z = 12. 46. x:y:z = 2:5:1, and x* + y 2 + 2 = 120. 47. x:y:z = 3: - 1:2, and x 2 + y 2 + *> = 56. 48. Divide 2800 into four parts proportional to 5 : 3 : 4 : 2. 49. Divide 1250 into four parts proportional to 3:5:11:6. CHAPTER 14 PROGRESSIONS 1 52. Arithmetic progressions A seqwnce of things is a set of things arranged in a definite order. An arithmetic progression (abbreviated A.P.) is a sequence of numbers called terms, each of which, after the first, is derived from the pre- ceding one by adding to it a fixed number called the common differ- ence. The common difference can be found by subtracting any term from the one following it. ILLUSTRATION 1. In the arithmetic progression 9, 6, 3, 0, 3, , the common difference is 3. The 6th term would be 6. 153. The nth term in an arithmetic progression Let a be the first term and d be the common difference. Then, the second term is a + d; the third term is a -f 2d; the fourth term is a -f 3d. In each of these terms, the coefficient of d is 1 less than the number of the term. Similarly, the tenth term is a + 9d. The nth term is the (n l)th after the first term, and is obtained after d has been added (n 1) times, in succession. Hence, if I represents the nth term, / = a + (n - l)d. (1) ILLUSTRATION 1. If a = 3 and d 4, the 18th term is 3 -f- 17(4) = 71. 1 54. Sum of an arithmetic progression Let S be the sum of the first n terms of an A.P. The first term is a; the common difference is d', the last term is I; the next to the last term is I d, etc. On writing the sum of the n terms, forward and backward, we obtain 222 PROGRESS/ONS l; (1) S~l + (l -d) + (l -2d) + --- + (a + 2d) + (a + d) + a. (2) On adding corresponding sides of (1) and (2) we obtain where there are n terms (a + 1). Hence, 2S = n(a + 0, or S = 5 ( a + I). (3) EXAMPLE 1. Find the sum of the A.P. 8 + 5 4- 2 4- to twelve terms. SOLUTION. 1. First obtain I from I = a + (n l}d. We have a 8, d - - 3, and n = 12: I = 8 + 11(- 3) = - 25. 2. From (3), S = 6(8 - 25) = - 102. If we rewrite (3) in the form S = np-J)> (4) we observe that the sum of an A.P. of n terms equals n times the average of the first and last terms. On substituting I * a + (n l)dm (3), we obtain S = % [2a 4- (n - l)d]. (5) & The quantities a, d, I, n, and S are called the elements of the general arithmetic progression. When three of the elements are given, we may obtain the other two by use of I = a 4- (n l)d and formulas 3 and 5. EXAMPLE 2. Find the remaining elements in an A.P. for which a = 2, I 402, and n = 26. SOLUTION. 1. We wish to find d and S. From (3), S = 13(404) = 5252. 2.' From I = a + (n - l)d, 402 = 2 + 25d; hence, rf = 16. If a sequence of three numbers a, 6, and c forms an A.P., then b a c 6, because each side of this equation is equal to the common difference. EXAMPLE 3. Find the value of k if (17, k, 29) form an A.P. SOLUTION, k - 17 = 29 - k', 2k = 46; hence, k = 23. PROGRESSIONS 223 t EXAMPLE 4. Find the sum of the A.P. 6 + 9 + 12 H h 171. SOLUTION. 1. We have given a - 6, d 3, and I = 171. 2. To find n, use Z = a + (n - l)d: 171 = 6 + 3(n- 1); 171 = 6 + 3n-3; n = 56. KA 3. To find 5, use (3) : 8 = ^ (6 + 171) - 4956. & EXAMPLE 5. Find the 3&th term in an A.P. where the 4th term is 8 and the common difference is 5. SOLUTION. 1. Think of a new A.P. where 8 is the 1st term; the former 39th term is the 36th term of the new progression. 2. Use I = a + (n - l)d with a = - 8, d = 5, and n = 36: desired 39th term = - 8 + 35(5) - 167. EXERCISE 82 Write the first six terms of an A.P. from the given data. 1. a = 15; d = 3. 2. a = 17; d = - 3. 3. a = 18; d = 2. v / 4. o = &; d = A. Which sequences do not form arithmetic progressions? 6. 3, 7, 11, 15. 6. 15, 17, 20, 22. 7. 23, 20, 17. 8. 35, 32, 30, 28. Find the value of b for which the sequence forms an A.P. 9. 3, 8, b. 10. 25, 21, b. 11. 15, b, 13. * 12. b, 17, 23. Find the specified term of the A.P. by use of a formula. 13. Given terms: 4, 7, 10; find the 50th term. 14. Given terms: 5, 8, 11; find the 29th term. 15. Given terms: 4, 2, 0; find the 41st term. 16. Given terms: 3, 3J, 3J; find the 83d term. 17. Given terms: 2.4, 2.6, 2.8; find the 39th term. 18. Given terms: 3, 2.95, 2.9; find the 201st term. Find the last term and the sum of the A.P. by use of formulas. 19. 8, 13, 18, to 15 terms. 20. 3, 5, 7, to 41 terms. 21. 9, 6, 3, to 28 terms. 22. 13, 8, 3, to 17 terms. t 28. 2.06, 2.02, 1.98, to 33 terms. 24. 5, 4i, 4, to 81 terms. 224 PROGRESSIONS In each problem, certain of the elements a, d, I, n, and S are given. Find the missing elements. 25. a = 10, 1 = 410, n - 26. 26. a = 4, 1 - 72, n = 18. 27. a - 17, 1 - 381, d - 4. 28. i - 53, d = 4, n = 19. 29. I - 87, d - - 3, n - 18. 30. a = 27, Z = 11, d = - J. 31. a - 60, Z = 0, d = - f. 32. S = - 2496, n = 52, a = 3. 33. S = 2337, n - 38, d - J. 34. n = 26, S = 5278, d = 16. Ftnd the value of k for which the sequence of three terms forms an A.P. 36. (3 - 2*); (2 - *); (4 + 3*). 36. (2 + *); (2 + 4fc); (6fc - 1). 37. Find the 45th term in an A.P. where the 3d term is 7 and the common difference is J. t 38. Find the 59th term in an A.P. where the 4th term is 9 and the common difference is .4. 39. In the A.P. .97, 1.00, 1.03, -, which term is 5.02? 40. In the A.P. 16, 13.5, 11, , which term is - 129? 41. Find the common difference of an A.P. whose 6th term is 9 and 37th term is 54. 155. Arithmetic means The first term, 'a, and the last term, l } in an arithmetic progression are called the extremes of the progression. The other terms are called arithmetic means between a and /. To insert k arithmetic means be- tween two numbers, a and J, means to find a sequence of k numbers which, when placed between a and Z, give rise to an A.P. with a and / as its extremes. EXAMPLE 1. Insert five arithmetic means between 13 and 11. SOLUTION. 1. After the means are inserted, they will complete an A.P. of seven terms, with a = 13 and I = 11. We shall find d for the progres- sion and then form the terms. 2. From I = a + (n - l)d, - 11 - 13 + 6d; d - - 4. 3. Hence, the missing terms are (13 4), or 9; (9 4), or 5; etc. The A.P. is (13, 9, 5, 1, 3, 7, 11). Therefore the arithmetic means are (9, 5, 1, - 3, - 7). PROGRESSIONS 225 When a single arithmetic mean is inserted between two numbers, it is called the arithmetic mean of the numbers. Thus, if (6, A, c) form an A.P., then A is called the arithmetic mean of 6 and c. Then, A - b = c - A or 2 1 A = c + 6. Hence, x _ A or the arithmetic mean of two numbers is one half of their sum. Thus, the arithmetic mean of b and c is the number which is frequently called the average of 6 and c. ILLUSTRATION 1. The arithmetic mean of 7 and 15 is J(7 + 15) = 11. Note 1. The average of k numbers is denned as their sum divided by k. As a generalization of equation 1, the average of k numbers is frequently called the arithmetic mean of the numbers. Unless we are dealing with just two numbers, so that k 2, the arithmetic mean of k numbers has no con- nection with the notion of arithmetic means as they occur in arithmetic pro- gressions. 1 56. Applications of arithmetic progressions In a problem dealing with an A.P., write down the first few terms of the progression and describe them in the language of the problem. Then, decide which elements are known and which you wish to find. EXAMPLE 1. A man invests $1000 at the end of each year for 30 years at 6% simple interest. Find the accumulated value of his investments at the end of 30 years, if no interest is withdrawn until then. SOLUTION. 1. The first $1000 invested will draw interest at 6% for 29 years, or a total of $1740 interest; the resulting amount at the end of 30 years is $2740. 2. The second $1000 invested will draw interest for 28 years; the re- sulting interest is $1680 and the amount at the end of 30 years is $2680. 3. Etc.; the $1000 invested at the end of 29 years will draw interest for just one year; the resulting amount is $1060. The last $1000 is invested at the end of 30 years and receives no interest. 4. The total amount at the end of 30 years is 274Q + 2680 + 2620 + + 1060 + 1000. We wish S for an A.P. in which a = 2740, n = 30, and d = - 60. 5. From S - 5 (a + !), S =.^(2740 + 1000) * $56,100. 226 PROGRESSIONS EXAMPLE 2. A contractor has agreed to pay a penalty if he uses more than a speckled length of time to finish a certain job. The penalties for excess time are $25 for the 1st day and, thereafter, $5 more for each day than for the preceding day. If he pays a total penalty of $4050, how many excess days did he need to finish the work? SOLUTION. 1. The penalties are $25, $30, $35, , which form an A.P. where a 25, d = 5, and S = 4050. We wish to find the number of terms, n. 2. From S = Jn[2a + (n - 1)<T|, 4050 - -[50 + 5(w - 1) J (1) * / 3. To solve (1), multiply both sides by 2: 8100 - 50n + 5n 2 - 5n; w 2 + 9n - 1620 - 0. (2) On solving (2) by factoring, or the quadratic formula, we find n = 36 and n = 45. The negative root has no application in the problem. Hence, there are 36 excess days. CHECK. The student should compute the sum of 25 + 30 + 35 H to 36 terms. EXERCISE 83 1. Insert four arithmetic means between 2 and 17. 2. Insert five arithmetic means between 2 and 40. 3. Insert five arithmetic means between 7 and 17. 4. Insert four arithmetic means between 19 and 12. 6. Insert six arithmetic means between 15 and 16.5. 6. Insert seven arithmetic means between f and 7. 7. Insert five arithmetic means between f and 6. / m Find the arithmetic mean of the numbers. 8. 6; 38. 9. 15; 37. 10. - 13; 27. 11. - 15; - 23. 12. x; y. 13. Find the sum of all even integers from 10 to 380 inclusive. 14. Find the sum of all odd integers from 15 to 361 inclusive. 15. Find the sum of the first 38 positive integral multiples of 3. 16. Find the sum of all positive integral multiples of 5 which are less than 498, PKOGKESS/ONS 227 17. There are 16 rows of billiard balls in a symmetrical triangular arrangement on a table, with 46 balls in the first row and 3 less balls in each other row than hi the one preceding it. How many balls are on the table? 18. Find the sum of all positive and negative integral multiples of 6 between 55 and 357. 19. The horizontal base of a right triangle is 15 feet long and the side perpendicular to this base is 45 feet long. At intervals of 1 foot on the base, a perpendicular is drawn to the base and reaches to the hypotenuse. Find the sum of the lengths of all perpendiculars, including the vertical leg of the triangle. 20. A man invests $1000 at the end of each year for 12 years at 6% simple interest. What is the accumulated value of his investments at the end of 12 years? Find the total sum of money paid by the debtor in discharging his debt. 21. Debtor borrows $10,000. Agrees to pay: at the end of each year for 10 years, $1000 principal and simple interest at 3% on all principal outstanding during the year. 22. Debtor borrows $20,000. Agrees to pay: at the end of each year for 20 years, $1000 principal and simple interest at 5% on all principal outstanding during the year. 23. A man invests $1000 at the beginning of each year for 20 years at 5% simple interest. Find the accumulated value of his investments at the end of 20 years. 24. The 4th term of *an A.P. is 215 and the 44th term is 55. Find the sum of the first 20 terms. 25. If y = 5x -j- 8, find the sum of the values of y corresponding to the successive integral values x = 1, 2, 3, , 30. 26. The bottom rung of a ladder is 28 inches long and each other rung is one half inch shorter than the rung below it. If the ladder has 18 rungs, how many feet of wood were used in making the rungs? *. 157. Geometric progressions A geometric progression (abbreviated G.P.) is a sequence of num- bers called terms, each of which, after the first, is obtained by multi- plying the preceding term by a fixed number called the common ratio. The common ratio equals the ratio of any term, after ihe first, to the one preceding it. 223 PROGRESSIONS ILLUSTRATION 1. In the G.P. 16, 8, -f 4, 2, , the common ratio is - J; the 5th term would be (- J)(- 2) = + 1. To determine whether or not a sequence of numbers forms a geometric progression, we divide each number by the one which pre- cedes it. All of these ratios are equal if the terms form a G.P. 8 x 64 ILLUSTRATION 2. If 3, 8, and x form a G.P., then = or x = -=- O O O If the terms of a G.P. are reversed, the terms will form a G.P. whose common ratio is the reciprocal of the ratio for the given G.P. ILLUSTRATION 3. In the G.P. (4, 8, 16, 32), the common ratio is 2. When the terms are reversed, we have (32, 16, 8, 4), where the ratio is ILLUSTRATION 4. The G.P. (a, ar, ar 2 , ar 3 ) has the common ratio r whereas the G.P. (ar 8 , ar 2 , ar, a) has the common ratio arVar 8 or 1/r. 1 58. The nth term of a geometric progression Let a be the first term and r be the common ratio. Then, the second term is ar; the third term is ar 2 . In each of these terms the exponent of r is 1 less than the number of the term. Similarly, the eighth term is or 7 . The nth term is the (n l)th after the 1st and hence is found by multiplying a by (n 1) factors r, or by r n-1 . Hence, if I represents the nth term, (1) ILLUSTRATION 1. If a = 3 and r = 2, the 7th term is 3(2 6 ) = 192. 1 59. Sum of a geometric progression Let S be the sum of the first n terms of a G.P. The terms are (a, ar, ar 2 , *, ar n " 2 , ar"- 1 ), where ar n " 2 is the (n l)th term. Hence, S =* a + ar 4- ar 2 + + ar n ~* -f ar n-1 , (1) and Sr ar -f ar 2 + ar 8 -f -f- ar n-1 + ar n ; (2) in (2) we multiplied both sides of (1) by r. On subtracting each side of (2) from the corresponding side of (1), we obtain S - Sr = a - ar n , (3) because each term, except ar n , on the right in (2) cancels a correspond- ing term in (1). From (3), S(l r) - a ar n , or PROGRESSIONS 229 Since I - ar n ~ l y then rl ar n . Hence, from (4), In using (4), it is sometimes convenient to rewrite it as 1 - r n S = ai L. (6) EXAMPLE 1. Find the sum of the G.P. 2, 6, 18, to six terms. SOLUTION. n = 6; a = 2; r = 3. From (4), 2 - 2-3 2 - 1458 79R S== 1-3 = -2 = 728 ' . Formula 5 is convenient when Z is explicitly given. EXAMPLE 2. Find the sum of the geometric progression (1.05) 2 + (1.05) 6 + (1.05) 8 + - + (1.05) 86 . SOLUTION, a = (1.05) 2 ; r = (1.05) 3 ; I = (1.05) 35 . From formula 5, o = (1.05) 2 - (1.05) 8 (1.05) 36 (L05) 2 - (1.05) 88 6 1 - (1.05) 3 1 - (1.05) 3 Note L When a sufficient number of the elements (a, r, n, Z, S) are given, we find the others by use of I = ar n-1 , (4), and (5). EXAMPLE 3. If S = 750, r = 2, and Z = 400, find n and a. a rl SOLUTION. 1. From S l t-rrr\ < K/ % 750 = -= =~; hence, a = 50. J. ~~ & 2. From I = ar n ~\ 400 = 50(2*-'); 2r~ l = = 8; 2n-i _ 23; hence, n 1 = 3, or n - 4. If three numbers (a, 6, c) form a G.P.. then - = T- a o 10 50 ILLUSTRATION 1. If (a, 10, 50) form a G.P. then ~ Tn or a 230 PROGRESSIONS 1 60. Geometric means The first term, a, and the last term, I, in a G.P. are called the extremes of the progression. The other terms are called geometric means between a and I. To insert k geometric means between two numbers, a and I, means to find a sequence of k numbers which, when placed between a and Z, give rise to a G.P. with a and I as its extremes. In asking for geometric means, we shall desire only real valued means. EXAMPLE 1. Insert two geometric means between 6 and ^. SOLUTION. After the means are inserted, they will complete a G.P. of four terms with a = 6 and I = *-. We shall find the common ratio of the progression, and then its two middle terms. From I = ar n ~ l t with n = 4, we obtain i? = fir- rsA. 'OE..? 9 or; 27 ; r ^27 3' The G.P. is (6, 4, f , *). The geometric means are 4 and f . EXERCISE 84 Write the first four terms of a G.P. for the given data. 1. a = 5; r = 3. 2. a = 16; r = J. 3. a = 4; r - 2. 4. a = 27; r = - J. In case the numbers form a G.P., state its common ratio and write two more terms of the G.P. 6. 4, - 12, 36, 108. 6. 10, 5, J, f . 7. 8, f , f , ft. 8. 4, 2, 1, 0. \ 9. a, ax, ax 2 , ax 8 . 10. (1.02) 4 , (1.02), (1.02) 8 . 11. (1.01)- 5 , (1.01)- 8 , (1.01)-*. Find the value of x so that the three numbers form a G.P. 12. 5, 20, x. 13. x, 12, 36. 14. 4, x, 16. 15. x, - 4, ia 16. If the G.P. (81, 27, 9, 3) is reversed, what fact do you observe? By use of I = ar""" 1 , find the specified term of the given G.P. without finding the intermediate terms. 17. 3, 9, 27; find the 6th term. 18. 4, - 12, 36; find the 9th term. 19. 12, 6, 3; find the 8th term. 20. 6, - 3, f ; find the 9th term. PROGRESSIONS 237 Find the last term and the sum of the G.P. 21. 4, 12, 36, to 7 terms. 22. 12, 6, 3, to 6 terms. 23. 5, - 15, 45, to 6 terms. 24. 25, 2.5, .25, to 7 termu. 25. 3, - 6, 12, to 7 terms. 26. jfc, - }, 1, to 6 terms. 27. 3, 66, 126 2 , to 8 terms. 28. 4, 8z 2 , 16s 4 , to 7 terms. Employ formula 5 on page 229 to find the sum of the G.P. 29. 4 + 2 + 4- ife. 30. 5 + 15 + + 3645. Find tfie missing elements of the G.P. 31. a = 5; r - 2; Z = 640. 32. a 2; r = 3; Z = 486. 33. r = 10; a - .001; I - 100. 34. 5 - 2186; I = 1458; a = 2. 35. S = 275; r = - 2; Z = 400. 36. S = *#*; a - - f ; Z - 135. 37. a = 256; r - i; 2 = J. 38. a = 1458; r = J; J - . Find Me specified term without finding the first term of the G.P. 39. The 10th term, if the 6th term is 5 and common ratio is 2. 40. The 12th term, if the 8th term is 25 and common ratio is .1. 41. The 4th term, if the 8th term is 40 and common ratio is 2. 42. The 5th term, if the 9th term is 80 and common ratio is J. * Insert the specified number of geometric means. 43. Five, between 2 and 128. 44. Five, between 128 and 2. 45. Three, between 4 and 324. 46. Four, between J and 81. j 47. Six, between .1 and 1,000,000. 48. Three, between 16 and .0001. // x and y are of the same sign, and if a single geometric mean G of the same sign is inserted between x and y, then G is called the geometric mean of x and y; (x, G, y) form a G.P. Find the geometric mean of the numbers 49. J; 16. 60. i; 36. 51. 4; 25. 52. - 9; 53. Find the geometric mean of x and y. S*wte the result in words Find an expression for the sum and simplify by use of the laws of er but do not compute. Use formula 5 on page 229 when convenient. 54. 1 + (1.03) + (1.03) 2 + - + (1.03). 55. 1 -I- (1.05) + (LOS) 2 + + (1.05) 56. (1.02)* + (1.02) + (1.02)< + + (1.02)". 232 PROGRESSIONS 67. (1.06) 4 -I- (1.06) 6 + (1.06) + - - - + (1.06); 58. 1 + (1.02) 3 + (1.02) H ---- to 21 terms. 69. (1.02)- 16 + (1.02)-" + (1.02)- 18 + + (1.02)-*. 60. (1.03)-" + (1.03)-" + (1.03)~ 12 + + (1.03)- 4 . 61. 1 + (1.02)* + (1-02) + (1.02)* + - - -f ( 62. If the 7th term of a G.P. is 5 and the llth term is 3^, find the inter- mediate terms. 63. For what values of k. do the three quantities (k + 3), (6k -f 3), and (20k + 5) form a G.P.? - . 64. Find the sum of a G.P. of 7 terms whose 3d term is j and 6th is -fo. -66. How many ancestors have you had in the twelve preceding gener- ations if no ancestor appears in more than one line of descent? 5. An investment paid a man, in each year after the first, twice as much as in the preceding year. If his investment paid him $13,500 in the first four years, how much did it pay the investor in the first and the fourth years? 67. In a lottery, it is agreed that the first ticket drawn will pay its owner $.10 and each succeeding ticket twice as much as the preceding one. Find the total amount paid on the first 10 tickets drawn. 68. Find the sum of the first 19 positive integral powers of 1.03, given that (1.03) 10 = 1.344. 1 61 . Applications of geor #r. progressions >.. , .1; .. . i ./,.'.' *W^\en a proHe*^ is met where a sequence of terms is suspected of forming ^ A.P., generally it is an advantage to compute the explicit values of the first few terms in simplest form in order to verify the 'istence of a common difference between the terms. On the other % d, if a sequence of terms is suspected of forming a G.P., it is best 4te the first few terms, without actually computing them, in a r hich will exhibit clearly any constant factor which appears to T e powers. * 1. A rubber ball is dropped from a height of 100 feet. On i, the ball rises one half of the height from which it last fell. *, has the ball traveled up to the instant it hits the ground for PKOGRESS/ONS 233 SOLUTION. 1. We list the first few distances traveled by the bouncing ball: 1st fall = 100ft. 1st rise = i(100)ft. \ =i(100)ft. =100 ft. 2d rise = i(i)(100) ft. 4(100) ft. \ 3d fall = i(100)ft. )-m-100)ft. 3d me = i(i)(100) ft. - 10 ft. efc. 2. The 1st fall brings in an unsymmetrical term. On neglecting it tem- porarily, the total distance, in feet, traveled otherwise up to the time of the 12th fall is the sum of 100, i(100), J(100), - . to eleven terms. (1) In (1) we have a G.P. with a = 100, r = J, and n = 11. The sum S is obtained from ~ rn - ion 1 " (i) 11 _ ininsik! - 100(2047). ~ ~ ~ " ~ 1024 ' _ 25(2047) _ 256 ~ 3. On adding the 1st fall to S, we find that the total distance traveled by the ball is 299|f J feet. MISCELLANEOUS EXERCISE 85 Solve by methods involving progressions. 1. If $500 is to be divided between 10 men so that the first one re- ceives $5 and each succeeding man obtains a fixed amount more than the preceding man, how much will the 10th man receive? 2. At a bazaar, tickets are marked with the consecutive even integers 2, 4, 6, and are drawn at random by those entering. If each person pays as many cents as the number on his ticket, how much money is received if 1000 tickets are sold? 3. The path of each swing, after the first, of a pendulum bob is .9 as long as the preceding swing. If the first swing is 40 inches long, how far does the pendulum travel on the first 8 swings? 4. A man piles 152 logs in layers so that the top layer contains 2 logs and each lower layer has one more log than the layer above. How many logs will be in the lowest layer? 234 PROGRESSIONS 6. In a professional golf tournament, the total prize money of $5187 is divided among the six players with lowest scores, so that each man above the lowest receives as much as the man below him. How much does the man with the lowest score receive? 6. A body, dropped from a position of rest in a vacuum near the earth's surface, will fall approximately 32 feet farther in each second, after the first, than in the preceding second. If a body -falls 10,000 feet in 25 seconds, how far does it fall in the first second? 7. Find an expression for the sum of the first n positive integers. 8. Find an expression for the sum of the first n positive even integers. 9. At the beginning of each year, a man invests $300 at simple interest at the rate 7%. At the end of 15 years, what is the total value of his in- vestments if none of them have been disturbed, and if all required interest is paid on that date? 10. The radiator of a motor truck contains 10 gallons of water. We draw off 1 gallon and replace it with alcohol; then, we draw off 1 gallon of the mixture and replace it by alcohol; etc., until 9 drawings and replacements have been made. How much alcohol is in the final mixture? 11. Find the sum of the first 40 positive integral powers of x. 12. In creating a vacuum in a container, a pump draws out J of the remaining air at each stroke. What part of the original- air has been re- moved by the end of the 7th stroke? - 13. A pendulum bob moves over a path 15 inches long on its first swing. In each succeeding swing the bob travels fo^ur fifths of the distance of the preceding swing. How far does the bob travel during the first six swings? 14. In a potato race, tw^njiy potatoes are placed at intervals of 5 feet in a line from the starting poih^with the nearest potato 25 feet away. A runner is required to bring the potatoes back to the starting place one at a time. How far would he run in bringing in all the potatoes? 15. A speculator will make $1200 during the first month and, there- after, in each month, $100 less than in the preceding month. If his original capital is $2700, when will he become bankrupt? 16. Two men start in a distance run. One man proceeds at a uniform speed of 300 yards per minute. The second man travels 435 yards hi the first minute, but, thereafter, in each minute he goes 30 yards less than in the preceding minute. When will the first man overtake the second? 17. Prove that the squares of the terms of a G.P. also form a G.P. Then state a more general theorem of this nature. 18. Prove that the reciprocals of the terms of a G.P. also form a G.P. PROGRESSIONS 235 19. A rubber ball is dropped from a height of 300 feet. On each re- bound, the ball rises one third of the height from which it last fell. What dis- tance has the ball traveled up to the instant the ball hits the ground for the 7th tune? 20. In a certain positive integral number of three digits, the digits form an A.P. and their sum is 15. If the digits are reversed, the new number is 594 less than the original number. Find the original number. Note 1. If P is the value of a certain quantity now, and if its value in- creases at the rate i (expressed as a decimal) per year, then the new value at the end of one year is (P -f Pi), or P(l 4- i). That is, the value at the end of any year is (1 -f t) times the value at the end of the last year. The values at the ends of the years form a G.P. whose common ratio is (1 + i). If A represents the value at the end of n years, then This formula is referred to as the compound interest law because, if a principal P is invested now at the rate t, compounded annually, the amount A at the end of n years will be P(l + i) n > Applications involving compound interest will be treated later. In all of the following problems, it will be assumed that any rate is constant. 21. If 300 units of a commodity are consumed in a first year, and if the annual rate of increase of consumption is 6%, (a) give an expression for the amount consumed in the 7th year; (6) find the total consumption in the first 12 years, given that (1.06) 12 = 2.012. 22. A corporation will sell $1,000,000 worth of its products this year and the sales will increase at the rate of 5% per year. Find the total sales during the first 25 years, given that (1.05) 26 = 3.38635494. 23. The population of a city increased from 131,220 to 200,000 in 4 years. Find the rate of increase per year. 24. A piece of property was purchased 4 years ago for $4860 and its value now is $15,360. Find the annual rate at which the value increased. 25. The value of a certain quantity decreases at the rate w (expressed as a decimal) per year. If // is the value now, and K is the value at the end of n years, prove that K = H(l w} n . (This formula is the basis for computing depreciation charges in business under the so-called constant- percentage .nethod.) 26. A n-.otor truck was purchased for $2500, and its value 4 years later is $1024. .Find the rate, per year at which the value has depreciated. 27. A tytel, purchased 3 years ago for $512,000, is sold for $343,000. Find the n te per year at which its value has depreciated. 236 PROGRESSIONS *162. Harmonic progressions A sequence of numbers is said to form a harmonic * progression if their reciprocals form an arithmetic progression. ILLUSTRATION 1. The sequence (1, J, , ^, J) is a harmonic progression because the reciprocals (1, 3, 5, 7, 9) form an A.P. To insert k harmonic means between two numbers, we first insert k arithmetic means between the reciprocals of the numbers. The re- ciprocals of the arithmetic means are the harmonic means. EXAMPLE 1. Insert five harmonic means between 4 and 16. SOLUTION. 1. First, we insert 5 arithmetic means between J and $. 2. From I = a -f (n l)d, with a = J, I = -j^, and n 7, we find A = 4 3. Hence, the A.P. is (J, & 4. The corresponding harmonic progression is (4, 3ft -^, 3?, 8, ^, 16). Hence, the harmonic means are (3ft ^ , ^, 8, *EXERCISE 86 Insert the specified number of harmonic means. 1. Four, between and ^. 2. Five, between J and 3. Four, between & and . 4. Four, between 4 and 24. 5. Five, between f and J. 6. Four, between J and 3. // (c, .ff, d) /orw a harmonic progression, then H is called the harmonic mean of c and d. Find the harmonic mean of the numbers. 7. 4; J. 8. 9; 6. 9. 4; - 8. 10. 12; 36. 11. x and y. *163. Geometric progressions with infinitely many terms Let S n represent the sum of the progression o, ar, ar 2 , , ar"" 1 . Then, by (4), page 229, a + ar + ar 2 H ---- * Suppose that a set of strings of the same diameter and substance are stretched to uniform tension. If the lengths of the strings form a harmonic progression, a harmonious sound results if two or more strings are caused to vibrate at one time. This fact accounts for the name harmonic prop ession. PROGRESSIONS 237 ILLUSTRATION 1. Consider the endless geometric progression 1, - -> , =jj^j to infinitely many terms. . (2) In (2), r = -; the nth term is ^; 1 - r = ^; ar" = ^- By(l), i + l + 1 + 1 + ...+^ ^2--^. (3) If n grows larger, without limit, the term ^ grows smaller, and is as near to zero as we please, if n is sufficiently large. Thus, if n 65, J_ = JL _ 1 _ 2 n-i 2 64 18,446,744,073,709,551,616* which is practically zero. Hence, in (3), 8 n will be as near to (2 0) as we please for all values of n which are sufficiently large. To summarize this statement we say that as n becomes infinite S n approaches the limit 2, and we call 2 the sum of the progression 1, 4, 4, 4, to infinitely many terms. We sometimes use "n >oo " to abbreviate "n becomes infinite." Then, our conclusion can be briefly written limit S n = 2. n Now, consider (a, ar, ar 2 , to infinitely many terms), under the condition that r is a number between 1 and -f 1. Then, as n > oo , the absolute value of the numerator ar n in (1) grows smaller, and is as near to zero as we please for all values of n sufficiently large. Hence, from (1) we see that, as n grows large without limit, the value of S n approaches / a 0_\ a \r=r r^7/' or r^"r" That is, limit S n = =-2 (4) n too 1 r This limit of the sum of n terms, as n becomes infinite, is callejd the sum of the geometric progression with infinitely many terms. If S represents this sum, then - r Thus, if | r | < 1, (5) (a + or + <n* + to infinitely many terms) = * _ (6) 238 PROGRESSIONS Note 1. Recognize that S in (5) is not a sum in the ordinary sense of the word, but is the limit of the sum of n terms as n grows large without bound. ILLUSTRATION 2. By use of (6), with a = 5 and r }, 4- 4- T + to infinitely many terms] 7-377 = 10. Practically, this means that, if we should add a relatively large number of terms, we would obtain approximately 10, and that by adding enough terms we can obtain as close to 10 as we may desire. Thus, Su = The indicated sum of a sequence of numbers is frequently called a series. An expression of the form + u% + u* + to infinitely many terms (7) is called an infinite series. Accordingly, the expression on the left in (6) is referred to as the infinite geometric series. EXAMPLE 1 . Find a rational number equal to the endless repeating decimal .5818181 SOLUTION. Because of the meaning of the decimal notation, .5818181 - = .5 + .081 4- .00081 H ---- to infinitely many terms, where we notice that (.081 + .00081 + ) is an infinite geometric series with a .081, and r = .01. By (6), the sum of this series is .081 = .081 = 9 1 - .01 .99 110* Hence, .5818181 ---- .5 + rf^ - & + rf * - f? - Comment. By use of the method of Example 1, we can show that any endless repeating decimal is a rational number. In any infinite series such as (7), let S n represent the sum of the first n terms. Then we say that the series has a sum S, and call the series a convergent infinite series which converges to S in case the limit of S n is S as n becomes infinite. If S n has no limit as n becomes infinite, we say that the infinite series is divergent, or diverges. Note 2. In this section we have proved that the infinite geometric series in parentheses in (6) has a sum, or converges, when | r \ < 1. When | r \ ^ 1, the series is divergent, or does not have a sum, because in this case S n in (1) does not approach a limit as n > oo . Thus, for the G.P. (1, 2, 4, ) where r = 2, we find that S n increases beyond all bounds as n - oo . PROGRESSIONS 239 *EXERCISE 87 Find the sum of each of the followng infinite geometric series by use of the established formula. I- 7 + J + I + ' ' 2. 12 + 3 + J + - -. 3. 15 + 5 -f f + . . 4. 10 - 5 + f + 6. 1 - J + t 6. 1 - J + J . 7. 1 - .01 + .0001 . 8. .8 + .08 -f .008 H . Find a rational number equal to the given endless repeating decimal, where repeating parts are written three times. 9. .333 . 10. .444 . 11. .666 . 12. .0999 . 13. .8333 . 14. .1666 . 15. .212121 . 16. .050505 . 17. .030303 . 18. .838383 . 19. .454545 . 20. 4.222 .. 21. .2111 . 22. .345345345 . 23. .210210210 - . 24. 252.525.- -. 25. 16.7167167 - -. 26. 25.05050 -. 27. .153846153846153846 . 28. .076923076923076923 . 29. A pendulum is being brought to rest by air resistance. The path of each swing, after the first, of the pendulum bob is .98 as long as the path of the previous swing (from one side to the other). If the path of the first swing is 30 inches long, how- far does the bob travel in coming to a position of rest? 30. A rubber ball is dropped from a height of 100 inches. On each rebound the ball rises to f of the height from which it last fell. Find the distance traveled by the ball in coming to rest. 31. The side of a certain square is 10 inches long. A second square is drawn by connecting the mid-points of the sides of the 1st square; a 3d square is drawn by connecting the mid-points of the sides of the 2d square; etc., without end. Find the sum of the areas of all the squares. Note 1. If | r \ < 1, we know that S, of (5) on page 237, is approximately equal to S n if n is large, and our confidence in this approximation increases as n increases. When n is large, it is decidedly easier to compute S than /Sn, and hence it is convenient at tunes to use S in place of S n < 32. (I) Find Sn for the G.P. 3, f , f , . (II) Find the sum of this pro- gression extended to infinitely many terms. 33. Find &, approximately, for the G.P, 8 -f J + f H . CHAPTER 15 LOGARITHMS 164. Logarithms Logarithms are auxiliary numbers which are exponents, and which permit us to simplify the operations of multiplication, division, rais- ing to powers, and extraction of roots, applied to explicit real numbers. Previously, we have introduced exponents only under the assump- tion that they are rational numbers. In connection with logarithms, however, when we mention an exponent, it may be any real number, rational or irrational. A logical foundation for the use of irrational exponents is beyond the scope of this text. Hence, without dis- cussion, we shall assume the fact that irrational powers have meaning and that the laws of exponents hold if the exponents involved are real numbers, either rational or irrational, provided that the base is positive. ILLUSTRATION 1. The student may safely use his intuition in connection with the symbol 10^ = 10 1 - 414 ---. Closer and closer approximations to 10^ are obtained if the successive decimal approximations to v^ are used as exponents. 165. Logarithms to any base In the following definition, b represents any positive number, not 1, and N is any positive number. DEFINITION I. The logarithm of a number N to the base b is the exponent of the power to which b must be raised to obtain N. In other words, if 6* N then x is the logarithm of N to the base 6. To abbreviate "the logarithm of N to the base b," we write "log*, i/V." Then, by Definition I, the following equations state the same fact, the first equation in exponential form and the second in LOGARITHMS 247 logarithmic form: N = b* and x = log* N. , (1) ILLUSTRATION 1. If N - 4 5 , then 5 is the logarithm of N to the base 4. ILLUSTRATION 2. "Iog 2 64" is read "the logarithm of 64 to the base 2": since 64 = 2*, hence Iog 2 64 = 6. ILLUSTRATION 3. Since v^5 = 5$, hence Iog 5 \^5 = j = .333 . ILLUSTRATION 4. To find Iog 2 J-, we express J as a power of 2: since 5 = r = 2~ 3 , Aence Iog 2 = 3. o & o ILLUSTRATION 5. If logb 16 = 4, then b 4 = 16; 6 = v'Te = 2. ILLUSTRATION 6. If log 2 = J, then cri = 2. Hence, 1 For any base b, we have 6 = 1 and 6 1 = b. Hence, log & 1 = 0; log b 6 = 1. (2) Note 1. We do not use b = 1 as a base for logarithms because every power of 1 is 1 and hence no number except 1 could have a logarithm to the base 1. 'x In the definition of log& N, we stated that N was a positive number. That is, in this book, if we speak of the logarithm of a number N we shall mean a positive number N. Also, we stated that the base b is positive. These agreements were made to avoid meeting im- aginary numbers as logarithms. In advanced mathematics it is proved that, if N and b are positive, there exists just one real logarithm of N to the base b. EXERCISE 88 1. Since 27 = 3 s , what is the logarithm of 27 to the base 3? 2. Since 625 = 5 4 , what is the logarithm of 625 to the base 5? 3. Since J = 3" 1 , what is the logarithm of J to the base 3? 4. Since 1 = 6, what is the logarithm of 1 to the base 6? 242 LOGARITHMS 5. If the logarithm of N to the base 4 is 3, find N. 6. What number has 2 as its logarithm to the base 4? Find the number N whose logarithm is given. 7. log, N - 4. 8. log. N = 2. 9. logio N - 1. 10. logio N - 2. * 11. Iog 7 N = 0. 12. log* JV - - 1. 13. log w AT = - 1. 14. Iog 2 'AT - - 3. 16. Iog 4 AT = - 3. 16. Iog 10 N - - 4. 17. Iog 4 AT = J. 18. Iog 26 N = J. 19. logs AT - |. 20. logioo N - 2. 21. logioo AT = f . 22. Find the logarithm of 125 to the base 5. 23. Find the logarithm of 1,000,000 to the base 10. Find the specified logarithm. 24. logs 25. 25. log* 16. 26. Iog 2 8. 27. log? 49. 28. log, 27. 29. Iog 4 64. 30. logio 1000. 31. Iog 5 625. 32. logu 5. 33. Iog 9 3. 34. logioo 10. 35. log^ 3. 36. Iog 4 J. 37. loge J. 38. Iog 2 J. 39. log, &. kFind the value of the unknown letter in the problem. 40. Iog6 16 - 2. 41. log* 125 - 3. 42. log* 625 = 4. 43. Iog5 1000 = 3. 44. log, 9 - J. 46. Iog 4 = - 1. 46. loga 7 - - 1. 47. log* J - - 2. 48. log y = 2. 49. Iog6 2 - J. 60. logb -fs - 2. 61. Iog6 .0001 = - 2. 62. logj 16 = x. 63. log^ N == - 3. 64. logs N = - f . 166. Common logarithms Logarithms to the base 10 are called common logarithms and are the most useful variety for computational purposes. Hereafter, un- less otherwise stated, when we mention a logarithm we shall mean a common logarithm. For abbreviation, we shall write log N, instead 3! logw N, for the common logarithm of N and read log N as the logarithm of N. Then, from the definition of a logarithm, the follow- ing equations are equivalent: N = 10 and x = log N. (1) LOGARITHMS 243 ILLUSTRATION 1. Since 10,000 - 10 4 , hence log 10,000 = 4. Since ^Io = 10*, hence log ^Io .333 Since .01 = 1Q- 2 , hence log .01 - 2. We have seen that log N may be either positive, negative, or zero, depending on the value assigned to N. Also, we notice that log N is an integer when and only when N is an integral power of 10. ILLUSTRATION 2. For future reference, the student should verify the following logarithms by use of the definition of a logarithm. N = .0001 .001 .01 .1 i 10 100 1000 10,000 100,000 log AT- -4 -3 2 i 1 2 3 4 5 167. Some properties of logarithms The following properties hold if the logarithms are taken to any base 6, but we shall write proofs only for the case where b = 10. 1. The logarithm of a product equals the sum of the logarithms of its factors. For instance, log MN = log M + log N. (1) ILLUSTRATION 1. If M = 10 3 and N = 10 8 , then MN = 1010 = 10 8 . Also, by the definition of a logarithm, log M = 3, log N = 5, and log MN = 8. Hence, log MN log M + log N in this special case. Proof of Property I. 1. Let log M = x and log N y. Then, M = 10*, N - 10", and MN - 10*10* = 10*+. 2. Since MN = 10*+", hence log MN - x + y - log Af -f log II. !T/ie logarithm of a quotient equals the logarithm of the dividend minus the logarithm of the divisor. That is, log 7 = log M - log N. (2) Proof. 1. Let log M x and log N * y. Then, (By a law of exponents) 2. Hence, log -^ =* x - y - log Af - log AT. 244 LOGARITHMS ILLUSTRATION 2. If we are given log 3 .4771, then log 300 = log 3(100) - log 3 + log 100 = .4771 + 2 - 2.4771; log .003 = log Ttffof = log 3 - log 1000 = .4771 - 3 - 2.5229. EXERCISE 89 Express each number as a product or quotient, and find its logarithm by use of the given logarithms and the logarithms of integral powers of 10. log 2 = .3010; log 3 - .4771; log 7 = .8451; log 17 = 1.2304. 1. 6. 2. 21. 3. 34. 4. 51. 5. 30. 6. 70. 7. 1700. 8. 2000. 9. 42. 10. | . 11. J. 12. Jf 13. 14. f 16. A. 1& rbs- 17. A- 18. .17. 19. .007. 20. .0003. 21. .017. 22. .0119. 23. .042. 24. 10.2. 168. Characteristic and mantissa Every number, and hence every logarithm, can be written as the sum of an integer and a decimal fraction which is positive or zero and less than 1. When log TV is written in this way, we call the integer the characteristic and the fraction the mantissa of log N. log N = (an integer) -f (a fraction, ^ 0, < 1) ; log N = characteristic + mantissa. (1) ILLUSTRATION 1. If log N = 4.6832 = 4 -f- .6832, then .6832 is the man- tissa and 4 is the characteristic of log N. ILLUSTRATION 2. If log N = 3.75, then log N lies between 4 and 3. Hence, log N 4 -f- (a fraction). To find the fraction, subtract: 4 - 3.75 = .25. Hence, log TV = - 3.75 = - 4 + .25. ILLUSTRATION 3. The following logarithms were obtained by later meth- ods. The student should verify the three columns at the right. LOGARITHM CHARACTERISTIC MANTISSA log 300 = 2.4771 = 2 + .4771 2 .4771 log 50 - 1.6990 = 1 -f .6990 1 .6990 log .001 - - 3 = - 3 -f .0000 -3 .0000 log 6.5 0.8129 = + .8129 .8129 log .0385 - 1.4145 = - 2 + -5855 -2 . .5855 log .005 = - 2.3010 = - 3 + .6990 Q .6990 LOGARITHMS 245 1 69. Properties of the characteristic and mantissa ILLUSTRATION 1. All numbers whose logarithms are given below have the same significant digits (3, 8, 0, 4). To obtain the logarithms, log 3.804 was first found from a table to be discussed later; the other logarithms were then obtained by the use of Properties I and II. log 380.4 = log 100(3.804) = log 100 + log 3.804 - 2 + .5802; log 38.04' = log 10(3.804) - log 10 + log 3.804 - 1 + .5802; log 3.804 =.5802 - + .5802; Q 804 log .3804 =log~ = log 3.804 - log 10 = - 1 + .5802; log .03804 = log - = log 3.804 - log 100 = - 2 + .5802. Similarly, if N is any number whose significant digits are (3, 8, 0, 4), then N equals 3.804 multiplied, or else divided, by a positive integral power of 10; hence, it follows as before that .5802 is the mantissa of log N. In Illustration 1, the characteristic of log 380.4 is 2, of log 38.04 is 1, etc. These facts could have been learned as follows. ILLUSTRATION 2. To find the characteristic of log 380.4, notice the two successive integral powers of 10 between which 380.4 lies: 100 < 380.4 < 1000. Hence, - log 100 < log 380.4 < log 1000; or, 2 < log 380.4 < 3. Therefore, log 380.4 = 2 + (a fraction, > 0, < 1); or, by definition, the characteristic of log 380.4 is 2. \ In Illustration 1 we met special cases of the following theorems. THEOREM I. The mantissa of log N depends only on the sequence of significant digits in N. That is, if two numbers differ only in the position of the decimal point, their logarithms have the same mantissa. THEOREM II. When N > 1, the characteristic of log N is an integer , positive or zero, which is one less than the number of digits in N to the left of the decimal point. THEOREM III. If N < 1, the characteristic of log N is a negative integer; if the first significant digit of N is in the fcth decimal pkce, then 'k is the characteristic of log N. 246 LOGARITHMS ILLUSTRATION 3. By use of Theorems II and III, we find the character- istic of log N by merely inspecting N. Thus, by Theorem III, the char- acteristic of log .00039 is 4 because "3" is in the 4th decimal place. By Theorem II, the characteristic of log 1578.6 is 3. 1 70. Standard form for a negative logarithm Hereafter, for convenience in computation, if the characteristic of log N is negative, - ft, change it to the equivalent value [(10 - k) - 10], or [(20 - *) - 20], etc. ILLUSTRATION 1. Given that log .000843 = 4 + .9258, we write log .000843 - - 4 + .9258 = (6 - 10) + .9258 = 6.9258 - 10. The characteristics of the following logarithms are obtained by use of Theorem III; the mantissas are identical, by Theorem I. IST SIGNIP. DIGIT IN ILLUSTRATION Loo AT STANDARD FORM 1st decimal place 2d decimal place 6th decimal place N = .843 N - .0843 N = .00000843 - 1 + .9258 = 9.9258 - 10 - 2 + .9258 - 8.9258 - 10 - 6 + .9258 = 4.9258 - 10 1 71 . Four-place table of logarithms Mantissas can be computed by use of advanced mathematics, and, except in special cases, are endless decimal fractions. Computed mantissas are found in tables of logarithms, also called tables of man- tissas. ILLUSTRATION 1. The mantissa for log 10705 is .029586671630457, to fifteen decimal places. Table II gives the mantissa of log N correct to four decimal places if N has at most three. significant digits; a decimal point is under- stood in front of each mantissa in the table. If N lies between 1 and 10, the characteristic of log N is zero so that log N is the same as its mantissa. Hence, a four-place table of mantissas is also a table of the actual logarithms of all numbers with at most three sig- nificant digits, from N , 1.00 to N 10.00. In case N is less than 1 or greater than 10, we must supply the characteristic of log N by use of Theorems II and III besides obtaining the mantissa of log N by use of Table II. LOGARITHMS 247 EXAMPLE 1. Find log .0316 from Table II. SOLUTION. 1. To obtain the mantissa: find "31" in the column headed N in the table; in the row for "31," read the entry in the column headed "6." The mantissa is .4997. 2. By Theorem III, the characteristic of log .0316 is - 2, or (8 - 10): log .0316 = - 2 + .4997 - 8.4997 - 10. ILLUSTRATION 2. From Table II and Theorem II, log 31,600 4.4997. EXAMPLE 2. Find N if log N = 7.6064 - 10. SOLUTION. 1. To find the significant digits of N: the mantissa of log N is .6064; this is found in Table II as the mantissa for the digits "404." 2. To locate the decimal point in N: the characteristic of log N is (7 10) or - 3; hence, by Theorem III, N = .00404. ILLUSTRATION 3. If log N 3.6064, the characteristic is 3 and, by Theorem II, N has 4 figures to the left of the decimal point: the mantissa is the same as in Example 2. Hence, N 4040. DEFINITION I. A number N is catted the antilogarithm of L in case log N ~ L, and for abbreviation we write N = antilog L. ILLUSTRATION 4. Since log 1000 = 3, hence 1000 = antilog 3. ILLUSTRATION 5. In Example 2 we found antilog (7.6064 10). EXERCISE 90 In Probkms 1 to 8, each number is the logarithm of some number N. State the characteristic and the mantissa of log N. 1. 2.9356. 2. 15.2162. 3. - 1.300. 4. - 2 + .3561. 6. 3.5473. 6. 7.2356 - 10. 7. - 5.675. 8. 5.1942 - 10. Write the fottowng negative logarithms in standard form. 9. - 1 + .2562. 10. .3267 - 3. 11. .4932 - 6. 12. - 3.4675. State the characteristic of the logarithm of each number. 13. 637,500. 14. 368. 16. .000673. 16. .00897. 17. .000007. Use Table II to find the four-place logarithm of the number. 18. 65.4. 19. 43.2. 20. 178. 21. .00785. 22. .0346. 23. 9.46. 24. 6530. 25. 17,800. 26. .00005. 27. .086. 28. .000358. 29. 101,000. 30. .00089. 31. 157,000. 32. .0000002. 248 LOGARIJHMS Find the antilogarithm of the given logarithm by use of Table II. 83. 2.3856. 34. 3.3927. 36. 3.6684. 36. 1.8785. 37. 0.1553. 38. 2.1461. 39. 1.8692. 40. 0.9727. 41. 2.4800. 42. 0.5611. 43. 7.7701 - 10. 44. Q.8041 - 10. 46. 8.9823 - 10. 46. 4.8915 - 10. 47. 4.9542 - 10. 48. 2.9340 - 10. 49. 9.4216 - 10. 60. 8.7284 - 10. 61. - 2.3010. 1 72. Interpolation for a mantissa Interpolation in a table of mantissas is based on the assumption that, for small changes in N, the corresponding changes in log N are proportional to the changes in N. This principle of proportional parts is merely an approximation to the truth but leads to results which are sufficiently accurate for our purposes. We agree that, whenever a mantissa is found by interpolation from a table, we shall express the result only to the number of decimal places given in table entries. Also, in finding N by interpolation in a table of mantissas when log N is given, we agree to specify just four or just five significant digits according as we are using a four-place or a five-place table. No greater refinement in the result is justified because the unavoidable error, which may arise, frequently will be as large as 1 unit in the last significant digit which we have agreed to specify, although the error is rarely larger. t EXAMPLE 1. Find log 13.86 by interpolation in Table II. SOLUTION. 1. We notice that 13.80 < 13.86 < 13.90. Hence, by the principle of proportional parts, we assume that, since 13.86 is 3% of the way from 13.80 to 13.90, log 13.86 & & of the way from log 13.80 to log 13.90, or log 13.86 log 13.80 + .6(log 13.90 - log 13.80). 2. Each logarithm below has 1 for its characteristic, by Theorem II. From table: log 13.80 = 1.1399" log 13.86 = ? From table: log 13.90 = 1.1430J Tabular difference is 31 .1430 - .1399 = .0031. .6(.0031) = .00186, or .0019. log 13.86 = 1.1399 + .6(.0031) = 1.1399 + .0019 = 1.1418. Comment. We found .6(31) = 18.6 by use of the table headed " 31 " under the column of proportional parts in Table II. LOGARITHMS 249 ILLUSTRATION 1. To find log .002914: 10 ~ T2910: mantissa is -46391 ' L2914: mantissa is ? J _ 2920: mantissa is .4654 _ Tabular difference is 15 .4654 - .4639 - .0015. x - .4(15) = 6. Hence, the mantissa for 2914 is .4639 4- .0006 = .4645. Hence, by Theorem III, log .002914 = - 3 + .4645 7.4645 - 10. EXAMPLE 2. Find N if log N = 1.6187. SOLUTION. 1. The mantissa .6187 is not in Table II but lies between the consecutive entries .6180 and .6191, the mantissas for 415 and 416. 2. Since .6187 is -fa of the way from .6180 to .6191, we assume that N is of the way from 41.50 to 41.60. 11 [1.6180 = log 41.501 Ll.6187 = log N J* . ^1.6191 = log 41.60 J 41.60 - 41.50 = .10 .10 x = A(-IO) = .064, or approximately .06. N = 41.50 + -&C10) = 41.50 + -06 - 41.56. ILLUSTRATION 2. To find N if log N = 6.1053 - 10: 34 15 r.1038, mantissa for 1270H L.1053, mantissa for ? J .1072, mantissa for 1280 J 10 = .4. Hence, x = .4(10) = 4. 1270 + 4 = 1274. Hence, .1053 is the mantissa for 1274 and N = .0001274. Comment. We obtain Jj = .4 by inspection of the tenths of 34 in the columns of proportional parts. We read 13.6 = .4(34) or ^ = .4, and ^ = .5. Since 15 is nearer to 13.6 than to 17, hence J| is nearer to .4 than to .5. Note 1 . When interpolating in a table of mantissas, if there is equal reason for choosing either of two successive digits, for uniformity we agree to make that choice which gives an even digit in the last significant place of the final result of the interpolation. 4 1 73. Scientific notation for a number Any positive number N can be written in the form . (1) 250 LOGARITHMS where P is a number greater than or equal to 1 but less than 10, and fc is an integer, either zero or positive or negative. We refer to the right-hand side of (1) as the scientific notation for N. ILLUSTRATION 1. 5,832,900 = 5.8329(10*). .00000058329 = 5.8329(.0000001) - 5.8329(10~ 7 ). The scientific notation gives a brief and easily appreciated form for writing very large or very small numbers. ILLUSTRATION 2. The nucleus of an atom has a diameter which is esti- mated as less than 3(10~ 12 ) centimeters. The mean distance from the sun to the outermost planet Pluto is approximately 3.67 X 10 9 miles. One light-year, the distance which light will travel in one year hi interstellar space, is approximately 6 X 10 12 miles. In equation 1, N and P have the same significant digits because the factor 10* merely alters the position of the decimal point to change P into N. Hence, the scientific notation is very useful hi writing a number N, particularly if it is very large, when we wish to show how many digits in N are significant. This feature was referred to in Section 48, page 53, for the case where k of (1) is positive. ILLUSTRATION 3. If 68,820,000 is the approximate value of some quantity and if just five digits are significant, this is not indicated by the usual form of the number. We write it as 6.8820(10 7 ) to show that one of the zeros is significant. ILLUSTRATION 4. If N = 1.352(10*), then, by Property I, page 243, log N = log 1.352 + log 10 8 = 0.1309 + 8 = 8.1309. Thus, 8 is the characteristic and log 1.352 is the mantissa of log N. \ Consider any number N, where N = P(10*) where k is an integer and 1 ^ P < 10. Then k is the characteristic and log P is the mantissa of log N, because log N - log P-f log 10* - log P + k, where log P < 1, since 1 ^ P < 10. ILLUSTRATION 5. If log N = 9.7419, and if we use the form N => P(10*), we have k = 9 and log P = .7419: P - 5.520 and 2V * 5.520(10*). (Four digits significant.) LOGARITHMS 251 Note 1. Besides common logarithms, the only other variety used ap- preciably is the system of natural, or Naperian logarithms, for which the base is a certain irrational number denoted by e where e = 2.71828 Natural logarithms are useful for theoretical purposes. Note 2. logarithms were invented by a Scotchman, JOHN NAPIER, Baron of Merchiston (1550-1617). His original logarithms were not the same as those now called Naperian logarithms, in his honor. Common logarithms, also called Briggs logarithms,' were invented by an Englishman, HENRY BRIQQS (1556-1631), who was aided by Napier. EXERCISE 91 ^i Find the four-place logarithm of each number from Table II. 1. 1826. 2. 25.63. 3. 532.2. 4. 12.67. 5. 35.94. 6. 1.293. 7. .3013. 8. .4213. 9. .5627. 10. .03147. 11. .01563. 12. .001139. 13. 90,090. 14. 203,500. 15. .001439. 16. .05626. 17. 1.233(10- 4 ). 18. 1.417(10'). 19. 3.126(10 3 ). 20. 2.438(10-*). Find the antilogarithm of each four-place logarithm from Table II. 21. 3.2367. 22. 7.1247 - 10. 23. 6.1640. 24. 8.9935 - 10. 26. 3.1395. 26. 2.9276. 27. 1.6016. 28. 0.4906. 29. 6.3350 - 10. 30. 4.1436 - 10. 31. 9.6715 - 10. 32. 8.0255 - 10. 33.- 8.8862. 34. 2.1952. 35. 0.0130. 36. 5.5511. 37. 5.9885 - 10. 38. 8.3358 - 20. 39. 9.6270 - 10. 40. 6.4228. 1 74. Computation of products and quotients Unless otherwise specified, we shall assume that the data of any given problem are exact. Under this assumption, the accuracy of a product, quotient, or power computed by use of logarithms depends on the number of places in the table being used. The result is fre- quently subject to an unavoidable error which usually is at most a few units in the last significant place given by interpolation. Hence, usually we must compute with at least five-place logarithms to obtain four-place .accuracy, and with at least four-place logarithms to obtain three-ylace accuracy. As a general custom, in any result we shall give all digits obtainable by interpolation in the specified table. 252 LOGARITHMS EXAMPLE 1. Compute .0631(7.208) (.5127) by use of Table II. SOLUTION. Let P represent the product. By Property I, we obtain log P by adding the logarithms of the factors. We obtain the logarithms of the factors from Table II, add to obtain log P, and then finally obtain P from Tkble II. The computing form, given in blackface type, was made up completely as the first step in the solution. log .0631 = 8.8000 - 10 (Table II) log 7.208= 0.8578 (Table II) log .6127 = 9.7099 - 10 (Table 'II) (add) log P = 19.3677 - 20 - 9.3677 - 10. Hence, P - .2332. [ = antilog (9.3677 - 10), Table II] 431.91 r> or.* EXAMPLE 2. Compute q , SOLUTION. 1. By Property II, log q equals the logarithm of the numer- ator minus the logarithm of the denominator. 2. Before computing, we round off each given number to four significant digits because we are using a four-place table. For instance, 15.6873 be- comes 15.69. log 431.9 - 2.6354 (Table II) _(-) log 15.69 = 1.1956 (Table II) log q = 1.4398. Hence, q = 27.53. (Table II) 257 EXAMPLE 3. Compute q = Sn^s' SOLUTION. We employ Property II. log 267 = 2.4099 = 12.4099 - 10 (-) log 8966 - 3.9521 = 3.9521 log q = ??? = 8.4578 - 10. Hence, q = .02869. Comment. When we first tried to subtract log 8956 from log 257, we saw that the result would be negative because log 8956 is greater than log 257. In order that log q should appear immediately in the standard form for a negative logarithm, we changed log 257 by adding 10 and then subtracting 10 to compensate for the first change. Actually, log q = 2.4099 - 3.9521 - - 1.5422 = 8.4578 - 10. Whenever it is necessary to subtract a larger logarithm from a smaller one in computing a quotient, add 10 to the characteristic of the smaller logarithm and then subtract 10 to compensate for the change. LOGARITHMS 253 A * * (4.803) (269.9) (1.636) EXAMPLE^ Compute g= v (7880)(253.6) ' INCOMPLETE SOLUTION. First make a computing form. (log 4.803 - . . flog 7880 = (+) log 269.9 = *" N \log 263.6 = [log 1.636 = log denom. = log numer. = ( ) log denom. = log q = Hence, q = EXAMPLE 5. Compute the reciprocal of 189 by use of Table II. 1 SOLUTION. Let R = 189 log 1 = 0.0000 = 10.0000 - 10 (-) log 189 - 2.2765 = 2.2765 log R = ? = 7.7235 - 10. Hence, R = .005290. Comment. In writing any approximate value, it is essential to indicate all final zeros which are significant. Hence, in writing R = .005290 in Example 5, the final zero was essential. It would have been wrong to write R .00529, because this would not show that we had reasonably ac- curate information concerning the next digit, zero. Note 1. Before finding the four-pl&ce log NUN has more than four significant digits, round off AT to four significant digits. EXERCISE 92 Compute by use of four-place logarithms. 1. 31.57 X .789. 2. .8475 X .0937. 3. 925.618 X .000217. 4. 925.6 X .137. 6. .0179 X .35641. 6. 3.41379 X .0142. 7. (- 84.75) (.00368) (.02458). 8. (- 16.8) (136.943) (.00038). HINT. Only positive numbers have real logarithms. First compute as if all factors were positive; then determine the sign by inspection. 675 728.72 .0894 ** 13.21 895 .6358 325.932 568.5 1A 753.166 1K .0421 1 lv ^rfc ^ ^ * JL4 /\rti*n o * JLO 23.14 9273.8 .53908 w 100,935 254 LOGARITHMS 16.083 X 256 9.32X531 17 JL f 47 + .0158 .8319 X .5685 .53819 X .0673 .42173 X .217 A 5.4171 X .429 ' .3852 X .956 18.1167 X 37 .00073 X .965 (- .29)(.038)(- .0065) n . (5.6) (- 3.9078) (- .00031) (- 1006.332) (2.71) (132) (- 1.93) Compute the reciprocal of the number. 25. 63283. 26. .00382. 27. .02567. 28. .0683(.52831). 29. (a) Compute 652(735); (6) compute (log 652) (log 735). 80. (a) Compute .351 * 625; (6) compute (log .351) -5- (log 625). *175. Cologarithms * The logarithm of the reciprocal of N is called the cologarithm of N and is written colog N. Since log 1=0, colog N = log - = - log N. (1) T i n i noi i ! log 1 = 10.0000 - 10 ILLUSTRATION 1. Colog .031 - log -ggjl ( , )log .Q* 31= 8.4914-10 colog .031 = 1.5086. The positive part of colog N can be quickly obtained by inspec- tion of log N: subtract each digit (except the last) in the positive part of log N from 9, and subtract the last digit from 10. ,, , n * 16.083 X 256 , r . . ., EXAMPLE 1. Compute q = 47 v AICQ "7 use ^ coioganthms. SOLUTION. To divide by N is the same as to multiply by 1/N. Hence, instead of subtracting the logarithm of each factor of the denominator, we add the cologariihm of the factor: 16.083 X 256 _ , . / 1 \ / 1 " (16 ' 083 X 47 X .0158 log 16.08 = 1.2063 log 266 = . 2.4082 log 47 - 1.6721; hence, colog 47 - 8.3279-10 log .0158 - 8.1987 - 10; hence, colog .0168 - 1.8013 q = 5542. < - (add) log q - 13.7437 - 10 - 3.7437. * The instructor may wish to direct occasional use of cologarithms. LOGARITHMS 255 1 76. Computation of powers and roots We establish the following property of logarithms as an aid to computing powers. III. The logarithm of the kth power of a number N equals k times the logarithm of N: log N* = k log N. (1) Proof. Let x log N. Then, by the definition of a logarithm, N - 10*. Hence, N k - (10*)* - 10**. (A law of exponents) i Therefore, by the definition of a logarithm, log N k kx - k log N. (Using x = log N) ILLUSTRATION 1. Log 7 6 = 5 log 7. Log N* = 3 log N. EXAMPLE 1. Compute (.3156) 4 . SOLUTION. By Property III, ' log (.3156)* = 4 log (.3156) - 4(9.4991 - 10). log (.3166)* - 37.9964 - 40 = 7.9964 - 10. Therefore, (.3156) 4 = .009918. Recall that any root of a number is expressible as a fractional power. Hence, as a special case of Property III we obtain IV. log Proof. Since 3/N = W*, we use Property III with k T- log y/Jf log JV - T log N. ILLUSTRATION 2. Since VJ? = Ni and log VN = ilog N\ log */N = \ log N. 2 3 EXAMPLE 2. Compute ^.08351. SOLUTION. By Property IV, log 4/N * J log AT. Hence, 256 LOGARITHMS log v^.08361 = 58.9218 - 60 6 8.9218 - 10 _ * 6 = 9.8203 - 10. (2) Therefore, \/. 08361 .6611. Comment. Before dividing a negative logarithm by a positive integer, usually it is best to write the logarithm in such a way that the negative part after division, will be 10. Thus, in (2), we altered (8.9218 10) by subtracting 50 from 10 to make it 60, and by adding 50 to 8.9218 to compensate for the subtraction; the result after division by 6 is in the standard form for a negative logarithm. T7 on*/ 05831) 8 \* EXAMPLE 3. Compute q = [ ~= ] \65.3VT46/ SOLUTION. 1. Let F represent the fraction. Then log q = log F. 2. Notice that log (.5831) 3 = 3 log .5831; log Vl46 - i log 146. log .6831 = 9.7658 - 10 log 146 = 2.1644 3 log .6831 = 9.2974 - 10 \ ( -) log denom. = 2.8971 j log F = 6.4003 - 10; 21ogF - 2.8006 - 10 = 42.8006 - 50. , 21ogF 42.8006 - 50 c , Am log q = jp = g = 8.5601 10. Hence, q = .03632. / log 66.3 = 1.8149 1 j log 146 - 1.0822 log denom. = 2.8971. EXERCISE 93 Compute by use of four-place logarithms. 1. (17.5) 8 . 2. (3.1279) 4 . 3. (.837)*. 6. VT09. 9. (1.04) 7 . 13. (700,928)*. 17. (- 1.03)*. 21. (143.54)*. 6. -v/2795. 10. (10,000)*. 18. (- 1796)*. 22. 4. (.0315) 8 . 8. 12. (.0138273)*. 16. 20. (- .00831) 8 . 23. (- .0057)*. 24. (157)" 8 . HINT for Problem 24. Recall that (1$7)~ 8 - 1/(157 8 ). 26. (13.67)*. 26. (3.035)-*. 27. (.98)-*. 28. (.831447)- 6 . LOGARITHMS 257 29. (1.03)-. 30. (1.05)-. 31. (1.04) 100 . 32. (1.04)-. 1. Given the seven-place log 1.04 = 0.0170333, compute Problems 31 and 32 and compare with the less accurate former answers. Compute by use of four-place logarithms. 33. .958(12.167) 2 . 34. 10'- 66 V265. 35. 10 2 W1J8147. 36. 37. 5 3 * 4 ; 31 !.- 38. (25 ;I^ 2 >'. 39. - 0198 21.4V521.923 * 1893.32 "" (3.82616) 40 758>32 41 * / 89-1 A0 / 47.5317 ' (46.3) 3 " \163 X .62* ^ \.031 X .964* 10-*V?78 10-i"v387 .. ^- 463.19 (.983174)' ** (57)(8.64)' ' 46 /(- 316)(.198) /54.2VT89\ * \ .756392 ' 47 ' V .157386 ) 5731.84 ATofe ^. Observe that no property of logarithms is available to simplify the computation of a sum. Use logarithms below wherever possible. 49. . + 89.532 ^45 - 364.1 R1 V57 + 2.513 ' U ' (.9873) 2 + 16.3* 5L "453 X .110173 62 - + 1 B3 log 86 log 567 - 20 " (1.03)* + 1 log 53.8* * log 235 66. (2.67) 1 -* 2 . 66. (53.17)- 4 . 67. (60.2)--. 58. (.065) - M2 . HINT for Problem 57. - .43 log 59.2 = - .7621 = 9.2379 - 10. 69. Compute (a) (antilog 2.6731) 2 ; (ft) [antilog (- 1.4973)] 2 . DEFINITION I. The geometric mean of n numbers is defined as the nth root of the product of the numbers. Thus, the geometric mean of M, N, P, Q, and R is ^MNPQR. In each problem, find the geometric mean of the given numbers. 60. 138; 395; 426; 537; 612. 61. .00138; .19276; .08356; .0131. // a, 6, and c are the three sides of a triangle t it is proved in trigonometry that A, the area of the triangle, is given by A = VS(S - a)(8 - b)(8 - c), where S = J(a + 6 + c). Find the area of a triangle whose sides are as follows. 62, 375.40; 141.37; 451.20. 63. .089312; .0739168; .024853. 258 LOGARITHMS The time t in seconds for one oscillation of a simple pendulum whose length is I centimeters, is given byt wv-> where g = 980 and v = 3.1416. 64. (a) Find the time for one oscillation of a simple pendulum .985 centi- meters long. (6) Find I if the time for one oscillation is 3.75 seconds. 65. Let d be the diameter in inches of a short solid circular steel shaft which is designed to transmit safely H horse power when revolving at R revolutions per minute. A safe value for d is Find the number of horse power which can be safely transmitted at 1150 revolutions per minute if d - 1.9834. 66. The weight w, in pounds of steam per second, which will flow through a hole whose cross-section area is A square inches, if the steam approaches the hole under a pressure of P pounds per square inch, is approximately w .0165AP-* 7 . How much steam at a pressure of 83.85 pounds per square inch will flow through a hole 12.369 inches in diameter? *1 77. Exponential and logarithmic equations A logarithmic equation is one in which there appears the logarithm of some expression involving the unknown quantity. EXAMPLE 1. Solve for x: log x + log -^- 6. SOLUTION. By use of Properties I and II of page 243, log x + log 2 -f log x log 5 = 6. 2 log x - 6 + log 5 - log 2 - 6.3980. (Table II) log x - 3.1990; x - antilog 3.1990 - 1581. (Table II) An equation where the unknown quantity appears in an exponent is called an exponential equation. Sometimes, an exponential equa- tion can be solved by taking the logarithm of each member. EXAMPLE 2. Solve 16* * 74. SOLUTION. Equate the logarithms of the two sides: x log 16 = log 74; log 1.869 - 0.2716 log 74 1.8692 (-) log 1.204 0.0806 * * log 16 " 1.2041* log x - 0.1910; hence x - 1.552, LOGARITHMS 259 *1 78. Logarithms to various bases The base 10 is convenient for a system of logarithms when they are being used to simplify computation. The only base other than 10 which is used appreciably is the irrational number e = 2.71828- , which is fully as important a constant hi mathematics as the familiar number IT. Logarithms to the base e are called natural, or Naperian, logarithms. Natural logarithms have many advantages over com- mon logarithms for advanced theoretical purposes. Recall that the equations N = 6* and x = log& N are equivalent. Hence, if N and b are given, we can find log$ N by solving the exponen- tial equation N b* by use of common logarithms. In particular, the natural logarithm of N can be found by solving N e* for x. EXAMPLE 1. Find log, 35. SOLUTION. Let x = log, 35; then, 35 = e*. On taking the common logarithm of both sides we obtain x logio e = logio 35. logio 35 1.5441. log 1.544 = 10.1886 - 10 X logio e 0.4343' (-) log .4343 - 9.6378 - 10 * x . 3.555 = log. 35. log * - 0.5508. THEOREM I. If a and b are any two bases, then log N = (toga &) (log* N). (1) Proof. Let y = lo& N; then N - If. (2) Hence, logo N = logo If - y logo b (logo 6)(log& N). The number log a b is called the modulus of the system of base a with respect to the system of base b. Given a table of logarithms to the base 6, we could form a table of logarithms to the base a by multiplying each entry of the given table by logo 6. *EXERCISE 94 Solve for x or for n, or compute the specified logarithm. 1. 12' - 28. 2. 51* - 569. 3. &* - 28(2'). 4. 15 s * - 85(3*). 6. .67* = 8. 6. .093* - 12. 7. (1.03)-" - .587. 8. (1.04)" - 1.562. 9. (1 ' 05 iT e "" 1 - 6.3282. 10. log W + log - - 5.673. .UO * 11. log. 75. 12. log. 1360. 13. log, 10. 14. logu 33. 15. log.s 23.8. 260 LOGARITHMS 16. Find the natural logarithm of (a) 4368.1; (6) 4.3681. (Notice that the results do not differ by an integer, so that Theorem I of page 245 does not hold for natural logarithms.) *1 79. Graphs of logarithmic and exponential functions We recall that y = log a x and x = a v are equivalent relations. We call logo x a logarithmic func- tion of x and a v an exponential function of y. ILLUSTRATION 1. In Figure 18, we have the graph of y = log, x. For any base a > 1, the graph of logo x would be similar. This graph assists us hi remembering the following facts. I. ' If x is negative, Iog x is not defined. II. If < x < 1, loga x is negative, and log a 1 = 0. III. // x increases without limit, log a x increases without limit; if x approaches zero, log a x decreases without limit. Since y = log a x is equivalent to x = a v , these equations have the same graph. Thus, in Figure 18 we have a graph of x *= e v . *EXERCISE 95 1. Graph y = logio x for < x f 30. From the graph, read the value of .3. 10 -.e. io.s. 2. Graph y = 2* from x = 6 to x = 4. From the graph, read log z 6. *1 80. Applications to compound interest On page 235 we saw that, if $P is invested at the interest rate i com- pounded annually, the amount $A on hand at the end of n years is given by the formula A = P(l + i) n - ILLUSTRATION 1. If $1000 is invested for 20 years at 5% compounded annually, the amount at the end of the time is A = 1000(1.05)*> = 1000(2.653) - $2653. (From Table III) LOGARITHMS 261 For unusual values of t, it is impossible to employ an interest table as in Illustration 1, but then logarithms can be used. EXAMPLE 1. Compute A = 2000(1. 036) 26 . SOLUTION. log 1.036 = 0.0153; 25 log 1.036 = 0.3825 log 2000 = 3.3010 (+) Hence, A = $4826. log A = 3.6835. We claim only three significant digits in the result because accuracy is lost in the multiplication 25 log 1.036. The 4th digit in 4826 is unreliable. EXAMPLE 2. Solve for the rate i: 2500 = 2000(1 + i) 8 . SOLUTION. 1. (1 + t) = $$& or (1 + i) 8 = 1-250. 2. Hence, 1 4- i = v' 1.250; we compute this root by use of Table II. log 1.250 = 0.0969; J log 1.250 = 0.0121; hence, v / L250 = 1.028. 3. Therefore, 1 + i = 1.028; or, i = 1.028 - 1 = .028 = 2.8%. ^EXERCISE 96 By use of Table III, find the compound amount at the end of the time if the money is invested at the specified rate compounded annually. 1. $2500; at 4%, for 16 years. 2. $1200; at 6%, for 13 years. 3. $1600; at 3%, for 35 years. 4. $400; at 5%, for 42 years. From A = P(l + i) n , we obtain P = A(l + i)-*. Use this result and Table IV to solve Problems 5 and 6. 5. What principal should be invested now at 5% compounded annually to create $2500 as the amount at the end of 15 years? 6. What principal should be invested now at 6% compounded annually to create $1000 as the amount at the end of 26 years? 7. At what interest rate compounded annually will a $2000 principal grow to the amount $3500 at the end of 10 years? 8. At what interest rate compounded annually will a $3000 principal double itself by the end of 15 years? 9. How long will it take $300 to grow to the amount $750 if invested at 5% compounded annually? (Recall Section 177.) 10. How long does it take money to double if invested at 6% compounded annually? 11. How long does it take money to double if invested at 6% simple interest? Compare with the result of Problem 10. CHAPTER 16 SYSTEMS INVOLVING QUADRATICS 1 81 . Graph of a quadratic equation in two variables A solution of an equation in two variables x and y is a pair of values of the variables which satisfies the equation. The graph or locus of the equation is the set of all points whose coordinates, (z, y), form real-valued solutions of the equation. EXAMPLE 1. Graph: z 2 -f- j/ 2 = 25. SOLUTION. 1. When x = 0, y 2 = 25; y = 5. Two solutions of the equation are (0, 5) and (0, 5). 2. When y - 0, x 2 = 25; x - 5. Two solutions of the equation are (5, 0) and (- 5, 0). 3. We plot the four points just found, with the same unit on OX and OY in Figure 19, and verify an advance inference that the graph is a circle whose center is the origin and radius is 5. Comment. Let P, with the coordinates (x, y), be any point in the coordinate plane with origin at 0, in a system where the same Fig. 19 unit is used in measuring all lengths. Then x* + y* - (OP) 2 . Hence, if x 2 + j/ 2 = 25, the point P must lie on a circle about as center with (OP) 2 = 25, or with radius 5. EXAMPLE 2. Graph: 9x* 4y* * 36. SOLUTION. 1. Solve for x in terms of y: 9z 2 ~ 36 + (1) SYSTEMS INVOLVING QUADRATICS 263 - x - g 9 x = 9+ 2 . (2) 2. We assign values to y and compute the corresponding values of x. Thus, if y = 0, then z = =t \/9 = =t 2. If y = 3, then * = =fc vl8 - 2V2 = 2.8. We tabulate the corresponding values of x and y in the following table. (a) y = -6 -3 3 6 x - V9 + y 2 x 4.5 2.8 2 2.8 4.5 (6) y = -6 -3 3 6 x = - V9 + y 2 X = -4.5 -2.8 -2 -2.8 -4.5 We plot the points given by the pairs of values of x and y in the table. In Figure 20, the points listed for (a) in the table give the open curve FDE; the points for (6) give HBO. These two open curves, together, are called a hyperbola, and it is the graph of equation 1. Each piece of the hyperbola is called a branch of it. Comment. The equation 9x 2 4y 2 = 36 defines z as a tow-valued function of y, as shown in (2), or y as a tow-valued function of x. The graph of the equation consists of the graphs of the two single-valued irra- tional functions x = and x The graph of the first of these is the branch FDE and the graph of the second is HBO. The two branches together make up the graph of equation 1. The branches are symmetr- Fig. 20 rical with respect to the y-axis. Note 1. To every hyperbola there correspond two characteristic lines, called asymptotes, which are indicated by dotted lines in Figure 20. As we recede out on any branch of the hyperbola, the curve approaches the corresponding asymptote but never reaches it. By moving far enough out on the branch, we may approach the asymptote as closely as we please. It is proved in analytic geometry that the equations of the asymptotes for equa- tion 1 are obtainable as follows: . 264 SYSTEMS INVOLVING QUADRATICS 1. Replace the constant term in the equation by 0, and factor the left member: 9x 2 - 4t/ 2 = 0; (3x - 2y)(3x + 2y) = 0. 2. Equate each factor separately to zero: 3x - 2y = and 3x + 2y = 0. These are the equations of the asymptotes. EXAMPLE 3. Graph: x 2 -f 4y 2 = 25. (3) SOLUTION. 1. Solve for y: y 2 = i(25 - x 2 ) ; y = JV25 - x 2 . (4) 2. To obtain real values for y, the numerical value of x may not be allowed to exceed 5. Thus, if x = 8, V25 x 2 = V 39, which is imaginary. 3. Place x = in (3) to determine the t/-intercepts: - 25; f . Hence, two points on the graph are (0, and (0, f ), labeled A and C in Figure 21. 4. Place y in (3) to determine the x-intercepts: x 2 = 25; x = 5. Hence, two points on the graph are (5, 0) and (- 5, 0), labeled B and D in Figure 21. Fi 9 . 21 5. As many more points as desired can be found by substituting values of x in (4) and computing the corresponding values of y. When the points are joined by a smooth curve we obtain the oval ABCD in Figure 21. The curve is called an ellipse. The graph of the positive valued function y = JV25 x 2 , from (4), is the half of the ellipse above the x-axis. The graph of y = JV25 x 2 is the lower half, which is symmetrical to the upper half. The whole ellipse is the graph of (3). The facts stated in the following summary are proved in more advanced mathematics. SUMMARY. The graph of , any quadratic equation in two variables x and y with real solutions is either an ellipse, a hyperbola, a circle, a parabola, a pair of straight lines, or a single point. 1. J/fc is positive, the graph of jc a H- y a = c is a circle whose radius is \/c and center is the origin, provided that the same unit is used on the scales of the x-axis and y-axis. SYSTEMS INVOLVING QUADRATICS 265 2. // a, 6, and c have the same sign, the graph of ax 9 + &y* = c is an ellipse, with center at the origin; if a b, the ellipse is a circle, pro- vided that the same unit is used on the scales of the x-axis and y-axis. 3. // a and b have opposite signs and if c is not zero, the graph of ax* -f by 2 = c is a hyperbola. 4. If c j 0, the graph of xy = c is a hyperbola; ifc>Q, one branch of the hyperbola lies wholly in quadrant I, and the other in quadrant III; if c < 0, the branches are in quadrants II and IV, respectively. The coordinate axes are the asymptotes of the hyperbola. 5. // a quadratic equation in x and y does not involve y 2 or xy, the graph of the equation is a parabola whose axis is parallel to the y-axis; if the equation does not involve x 2 or xy, the graph is a parabola whose axis is parallel to the x-axis. ILLUSTRATION 1. The graph of the equation 3# 2 + 7y* = 8 is an ellipse, of 5x 2 y 2 = 7 is a hyperbola, and of 4z 2 -f- 4y* = 25 is a circle, whose radius is 5/2. EXAMPLE 4. Determine the nature of the graph of 2x* - xy - 3y* = 0. (5) SOLUTION. 1. Factor: (2x 3y)(x 4- y) = 0. Hence, (5) is satisfied by values (x, y) in case (a) 2x - 3y = 0, or (6) x + y = 0. Therefore, the set of all points (x, y) satisfying (5) consists of those satisfying (a) and those satisfying (6). Or, in other words, the graph of (5) consists of the graph of (a) and the graph of (6). 2. The graphs of (a) and (6) are straight lines through the origin. Hence, the graph of (5) consists of these two straight lines. Comment. Another case similar to Example 4 was met in finding the asymptotes in Example 2. They were the two straight lines which are the graph of the equation 9z 2 4y 2 = 0. EXAMPLE 5. Determine the nature of the graph of 5x - 7 = 0. SOLUTION. 1. Solve for y: y = |z 2 f x + J. 2. Thus, y is a quadratic function of x, and therefore the graph of the given equation is a parabola whose axis is parallel to the y-axis. To graph the equation, we would compute the coordinates of the vertex of the parabola and proceed as in Section 132, page 186. 266 SYSTEMS INVOLVING QUADRATICS 182. Routine for graphing It is important to be able to construct reasonably good graphs quickly. Beyond this, it is also essential to have a procedure for im- proving on such graphs when the necessity arises. The following suggestions are of aid hi constructing graphs quickly for equations of the second degree hi x and y. 1. Refer to the summary of Section 181 and if possible decide on the nature of the graph before carrying out details of the work. 2. When the graph of ax z H- by 2 = c is a circle, find its radius, 17 -> and construct the circle with compasses. 3. When the graph of ax* 4- by* = c is an ellipse, find the x-intercepts by placing y ~ and solving for x, and find the y-4ntercepts by placing x = in the given equation. Then, sketch the ellipse through the four intercept points thus obtained* 4. When the graph of ax 2 4- by 2 = c is a hyperbola: Find its asymptotes by replacing c by and constructing the two straight lines which are the graph of ax 2 -f- by* - 0. Find the x-intercepts or the y-intercepts. (One set of intercepts witt be imaginary because the hyperbola will cut just one of the coordinate axes.) Sketch the hyperbola through the real intercepts thus found, with each branch of the curve approaching the asymptotes smoothly. 5. When a quadratic equation in (x, y) is linear in one variable, solve for it in terms of the other variable and then graph the resulting parabola by the method of Section 132. ILLUSTRATION 1. To graph 9x 2 4y 2 = 36 quickly, we first note that the graph will be a hyperbola. We substitute y = and find that the ^-intercepts are real, x = 2; we thus obtain points B and D in Figure 20, page 263. We obtain the asymptotes, as in Note 1, page 263. Then we sketch branches EDF and GBH through points B and D in Figure 20. To improve on a graph as obtained through the preceding sug- gestions, or when doubt arises as to the nature of a graph, solve the given equation for one variable in terms of the other and compute as many points as needed, with Example 2 of Section 181 as a model. * Illustrated in Example 3, page 264. SYSTEMS INVOLVING QUADRATICS 267 1. x* + y* = 9. 4. 4z 2 + 4y 2 = 9. 7. 9z 2 - I 2 - 0. 2. 5. 8. = 4. EXERCISE 97 Graph each equation on cross-section paper. - 36. 3. 4x 2 - j/ 2 - 16. 6. xy - 6. 0. 9. 4y 2 9x 2 = 36. 11. y - a; 2 - 4x + 7. \ 13. 3s 2 + 4xy 4y 2 * 0. 16. (a; 2y)(3z 2y 6) = 0. 16. Rewrite statements I and II of page 187 for the graph of x = ay 2 + by + c, where it is understood that the o>axis is to be horizontal as usual. 10. (Qx - 3/)(3s + 2y) - 0. 12. 2y 4s -f 6z 2 - 9. 14. 36 - * 2 - 9w 2 - 0. Graph each equation, with the aid of Problem 16. 17. x = 4y 2 . 18. x - 2y 2 + Sy - 6. 19. 9 - 0. 183. Graphical solution of systems involving quadratics EXAMPLE 1. Solve the following system graphically: = 1 (1) (2) SOLUTION. 1. We graph each equation, on one coordinate system. The graph of (1) is the hyperbola and the graph of (2) is the ellipse hi Figure 22. 2. Any point on the hyperbola has coordinates which satisfy (1), and any point on the ellipse has coordinates which satisfy (2). Hence, both equations are satisfied by the coordinates of A, B, C, and Z>, which are the points of intersection of the ellipse and the hyperbola: A: (x = 9, y = 2). B: (x - 3, y - 2). C: (x 3, y - - 2). D: (x - 3, y - - 2). These pairs of values are the solutions of the system [(1), (2)] and can be checked by substitution in the given equa- tions. Only real solutions can be found by the preceding graphical method and, usually, solutions can be read only approximately from a graph. Fig. 22 268 SYSTEMS INVOLVING QUADRATICS EXERCISE 98 Solve graphically. - (* 2 + 2/ 2 = 16, (2* + y = 3, o ** -2 = 3. * tf + f -9. * 9t/ 2 = 36, / 25x 2 + y 2 = 25, *i < /y.2 Q V """ i/ f y - 2z 2 - Sx + 9, f 2z 2 - z?/ - 6y 2 = 0, = 12. 2 + ? 2 = 4. 4t/ 2 = 36, \25z 2 4- 4y 2 - 100. \z 2 + t/ 2 = 9. 1 84. Solution of a simple system A system consisting of one linear and one quadratic equation in two unknowns x and y usually has either (a) two different real solu- tions, or (6) two real solutions which are the same, or (c) two im- aginary solutions. These possibilities correspond, respectively, to the following geometrical possibilities: the straight line, which is the graph of the linear equation, (a) may cut the graph of the quadratic in two points, or (b) may be tangent to, or (c) may not touch the graph of the quadratic. EXAMPLE 1 Solve- f 4x 2 - 6^ + 9t/ 2 = 63, (1) EXAMPLE 1. bolve. \ 2* - 3y = - 3. (2) SOLUTION. 1. Solve (2) for x: x = 3y " 3 - (3) Zi 2. Substitute (3) in (1) : ' . 63. 2,2 _ y _ 6 . . (y _ 3 )(y + 2 ) = 0; y = 3 and y = - 2. 3. In (3), if y = 3, then x = 3; if y = - 2, then x = - 9/2. 4. The solutions are x = 3, y = 3 and x =_i v =_2 ./. Since a solution of a pair of equations in x and y is a pair of related values of x and y, it is very essential that each solution should be plainly indicated as a pair of values, as in Example 1. Note %. A system of the type considered in this section will hereafter be called a simple system, SYSTEMS INVOLVING QUADRATICS 269 SUMMARY. To solve a system of one linear and one quadratic equation in x and y algebraically: 1. Solve the linear equation for one unknown in terms of the other, say for y in terms of x, and substitute the result in the quadratic equation; this eliminates one unknown. 2. Solve the quadratic equation obtained in Step I and, for each value of the unknown obtained, find the corresponding value of the other unknown by substitution in the given linear equation. EXERCISE 99 Solve, (a) graphically and (b) algebraically. , , <> - -16, , , \4d + 3c = 50. * \u + 2v - 6. a + 2b - 4. Solve algebraically. (2x + 7/ + 3 = 0, (z 2 + 92/ 2 = 25, \ 2z 2 + y 2 - Qy = 9. \x - 7 + 3y = 0. -46 = 12, \ a 2 + 2a - - 46 - 12. z 2 + 2z - 4y = 23 - y, f 2x2/ + 3?/ + 4x - 1 - 0, + 2y = 5. \2aj + y + 3 = 0. / y - 2x + 1 = 0, t / 2x + z/ = 2o - 1, + 24i/ = 36. \ 4a; 2 + 4x + y = 2a 2 ' - 2xy + 2/ 2 + 8z - 2y = 3. + y z - 2by = 2a 2 -f & 2 . 185. Solution of a system of two quadratic equations When both equations of a system are quadratic, the system usually has/owr different solutions, all or two of which may involve imaginary numbers. The student should recall his graphical solutions of sys- tems of this type where four solutions were obtained. Note 1 . The fact stated in the preceding paragraph is a special case of the following theorem which is proved in a later course in algebra: 270 SYSTEMS INVOLVING QUADRATICS A system of two integral rational equations in x and y, in which one equation is of degree m and the other is of degree n hi x and y, usually has mn solutions. Thus, a system consisting of an equation of the third degree and a quadratic usually has 3 X 2 or 6 solutions. Usually, the algebraic solution of two simultaneous quadratics brings hi the solution of a fourth degree equation hi one variable. At this stage in algebra, the student is able to solve only very simple fourth degree equations. Hence he is not prepared to consider the solution of all systems of simultaneous quadratics. Therefore, in this chapter we consider only special elementary types of systems. 1 86. Systems linear in the squares of the variables When both equations have the form ax 2 -f 6t/ 2 = c, the system is linear in x 2 and y 2 and can be solved for x 2 and y 2 by the methods applicable to systems of linear equations. EXAMPLE 1. Solve: } * ' '* ~ ""> (1) (* 2 \ x* x* + 2y 2 = 34. (2) SOLUTION. 1. Multiply by 2 in (1) : 2s a -f 2y 2 50. (3) 2. Subtract, (3) - (2): z 2 = 16; x = db 4. 3. Substitute x 2 = 16 in (1): 16 -f 2/ 2 = 25; y 2 = 9; y - 3. 4. Hence, if x is either 4* 4 or 4, we obtain as corresponding values y -f 3 and y 3, and there are four solutions of the system. -4,y x = 4, y = - 3 x = 4, y = EXERCISE 100 Solve each system, (a) graphically and (b) algebraically. *' ^ - -- . 9. 2 ' is 2 + t/ 2 = 36. " \9z 2 -f W - 16. Solve algebraically. , / z 2 + 4s/ 2 = 14, / re 2 - t/ 2 = 4, A / 2* 2 - 32/ 2 - 3, L i * = S*y2 - Ifi. ' ^ rt " - - 16. 1 2 2 -f v 2 - 11. ' I 5ar 2 -h 2w 2 - 17. f 9s* - \ &C 2 - 9i 2 - %* - 6, ft / ISc 2 - 8 -h w , ** ^ i <n x r^/v A V 7. 1 15 - 12d - 20c*. 1 6 - 3^ + 5r*. f 7r* -1- Ait 2 = 3ft ( 7r* ft?/ 2 = AQ 10 <* 11 V ^ ' 12 < ' \ llr* + 5a* - - 4. ' \ 9x 2 + 2y 2 = 13. \ 4x 2 -h 9y 2 = - 4. EXAMPLE 1. Solve: ( * + * ~ 14 J , ft \ x 2 - 3xy + 2y 2 - 0. SYSTEMS INVOLVING QUADRATICS 271 1 87. Reduction to simpler systems <J> (2) SOLUTION. 1. Factor (2): (x - 2y)(x - y) = 0. (3) 2. Therefore, (2) is satisfied if either z 2y = 0, or x y = 0. 3. Hence, (1) and (2) are satisfied if and only if x and y satisfy one of the folio wdng systems: ( & + yt = 14, T / x* + 2/ 2 - 14, l 4. On solving (I) by the method of Section 184, we obtain two solutions: (x = \/7, y = V7) and (x = V7, 2/ = V?). From (II) we obtaui (a? = fVTO, y - JVTO); (a; - - fVTO, y - - We say that the given system in Example 1 is equivalent to the systems I and II because the solutions of the given system consist of the solutions of (I) together with those of (II). The preceding method applies if, after writing each equation with one member zero, we can factor at least one of the other members. 1 88. Elimination of constants A system in which all terms involving the variables are of the second degree can sometimes be solved by use of the equation we obtain on eliminating the constant terms from the original system. EXAMPLE 1. Solve: INCOMPLETE SOLUTION. 1. Eliminate the constants. Multiply (1) by 2: 2z* + fay - 56. (3) Multiply (2) by 7: 7xy + 28y 2 = 56. (4) Subtract, (3) - (4): 2z 2 - xy - 28y 2 = 0; or, (2x + 7y)(x - 4y) - 0. (5) 2. To solve [(1), (2)] we may now solve [(2), (5)]. This system is equiva- lent to the following simpler systems: 4t/ 2 = 8, / xy + 4y 2 = 8, - 0. (6) Instead of using (2) in (6) we could equally well have used (1). Each system in (6) has two solutions and thus [(1), (2)] has four solutions. 272 SYSTEMS INVOLVING QUADRATICS EXERCISE 101 Solve algebraically and graphically. 1i I y t {% j **' \ y i / vj / ' H" y)( x "" 2y) = 0. \ (x 2/)(x 4- 3y) = 0. . Hereafter in this chapter, leave all surd values in radical form. Moreover, unless otherwise stated, to solve a system will mean to solve al- gebraicatty. f Solve by reducing to simpler systems. 3 ( 2z 2 + 5*s/ - 3y 2 = 0, (3x* + 5xy r \2s 2 -f Zxy = 2. \ x 2 + sy = 4. , = 0, * (a; 2 -f 3y 2 = 7. \ 2z 2 - xy - 2y* = 8. Solve by eliminating the constant terms. I x* + 3xy = 28, 8 f x 2 - 5o;y + 6t/ 2 = 10, \xy + 4y* = 8. ' \x*-xy = 4. ' 2c 2 - 2cd = 15. * 2a: 2 + 2/ 2 = 5. = 11, fx 2 \t - 2 2 + 3 0. \ x 2 - a^ + 4y z = 40. 6 = 0, f 2w 2 -f 3mn = 1, 1 y + 2/ 2 - 35. \9w 2 -f 8n 2 = 9. , t/ 2 + 7 = 0, \ 2mn -f n 2 = 64. 1D * \x 2 - Zxy - If + 4 - 0. *189. Additional devices for reducing to simpler systems ( * + * = , 27 ' I x + y = 3. (2) EXAMPLE 1. Solve: * + * = 27 ' SOLUTION. 1. Factor (1): (* + y)(x* - xy + y*) = 27. (3) 2. Divide, (3) by (2) : x 2 - xy + y* = 9. (4) 3. Hence, (z, #) satisfies [(1), (2)] if and only if (x, y} satisfies f x + y - 3, (5) \ x* - xy + y* = 9. (6) The student should complete the solution by solving [(5), (6)] by the method of Section 184. SYSTEMS INVOLVING QUADRATICS 273 EXAMPLE 2. Solve: ( * + ^ + * - 20, (7) I xy = 5. (8) INCOMPLETE SOLUTION. 1. Add, (7) + (8): x* + 2xy + y* 25. (9) 2. From (9), (x + !/) 2 = 25; hence z + y = 5, or a; + y = 5. 3. To solve [(7), (8)], we would solve each of the following systems: (x + y = 5, ! X +.V = - 5, 1 xy - 5. \ xy = 5. *190. Determination of tangents to curves EXAMPLE 1. Find the value of the constant k so that the graphs of the equations in the following system will be tangent: f x 2 + ?/ 2 = k 2 , (1) \x + y = l. (2) SOLUTION. 1. If the graph of (2) is tangent to the graph of (1), then the two solutions of the system [(1), (2)] must be identical. 2. Substitute y = 1 - x in (1) : 2z 2 - 2x + (1 - fc 2 ) = 0. (3) 3. From Step 1, we notice that (3) must have equal roofa. Hence, Us discriminant must be zero, or (- 2) 2 - 4(2)(1 - fc 2 ) = 0; 8fc 2 - 4 = 0; k = =fc i>/2. Thus, the straight line is tangent to the circle if its radius is .707. *191. Equations symmetrical in x and y. An equation in x and y is said to be symmetrical in x and y in case the equation is unaltered when x and y are interchanged. A quadratic equation in x and y is symmetrical in x and y if the coefficients of x 2 and y 2 are equal and those of x and y are equal. The method of the next example applies where each equation is symmetrical in the unknowns. - ! <a i 2y = 8, (1) EXAMPLE 1. Solve: < . , * ) rt \ I 2xy -f x + y = - 4. (2) INCOMPLETE SOLUTION. 1. Substitute x u + v; y = u v. (3) From (1) : 2w 2 -f 2t^ + 4u = 8, (4) From (2) : 2w 2 - W + 2u - - 4. (5) * 2. Solve the system [(4), (5)3 for u and v: Eliminate y 2 , [(4) + (5)]: 4w 2 + 6u - 4 = 0. (6) 3. Solve (6) for u; then obtain v from (4). Each pair of values (u, v) when placed in (3) gives a solution of [(1), (2)]. 274 SYSTEMS INVOLVING QUADRATICS 4. 6. 8. *EXERCISE 102 Solve by any convenient method. \8a 8 + 6 s - 98. 2 * ( a - 6 = 3. 3 * ( y(x ( a; 2 H- xy H- y 2 = 7, + y) + y) 40, 20. f * \ }- 4, - 24, - 4 - 0. 4t^ = 9, uv + v = 3. f z 2 + 3,2 . 13, I OPfl/ =r ft ^ **y ^^ " f4c 2 -h3^-f2/ 2 = 8, / \ xy = 1. = 2. 11. = 5. 12. 14. \ 3j/ icy y* = 4. ' x + 2y -f 2 = 3, 13 f * + xy + 2/ 2 - 61, " \ icy = 29 - x - y. y = 6, + ^ -f = 14. 16. - 2?/ 2 - 1, 8. Find the values of k for which the graphs are tangent. Then, if k is real, graph the equations of the resulting system. - y = 5. ' 17 ' \Sy + 3x - *. ' 18 * 1 4 + 2y + kx = 0. 16 an expression for c in terms of the other constants in case the graphs of the two equations in the variables (x t y) are tangent. 1ft /!/ - mx + c, ^ (y = mx + c, M ( x - my -f c, i. < . + 4y2 ^ 36 -w. | fl2a . 2 _ ^2 = (W> 22. the method applying to symmetrical equations. 4x + y 2 4y = 17, 22 / ^ "~ 3a ^ + 2/ 2 == 1, 6 = 0. . 1 2x 2 - xy -h 2t/ 2 - 17. 25. 24. without first clearing of fractions. 11, ^ =: H- 28 - 0. 26. 27. - - 7. SYSTEMS INVOLVING QUADRATICS 275 * MISCELLANEOUS EXERCISE 103 Solve graphically. f4* 2 + y 2 = 25, 2 fa* + 42,*= 16, f* + y'20, L '\2x + y = T. A \* 2 -s/ 2 = 9. * \ z* - 4y 2 = 4. ^ f (a - y _ 2)(* - y - 1) - 0, fi y = 1 - s*, 6-8. Solve Problems 1, 2, and 4 algebraically and compare the results with the solutions previously obtained. Graph each equation. fc 4z 2 + V - 0. 10. 4z - ty* - 0. 11. 4x - 9y* - 36. algebraically. -2*+ 1=0,, Sx - 4y - 3y*. 4x* - xy - y* - 4. 17 110 - t> 2 - 9, '* , ' 3. V * r' - rA = 75. or a; , - 4 2 = a 2 - 2o6 - 6 2 . Solve each problem by introducing two unknowns. 22. The sum of two numbers is 28 and the sum of their squares is 634. Find the numbers. 23. Find the dimensions of a rectangle whose area is 60 square feet and whose diagonal is 13 feet long. 24. The area of a rectangle is 55 square feet and its perimeter is 49 feet. Find the lengths of the sides of the rectangle. 25. The sum of the reciprocals of two numbers is 3 and the product of the numbers is . Find the numbers. 26. A man divides $500 between two investments at simple interest, a first part at twice the interest rate obtained on the second part. The first investment grows to the amount $345 in 3 years, and the second to the amount $220 hi 4 years. Find the interest rates and the sums invested. 276 SYSTEMS INVOLVING QUADRATICS 27. The sum of the squares of the two digits of a positive integral number is 65 and the number is 9 times the sum of its digits. Find the original number. 28. Some men row 15 miles downstream on a river to a mountain and then climb 12 miles to its summit. They take 9 hours for the journey and, the next day, 9 hours to return. Find the rate at which they row in still water and their speed in ascending the mountain, if they descended 1 mile per hour faster than in ascending, and if the rate of the current of the river is 1 mile per hour. 29. 'A weight on one side of a lever balances a weight of 6 pounds placed 4 feet from the fulcrum on the other side. If the unknown weight is moved 2 feet nearer the fulcrum, the weight balances 2 pounds placed 9 feet from the fulcrum on the other side. Find the unknown weight. 30. A farmer has 3 fields of equal size and pays each of his workmen $4 per day. He paid a total of $42 to 2 Workmen after each, working alone, plowed one of the fields. These men took 2f days to plow the third field when working together. How many days did it take each man to plow a field alone? 31. Two towns on opposite sides of a lake are 33 miles apart by water. At 6 A.M., from each town a boat starts for the other town, traveling at uni- form speed. The boats pass each other at 9 A.M. One boat arrives at its destination 1 hour and 6 minutes earlier than the other. Find the time it takes each boat to make the trip across. 32. A wheel of one automobile makes 96 more revolutions per mile than a wheel of a second automobile. If 20 inches were added to the length of the radius of a wheel of the first automobile, the result would be the diameter of a wheel of the second automobile. Find the diameter of a wheel of each automobile, using 22/7 as the approximate value of TT. APPENDIX NOTE 1. THE IRRATIONALITY OF V2 If there exists a rational number which is a square root of 2, then there exist two positive integers m and n, such that (l) n m where is a fraction in lowest terms. In other words, if V2 is rational 7T there exist two integers m and n, without a common factor, such that (1) is true. Let us show that this assumption leads to a contradiction. 1. Square both sides of (1): = - ; or > 2n 2 = m 2 . (2) We see that 2 is a factor of the left member of 2n 2 = m 2 ; hence 2 is a factor of the right member. Therefore 2 is a factor of m because otherwise 2 could not be a factor of m 2 . That is, m = 2k, where k is some positive integer. 2. Place m = 2k in (2) : 2n 2 = (2fc) 2 = 4& 2 ; w 2 = 2k 2 . (3) Consider n 2 = 2& 2 ; since 2 is a factor of the right member, hence 2 is a factor of n. 3. We have shown in Steps 1 and 2 that m and n have 2 as a factor. This contradicts our original assumption that m and n had no common factor. Hence, the assumed equation 1 has led us to a contradiction, and it follows that (1) itself must be false. Therefore no rational number exists which is a square root of 2, or, \/2 is an irrational number. Comment. We easily verify that (1.4) 2 = 1.96; (1.41)* - 1.9881; (1.414) 2 = 1.999396; etc. On considering the sequence of numbers 1.4, 1.41, 1.414, 1.4142, 1.41421, -, (4) we see that the square of each number in (4) is less than 2 but that, on pro- ceeding to the right in (4), the squares of the numbers approach 2 as a limit. Each number in (4) is a rational number; we refer to these numbers in (4) as the successive decimal approximations to A/2. 278 APPENDIX NOTE 2. EXTENSION OF THE INDEX LAWS TO RATIONAL EXPONENTS A complete proof that the index laws hold for any rational exponents could be constructed by showing, hi succession, that the laws hold if the exponents are (1) any positive rational numbers and (2) zero, or positive or negative rational numbers. Without giving a complete discussion, we shall indicate the nature of the methods involved by proving some of the necessary theorems. For convenience in details, we shall assume tha$ the base is positive. In our proofs, we use the index laws for positive integral exponents and the definitions of Sections 113, 114, and 115. (m\p mp a n ) = a n . Proof, (a*)* = [(o) W ] P = (a) mP ; [(3), page 151; (II), page 142] (!\p tap a n ) = a n . [(3), page 151] THEOREM II. // m, n, p, and q are positive integer s t then m j> m j mq+np I * 2\ n fl / m \*tf/ 2\*m Proof. [a* an) = (a*) Uv [(IV), page 143] _ a m a pn < (Theorem I) (m J>\nq a oJ a m *+*. [(I), page 142] Therefore, by the definition of an ngth root, / . \A. ao = ^a mfl+pn ; n = a (m\ o> r/ !\ Suggestion for proof. Compute |_ V */ fl In the remainder of this note we shall assume that the index laws have been completely established for all positive rational exponents. THEOREM IV. Law I of Section 105 holds if the exponents are any positive or negative rational numbers. Comment. We are assuming that Law I has been established if both exponents are positive. Hence, it remains to show that, if h and k are any positive rational numbers, then a~ h a~ k = o~*~*, and o*a~* a*~*. Incompkte proof. By the definition of a negative power, -* a -* =*.--= -; or, a* a* a*" 1 "* APPENDIX 279 NOTE 3. ABRIDGED MULTIPLICATION ^The following example illustrates a method for abbreviating multiplication of numbers with many significant digits when the result is desired with accuracy only to a specified number of places. EXAMPLE 1. places. Compute 11.132157(893.214), accurate to two decimal SOLUTION. 1. To multiply by 893.214, multiply in succession by 800, 90, 3, .2, .01, and .004 and add the results (in ordinary multiplication these operations are in reverse^ order). Since we desire accuracy in the second decimal place, we carry two extra places, or four decimal places, in all items. 2, In the abridged method, to multiply by 800 we multiply by 8 and move the decimal point; all digits of 11.132157 are used in order to obtain four significant decimal places. This first operation accurately locates the decimal point for the rest of the items. ORDINARY METHOD ABRIDGED METHOD 11.132157 893.214 vWvV 11.132157 893.214 Multiply by 44528628 11132157 2 2264314 33 396471 1001 89413 8905 7256 8905.7256 1001.8935 33.3963 2.2264 .1113 .0444 800 90 3 .2 .01 .004 9943.398482598 9943.3975 Add ^Result = 9943.40 Result = 9943.40 3. To obtain four decimal places when multiplying by 90 we do not need the last digit of 11.132157; to indicate this place 'V" over *'7" and multiply 11.13215 by 90. Next, place ' V" over "5" and multiply 11.1321 by 3; then " >/" over "1" at the right and multiply 11.132 by .2; then " >/" over "2" and multiply 11.13 by .01; then " V" over "3" and multiply II. 1 by .004. Then add and round off to two decimal places, obtaining 9943.40. Note 1. The advantages of the preceding abridged method are obvious. As compared with the ordinary method, less labor is involved, the decimal point is accurately located, and fewer mistakes will occur in the final addi- tion. Note g. An abridged method of division can be developed similar in prin- ciole to the method of abridged multinlicatioiv abova. . . 280 APPENDIX NOTE 4. A FALLACIOUS PROOF THAT 2 - 1 The following absurd result that 2=1 illustrates the contradictions that arise if division by zero occurs. 1. Suppose that y 6. 2. Multiply by y: y 2 = by. 3. Subtract 6 s : y* - 6> - by - 6*. 4. Factor: (y - 6)(y + 6) = b(y - 6). 5. Divide by (y 6) : y 4- 6 = 6. 6. Since y = 6 (Step 1), 6 + 6 = 6, or 26 =* 6. 7. On dividing both sides by 6, we obtain 2=1. Discussion. In Step 5 we divided by zero, because y 6 = if y 6. Hence, Steps 5, 6, and 7 are not valid, because division by zero is not allowed. NOTE 5. THE SQUARE ROOT PROCESS OF ARITHMETIC Consider finding V7569. Since the radicand has four digits to the left of the decimal point, the square root must have two digits to the left of the decimal point, because the square of any number of units is less than 100, and the square of any number of hundreds is greater than 10,000. We observe that 80 is the largest whole number of tens whose square is kss than 7569. Hence, we consider finding x, a number of units, so that <80 + z) 2 = 7569. (1) By the formula for (a -f 6) 2 , from (1) we obtain 6400 + 2-80- x + a* = 7569; ,(2) 160* + z 2 = 7569 - 6400; 160z + z 2 = 1169. (3) From (3), z(160 + x) = 1169, 1169 x = ieo+T An approximation to (4) is obtained if we use the trial divisor 160 in place of (160 -f x) ; this gives x - 115? _ 7+ / x ~ 160 ~ 7 ' (5) Then, we take x * 7 and verify that the complete divisor, (160 -f- x) or 167, gives 1169 - ,, 7, exactly. Hence, V7669 = 80 -h 7 - 87. We verify by squaring that 87* - 7669. APPENDIX 28? In the following examples, the student will observe, in brief form, steps corresponding to those just explained in detail by reference to the formula for (a + &) 2 . Hereafter, the details will be carried out without making the possible contacts with the square of a binomial. Essentially, at each stage of the following arithmetical process, we have knowledge of a in a binomial (a -f 6) and we obtain an approximation to b so that (a + 6) 2 will be as nearly equal as convenient^ at that stage, to the number whose square root is being obtained. EXAMPLE 1. Find V7569. SOLUTION. Arrange 7569 into groups of two figures each, starting at the decimal point. After each of the following steps, read the corresponding explanation below. Step 3 Step 4 8 87 Step 1 8 Step 2 8 75 69 64 75 69 64 11 69 160 75 69 64 11 69 Ififi A v/vr 167 75 69 64 11 69 ^ t//\ XvJU 167 StepS 8 7 7569 64 11 69 11 69 Explanation. 1. The largest perfect square less than 75 is 64. Write 64 below 75. Write V64, or 8, above 5 of 75. 2. Subtract 64 from 75. Bring down the next group, 69. 3. Form the trial divisor: 2X8= 16; annex (160). 4. Obtain the complete divisor: 1169 -f- 160 = 7 + . Add 7 to 160, forming 167 as the complete divisor. Write 7 over 9 of 1169. 5. Find 7 X 167, or 1169. Subtract. Since the remainder is zero, V7569 = 87. The complete solution appears as Step 5. It alone would be written in an actual solution. EXAMPLE 2. Find V1866.24. SOLUTION. 1. First trial divisor: 2X4 8. Annex 0, giving 80. 2. First complete divisor: 266 -* 80 - 3+; 80 + 3 83, the complete divisor; write 3 over the right-hand 6 of 66. 43. 2 3. Place a decimal point in the square root, above that of 1866.24. 80- 83 4. Second trial divisor. 2 X 43 = 86. Annex (860). 5. Second complete divisor. 1724 -f- 860 - 2 + ; 860 + 2 - 862. Place 2 above 4 of 24. 2 X 862 - 1724. 18 66.24 16 266 249 OftA ouu 862 1724 1724 Check. We find that (43.2)* - 1866.24, or V1866.24 - 43.2. 282 APPENDIX EXAMPLE 3. Find V645.16. SOLUTION. 1. The largest perfect square less than 6 is 4 2 5. 4 or 2*. Hence, 2 is the first digit of the square root. 2. After forming the trial divisor, 40, it appears that the next figure may be 6 since 245 -*- 40 = 6 + . But 6 X 46 = 276, and this is more than 245. Therefore, we must use 5 as the second figure of the square root. 3. To form 500, we take 2 X 25 and annex 0. 2016 -f- 500 is 4+. Then 500 + 4 = 504. The result is 25.4. 6 45.16 4 245 225 504 2016 2016 SUMMABY. To find the square root of a number, written in decimal notation: 1. Separate the number into groups (or periods) of two figures each, both ways from the decimal point. 2. Below the first group, write the largest perfect square less than that group. Above the group, write the square root of this perfect square. 3. Subtract the perfect square from the first group; bring down the next group, thus forming the first remainder. 4. Form the trial divisor by doubling the part of the root now found, and annexing zero. . Divide the first remainder by this trial divisor, taking as the quotient only the whole number obtained. (Possibly reduce the number by I.) Write this quotient above the next group. 5. Form the compkte divisor by adding to the trial divisor the new figure of the square root found in Step 4. 6. Multiply the compkte divisor by the new figure of the square root. Write the product under the remainder. Subtract. 7. Continue in this way, following Steps 4 to 6, until the remainder is zero, or until you have as many places in the square root as are requested. Note 1. As a rule, for a random number N, VN will not be a terminating decimal. Then, in finding VAT, we annex zeros at the right in N and carry out the square root process to as many decimal places as desired. EXERCISE 104 Find the square root of each number. Obtain the result correct to hundredths by carrying out the process to thousandths. 1. 3969. 2. 134.56. 3. 273,529. 4. 8299.21. 5. 105,625. 6. 936.36. 7. 40.8321. 8. 2.1904. 9. 78.354. 10. 15,765. 11. 1643.8. 12. 7.809, ANSWERS TO EXERCISES Note. Answers to odd-numbered problems are given here. Answers to even-numbered problems are furnished free in a separate pamphlet when requested by the instructor. ' Exercise 2. Page 7 1. 56. 3. - 12. 5. 36. 7. 0. 9. - 8. 11. 56. 13. 5. 16. - 8. 17. 4. 19. 3. 21. - 9. 23. - 168. 26. 120. 27. 120. 29. - 360. 31. - 2. 33. - 4. 36. 2. 37. - 13. 39. 5. 41. 52. 43. f. 46. - 60. 47. 14.4. 49. 4. 61. 31. 63. - 96. 66. 360. Exercise 3. Page 12 1. 27. 3. - 18. 6. 13. 7. - 16. 9. - 32. 11. - 35. 13. 0. 16. - 13. 17. - 18.2. 19. - 7. 21. 13. 23. 61; 29. 26. - 30; 4. 27. - 36; - 70. 29. 17; - 17. 31. 3.3; - 11.9. 33. - 5. 36. - 7. 37. 10. 39. - 10. 41. 22. 43. 3. 46. 13. 47. 42. \ 49. 12. 61. 28; 4. 63. - 40; - 26. 66. 44; - 32. 67. 100. 69. - 53. 61. - 36.3. 63. - 14; 14; 0; 0. 66. 3; 9; - 18; - 2. 67. 23; - 23; 0; 0. Exercise 4. Page 15 11. <. 13. <. 16. >. 17. >. 19. <. 21. >. 29. - 5 < - 3; | - 5 | > | - 3 |. 31. > - 3; | - 3 | > | |. 33. | - 2 | < 7; - 2 < 7. 36. 2 > - 6; | 2 | < | - 6 |. Exercise 5. Page 17 1. 10. 3. - 24. 6. 5. 7. 8. 9. 44. 11. 36. 13. - 2. 16. 0. 17. - 24. 19. - 13. 21. - 2o -f 56 - c. 23. 31 - 5a + y. 26. - 8a + 36 + c. 27. 15o. 29. 15a. 31. 8a - 12. 33. - 15 + 5a + 30c. 35. 18 - 12a 4- 156. 37. - (5 - 7a + 46). 39. - (- 6 + 3x + 4y). 41. 16 - (4a + 6 - 3c). 43. 2ac - (- 3 + 5a - 4c). Exercise 6. Page 20 1. lla. 3. l&c. 6. 6cd. 7. 5x 6a. 9- 3a - lie. 11. 19c - IScd. 13. - 2a - 26 -f 4; - 60 + 166 - 10. 16. - x + db - 4c; 7x - lldb + 2c. 17. - 3m - k - 6h; 9m - 9fc + 4A. 19. - 14x - 3y. 21. - 9ac - 7xy + 46. 23. 9o - 206. 290 ANSWERS ^ 25. 3fl + 14A - 23*. 27. 3a -f 10y - 3. 29. 21a - Sly + 9. 31. 2t - 3. 33. 2a. 35. - r - s. 37. 2. 39. 13 - 4x. 41. - 56 + 10. 43. - h + 12* - 36. Exercise 7. Page 24 1-1. 3. 5. f 7. i 9. f 11. - i 13. f. 15. 3. 17. 3/4y. 19. 9/26. 21. - |. 23. - 5a/3. 25. tV 27. 46/d. 29. & 31. f . 33. 6. N 35. 3c/5. 37. ^. **. 12 c 5bc 10 vJL* "ZTT" * 43. ~ 45. ^- 47. f. 49. 51. A. 76 7 36 5 9d h 9W vtffc 55. 4. 57. Af. 59. 1 61. 63. yfy. 65. - |. ' 2k' c Exercise 8. Page 27. 1. 16. 3. 100. 6. 10,000. 7. 1. 9. 1. 11. - 27. 13. - 125. 15. - 243. 17. - 1000. 19. H. 21. - i 23. - 16. 25. -216. 27. -75. 29. 320. 31. n odd, neg. ; n even, pos. 33. - 120. 37. J. 39. y w . 41. x\ 43. 6 18 45 /i*+* W Cv 47. a 8 . 49. *y. W(W2 </!/ 53. h*. 55. 816 4 . 67. 59. g- 61. ft. 63. -y^S. .!5i 4 . ._ a 4 c 4 Of _- - * 69. 27c. 71. 1 j JL C4/ */ fcHo 4 166 4 73. Six 8 . 7K /r^^-w I V t C/ 1 f*^ * 79. r^t/^ 1 / ^,4 125a Exercise 9. Page 28 3. 3fl 2 6. 6. 152 8 . 7. 9. - 4s 2 *. 11. - 2x 2 j/ 2 . 13. 2as 6 . 15. - 8o 8 6>. 17. 24rW. 19. 12a; - 3y. 21. 20a - 15. 23. 15z + 12y. 25. 6s 2 lOa? 4 . 27. 6x 2 2x*. 29. 3^ 2 fc 3hk. 31. 15t^ 20to* 10w>. 33. 6a 4 6. 35. 24m 6 n 4 . 37. Qx h+n y l+k . 39. 3^ 44r A; 34 '*. 41. 14x- 10x 4 + &c - 12x 2 . 43. 3o 4 6 4 -f 15o6 + 18o 2 6 2 - 9o6. 46. 3 - 2y + 4y 2 . 47. 4a 2 - lOa - 14. 49. 6x - 24x* - 12. Exercise 10. Page 30 1. 3* + x - 12. 3. 2z 2 - 3x - 35. 5. 20a 8 - 43o + 21 7. 4# - 9* 2 . 9. 2o + ab - 156*. 11. 9r - 25a*. 13. 2a 2 6 - ab - 15. 15. c 2 ^ - x 2 . 17. a 2 + 60 + 9. 19. W - 8Wfc -H 16A J . 21. 9a* - 12o6 + 46. 23. oV - 2a6x -f V ANSWERS 297 25. y + 3y* - 10. 27. 3a + a'6 2 - 46*. 29. y - 8. 31. x - x 2 - llx + 15. 33. 6 - 5x - 6x 2 - x 8 . 35. 2x - 5x 2 - 8x + 5. 37. 20 - 146 - 36* + 6. 39. 6s 4 - 7x + 12x 2 - 19x + 7. 41. 2^ - 20y - 6j/ 2 f 25y - 25. 43. 15s 4 - 17x + 12x 2 + 17* - 15. 45. 25x 2 -f 4y* + 9 + 30* - 12y - 2Qzy. 47. a* -f 6. 49. 2x - x 2 - 16x +"l5. 51. 6a - llo 8 - 17o + 30. 53. x* - 27y*. Exrc!$ 11. Pag* 33 1. y 2 . 3. l/x. 6. x. 7. 9. 9. 1/x. z 4 11. 2y*. 13. 4x. 15. 7a. 17. l/5r. - ^^ ^ - . Jfc 3 66 21 23. -- 25. r 27. tot. 29 y> /i c 2 Li- 4c 33. 7x. 35. - J. 07 __ , . 87 ' S6 2 89. 6xy. 41. fa + 56. 43. - a - 46. 45. 3a - 2a 2 . 47. - 2 + 5a*. 49. r - ! 2 ~ 2x + 3. 53. y 2 - y + 5. 55. A-^i + r.- 57. 66-3a. 59. - ^ -h 2x + ? - - 15 5x* 3x 2 2 a; 61. - 4y + -'- 63. # - a - Exercise 12. Ps 36 1. x + 4. 3. c - 3. 6. 8 - 3. 7. y - 4. 15. o-6-^--r- 17. x + 2. 19. 3x+ll + - 2o 4- 6 x 21. x - 2. 23. 2x - x - 6 - r-^-r- 25. 2y - 3 - 2x - 3 * 4y 2 - Zy + 2 27. x - x 2 - 4 H ?_^ 29. x 2 + 3xy + 4y 2 . 81. x 2 - 3x + 9. x 3 33. x 2 + xy + y*. 35. 4w^ + 6u> + 9. 37. a 4 + a^ + 6*. 39. 2x 2 - 3. Exercise 13. Page 39 ^ 15. o 21. 36c. 23. 2 + 6- - a 6. 5 * a S< 3 Oil ^ /Y* JU U ^^^ w(f -7 ' 8 I 17. 6. 25. 28fec*. U * 6 19. 34. O7 80 Ale $f. 2o 2 6 2 . 292 ANSWERS 29. - 8L. jt. 35.. 37.48. 35 Q 39. 300. 41. 18,900. 43. 24a 3 6. 45. 36az 4 . 47. 80/W. Exercise 14. Page 41 1. V- 3. i 6. &. 7. ft. ' ^' 11.^3*. 13. H. 16. -|. 17. J|. 21. 2 -3* + 6 */. 23. ^J>. 26 . . . 31. 13 5r-3fe 106* - 3a 206 - 15ay + 3a .< 19o -f 94 6 - ' 20 ._ 12 Art 5x - 9 ._ 2y 2 2 - 4yz - 6y + 9 45. - 7; - 47. - -- 49. 4* Ox . 3 - 13z _. 27y - 10^ + 20 __ 3 - Sab 3 - 4a 2 6* Exercise 15. Page 44 9. . 11. H. 13. 2 - 3a 5 - 3bc 5 + 4a' 46c + 3' ' 5a 2 fe - 3 3 * ' - 2 4x 2 - Zxy* 20x 2 - 6 + 156 2o + 3 2 - 3z 36x - 24 33. ^. 35. f . 37. - I 39. &. 41. - &. 43 ! 45 2a-3fe 36 - 2a* ' 3o + 5/i* Exercise 16. Page 46 1. - 60. 3. 0. 6. f . 7. - 4. 9. 1. 11. 6. 13. 17. 15. - 32; - 14. 17. 42; - 8. 19. <. 21. >. 23. 26 -}- c - 3a. 25. a'6 2 - 3a'6. 27. foy - 12zV. 29. 9Jk - 11 A. 31. 15 - 2o. 33. - f 35. &. 37. ^. 39. ^f. 41. 24 W. 43. ANSWERS 293 a 2 *) 2 45. 625c 8 d 12 t/ 4 . A7 A 7 _ Tile J2 5 AQ . TBV* 53L 63. 4 4x* 55. 4z 2 - Sx - 21. 67. 62;* - l; 3x|/ + 15y. 69. 2s 3 - 5s 2 - 8z 4 - 13s + 15. 61. a*. 68. - 3 4 -1 - 8w 65. 2z 2 + 4x 3 + \j A7 ^^y 4^^^ K cm ff 2z -5 VI. ^, to 4 71. 31 - 12a 73 Qy '~ { toy* - 20x + 12: rt/ . 7fg 10 12 Id. 12x 2 y 3 ID. %%. 77 9y 2 - 30z 2 i/ 70 2 **>- 3x 2 i/ 01 5. ! 4c 3y - 5x Exercise 17. Pase 50 1. 3.25. 3. 100,000. 5. .0001. 11. + - 2 + , 9. 3(10 3 ) + 10 2 + 4(10) -f 9. 15. 5735.35. 17. 14.1192. 23. - 103.7698; 16.0762. 29. 326,530. 31. .000317. 13. 536.437. 21. 6.64; 3. 27. 5.32. 1. 4.914. 9. .000054322. 1. 15.326; 15.3. 7. .034564; .0346. 13. 31.54; .586. 19. 2,056,000. Exercise 18. Pase 52 3. 5.993. 6. .51312. 11. 2.1435402. a-5 7. .0000001. 19. .0681. 25. 1.178. 33. 5.738. 7. 13.62528. Exercise 19. Pase 55 3. .31486; .315. 9. 566.5 and 567.5. 16. 11.4034; .054. 21. 10 2 (6.7538). 5. 195.64; 196. 11. 567.35 and 567.45. 17. 2738. 23. 4.5726(10 4 ). 25. 4.5312(10 6 ); 4.53(10). 27. 7.2200(100; 7.22(10'). 29. 2.6(10) cu. ft Exercise 20. Pase 57 1. 1.37. 13. 3. 57.2. 6- 6. .263. 17. 7. 150. 19. .625. Exercise 21. Pase 62 9. .02981. 21. .15. 11. .286. 23. .4375. 1. 11. - 21. J. 31. 8. 41. 3. 61. f 63. - 3. - 3. 13. 1. 23. - 3. 33. 6. 43. .36. 7. 55. 2. 5. i 7. 0. 9. $ 16. f. 17. f . 19. - f . 26. V- 27. 15. 29. . 35. 2. 87. 4. 39. 17. 46. f. 47. 4. 49. f. 57. - f . 59. 283.46. 61. 515.02. 294 ANSWERS Excreta 22. Page 66 1 3 + C ^ , , , . K .-, ft 1 c 11. ^ 5 '3~a **6- 17 -. 2a 3 9a6 15 IL 2 1S * 2 9L ^ 1T *3fc L * 12 27 9 41 ^ r^ n ^M ir 2L 36 - a 9ft . - . 6a-5c _-. 29> 35 10. M 9.ft7 r/f 23. * 2a-36 2a6 3c ' o6c) -6d n . . 33. a - 9 m k / <\j I a -\- d . I a 35. a - I (n l)d: n - 3 - : d = - - d n 1 5 o tt >. M + N + P ^^,0,0 37. a - - - 39. A - - r - -- 4t C - .12n + 6. r* 1 3 Extrcta 23. Page 69 1. 32.5* and 35.5'. 3. 22.5' and 5'. 5. 3f. 7. 15; 16; 17. 9. 8'. 11. 40'; 120'. 13. 13 nickels; 39 dimes; 36 quarters. 16. 80 bu. 17. 8f hr. 19. 3H da. 21. 2$ hr. Excrcist 24. Page 72 L .05. 3. .0375. 6. 1.263. 7. 7%. 9. 2%. 11. 135%. 13. 8.32. 16. 37*% of 200. 17. 175% of 200. 19. 452.9, approximately. 21. 560 dimes. 23. $22,000. 26. 75 Ib. at 70 ff; 25 Ib. at 50*f. 27. 20 gal. 29. Approximately 88.9 bu. at $1.25 and 111.1 bu. at $.80. 31. 3 gal. 33. 58&%. Exercise 25. Page 74 1. 21^r ft. from fulcrum on other side. 3. 63& Ib. 6. 8 ft. from fulcrum on side of 40 Ib. weight. 7. 69& Ib. Exercise 26. Page 77 1. 50 m.p.h. 3. At end 7| hr. 6. 16$ sec. til 7. ~ sec. 9. 310 m.p.h. 11. At end 10 yr. x > 13. 1792 mi.; 7 hr. and 28 min. 16. 1306$ mi.; 7 hr. and 28 min. 17. Approximately 8.13 hr. 19. At lOft min. after 2 P.M. Exercise 27. Page 80 L $180.00; $5180.00. 3. $48.00; $3048.00. ft. $159.00. 7. $2914.98. 9. $42,857.14. ANSWERS 295 It $1000.00. 18. $5000.00. 18. At 5%; gains $17.86. 17. $4000 at 5%; $3000 at 4%. Excreltt 28. Past 83 1. rfc 5. 3. 11. 5. db $. 7. 3. 9. 9. 11. 14. 18. |. 16. i. 17. |. 19. x*. 21. a. 28. a. 25. 2a*. 27. 2^. 29. 7*. 81. 8*c*. 83. 7*A 36.?. 37. |- 39. ~- 41. Extreist 29. Past 86 1. 15a - 20t>. 3. 4a6x - a*6x. 6. c* ^ 7. a* + 2ay + y*. 9. 16 - y. 11. 9 - 13. a 4 - 96*. 16. a* - 60 + 8. 17. x* + lOx + 25. 19. 4a* - 20a + 25. 2L 4z - 4u* + M* 23. 4a* -f 4a6 + #. 25. 6 + 5x + x*. 27. x* + 4z - 45. 29. o + 5a6 + 66. 81. 6x* + 17 + 12. 33. 8y - IQxy + 3a?. 36. 6y -f y - 15. 37. 21u* + 29u> - 10. 39. 8x + 6xy - 9j/. 41. - 12 + Ifo - 5x*. 43. - 3aj* + 19x - 20. 46. x* + 4a* + 4. t 47. 4*V - 12xy -f 9y*> 49. 9 + 246x + 16Wx*. 61. a;* - z + i 63. i - ^ + 4. 66. <AP - 9z*. 67. a? + .3x - .1. 69. 6 + 1.1* - .1*. 61. io - #*. 63. .08x* - .2&r - .15. 66. 16 - 8a* + x. 67. - 14oz + 21x + 7x*. 69. 6 - 7x - 20x*. 71. 4s* -f 12xy + 90*. 73. 4s* - Sxy + 4y*.' 76. lOOc* - 300cd + 225(P. 77. 12x* - x* - 6. 79. 2a* - a*6* - 15M. 8L 21a + a6 - 10M. 83. 12x - sV - 6. 86. 9u - 15w*e - 14. Extrclst 30. Past 88 L x* + y* H- 2xy + 4 + 4x + 4y. 3. 9 - 12* -f 6y + 4x* - 4xy + &. 6. 9x + !/* + 25 + 6xy -f lOy + 30x. 7. 16o* -f 6* + c* - 806 - 8oc + 26c. 9. 4z* - 12ox + 12M* + 9o* + 96* - 18o6*. 11. x* + 2xy -f y* - 9. 13. 16 - 4a* - 4o6 - 6*. 16. a + 2ab + 6* - x*. 17. 9a* - Gay + i/* - 16. 19. x* - y -f 2y* - *. 21. 4x* + 4xy + y* + a* - 60 + 9 + 4ax - 12x + 2ay - 6y. 23. 4x* + z* 4x* + y* 4y + 4 + 4xy 8x 2yi + 4z. 26. 4a + 4xy + j/* - * -f 6 - 9. 27. c* - 4cd + 4d* - a* - 2ax - x*. 29. 4a* + 96* + 16c* + 12o6 + 16ac + 246c. 31. to* + 25x* + 9a* - lOua + taw - 30ax. Exttcist 31. Past 90 t x(3 + 6). 3. 2x(3y + a). 6. y(2c + <P + 1) 7. x(36 - a + c). 9. (<-<#- 4a). 296 ANSWERS 11. ay*(3ay - 2 + !/*). 13. vto*(Wx - 6 + 5wx 2 ). 15. (w - z)(w + 2). 17. (8 - xy)(8 + xy). 19. (2x - y)(2x + y). 21. (6d + ll)(ed - 11). 23. (2a + 36)(2a - 3b). 25. (16a + l)(16a - 1). 27. ($ + u>)(i - u>). 29. (5w + cd)(5w> - cd). 31. (6a6 + 8x)(6a6 - 8x). 33. a(x - y)(x + y)(x 2 + y 2 ). 41. (x + 6) 2 . 43. (a - I) 2 . 45. 12>; (w - 6) a . 47. (x - 9) 2 . 49. (7x + a)'. 51. (8 - a6) 2 . 53. 12x2; (2x + &)'. 65. 20acd; (2cd - So) 2 . 57. (3x - 5y)*. 59. (2x 2 - 7) 2 . 61. (2a 2 - St 2 ) 2 . 63. (72 - 26)(72 + 26). 65. 4w(3t; - w)(3 + w). .67. (x - 5j/ 2 ) 2 . 69. x(2o - I) 2 71. 2(7u 2 - 5)(7w 2 + 5) 73. 25(x - 26 s ) (x + 26 s ). 75. (4x 2 + 25^) (2x - 5t>)(2z + 5t>). 77. 2(3w - 5). 79. 3(7x - 5y^)(7x + Sw 2 ^). 81. 4(100). 83. 1600. 85. 280. Exercise 32. Page 93 L (x + 5)(x + 3). 3. (a - 6)(o - 2). 5. (x - 5)(* - 3). 7. + 7)(* - 3). 9. (x - 6)(x + 3). 11. (w - 6)(w + 8). 13. (5 + u>)(8 - W). 15. (6 - w) (4 + w). 17. (8 + y)(4 - y). 19. (9 + Jb)(6 - Jb). 21. (x 4 - 12) (a; + 6). 23. (5a + 7) (a + 1). 25. (50 - 3)(2x - 1). 27. X 2 (4x - 3)(2x - 1). 29. y(Zy + 5)(y - 1). 31. (3x - 5)(x + 2). 33. (4w^ + 3)(2w>' - 3). 35. (5a 2 - 7)(3a 2 -f 4). 37. (7 + 2s) (1 - 3x). 39. (1 - 3x)(9x + 2). 41. (3x + 2)(x + y). 43. \2w + 5z)(4w - 3). 46. (6u/ + u)(2w - 5u). 47. (3a - 56)(2<z - 6). 49. (lOa + x 2 )(10a - x 2 ). 61. (x - 2y)(x + 2y)(x 2 + 4y 2 ). 63. (8a - 3c) 2 . 66. (4 - 3x)(2x + 5). 67. 2x(x - y)(x + y). 59. (3x 2 + 2y) 2 . 6L (5x + 106)(5a: - 106 2 ). 63. r(2 - 5ft) (1 - 3ft). 65. (| - 2y)(* + 2y)(^ + 4j/ 2 ). 67. Prime. 69. (3x 2 + 5)(x 2 - 4). 71. 2 4 (5w - 2)(5to + 2)(25to + 4). 73. (2y + 2)(2y* - *). 75. - (3a - 66). 77. (2x 2 - 5)(x* + 3). 79. (3a 2 - 5s/ 2 )(a 2 + 3y 2 ). Exercise 33. Page 95 1. 2(x + 2y). J 3. (c + d)(x + y). 5. (2ft - 3&)(m - 2). 7. (2d - 5c)(r + ). 9. (3ft - l)(w - 2). 11. (3a + 26) (w - 2k). 13. (a H- 6)(3c + d). 16. (c + 3d)(r - ). 17. (2x + y)(c - d). 19. 4(x - 6)(ft - 2c). 21. (x - 2)(x - l)(x + 1). 23. (x + 2)(x 2 + 1). 25. (x - 3)(x 2 + 1). 27. (a - 3)(a 2 + 1). 29. (3x - 2)(x 2 -f 2). 31. (2 + x)(r - a). 33. (x - s - 3)(x + 8 + 3). 35. (22 + w - y)(22 + w + y). 37. (c - 3d - 2x - y)(c - 3d + 2x + y). 39. (2 + 1 - 3x)(2 + 1 + 3x). 41. (y + z + 2x)(y + z - 2x). 43. (2o - 3* - l)(2a + 32 + 1). 46. (3x - y + 2)(3x + y - 2). 47. (4o - 1 + 3x)(4a + 1 - 3x). 49. (6 + c)(x - y)(x + y)(x 2 + y 2 ). ANSWERS 297 51. (z 2 - w)(s 2 4 w - 1). 63. (r 4 3t - a - 6)(r 4 3< 4 a + 6). 65. (c + 2 - 3d - h)(c 4 2 4 3d + A). 67. (3x - y - 5o 4 5) (3* - y 4 5a - 6). 59. (a + 6 + 3x)(a + 6 - 3x). 61. (2o - 36 4 2x + y)(2a - 36 - 2x - y). 63. (2x - 3y)(2x + 3y)(4x 2 + 9y' + 1). Exercise 34. Page 98 1. x 9 - xy 4 y 2 . 3. a 2 - 3a6 + 96 s . 5. c 4 w>. 7. 27a 3 - c 8 . 9. 1 - 27x 8 . 11. 6" - 8x. 13. (d -y)(d? + dy + y 2 ). 15. (y - 3)fo + 3y + 9). 17. (1 - tO(l 4^4^). 19. ( 4 10)( 2 - 10 + 100). 21. (1 - 3)(1 + Zx + 9a; 2 ). 23. ( 26. (6x - i/z)(36x 2 + 6x2/2 + 2/ 2 2 ). 27. (7a - 29. A 8 - 3h?k 4- 3/iA; 2 - A; 8 . 31. u* -f 9u + 27u -f 27. 33. &c + 12m 2 + 6w*c + w*. 36. 64z + 48z 2 y + 12xj/ 2 -f 37. a 6 - 6a 4 x + 12a 2 x 2 - Sx*. 39. c - G^c 2 + 126c - 86. 41. 8c - 36c 4 z 4- 54c 2 z 2 - 27z 8 . 43. (x + 2)(x - l)(x 2 + * 4- l)(x 2 - 2x + 4). 46. (2x - 37/)(4x 2 + Qxy + 9y 2 )(x + y)(x 2 - xy + y 2 ). 47. (a - 49. (w - 3x) 3 . 61. (c - d - a)(c 2 - 2cd + ffi + <w - a^ + a 2 ). Exercise 35. Page 99 1. (a 2 + a + l)(a 2 - a + 1). 3. (3a 2 + 2a + l)(3a 2 - 2a + 1). 6. (z 2 + hz + /i 2 )(z 2 - ^z -f /i 2 ). 7. (2i0 2 42aw>43a 2 )(2t0 2 -2au>43a 2 ). 9. (5a 2 4 5ab 4 26 2 )<5 > a 2 - 5a6 4 26 2 ). 11. (x 2 - 2x 4 2)(x 2 4 2x 4 2). 13. (z 2 4 4Az 4 8fc 2 )(z 2 - 4^z 4 8A 2 ). 15. (9z 2 4 12xz48x 2 )(9z 2 - 12xz48x 2 ). 17. (3a 2 4 2oc - 2c 2 )(3a 2 - 2ac - 2C 2 ). 19. (5a 4 3y)(a - y)(5a - 3y)(a 4 y). 21. (3x 2 4 3xy - 5y 2 )(3x 2 - 3xy - 5y 2 ). Exercise 36. Page 101 1. (2ob) 3 . 3. (2a6) 4 . 5. (5x 2 y) 8 . 7. (4 9. (a - x)(a 4 x)(a 2 4 * 2 ). 11. (2 - w)(2 4 w)(4 4 t^ 2 ). 13. (x< 4 y 4 )(^ 2 4 y*}(x -)(* + y). 15. (3 - 2x)(3 4 Jto)(9 + 4x). 17. (u - l)(w 4 l)(u 2 4 w 4 l)(w 2 - u + 1). 19. (x - 2y)(x 4 2y)(x 2 4 2xy 4 4y 2 )(x 2 - 2xy 4 4y). 21. (x 2 4 l)(z* - x 2 4 1). 23. (x 2 + 9)(x -W + 81). 26. (4 4 o 2 )(2 - a)(2 4 a)(16 4 a<). 27. (a 4 b)(a 2 - ab 4 6 2 )(a - a 8 6 8 4 6). 29. (3x 2 - y)(3x 2 4 3/)(9x< 4 J/ 2 ). 31. (5 - 2x)(5 4- 2x)(25 -f 4x*). 33. (a - 26) (a 4 26) (a 2 4 2a6 4 46 s ) (o 2 - 2a6 4- 46 s ). 36. (2a - 3x 2 )(4a 2 4 6ax 2 4 9x). Exercise 37. Page 103 L x 6 4 x*y 4 sV + V 4 xy 4 4 y 6 . 3. x* - 2xy 4 4xV - 8xy 4 6. a 4 o^ 4 aV ay 8 4 y 4 . 298 . ANSWERS 7. xi 9. 2* + w + w^ + w*z + to 4 . It x 4 + z + x* + x + 1. 13. * - xty + *V - aty 4- *V ~ *Y + *V ~ V + *y* - V*. 15. a* + 2a + 4. 17. x 4 + y 4 . 19. a - a*b + a6 - 6. 21. x 4- xy 4- tf. 23. z - xV 4- *V - y f - 25. 4s> + 6x -f 9. 27. z 4 + 2V + 4x*z* + tote 4- 16x. 29. (a - t0)(a + w)(o + c). 8t (t* 88. (1 - y)(l -f y)d + y + i/)d - y 85. (w - )(ti* + v + )(ti + t*V + ). 87. (2o - l)(16a* + 8a + 4a + 2a + 1). 89. (2 + *)(64 - 32x + 16x - 8x + 4x* - 2x 41. (a - 3a?)(a' + 3aa; + 9x<). 48. Prime. 45. (ti* + t>)(w tt -!* + *). 47. Prime. 49. (2 - *)(4 + 2x + x)(64 + &r + x<). 5L (u* Extreist 38. Past 105 1. 3. f. 5. 7. ^4^- 9. 11. ~- a 13. ^-4-r- 15. ^-T^:- 17. 23. 2 2s - ' x - 2y 2 3x 4- 2y !*-*a. 27.--?- 29. ^. a 4- ^ * 4- y c 3 x ^ x4-3 5 2x 31. r-r-r-* 83. - . , 35. , . n -rr- w * 26 -x Excrcis* 39. Pag 107 ( + 3)(* - 2) 4z - 9 2o - 4 7 . . ,. * 1 ^r 4o6 15(o - 6) 15(x - y) 17. - 4y 6rf - 2c 3(4x - 13s - 2x + 10 - 19x + 4 6a - 6 -6x * (2x- l)(3x + 3)' 2o - 3 - 3xy + 8 * ' ' *"* 2x-2y 6(a-n) * x + x - 12 14 + 2n 6c-5c-H30 Iftc 8 + 36z -f 45 %FflV^ f^ f 4 \ r * * \ Vv^r ^^ / A ^^ \ X A ^V V ^FUi 3(1 - n)(n + 4) 2(3c - 2)(c* - 9) (3 - 2x)(8z - 27) 37. 43. ANSWERS , 299 12x + llx - 25s - 9 . 13* + 18x - 3 - 4)(2z - 3) 3(x + l)(2z - 3) 2(9 - 4x)(x - 3) 9or - t* - 60* -h 81o 4 - 9or* -j-'3or* - 27or (r - 3a)(r + 3a) Extrcist 40. Past 110 + 4) Li 3.-- 6. 9. (2a -f 36) (a* - 06 + 6*) v 'b-a 17 . Z 19 . . 2 L a + 36 12 2y + 3x 2 23. - lOsy 1 w 3x + 4 2y(y -r ~ y i ~ / _ 29. 3L S3. 2x(x + 4) 36. , ' ,- 37. * v * ""^ 39. ^ c(n v) y + 5 4L^ ac 36c a 8 6* 2 4a -6) jiT v***' i^ yy / y^v i **y / MQ * 7 - - _ y) 2 -x 3a-26 (3a - 2)(2a - 2a(5a - 1) (5o - 3)(3 - 3a) ExrcUt 41. Past 115 1. 14. 3. - 5. 6. i 7. - 11. 9. 3. 11. 2. 13. - 2. 16. 5. 17. 4. 19. - f . 21. &. 23. - 5. 26. 1. 27. 3. 29. tf 31. 4 hr. 33. 380 m.p.h. 36. 15 m.p.h. Extrcist 42. Paft 117 2M-Ja; 3. 3a -f 6. 6. ^^. 7. 2n. <5 * t _1_ M K C rf 13. 2a. a . 2o6 6 a 17. - 19. i 21. 26. 23. r Extreii* 43. Pg 118 1. 0s* - 250 1 . 3. 4x* 4- 12x H- 9. 6. y 4 - 6tcy* 7. o - 64. 9. (y + 5*)(y - 5). 11. (i - 4y). 300 ANSWERS 13. (o - 36) (a' + 3o6 + 96*). 15. (3y + 2 2 ) 2 . 17. (2 + 7)(f - 3). 19. (4x + 1)(2 - to). 21. 5(i 23. 2(a + 2)(a + l)(a - 1). 25. (x - a - 36) (x + a + 36). 36x-f 6 2a - 6 - 5a6 - 56 a Ol. . . - - 2y 3x x* a 2 6 2 83. * " * 36. -*. 37. -2. 39.-^-- o 3 c 1 a + 6 Exercise 44. Page 120 13. (5, 1); area = 40 sq. units. 16. 9 sq. units. 17. 10 sq. units. 23. All abscissas are 2. 25. 4 units. Exercise 45. Pase 124 1. (a) 8 and 4; (6) - i, - f, and 0. 15. (6) Equals if x = 4.4 or 1.6; equals 10 if x = 6.5 or .5. Exercise 46. Page 1ST L 7. 3. - 1. 6. f . 7. - 33. 9. |. 11. 4c - 12C 9 . 13. 9; 6* - b + 3; c< - c 2 + 3; s - 5x + 9. 15. 4; 4; H*; (x + 2y)/(x - y). 17. - 5; 27; c 2 + 66c. Exercise 47. Page 130 19. x 5; x = 4. 21. Cuts x-axis at (5, 0); y-axis at (0, 3). 23. Cuts x-axis at ( f, 0); #-axis at (0, f). 25. y = fx - Jgk Exercise 48. Page 132 ATote. In this answer book, in any solution of a system of equations, the values of the unknowns will be arranged in their alphabetical order. 1. (- , - If). 3. (2, 5). 5. (- 2, 3). 7. (- 2J, - f). 9. (1, f). if. No solution; parallel lines. 13. No solution; parallel lines. 15. Infinitely many solutions. Exercise 49. Page 134 1. (3, 2). 3. (- 1, - 3). 5. (0, - 4). 7. (2,2). 9. (- f, |). 11. (i |). 13. (2, f). 15. (5, 2). Exercise 50. Page 135 1. (5, 2). 3. (7, $). 6. (0, 0). 7. <0, 0). 9. (3, - 2). 17. (- ff, - V). 19. (.42, .19). 21. (- .35, .27). 23. (2, 3). 25. (5, - 3). 27. (3, 2). 29. (5, - 3). Exercise 51. Page 136 !(-!,- i). 3. (2, 5). 5. (i, - |). _ /2 6\ /a 6\ /3M-* M \ 7 ' la' 2 j ' *' W a) ' U ' UM^' " 9AT2fc/ ' 13. (26, -3o). 15. (a + 6, 6 -a). 17. (m - n, 2m "" 2n V ANSWERS 301 Exercise 52. Page 138 1. (1, 2, - 2). 3. (- i - f, |). 6. (f, - f, f). 7. (ft, - A, A)-' <- 2, 3, 3). 11. (i i - i). 13. (- 1, - 1, 3, 2) Exercise 53. Page 140 1. 30; 120. 3. 41i; 48*. 5. 5 gal. of 20%; 2 gal. of 50%. 7. 11' by 3'. 9. 1st, 3 lb.; 2d, 6 Ib. 11. 13, or 26, or 39. 13. 40 lb. silver; 80 lb. lead. 16. $3000 at 3%; $2500 at 4%; $4500 at 6%. 17. 465. 19. y = - 4x - 11. 21. y - - Js + 2. 23. ^ = 2. 26. Land, 90 mi; water, 48 mi. Exercise 54. Page 143 1. 32. 3. - 243. 6. &. 7. Minus. 9. x*+*. 11. z 16 . 13. 32a". 16. 625zV- 17. - 8z. 19. 16a. 21. a 2 *. 23. d 2 **. 25. c*cP*. 27. .09cd. 29. - Zr 1 <w <M w 31. 33. SO. 37. d 3 4 6 s a* -IS- - _ __ > a 2 " 64s 2 w*c* 200x 66. (a) 16; - 16: (6) n odd. Exercise 55. Page 147 1. 8. 3. 9. 6. =fc J. 7. =fc .1. 9. 12. 11. i 13. f 16. - 3. 17. 5. 19. - 6. 21. 3. 23. 5. 26. =fc . 27. d. 29. 3. 31. 3. 33. 57. 36. 4zy 3 . 37. 6. 39. - 2. 41. 2. 43. 4. 46. 2. 47. - 1. 49. 6. 61. 20. 63. 20. 66. i 67. - i 69. .1. 61. .1. 63. .2. Exercise 56. Page 150 1. 6. 3. a. 6. x*. 7. *. 9. y 9 . 11. x 9 . 13. z 8 . 16. 2y. 17. 2y. 19. J. 21. . 23. &. 26. f. 27. 3s*. 29. - 2z. 31. x*y*. 33. 2a 2 . 36. - .1. 37. 2xy*. 39. - 2. 2x 41. xz*. 43. .2rc. 46. .5x. 47. ^ w . ^-^ 2y Exercise 57. Page 153 1. 3. 3. 2. 6. J. 7. &. 9. 8. 11. i. 13. sV. 16. i 17. i 19. 1. 302 ANSWERS 21. i 23. 3. 26. .6. 27. i 29. &. 31. -^ r . 33. - 1. 35. - i 37. - 5. 39. 10. 41. 125. 43. 216. 45. 625. 47. i 49. 16. 1 y KR * K* 3 65. 57. TT d 1 A 4 ML*. 6L a ^J^f9 * ^ 65. 6. 67. ac. 69. J-- y* 4x* 5a "-TO 78 C n. ^ vv 125 8 ^ M 18 J I / f o * * * I e? * a*c*d< (k* oa' 9 ay 81. r*. . 83. 5y-. 85. 5y~. 87. 4(3- 1 a*x-'y-'). 89. 8x*y -V. 91. 3(1. 04)-". 93. c(x - 5y)-. 95. </z. 97. <^6*. 99. 5^c. 101. b#x*. 103. 6*. 105. ^6c. 107. V5x\ i/. 109. ^49a 4 . 111. ai 113. 5*ci 116. (a* - 36)*. 117. (c - 3d)^. 119. (a - 6)i 121. (4 - x)i. Exercise 58. Page 155 1. X*. 3. x 4 . 6. o. 7. 16. 9. 8x. 11.^' 13. 25. 15. a*. 17. - 19 - a 9 x* * 6* 8 o 9y xV * x* * X*' * ~x*' 2? * "36" 29. 8. 31. 625. S3. 35. r- 16 37. xl 39. i- a* 4L xi* 43. ~ a* aiV , 45. ^r- 47. 49. ao*. 61. - fei 2y y >- 66 ol6> 67 3 . 59. 9m. 27 " Six 61. 25x*. 63 a AfC , r . 7 6 v , ab fi h r~j~~~r i ab Oifw^ , fid 71 , iy TK * 1 _ I JL _ c 1 o * i i i . ^ a 6* -f- oo -f- a* o t* 3a6 77 ^ - 79. x- - - y" 1 . 8t 16x* - yf . 88. x* - y* "' ., _L 1 a + c >* 85. 15aH - 14a* - 8. 87. a" 4 + 2a-*& + &. 89. at + 2ai6* -H 6*. 91. o" 1 + 2a~V 4-y 4 . 93. a- + 3a"*6 + Sa'V + 6. 96. 27 - 27y l - f 9y* IT*. 97. 6~a irut ^^^ ;-* - 15x- 4 . 11 99. 3"x5. nK at fciv mir <A< >lm^.^fcMk ^L^^^Ltt *f w * w tMt w*^jWwfp /.*. . ^ ^^-1\/^ i . ._i\ 109. (3ar> - 6")(3ar + 6-). 111. (2x* - yi)(2xi -f yi). ANSWBRS 303 113. (3x* - 5y*)(3a* + 5y). 115. (2a* - 36*)(2a* + 36*). 117. (z - 3ar)*. 119- (Sa- 1 - 6-). 121. (6x* + yty. 123. (Sar 1 - 2y)(or l + y). 125. (2a* + 36*)(4a* - 6a*6* + 96t). 127. (6 - x*)(36 + 60;* + x*). 120. Sar 1 + jr. Extrclst 59. Pagt 158 1. 3^2; 4.242. 3. 2VS; 4.472. 5. 10V; 14.14. 7. 3V3; 5.196. 0. 6V2; 8.484. IL .3V6; .6708. 13. 2^2; 2.520. ^ 15. 3^4; 4.761. 17. - ^5; - 1.710. 10. - 3^2; - 3.780. 21. tf^'x. 23. y<^P. 25. x*V&- 27. 3a. 20. 2a^. 31. 33. 2ayV^. 35. ayi/lV. 37. Sy^^. 30. 41. - o^. 43. xy*V3ij*. 45. 2d^cd. 47. - 40. .5xV;. 51. V^. 53. - >^. 55. y* 2y* 57, - 4; ^2T 50. 3Vl + y . 81. aVl + 56. 63. atr 66. x*. 67. 2x^2T 60. ( + ^ ** 71. 8^/2? 73. ab 75. 3V. ' 77. (a - 5fe)V2. 70. (3 - x)^3x. 81. (2x - Exercist 60. Pigt 160 1. Vl5; 3.873. 3. 5V2; 7.070. 5. 6V^ ; 8.484. 7. 40. 0. 18\/2; 25.45. 11. 5>/42; 32.40. 13. 54. 15. 9^60; 35.24. 17. - 2^; - 2.884. 10. V7; 2.646. 21. ^9; 2.080. 23. VH. 25. ? i ^O _ _ 27. - 20. 3xV5. 31. 3*V2x. 33. 3a6< / 2a6. a 35. 375a. 37. 54s. 30. 6 s ** + 6. 41. - 9 + 7<s/5. 43. 1. 45. 18 + 13V6. 47. 5. 40. 2>/6 + VlO + 8>/3 + 51. 27 + 10>/2. 53. 14 - 4>/6. 55. a - 9fy. 57. 6x - 15y + 50. a + tor + 26Vax. 61. - xyzV*. 63. Vl8a. 65. V^te. 67. ^27V. 60. ^486. Exercist 61. Pgt 162 1. ^V2; .707. 3. i\/10; .632. 5. iVlO; .7905. 7. J^; .630. 0. i<^; .6785. 11. - ^^7; - .2646. 13. rb"^; -05477. 16. - A^30; - .3107. 17. ri^iKJ; .1095. 304 ANSWERS 19. ' 26 35. 2c 43. - - 53. 1. $; .577. 7. ^v^; 1.291. 9 ~ 5 3 13. o ; .057. 25. 8 ~ 5 ; -.465. 2 31. i^5; .342. 87.- 2o& A 21. 29. a 23. 26 V 3z ' 5to5 25. - 3a 45. a 39. 47. ax 49. _ , 5ao 3) 61. 4V5. 63. 0. Exercise 62. Page 163 3. fV5; 2.683. 9. |Vl5; 1.549. 65. a a. 5. 11. ; 1.155. ; 1.890. 15. 3 - 2V2; .172. 17. 7 ~ e 3 6 ; - .0694. ; .689. 17 27. - i^l8; - .437. 29. - ^100; - .4642. 33. 89. 26c 35. - 41. 2V5 + 4V-3 V6 14 45. 4.900. ax* 9. 17. 25. 33. 41. 49. 67. Exercise 63. Page 166 5. 13. Vs. 21. 29. 37. 9. 45. 63. ^3. 61. 3. It 19. 27. 35. 43. 9a. 51. 59. 7. 15. V^. 23. 31. 39. 47. 56. 63. ANSWERS 305 65. v^27. 73. 81. 67. o< 75. 1. 9L 97. 105. v'a. 113. 2vl2. 119. 2(V + ^2). 121. 26 107. #Zy. 123. Vtf a 69. 77. * 3* 93. 6^ 101. 26 109. 4- 6 -2a I /-| y* 95. 3^. 103. a 111. Exercise 64. Page 168 3. 1. 5. 3>/3; 5.196. 9. . 11. fVH; 1.342. 13. 239. 17. fcVg; .306. 19. v^2; 1.260. 23. &VI6; .1265. 26. - 29. 35. 3zi 43. - a; 8 61. a 67. 63. *4 79. yVx*y. 87. - : i. (a- 31. 37. 45. 81. 2z J 89. 39. 47. 59. - 65. 71. Z</x. 73. 6v y 6. 83. -f a 6 7. 125. 15. 1^12; .572. 21. 3V2; 4.242. 33. 41. 36. 65. 4; - 3. 61. 67. 75. v*6. 77. . 8R (31 - 180)^3 91. 97. 06 306 ANSWERS Exercise 65. Pa 9 172 L 3i. 3, 6e. 5. St. 7. 5*V2. 9. 11. #. 13. IT'- 15. it. 17. .3*. 19. ft. at .6i. 23. itV5. 25. &iVll. 27. 2W. 29. obi. 31. 2isV2. 33. 5tfc\/3. 35. SisVVsty. 37. 39. frdi. 41. ~-- 43. d=fi. 46. 47. t. 49. - 1. 51. - 1. 53. 10. 55. 13. 57. - 29 4- 11*. 59. 21 - 20t. 61. 9 + 40*. 63. Si - 40. 65. 4 + 19i. 67. - 15. 69. 3*V + 10V2. 71. - 20*V - 17. 73. (4* - 19); (- 34 - 6t); (- 70t - 66). 75. .- 77. 1 + 2*. 41 Exercise 66. Page 174 Note. In simplifying radicals in the solution of an equation, it will be assumed that any literal factor of a radicand is positive if this adds to our convenience in the reduction. 1. =fc 5. 3. 3t. 5. =fc ft*. 7. $36; d= 1.183. 9. iv IL -. 13. ijl; .829. 6a 15. 30 1.826. 17. i2; .707. 19. * . . 2(1 + c) m T 33. db *V2; db .707. 35. 14. 37. iVl3; 1.803. Exercise 67. Page 176 1. 5; - 2. 3. - 4; 3. 5. 0; f. 7. 0; . 9. 0; f. 11. f 13. - 3; i 15. *; . 17. f ; I 19. - i; - i 21. - 2; - f. 23. f ; - 4. 25. - J; - f. 27. f ; - 3. 29. f ; - 2. 31. - |; 1. 33. - ; |. 35. 0; - 37. - 3&; - 26. 39.' - 6; 6. -5 ; S' ^ 'I 5 -5' 49. i 51. - 3; - i 53. 0; ; - f . 55. - 3; |. Exercise 68. Page 180 L 16; (x - 4). 3. c; (x - c). 5. |; ( - ). 7. f|; ( + *)* ^ - 7; 1. 11. 3; - 7. 13. - 2; - 2, ANSWERS 307 2 :t Vl4 1C. - ~ - : - 2.871; .871. 17. (2 + t); (2 - t). : o 2 4-*- \/TQ 23. -~-^= -: - 2.786; .120. 25. |; - 1. o 29. 6a; - 3a. 3t 6; - $6. 38. - 2 d= V4 + oc - tf - 4HP 86. - 37. 9 Extrcisc 69. P*gt 182 1.}; - f . 3. f; - J. 5. 1 3t. 7. f ; f. 2 dr V 1 4- \/2 - 1<U- 1 Rfift -- . -~ x oi v ^ v. ^ . 107 1 l.OOO. ttll:i 2 5V2. OO. ^ . ^.loo, 1. - 12ac 2a IT - - . .CMJU, .100. 19. t*. l2*Vl3 "' 4 23. .3; .5. 20 32i o 31. t; f. 37. fd; - id. JO ,7. 0^ 25 ' 2 ~26i 2 A ^ OA/ zt ^ y/c "* A^&UAC ^ii , _-. . u> 10* r. ?; - *O. a , ^C. 4ft 2 - 5 1+ V ' 51. y - x -f 2; y - ^(1 - x). 53. a; = y 2; x * 1 2y. Extrcist 70. Pag 184 L - 3; t 3. |; - f. 5. 9; - 5. 7. - f ; * 9. *; f . ' 11. i; i. 13. 2 ^ 5 ; .847; - .047. 15. 0; f- o 17. iA'; .905. 19. =fcfc6. 21. |; - 1. 23. 6 \/41: 12.403; - .403. 25. 4; - f. 27. 6; i - h =* oo i wi A 01 L 11 oo 29. i; ^(1 b). 31. fc; JA. 33. C 35. . 37. 13' by 17'. 39. ^; - tf. 41. 14.928'. 43. 6.48 yd. ' 46. 20 m.p.h. 47. 6 m.p.h. 308 ANSWERS 49. (a) t - ; (6) s - 500' at t - 3.46 sec. and 9.04 sec.; 8 - 0' <7 at t = sec. and t = 12.5 sec. Exercise 71. Page 188 L Vertex (0, 0); axis x * 0; min. = 0. 3. Vertex (0, 0); axis x 0; max. = 0. 5. Vertex (0, 5); axis a; = 0; min. =5. 7. Vertex ( 3, 4); axis x 3; min. = 4. 9. Vertex (1, 5); axis x = 1; max. = 5. 11. Vertex (2, 5); axis x = 2; min. = 5. 13. Vertex (f, 9); axis x = f ; min. = 9. 15. Min. = - 13. 17. Max. 8. 19. At end 2$ sec. 27. 30; 30. 29. 7*" by 15". Exercise 72, Page 190 1. f. 3. 3; 3. 5. Roots imag. 7. 1.6; 4.1. 9. Roots imag. 11. 2.6; - .6. 13. - 3; - 1. 15. (0, 2|); (1, 1*); (3*; - 1). Exercise 73. Page 192 1. Disc. = 9; real, unequal, and rational. 3. Disc. = 12; real, unequal, and irrational. 5. Disc. = 0; real, equal, and rational. 7. Disc. = 1705; real, unequal, and irrational. 9. Disc. = 0; real, equal, and rational. 11. Disc. a 16; imaginary and unequal. 19. 5.2; 1.2. 21. 1.2; .2. 23. Disc. 49; graph is a parabola concave upward, with its axis perpendicular to the 3-axis, cutting z-axis in two points, and hence the vertex is below that axis. 26. Disc. = 59; graph is a parabola concave downward, with its axis per- pendicular to x-axis, which does not meet that axis and hence lies entirely below it. 27. Disc. = 52; etc. 29. - 2 + 5i. 31. - 6*. Exercise 74. Page 195 1. - 5; - 3. 3. f; f 5. - f ; - |. 7. 0; - tf. 9. - |; - f. 11. f ; - f 13. - ; - |. 15. y - |. 17. - r-|-; ^4-- 19 - fir; fr^- 21 - te ' + 1* - 3 - . 5 + o 5 -h a 1-fcl+c 23. x* + 2x - 7. 25. x 2 + 5z -f 6 = 0. 27. 3z - 7x + 2 - 0. 29. 3z - x - 10 - 0. 31. z 8 - 2 = 0. 33. x - 18 - 0. 35. 9z 2 4- 4 - 0. 37. x 8 -f 4z - 1 = 0. 39. 2a - 2x - 13 = 0. ANSWERS 309 41. x 9 - Ox + 34 0. 43. a;* - 8z + 20 - 0. 45. x* - 4x + 24 0. 47. 3z + 4a + 2 - 0. 49. (ftc - 8)(3x + 5). 51. (8z - 15)(3z 4- 4). 53. 27x a + 12x 32 =. 0. 56. No. (Disc, is not a perfect square.) 57. (x + 3 + i)(x + 3 - i). 59. (x - $ + ft) (a; -*- Exercise 75. Page 197 1. 1; 2. 3. fc2; 2. 5. 2t; =fc . 7. 3i; =t 9. |; db i. 11. i? 1. 13. d= 2i; =fc iV. 15. 5; 1. 17. - 1; - f. 19. 3; 2; - 1. 21. - 1; f; 1(1 =t t 23. - *; - |. 25. 1; - 4; *(- 3 =t V5). 27. 1; - 3; *(3 V5l). 29. 1; 3; - 2; - 4. 31. 2; - ^2. 33. ; - 1; *(1 tVf). 35. d= 2Va; =b i^ea. 37. ; J(~ 1 tV5). 39. 3; *(- 3 db 3tV3). 41. f; fi. 43. - ; ^(3 3iVi). 45. 4; (- 2 VS). 47. - 1; J(l db iV). 49. i; i(- 1 =b *V). 51. 1; =t . 53. 3; 3i. 55. 2; 2t. 57. ; d= ft. Exercise 76. Page 200 1. 7. 3. No sol. 5. 12. 7. - 13. 9. 14. 11. f >/2. 13. No sol. 15. 4. 17. 9. 19. 4; - 2. 21. 0; f V5. 23. 0. 25. No sol. 27. 3; - 1. 29. i 31. 4; |. 33. 1; 2; - 3. 35. o. 37. 0; 46. 39. ; - 41. 7T t 43. 1; &. 45. 8; - ^-. 47. 16; if. 49. 16. 61. =t ^V2. 53. 4. 65. No solution (any principal root is positive). 57. 243. 59. - 243. 61. - 25$. 63. ^4 (x~% = 8 has no real solution). Exercise 77. Page 203 1. db 1 3. 2. 6. 0; - J. 7. H (& = 1 is not a solution because it does not give a quadratic equation). 9. 10; 2. 11. ^; k does not apply. 13. - .268; - 3.732. 15. - . 17. &. 19. - i 21. if. 23. - &. 25. =fc 3V5. 27. - f . 29. - f. . 31. =fc .816. 33. 1; - 2. Exercise 78. Page 206 1. a 6 + 5o6 + 10a6 J + 10a*& + 5a& + 6 s . 3. x 8 - 8ofy + 28zV - 56x 5 y + 70x 4 y 4 - 56xV + 28a;V - 8xy -f y. 5. 16 + 32a 4- 24a 8 -f 8a + a. 7. 7296 - 14586 B y + 12156y - 540&V + 1356V - 18^ + V*. 9. a -f So^ + 3a6* + 6. IL o - 6a w 6 4- 15a6* - 20a6 -f 15a6 - 6a'6 l <> 13. x - 310 ANSWERS 15. a* - 17. a* - 4oir H- . , 8x* , 24x 32x* 19. x - + --- ~ a* a* a* a 25. a" -f 20a& -f 190a6. 81. 2x* - 30 2ox + 435 / _ ^ ^ \ 85. * n - 89. 720. I5xy* - - 40IT 6 + IT 8 . 16 27. 1 - + 2.31. 87. a*" - Ig0ou lg ^ 29. 1 - 12 >/2 -f- 132. + / ^^ -j \ 41. 39,916,800. Exercise 79. P3t 208 8. 35z- 5. - 48. 126. 7. 9. - n(n-l)...(n~5) 15. 21. 6! 11. .00056. ^ 18. 15x*. ; 2016xy 10 . 28. 3/ 25. 27. 126o>6; 29. 10,000 - 4000o + 600a - 40a + o. 3t 96,059,601. 83. 132,651. 36. 1.127. 87. 1.230. 39. .904. 41. 1.243. 48. .002. 45. 14,776,336. Extrcls* 80. Page 211 1. f . 8. f . 5. s/a. , 7. 11. *. 18. 30. 15. f. 17. 10"; 8'. 2L 32'; 21$". 28. 5500 sq. in. 25. 5.366'. 27. 28.7'. 81. d= 1. 88. 25. 85. db 9i. 87. $ 9. 19. 20; 70. 29. db v^6. 39. d= (y + 3). ^ oe 41. 35. A 48. -T- 3 Am 45. 27 77 175 . 47. 9m 4 5n 7. Exercise 81. Ps 217 8.Z- . V 1 9. T a* 15. u varies directly as x 9 and y. 25. ^. 27. 784'. 29. 2700 Ib. 18. z is proportional to x*. 12t/ 21. - -r^- 28. H - - ox 81. 4| ft. 88. (a) 71f Ib.; (6) 7 sq. in. 87. 5* ft. in diameter. 89. 1824 r.p.m. 4L /i : /, 1:16. 48. 56%. 45. (10, - 6, 4). 47. (6, - 2, 4) or (- 6, 2, - 4). 49. (150, 250, 550, 300). ANSWERS 3?T Exercise 82. Page 223 L 15; 18; 21; 24; 27; 30. 3. - 18; - 16; - 14; - 12; - 10; - 8. 9. 13. 11. 14. 13. 151. 16. - 76. 17. 10. 19. I - 78; S - 645- 21. i - - 72; S - - 882. 23. I - .78; S - 46.86. 25. d - 16; 8 - 5460. 27. n - 92; 5 - 18,308. 29. a - 138; S = 2025. ^ 31. n - 21; 5 = 525. 33. a - - *', I" 126i . ' 35. k - - 1. 37. 21. 39. 136th. 41. |f. Exercise 83. Page 226 1. 5; 8; 11; 14. 3. 8; 10; 12; 13; 15$. 5. 10.5; 6; 1.5; - 3; - 7.5; - 12. 7. - i; 1; f ; ; ^. 9. 26. 11. - 19. 13. 36,270. 15. 2223. 17. 376. 19. 360'. 21. $11,650. 23. $30,500. 25. 2565 Exercise 84. Page 230 1. 5; 15; 45; 135. 3. 4; - 8; 16; - 32. 7. &; ^. 9. or 4 ; or*. 11. 1.01; (1.01) 8 . 13. 4. 15. f. 17. 729. 19. &. 21. I - 2916; S = 4372. 23. I = - 1215; S = - 910. 25. I - 192; -S - 129. 27. Z*>= 3846'; 8 - 29. W. 81. n - 8; 1275. 33. n = 6; 5 - 111.111. 35. a - 25; n - 5. 37. n = 11; S = ^^i. 39. i = 80. 41. f. 43. (4; 8; 16; 32; 64) or (- 4; 8; - 16; 32; - 64). 46. (12; 36; 108) or (- 12; 36; - 108). 47. 1; 10; 100; 1000; 10,000; 100,000. 49. 2. 61. 10. . _ . M 05^ 7 1 63. VJJ, if x > 0; - v^ if x < 0. 66. ^^- -- _ (1.06)> - (1.06) 4 eA . 1 - (1.02)-" (1.02)" - 67. - :rr - 59. - -- Ol. . . .06 .02 ( 1>0 2)i 63. - A; 2. 66. 8190. 67. $102.30. Exercise 85. Pase 233 1. $95. 3. 227.8" approximately. 6. $3125. 7. *n(n - 1). 9. $7020. U. 1 x 13. 55.339" approximately. 16. At end 2 yr. 19. 599.59' approximately. 21. (a) 300(1.06)*; (6) 5060 units. . 23. Approximately 11.1% per year. 27. 12i%. Exercise 86. Page 236 1. i; t; il A- 3. A; i; f; f. 6. 1; f; f; A; 7. 2. 9. 16. 11. 2*y/(* + y). 372 ANSWERS 1. 14. 13. * . 25. 1. 3. 15. &. 29. 3. 43. 10. Exercise 87. Page 239 3. 22i 6. I 7. flft. 9. i iL 1K _7_ r 1T JL 1O 5 Q1 19 QQ * w 33 * 33 ** TT * IJL * 90 *^ 27. 3^. 29. 1500". 31 200 sq. in. 33. 12. Exercise 88. Page 241 3. - 1. 6. 64. 7. 81. 9. 10. 1L 1. 13. A 17. 2. 19. 2. 21. 1000. 23. 6. 26. 2. 27. 2. 31. 4. 33. i. 35. $. 37. - 1. 39. - 3. 41. 5. 45. 1. 47. 2. 49. 8. 61. 100. 63. 64. 1. .7781. 9. 1.6232. 17. - .5229. Exercise 89. Page 244 3. 1.5314. 6. 1.4771. 11. .3680. 13. .7533. 19. - 2.1549. 21. - 1.7696. 7. 3.2304. 16. - .3853. 23. - 1.3768. Exercise 90. Page 247 1. Ch. = 2; man. = .9356. 3. Ch. = -2; man. = .700. 7. Ch. = 6; man. = .325. 11. 4.4932 - 10. 13. 5. 19. 1.6355. 21. 7.8949 - 10. 27. 8.9345 - 10. 29. 5.0043. 35. 4660. 37. 1.43. 39. 74.0. 45. .0960. 47. .000900. 6. Ch. = 3; man. = .5473. 9. 9.2562 - 10. 15. - 4. 23. 0.9759. 31. 5.1959. 41. 302. 49. .264. 17. -6. 26. 4.2504. 33. 243. 43. .00589. 51. .00500. Exercise 91. Page 251 1. 3.2615. 9. 9.7503 - 10. 17. 6.0910 - 10. 25. 1379. 33. 7.695(10 8 ). 3. 2.7261. 11. 8.1939 - 10. 19. 3.4950. 27. 39.95. 36. 1.030. 6. 1.5556. 13. 4.9546. 21. 1725. 29. .0002162. 37. .00009738. 7. 9.4790 - 10. 16. 7.1581 - 10. 23. 1.459(10). 31. .4693. 39. .4236. Exercise 92. Page 253 Note. In some classes, the teacher may desire to teach the use of 5-place log- arithms. For the advantage of such classes, in the case of each computation problem in the remainder of this chapter, the result obtained by use of 5-place logarithms is given in black face type beside the result found with 4-place log- arithms. 1. 24.91; 24.909. 3. .2009; .20086. 5. .006380; .0063797. 7. - .007667; - .0076660. 9. 51.10; 61.098. 11. .1406; .14061. 13. 24.56; 24.668. 16. .07808; .078096. 17. 5542; 6644.4. 19. 27.61; 27.609. 21. .003467; .0034669. ANSWERS 313 23. - 2.627(10-'); - 2.6266(10-*). 26. 1.580(10-'); L6802(10" 6 ). 27. 38.96; 38.966. 29. (a) 4.792(10 6 ); 4.7922(10 8 ): (6) 8.065; 8.0662. Exercise 93. Page .256 L 5358; 6369.6. 3. .4107; .41082. 6. 1.044; 1.0440. 7. .9500; .94986. 9. 1.315; L3158. 11. .6030; .60296. 13. 28.93; 28.936. 16. .1585; .16849. 17. - 1.010; - 1.0099. 19. 50.32; 60.324. 21. 41.47; 41.470. 23. .1266; .12668. 26. 2.111; 2.1111. 27. 1.041; 1.0412. 29. .8630; .86268. 31. 50.12; 60.466. By preliminary use of 7-place table, the results are .5050; .60604. 33. 141.9; 141.82. 36. 215.1; 216.08. 37. .4971; .49714. 39. .001352; .0013626. 41. .9388; .93896. 43. .3986; .39882. 46. - 1.916; - 1.9166. 47. 21.76; 21.768. 49. 134.9; 134.84. 61. - .136;* - .1366.* 63. 1.118; 1.1177. 66. 4.908; 4.9086. 67. .1730; .17294. 69. By 4-place table: (a) 2.219(10 4 ); (6) 3.222(10- B ). 61. .02323; .023229. 63. .0007867; .0007869.* ' 66. 236.1; 236.13. Exercise 94. Page 259 1. 1.341; 1.3410. 3. 1.319; 1.3194. 6. - 5.195; - 6.1923. 7. 18.1;* 18.02.* 9. 5.63;* 6.634.* IL 4.317; 4.3176. 13. 2.303; 2.3026. 16. - 14.2;* - 14.20.* Exercise 96. Page 261 1. $4682. 3. $4502. 6. $1203. 7. 5.8%. 9. 18.8 yr. 11. 16 yr. Exercise 98. Page 268 1. (4, .5); (- 2.8, - 2.9). 3. (5, - 3). 6. (2.1, 1.5); (- 2.1, 1.5). 7. (3.2, 3.7). 9. No real solutions. Exercise 99. Page 269 1. (3, - 4); (- 4, 3). _ 3. (5, - 3); (5, - 3). /6 2*W 6 + *V6\ /6 + 2t'v / 6 t v 5. ^ _ , _ Y ^ > ; 7. (- 1, '- 1); (- 3, 3). 9. (4, 2); (4, 2). 11. (1, *); (i 2). 15. (il); (-i3).' Exercise 100. Page 270 1. (1.837, .790); (- 1.837, db .790). 3. (|V2, ); (- |V^, ffl. 5. (V5, 1); (- V5, 1). 7. (db V2, V); ( \/2, - V5). 9. (^v^, iv^S); ( |V, iv^S). 11. (V, i^/7)', ( v^, * The result is not reliable beyond the last digit given in the answer. 314 ANSWERS Exercise 101. Page 279 1. (V2, - V5); (- 3. (i, 1); (- ft, - D;_ (- V2, ftV5); (V2, - 6. (- V3, V3); (V3, - V); (- 2, 1); (2, - 1). 7. (14, - 4); (- 4, - 1); (- 14, 4); (4, 1). . (I, - ft);- (ft, - *), (- f, i); (- ft, ft). (- ft, ft); (i - ft); (_- 2, i); (2, - i). 13. (ftV2, JV5); (- ftV2, - JV5); (2, 5); (- 2, - 5). 15. (6, 4); (- 6, - 4); (- Exercise 102. Page 274 X 1. (- ft, 5); (ft, - 3). 3. <|Vl5, f Vl6); (- |Vl5, - Vl5). 5. (- 2, 3); (6, - 1). 7. (f, - f); (- 1, - 2); (f, V); (- 1, 1). 9. (i, 2); (- i - 2); (1, 1); (- 1, - 1). 11. (2, 1); (1, 2); [ft(- 4 + ,V6), ft(- 4 - iVe)]; Cft(- 4 - Vg), J(- 4 + t-s/6)]. 18. (- 5 + *Vl4, - 5 - iVII); (- 5 - Vl4, - 5 + tVl4); (5, 4); (4, 5). 15. (ft, 2, - 1); (ft, 2, 1); (- ft, 2, - 1); (- ft, 2, 1); (ft, - 2, - 1); (ft, - 2, 1); (- ft, - 2, - 1); (- ft, - 2, 1). 17. 25. 19. c - V9 + 4m 3 . 21. c = d= V o s + 6m 8 . 23. (3, 1); (- 3, - 1); (1, 3); (- 1, - 3). 25. (i, - |); (i - i). 27. (i, 1); (- |, 1). Exercise 103. Page 275 1. (2, 3); (ft, 4). 3. (4.1, 1.8); (- 4.1, =fc 1.8). 5. (- 1.8, -2.1); (2.5, -5.2). 7. (|V65, *V35); (- f V65, ftV5). 13. (f, - 2); (4, - 7). 16. (- 1, 0); (1, 0); (V5, - 3*V2); (- *V2, 3tV^). 17. (3, 6); (- 3, - 6); (- 4^3, 5^3); (4V3, - 5V). 19. (10, - 5); (- 10, 5); (ftV, J^V2); (- ftV2, - J^V2). 21. [ft(a + 1), ft(a - 1)]; [ft (a - 1), ft (a + 1)]. 23. 12' by 5'. 26. ft; f 27. 81. 29. 3 Ib. 31. 6f hr.; 5ft hr. Exercise 104. Page 282 1. 63 (exact). 3. 523 (exact). 5. 325 (exact). , 7. 6.39 (exact). 9. 8.85. 11. 40.54. INDEX Numbers refer to pages. Abscissa, 119. Absolute value, 4. Addition, 8. Antilogarithm, 247. Approximate values, 52. Arithmetic means, 224. Arithmetic progression, 221. Asymptote, 263. Base, of a logarithm, 240. for a power, 25. Binomial, 28. Binomial formula, 207. Briggs, 251. Characteristic, 244. Coefficient, 18. Cologarithm, 254. Common logarithm, 242. Completing a square, 177. Complex fraction, 43. Complex number, 171. Compound interest, 260. Conditional equation, 58. Conjugate imaginaries, 192. Constant, 64. Coordinates, 119. Decimals, 48. terminating, 49. Degree, of a polynomial, 61. of a term, 61. Denominator, 6. Dependent equations, 132. Dependent variable, 121. Difference of numbers, 10. Difference of squares, 85. Discriminant, 191. Dividend, 6. Division, 6. Divisor, 6. Ellipse, 264. Equation, 68. of a curve, t29. of a line, 129. Equivalent equations, 59. Exponential equation, 258. Exponential function, 260. ' Exponents, general, 150. laws of, 25, 142. positive integral, 25. Extraneous roots, 114, 198. Factor, definition of a, 3. Factorial symbol, 206. Factoring, 88. Fractions, 22, 103. Function, definition of a, 121. Functional notati6n, 126. Fundamental operations of algebra, 3. Geometric means, 230. Geometric progression, 227. infinite, 236. Graph, of an equation, 128. of a function, 122. of a quadratic equation in two variables, 262. of a quadratic function, 186. Harmonic means, 236. Harmonic progression, 236. Highest common factor, 106. Hyperbola, 263. Identical equation, 58. Imaginary number, 144, 170. pure, 171. Inconsistent equations, 132. Independent variable, 121. Index laws, 25, 31, 142, 154, 278. Index of a radical, 146. Inequalities, 14, 15. Infinite series, 238. Integral rational polynomial, 28. Integral rational term, 28. Intercepts of a graph, 128. Interpolation, for logarithms, 248. Irrational equation, 199. Irrational function, 148. Irrational number, 147. 376 INDEX Linear equation, 61. Linear function, 122. Logarithm, base of a, 240. characteristic of a, 244. definition of a, 240. mantissa of a, 244. Logarithmic equation, 258. Logarithmic function, 260. Logarithms, properties of, 243, 255. Lowest common denominator, 40, 106. Lowest common multiple, 38, 106. Lowest terms, for a fraction, 22, 104. Mantissa, 244. Maximum value, 187. Mean proportional, 212. Minimum value, 186. Monomial, 28. Naperian logarithms, 259. Napier, 251. Natural logarithms, 251, 259. Negative of a number, 9. Negative numbers, 4. Numerator, 6. Numerical value, 4. Ordinate, 119. Origin of coordinates, 119. Parabola, 186. Pascal's triangle, 205. Percentage, 70. Perfect nth power, 100, 148. Perfect square, 82, 90. Polynomial, 28. integral rational, 28. Power of a base, 25. Prime factor, 88. Prime integer, 38. Principal root, 145. Progressions, arithmetic, 221. geometric, 227. harmonic, 236. Proportion, 210. Proportional parts, principle of, 248. Pure quadratic equation, 173. Quadratic form, 196. Quadratic formula, 180. Quadratic function, 186. Quadratic in one unknown, complete, 173. discriminant of a, 191. graphical solution of a, 189. pure, 173. Quadratic in two variables, 262. Quotient, 6. Radicals, 82, 146. properties of, 149. simplification of, 157, 166. Radicand, 82, 146. Radius of action, 76. Ratio, 6, 210. Rational function, 148. Rational number, 147. Rationalizing a denominator, 161. Real number, 1. Reciprocal, 44. Remainder, in division, 34. Repeating decimal, 238. Root, of an equation, 59. of a number, 145. Rounding off numbers, 53. Scientific notation for a number, 249. Significant digits, 52. Signs, laws of, 6. Similar terms, 18. Simple interest, 79. Solution, defined for an equation, in one variable, 59. in two variables, 127. Solution of a system of two equations, 131, 267. Square root, 82, 280. Subtraction, 9. Surd, 148. Systems of equations involving quad- ratics, 262. Systems of linear equations, in three unknowns, 137. in two unknowns, 131. Terminating decimal, 49. Transposing terms, 60. Trinomial, 28. Uniform motion, 75. Unknowns, 58. Variables, 64, 121. Variation, 213. constant of, 213.