AN ALGEBRA FOR SECONDARY SCHOOLS BY E. R. HEDRICK PROFESSOR OF MATHEMATICS THE IMVI'.KSHY OF MISSOURI (? NEW YORK •:• CINCINNATI •:• CHICAGO AMERICAN BOOK COMPANY OOPYRIGHT, 1908, BY E. R. HEDRICK. Entered at Stationers' Hall, London. hedrick's algebra. W- P. i i . . . > <3A isa fo MY SISTER AND TEACHER MRS. ERNST VOSS nee AURIE VAIL HEDRICK THIS BOOK IS DEDICATED IN GRATEFUL RECOGNITION OF HER LONG DEVOTION AND ASSISTANCE 348583 PREFACE The presentation of a new text on elementary algebra to the public demands justification. It is the author's earnest hope that this book will fill a need felt by many teachers for a book that is at once thoroughly modern, yet conservative of what was good in the older text-books. The drill afforded by the older texts will not be found wanting. The exercises in all instances are numerous, especially in those topics where drill work is needed to develop technique. It is strongly recommended th4-:i!) Summary IV. Multiplication and Division; Factoring; Appli cations. §§ 40-67 Part I. Numbers and Monomials. §§ 40-48 Part II. Longer Expressions. §§ 49-55 Part III. Special Multiplications; Factors; Type forms. §§ 56-63 Part IV. Applications; English translated into Alge bra. §§ 64-67 . . . . Summary V. Fractions (Common Factors; Reduction; Operations Proportion; Fractional Equations). §§68-85 . Part I. Common Factors; Reduction of Fractions §§ 68-71 Part II. Rules of Operation. §§ 72-76 Part III. Proportion. §§ 77-80 .... Part IV. Fractional Equations. §§ 81-85 . Summary ........ VI. Simultaneous Linear Equations. §§ 86-93 Summary ........ VII. Simple Powers and Roots. §§ 94-107 Part T. Powers and Roots of Numbers. §§ 94-101 ix x CONTENTS CHAPTER PAGF Part II. Powers and Roots of Monomials. §§ 102- 107 102 Summary 201 VIII. Quadratic Equations. §§ 108-121 . . . .203 Part I. Methods of Solution ; Character of the Roots. §§ 108-115 203 Part II. Practical Applications; Problems. §110 '. 216 Part III. Properties of Quadratic Equations. §§117-121 223 Summary 236 IX. Variation; Indeterminate Equations. §§ 122-126 237 Summary 252 X. Simultaneous Equations involving Quadratics. §§ 127-134 253 Part I. One Linear and One Quadratic. §§ 127-130 . 253 Part II. Simultaneous Quadratics. §§ 131-134 . . 267 Summary 283 XI. Radicals ; Fractional and Negative Exponents. §§ 135-149 284 Part I. Operations; Fractional and Negative Expo- nents. §§ 135-144 .284 Part II. Applications: Radical Equations. §§145-149 300 Summary ......... 312 XII. Equations solved by Substitution. §§ 150-151 . 313 Summary 322 XIII. Progressions or Sequences. §§ 155-160 . . . 323 Part I. Arithmetic Sequences. §§ 155-158 . . 323 Part II. Geometric Sequences. §§ 159-161 . . 328 Summary ......... 333 XIV. Logarithms. §§ 162-167 334 Summary ......... 353 Appendix — Detached Coefficients — Remainder Theorem ; Fac- toring — Choice and Chance; Permutations and Combi- nations — Inequalities — Binomial Theorem — Euclidian Method H.C.F. and L.C.M. — Cube Root and Higher Roots — Limits and Infinite Scries; Irrational Numbers — Imagi- nary and Complex Numbers — Simultaneous Quadratics . 354 Tables . 407 Index -417 ALGEBRA FOR SECONDARY SCHOOLS CHAPTER I. INTRODUCTION PART I. NUMBERS AND SIGNS 1. Algebra. The word algebra is used as a convenient name for the continuation of arithmetic ; there is no defi- nite line between the two subjects, but symbols are used more freely in algebra than in elementary arithmetic. 2. Numbers. The first numbers we learn are the in- tegers, 1, 2, 3, 4, etc. The students know also another kind of numbers, fractions, which are useful in measuring quantities. Later we shall extend these ideas. 3. Signs and Marks. Various signs and marks, some- times called symbols, or characters, are used as in arith- metic, and new ones are introduced as they are needed. A table of these is found at the end of this book (p. 407). Addition, subtraction, multiplication, and division are indicated by the signs, +, — , x, -*-, respectively, as in elementary arithmetic. Thus, 3 + 2, 3 - 2, 3 x 2, 3 - 2, are read " 3 plus 2," " 3 minus 2," "3 multiplied by 2," and " 3 divided by 2,"' respectively. The product 3 x 2 is also often written 3 • 2, the dot • being a simplified cross, x ; and we may say "3 times 2" instead of "3 multiplied by 2." The quotient 3 -=- 2 is sometimes written 3:2, also f, also 3/2; the form 3:2 is often called the ratio form ; the form § (as well as the form 3/2) is often called the fraction form; but they all mean the same thing. The form § may also be read "3 over 2." 1 2 INTRODUCTION [Ch. J One important sign is =, read " equals," or " is equal to." A statement that one thing is equal to another is called an equation or an equality ; e.g. 2 + 4 = 6. Two num- bers are equal if, and only if, they are the same number, though they may be written differently. Thus, 2 + 4 is the same number as 6, but it is written differently. An equality of two fractions (or ratios) is called a pro- portion. Thus, | = f (or 2:4 = 3:6) is a proportion. The radical sign V is used to indicate square root ; thus Vl6 = 4. For further uses of this sign, see p. 9. 4. Abbreviations. In arithmetic we use such abbrevia- tions as ft. for 1 foot, in. for 1 inch, hr. for 1 hour, etc.; and we write 36 in. = 3 ft., etc. In algebra we abbreviate still more. Thus, we may as well write merely/ for one foot and i for one inch, then 36 i = Sf. Such abbreviations often simplify problems and shorten calculations ; they make general statements possible. Thus, instead of saying interest =(rate) x (principal) x (time in years), we may say i = r x p X t with the same advantage that we gain in writing Mo. for Missouri ; namely, we save time and space. There is nothing mysterious or difficult about such abbreviations if they are once explained. Tt should be noticed that the same letter may be used for different things in different problems ; thus i was used to mean one inch in one connection above, and again to mean interest. Such abbreviations are simply temporary, for a single problem; the student will not be confused if he clearly understands what each letter means. It is very important, however, to avoid using the same letters for two different things in the same problem. In using letters as above, the sign of multiplication is often omitted; thus rpt means rxpx U and 5 pr means 5 x p x /• ; but it is not safe to omit the sign when two factors are ordinary Arabic numbers, since, for example, 3-4] NUMBERS AND SIGNS 3 52 means 50 + 2, not 5x2. However, this may be done in combining numbers and letters, as in arithmetic, when we say that if in. means 1 inch, 3 in. means 3 inches. EXERCISES I: CHAPTER I [Tables of .symbols, weights, measures, etc., will be found at the back of the book.] 1. Express 4/+3i ia inches if / means one foot and i means one inch. 2. Express 2h + 4 m — 14 s in seconds if h means one hour, m one minute, and s one second. 3. If b denotes one bushel, p one peck, and q one quart, ex- press 3b—p — 2q in quarts. Also express 3b— p — 2q in pecks ; in bushels. 4. Tf t stands for one tenth and h for one hundredth, ex- press 1 — 3 1 -4- 5 h in per cent. Express 67 c / c in terms of t and h. 5. Let c denote one cubic centimeter, i one cubic inch, and /one cubic foot. Express T V/ — 132 i in terms of c, if we let i equal 16 c (which is very nearly correct). 6. Express 2 m + 30 y + 3/ in terms of/ if m = 5280/ and 2/ = 3/ [Xote that figures given are actually applicable to the case in which /denotes one foot, y one yard, and m one mile.] 7. If g = 4 q, express 2 g + 3 q in terms of q ; in terms of g. [Can you mention a case of measurement to which this problem would be applicable ?] 8. If y denotes one yard and i one inch, express »a = y + 3|i in terms of i. What unit of length does m denote ? 9. If x denotes a single thing, d a dozen, and m a gross, express 2 m + 11 d — 6 x in terms of d. 4 INTRODUCTION [Ch. I 10. Express a — 17 b + 8 c — 50 d in terms of d if a = 24 6, b = 60 c, c = 60 d. To what measures will this apply ? 11. What is the value of 4 a + 3 b — 3 c if a = 2, b = 5, c = 3? 12. What is the value of 6 xt - 3 ?/ - 5 1 if a; = 3, y = 0, <=2 ? 13. Express 4 m — 3n + 10p in terms of n if m = 2n, n = 5 p. What is the value of the result if n = 52? if a = 32? if n = ? 14. What is the value of 2q s + 6 ?/ + c if a = 3, 6 = 4 j c = g, ^ = 5,^ = 2? 3a; ~ 6 15. If ^4 = 5 «, jB = 3 &, express — in terms of a and b. Ab — Ba Find the value of this result if a = 2 and b = 4. 16. If # denotes the area of a square, and y the length of a side, then x = y • y. What is the area of a square whose side is 4 feet long? (The product y-y is often written ^ is called the square of y because y • y is the area of the squ«,?e.) 17. Express X y in terms of a; alone if y = 2 a?. Then 1 — ay find its value if x = \. 18. What is the area of a rectangle whose length is a and whose breadth is b ? What is the area of a rectangle 5 inches long and 3 inches wide ? 19. Let m denote the side of a cube, n its volume. Express n in terms of m. If m = 2 feet, what is n ? If n = 27 cubic inches, what is m? 20. Let the dimensions of a room, measured in feet, be de- noted by vo (the width), I (the length), and h (the height of the ceiling); and let c be the cost (in cents) per square yard of plastering. How much will it cost to plaster the walls and ceiling of the room, neglecting doors and windows? What will the cost be if I = 16 feet, w = 15 feet, h = 10 feet, and c = 15 cents ? _r,] NUMBERS AND SIGNS 5 5. Problem Solving. Abbreviations are especially useful 11 the solution of problems. Ex. 1. Find a number such that the sum of that number Mid half the number is 18. Solution Solution (not abbreviated) (abbreviated) Consider Me number to be found. Let n = = the number to be found Then (that number) + 1 (that Then n + I n = 18, number) = 18, or, | (that number) = 18, or, | n = 18, or, I (that number) = 6, or, \ » = 6, or. (that number) = 12 (answer). or, n = 12. \Check: Check: 12 + I • 12 = IS. (Correct.) 12 + \ . 12 = 18. (Corret A check is any process for testing an answer. If, as here, the answer can be tried directly in the given problem, the check is complete ; a complete check shows that the answer is surely correct. [The teacher should explain carefully the check on the answer in this problem, and the students should be required to check their answers in all cases.] Ex. 2. A merchant sold tea for 35 cents per pound. What was the cost to him if he made 25 % profit on the cost? Solution Solution (not abbreviated) (abbreviated) Consider the cost to merchant. Let c — the cost to merchant. Then the profit to merchant is 25% Then profit to merchant = 25 %c. of cost to merchant, or I of cost to merchant. Then profit to merchant = \c. Hence, the total price is the cost Hence total price is c + $ c. to merchant added to I of cost to merchant, or | of cost to merchant. Hence total price = f c. Then 35 cents is | of cost to mer- Then 35 cents = f c, chant, or 28 cents is the cost to merchant. 28 cents = c. Check: 28 + J • 28 = 35. Check: 28 + \ • 28 = 35. t> INTRODUCTION EXERCISES II: CHAPTER I [Check each result as indicated in the preceding examples.] 1. If the sum of three times a certain number and half that number is 14, what is the number ? [Let the students invent and i:>ropose to each other such problems.] . 2. What number yields a remainder 5 if diminished by the sum of one half of itself and one third of itself ? 3. What must be the amount of money invested in an enter- prise yielding 25 % profit, in order that the money in hand at its conclusion may be $ 1000 ? 4. A man attempted to charge 8 % interest on money. Being able to collect only the legal rate (6 %) he made $20 less than he expected. What was the amount of money loaned? 5. A merchant buys butter for 30 ^ a pound and sells it for 36^ a pound. What is his per cent of profit on the cost '.' 6. A fruit dealer sold oranges at 5 cents apiece or 50 cents per dozen. He found that he received $ 54.00 for 100 dozen. How many w r ere sold singly and how many by the dozen ? 7. What must be the per cent of profit on an investment if $525 is to produce $600? 8. A shoe dealer buys 100 pairs of shoes at $2.00 each and sells 75 pairs at $2.50 each. To sell the remaining 25 pairs he marks them down so as to make 20% profit on the whole. AYhat is the price per pair that will give this result? 9. I propose to a friend the following puzzle : " Think of any number, add 5 to it, multiply the result by 3, and subtract 4." He gives the result as 17. What was his number ? 10. I have an opportunity to lend money at 5% simple interest for a period of 8 years. How much must I lend in order that the amount may be $3500? 11. How large a cubical box may be covered with 24 square • inches of paper ? PART II. PRELIMINARY DEFINITIONS 6. Expressions. All single groups of numbers and char- acters of the kind already used are called expressions; in order to deserve this name, the group of numbers and characters must have some meaning. Thus, 3/+ 2 i has a meaning if f means 1 ft. and i means 1 in.; then 3/'+ 2i is an expression. Again, the group of characters p + rpt is an expression, although the meaning is not wholly clear. It is at least clear that some num- ber^) is to be increased by the product of three quantities, r, p, and /. The meaning becomes clearer if we are told that /j means the principal, /• the rate, and t the time in years in a certain interest problem; then the above expression clearly means the amount. The meaning be- comes still clearer when the numerical values of r, p, and t are given, say r = 5%,/j = 8 125, and / = 3 years. These various stages in clear- ness do not contradict the fact thatyj + rpt has in itself (without any further explanation) a certain meaning as expressed above. 7. Terms. An expression may contain one or more + or — signs, which separate it into parts ; these parts are called the terms of the expression. Thus in p + rpt the terms are p and rpt. In 2 ax + ■' " + ic the terms are 2 ax and °' f " and 4 c. To be sure, the term ° - 1 ~*~ " itself 3 3 contains a + sign, but it is not separated into parts by that sign as it stands. On the whole, the word term is used rather loosely, the inten- tion being to distinguish those parts which stand in rather compact groups as compared with the rest. The terms of an expression are calculated separately, and these results are added or subtracted in their order. Thus, as a general rule, the multiplications and divisions are carried out first, after which the terms thus formed are added or subtracted. If anything else is intended, parentheses are used to show that the expression inside the parentheses is to be calculated first. Thus, the expression 2 ax + m (."> y + 2) + 4 c has three terms: 2 ax, m (5 y + 2), and 4 c ; the term m (5 y + 2) means m times the sum of by and 2. See also §11, p. 10. 7 8 INTRODUCTION [Ch. I To distinguish expressions the special names monomial, binomial, and trinomial are often used for expressions that have one term, two terms, or three terms, respectively. 5 v + 2 Thus rpt and -\ " are monomials, p + rpt is a binomial, and 5 y -L o " 2 ax -\ — J — 4 c is a trinomial, etc. 3 When we wish to refer to a complicated expression of more than three terms, we may simply call it an expression. The word polyno- mial may also be used in certain simple cases defined in § 9 below. 8. Factors. When several numbers or expressions are multiplied together, any one of them is called a factor and the result is called the products Thus the factors of the product rpt are r, p, and t. Any factor is called the coefficient of the rest of the product ; usually, however, the word coefficient is under- stood to be that factor which is represented by Arabic numerals, or which is supposed to be a known number, un- less the coefficient of a special part is required. Thus, in 2 axy the coefficient is 2 unless a specific part is mentioned, but if we are asked for the coefficient of xy, the answer is 2 a ; the co- efficient of 2 ax is y, and so on. Coefficient is but another name for multiplier. Thus, in 2 axy, 2 is the multiplier of axy and 2 a is the multiplier of xy; but coefficient is the word generally used. If the coefficient is 1, it is not written, since multiplying a number by 1 does not change it. Thus, in axy the coefficient is 1 unless the coefficient of a special part is required. Terms that are precisely the same or that differ only in their coefficients are called similar terms or like terms. Thus, 2x and Sx are similar; "dm 2 n and 4m 2 n are similar. Terms that are not similar are called dissimilar or unlike. 9. Powers. The product of two equal factors is called the square of that factor ; it is indicated by a small figure 2 at the upper right-hand corner. Thus, x • x = x 2 . The product of three equal factors is called the cube of that factor. Thus, x • x • x = x 3 . 7-10] PRELIMINARY DEFINITIONS 9 The product of any number of equal factors is called a positive integral power or a simple power of that factor ; it is indicated by writing the number of the factors at the upper right-hand corner of the factor. Thus, we write x ■ x- x • x = z* ; x - x- x • x • x = t > \ etc. The number of factors is called the exponent of the power. Thus, in x* the exponent is 4, and x* is called the fourth poiver of x, or simply x fourth power. We shall later extend these definitions of power and exponent so as to give a meaning to such forms as a;*, x~\ x°, etc., which are at present meaningless to the student. See pp. 286, 296. Since any number x may be considered as taken once as a factor to make itself, x 1 means the same as x. Hence, it is useless to write the exponent 1, and when no exponent is written, 1 is understood. A polynomial in certain given letters is an expression whose terms, if they contain one of those letters at all, con- tain it as a factor which is a simple power.* 10. Roots. If a number is the product of two equal factors, one of the equal factors is called the square root of the number; if a number is the product of three equal factors, one of the equal factors is called the cube root. In general, if a quantity is the product of a number of equal factors, one of them is called a root of that quantity ; and the number of the factors is called the index of the root. Thus, 4 has two equal factors, 2 and 2. Hence, the square root of 4 is 2, written Vi = 2 ; likewise, since 3 • 3 • 3 = 27, the cube root of 27 is 3, written V27 = 3. Again, since a • a — a 2 , Va? = a. So also, yfa? = a ; Va* = a. etc. The index is always written at the upper left-hand corner of the sign V , except for square roots, which are in- dicated by v 7 instead of v . These ideas will be extended later. *A broader use of polynomial is common in many text-books, but it is l\ot good usage. Standard mathematical works otlier than text-books agree on the definition used above. The word is not used extensively in this book until alter p. 88, where a more detailed explanation is given. 10 INTRODUCTION [Ch. I 11. Parentheses. When we wish to group terms to- gether, we use ordinary parentheses (see § 7, p. 7). Thus, ( a _|_ J)2 means the square of the expression a + b. In order to avoid confusion when one pair of paren- theses occurs within another, we use different forms, as follows: ], called brackets; | $, called braces; and" ~~, called the vinculum. These names are convenient to dis- tinguish the shapes, but they are all used alike, and they are all called parentheses except when some confusion would result. Thus (2 + 3) • 4 means 5 • 4, or 20; (a + h)c means the sum of a and b multiplied by c; 2Va + b means twice the square root of the sum (a + b). We may now use more complicated expressions, the con- venience of which will be seen in the next chapter. For instance, 2(a+6)(c + rf) + (a+6) 2 means twice the product of (a plus b) and (c plus d) added to the square of (a plus b). If the letters used mean certain numbers, say a = 2, b — 1, c = 3, d = 4, then a + b = 3, c+d = 7, and 2(a + b)(c + d) + (a + by 2 = 2 . 3 • 7 + 3 2 =42 + 9 = 51. EXERCISES III : CHAPTER I Find the value of : 1. 2 a 8 if a = 2 ; if a = 3 ; if a = 5. 2. 4 d 2 b 3 if a = 2, b = 3 ; if a = 1, & = 2 ; if a = 5, 6 = ? . 3- Vpr if i> = 2, r = 8 ; if p = 3, r = 12 ; if p = 18, r = 8. 4. ^ + 4 -4 a? -a; 2 if x = l; if « = 2; if a; = 3 ; if a; = 4. 5. V5W + 4a 2 (& + c) if a = 5j & = 2 , c = 3; if a = 20, & = 1, e = 2. 6. Find the value of x^/1 -f - yy/ l-x 2 [ix = ^ y= * xy + Vl — ar' • VI — y' 1 7. ft V^-^ + yVq 2 -^ jf a = 5, 6 = 3^ = 13,^ = 5. V(ar'- ? />r'-& 2 )-ty 11] PRELIMINARY DEFINITIONS 11 8. 3" if n=2; if n = 3; if n = l. 9. ./•" if x = 2, if = 3; if a; = 4, y = l; if aj=l, 2/ = 2. 10. tftft if a = 64, 6 = 3, c = 2 ; if a = 256, b = 4, c = 2. Express : 11. (2/+ 3 £) 2 in square inches if /= one foot and i = one inch. . 12. (2 t + uf in units if t denotes one ten and u one unit. Then write the result in terms of f 2 , t, and u. Show that : 13. (.i- + ?/) 2 = x 2 + 2 xy + >f if x = 3, ?/ = 2; if a? = 5, y = l; if x- = 4,, y = f. 14. ( 3 / 2 + 1) 2 = (% 2 - l) 2 + (2 ra) 2 if n = 1 ; if n = 2 ; if n = 3 ; if n = 4. 15 = 3, q = 2. 18. In Ex. 1 name the coefficient of a 3 ; the exponent of a. 19. In Ex. 2 name the coefficient of a 2 b n ; of >(4 - 1- (14 U-f- -8- 9- 10 Iff -—1-2 -3-4 -5-(i -1--8 -y- ■ '■ z w'~ Fig. 6. 18 MEASUREMENT AND EXPRESSION [Ch. II On a certain day the temperatures were: Time of Day 8 -10 9 - C, 10 11 4 12 Id 1 12 2 20 3 18 4 10 •5 15 6 12 7 10 8 5 9 _ 2 Temperature The diagram that corresponds to the preceding table is shown in Fig. 7 : -&£- T — 5«— > (. 1 0° X r Mil ! 3 l 1 1 s 1 1 l 2 i a • ; - * i 7 8 * 1 - } c 2 -io J ? a, &x Fig. 7. If the temperature is read oftener, say every fifteen minutes, a much better figure results. Such a figure is shown in Fig. 8. . • • •. • • i . Ill • . •, . • • i ■ • i • • • n mt 1 8 l i 10 l i 1 2 i 2 3 1 , 6 i ■i • • i V • * < • a l • - ^. r h l''io. 8. 1«] RELATIONS — GRAPHS 19 Let the student judge as nearly as he can from this figure the temperature at 8.30 a.m., at 11.15 a.m., at 2 p.m., at 9.15 p.m. Notice that this cannot be told accurately, but only approximately. Notice also that the temperature can be estimated at times when it was not really measured. Thus the temperature at 11.40 a.m., though not actu- ally measured, may be judged by the figure to have been about 8°. A smooth curve drawn through the points, as shown in Fig. 9, gives a very accurate idea of what the temperature was at any moment. >ri fj, T nn 8 9 »l 1 I 1 l i 1 ;> 3 . ) ' i Sa t u / ^ / a / -io u • : Fig. 9. The figure may be used also in another way. We may ask, "At what time was the temperature 14°?" By fol- lowing the horizontal line through the mark " 14 " till it strikes the curve, we see that the temperature was 14° at about 1.20 p.m. and again at about 5.15 p.m. It should be noticed that there are two answers. Let the student select the time of day from the figure when the temperature was 18°; 10°; 0°. Notice that it was never 30° on that day. Let the student actually use a thermometer and draw such a figure as that just given, and let him then show what the temperature must have been (approximately) at some time when he did not meas- ure it. Let him also show (about) when the temperature was (say) 40°. Such figures as those just drawn are often called diagrams, or graphs. Drawing such a diagram or graph is called plotting it ; locating a point is called plotting the point. 20 MEASUREMENT AND EXPRESSION [Ch. ii 17. Scales of Prices. Many quantities may be drawn in figures like the preceding. Following are examples : Ex. 1. If butter is 30 cents per pound, draw a figure to represent the cost of any number of pounds. We first make a table as shown below : No. of lb. 1 2 3 4 5 150 6 7 8 10 11 12 13 14 15 etc. Price in f- 30 CO 00 120 180 [Let the student fill in the blank spaces.] Then, representing the number of pounds of butter on a horizontal line, as shown, and the cost on a vertical line, as shown, we have as before a set of points, A, B, C, D, E, ■•-. Joining these by a smooth curve, we get a close idea of the cost of any number of pounds, even a fractional number. For example, the cost of 3£ pounds is $1.05, as the figure indicates. The " curve " in this case is really a straight line ; this may happen in any example, but the word curve is used at least until we are sure the line is really straight. Most price curves are straight lines. See §§ 18, 19, pp. 23, 24. Notice that no butter cost no money. Thus at the point marked "0" on the horizontal line we should draw the poiutai the height 0. Notice that buying - 2 pounds of butter means selling 2 pounds. The cost of — 2 pounds is, therefore, less than nothing; in fact, a man is, of course, paid for the butter if he really sells it. Instead of saying that he is paid 00 cents for selling 2 pounds, we may say that the cost. is - 60 cents if he buys - 2 pounds. This would be represented in Cost 200 rf=!J3 -i 1 L _ 1 _ 7 r - 4 E 7 T -4 s 100 ^=!M .7 v M 1 i n/ 50 c' fl ; /: 1 J A to g T . / lb. ; o * 5* j 7 j 7 f 1 Fig. in. 17] RELATIONS— GRAPHS 21 the figure by a point corresponding to — 2 on the horizontal line and to — (iOtf on the vertical line. With this understanding the curve of prices just above is an unbroken straight line. EXERCISES II: CHAPTER II 1. Draw a figure to represent the temperature at various times of the day from the following table : A.M. M. 12 P.M. Time . . . 8 30.5 9 10 11 1 2 3 4 5 6 7 8 9 etc. Temperature 31.5 31 30.5 29 28.5 29.5 31 31.5 31 30 29.5 29 28 etc. 2. From the figure in Ex. 1 estimate the temperature at at 11.30 a.m. ; at 2.30 p.m. ; at 6.15 p.m. 3. At what time of day was the temperature about 30.5° ? 29° ? Is there more than one answer in each case ? Find all the answers. 4. At what times (about) during the day did the tempera- ture change from rising to falling ? from falling to rising ? . 5. If the price of tea is 35 cents per pound, draw a figure to show the cost of any number of pounds. 6. From the figures give the cost of 3\ pounds; of -Im- pounds; of — lg- pounds. [Hint. — Continue the line below and to the left.] 7. How much tea can be bought for 81 cents? for $1.96? for — $1.05 ? What does the last statement mean ? 8. The population of the U.S., as determined by the decen- nial census, is approximately given in the following table: Year . . 1790 1800 1810 1820 1830 1840 17 1850 23 1860 31 1870 39 1880 50 1890 63 1900 76 Millions 4 5 7 10 13 Draw a graph showing the population at various times. 9. Estimate the population in 1S05 ; in 1885 ; in 1895. 10. When was the population about 15,000,000 ? 60,000,000? 70,000,000 ? 22 MEASUREMENT AND EXPRESSION [Ch. II 11. Represent by a graph the simple interest at 6 % for one year on any amount of money. Choose sums at intervals of $ 1000 from - $5000 to + $ 5000. 12. What interest shall I have to pay at 6 % for a year's use of $2500? $4500? —$2500? Interpret the last statement in the light of Ex. 11. 13. What principal yields as interest for one year at G% $195? -$195? -$105? 14. The number of inhabitants of the United States of school age from 1870 to 1904 is shown in the Fig. 11. What was the number in 1870 ? 1880? 1885? 1890? Number of School Age 24000000 23000000 22000000 21000OC0 20000000 19000000 18000000 17000000 16000000 15000000 14000000 13000000 12000000 }'. iirt, Fig. 11. 15. When was the number of inhabitants of school age about 12,000,000 ? 14,000,000 ? 10,000,000 ? 20,000,000 ? 17-18] RELATIONS — GRAPHS 23 16. From the figure construct a complete table of the num- ber of inhabitants of school age from 1870 to 1905 (every year). 17. Draw a straight line graph to represent the correspond- ing readings of a Centigrade and a Fahrenheit thermometer. (See tables at back of book.) 18. Give the Centigrade temperatures corresponding to Fahrenheit temperatures 22° ; 35°; -17°; -32°. Give the Fahrenheit temperatures corresponding to Centigrade tempera- ture -25°; -12°; -5°; 20°; 85°. 19. What temperature has the same reading on both scales? The teacher should also have each student take some problem in his own experience : the number of inhabitants in his town ; the number of pupils in his school ; the earnings in his father's store; the price of wheat or of cattle; the amount of rainfall in various years; or any other familiar instance of varying quantity. This problem should be carefully investigated by the student, and the figure should be drawn. The World Almanac, which can be had by addressing The World, New York City, contains much valuable information of the kind needed in such problems. 18. Equation and Graph. In the case of prices of a commodity, such as in the example in § 17, we may also represent the cost of any number of pounds by an equa- | lion. Let c be the cost in cents, p the number of pounds ; ' then if butter costs 30^ per pound, c = 30 jo. This equation represents the same thing (cost of an amount of butter) as Fig. 10 ; hence, we say that this equation belongs to that figure, or, briefly, this is the equation of that figure; and that figure is the graph, of this equation. Notice that the figure may easily be drawn from the equation. Taking various simple values of p, we get the following; table : 1 30 2 60 3 10 etc. - 1 -2 etc. Cost * 90 300 etc. -30 -60 etc. 24 MEASUREMENT AND EXPRESSION [Ch. TI If we plot these points, as before, we get a figure like Fig. 12. It will be evident to the student from the fig- ure that this graph is a straight line through the starting point, as men- tioned in § 17. The proof of this fact is given in § 80, p. 110. 19. Equations of Prices. We proceed to extend the sugges- tion of § IT that most price curves are straight lines. If the cost of one article (or of one unit of meas- ure of a measurable commodity) is known, the cost of an} 7 num- ber is given by the following equation, where c is the total cost, h the cost of one, and n the num- ber bought: total cost = (known cost of one) x (number), or, c—k-n. As in the example of § 18, the graph of such an equa- tion is always a straight line through the starting point. (See also § SO.) — 1 — 1 — 1 — 1 — 1 — [ — 1 — 1 — 1 — 1 — I — 1 — 1 — I 7 300^=! 5-3 r _/ 7 L '/> / 6" 7 L ^7 7 L 200 d= $2 / 7 L ~1 7 / l 1 1 1 100<><• / _^_ * 7 r !\ 10g( 7 7 1 lb. /0 5* 10* t _f 1 7 r / i Fig. 12. 18-20] RELATIONS — GRAPHS 25 20. Linear Equations. Straight Lines. The student should notice that ordinary proportion* might be used in the preceding examples. Other examples in which proportion might be used lead to equations similar to those above, and to straight line graphs. Ex. 1. If a man walks 3 miles per hour, and if d denotes the total distance (in miles), and n the total number of hours he walks, evidently d = 3 n. Ex. 2. If x denotes the number of feet in a certain length, and y denotes the number of inches in the same length, y=\2x. [Let the student make a table of values and draw a graph for each of these examples.] Many examples arise in business and in science in which ordinary proportion could not be used directly. Such examples may lead to straight line graphs. Ex. 3. If the cost of setting the type for a circular is $2.00 and the cost of paper and printing is \$ per copy, find the cost of any number of copies. Let c mean the total cost in cents ; let n be the number of copies. Then, c - i „ + 200. v In order to plot the graph of the equa- tion c = J»+ 200, let us call c x the cost of printing and paper alone, then c^^n and the figure is a straight line, as above. Now the effect of the cost of setting the type is to increase the price by 200)*, no matter what the number of copies. Hence, the figure for the total cost is found by simply raising the whole figure by an amount that denotes 200 f on the vertical scale. Hence, the final figure is also a straight line. " — — j- -■' 1 i ~ '" " 1 _/■» ; s ■ "^vb-^T t ± <.2L* 'dz o _l ^tPTJl § ' ^>"n A ' 1 >*" n i 1 ° . j^ 1 o 1 Jl^ 100 '> 1* ' ,> "\ 1 J^-- J J^ o 4-im • aoo I p \ i TTl 1 I i 1 I 1 II 1 \ Fig. 13. * It is assumed that the student is familiar with ordinary proportion as treated in all arithmetics. If not, the teacher may well recall it to his mind. 26 MEASUREMENT AND EXPRESSION [Ch. All of the equations of §§ 18, 19, 20 are of the form y — kx + / where k and I are fixed numbers. The graph for y and x is a straight line. For this reason any equation of the form y = kx + I, or any equation that can be reduced to this form, is called a linear equation or a simple equation. EXERCISES III: CHAPTER II 1. If 11 tickets to an entertainment cost $2.75, how much do 5 such tickets cost? Let n be the number of tickets bought, and c the total cost (in cents) ; then c = kn, where k is the price of 1 ticket (in cents). On a figure, as before, plot the values n = 11, c = 275 (cents), given in the problem, at P. Join P and O; the line PC) represents the cost of any number of tickets. Go 5 units to the right of O and then up to the point Q ; opposite Q is the point which represents the number 125 on the vertical line. Hence, 5 tickets cost 125 cents. ' We might have found this from the equa- tion. For c — kn, and c — 275 when n = 11 ; hence, 275 = 11 k, or, k = 25. If now n — 5, we have c = 25 n = 25 . 5 = 125, which is the result just found. The advantage of the graph is that it solves all such problems (approximately) at once. Again, the figure quickly shows the answer to many reverse problems. Thus, if a man has $2.10, how many tickets can he buy ? Look for the point for 210 on the verti- cal line, pass over to the line OP, then look down at the horizontal. It is clear that he can buy only 9, for the value found is less than 1<». 2. Draw a graph to show the relation between kilograms and avoirdupois pounds. What is the equation expressing the former in terms of the latter? Compare Ex. 2, p. 25. Fig. 14. 20] RELATIONS — GRAPHS 27 [Suggestion. Let z denote the number of kilograms in a given weight, and let y denote the number of pounds in the same weight. Then y = 2.2 x x (nearly), since 1 Kg. = 2.2 lb. See Tables.] 3. Express approximately in kilograms 3 lb., 7 lb., 11 lb. Express approximately in pounds 0.5 Kg., 3 Kg., 4.5 Kg. 4. What is the equation representing the area A of a rec- tangle whose base is 5 and altitude a ? Draw the graph. 5. Find A in Ex. 4 if a = 2 ; if a = 5 ; if a = 3|. Find a if .4 = 15; if A = 27%. 6. Express by an equation the relation between United States dollars and English pounds sterling. Compare with Ex. 2. Draw the graph. 7. Express by an equation the relation between the differ- ence in time of two places on the earth's surface and the differ- ence of longitude. Draw the graph. Interpret the meaning of positive and negative values of each difference. [Suggestion. A difference of 15° in longitude makes a difference of 1 hour in time.] 8. Expressing longitude east of Washington, D. C, as positive, and longitude west as negative, find the time corre- sponding to noon at Washington at a place whose longitude is -f 30°; —30°; +25°; —40°. Find the longitude of a place whose time differs from that at Washington by -f- 3 hr. ; -3hr.; + 21 hr.; -5|hr. 9. Represent by a graph the distance traversed by a bicy- clist at the rate of 8 miles an hour. What is the equation connecting the distance and the time ? 10. In what time will the bicyclist cover 2.3 miles? 3.2 miles? How far will he go in H hours? in o\ hours? 11. The cost of setting type for an order of business cards is $0.75; the cost of cards and printing is \$ apiece. What is the equation for the cost of any number of printed cards ? Draw the graph. Estimate the price of 150 cards ; of 500 cards ; of 650 cards. 28 MEASUREMENT AND EXPRESSION [Ch. II 12. Another firm offers to disregard the initial cost of type- setting and to print the cards for \$ apiece. What is the cost of any number of copies ? Draw the graph on the same diagram as that for Ex. 9. 13. Which firm will most cheaply print 100 cards ? 500 cards ? Which firm will do the most printing for $ 1.25 ? for $ 2.00 ? How many cards will be printed just as cheaply by one firm as by the other, and what will be the cost of the printing? 14. The initiation fee in a certain club is $ 3.00 ; the yearly dues are $2.00. What is the cost of membership in the organ- ization for any number of years ? Draw the graph. 15. Find the cost of membership in Ex. 14 for 5 years; for 12 years ; for 15 years. Find the length of membership of a member who has paid $11; $29. 16. The bicyclist of Ex. 9 has a distance of 20 miles to go. Represent by an equation the relation between the time he rides and the distance that remains for him to travel. Draw the graph. 17. How far has he left to go after traveling .82 hour? 1.77 hours? How long must he travel in order to be within 12 miles of his destination ? 4 miles ? 18. Draw in the graph whatever should correspond to the bicyclist's riding past his destination. How long after start- ing will he have left —4 miles to travel to his destination? — 12 miles ? 19. Two trains are running at the same rate in the same direction. If x is the distance passed over by one train in any length of time, and y the distance passed over by the other in the same time, what is the relation between y and x? Draw the graph showing this relation. 20. Two trains start from Chicago and run at the same rate in opposite directions. If x and y represent the distances of the two trains from Chicago at any time, express the relation between y and x by an equation and by a graph. 20-l' 1 J RELATIONS — GRAPHS 29 21. Suppose that one train starts from Boston when another is 5 miles ahead, and they travel at the same rate in the same direction. When the first train is x miles from Boston, how far is the other from Boston ? If y is the distance of tin- second train from Boston, what is the relation between y and x? Draw the graph. 22. A problem similar to the preceding leads to the equa- tion t/ = x + 9. Plot the graph. When x=3, what is y? When ?/ = 17, what is x? 23. From a time-table of some railroad between two im- portant cities, construct a figure to show the movements of trains ; mark the distances, starting from one city along the main horizontal line, and mark the times of day from 12 o'clock of one day to 12 o'clock the next day on the main vertical line. Then plot the position of each train at the time shown in the time-table. Plot graphs corresponding to the following equations (plot only positive values unless you know how to perform the necessary operations with negative values) : 24. y =x — 2. When y = 7, what is x ? When y = 5, what is x? 25. ?/ = 4.r— 2. Plot only points for which x is positive. What is y when x = 2-L ? when x= o\ '.' 26. 2y = x + 3. (This may also be written, y = ^x + %.) What is y when x=l? 3? 5? What is x when y =2 ? 3? 4? 27. 5 y = 2 x + 10. What is y when x = 0? 2? 5? What is x when y = 4 ? 6 ? 21. Other Examples. Some examples do not lead to straight line graphs. Those which give a linear equation (§ 20) do, of course ; but it is easy to make examples which lead to other kinds of equations and to other kinds of figures. The following example will dispel the false idea that all graphs are straight lines. 30 MEASUREMENT AND EXPRESSION [Ch. \l Ex. 1. What is the area, A, of a square in terms of its side, s ? Draw the graph. Solution. If A is the area (in sq. ft.), and .s is the length of one side (in ft.), we have A = s' 2 . Giving .s- various values, we find: Length of side (s) : 1 2 3 4 5 etc. i 2 3 2 etc. Area of square (A) : 1 4 9 1 % Point in Fig. 15 : A B C D E [Let the student fill in the blank spaces and extend the table.] Y ^ -j» p.. 1 J 4 4 -i (orecO 1 20 + f L wt 1 -C 4 7 I -> If io t£. d \ % -4£ / - -<-tffi ^F^ gr "2f -i.cli it y 5 S («<<*«) Si i tie: 1 //«, horizontally 1 \q.ft. vertically X Fig. 15. Plotting these values of .s and A, we obtain a graph that is not a straight line (Fig. 15). This fig- ure may be used as above ; thus, if the side is 2.3, we may find the area by going out 2.3 on the hori- zontal line to the right, then up to the curve (R in Fig. 15), then over to the vertical line ; the value of A is seen to be a little over 5 ; really, A = 5.29. Again, if A = 11, we can find s = 3.3 (about) ; the point is T in Fig. 15. 22. Review Exercises. The following exercises lead to equations like those in §§ 18, 19, 20, that is, the graphs are straight lines. When one of the quantities that vary is given, the other can be found, either from the equation or (approximately) from the figure. The two answers should be the same, except for the slight inac- curacy of the figure ; thus the figure serves as a check. 21-22] REVIEW 31 REVIEW EXERCISES IV: CHAPTER II 1. Represent by a graph the multiplication table for mul- tiplier 7, from 7 x to 7 x 12. Read off 7 x 4.5; 7 x :'.."» ; 7x31 Determine a number, n, such that 7 n = 85.5 ; 7 n = 3.5. 2. A certain grade of cloth costs 20^ a yard. Represent in a figure the cost of any number of yards. Read from the figure the cost of 3.5 yards; — 5.5 yards; 1.4 yards. (Write equations.) How many yards cost 64 ^? 86^ ? — 50^? 3. The value of farm property in the United States is given at intervals of ten years from 1850 to 1900 in the following table : Year .... 1850 1860 1870 1880 1800 1000 Billions of Dollars 3.08 7.08 S.-.lll 12.18 16.08 20.44 Draw the graph; estimate the values for intermediate years, recording your results in the form of a table. 4. Represent graphically the relation between the circum- ference of a circle and its radius. Use the figure to determine approximately the radius of a circle whose circumference is 3 ft. ; 22 cm. ; 35.5 units. (See Table.) 5. Letting d represent one degree, what is the circumfer- ence of a circle in terms of d ? If r is the radius of the circle, what is the circumference in terms of r? What relation there- fore holds between d and r? Draw the picture. Express as a multiple of the radius of the circle an arc of 57.3°; 100°; 172°. How many degrees are there in an arc twice as long as the radius ? 6. Represent by a picture the relation between pounds and ounces. Construct and solve problems in the reduction of pounds to ounces and ounces to pounds. Compare Ex. 2, p. 28. 7. Find out the rate per $ 100 valuation of your city, county, and state taxes. Plot a figure to represent the tax on 32 MEASUREMENT AND EXPRESSION [Ch. II any valuation by representing the valuation along the horizon- tal line ($100 to one large space) and the tax along the vertical line ($ 1 to one large space). Draw a line to indicate the total tax. Express the same facts by equations. Estimate the taxes on a house and lot valued at $2750. Find the value of property taxed at $ 6.25. 8. A man rides horseback for an hour and a half at the rate of 8 miles an hour; then, leaving his horse, he walks back at the rate of 4 miles an hour. Draw a picture showing the distance, d, of the man from his starting point at any time, t. How long does the trip take? 9. Another man starts at the same time from the same place as the man in Ex. 8, walking at the rate of three miles an hour in the same direction. How long after the start will the first man, returning, meet the second, and how far from the starting point will the meeting take place? Solve only graphically. 10. A train that leaves Kansas City at 10 a.m. arrives in St. Louis at 6 p.m. The distance is 280 miles. Find the aver- age speed. If t denotes the time (in hours) after leaving Kansas City, and d denotes the distance (in miles) from Kansas City, express by an equation and by a graph the relation between d and t. Find d when t = 2^. Find t when rf = 105; when d = 200. 11. A train leaves New York at 2 p.m. and arrives in Chicago (940 mi.) at 5 p.m. the next day. Express by an equation and by a graph the relation between distance and time. When should the train reach Buffalo (430 mi. from New York) if its speed is never changed? 12. Plot the graphs of the equations y = 2 x — 9, y = x — 3, on the same diagram. What pair of values of x and y satisfies both equations, i.e. belongs to both figures? 13. Proceed as in Ex. 12 in case of the equations: y = 10 — x, y — x—1. 22] SUMMARY 33 SUMMARY OF CHAPTER II: MEASUREMENT AND AIDS IN EXPRESSION, pp. 13-32 Part I. Simple Measurement; Negative Quantities, pp. 13-16. Simple Scale of Positive Numbers: measurement by rulers. § 12, p. 13. Introduction of Negative Numbers : thermometers ; readings below zero; other opposites. §§ 13-11, pp. 13-14. Representation of Negatives: vertical scale like thermometers; horizontal scale, positive to the right. Addition and Subtraction if a Positive Number: motion up or down on vertical scale, forward or backward on horizontal scale. Exercises I. § 15, pp. 15-16. Part II. Relation Between Two Quantities ; Graphs, pp. 17-32. Temperature Curve: development by plotting points, estimation of temperature at given time; time for given temperature ; graphs. § 16, pp. 17-20. Price Curves: usually straight lines. Exercises II, § 17, pp. 20-23. Graph of an Equation: each pair of numbers from the curve satis- fies the equation. § 18, p. 23. Equations of Prices : c= kn; straight line graph through starting point, § 19, p. 24. Linear Equation : ordinary proportion, y = kx + I ; straight line graph. Exercises III. § 20, pp. 25-29. Other examples: equations leading to figures not straight lines; graph for area of square. § 21, pp. 29-30. Review Exercises for Chapter 11: relations between quantities. Exercises IV. § 22, pp. 30-32. CHAPTER III. ADDITION AND SUBTRAC- TION; SIMPLE EQUATIONS PART I. GENERAL RULES FOR OPERATION ; PARENTHESES 23. Extension of the Operations. The student has seen C§ 15, p. 15) how to add and subtract in several new eases. -.ientary addition remain true; these properties are mentioned below. 24. Properties. One important characteristic of addi- tion is the fact that the numbers added may be taken in any order without changing the result. Thus, 3 + 4 = 4 + 3, for both equal 7. The same thing is true in all additions of elementary arithmetic. Instead of writing down every possible example, we may say that a + b = b + a, if a and b mean any two numbers whatever. 34 GENERAL RULES 35 This fact is often used to check the correctness of additions, for if a problem is solved by adding the numbers in two different orders, the results should be the same. Thus, one often adds a column from the bottom upward, and then also from the top downward. The most fundamental properties, including, for con- venience, the properties of multiplication, are the fol- lowing : I. a + b = b + a. This is mentioned just above. Example : 3 + 4 = 4 + 3. II. a x 6 = 6 x a. This has to do with multiplication. Example : 5x7 = 7x5. III. a+(6+c) = (a + 6) + c Example : 2 + (7 + 4) = (2 + 7) + 4, that is, 2 + (1 1) = (9) + 4. This rule, also, is often used in adding columns of figures; thus, instead of adding the numbers one by one, we may add groups, if convenient ; for example, if a figure 7 follows a figure 3 we add 10, instead of adding 7 and then 3. IV. a x(6 x c) = (a x 6) x c Example : 3 x (5 xl) = (3 x 5) x 4, that is, 3 x 20 = 15 x 4. This is often used; thus, (5 x 8) x 6 is easier than 5 x (8 x 6). V. a(b + c) = ab + ac Example : 5 (3 + 7) =5 x 3 + 5 x 7, that is, 5(10) = 15 + 35. This is often used; thus, 52 x 7 = 50 • 7 + 2 • 7, which is easier. This is really the principle of multiplication taught in arithmetic. These rules are called axioms, because tliey are merely- assumed to be true ; there is no pretense of proving them. We take these rules in preference to any others because experience has shown that they are the simplest and most useful. See also the footnote on p. 56. These rules are frequently named as follows : T. Commutative Law of Addition; II. Commutative Law of Multiplication ; TIL Associative law of Addition; IV. Associative Law of Multiplication; V. Dis- tributive Law. We shall not often use these names. 36 ADDITION AND SUBTRACTION [Ch. Ill EXERCISES I : CHAPTER III Show that the above rules hold if : By means of Rule V : 1. a = 7, 6 = 3, c = 5. 5. Multiply 28 by 5. 2. a = 0, 6 = 2, c = 7. 6. Multiply 63 by 8. 3. a = I, & = |, c = f. 7. Multiply 75 by fc. 4. a = 1, 6 = 2, c = 3. 8. Multiply 6| by 7. Use the above rules in order to perform the operations indicated : 9. 3 x (4x31). Solution. 3 x (4 x 3f) = 3 x (3| x 4) = (3 x 3£) x 4 = 10 x 4 = 40. 10. 12i% f $200. Solution. 12J% of |200 = (12i x r i 5 ) x $200= 12 J x ( T fo x $200) = $25. 11. 2 + (8 + 7). 14. [8 + (12 + 4)] x 5. 12. 5 x 7 + 5 x 3. 15. 4i x H. 13. 16f % of $ 300. 16. 53 x 53. 25. Addition and Subtraction: Negatives. As mentioned in § 23 : (1) To add a positive number we go forwards on the scale. (2) To subtract a positive number we go backwards on the scale. We now add to these the following : ( 3) To add a negative number we go backwards on the scale, by the amount indicated by the number. Thus, 9 dollars + 3 dollars debt = 9 dollars + ( - 3 dollars) = 6 dollars, which is the same as 9 dollars — 3 dollars. Likewise, 9tf + ( — 3rf)= 6rf =,9rf — Bd, no matter what d means. This scheme is useful. The student, will see also that in this process the rules mentioned above hold good. For example, (9 ~ _ 4 _ 3 _ 2 -1 1 2 3 4 5 [ ■ i i | i i i | i i ■ f i i i | i I I | i i ' | ' ■ ' I ' ' -i I ' ' ' I ;a»> >- To add a positfue number ) f nr1l ,„ rf j<. 0r to subtract a negative number \ Go fomardS. -4—3-2-1 1 2 3 4 5 I | i i I ) i i I i i i I i i i I i i i I l i i | i I i 1 i ' ' I I '- 1 I To subtract a positive number I T or 4 ^ f. . v Go backwards. to add a negative number j Fig. 17. (4) To subtract a negative number we go fonvards on the scale, by the amount indicated by the number. Thus, subtracting (or removing) a debt amounts to increasing the wealth of the man who owed it. If you have ten dollars and owe four dollars, your total wealth is six dollars. But if some one pays the debt for you, you will really have ten dollars ; i.e. 6d— (— 4 d) = 10 d. (5) To subtract any number, change its sign and then add the resulting number. This rule is very important, as will be seen later. 26. Addition and Subtraction of Several Numbers. In business and ordinary life we often need to add several numbers some of which are positive and some negative. Thus, if a merchant has several debts and also several different amounts of money and valuable goods, all the debts are added together to find the total debts, and the values of all the money and goods are added together to find the total "assets." The difference between these amounts represents the real wealth : it is a debt (i.e. negative wealth) if the total debts are greater than the total assets; it is real wealth (i.e. positive), if the total assets are the greater. Some examples are given below in which this process proves useful. When several numbers are to be added together, we add all the positive numbers to make a positive total, then all the 348583 38 ADDITION AND SUBTRACTION [Ch. Ill negative numbers to make a negative total; the sum sought is the difference of the amounts of these totals with the sign of the greater one prefixed to it. Thus, the sura of 6 d, — 7 d, 4 d, - 10 d, 25 d, is found as follows : 6 d + 4 d + 25 d = 35 d, the total of positive numbers. — 7 d — 10 d — — 17 d, the total of negative numbers. 35 d + ( - 17 r/) = 35 d - 17 d = 18 d, the final answer. Again, 3 d + (- 5 rf) + 8 d + (- 12 d), is found as follows : '.id +8 d = lid, total + - 5 ,/ - 12 d = - 17 tf, total - 11 r/ + (- 17 d) = -(17 - 11) d = - 6 miles an hour. l>y hoisting sails, use is made of the wind blowing in the same direction at 8 miles an hour. What is the speed at which the vessel moves ? Express in terms of positive and negative numbers. 20. A vessel is making use of both sails and steam; the latter alone will propel it at 15 miles an hour, the wind opposes the motion at the rate of 10 miles an hour, and an opposing current is flowing at 2 miles an hour. What is the speed of the vessel ? The sails are lowered. What now is the speed ? Express in terms of positive and negative numbers. 27. Addition and Subtraction of Negative Numbers. By the rules above : (1) Adding a negative number is equivalent to subtract- \ing a positive number of equal amount; they both mean going backward 071 the scale by the same amount. (2) Subtracting a negative number is equivalent to add- ing a positive number of equal amount ; they both mean going forward on the scale. Thus, 5+ (-2) = 5 - 2 = 3, ; 5- (-2) = 5 + 2 = 7, 'and so on. This holds equally well for dollars (d is used below) or for any other unit quantity : 5rf + (-2d) = 5d- 2d= 'id. 5rf-(-2rf) = 5rf+ 2 d = 7 (I. It follows that additions and subtractions of nega- tive numbers may be turned into subtractions and addi- tions of positive numbers by changing the sign of the [given quantity. Hence, we never really distinguish betiveen such expres- sions as 5 + ( — 2) and 5 — 2 ; or, in general, a -f ( — b <~) = a — b ; 40 ADDITION AND SUBTRACTION [Ch. Ill nor betiveen 5 — ( — 2) and 5 + 2 ; or, in general, a — ( — o) = a + o. TFe a/so remark that 5 — ( + 2) means 5 — 2 ; or, m general, a — ( + 6 ) = a — o. With this understanding, subtractions or additions of either positive numbers or negative numbers may be performed as above, by first thinking of all subtractions as turned into additions. EXERCISES III : CHAPTER III 1. A man has $215 in one bank, $134 in another, and goods worth $1250; there stand against him a debt of $75, and a mortgage of $750. Using d for one dollar, find his total wealth. 2. Letting d stand for one dollar, complete the following table, which represents increases ( + ) and diminutions (— ) of the wealth of a family of three brothers : Jan. Feb. Mar. Total for Tiirek Months A. . . 912 d - Ud 36 d B. . . — 721 d 255 d - 1000 d C. . . 2200 d - 133 d 756 d Total . 3. A weight of 20 g. (grams) in a certain piece of mechanism is to be raised by two wires pulling upward by 6 g. and 8 g. respectively, and a spring exerting an upward pull of 10 g. With what upward force will the weight rise if the smaller wire has been broken ? Will the total force be exerted up or down, and what will be its magnitude ? 4. If p denotes one pound, the weight of the basket of a balloon is 320 p, of the instruments 68 p, of each of 9 sand 27] GENERAL RULES 41 bags 100 p; the weight of the balloon itself is— 1500 p. "What is the total weight of balloon and contents ? 5. Let m denote miles an hour. A boat using both sails and steam is urged forward by its steam power at the rate of 12 m, and by the wind at the rate of 1 m. It sails against a current flowing at the rate of 5 m. What is its rate of progress? Perform the following additions : L 6. 12 + 9 + (-7) + (-13) + (-2). [ 7. (_7) + (-9) + 5 + (-2) + 20. 8. 8 d + 7 d + (-20 d) -f 6 d + (- 3 d). 9. (_ 7 a) +4 x + 10 x + (- 5 x) + 12 x. 10. (a) 15 (6) -5 («) lb (d) 2 2 -18 -4 -46 -82 32 -3 86 5z -17 -2 12 6 9z - 5 (/) 10 4.3 <0 -36 16 p -17 2 (e) 6 (A) -12/ -7 -7.25 -18;) -10/ -9 -6.8 12* 75/ -15 10.53 3* -68/ 20 -8.2 -15* - / 11. The average of two numbers is their sum divided by 2, I Find the average of 4 and 6 ; of — 4 and 6 ; of 10 and 15 ; of j - 10 and 15; of - 4 and - 6 ; of - 10 and - 2 ; of 5 d and 7 d\ i of 7 d and — 5 d ; of —7 c? and —5d. 12. The average of several numbers is their sum divided by the number of numbers added. Find the average of 2, 3, 4, and I ; of - 2 , 3, - 4, 5 ; of 5 d, - 7 d, and 3d; of - 10 d, - 12 d, and — 4 d. 13. In Ex. 2, find the average gain per month of A during the three months given ; of B ; of C ; of the whole family. 42 ADDITION AND SUBTRACTION [Ch. Ill 14. In Ex. 2, find the average gain of the three men during January ; during February ; during March. Find the average gain for the three months. 15. Find the average of the numbers added in Ex. 6; in Ex. 7; Ex. 8 ; Ex. 9 ; in each part of Ex. 10. 16. In Ex. 2, how much above the average earnings during January were A's earnings during that month ? How much were C's above the average ? How much were B's below the average ? Can you state that B's earnings were a certain amount above the average by using a negative number ? 17. A business house gains $5000 one year, $2000 the next; it loses $1500 the third year, and $3000 the fourth; the fifth year it gains $ 1000. What is the total gain for the five years? Find the average yearly gain for the five years. 18. In Ex. 17, how much above the average gain per year was the gain during the first year ? during the second ? third? fourth? fifth? 19. A business house during three successive years loses $500, $300, and $350. What is the total loss? the average loss ? What is the total gain, if the results are expressed as gain? the average gain? State the result as a problem in averaging negative numbers. 20. The results of a business for five years are as follows : Fik8t Teas Second Year Third Year Fourth Tear Fikth Year Gain $ 1732 -f 3000 Loss $ 5251 $ 2572 1135 What is the total net gain ? the average gain? State the result as a problem in averaging positive and negative numbers. 21. Find the average of the following temperatures: 3.5°, - 2.5°, - 1°, - 2°, — -• ' • 27] GENERAL RULES 43 22. What is the average temperature in the example of § 16, p. 17 '.' in the example on p. 18 ? 23. A rifle gives the ball the muzzle velocity of 1200 feet per second. Fired directly backwards from a moving car, the actual velocity of the ball is 1150 feet per second; how fast is the car moving ? 24. Fill in the spaces left blank in the following account of the gains of a merchant at his three stores : First Store Second Store Third Store Total 1902 1903 1904 1905 1906 $5,000 $3,000 - $ 50 -$2,250 $ 10,500 $ 8,000 $ 6,250 - $ 1,050 $ $ $ $ $ $16,(101) $11,525 $ - $ 4,500 $ 1,000 Total gain of five years $8,700 $ $28,225 Average gain per year $4,540 25. The total weight of a balloon and basket containing a man, apparatus, sand bags, etc., and small balloons, is — 25,000 p, where "p" denotes one pound. The man weighs 150 p, the basket and apparatus 50 p, the ten sand bags each weigh 75^: and the ten balloons each —bp. What is the weight of the large balloon ? 26. Carry out the following subtractions (a) 15 (c) -25 (e) -30 -17 (6) -25 (d) 25 17 -17 (/) 17 (9) 17 (0 5x -25 (*) 25 -17 (./) — 6 a; -17 -lop -25 25 — 25 p 44 ADDITION AND SUBTRACTION [Ch. Ill 27. Reduce the following expressions to simpler forms : (a) 7x + (-llx)-(-12x) + (-5x)-Qx. (5) _256-(-306) + 126-166-(-6). (c) 10d + 7 d-(-5d) + (-30d). <*) (-5) + (-5) + (-5) + (-5) + (-5). (e) (_3fc) + (-3fc) + (- 3 ^) + (-3^)- (/) 125 2 - ( - 375 z) + ( - 75 J 2) - 502 2. (g) 13 r - 50 r - (- 65 r) + 73 r - 11 r - (- 12 r). [Check each exercise by substituting some number for the letter that occurs both in the exercise as given and in your answer.] 28. Addition of Monomials. We have done one thing above which should be stated carefully. Having given 9d+3d, we write as answer 12 d ; and this is correct whether d means dollars, dimes, or anything else. Thus, 9 d + 3 d = (? + 3)rf = 12 d, 14 a; +-113= [14 + (-11)] 2: =3 as, and so on. Rule. To add terms that have a common factor, add the coefficients of that factor ; the sum is the sum of the co- efficients times the common factor. This is, in fact, merely another statement of V, p. 35 ; e.g. 9d + Sd = (9 +3)rf= 12 d. The same rule applies for the sum of any number of terms. Thus, (6(f) + (-Id) + (4 + r) 10. 1.35^ -2.7^ T_ &s¥ 11. 6 (ax 2 + by 2 ) 12. -3(/aj+gy+Jte) 13. 4(-ap + bq) - 8 (as 2 + by 2 ) - 8 O + gy + &») - 5 (- qp + &g) 3 (ax 2 + by 2 ) 7 (fx + r/.y + 7iz) - ( - ap + bq) Perform the following subtractions : 14. 12 bcyz 15. — 10 bcyz 18 21. 4 (# + // — 2) -3(x + y-2) t aryz 3 x s yz 2 16. 3.81 /- 2 17. im(m-l) -4.21V £m(m-l) J - - 5f, 19. -1. it 20. -10 (-- -1) R 2 b 2 V ~ nJ h -5 at -12 f l - - L ) R 2 lr V It) 22. 6 y^/vr + !f — 4 y^/ni' ' + f 23. - 1.3 VI- -y 1 -1.5V1- -y 2 29-30] GENERAL RULES— PARENTHESES 47 24. Subtract the last quantity in each of the first 13 exercises from the sum of all that precede it in the same exercise. 25. Average the quantities, 5 abx, —Tabx, 10 abx, — 12 abx. 26. Average the quantities in each of the first 13 exercises. 27. Average the quantities, 7 (Irx 2 + cry 2 ), — 10 (b-x 2 + a 2 y 2 ), tfx 2 + cry 2 , an d 6(6 2 .r + a 2 y 2 ) . 28. Subtract the first quantity in each of the first 13 exercises from the average of the rest of the quantities in the same ex- ercise. 30. Grouping in Addition. If longer expressions are to be added together, we use the fact that the order in which terms are added is immaterial, stated in Axioms 1 and III, p. 35 : I a + b = b + a; III a + (& + + 5) + e. Thus, we may inclose any number of terms in parentheses preceded by the sign + (plus); likewise ive may remove parentheses preceded by the sign + . This means we may add the terms inside a pair of parentheses together before we add their sum to the others, or we may group the terms in any other way we please. This is done frequently in everyday life. Thus, if a merchant owns three stores, as in the example on p. 18, he may count his liabili- ties and his assets for all these stores, or he may count them separately for each store. His total wealth will he the result of adding all these, and will he the same hy either method of working. This fact is often used to check the work. Ex. 1. To calculate 44 + 36 + 17 + 33 + 57 we may add all together or we may group in parentheses, thus : (41 + 36) + (17 + 33) + 57 = 80 + 50 + 57 = 187, or, (44 + 36) + 17 + (33 + 57) = SO + 17 + 90 = 187, ■ or. (44 + 33) + 36 + (17 + 57) =77 + 36 + 74 = 187. j Which of these is the easiest way ? 48 ADDITION AND SUBTRACTION [Ch. Ill Ex. 2. A merchant has three stores. These have the follow- ing amounts ; the debts due and mortgages to be counted as negative : Cash Fixtures Stock Acc'ts Due Debts Due Mortgage Total 1st Store .$327 $ 56 $125 $250 $ 1200 $463 $651 $ 500 2d Store $515 $2100 $590 $980 $2000 3d Store $357 $1570 $ 25 $415 $1400 Total Find (a) the total value of each store ; (b) the sum of these three to get the merchant's wealth; (c) the total cash in the three stores and the total of each of the other columns; (d) the sum of these totals to get the merchant's wealth. Do your total answers agree ? EXERCISES V: CHAPTER III Note. The student should check all answers in which letters are used, by substituting numbers for the letters at random, as suggested above. 1. The gains of a merchant's two stores are as follows for a period of five years. Fill in the spaces left blank : First Store Second Stoke Tota i. A VEIIAGE 1902 $1525 -$7530 1903 $ 1035 - $ 5000 1904 $ 105 $ 565 1905 - $ 355 $3325 1906 $ 870 $ 9565 Total for five years Average per year 30-31] GENERAL RULES — PARENTHESES 49 Perform mentally the following additions, abbreviating by grouping terms as often as possible : 2. 6 + 8 + 4 + 3 + 2. 3. 12 + 17 + 28+35 + 30 + 5. 4. lld + 6d + 9d + 3d + d + 4<*. 5. 1 7 xy + 8 xy + 12 xy + 23 xy. 6. 3 z"t + 9 z*t + 12 zH + z*t + 7 z*t. Collect into single terms : 7. (12 6c + 9 be + 8 6c) + (7 be - 3 6c + 2 6c). 8. (_ r-\' - 8 rv + 13 r 2 v) + 8 j^y + (7 >~r - 12 rV). 10. Average the expressions: (7pq -5pq +pq), (Spq - 3pq -9pq), (7 pq +pq - 3pq). 11. Average the expressions : 2 Imn — 5 Imn — 8 Imn, — 9 Imn + Z»im — 2 /m», 6 ?mw + 4 hnu — Imn. 31. Addition of Longer Expressions. We may now add longer expressions by grouping - together those terms which are " similar " (or " like ") (see § 8, p. 8). Ex. 1. One farmer has 7 horses, 12 cows, and 4 sheep, another farmer has 6 horses, 5 sheep, G cows. Find their total posses- sions. We say : (7 horses + 12 cows + 4 sheep) + (6 horses -f 5 sheep + 6 cows) = (7 horses + 6 horses) + (12 cows + 6 cows) 4- (1 sheep + 5 sheep) = 13 horses + 18 cows + 9 sheep, but we do not try to add cows to sheep or to horses. Likewise (7 h + 12 c+4s) + (6 ft + 5.s+6 c) = (7 A+6 A) + (12 c + Q c) + (4 s + 5 s) = 13 h + 18 e + 9 s, no matter what h, c, and s mean. 50 ADDITION AND SUBTRACTION [Ch. Ill Again (6 ax + 20 b 2 x + 5 ab) + (6 h 2 x + 30 ax + 7 ah) = (6 ax + 30 ax) + ('20 h 2 x + G h-x) + (5 aft + 7 a&) = 36 ox + 26 Irx + 12 ah, no matter what o, h, and x mean. Check this addition by trying a = 2, b — 3, x= 1. Try some numbers of your own choice. Notice that there is no attempt to add together the final terms 36 ax and 26 Irx and V2ab, just as there is no attempt to add cows to horses or to sheep in the first example. Rule. To add long expressions, ivrite them in columns with the similar terms in the same column ; add the columns separately ; write the total sum as the sum of these separate results. This was done in Ex. 2, p. 48. Likewise to add the expressions (2 abc + 3 a 2 x - 7 ab - 4 be + 5 ex 2 ) and (6 ab - 4 a 2 x + 5 be - 2 ex- - 7) and (4 be - 3 ab - 3 ex 2 + 2 a 2 x + 4) we write Given 2 abc + 3 a 2 x - 7 ab - 4 be + 5 ex 2 - 4 a 2 x + 6 aft + 5 be - 2 ex 2 - 7 2 « 2 x - 3 ab + 4 be - 3 ex 2 + 4 Sum 2 ate + a 2 x -4aH56c — 3 EXERCISES VI : CHAPTER III Add the expressions given in each of the following exercises: [Check each result by substituting random numbers for the letters.] 1. 5a + 76-3c, -12a + b-2c. 2. 3 x — 4 y, 7 x 4- 2 y, — 2 x — y. 3. s fx 2 _ 4 gy 2 + 7 ^ /aj2 + r/// 2 + ^ 2 /ar> + ^ _ 3 to?> 4. aZ — 8 bm, 6>u — 2 en, 4 cw — 5 al. 5. 1 (') aiV — 19 xyz, — 25 abc — 11 07/2, 15 xyz. 6. 19ap+13&g— 15cr, — 176g+cr— hap, Sbq— 11 cr— 2 ap. 7. 7 &C2/3 — 9 cazx — 3 «&#?/, — 2 &c#2, —3 caza + 9 abxy, 2 cazx — bcyz. 8. 6 x 2 - 3 x + 7, - 5 ./-' + 2 a; - 3, 2 x 2 - 9x - 5. 9. x 2 + 2 ftx + lr, x 2 - Ir, h 2 - hx - 2 x 2 . 31-32] GENERAL RULES — PARENTHESES 51 10 . 7 + 3t-%gi?, -5t + 3gfi } e-2^gt s . 11. 5(a + /o-"C'+rt)+ 4 (- i '+^), -(a+b)-(c + d), 3(a+b) + 2(c + d)-5(x + y). 12. 3 (7 x - 5 y) - 8 (2 x - 3 y) + 10 (a; - ?/), - (7 a; - 5 //) + 7 (2 x - 3 y) - 4 (a- - >y), _(7a;-5y) + 2(2a;-3j/)-5(a;-2/). [Also collect your result into only two terms.] 13. 2 (- x - 3 y) - 5 (a; + 3 y), -(- x - 3 */) + 6 (a + 3 >/). [Collect your result into as simple a form as possible.] 14. (6 ax + 9 by -3 cz) + (6 ax + 9 by - 3 cz) + (6 ax + 9 &?/ — 3 cz) + ((5 ax -f 9 &# - 3 cz). 15. Add a,.r -f &#, n,x + &22/j ":;■'' + %i by collecting the coeffi- cients of x into one term and the coefficients of y into one term. The use of letters with subscripts, as here, is very common; thus, a, (read " a, sub-one "), a,. a s , />,, h., (read " a, sub-two," " a, sub-three," " b, sub- two," "&, sub-three," etc.), are very often used in the place of separate letters. They should be treated simply as separate letters, psed to indicate separate numbers, and no meaning should be attached to the small figures except to distinguish a 2 from a 3 , for example. 32. Subtraction of Longer Expressions. Just as before, to subtract, we proceed as in addition after changing the bign of the quantity or quantities to be subtracted. If \here are several quantities to he subtracted, we must be care- ful to change the sign of each one before proceeding to the addition mentioned. Ex. 1. A firm of stockmen have $2500 cash, 600 sheep, 325 cows, a mortgage (debt) of $1500, and they owe 25 horses to another firm. What is their total wealth ? The total wealth is 2500 d + 600 s + 325 c - 1500 d - 25 h, where the letters stand for the article having the initial. 52 ADDITION AND SUBTRACTION [Ch. Ill Ex. 2. If one partner, wishing to set up an independent busi- ness, takes part of the debts as well as part of the assets : $ 500 cash, 125 sheep, 150 cows, a mortgage (debt) of $300, what is left for the others ? We must subtract (500 d + 125 s + 150 c - 300 d) from (2500 d + 600 s + 325 c — 1500 d — 25 h). To do this we change the sign of each of the quantities to be subtracted; then we add the result to the original amount. Original, 2500 d + 600 a- + 325 c - 1500 d - 25 h Add _ 50Q a _ 125 s - 150 c + 300 d Answer, 2000 d + 475 s + 175 c - 1200 d-25h Ex. 3. (4 a - 7 bx + 3 c 2 )-(6 a + 5 bx - 2 c 2 ). Original, 4 a — 7 bx + 3 c 2 6a + 5bx — 2c' 2 (Change signs mentally and add.) ^4 nswer, — 2a — 12bx + 5 c 2 This holds no matter what a, b, x, c mean. Try it if a = 4, b — 3, x = 5, c = 1. EXERCISES VII: CHAPTER III [Check each result by substituting random numbers for the letters.] Subtract the second expression from the first : 1. a - b, 2 a - 3 b. 2. 31-5 n, -l-4n. 3. 6 xy — 4 ab, —5xy-\-7 ab. 4.-3 kx 2 - 4 by 2 + 7 mz 2 , - fc.x 2 - 5 6y 2 + 2 mz 2 . 5. - 1 1 xy 2 + 2 yz 2 , - 1 a-// 2 + 3 yz 2 + za; 2 . 6. .r//z 4- a&c + Imn, av/z + 2 «6c + 3 hnn. 7. x + 2 V#2/ + y, x — 2 ~\/xy + y. 8. a 2 - 2 ab + b 2 , b 2 - 3 ab - 4 a 2 . 9. r/ryz — rpzx — pqxy, rpzx -)- pqxy — gryz. 10. cr 2 — ?/ 2 — 2 ,yz — z 2 , z 2 + x 2 — 2 yz + y 2 . 11. 2 + llt- 13 ^ 2 , 1 + 12 < - 10 ^ 2 . 12. 6 (5 a; -3?/)- 2(* + ?/), 5(5 x-3y) -3(x + y). Then collect your result into two terras. 32-33] GENERAL RULES — PARENTHESES 58* Perform the operations indicated : 13. (—9 ax- + 7 by — cz) - ( - 8 ax — by + 2 cz). 14. (2x-y)-(x-2y) + (-x-y). 15. (2x-y)-(x-2y)-(x + y). 16. O 2 - 2 hx + If) - O' 2 - 6 hx + 9 Jr) - (V 9 - ft 2 ) + (h s - /*.r). 17. (x 2 - 2 av/ + if) - (3 a 2 - 5 ag + 2 ? /) + (2 x* -3xy + f). 18. Subtract a v v + a 2 y from b^x + o 2 ?/ by collecting the coefficients of x into one term, and the coefficients of y into one term. 33. Removal and Insertion of Parentheses. We saw that parentheses preceded by the sign + may be inserted or removed as we wish, without any other change. But if a pair of parentheses is preceded by the sign — , the whole interior is to be subtracted from what goes before it. We must therefore change the sign of each term inside the parentheses when we take them away. Thus, (4 a - 7 bx + 3 c 2 ) - (6 a + 5 hx - 2 c 2 ) = 4 a - 7 bx + 3 c 2 - 6 a — 5 hx + 2 c' 2 = - 2 a — 12 hx + 5 c 2 , which is the same as the result obtained in § 32. Notice that the first term inside a pair of parentheses has the sign plus if no sign is written ; in removing pa- rentheses preceded by the sign — , care should be taken not to over- look the first term. Similarly, if we insert parentheses preceded by the sign — , we must change the sign of each term we inclose. Thus, 4 a - 7 bx + 3 c 2 - 6 a - 5 bx + 2 e 2 = (4a - 6 a) - (7 bx + ohx) + (3 c 2 + 2 c 2 ) = -2 a - 12 bx + 5 c 2 , which is, of course, the same result obtained before. Notice that the middle pair of parentheses is preceded by the sign — , and that the signs of the terms inside of it are changed as they are put inside. To remove or to insert parentheses, change the sign of I none] j? t j lg term8 if the sign \ precedes the pare n- [ each J I J theses. (Read " none " with -f ; read tl each " with - .) 54 ADDITION AND SUBTRACTION If several parentheses are used, one inside the other, remove the one farthest inside first, and the others in their order. The student should read § 11, p. 10, and should state the forms of parentheses used in algebra. After considerable practice the student should be expert enough to remove any one pair of parentheses -without touching the others; he may then find it more convenient not to use the order, specified above. EXERCISES VIII: CHAPTER III Remove the parentheses in the following exercises, and sim- plify as much as possible : 1. 7 + (-3 +2). 5. -116+[8 6-(26+6)-36]. 2 . 7_(3_2). 6. Skz-[7kz-(3kz-57czyj. 3. - G + [5 - (7 + 3) + 12]. 7. 8 kz - [(7 kz - 3 kz) - 5 kz]. 4. 13 _ [7 _ (2 _ 5)]. 8. 8 kz - (7 kz -3kz-5 kz). 9. [8 kz - 7 kz] - [3 kz - 5 kz]. 10. [8 kz - (Ikz - 3kz)] - 5 kz. 11. [G mn 2 -(SP- mrr + 3 >P - mn 2 ) - (22 mir - 8 I 3 )]. 12. 13 x>/z — [(5 abc — xyz) — (3 x\jz + 7 abc — xyz)]. In the following, remove the parentheses, beginning with the outermost : 13. —6 +[5 — (7 + 3) + 12]. Does your result agree with Ex. 3? 14. 13 — [7 — (2 — 5)]. Does your result agree with Ex. 4? 15. [G mn*- (8 P-vur + 3 n s - mir) - (22 mn 2 -8 P)]. Does your result agree with Ex. 11 ? 16. a -[_(--&)]. In the following examples rewrite the expressions, inclosing the last two terms first in parentheses, preceded by the plus sign, and then in parentheses preceded by the minus sign : 17. a + b + c. 19. a + 6 — c. 21. a— 6+c. 18. a — b — c. 20. b — a — c. 22. xy — xz + z 2 . PART II. APPLICATIONS. LINEAR EQUATIONS 34. Equations. We may apply the knowledge gained to solve many problems similar to those of Chapter II. In Chapter II, § 20, we had an equation c = £ n + 200, where c means the total cost in cents, and n the number of copies of a certain printed pamphlet. If the cost were $4.25, or 425 cents, we could find the number of copies. (1) c= l ,n + 200, or, since c = 4:25, (2) 425 = i n + 200. Subtract 200 from each side of the equation (2), (3) 225 = \ n, since, if 425 and -J n + 200 are the same number, it follows that 425 less 200 is the same number as h n + 200 less 200. Now multiply each side of the equation (;5) by 2, 450 = n, since, if 225 and \n are the same number, twice one of them is the same as twice the other. We now check this by putting n = 450 in the original equation (2) ; this gives 425 =i (450) + 200, which is evidently correct. The check used above is complete (p. 5); i.e. the check leaves no doubt whatever concerning the correctness of the answer. Whenever the answer can be tried directly in the given problem, as was done above, the check is complete. Hence, this should be done whenever possible. 35. Operations on Equations. The work just done is nothing more than we could have done in Chapter II, but it is carefully stated. In particular, we consciously subtracted 200 from each side of equation (2), and we multiplied each side of equation (3) by 2. The student will see that we may always do any one of the following things with- out disturbing the equality : 56 ADDITION AND SUBTRACTION [In. Ill I. We may add the same number to each side of an equation. II. We may subtract the same number from each side of an equation. III. We may multiply each side of an equation b} r the same number. IV. We may divide each side of an equation by the same number, when the division is possible (see p. 106). For an equation indicates that the two sides each repre' sent the same number, hence the result of adding, subtract- ing, multiplying, or dividing by the same number on both sides must be the same.* Always perform the operation upon each side as a whole, not upon a part of it. 36. Definition of Linear Equations. An equation with one unknown letter which contains the unknown number only in its first power, in the form : Ax + B = 0, where A and B are known numbers, or which can be reduced to that form by the operations of § 35, is called a simple or linear equation. Most of the equations we have studied up to this time are linear. (§ 20, p. 25.) 37. Examples. In this Chapter we have learned how to remove parentheses, how to add and subtract negative quantities as well as positive quantities. These operations assist in solving equations. Ex. 1. Given the equation 2 x — 4 = 5, where x is an un- known number ; to find x. {\) 2x-4 = 5. Add 4 to each side : 2 x = 5 + 4, or, 2x = 9. *These rules really state the axiom that there can he only one result for any addition, or subtraction, or multiplication, or (possible) division. Notice that each of these operations is performed by means of a number ; if any expression other than a simple number is used in any of them, care is necessary to insure that it does represent a Dumber. See also p. 1(><>. B5-37] LINEAR EQUATIONS 57 Divide each side by 2i x — § = 4^. Check: Tut \\ for x in (1): 2 (U) -4 = 5, or 9-4 = 5; since this is correct, we see that the value of x found is correct. Ex. 2. Given the equation (1) 2(3*-4)-12 = 2[>-l)-(2z-3)], where x is unknown ; to find x. Remove the parentheses as in § 33, p. 53. (2) 6x-8-12=2(x-l-2z+ 3), or, (3) 6x-20 = 2(-x+2), or, ( J ) 6x-20=-2a: + 4. Add 2x to each side: (5) 6 x + 2x- 20 = 4. Add 20 to each side : 6 x + 2 x = 4 + 20, or, 8 x = 24. Divide each side by 8 : x = 3. Check: Put 3 in the place of x in (1), 2 (3 • 3 - 4) - 12 = 2 [(3 - 1) - ( 2 • 3 - 3)], or, 2 (5) - 12 = 2 (2 - 3), or, 10-12=2(-1), or, -2 =-2, which is seen to be correct ; the answer x = 3 is therefore correct. (The general plan of solution is as follows : (1) Perform all indicated operations. If necessary, re- move fractional coefficients by III, § 35. (2) Transpose all terms that contain the unknown let- ! ter whose value is required to one side ; all other terms to the other side. (3) Combine the terms on each side, and collect the co- efficient of the unknown letter (§ 28). (4) Divide both sides by the coefficient of the unknown letter. (5) Check the answer thus found by trying it in the .given equation. 58 ADDITION AND SUBTRACTION [Ch. Ill EXERCISES IX: CHAPTER III Solve the following equations to find the value of the un- known letter. If there are several, solve for the letter specified. 1. x + 3 = 7. 4. x — 3 = 7. 7. x + 3 = — 7. 2. a;— 3 =—7. 5. — .r + 3 = 7. 8. — x -3 =—7. 3. 2^ + 5 = 17. 6. 9/- — x?y* 6. 2y + 7y] b j 3. lx — my n _ Ibz mx-t-ly * t 7. How is the average of two numbers x and y found ? If we know the average of x and y, how may we find the sum of x and y ? If the average of x and y is 5, what is x + y? Plot the figure, showing all points, the average of whose coordinates is 5. 8. Plot the figure of all points such that the average of x and y is — 3. 9. The average of — x and y is — 5. Plot the figure. 10. The average of x and — 3 y is 2. Plot the figure. 11. The average of y and 2 is x. Plot the figure. 12. The average of —5y and 10 is x. Plot the figure. 13. Denote by y the average of x and —10 for various values of x. What equation connects x and y? Make a table of values of x and y, and plot the figure. 14. Denote by y the average of — 3 x and 6. Plot the figure. 15. Denote by y the average of — 2 x and — 3. Plot the figure. 64 ADDITION AND SUBTRACTION [In. Ill Perform the operations indicated : 16. (ax — 2 by + 3 cz) + (by - 2 cz + 3 ax) + (cz — 2 ax -f 3 fa/). 17. (6 ar 3 ?/ 2 + 3 x 5 - 7 a-?/ 4 ) - (3 ay 4 - 4 if - 7 a; 2 / + 2 a; 3 ?/ 2 ) + (.x 5 - xhf + a; 4 ?/) - (4 x' + 4 /) + (10 xy* - 7 aft/ 3 ). 18. (a; 2 + i/ 2 + z 2 - 2 .y^ - 2 z.c - 2 a*/) - [(.r + z 2 - 2 yz) + (z 2 + a; 2 - 2 za-) + (x 2 + y 2 - 2 a*/) - (a* + y 2 4- « 2 )]. 19. (af + ?/ + z 3 - 3 a-?/z) - [(f - 3 ?/ 2 z + 3 ? /z 2 - z 3 ) + (z 3 - 3 z 2 x + 3 zx 2 - .r 3 ) + (as" - 3 afy + 3 xy* - i/ 3 )] + [(3 ?/ 2 z - 3 ?/z 2 ) + (3 z 2 x - 3 zx- 2 ) + (3 x 2 y - 3 ax/)]. Solve the following equations for the letters denoting un- known quantities : 20. 2(aj-5)+3(6-a;) = 0. 22. i(z-5)+(10-z)=£z. 2i. 6^+4-p=19/, + 25-7i>. 23. 7-[5 + (3-2n)]=15. 24. (1 - w) - [«r - (3 to - 5)] + [3 - (5 - w 2 )] = 0. 25. (6ft-9)+[5-(3fc-2)]=0. 26. (2x+7)-(5+a-) = -3. 27. 2 ax — x 2 — 2xy. (Solve for a ; then also solve for y.) Simplify the expressions : 28. [(6 xy - 7 ab) - 2 (5 xy - 3 aft)] + [(5 xy - 3 ab) - 2 (6 an/ - 7 oft)]. 29. [2 (10 p 2 + 9pg - q 2 ) + 3 (2 /> 2 - 3;>r/ + 2 r/)] - [5 (10p 2 + 9 j pr / _r/)-(2 i r-3 i >r / + 2r/-')] + [2(10 i r + 9^-^-3(2 p 2 - 30. The sum of two numbers is 3 a + b; their difference is a — b; what are the numbers? 31. The sum of two consecutive integers is 4 A: — 3; what are the numbers ? 32. How many pounds each of spice worth 20 and 50 cents a pound should be taken to form 12 pounds of a mixture worth 30 cents a pound ? 33. What is the amount at compound interest at rate r on principal p for two years ? What principal will yield $605 at 10 per cent at compound interest for 2 years? 39] SUMMARY 65 SUMMARY OF CHAPTER III: ADDITION AND SUBTRACTION; SIMPLE EQUATIONS, pp. 35-65 Part I. General Rules for Operation ; Parentheses. pp. :U-.">4. Extension of the Operations: Addition of a positive, forward motion ; subtraction of a positive, backward motion. § 23, p. 34. Fundamental properties of addition and multiplication : five rules: I. a -\-b =b + a (Commutative Law of Addition) ; II. axb = b x a (Commutative Law of Multiplication); III. a + (b + c) = (a + 6) + c (Associative Law of Addition) ; IV. a x (b x c) = (a x b) x c (Associative Law of Multiplication) ; V. a(b + c)=ab+ ac (Distributive Law). These five are axiom*. Exercises I. § 24, pp. 34—36. To add a negative number ; backward motion. To subtract a negative number: forward motion. (Adding 1 , (subtracting] \ > a negative number: equivalence to ] ,,. \ a } Subtracting ) J [ adding J positive number of the same amount. To subtract any number : change its sign and add. § 25; pp. 36-37. To add several numbers: the difference of negative total and positive total amounts with sign of greater total amounts. Exercises II. § 26, pp. 37-39. Equivalent Expressions : a +(- b) and a - 6; a - (- b) and a + b. Exercises III. § 27, pp. 39-14. To add similar monomial terms: sum of coefficients x common factor; to subtract : change sign and add. Exercises IV. §§ 28, 29, pp. 44-4:,. Use of parentheses : grouping of terms in addition. Exercises V. § 30, pp. 47-4!t. To add longer expressions: add similar terms in columns. Exercises VI. §31, pp. 49-51. To subtract a longer expression : change the sign of each term and add. Exercises VII. § 32, pp. 51 -53 . f no } . f + 1 To remove or insert parentheses: change \ \ signs if \ \ pre- cedes the parentheses. Exercises VIII. § 33, pp. 53-54. 60 ADDITION AND SUBTRACTION / Part II. Applications ; Linear Equations. pp. 55-64. Equations : typical solution ; complete check. § 84, p. 55. Permissible Operations: add, multiply, divide, subtract; operate on each side as a whole. § 35, pp. 55-56. Linear or Simple Equations : first power of unknown. § 36, p. 56. Solution of Examples: typical solutions; plan of solution. Ex- ercises IX. § 37, pp. 56-58. Transposition : change of sign ; transpose terms, not factors. § 38, pp. 58-59- . Problems stated in English : typical solutions ; directions for student. Exercises X. § 89, pp. 58-62. Review Exercises XI. PP- 63-64. CHAPTER IV. MULTIPLICATION AND DIVI- SION; FACTORING; APPLICATIONS PART I. MULTIPLICATION AND DIVISION OF NUM- BERS AND MONOMIALS 40. Multiplication. As in the case of addition, we wish to extend the idea of multiplication so that we can mul- tiply any kinds of numbers together. We shall always keep the rules given on p. 35, and in every other way we shall try to preserve the spirit of what was known as mul- tiplication in elementary arithmetic. Multiplication was originally applied to integers only. In multiplying 5 by 4 we take 5 four times and add ; thus, 5 x 4 = 5 + 5+5 + 5 = 20. This notion very clearly does not hold for fractions. In multiplying 12 by | we do not take 12 two thirds times, for that is absurd. To multiply 12 by | we may note that 12 x | = | x 12 = | + | + -..(12 times) =8, by virtue of our Rule II, p. 35. In general, if a, b, c, are any positive integers, b b b , b , , N ab ax- = -xa = -H \- ■■•(a times) = — c c c c e If we now take, say f x |-, we note that (f x f) x 35 = f x (I- x 35) = f x 20 = 12, by IV, p. 35. Hence, (f x f) x 35 = 12, or, by IV, p. 56, | x 4=i2. 07 68 MULTIPLICATION AND DIVISION [Ch. IV Likewise, in general, for (^x-)xbd = -xt-xbd) \b dl b \d I a c _ac b d~bd' = - x cb. b = ac. Thus,we arrive at the usual rule for multiplying fractions by means of the Rules II and IV, p. 35. The elementary definition of multiplication (for in- tegers) is: The product of two integers is found by taking the multiplicand as many times as is indicated by the multi- plier and adding. We extend this definition for other numbers by saying : The product of any two numbers shall be such that the rules of p. 35 shall remain true. Thus, the rule for fractions results in this way, as shown above. We shall use this principle to find the product in all new cases. As another example, consider the product of any num- ber and zero ; say 4x0. We say 0x4 = + + + = 0. But, 4x0 = 0x4 by Rule II ; hence, 4 x = 0. The product of any number and zero is zero. The proof shows this to be true for any integer only. The follow- ing argument shows that it is true also for any fraction : To find Ox-; consider (c — c)x-=0x-- b v ' b b b b b b This proof need not be learned at this time. 41. Products of Negatives and Positives. It frequently becomes necessary to multiply a negative number by a positive number. 40-41] NUMBERS AND .MONOMIALS 69 Thus, if a man makes ten debts of $50 each, his total debt is $500, i.e. (- 50 d) x 10 = - 500 d. This holds whether d means dollars or anything else. In general, ( — a) x b = — ab. To multiply a negative quantity by a positive quantity, multiply the two amounts as if both were positive and prefix the sign — to the answer. This is convenient also in solving equations. Thus, the equation 3 x — 6 = gives x = 2 ; cheek : 3(2) —6=0 (correct). But if we first transpose K 6x, we get — .6 = — 3a;; an attempt to put 2 in place of x here gives — 6= — 3(2), which must be correct in order to avoid contradicting the previous check. Since the factors should give the same result when multiplied in any order, the same rule holds for multiply- ing a positive number by a negative number. In formula : a x( — by = —ab. We can easily prove that these results must be true if the rules on p. 35 hold true, but it is not necessary for the student to learn this proof at this time. For a x [ft + ( - ft)] = a x (ft) + a x (- ft), by Rule V. But, « x [ft + ( - ft)] = since 6-6 = 0. Hence, ah + a(— ft)= 0, or, a x ( — ft) = — ab, which is one of our rules. Moreover, ( — ft) x a — a x ( — 6) by Rule II, = — aft, which is the same as our rule above. EXERCISES I: CHAPTER IV 1. A man owes twelve debts of S30 each. What is his total indebtedness? State the problem in terms of negative numbers. 2. Each inhabitant of a city pays a certain tax of $ 10. What is the increase in the wealth of the city due to the de- parture of fifteen residents? 3. How much is the downward pressure of a bar of iron lessened by the attachment of twelve springs, each pulling up 4. 5x-2. 9. 5. 7x-3. 10. 6. 4X-1. 11. 7. -4x1. 12. 8. -2x5. 13. 19. — 5x6(6c4 -ca + aV 70 MULTIPLICATION AND DIVISION [Ch. IV with a force of 75 pounds ? Express in terms of negative numbers. Perform the following multiplications : 3x(-lx7). 14. 2 x -(3^x2!) 7 x - 15 d. 15. x - 3.* G x -13ft. 16. -3x0. 6 x -5 x 9 M. 17. -2i x x 3 x. 9x3 x 2 i/z 3 . 18. - 7.2 x 2.3 x*y. 20. -3(ar+r-2 xy) x 10. 21. -13(j£- + -2- + -*2-)x0. \q — r r—p p-qj 22. In arithmetic we multiply by taking the multiplier as many times as the number of units in the multiplicand and adding. Supposing this to be a proper method whenever the multiplier is a positive integer, show that — 5 X 12 = — GO ; 0x7=0. 42. Negatives multiplied. Tt is frequently necessary in algebra to multiply one negative number by another. It is convenient, and, in fact, necessary if we are to avoid contradictions, to say that the product is positive. Thus, given the equation 3x4-6 = 0, we find x = - 2 ; check : 3 ( _ o) + 6 = (correct) . I f we first transpose 3 x, we have 6 = - 3 x ; an attempt to put - 2 in place of x in this form of the equation gives 6 =- 3(- 2), which must be correct if we are to avoid contradict- ing the previous check. Hence, we say (— 3)(— 2)= 6. In general, (— a) X (— b) = + ab, or, To multiply one negative number by another^ multiply their amounts as if both were positive and prefix the sign 4- to the answer. * Observe that +0 = — 0, since each means no motion al all on our scale. Hence, in the answers, write merely whenever either + ui — would result. 41-43] NUMBERS AND MONOMIALS 71 This is easily proved by the rules of p. 35. For (-a)x(ft-6) = (-a)x 6+(_ a )(-6), or, = - ab +(- a)(- 6), or, + af> = (- r/)(- /;)• [The student need not learn this proof at this time.] The preceding rules may be stated as follows: — a x + b gives — a£>, + a x — b gives — aJ, — a x — b gives + ab, and, of course, + a x. + b gives + aJ, or, Rule of Signs : 7// multiplying, like signs give + and unlike signs give — . EXERCISES II : CHAPTER IV Perform the following multiplications : 1. - 2 x - 3. 5. - 7 x - 5. 9. - 13 x - 5 d. 2. — 3 x 7.3. 6. 5 x — 7. 10. — 4 x — 7 xyz. 3. 7x — 5. 7. — lx— 1. 11. 7x— 5x— .2. 4. — 7 x 5. 8. — 13 x 0. 12. — 1 x — 1 x — 1. 13. 2 x - 3 bhjz x - 13. 14. - kz x - 3 x - 5. is. — I x || (/r + «•■* + «y) • 16. - 7 afy x x - 3|. 2 s, _ 3 A/'- - *Y no _ qv 5 *L 17. _^ X -x ^ • 18. -9X 9 8V,(/s + /V 3 m 3 — n 2 19. — lX— lx— lX— 1. 20. -lx -lx— lx-lX — 1. 21. (-3) 3 . 22. (-3)\ 23. (-3) 5 . 43. Multiplication of Monomials by Rearrangement. Mo- nomials mug be multiplied by rearranging their factors. Thus, 4ulj x '■! ab =(4 x a x b)x (3 x a x ft) = (4 x 3) x (« x a) x (/> x fr) = 12 x « 2 x b\ since, by Rule II, p. 35, the order of multiplication is immaterial. 72 MULTIPLICATION AND DIVISION [Ch. IV Check: If we put a = 2, b = 3, we find 4 • 2 • 3 x 3 • 2 • 3 = 12 • 2 2 . 3 2 = 12 • 4 ■ 9 = 432 (correct). Likewise, ( - 4 ab) x (3 ab) = - (4 ab) (3 aft) = - 12a 2 6 2 , by the rule for signs given just above. Check: If we put a = 2, b = 3, we find ( _ 4 . o • 3) x (3 • 2 • 3) = - 12 • 2' 2 • 3- = - 12 x 4 x 9 = - 432 (correct). EXERCISES III: CHAPTER IV Perforin the following multiplications by the method just given. Check each result by substituting numerical values for the letters. 1. —5a 3. — 12x 2 y 5. Ip 6 b - 3 xy 2 -&(k + m + n) 2. — 2Jcr 4. 9m? 2 MJ 6. — 2(x + y + z — 1) - 3 x- - 7 ;to 3 m 2 - 3 (as + y + z- If 7. — 2 a X — 2ax— 2 a. 8. — 7 /-y x 3 cc^a x — to 3 . 9. (-3 xf. 10. (2m 2 »t) 2 . 11. (-3a 2 ?/z) 4 . 12. - (a + 6)(a; 2 + y-f x 3(.r + //->(„ + hf x - 2 (a + />)(.» 2 +?/ 2 ). 13. l 2 m s n 4 p 5 c 2 . 4. 9 (2 2 x 3 3 ) x 3 (2 x 3 2 ) . 8. 2. 5 m?n s x - 8 i»y x + £ rip\ 9. 5/V% 3 x - 3fffh x - 2/ 2 7i 3 . 10. 7.2 a?bc 2 x - 4.1 ab 3 c x 1.3 a 2 6c 3 . 11. -S(y - z)(z - x) x2(z -x)(x- y) 2 x - (y -z). 12. -2(-3 &c)(-2ca) x -3(-2ca)(-a&) x (- a&)(- 3 6c). 13. 5V%x-3(Vi) 3 x-2(V%) 5 . "•- i8 (?)(^) Sx I(tJ(^) x -K^ Simplify the result as much as possible. 15. a m x a". 16. — a m x — a\ 17. (— a) m x (- a) n . 18. (-a) 2 ; (-a) 3 ; (-a) 4 ; (-a) 5 ; (-a) 6 ; (-a) 7 . 19. (— a)'", if m is even. 20. (— a) m , if m is odd. 21. (-2 a) 3 . 22. (3rsH 3 ) 2 . 23. (- 4 x 2 yf(2 xy 2 )\ 46. Division. Division is the reverse of multiplication ; the result of division is called the quotient ; that is, dividend -J- divisor = quotient, if quotient x divisor = divi- dend. For example : 12 ^3 = 4, for 4x3 = 12. Again, f h- f = - 1 /-, for V x § = |. a C w/ r nil - -f- - = — , tor — - h d be be We shall consider this again later (Chap. V, p. 134). T , n c nil <■ nd c a In general, --=-- = — , for — »x- = — ■ h d be be d b 45-46] NUMBERS AND MONOMIALS 75 A negative number divided by a positive number gives a negative result ; thus, - 12 -*- 3 = - 4, for - 4 x 3 = - 12. A positive number divided by a negative number gives a negative result ; thus, ( + 12)-h(_3)=-4, for -4 x- 3 = 12. A negative number divided by a negative number gives a positive result ; thus, (-12) -j- (-3) =4-4, fori x -3= -12. The rule may be stated in short as follows : Rule of Signs. In division, like signs (jive •plus ; un- like signs give minus. Notice that this rule reads precisely like the rule for signs in multi- plication, p. 71. Let the student show why this should be the case. Zero divided by any number, not zero, gives zero ; thus, 0-3 = 0, for 0x3 = (p. 68). The quotient of any number divided by zero does not exist ; thus, 3 — = ? (does not exist), for the question, ? X = 3, has no answer. EXERCISES V: CHAPTER IV Perform the following divisions : i. _i8-5--G. 3. 18 -s- -G. 5. _8h--2. 7.-5-2. 2. -18-6. 4. 15 -5- -3. 6. 7 -5- -3. 8. -13 -5- -5. 9. 6 a -*- — 2 a. 10. 15 bcp — - 3. 11. - 17 xyz + —2xyz 12. -3(6c + ca + a6)-4--2. 13. _^--^. 2 " 2 Rw J \ Rw 15. — abcxyz -. — abc. 16. Oh — G jryz\ 17. — 51 m -*- 0. 14. uf*- 1P \+-7" 1 * ^ 70 MULTIPLICATION AND DIVISION [Ch. IV 47. Division of Simple Powers. We may rind out how to divide many other expressions by the same reasoning, when the exponents are positive integers. Thus, a 5 -s- a 3 = a 2 , because a 2 x a 3 = a 5 , and a" -h a 4 = a 3 , because a 3 x a 4 = a 7 , and a m -r- a" = a m ~ n because a m ~" x a" = a m . This rule is proved now only when m and n are positive integers, and m is greater than n. Later we shall prove that it holds for other values of m and n. See Chapters VII and XI. Hence, in dividing powers of the same quantity, the ex- ponent in the quotient is the difference of the exponents, the exponent in the divisor being subtracted from that in the dividend. The rule is seen also by writing out the factors in full : ( f + a* = «•»•";*•*•*•* = a.a.a = a% V/- 12 -0. - 17 Mv -4- - 5 HV. 16. (a?+.vWa+6)*+(s - 6 a& - 21 afy-V t» 2 y> 3 X — 5 m/? a.rhi m z n X b x"~ l y l ~ 10 wV abx ,, - m y'- m x -ffVfr" 22 Qr +y) 8 (a + 6) g .4#a& 2(a; + ?/) 2 (a-f-&) 2 ,_ — 13 mnV x — 5 m/> __ a.r l >/ m z n x bx n ~ l y l ~ m z 18. • 21. 10 wV ,_ ^l 8 a'6» x --BV5" „„ (.r+y) 3 (a + &) 2 x(a-+y) 2 (a+&) 3 PART II. MULTIPLICATION AND DIVISION OF LONGER EXPRESSIONS 49. Monomial x Binomial. In § 28, p. 44, we saw that 9d+3d = (9+3)f expressions. Thus, given the expression 8 a% — 12 a 2 b 2 + 20 ab s , we can see by inspection that 4 ah is a factor, and we write 8 a% - 12 aW + 20 ah* = 4 ab(2 « 2 - 3 ab + 5 P ) . [Let the student read again and state the definition of factor, § 8,p.8.] EXERCISES VIII: CHAPTER IV Perform the following divisions ; check each result by substitution of numerical values : kx — 3 k „ 6 a 2 — 9 ab _ — 12 xy — 16 ax + 3 x 1. 2. • 3. — • k —3 a — 4 x a 2 b 3 — 5 a s b 2 — a 2 b 2 „ s 2 — as — bs — cs 4. • 7. • - a 2 b 2 3 s 2fx 2 + 3gxy-7hxz 9 ct-x- 3 - 12 a¥ - 27 ofa 2 2 x — o ax~ i^ 2 -7^ 12 afyV - 9 ays 8 +15 aryV i gr 3 x->/-z- Factor the following expressions as the product of a mono- mial factor and another expression : 10. ab + ac. 16. 2 ar? — 6 arx 2 + 8 a 8 x. 11. ±xy + 6xz. 17. 15 m 3 n 2 — 20 mV +25 »;V. 12. 5 a 2 b s - 10 a s b 2 . 18. 8 a&V - 12 a 2 ffc + 10 «. s 6 2 c. 13. 6 m 3 n 5 - 8 mV. 19. 6 rW + 1 2 rW - 18 rW, 14. _ 9 ar fi - 12 a 2 r 2 . 20. 8 apY + 24 6pY — 32j»y. 15. — 1 1 «pr/ 2 + 13 a/) 2 g. 21. — 10 x 2 yz — 5 xy 2 z — 15 scyz 2 . 22. Factor the numerator in each of the exercises 1-9. 23. Express the answer in each of the exercises in Ex. VII as the product of two factors. 24. If twice a number is known, how is the number found? If 2 a; = 12, »=? If 2x = -8,.x= ? If ax = a 2 , x = ? 25. I f ax = a 2 — 2 ay, x = ? If ax = a 2 + ab + ay, x = ? 51-52] LONGER EXPRESSIONS 81 26. If 2 x - 3 = 5,x=? If ax - ab = ay, x=? 27. 12 x + 16 // = 48. Reduce to simpler form and plot the figure both before aud after this reduction. Simplify : 28. (ax'" + bx") -=- af. 30. (cCy — 2 aSy) -=- a;?/. 29. (a./""//" + &»V) -s- x r y s . 31. (5 x'-" - 15 x m ~ H ) + - 5 a"-". 52. Product : Two Binomials. In the product of two binomials, for example, (a + b)(c + d), we may write first ( a + j^ ^ + ^ _ a (c + c i ) + i, {e + #) by what precedes, if we regard (e + d) as a simple quantity. Now, a(e + d) = ac + ad, and b(c + d) = be + bd. 1 1 cnce, (a + b) (c + d) = ac +- ad +- be + bd. Notice that this is true for numbers of various kinds : (4 + 2) (3 + 6) = (4 + 2) • 9 = 4 ■ 9 + 2 • 9. But, 4-9 = 4(3 + 6) =4-3 + 4.6, 2-9 = 2(3 + 6) = 2.3 + 2-6. Hence, (4 + 2) (3 + 6) = 4 • 3 + 4 • 6 + 2 • 3 + 2 • 6 (correct). As an example, consider (2 ab + 3 ft 2 ) (5 a — 4 6). We write this in the form : Multiplicand : 2 ab + 3 b 2 Multiplier : 5 a — 4 b First partial product : 10 a' 2 b + 15 aft 2 Second partial product : — 8 ab' 2 — 12 b a Total product : 10 a 2 b + 7 aft 2 - 12 6 8 Care must be taken not to overlook negative signs. In writing down the partial products we write in one line the product (2 ab + 3 fe 2 )(5 a), and then the product (2 ab + 3 6 2 )( - 4 6). The student should he careful to write similar terms in these products underneath one another, so that they shall be conveniently placed Eor addition. 82 MULTIPLICATION AND DIVISION [Cu. IV 53. Product of any Expressions. The product of longer expressions may be found by a similar rule. In any case we multiply each term of one of the expressions by each term of the other in some convenient order and add the resulting partial products. The work will be easier if the terms in each of the given expressions are arranged first in a definite order ; thus, if some letter is selected that occurs in most of the terms, the terms may be arranged so that the exponents of that letter increase as we go toward the rigid or left. Thus, to multiply, + 4 x - 3 x* + 2 by 2 + 5 x 2 - 3 x, we rearrange with respect to the letter x and write: Multiplicand : 2 + 4 x - 3 x 2 Multiplier : 2 - 3 x + 5 x 2 First partial product : 4 + 8 x — 6 x 2 Second partial product : — x — 12 x 2 + 9 a; 3 77/ ird partial product : + 10 x 2 + 20 x s - 15 x* Product : 4 + 2 x - 8 x 2 + 29 X s - 15 a; 4 Check: If x = 1, multiplicand = 3; multiplier = 4; product = 12 (correct). An arrangement in which the exponents of the chosen letter in- crease as we go toward the left is equally good. In either arrange- ment the similar terms fall under one another if we shift the partial products to the right by one term each, provided all positive integral powers up to the highest are present in both factors. If any terms we should naturally expect are absent, care should be taken to put similar terms in the partial products under one another. EXERCISES IX : CHAPTER IV Perform the following multiplications, and check as usual: 1. X + y 4. X ~ y 7. a + b 10. a ~- uh + h ~ a + 6 • — a — b a — b a + b 5 <(+b a + b 6. a-b a — b 2x-S y -x + 5 y x + 2y 2 x - y a -" y ii. — x + 5y 9. * + -" 12. a 2 — ab + b- a—b a 2 + ab + lr a + b 93] LONGER EXPRESSIONS 83 13. a s +db + b 2 22. 2ar i -3a 8 -ar ! + 7a;-3 a-b 2 x - 1 14. .r + 3.e-4 23. aW — 2 a?x* — 3 ax 4 + X s 2.>;4-3 - a 2 4- a.r - 2 x 2 15 . _ m 4 + 2m 2 -7 24. 2.r>-7.r-4.r + 2 2 m 3 -3 ^_5,v + i 16. r > + .- )r _i 25. 4r 3 -6r 2 + r4-l ,■8 + 3 r + 4 -r 2 4-3r + 2 17. .,.-y _ 3 ., 7/ + 4 26. 3m 4 -2m 8 +6m 2 +4ra-3 _ 3 .,• ,/ _|_ 5 2 m 3 — 2 m 4- 3 18. l_4. r + x 2 27. _2.r 4 4-3.r-5 3 + 3 .« - 2 .r a^ + g^ + g + l 19. a?-3xy + 2y 2 28. l-2x i + 3x s —4:X* 3 a; -2 ?/ l-*- 2 + 2.r 20. ^ + 5-305 29. a? + 2-3si? + x — 4rf 2 .)• - 6 + a? .r ' 4-1-3 a 21. 1 >*-2pq- + 7rf 30. >' 4 - 3 ft + 27^ 2 - rt 3 4- 1* if+yq + 'f l^-f + rt Find the following products and powers : 31. (x-l)(x-2)(x-3). 32. (x-a)(x + a)(x 2 +a 2 ). 33. (a — 5 b) (a 4-3 b) (a— b). 34. (a + b)\ 36. (a + b)*. 38. (a-&) 8 . 35. (a4-&) 8 . 37. (a-6) 2 . 39. (a - b)\ 40. (a 4-6 4- c) (a 2 + b 2 + c 2 — be — ca — a&). 41. (s 2 -4)(a 2 4-4). 42. (^ + y m )(^-2r> Show that : 43. (aj 2 4 ■>•.!-) = x* + .r>r 4 !/*■ 44. (a 2 + b-) (x 2 + .v 2 ) - (ax + &y) 2 = (ay - fcc) 2 . 84 MULTIPLICATION AND DIVISION [Cii. IV 54. Division of Longer Expressions. Just now we shall not attempt to divide one expression by another except in a few easy cases in which the quotient is known to be simple. This is the case if the dividend is really known to be the product of the divisor times some simple expression. Thus, taking the example worked out in § 53, we know that we can divide 4 + 2 x - 8 x 2 + 29 x 3 - 15 x* by 2 + 4 x - 3 x 2 , and we know that the quotient is 2 — 3x + 5 a; 2 because we just multiplied the latter two expressions and found their product to be the first expression here mentioned. If the quotient were unknown, we should write down the following scheme (see explanation below) : „. . Dividend : 4 + 2 x - 8 x 2 + 29 x 3 - 15x 4 2 + 4 x - 3x a 3 x + 5 x r\ t2 1st Partial Product: 4 + 8 x - 6 x 2 (subtract) Quotient 1st Remainder : - 6 x - 2 x 2 + 29 x 3 - 15 .r 4 2d Partial Product: - 6 x - 12 x 2 4- 9 x 3 (subtract) 2r/ Remainder : 10 x 2 + 20 x 3 - 1 5 x* M Partial Product: 10a a + 20 x 3 - 15-r 4 (subtract) Final Remainder : Check: If x = l, dividend = 12 ; divisor = 3 ; quotient = 4 (correct). The final remainder being zero, the division is said to be exact. This scheme is useful in rediscovering the partial products in the work in the example of § 53. The explanation is as follows : (1) The first term of the dividend divided by the first term of the divisor gives the first term of the quotient. Otherwise the work of § 53 would not give the first term of the product as shown. (2) This first term of the quotient x the divisor is the first partial prod- uct in i 53 : we place it underneath the dividend. (3) The difference between the dividend and the first partial product must be all the rest of the. whole product in § 53 ; ice therefore subtract them (result called " 1st remainder"). The next steps are all taken for similar reasons and are really repe- titions of the above steps ; they are : (4) (First term of first remainder) -f (first term of divisor) = (second term of quotient). (5) (Second term of quotient) x (divisor) = (second partial product). 54] LONGER EXPRESSIONS 85 (6) (First remainder) — (second partial product) = (second re- mainder). (7) (First term of second remainder) -*- (first term of divisor) = (third term of quotient, last in this example). (8) (Third term of quotient) x (divisor) = (third partial product). (9) (Third remainder) — (third partial product) = final remainder (= in this example). Steps (1), (2), (3) are simply repeated as many times as necessary. Thus, the steps (4), (5), (6) and the steps (7), (S), (9) are the same kind of steps. If the final remainder is zero, the division is said to be exact; this always happens if the dividend is really the quotient multiplied by another simple expression. If the remainder is not zero, the division cannot be entirely carried out ; in that case what is left over is called the final remainder, or simply the remainder. This will usually happen if the example has not been carefully selected to avoid it. Thus, if we try to divide 6 - 10 x + 3 x 2 - 7 x 3 - 24 x 4 by 2 + 4 x — 3 x' 2 , the work is as follows : Dividend : 6 - 10 x + 3 x 2 - 7 x 3 - 24 x 4 1 2+ 4.r- 3x 2 Divisor 1st p. p. 6 + 12 a;- 9x 2 (subtract) [3-11 x + 28a:' 2 Quotient 1st rem. - 22 x + 12 x 2 - 7 x 8 -24 x* 2d p.p. -22x-44x 2 + 33 x 3 (subtract) 2d rem. 56 x 2 - 40a; 3 -24x 4 Zdp.p. mx 2 +U2x 3 -Ux i (subtract) — 152 x 3 + 60 x 4 Final Remainder In this case the division is " not exact " since the final remainder is not zero. Just as in arithmetic, we may still express the result; thus 31 -f- 7 gives quotient 4 and remainder 3 ; we say .... , ... .. . . remainder dividend -r- divisor = ouottent + ■ — — , divisor i.e. 31 -^ 7 = 4+ f. Similarly, here we say (6 - 10 x + 3 x 2 - 7 x 3 - 24 x*) + (2 + 4 x- 3 a; 2 ) (dividend) (divisor) (remainder) — 152i 8 4- 60 r 4 =V-iix + 28x*)+ 2 7 4g !aJ[ -- (quotient) - ^.^ 86 MULTIPLICATION AND DIVISION [Ch. IV It is absolutely necessary for all this ivork that both divi- dend and divisor be carefully arranged in the same order (either exponents increasing to the left or to the right) ■with regard to the same letter. In the examples just worked the exponents increase to tlw- right ; we may work the same examples equally well with exponents increasing to the left. The work for the example of p. 83 follows : Dividend : - 15x 4 -f 29 x s - 8x 2 + 2x + 4 - 15x 4 + 20x 3 + 10 x 2 9 x 3 - 18 x 2 + 2 x + 4 -3x 2 + 4x + 2 Divisor 5 x 2 - 3 x + 2 Quotient 9 a: 3 - 12 x- 2 - 6x - 6 x* + 8 x + 4 - 6 x* + 8 x + 4 Final Remainder Notice the work is really the same as before ; the answers also are in this case the same, but with the terms in reverse order. But if the division is not "exact," the work and the answers change when we change the arrangement. Thus, in the first example on p. 85 : Dividend: -24x 4 - 7x 3 + 3 a? -10 x + fi -24s 4 +32x 3 + 16 x 2 -39x 3 -13x 2 -10x- + G -39 a: 8 + 52 a: 2 + 26 a; 3 .<" 2 + 4 x + 2 Divisor h - B / Quotient - 65x 2 -3ox + G -65x 2 + *fax-+ 4 a — *|& x — 1 l' i Final Remainder This work is totally different from the work done before on the same example. It is for this reason that very few examples of this kind are given below. However, the student need not be surprised; for example, 31 -*- 7 = not only 4+^; but also 3 + ! J\ and 5— f Moreover, in the form on p. 85, the division maybe carried to further terms, if desired, just as 31 -f- 7 = 4.42 +'—> >» arithmetic. If several letters occur in a problem, one of them — usually the* most prominent one — is used for the purpose of arrangement. If another letter is used for arranging the expressions, the work may be wholly different, but the results will always be precisely the same if the division is exact. Thus: 54] LONGER EXPRESSIONS 87 Dividend : 2 x — 3 y Divisor x 3 — 7 x-y + 5 xy 2 — 4 y 3 Quotient 2 x 4 - 17 x*y + 31 xhf - 23 xif + 12 # 4 2x 4 - 3x 3 # - 14 x 3 y + 31 x 2 y 2 - 23 xy 3 + 12 # 4 - 14 x a y + 21 j-y 10 xy - 23 x# 3 + 12 ^ 4 lPx'ty 2 — 15x + /,). 3. (3x 2 +5.«-2)-(.r + 2). 11. {a?-tf)+(x-y). 4. (a 2 -12 ab -13 6 2 )-=-(a -13 ft). 12. (r» + l)-=-(r + l). 5. (ab-ac-bd + ed) + (a-d). 13. (m 8 + 8)-5-(m + 2). 6. (A;«-8P + 16)-(^-4). 14. (.r 4 -l)-=-0-l). 7. (a 8 -3a 2 & + 3a& 2 -& 8 )-s-(a-&). 15. (m* - w 4 ) -s- (m + n). 8. (^ + 4a-7/ 2 + 4^)-(.f 2 + 2/). 16. (a*-tf)+(x-y). 88 MULTIPLICATION AND DIVISION [Ch. IV 17. (A* + 3 A-B + 3 AB 2 + B*) -=- (A 2 + 2AB + B 2 ). 18. (2-x- 4x 2 + 17 ar* - 12 a; 4 ) -*- (2 + 3 a; - 4 af). 19. (6 x- 4 - 2 a- 3 - 8 a,- 2 + 4 a> - 8) -=- (3 a; 2 - a? + 2). CO. (6 - 2 wi - 8 m 2 + 4 m 3 - 8 m 4 ) -s- (2 — 4 m 2 ). 21. (8a^+10a ! 4 + 5ar + loa: 2 -12a- + 2)-(2a; 2 + 3a:-l). 22. (4 r 6 + r 5 + 2 /- 4 - 4 r 2 - r - 2) + (4 r - 3 r 2 + r - 2). 23. (2-17 r+46 ^_43 1^+20 ^Vl8 r^+8 r")-(l-4 r+2 r 2 ). 24. (24 ?/ 6 - 28 / + 14 f - 10 y" + 5 ?/ 2 - 1) -- (4 / - 2 y - 1). 25. (4 + Ap +p 2 - 3 p s - 2p 4 + p 3 +/') -s-(4 - 3p 2 + p 4 ). 26. (2 a,- 6 + afy+4 rttf- 2 tftf-xtf-ltf) -*- (ar 5 +afy+an, 2 + /). 27. (a; 5 + 7 ar 3 + 14 a? - 15 - 11 x- 2 - 2 **)-*-(- a; + 3 + x 2 ). * 28. (a- 4 - X s - 7 x 2 + 28 * - 11) -5- (a- 2 - 4 a- + 7). 29. (aj 2 — a + &- a; + <*&)-=- (a: — a). * 33. a* + (a» — x — 2). *30. (a; 4 - 2 a; 8 -10 a;)-*- (a: 2 - 3 a; +1). 34. (/- l)-*-(y 2 - 1). 31. (a 4 +& 4 +.a 2 & 2 )-s-(a 2 + & 2 -a&). 35. («n e — n 8 )-*.^ 8 — n»). 32. (m 5 + 32rc 5 )-*-(m + 2w). * 36. (# - 4)-=- (A; 2 - 1). 37. (72 3 - 6 # 2 + 11 fl - 6) + (R 2 - 5 fl + G). 38. (./' + x + 5 - 2 a? - 7 a,- 2 -f 2 a: 4 ) -=- (2 a; 2 - 5 - x + ar 3 ). * 39. (r 5 - 4 >- 4 + 5 r 3 -8 v 2 + 8 r - 6)^(^-3^ + 2 r- 7). 55. Polynomials. The expressions in the exercises given in this chapter are all very simple. They are in fact poly- nomials. (See § 9, p. 9.) The most general polynomial in the single letter x is a + bx+cx* + dx z + ••• +lx n where a, b, c, •••, I are constant numbers, possibly zero, and where n is some positive integer. 54-55] LONGER EXPRESSIONS 80 The most general polynomial in two letters, x, y, is a + bx+ cy + dx? + exy +fy 2 + (a finite number of terms), where a, 6, c, ••• are constant numbers, possibly zero. Each term of a polynomial contains each letter, if at all, only as a factor which is a simple power. If a letter oc- curs under a radical sign, or in the denominator of a frac- tion, the whole expression is not a polynomial in that letter.* Thus, 3 — 2 x + 5 x 2 is a polynomial in x, 4 my 2 + 6 m 2 y — 7 m 3 is a polynomial in y and m. 5 x V# + 3 x 2 y is a poly- nomial in x, but is not a polynomial in y (because \/y is not a simple power). The expression Ix 2 — 3 + - is a polynomial in x, but is not a polynomial in y. All the expressions in this chapter are polynomials in all the letters in them, except some few exercises that contain fractions or contain letters as exponents. There will later be many examples that are not polynomials, in the Chapters on Radicals (Chapters VII, XI) and in the Chapter on Fractions (Chapter V). The distinction is of especial importance in Factoring. (See p. 91.) REVIEW EXERCISES XI: CHAPTER IV Perform the indicated operations ; check as usual : 1. (a 2 -a + l)(a 2 + a + l). 3. [x 2 -(a + b)x+ab](x - c). 2. ( v _5)(?j + 3)<>-2). 4. (a 2 + 6 2 -a&) 2 . 5. (m 3 - 2 m 2 + m - 3) (m 2 - 3 m - 2) . 6. (2x-3y)(3y-5z)(5z-2x). 7. (Sx 2 - 4 x + 5)(2 .r 3 - 3 x 2 - 4a- 2)(1 - 2 *- 3 x 2 ). * Although many elementary text-books define this word no as to in- clude any kind of terms whatever, the author of this book can find no standard authority for any other definition than that here given. See, e.g., Bocher, Introduction to Higher Algebra, pp. 1-4. 90 MULTIPLICATION AND DIVISION 9. (f + 2 rs + 4 1-) (r 2 - 4 f) (r 2 - 2 rs + 4 « 2 ). 10. (V - 5 x 4 y - 6 xhf + x 2 f-2 xy 4 + f) (3 x 2 - 2 /> 11. (2 x - a 2 + 1) (s 3 + 2x-x* + 4) (as + 3). 12. O/ + x 2 + 23 a; - 15 t — sr Show that : 22 a>* + l x 2 — 6 x + 5 (ar 3 -3afy4-3a?/ 2 -y 3 )(^4-ay4-?/ 2 ) = (a . _ , /); a?--- 3 (9 ra 2 + 3 ran + w 2 )(9 m 2 - 3 win + n 3 )(9 m 2 - n 2 ) _ -, (27ra 3 -n 3 )(27ra 3 4-n 3 ) 24 (a 2 + V + c 2 )(-r 2 4- y 2 + z 2 ) - («* + by + c*) 2 _ L (ay — &a;) 2 + (&2 — cy)' 2 + (ex — az) 2 25. Divide a,- 3 - 3 a 2 4 5 x — 7 hy x — a, keeping coefficients of like powers of x together. Arrange your remainder as a polynomial in a with exponents decreasing to the right. What do you note about the remainder ? 26. What would be the remainder on dividing ar 3 — 3 a; 2 4- 5 x— 7 by x - 1? by a; - 2 ? by x + 1 = x- (- 1)? 27. Divide a: 4 — a^ 4- 3 x — ra by x - 1. What is the remain- der? How may ra be chosen so that the division may be exact ? 28 * 4 -(*+!)* ' ** + (» + !) 29 l-3g-aQ42(l-.rV l-(l-s) PART III. SPECIAL MULTIPLICATIONS; FACTORS; TYPE FORMS 56. Introduction. There are some examples that occur so often in applications of algebra that it is desirable to commit them to memory. As an example of what is to be done here, consider (x + y)' 2 , which means (x + y)(x + y). Actually multiplying, we get, x + y x + y x 2 + xy xy + y 2 x 2 + 2xy + y 2 Then (x + y) 2 — x 2 + 2 xy + y 2 . This holds whatever x and y may mean. Thus (4 + «) 2 = 4 2 + 2 (4 • 0) + 6 2 = 100. Again (■/« + u) 2 = in' 1 + 2 mn + w 2 . To use this (or any other forms in this chapter), we try to see whether a given example can be made to look exactly like the known example by pairing off the parts. Thus, (3a + 2b) 2 is exactly like (x + y) 2 if z = 3a and y = 2b. Hence, since (x + y) 2 = x 2 + 2xy + y' 2 , (3 a + 2 b) 2 = (3 a) 2 + 2 (3 a) (2 b) + (2 b) 2 - 9 a 2 + 12 ab + 4 b 2 . Check: If a = 1 and 6 = 1, (3 + 2) 2 = 25 = 9 + 12 + 4 (correct). We try to remember the answer for another reason ; for (z 2 + 2 xy + «/ 2 ) -j- (a; + #) = x + y ; if this example in divi- sion is given, it is convenient to know the answer. Finally, we say the factors of x 2 + 2 x y + y 2 are (x + y) and (# + //), for the product of these factors is the given expression. (See p. 9.) The expressions, x 2 + 2 xy + y 2 , x + y, x + y are all poly- nomials in the letters x and y. (See p. 88.) In general, in this chapter and throughout the book, we shall seek only for polynomial factors of polynomials. 91 92 .MULTIPLICATION AND DIVISION [In. IV This is similar to the custom in arithmetic, where we usually seek only for integral factors of integers; thus, we say the factors of 10 are 2 and 5 ; we do not say the factors of 10 are | and 6, for example, though f x 6 = 10. Ex. 1. Find the factors of 4 m 2 + 20 mn + 25 n\ Given 4 m' 2 + 20 mn + 25 m' 2 , we may possibly notice that this is the same as (2 in) 2 + 2 (2 in) (5 n) + (5 n) 2 . Comparing with x 2 + 2x*y + y 2 , we see that 4 m 2 + 20 mn + 25 n 2 = (2 m + 5 u) 2 . The factors of 4 m 2 + 20 mn + 5 n 2 are therefore (2 in + 5 n) and (2 /« + 5 n) ; in other words, 4 ;n 2 + 20 mn + 25 n' 2 ) 2 - I*- OR 2 +16/2 + 04) -OR + 8). 11. (ic 2 + 3a ! + 2) 2 . 15. (p 2 + 3i>q + 2\ 28. (a 2 -f 2ab + b*)+6(a + b)c + 9c-. 57. Square of Sum. The result of § oti is: I. ( /+/ )^r , + 2//+/ 2 . The square of the sum of two terms equals the square of the first term, phis twice the product of the two terms, plus the square of the second term. 58. Square of Difference. Likewise II. ( x -y? = x*-2xy+yK Actually multiply (x - y)(x - y) to get this result. The square of the difference of two terms equals the square of the first term, minus twice the product of the two terms, plus the square of the second term. Ex.1. (4 k - 3 .s) 2 = (4 A') 2 - 2 (4 k) (3 s) + (3 s) 2 = 16 k 2 - 24 ks + 9 s 2 . Ex. 2. (18) 2 = (20 - 2) 2 = (20) 2 = 2 (20) (2) + 2 2 = 400 - 80 4 4 = 324. This will he found an easy way to square certain numbers. Ex. 3. (mr — 2 mn 4- ?r) h- (m — n) = m — n. This results from the formula above. Ex. 4. (a + b - c) 2 = [(a 4- 6) - cj = ( a + !,)-- 2 (a 4- 6) • c 4- c 2 - a 2 + 2 a& + 6 2 - 2 «c - 2 6c + c 2 — a 2 + 6 2 4 c- 4- 2 a6 - 2 ac - 2 6c. Ex. 5. Find the factors of 4 m 2 — 20 mn 4-25 ?r. As in Ex. 1, p. 92, we notice that the given expression may be written (2 m)-— 2 (2 m) (o ra) + (5 n , 2 . Comparing with II, we see that 4 m- - 20 wm + 2.") n 2 = (2 m - 5 ») 2 . Hence, the factors of 4 to 2 — 20 rnn + 25 n 2 are 2 ?« — 5n and 2 m — 5 n. Check by actual multiplication. 94 MULTIPLICATION AND DIVISION [Ch. IV EXERCISES XIII: CHAPTER IV Perform the following multiplications and divisions ; check as usual : 1. (5r-2s) 2 . 3. (-a+by. 5. (v-2 tuf. 7. (x-2?/+2) 2 . 2 . (9 6-2) 2 . 4. (2v-13«) 2 - 6. {x 2 -nx) 2 . 8. (x-y-zf. 9. (62_io& + 25)-5-(6-5). 14. [(x + y)+(x-y)y. 10. (16-8fc + fc 2 )-=-(4-A-). 15. [(x+y)-(x-y)¥. 11. (64a; 2 -16.r?/ + ?/ 2 )--(8a;-?/). 16. (a - b - c + d) 2 . 12. (m 2 -9m + 20J)-*-(m-4£). 17. ( a + &) 2 = [a-(-&)] 2 . 13. (V- 8 a- 2 +16) -(a 2 -4). 18. (19)-'=(20-l) 2 ; (29) 2 ;(9) 2 . Resolve into factors : 19. fc»-6* + 9. 24. 4e 2 -96e + 576. 20. Z 4 -12z 2 + 36. 25. 64 p 2 (f- Wpq + 25. 21. r 2 - 14 £ + 49. 26. 16 a 6 -40 00* + 25 a 2 . 22. v 2 + 9t 2 — 6vt. 27. 9 a 2 z 2 - 42 abxy + 49 & 2 ?/ 2 . 23. m 2 — 24 m>* + 144 n 2 . 28. (m 2 — 2 mn + n 2 ) —2{m—n)p+ p 2 . 59. Product of the Sum and Difference. Likewise, III. (x+y){x-y)=x i -y*. Actually multiply (x + y~) by (x — y) to get this result. The product of the sum and the difference of the same two terms equals the square of the first term, minus the square of the second term. Ex. 1. (3« + 2&)(3a-26) = (3a) 2 -(2//) 2 = 9a 2 -4& 2 . Ex. 2. (a + b + c) (a + b - c) = [(a + 6) + c] [(a + 6) — c] = (a + &) 2 - c 2 = a 2 + 2 a& + 6 2 - c 2 . Check by actually multiplying. Ex. 3. Find the factors of 4 a 2 - b'\ We write 4 a 2 - b 2 = (2 a) 2 - ft 2 =(2 a + 6) (2 a- 6). Hence, 2 a + b and 2 a — b are the factors. 58-60] FACTORS; TYPE FORMS 95 EXERCISES XIV: CHAPTER IV Perform the following multiplications and divisions; check as usual : 1. (4 a + &) (4 a - &). 8. (a - b - e)(a + b + c). 2. (2x + 3y)(2x-3y). 9. (a + 2&-3c)(a + 2& + 3c). 3. (a 2 +5) (a 2 -5). 10. ( z 2 -16r 2 ) + (z-4r). 4. ( :B 2 + 5)(-s 2 + 5). 11. (a*-b n ) + (a 2 + b*). 5. (mi — 10 kz) (711 + 10A*z). 12. (m *n* — m*nF) -5- (mn 2 + vnNi). 6. ( a _6-c)(a + 6-c). 13. (25n s — 49m*)-i-(5n— 7m). 7 . ( a -6-c)(a-6 + c). 14. (64;>y- 25 ) -s-(8pg+5). 15. 99 -s-11 =(100-1) -=-(10 + 1). 16. 99-5-9. 17. 624-5-26. 18. 24- 26 = (25-1) (25 + 1). 19. 15-17. 20. 35-37. 21. (a 2 + 2 a& + & 2 - 16 c 2 ) -5- (a + 6 - 4 c). 22. (a 2 + 6 ah + 9 & 2 - 4 x 2 + 4 a# — ;/ 2 ) -5- (a + 3 6 - 2x + y). Resolve into factors: 23. 100 r- 169 s 2 . 27. a 2 - 2 a& + & 2 - 9 c 2 . 24. 4.1 2 -9^r. 28. .r 6 -.7r. 25. a 4 — 64a 2 . 29. i y 4 -& 2 + 2&c-c*. 26. a 2 — 2 as + .« 2 — 4 ?/ 2 . 30. 1 6 — x-y • 31. z 4 + z- + 1 = (2 4 + 2 z 2 + 1) - z 2 . 32. a 2 + 2 a& + If - c 2 + 6 cy - 9 >/ 2 . 33. a 2 - 2 a& + b 2 - 4 Z 2 + 12 Im - 9 m 2 60. Likewise, IV. (jr + a)(jr+6)=jr2+ (o+6)jr + a6. Actually multiply (x + a) by (x + J) to get this result. The product of tivo binomials whose first term is the same equals the square of the common term, plus the sum of the second terms times the com- mon term, plus the product of the second terms. 90 MULTIPLICATION AND DIVISION [Ch. IV Ex.1. (x + 2)(x + 4) = x 2 + 6x + 8. Ex. 2. (as - 3) (a? -f 5) = [> + (- 3)] [.*• + 5] = a 2 + [(-3) + 5]sb+(-3)(5) = . c 2 + 2 x _ 15. Notice that the rule really applies to differences as well as to sums, as in example2. We need only express the difference x — 3 as the sum of x and — 3. On the same principle (x — y) 2 can be worked out by the rule for (x + y)*. Thus, (a - y)« = [.r + (- y)f = x* + 2 x(- y) + (- y)> = *» - 2 zy + y 2 , which is the same as the result in § 52. (See Ex. 18, List XII.) This rule is used chiefly to find the factors of given ex- pressions like the products above. Ex. 3. Find the factors of x 2 + 6 x + 8. We write x 2 + 6 x + 8 = (.r + ?) (./: + ?). Now the last terms have a product 8. Hence, we try such combina- tions as 1 and 8, and 2 and 4, etc. The pair 1 and 8 is not correct, for (.r 4- l)(x + 8) = x 2 + 9 x + 8, which is not the given expression. In fact, it is clear that the sum of the numbers of the correct pair must be 6. We want, then, a pair of numbers whose product is 8 and whose sum is 6. If we try a few pairs, we shall probably try 2 and 4 quite soon; this pair is correct, for 2 x 4 = 8 and 2 + 4 = (i. Checking the answer by multiplication, we find (x + 2)(x + i) = x' 2 + G x + 8. Hence, the required factors are x + 2 and x + 4. If, after trying all pairs whose product is 8, we find no pair that is correct, we must give up the problem just now ; later we shall solve such problems by a different method. Ex. 4. Consider x 2 -{-2x — 15, the resvdt of example 2. We must choose a pair of numbers whose product is — 15. Such pairs are — 1 and +15, +1 and — 15, + 3 and — 5, — 3 and + 5. But the sum of the pair must be 2; hence the last pair is the correct one since — 3 + 5 = 2. Check : (x — 3) (x + 5) = x 2 + 2 x — 15. The required factors are therefore (x — 3) and (x + 5). Ex. 5. Similarly, 1 + 2 x - 1 5 x 2 = (1 - 3 sr)(l + 5 x), and x 2 + 2 xy — 15 y 2 = (x — .°> y) (x +■ 5 y). The letters used should not confuse the student. 60] FACTORS; TYPE FORMS 97 EXERCISES XV: CHAPTER IV Perform the following multiplications : 1. ( (J+ 2)(q-3). 11. (M- 8)07 + 1). 2. (z + l)(z + 4). 12. (t-3a)(t-$a). 3. (m-2w)(m + n). 13. (xy- - 2){xy 2 + 10). 4. ( r -.v)(»-os). 14. (s + «-l)(s + <-3). 5. (>--3)0 + 5). 15. (xyz + 2 t)(xyz + 4: t). 6. (a + 6)(a + 2). 16. (1 - 2 e' ! )(l + 3 c 3 ). 7. (l_s)(l+3s). 17. (rc-i)^-!). 8. (/ + 2m)(/ + T4 18. (b -fa)(& + fa). 9. (r 2 -2)0 2 -9). 19. (2s-9)(2z + 8). 10. (.r 3 + 3)(ar J -4). 20. (Q pq - 2 rs)(6po-3 »•*). Resolve into factors: 21. z 2 + «z-6a 2 . 24. f 2 -2f-3. 27. a^-e.t + S. 22. a 2 +6 a +8. 25. .r 2 -5x- + 6. 28. ar + 6.i- + 5. 23. a 2 - 6 a + 8. 26. x 2 + 5 .1- + 6. 29. 1 - 3 Z - 18 z 2 . 30. m 2 -2 inn — Ion 2 . 33. ?/ 2 — 14 xy + 24 a; 2 . 31. r 2 - 12 7-6- + 35 s 2 . 34. TO 4 + 5m 3 w-24m¥. 32. 1 - 12 rs + 35 rV. 35. ? 2 - 15 fa + 56 f = (2 m + 3 n) (4 m + 5 n) . The letters used should not confuse the student. EXERCISES XVI : CHAPTER IV Perforin the following multiplications : 1. (2a-3)(a-4). 4. (5-x)(7 + 2z). 2. (3a + l)(5a-3). 5. (x + 5 y)(5x- y). 3. (z-6k){3z + k). 6. (3x-2az)(4:x + 3az). Resolve into factors : 7. 8z- + 8z-6. 15 6 /^V_^_ji2 8. 3 x 2 + xz - 52 z 2 . ^ w ' ?l 9 2 -a- 21 a 2 16 S^-* 2 " 45 - 17. 12.ry + 17^z + 6 2 2 . 10. 6 »V — 7 pa — 20. . _ , , _ 7 , 7 * 1 d 18. j2a 2 + a/t — 6/t 2 . 11. W-r-2r 2 . 19 2aar 9 + (4-3a)a;-6. 12. 2r a + r-15. 20. aft^ + Ca + ft^x-K 2 . 13. 7 .r 2 + z»/ - 6 /• 21. 12ar-(4ft-6)z-2ft. 14. 6m 2 -13mn + 6w 2 . 22. 4 e 2 + 960 e + 479. 62. Other Forms. There are many other general results which might be given here. Among them we mention the following : (1) (a + ft) (a 2 - aft + ft 2 )= a 3 + ft 3 . See Ex. 10, p. 82. (2) (a - ft) (a 2 + aft + ft 2 ) = a 3 - ft 3 . See Ex. 13, p. 82. (3) (a 2 - ft 2 ) (a 2 + ft 2 ) = a* - ft 4 , or (3 a) (a-ft)(a + ft)(a 2 +ft 2 )=a 4 -ft 4 . [This is really an application of § 59.] (4) (a + ft) 3 = a 3 + 3 a 2 ft + 3 aft 2 + ft 3 . See Ex. 35, p. 83. (5) (a + ft) 4 = a 4 + 4 a 3 ft + 6 a 2 ft 2 + 4 aft 3 + ft 4 . See Ex. 36, p. 83. 100 MULTIPLICATION AND DIVISION [Ch. IV (6) (a - ft) 3 = a? - 3 a 2 b + 3 aP - ft 3 . See Ex. 38, p. 83. (7) ( a _ by = a 4 - 4 a% + 6 a 2 ft 2 - 4 aft 3 + ft 4 . See Ex. 30, p. 83. These and many others may be found by the student by multiplying together the given factors. As before, if an expression can be put into a form exactly similar to one of these answers, its factors may be found by comparison with the factors given above on the left. EXERCISES XVII: CHAPTER IV Factor : 1. p3-8. 3. x 4 -81. 5. .16m 4 -?* 8 . 7. p 6 — q 6 . 2. p* + 8. 4. 27 -125 a 3 . 6. i*-f. 8. a 8 -ft 8 . 9. . r 3 +3a* + 3» + l. 14- a 3 - 30 x 2 + 300 x - 1000. 10. # - 6 f + 12 * - 8. 15. j> 4 - f p 3 + 1 p 2 - 2 4 T p 4- ¥ V- 11. i + 4£ + 6£ 2 + 4£ 3 +« 4 . 16. ^ + 3afy + 3a# 2 +Y-s 3 - 12. m 8 « 3 + 216ar 3 . 17. a 8 -3a 2 6 +3 aft- - ft 3 -8 c 3 . 13. (a + 6)» + (o-6)». 18. .r'-3x- + 3.r-l-125/ fi . Perform, by factoring, the operations indicated : q» + 3o a 6 + 3a6 2 + 6 !> 20 A; 4 - 4 A; 3 + 6A; 2 -4fc + l a 2 + 2aft + ft 2 A; 3 -3A: 2 + 3A;-1 Perform the following multiplications, and express the re- sult as a formula in each case : 21. ( + 3 qx + 3 ew/ +/>a?. 13. t 3 — ?/£'-' — xH + ar 2 */. 6. 3 + 2z — 6c — Icz. 14. « s + a« 2 + c« 2 + 6«4-ac«4-a6. 7. ab 2 — abc - &aj + ex. 15. ar m+ " — ax m - &a? + a&. 8. ar 5 — 4 a,- 2 + 3.x- -12. 16. 1— y — z+yz. 17. .,•-' _ a 2 [ = a; 2 + ax - ax - «-]. 18. .r + 2 a.*; + a-[ = .r 2 + + 9)-10. 12. a 2 + 2ab + b 2 -c 6 . 4. * 3 -64. 13. 3z 4 -8z 2 -35. 5. 4 a 2 - 20 ax + 25 x\ 14. 9 + 18a -7 a 2 , v 6. aV-3a 2 ft 2 a5 2 +3a& 4 a;-& 6 . 15. l-v-156v 2 . 7. x 4 -16. 16. 81 L 4 - 256. 8. p 2 -10pg + 25. ^ 17. 10 -a- 3 x\ §>10 - cz - 60 A 2 . / 18. i> 4 - 81 q\ 19. 63 = 64-1; 1001 = 1000 + 1; 65 = 64 + 1; 77 = 81-4. 20. m 4 — 4 m 3 n + 6 m 2 /i 2 — 4 »m 3 + n 4 — 16 jA 21. fc*-12# s + 54& 2 -108fc-175 = (A;-3) 4 -256. [Factor.] 22. a 6 - // = (a 3 - 6 3 )(a 3 + b s ). [Factor these factors.] 23. a 4 + o 2 6 2 + & 4 = (« 4 + 2 a 2 6 2 + b 4 ) - a 2 6 2 . [Factor.] 24. a 6 - b 6 = (a 2 - 6 2 ) (a 4 + d 2 b 2 + o 4 ). [Factor these factors.] 25. a s _ 6 8 _ ( a A _ b i^ a 4 + ^). [Factor these factors.] 2 6 . a w - 6 10 = (« 5 - b 5 )(a 5 + 6 5 ). [Factor these factors.] 27. a 10 - 6 10 = (a 2 - b-) (a 8 + a 6 6 2 + aW + a 2 6 e + //) . [ Factor.] 28. a 8 + a%- + « 4 6 4 + a 2 & fi + b 8 . [Make use of Exs. 26, 27 .] 29. a 2 - 2 a& + W - x 2 + 2 a# - ?/ 2 . 30. « 3 - 3 « 2 6 + 3 ab- - & 3 - x 3 + 3 a- 2 // - 3 xy- + 2/ 3 . 31. Ax 2 - 3 vlx + 2 .1 + 5a- 2 - 3 5.x + 2 B. 32. i>V + 2jfl + 7 vv> + 14 pt + 24 / + 12 r. 33. 1 — T — 8— t + rt + rs -\-st — rst. PART IV. APPLICATIONS. ENGLISH TRANSLATED INTO ALGEBRA 64. Expression of English in Formulas. The preceding chapters have given practice in writing formulas in the place of English. Thus, instead of the rule : " The square of the sum of two terms is equal to the square of the first term, plus twice the product of the two terms, plus the square of the second term," we merely write (x + y) 2 = x 2 + - xy + y 2 , which really says the same thing in much shorter form. Again, instead of saying, " The volume of a box in cubic inches is the product of its width times its length times its height, each measured in inches," we may simply say v = w • I • h, where v means volume in cubic inches, and w, I, h, stand for the width, length, and height, respectively, measured in inches. In many cases the student will find it easier to write down the formulas if he first tries the problem with known numerical values. In doing so it is advisable not to per- form the additions, multiplications, etc., but merely to indicate them as a guide in the case when no numbers are actually given. In this way the structure of the ex- ample is seen with simple numbers, and can be followed afterwards in the use of letters for unknown numbers. Consider the problem : Ex. 1. A tin box is to be made from a square piece of tin by cutting square pieces out of the corners ami then folding up the flaps. Find the size of the piece of tin which must be used to make a box 4 in. high that shall contain 100 cu. in. volume. Let us first become familiar with the problem by trying several numbers. Suppose the original plate were IS in. .square (in fig- 103 104 MULTIPLICATION AND DIVISION [Cii. IV D __ Q HH — I H E B Fig. 18. ure, AB = 7?C = 18 in.), and we cut out the shaded corners, each 4 in. square. Then HE = 18 in. — 2 x 4 in. = 10 in. Likewise FE = 10 in. If we now fold up the flaps, we get a box whose bottom is 10 in. square and whose height is 4 in. The volume of this box is 10 in. x 10 in. x 4 in. = 400 cu. in. It is now easier to try the given problem. To start with, we do not know how long AB must be. Let us call its length / in. and then try to find I. If we cut the corners out as before, we shall have HE = I in. — '2x4 in. = (/ — 8) in. just, as before. Likewise FE = (7 — 8) in. Hence, our box will have its bottom (7 — 8) in. square. Its height will be 4 in. Its volume is then (7 — 8) in. x (7 — 8) in. x 4 in. = 4(/ — 8) 2 cu. in. If this volume is to be 100 cu. in., we shall have 4(7 -8) 2 = 100. (7 - 8) 2 = 25, (7 — 8) = 5 or else — 5, (5) 2 = 25, and also (- 5) 2 = 25.* I — 8 + 5 or else 8-5. I = 13 or else I = 3. Divide both sides by 4 : hence, since Adding 8 to each side, we get It follows that The correct result is I = 13, for I = 3 would not really do at all not cut corners 4 in. square out of a piece of tin 3 in. square. we could As in this problem the student must always be careful to see which, if any, of several possible answers are the correct ones, for there may be answers that cannot pos- sibly mean anything, as is seen above. Ex. 2. If a printed book is to have a margin 2 in. wide, how large must the pages be cut if there is to be 70 sq. in. of print on them and the pages are to be twice as long as they are wide ? * There are no other numbers whose square is 25, for a positive num- ber less than 5 would give a square that is too small; a positive number greater than 5 would give a square that is too large ; and similarly no negative number except — 5 would produce precisely 25. It is very im- portant, in all problems, to make sure that the answers found are all that exist. If this is not done, some answer — perhaps the most important — may be overlooked. 64-65] EXPRESSION IX FORMULAS 105 Fig. 19. Let us first become familiar with the problem. Suppose the page is 10 in. x 20 in. (twice as long as wide). Then AB = 10 in.. EC = 2 x 10 in. = 20 in. If the margin (on each edge) is 2 in. wide, then EF = lo in. - 2 x 2 in. = 6 in., and FG = 20 in. -2x2 in. = 16 in. The printed portion is then (6 x 16) sq. in. = 96 sq. in. Suppose now that the width AB is not known in advance; call it win. Then . I ]', — w in. BC = 2 x w in. = 2 w in. If the margin (on each edge) is 2 in. wide, then EF=(w — 2)x 2 in. = (w — 4) in., and FG = [2 w - 2 x 2] in. = (2 w - 4) in. Hence, the printed portion is (w — 1)(2 w — 4) sq. in., or (2 w a — 12 iv + 16) sq. in. But the printed portion is to be 70 sq. in., hence, 2 io n - - 12 w + 16 = 70. Divide each side bv 2 : w- - 6 w + 8 = 35. Subtract 35 from each side : w i _ 6 w - 27 = 0. But this is O + 3)(w - 9) = (§60). Hence, w + 3 = 0, or else ic — 9 = 0, for the product (w + 3)(w — 9) would not be zero unless one of the factors were zero. If w + 3 — 0, then w = — 3 ; this is meaningless, for a piece of paper cannot be — 3 in. wide. If w — 9 = 0, then w = 9 ; and this is soon seen to be correct, for if w = 9, the actual printed area is 5 in. wide and 14 in. long; its area is therefore (5 x 14) sq. in. = 70 sq. in., as was required. 65. Simple Changes in Equations. We have made several changes in the equations above which we shall now review. Thus, we had in example 1, 4(7— 8) 2 =100 and we divided each side by 4. The justification for this is that 4(7— 8) 2 is the same number as 100; hence, |- of either of them is equal to ^ of the other. (Compare § 35, p. 55.) 106 MULTIPLICATION AND DIVISION [Cii. IV Likewise, we multiplied both sides by the same number, added the same number to both sides, subtracted the same number from both sides. The argument is the same. We see that in any equation, if it is a true equation, the two sides stand for the same number ; hence : Given any equation, ice may I. Add the same number to each side. II. Subtract the same number from each side. III. Multiply each side by the same number.* IV. Divide each side by the same number except zero.f V. Perform the same operation (of any kind) on the whole of each side, if the result is known to be a single num- ber in each case.$ (Take care to perform the operation on the whole of each side, not on a part of it.) These principles have been used frequently ; the first four were stated in another form in § 35, p. 55. 66. Product Equal to Zero. Another principle is the one used in example 2, p. 104. Since (w + 3)(w — 9) = 0, we knew that either (?#-j-3)=0, or that (iv — ( .)) = 0, for otherwise the product of the two could not be zero. * If both sides are multiplied by zero, the new equation is = 0, which is correct, but not useful. Avoid multiplying both sides by zero, and test any expression used as a multiplier to see if it is zero for the values finally found. t In dividing, great care is sometimes necessary to avoid dividing by zero. Notice that no number can be divided by zero. (See p. 75.) \ If the result of the operation performed is not a single number, great care is necessary. Thus, we said, in example 1, if (7 — 8) 2 = 25, we know that I — 8 is cither + 5 or — 5. To get this we take, the square root of em-It side. But 25 has two square runts, since (+ 5)' 2 =( + 5) x (+5) =25 and (—5)'- =(—5) x (— 5) = 25. Hence, we cannot be sure which one of these is equal to (1 — 8). Nevertheless, we may make a perfectly definite statement even in this case; namely, "(7 — 8) is equal either to 5 or else to — 5." 65-66] EXPRESSION IN FORMULAS 107 This principle is simply that two numbers, neither of which is zero, have a product which is not zero. This truth is quite evident after a little thought, and we shall state it as follows : VI. 7/' the product of two factors is zero, then at least one of the factors is zero. Let us solve additional problems to illustrate these prin- ciples : Ex. 1. Each member of a certain family gave each of the others a present, at Christmas, which cost 50 cents. If the total spent by the family was $10, how many persons are there in the f amily ? To become familiar with the problem, suppose there had been 6 memhers. Then eacli one would have given 5 presents, so that all together 6 x 5, or 30, presents would have been given. Since the number in the family, is not really known, let us call it n. Then each member gave n — 1 presents, so that all together there were n(n — 1) presents given. If each cost 50 cents, the cost in cents was 50 n(n — 1). But this is known to be 1000 cents: 50n(» - 1)= 1000. Divide both sides by 50 : »(»- 1) = 20. Multiply » — 1 by n, and subtract 20 from each side, n 2 _ „ _ 20 = 0, (n + 4)0 - 5)= 0. Hence, one or the other of these factors is zero : n + 4 = 0, or h — 5 = ; that is, subtracting 4 from each side or adding 5 to each side gives, n = — 4, or n = 5. Now n = — 4 is meaningless in this problem ; therefore n = 5 : there are 5 persons in the family. To check this result, we notice that there would then be 5 x 4 = 20 presents, which, at 50 cents each, would cost $10. This check is complete. Many practical problems arise in computing the effect of errors in measurements. In such problems the error 108 MULTIPLICATION AND DIVISION [Ch. IV in measurement is usually counted positive when the meas- urement is too large ; if the measurement is too small, the error is said to be negative. The following examples will illustrate the calculations of such error effects. Ex. 2. Let e denote the error (in inches) made in the meas- urement of the side of a square whose side is really 10 ft. long; and let E denote the error (in square inches) in the computed value of the area of the square. Express E in terms of e. If e = 3 (in.), find E. UE = 484 (sq. in.), find e. If E = - 119| (sq. in.) (i.e. the computed area is 119f sq. in. too small), find e. The side is really 10 ft., or 120 in. long. If the error is e, the meas- urement is 120 + e (in inches). Hence, the computed area is (120 + e)' ? (in square inches). Since the real area is (120)' 2 (in square inches), the error E in the computed area is E = (120 + e) 2 - (120)2 _ e 2 + 040 e. If e = 3 (in inches), the value of E is found l>y putting 3 in place of e in the preceding equation ; and we find E = 3 2 + (240) (3) = 9 + 720 = 729 ; hence, the effect of an error of 3 in. in measuring the length of the side of the square causes an error of 729 sq. in. in the computed area, or about 5 sq. ft. If E — 484 (in square inches), we have e 2 + 240 e = 484, or e 2 + 210 e - 484 = ; whence, (e - 2) (e + 242 ) = ; and either e - 2 = 0, or e + 242 = ; that is, either e = 2, or e = — 242. The answer e = — 242 is unreasonably large, since no one would con- ceivably make an error of the amount of 242 in. in measuring a length of 10 ft. The answer e = 2 is evidently the only one to which we need pay attention ; it means that the error of 484 sq. in. in the com- puted area would be caused by an error of 2 in. in the measurement of the side, the measurement being 2 in. too long, since e is positive. If E = — 1193 (in square inches), we have e 2 + 240 e = - 119f, or e 2 + 240 e + 119 3 = 0, or, multiplying by 4, we get 4 e 2 + 960 e + 479 = 0. 66-67] EXPRESSION IN FORMULAS 10!) Since 479 lias no factors except itself and 1, we would soon try the factors (2e + 1 ) and (2e -J- 47!)) ; these are correct, as will be seen by multiplying them together. Since their product is zero, we have either 2 e + 1 = 0, or 2e + 479 = 0, that is, either e = — i, or e = - 239J. The answer e — — \ means that the error made in measuring the side was negative, i.e. that the measurement was too short, and the amount of the error was } in. The other answer e = - 239J is un- reasonably large, and we need not consider it. Check: If e = — §, the measured length was 119J in.; hence the computed area was (119') 2 (sq. in.) or 14,2804 sq. in. The real area being 14,400 sq. in., the error E was - 119| sq. in. (correct). In the preceding problem, the answer e = — \ has a real meaning, though it is negative. In problems about actual things, always notice carefully whether or not a negative answer can be interpreted. Notice also whenever an answer is unreasonably large, or when it is unreason- able for any other cause, and be careful to say why any of several pos- sible answers is discarded. See also the footnote on p. 104. v 67. Solution of Problems. The following exercises will illustrate the uses of factoring and of multiplication and division in simple cases. They may all be solved by the principles of §§ G5-66 ; and the answers found will be correct if these principles are carefully followed. The equations solved above are quadratic equations; we shall study such equations in more detail in Chapter VIII, p. 203. Apparent answers maybe found which do not fulfill the conditions of the given problem. Thus, the equation (1) x — 3 = has only one answer, x = 3. But if both sides of equation (1) are multiplied by x - 2, the resulting equation (x - 2)(x - 3) = has two possible answers, x = 2 and x = 3, the first of which, x = 2, is not a possible answer for equation (1). No such false answer can arise if both sides are multiplied by a number, as required in III, § 65, if that number is not zero. In order to be sure that no mistakes have been made in the work, as well as to avoid the possible introduction of false answers, each answer should be checked, as above, by actually trying it in the given problem. Such a check is complete. 110 MULTIPLICATION AND DIVISION [Cn. IV EXERCISES XX: CHAPTER IV 1. In example 1, p. 103, we found that the given problem led to the formula v = 4 (l — 8) 2 . Find v when 1 = 8; when Z = 9; 10; 11; 13. Find v whenZ = 0; 1 ; 3 ; 5 ; 7 ; -- 1 ; - 5. Make a table of these values of v and /, and plot the graph for this equation. Does the shape of the graph resemble any you have drawn before? When v = 100, determine I from the graph. This procedure is called the graphical solution of the problem. Compare your results with the results on p. 104. 2. When v = 12\, l = ? What is v when 1 = 2? when Z = 3i? Solve both graphically and algebraically. For any value of /, how many values of v are there? For any value of v, how many values of I are there ? In the case of two results such as v = 100, I = 13 or 3, how is the correct answer to be distinguished from the incorrect one in the figure ? 3. In example 2, p. 104, the printed portion p = 2w 2 — 12 w + 16, if w = width. Find p if w = ; 1 ; 2 ; 5 ; 10 ; - - 1 ; - 5 ; — 10. Make a table and draw the graph. Solve graphically for w if p = 70, and compare your results with those on p. 105. How many values of iv are there? Point out why one is impossible as a true solution. 4. A certain rectangular yard is known to contain 050 square feet; it is found that a 150-foot rope exactly surrounds the fence. What are the length and breadth of the yard? Solve by letting I = length. 5. If A denotes the area of a field, as in Fx. 4, what rela- tion connects A and I? Draw the graph. Solve graphically for I if A =650. What can you say of the two values obtained for I in this case? 6. For what values of / in Ex. 5 is A = 0? What is the space between these values? If the number I represents the length of the field, what distance in the figure represents (IT] EXPRESSION IN FORMULAS 111 the breadth of the field? Point out on the figure the greatest possible value of the area A. For this value of A, what are the length and breadth of the field ? 7. By what amount must the length and breadth of a rec- tangular plot of ground 300 ft. long by 50 ft. wide be equally increased, in order that the area may be increased by 3000 sq. ft. ? 8. Represent by a graph the increase in area of the plot of ground corresponding to equal increase of length and breadth. Solve the problem graphically. Interpret negative values. What equal decrease of length and breadth will reduce the area 5025 sq. ft. ? 9. What error (e) in measuring the side would cause an error (E) of 241 sq. in., the computed area of the square men- tioned in Example 2, § 66 ? If e = 5, find E ; if e = -% find E; if # = -239, find e. 10. The sum of the squares of two consecutive integers is 113. What are the integers ? Have both positive and negative results a meaning in this problem ? 11. "Think of a number; add 3 to it; multiply the result by 2 ; from this result subtract 5, then multiply by the num- ber you first thought of. What is your result?" If the result given is 36, what is the number thought of ? If the re- sult given is 21, what is the number thought of? 12. Show that the difference of the squares of two consecu- tive integers n and n + 1 must be an odd number. If the dif- ference of the squares is given, how may the numbers be found ? Determine two consecutive integers such that the difference of their squares is 9, 15, 33. 13. Por the frame of a picture 10 by 15 in., 116 sq. in. of material are available. How wide may the frame be made if the frame is the same width on all sides, and if it exactly meets the picture ? Solve graphically, also. 112 MULTIPLICATION AND DIVISION [Cn. IV 14. The interest on a sum, p, of money at a given rate, r, for a given time, t, is i=prt. What is the amount, A ? What is the amount, A, for one year ? What is the amount, A', at annually compounded interest, for two years? Plot the annually compounded interest for two years on $200 at various rates. (Do not start the plotting of A' from 0, but (say) from $200. Choose rates from 0% to 12% at in- tervals of about \%.) For what rate will A' be $220.50? Solve first graphically, then by the equation. 15. A rectangular frame is so constructed that by means of a slide its width may be altered. When it is set at the width 5 centimeters, a rod 13 centimeters long just fits as diagonal. How much must the width be increased so that a rod 15 centi- meters long may fit as diagonal ? 16. A rectangular plot of ground 60 feet by 14 feet is to be doubled in area by equal increase of length and breadth. Draw a picture illustrating in general the alteration in area corresponding to equal alteration of length and breadth. Find the alteration necessary, both by use of the graph and by solution of the equation. 17. Let the error in the measurement of the side of a square whose side is really 15 ft. long be denoted by e (in inches), and the error in the computed area by E (in square inches). Express E in terms of e. If e = l, find E ; if e = 2, find E; if e — 3, find E ; etc.; if e= — 1, find E ; etc. Draw the graph, taking one small space on the horizontal line to- mean 1 in. in values of e, and one small space on the vertical axis to mean 1 square foot in values of E, i.e. 144 sq. in. Find e if E = 361 ; if E = 724 ; if E= - 359. 18. Let the error in measuring the radius of a circle whose radius is really 7 ft. long be denoted by e (in inches), and the error in the computed area by E (in square feet). Taking tt = 3\ (see table at back of book), show that E = 528 e + 3\ e 2 . Find E for various values of e ; plot the figure ; find e if E = - 524f . 67] REVIEW 113 REVIEW EXERCISES XXI: CHAPTER IV Perform the indicated multiplications ; check : 1. (a 3 - 3 a?b + 5ab 2 - b 3 ) (a 2 - ah + b 2 ). 2. (16 At - 4 k + 1) (16 k 2 + 4 k + 1). 3. (l- a -b)(l + a + b + a 2 + 2ab + b*). 4. (a + 2 6 + 3 c)(b + 2 c + 3 a)(c + 2 a + 3 6). 5. (y-z)(z-x){x-y). 6. (a 4 + a 3 + a 2 +a + l)(a-l). 7. (m 3n + m 2n + m" + 1) (m" — 1). 8. (ax 2 + bx-\-c)(lx 2 + mx + ri). 9. _ a ^v' x a e b d 2 ). 27. « c — b s . 28. 2 .i- 2 + 2 a; - 40. 36. p 5 — 5 p 4 ? + 10 p 3 q 2 - 37. v 4 - 12 ?- 3 + 54 r- - 108 r + 81 . 38. O 4 + 3 rr + 4) = (/ ( 4 + 4 n 2 + 4) - re. 39. a&ftj 2 ™ — (a 2 + tr)x m y + a^?/ 2 . 40. ft- fi - 14 ^ + 49 - tf+ 2 yz - z 2 . In the following exercises solve both from the graph ano 1 from the equation, unless otherwise directed : 41. A page is to have a margin of 1 inch, and to contain 35 square inches of printing. How large must the page be if the length is to exceed the width by 2 inches ? 42. How wide a path may be laid out inside the margin of a park area 100 x 250 feet, in order that the space taken up by the path may be 3400 square feet ? 43. Such a problem as Ex. 42 would arise as follows : how wide a path will take up an area not greater than, 4000 square feet, approximately. Solve this problem graphically. Note that we are not yet in a position to solve the problem even approximately by means of the equation. 44. For a kindergarten gift, a rectangular box with a square top is to be constructed 5 inches high. If exactly 2 square feet of colored paper is used to cover the box, including the lop and the bottom, what must be the length of one side? 67] REVIEW 115 45. In the preceding exercise, it is desired that not more than 5 square feet of paper be used to cover the box. How large may the side be chosen ? Solve only graphically. 46. " Think of a number, double it, add three, and square the result; from this result subtract seven, and halve the re- mainder." The result is 21 ; what was the number chosen? 47. Construct a puzzle similar to the above, leading to an easy equation, involving the square of the unknown number; propound it to a friend and find the number chosen by him. 48. Find two numbers whose sum is 14 and whose product is 48. 49. The hypotenuse of a certain right-angled triangle is ten feet long; a piece of string which is exactly one fourth as long as one of the perpendicular sides is found to be exactly one third as long as the other. How long is the string, and how long is each of the perpendicular sides ? (See Tables.) 50. What radius must a cone whose slant height is 5 inches have in order to be covered by 14 ?r square inches of tinfoil '.' 51. What is the error E (in square inches) in the computed value of the area of a square 50 ft. long, caused by an error e (in inches), in measuring the side ? Plot the graph as in Ex. 17, p. 112. Find e if E = 2404 ; if E = -3591. 52. How carefully must the side of the square in Ex. 51 be measured in order to be sure that the error in the computed area is not more than 2404 sq. in. ? How carefully must each mark be made if the side is measured with a foot rule in the ordinary way ? 53. Answer similar questions for E = 6025 in Exs. 51 and 52. Is it reasonable to suppose that the area can be computed (from measurement with a foot rule) to within 10 sq. ft. '.' 54. How carefully must one measure the radius of a circle whose radius is 21 ft. in order that the computed area may be correct to within 795} sq. in. ? Is it reasonable to suppose that the area can be computed (from measurement of the radius with a foot rule) to within 5 sq. ft.? (See Ex. 18, p. 112.) 116 MULTIPLICATION AND DIVISION SUMMARY OF CHAPTER IV: MULTIPLICATION AND DIVI- SION; FACTORING; APPLICATIONS; pp. 07-115 Part I. Multiplication and Division of Numbers and Mono- mials, pp. 67-77. Definition of Multiplication in General: fundamental properties to be preserved. § 40, pp. 67-68. Product of Negative and Positive : product of amounts preceded by - sign. Exercises I. § 41, pp. 68-70. Product of Tico Negatives : product of amounts preceded by + sign. Rule of Signs: like signs give + , unlike give — . Exercises II. §42, pp. 70-71. Multiplication of Monomials : rearrangement of factors. Exercises HI. § 43, pp. 71-72. Multiplication of Powers of same letter; add exponents. § 44, pp. 72-73. Final Rule for Multiplying Monomials : multiply coefficients, add exponents. Exercises IV. § 45, pp. 73-74. Division of Monomials : reverse multiplication. Exercises V. § 46, pp. 74-75. Division of Powers of the same Letter: subtract exponents. § 47, p. 76. Final Rule for Division of Monomials : divide coefficients, subtract exponents. Exercises VI. § 48, pp. 76-77. Part IT. Multiplication and Division of Longer Expressions. pp. 78-90. Monomial x Binomial: a (b + c)= ah + ac ; sum of products term- wise. § 49, p. 78. Monomial x Longer Expression : sum of products termwise. Exer- cises VII. § 50, pp. 78-79. L,onge~ Expressions -^ Monomial : sum of quotients termwise. Exer- cises VIII. § 51, pp. 79-81. Monomial Factors: application of rule; work by inspection. Product of any Expressions : sum of partial products. Exercises IX. §§ 52-53, pp. 82-83. Division of any Expressions : long division by reversal of partial products; principally exact divisions. Exercises X. § 54, pp. 84-88. SUMMARY 117 Polynomial : letters occur in simple powers ; a + bx + ex 2 + ■ • • + lx n ; a + bx + cy + ' which evidently means the same thing. [C is the common factor of numerator and denominator of the given fraction ; N is the numerator of the resulting fraction ; D is the denominator of the resulting fraction.] This principle may be proved by means of the rules of p. 35. For - = A , provided N = D • X, and VjJl^X, provided C • JV>(C . D) ■ X = C ■ (D • X). C ' D But by IV, p. 106, the two conditional equations preceded by "provided" are equivalent, for we may divide both sides of the N C • N last bv C, if C is not zero. Hence, — and ——- are the same number J D C • D X, prodded C is not zero. This proof need not be learned at this time. Notice particularly that C must not be zero. Thus, in substituting numbers in the place of letters in order to cheek an example, or for any other reason, if the numbers substituted would make one of these divisors zero, at any point in the work, they must not be used. Again, if both numerator and denominator are divided by an expression that contains unlnoton letters, we should be careful to say that the result is correct, provided the (unknown) value of that expression is not zero. Reversing the preceding rule, we may multiply both numerator and denominator by the same number, except zero. Notice, however, that adding the same number to both numerator and denominator is not allowable, in general. Thus, 3 ■ , , . 3 + 2 - is not euual to -• 4 [ i + 'J Likewise, subtracting the same number from both terms is not allowable. 69. Common Factors. In § 68 we used any common factor of both numerator and denominator as the divisor of both. These common factors may be found by inspec- tion in simple examples, as in § 51, p. 79. 120 FRACTIONS [Ch. V 20 a*V Ex. 1. Ex. 2. FRACTIONS 4b • 5 a?b* 4b 3a • 5 aW 3 a 15 a 4 6 3 a; 2 -! = (g-l)(y + l) ^ a;-l a: 2 +22- + l (»+l)(a; + l) # + 1 u o m 2 + 5 w + 4 (?n + l)(m + 4) m + 1 m 2 — 6 m — 40 (w — 10) (m + 4) w — 10 If all the common factors are removed, the fraction is said to be in its lowest terms. The rule is the same as in elementary arithmetic : To reduce a fraction to its lowest terms, divide both numer- ator and denominator by each of their common factors ; in the resulting fraction the numerator and the denominator should have no common factor. EXERCISE I: CHAPTER V Simplify the fractions : 18 96^ 2 3 . 3 2 . 5 2a% # 9 - 16 aW 20* ' 144* ' 2.3 3 -5 2 ' ' Sab 3 ' ' -24abx 4 ' „ 65 ,189 c 4 3 -5.6 2 _ XY 2 Z 3 ' 35 mVp 2 • 4 • 6. • 8. — — • J.U. — • '70 105 2.3-4 XY 3 Z 21m 3 wp 4 11. 16 m 3 n 2 -42 ^t 13 -lOafy* 3 14 p 2 q-pq\ 24 mp 2 ' ' - 30 xyzt 2 ' 2tfy 2 z p 2 q 2 27(a + b) 2 (x + y)V 65(g-r)(r-p)(j>-l) 8 ' 15(a + 6) 3 (a; + t/)r 6 ' 26(y-r) 2 (r-p) 3 (p-l)' 70. Highest Common Factor. As in elementary arith- metic, the product of all the common factors is called the highest common factor, usually abbreviated H. C. F. This is often called highest common divisor, H. C. D.; also, in elementary arithmetic, greatest common divisor, a. a d. 69-70] COMMON FACTORS; REDUCTION 121 In arithmetic we found a number as G. CD. Thus the G. C. D. of 60 and 72 is 12, for 60 = 2 • 2 • 3 ■ 5 and 72 = 2 • 2 • 2 • 3 • 3. The common factors are 2, 2, aud 3, and their product is 12. Just here we are working with polynomials ; that is, expressions each of whose terms contains only simple powers of the important letters. (See §§ 55, 56, pp. 88, 91.) We shall use as common factors of two such polynomials only factors which are themselves polynomials * The H. C. F. here is therefore a polynomial, though we shall later have examples of another kind. EXERCISES II: CHAPTER V Reduce to lowest terras after finding the H. C. F. of numera- tor and denominator : 1. 15 a?y— 10 xy 2 5 xy 2 12 a 3 6 2 c + 18 a&c 3 4- 15 6c 2 30 abc 3 8pq 5. z 2 2 2 -5z + ( + z-6 > „ 1 - 2 r - 8 r 2 16p s q 2 — 24j;g 3 1 _ 7 ,- _ 18 j- 2 4. m~n — mn 2 6. Z 2 Z' -5z + t ■-z-6 i a x 2 — 2 a?/ + ,'/ 2 O .1 m — ?r or + 4 «y — 5 y 2 9. m 6 — n 2 • 15. 6 v 2 _ v - 2 ra 6 + 2 m 8 w — 3 n 2 9 ir - 4 10. r 2 — 2 rs — 15 s 2 r 2 + 8 rs + 15 s 2 16. 3 z 2 _ 13 z + 12 9 z- - 24 2 + 16 11. u 2 + 9 uv - 10 V 2 2 u 2 — uv — v 2 17. 5 s 2 + st - 6 * 2 s 3 * - st 3 12. » 2 -2x- 35 2 x 2 - 5 x - 63 18. 12 x 2 — xy — 6 y 2 Sx 2 -2 xy -3y 2 13. 1 — x — Qx 2 3-X-24.X 2 19. 5 p 2 — 26pq + 5q 2 3p 2 — 5pq — 50 q 2 14. x 2 — y 2 x s — y 8 20. a 2 + 2 ab + b 2 - c 2 a 2 + b 2 + c 2 + 2bc + 2ca + 2ub * It should be noticed that an expression of even a single term may be a polynomial ; it is so, provided it contains only simple powers of the im- portant letters, or if it is a known constant. (See § 9, p. 9, and § 55, p. 88.) 122 FRACTIONS [Ch. V 71. Practical Work in Common Factors. The chief use of the H. C. F. is in the reduction of fractions. In cases where the factors cannot be found by inspection, the product of all those common factors which we can find is used in its place. Thus, practically speaking, we find the H. C. F. only in the simpler examples. Ex. 1. Find the H. C. F. of 36,285 and 44,895. We may notice that 5 and 3 are each common factors (by trial). Hence 15 is a common factor. The student might try for some time without rinding any other common factor. We should say, after a reasonable amount of work, that 15 is a common factor of 36,285 and 44,895, and that we cannot easily find any others. However, 41 is also a common factor; the H. C. F. is really 15 x 41 = 615. There is a long process that is sometimes used to make sure of the H. C F. ; this is hardly worth while at present, however, for the work required in it is longer than the value of the results justify. The process is found at the end of this book. (See Appendix, § 19.) Practically speaking, we wish to reduce a fraction only so far as i? convenient. For example, ||ff| = f £ \% after taking out the common factors 3 and 5 which are easy to find. If we wish really to divide numerator by denominator to find the value in decimals, it is worth while to find such easy common factors as 3 and 5 and remove them first. But it would be a waste of time to try past 11 or 13 in seeking common factors by trial. Certainly it would be a waste of time to try a number as high as 41 — which is really the next common factor, for the whole work required to divide 2419 by 2993 (to several places of decimals) is less than the work required to find the factor 41, either by trial or by the long process mentioned above (Appendix). We shall say that a fraction is reduced as low as is prac- ticable, or is in its loivest practicable terms, when all the com- mon factors of numerator and denominator that can be found by methods we know at present have been removed. Thus . 36^85 = 3-5-2419 = 2419 ^ a§ ticable) , 44895 3-5- 2993 2993 v although g» „*£»..» do-* terms). 71J COMMON FACTORS; REDUCTION 123 Likewise a ' b + a5/>8 + a%5 - = a3 ^ a * + a%2 + h *^ 1 ewlse ' a W-4a*b* + iaW-'Sa*b* a 2 6 3 (« 8 - 4 a 2 £ + 4 ai 2 - 3 6 s ) a(a* + a*b* + b*) ~ft 2 (a 8 -4a 2 6+4a& 2 -36 8 ) (as low as practicable), a (a* + aW + b*) = a(a 2 - ah + & 2 ) Q 2 + a h + /;-) i-(a a - 4 a 2 6 + 4 ai 2 - 3 b s ) ' b*(a*-ab + i 2 ) (a - 3 b) although = a ( a * +ab + b2 ) (lowest terms). Unless the factors happen to be known, the second operation for finding the result in lowest terms cannot be done without considerable work ; it is scarcely worth while, the result in lowest practicable terms being the one most often used. * EXERCISES III : CHAPTER V Reduce to the lowest terms practicable : 1 2.0 4 6 p 3315 o 824 6 4. 417 10 e 17384 -*■• 2418- *' 4845' °' 13206* *■ 434T0" - «'■ 18TJ40' 6 9636 7 «*-& 4 8 « 3 + 6 3 9 y + 7y-8& w 3 » + 3 m 2 w 2 — 4 mn 3 . re 3 — x*y — xy- + y 3 2 m 4 + 3 m 2 n 2 — 5 mw s x 3 + x l y — xy- — y 3 J-l- — o ^ ' !*• r 2 — 6 r + 5 x 3 — 2 a 2 ?/ + 2 an/ — t/ 3 a,- 3 ?/ - yy 3 3 a 5 & 2 + 6 a 3 6 4 + 27 a& 6 2^-5 a; 3 i/ 3 + 3 an/ 5 2 a 5 6 2 - 2 a A b 3 + 2 a 3 6 4 + 6 a 2 6 fi [Many of the above exercises will of course be reduced readily to the actual lowest terms.] * The question as to when any fraction really is reduced as low as is practicable is to some extent arbitrary, and depends on the breadth of the student's experience in working with literal expressions and recogniz- ing factors. If the factorization of a* + a 2 l>- + b* and a 3 — 4 d 2 b + 4 ab 2 — 3 b 3 , used above, were familiar, then the fraction would be in lowest terms practicable only when written in the form a ^ a ' a + — i • The 6 2 (a— 3 b) teacher will be able to judge what degree of reduction may be reasonably expected in the case of his own students. See also Appendix, § 34, et seq. PART II. RULES OF OPERATION 72. Rules of Signs. A fraction is merely an indicated quotient, as in elementary arithmetic. The rules for signs hold as in division. Thus, — a a -^=- rt ; and -b ft — a _ a ~b~ b' Likewise, - '(~l)'-^ -(r?)=+S' -&)- a "ft There are three important signs to be considered : the sign of the numerator, that of the denominator, that in front of the whole fraction. Of these three important signs any two may be changed at the same time, without changing the value of the fraction. The sign in front of the fraction affects the total result. Thus, in the fraction a , if a = 8 and b = 3, we get 2 a — b 3 • 8 + 5 ■ 3 _ 24 + 15 _ _ 39 _ _ g 2-8-3 " 16 - 3 " 13 To change the sign of the numerator (or denominator) be careful to change the sign of each term, including the first term '" _ 3a 4- ob _ - (3 a + 5 b) _ -3a -5b 2a- b 2a -b 2a -b Check : Let a = 8, b — 3 in this answer; -3-8-5.3 - 24 - 15 - 39 = -3. 2-8-3 16-3 13 . 3 a 4- 5 ft 3a + 5ft 3 a 4- 5 ft 3a + 5ft Atiain, ! = ■ = ■ = ■ • & 2s • r a; + 1 jc — 1 _ (x + l)-(x 2 -\) (x-l).(x-l) _ 2(x + 1) • (x + 1) l.(z»-l) (x + l).(*-l) (* + l)-(*-l) (x 3 + T 2_ a -_ l) + (a: 2 -23: + l)-2(x 2 + 2x+l) x 2 - 1 _ x 8 + x 2_a;_i + a; 2_2x + l-2j: 2 -4x-2 = x 3 - 7 a; - 2 x 2 - 1 x 2 - 1 In these examples one of the fractions has the sign — . This amounts to subtracting it from the others ; in any case the result is called the algebraic sum. As in ex- ample 1, if a term is inserted to be added to the others which itself is not written as a fraction (2xy in Ex. 1), such a term may be written in the form of a fraction rL^, for example) by inserting the denominator 1. Such a form as 2 xy + — + ^ is often called a mixed expression, y 2 2 x since part is in fractional form and part is not. f3-74] RULES OF OPERATION 127 As common denominator, any convenient expression may be taken of which each given denominator is a factor ; in any case the product of all the given denomi- nators will surely suffice. See also § 74, below. EXERCISES V : CHAPTER V _ x — 1 . 2-6) 2 ]. But a 2 - ft 2 = (a - 6)(a + ft), (a - ft) 2 = (a - 6) (a - ft), and 4(« + ft) = 40 + &). Hence, a common multiple is [(a-5)(a + &)][4]|>-&], where, just as before, we have omitted duplicate factors. This is the lowest common multiple* (L.C.M.), i.e. if any of the factors now remaining in the result were omitted, the result would not be a common multiple. The rule is precisely like that of elementary arithmetic : Factor each of the given expressions ; write as their L-CM. the product formed by omitting in each any of its factors * It is usual in arithmetic to use the word " least,'' in algebra the word "lowest," in this connection, 74] RULES OF OPERATION 129 that is already written down as many times as it occurs in the expression under consideration ; in short, omit the dupli- cate factors. Care must be taken to have a factor occur in the L. C M. as many times as it appears in any one factor. Practically it may not be convenient to factor the given expressions completely ; in this case we omit only those duplicate factors which can be found by the methods known at present. The lowest practicable multiple is thus found ; it is the simplest common multiple that can be found conveniently. Thus, given 44.895 and 36,285, we find, 44,895 = 3 • 5 • 2993. 36,295 = 3-5. 2419. Hence, the lowest practical multiple is (3 • 5 • 2993) x (2419) = 108,601,005. As a matter of fact (see p. 122) 41 is a factor of 2993 and also of 2419. Hence the L. C. M. is (3 • 5 • 41 • 73) x (59) = 2,648,803. In this example it is clear that t n ivr Product of the two given expres sions LCM = iLCuT ' This is true if only two numbers are given, for the H. C. F. is exactly the product of those duplicate factors that are to be omitted in forming the L.C.M. This remark leads (see Appendix, § 19) to a general method for finding L. C. M. in any case ; hut this process is usually long and is scarcely justified by the value of the result. In adding fractions by the rule of § 73, p. 125, choose as the common denominator the lowest practical multiple of the given denominators ; then reduce each fraction to this denominator by multiplying its numerator and its denomi- nator by the quotient formed by this common denominator divided by the given denominator, and proceed as on p. 125. hedrick's el. alg. — 9 130 FRACTIONS [Ch. V e x . i. 2x y - 2x ~y _ rf+zy 2 . x* — y 2 x-y (x + y) 2 We notice that the L. C. M. of the denominators is (x - y) (x + y) (x + y) = (x - y) (x + y) 2 . Hence, the expression given _ 2 xy(x + y) _ (2 x — y)(x + y) 2 _ (x 2 + 2y 2 )(x — y) ~ O - y) O + y)' 2 (*-y)(x + y)' 2 (* - y)(* + y) 2 _ (2 x 2 y + 2 xy 2 ) - (2 x 3 + 3 x 2 y - y a ) - (x 3 - x 2 y + 2 xy 2 - 2 y 3 ) (x -y)(x + y) 2 _ - 3 x 3 + 3 y 3 _ _ 3 (x 3 - y a ) (x-y)(x+y) 2 (z-yXz + y) 2 _ 3 (x - y)(x 2 + xy + y 2 ) _ 3 (x 2 + xy + y 2 ) (x + y)(x + y)(x-y) (x + y) 2 In this example the result is reduced to its lowest terms ; the student should try to do this in every example, and the result should be reduced to lowest terms or at least as far as is practicable. The student is advised to leave the common denominator in factored form, as in the preceding example, until the work is completed. EXERCISES VI: CHAPTER V Perform the indicated operations : 11 1 , 1 a + a 2 a — a 2 x 2 + 2 x — 15 x 2 — x — 6 x x — y x — 1 a? — 4 ' x 2 + xy x 2 +2xy + y 2 ' ' o; 2 + 2x--15 %*-x-& 5. X + V | 1 g .V + 2y' . x 2 + xy + y 2 x—y tf—tf c y-z y + z 4z y(y + z) y(y-z) y 2 -# •4-75] RULES OF OPERATION 131 7. , I + » • 1 8. - y) (.x- - z) (y- z) (y - x) (z -x)(z- y) 2 r - 1 r - 2 1 r 2 + 2 r - 3 r 2 - 1 r + 3 ^/ g .t — 1 .x— 1 4 ^-5a; + 6 ar-a;-2 a; 2 -2z-3' 1 , 1 10 - -= — ~ ; + x- 2 -3o;-4 a , -o» + 4 2x 4 a; 3 x' + y 2 x + y y — x x 4 — y* 12. -1^-1=1+ 7 13. 2 a; + 3 2-x2/l5. * + 1 2 * + 1 ■ 3 + 2a; 2 + 5x- + 2 4a: 2 + 5 a;- 6 8 a; 2 - 2 < -25c 2 ' u — v v + u - n a b c d 6 ~rv^ — r~* x "^ ' 10 - i x ~ x w x ~ • tr -f 2 wi> + V v — u o c a a ,. 1^^+ J^fe. 11. M x tf - fy 1 — r 1 — 16 r p + q 75] RULES OK OPERATION 133 x 2 — 5 x + 6 xr — 7 x +■ 12 ' ar 1 - 5 a; +4 * ar" - 7 a: + 10 ' 13. a- 6 cr + 2a + l a 2 - a -2 a 2 - 2 aft + ft 2 - 15 (s - a) 2 (y - b) s 77 (x -a)(z~ c) 14 (2 -c) 2 -25(y-6) 4 15 ax + or >: ar — cr ojb — ar a,- 2 + 3 ax + 2 a 2 16. — ~ y* v ^" + 4 a;y + 3 y 2 af' + 7 xy + 6 y'-' ar' + xy + y 2 17 xy + y 2 x pa; -py- qx + ?y p-g ar' + y 3 18 a 2 + 6 2 + c 2 + 2 6 c - 2 eg - 2 g& a 2 - fe 2 + 2 6c - c 2 a+6-c a 2 -b 2 -2bc-c i ' 19 a,- 3 + ary + a?y 2 + f x ^~ f ar 5 — 3 x*y + 3 ajy 2 — y 3 ar" + y 3 20 4a-- 2 + a,--3 3a,- 2 + 5a;-2 a^ + a;-2 8ar J -2a;-3' Q , 6 a 2 — a — 2 a 2 21. _ x - a 2 - a - 2 4 a 2 - 1 22 x 2 y + xy 2 ^ 2x 2 -xy-y 2 Gx 2 + xy — y 2 .ry- 23. a; 4 — y 4 X 3 ar* + 2 y 2 12 x 4 — x?y l — 6 y 4 ar 3 — ary — xy 2 + y 3 24. « 3 + 2 « 2 6 + 2 «6 2 + b 3 « 3 + 3 a 9 6 + 3 aft 8 + 6 s a 3 - 3 a 2 6 + 3 ab 2 - 3 6 s a 3 - 6 3 134 FRACTIONS [Ch. V 76. Division : Complex Fractions. We have also found and stated (see p. 74) the rule for dividing one fraction by another ; it is , a ^ c _aa b d be The proof consists in the fact that the quotient x the divisor — the dividend : ad c a — x - = -. be d b In words this rule reads as in elementary arithmetic: To divide one fraction by another, multiply the dividend by the divisor inverted. For ?^ = ^ = ^x-. b d be b c Ex 1 a 2 -! .x + l = x*-l x x-l = (x-lf | 3cc + 2 ' aj-1 3 a; + 2 x + 1 3a-|-2 Ex. 2. x* - 1 1= x 2 -! . a + l _. a 2 -! , 1 x-1 'Sx + 2' 3a + 2 ' 1 3^ + 2 x + 1 3 a; + 2 In this example the divisor is a fraction only after we insert the denominator 1. Though we should not usually insert this denomi- nator, it is often convenient to do so. The result of inverting a fraction is called its recipro- cal. Thus, the reciprocal of - is -. In general, the b a reciprocal of any number is 1 divided by that number. The reciprocal of ..1 e 1 . ., 1 , c a . -. a -, b b 4 is-; of - is 1 -^ - = 4; of-isl-s-- = lx- = -- 4 4 4 b baa The division of one fraction by another is often indicated by means of the fraction form. In this case the whole is called a complex fraction. 76] RULES OF OPERATION 135 a; 2 + l g--l = ^ + l , g + l = ie 8 + l a;-l = a^ + 1 Lx - 1 - x + j ~- tf-1'as-l z--l x + 1 ~(x + l) 2 ' x-1 ., a a + b a b Ex. 2. (i + 1) a + b a+b a + b -. b a — b b a a—b a—b a—b a—b a b s , a — b _ b(a — b) a+b a — b a + b a a(a + b) If there are more than two horizontals, great care should be taken to mark the main horizontal line heavily, for neglect to do so may lead to serious error. Thus, 2 , 5 10 , ., 3 2 1 2 _ = o x - = — , while - = - X - = — • 3 3 3 5 3 5 lo 5 Similarly, the position of the main horizontal line is essential in fractions in algebra, as in arithmetic; a mistake in recognizing which is the main horizontal line causes a mistake in the result. EXERCISES VIII: CHAPTER V What is the reciprocal of : l 0*>_9 _9? -_9 *■ *' 2' ^ 2" 2. a? 3 ? -a? --? a a n o> n — ct q a r> a ip ' b' ~b~' ^-b' ~b' Perform the following divisions : lo t o 16" "i" -64 . 40 -65 " -39 8. ^-5-100. 76 5 20 . 15 21 " 14' 144 . 126 25 ' 125 9 4 - 1 25 * 100 136 FRACTIONS f - 12 a 3 b 2 c . 2£a 2 b 50x?y 4 . 75arfy 35afy ' -7a;?/' ' 49z 2 " ' 56z 3 * 36 6c 2 32 aft 3 4 m 2 .- 12 Zm 25a ' 15 ' -9Zn 3 ' n 2 - 65 r 2 143 r.s 3 14. 64 s 2 -144 » ( 1+ £fcM 1+ 2# 16 ^ . a: 2 + 2 ay ' x* + 2xy + tf ' x 2 -f a *-2ab-3b 2 . a 2 ~b 2 ■ a 2 -2ab + b 2 '" a 2 b-ab 2 ' p« + J?g _6 g 2 . y-3pg + 2g 2 9 9 i p z — q l p+pq z 2 + 8z + 15 ^ ga + Sz-lQ • 2 2_6 2 + 8 " z 2 -5z + 4 ' z 4 -16 2-z-z 2 20. z 3 -! 1+z + z- Simplify the complex fractions : 6s 2 + x -12 4 + -| 2^ + 35-6 ■•ifFfs- 23 -3?Eir- 25 ' x 2 + 4 a-- + 4 a; 2 + a; — 6 a* - 2 x - 3 1 22. y 2 - 25 //-' + 7?/ + 12 v 2 + 10 y + 25 ^ - 2/ - 12 X — 1 , 05 + 1 a; + 1 ' a- — 1 a? 2 + 2 a; - 3 a; 2 - 5 a- + 6 2Q ^-3* 24> 1 ' ' «±i: x (*» + l) 2a5 2 -9a5 + 10 a?-l PART III. PROPORTION 77. Definitions and Introduction. If two fractions are equal, their four terms are said to be in proportion. Ex 1. f = |, for each is equal to \. This is sometimes written 8:6 = 4:8, and read "3 is to 6 as 4 is to 8," but there is no advantage in doing so. Ex.2. iL^L =: «±i,f„ 1 .*+i=*±lx a! " 1 - * -1 (x-1) 2 x-1 x-1 x-1 x-1 (x-1) 2 Likewise all equalities of two fractions given above are propor- tions. Any equation is a proportion if the denominator 1 is supplied in case no denominator is given. A fraction is often called a ratio, and - is written in the form a : b, and read '•'•the ratio of a to 6." 78. Standard Changes. We may make a new propor- tion (equation) from an old one by the operations of § 65, p. 106. Some of the forms derived have been given names. I. If - = -, we may add 1 to each side ; then, b d a , i__£ , i . a + b _ c-{- d b d b d This is called the process of composition. ^-,24, 2 + 3 4 + 6 nv 5 10 Ex.1. _=_, hence, —=—, or, - = -• Ex 2 x ~y= x2 ~ f • hence -**- = 2^ + 2xy , x + y x 2 + 2xy + y- i ' x + y x? + 2xy+y 2 II. If - = — , we may subtract 1 from each side; then, b d a ^ c -, a — b c — d 1= 1, or — : — = — — • b d b d 137 138 FRACTIONS [Ch. V This is often called the process of division, but this name is not well chosen, since "division'' means something else. ttt is- a c ,, a + b c+d -, a — b c — d III. If -=-, then —£— = — "—- and —7—= — — • b d b d b d Dividing the right-hand sides, and also the left-hand sides, we find, a+b c+d b d a+b c+d ,, or, a b c — d a—b c—d b d This is called the process of composition and division. IV. If - = -, we may divide 1 by each side, if neither is b d 11 b d zero; then — = -, or, - = — • a c a c b d This is called the process of inversion. V. If - = -, we may multiply each side by % bd; then b d - x bd = - X bd, or, ad = be. b d This is called the process of clearing of fractions or the process of cross-multiplication. VI. Finally, if ad= be, we may divide each side by bd, if bd is not zero, [or by dc; or by ab ; or by ac], then ad be ,i\ a c . bd bd b d r /AN a b /-on d c ^,v d b or, (2) - = - ; or, (3) - == -. ; or, (4) - = - c d b a c a. 78-79] PROPORTION 139 A comparison of these equivalent forms shows several permissible changes that are sometimes given names ; thus, the change from form (1) to form (2) is called alternation ; the change from form (1) to form (4) is called inversion ; but the simplest way to remember all of these is to remember the operation of clearing of frac- tions, No. V above, and the principle stated in VI, which amounts to dividing both sides by the same number. Besides these six, many other changes are allowable ; for example, we might add 2 to each side, or subtract 3 from each side, etc. In general we merely carry out operations that are allowable with any equation. These principles are often valuable in solving equations. EXERCISES IX : CHAPTER V In the following exercises, apply to each given proportion the process referred to by Roman numerals : l. ^±2 = 12 j n m 2 IV x-2 16' ' ' 3 1' x 8 4 x*-2xy + tf ■ 9 j Hj m IV 5 3 = 15. ^ yi (3 xr + 2 xy -f y 25 5 x 6- ^±\ = ^=Zl. I, II, HI, IV. 7. tf^ = ^-- I, II, III. a — bx + y xr + 4 13 8 P=M = \. I II in. 9 . a--56 = 2 T n ni IY y 2p + q 4' ' ' a + ob 1' ' 10 2a?-l 2. II V VI (2) 11 ™ a -n* = ™?-n s . I, II, VI (2). 79. Variables in Proportion. Proportions (equalities of fractions) arise whenever two varying quantities are so related that their quotient is always the same. (See § 20, p. 25.) 140 FRACTIONS [Ch. V Thus, if butter is 30 cents per pound, n pounds cost 30 n cents, or p — 30 n, where p is the cost in cents and n is the number of pounds. (See p. 23.) If n = 3, p = 90 ; if n = 4, p = 120 ; etc. ; and we have 90 120 , p , Qn , — = — = etc. = <- always = 30 always. 3 4 n In case two varying quantities y and x have a constant y quotient, i.e. - = &, where k is constant, any two pairs of JO values of y and x form a proportion, and we say that the variable quantity y is proportional to the variable quan- tity x; that is, any pair of values of y and x form a propor- tion with any other pair. We have already used such proportional quantities and have drawn corresponding figures, which we found to be straight lines. (See pp. 20-25.) We shall now make clear that the figure is always a straight line if ¥- = k, i.e. if y = k • x. x 80. Graph. In dealing with an equation of the form y = kx, we found always a straight-line figure. (See p. 25.) We see that if x=x ^ y = k . x=% y=2k; x= 3, y = 3 k; etc., and we wish to show that these points lie on a straight line. To plot the point z=l, y = k we go one unit to the right and k units upwards to the point marked A. To plot the point x=2, y=2k we go one unit beyond A to the right and k units above A (Fig. 20). The rectangle whose corners are and A is precisely the same shape and size as the rectangle whose corners are A and B. Hence, the diagonal OA has the same 7&-80] PROPORTION 141 direction as the diagonal AB ; consequently AB is an extension of the straight line OA. The argument is the same from B on to the next point C; from on to _Z), etc. Likewise backward from to L, thence to 31, etc. If smaller steps are taken, the argument is always the same. Thus, steps half as wide and half as high would again bring us to the same line. D/ k / Y C / k / B / / k A / 1 k / n / 1 1 X / 12 3 4 i. /. M / / / N m / V Fk 3. 2 3. The idea of steps — of actual stair steps — is a good illustration of this; the broken line one unit to the right and k up makes such a stairway. The edges of these steps make a straight line as on any ordinary stairway. Finally we may say, The equation ii = kx is represented by a straight line through the starting point, which rises k units in passing 1 unit to the right. 142 FRACTIONS [Ch. V Ex. 1. In the problem just mentioned p = 30 n, where p is the price in cents of n pounds of butter. We may now draw the corresponding figure much more easily than on p. 24 where the same example was given. If we draw only two points of the figure and connect these by a straight line, we know the figure is correct, since we know in advance that it is a straight line. If n = 0, p — (that is, no pounds costs no money) ; if n = 1, p = 30 ; if n = 5, p = 150. Plot the point n — 0, p = at A . Plot the point n — 5, p — 150 at B. Join .4 and B by a straight line. This straight line is the desired figure ; from it can be read off at once the answers (approximately) for a variety of problems : the cost of a given number of pounds ; the number of pounds that can be bought for a given amount of money, etc., as on p. 20. Thus, having worked the two simplest examples of which we can think (that is, if n = 0, p — and if n = 1, p = 30), we have in this figure the solutions (approxi- mately) for any problems in proportion which might be given under this example. It is well to plot a third point. Thus, if n = 10, p = 300. Plot this point at C. It lies on the straight line. If it did not, we should know that we had made some error. The value in plotting three points lies in this check on the accuracy of the work. Ex. 2. In the Fahrenheit and Centigrade thermometers the scales of temperature are made in divisions that are propor- tional. Thus 9 divisions on the Fahrenheit scale are equiva- lent to 5 divisions on the Centigrade scale. If the temperature rises 9 degrees by a Fahrenheit thermometer, it rises 5 degrees Fig. 21. SO] PROPORTION 143 by a Centigrade thermometer. The amount of rise or fall on either scale can be found by ordinary proportion, in case the rise or fall is known on the other. But the Fahrenheit scale is marked 32° at the freezing point, whereas the Centigrade is marked 0° at the freezing point. We must therefore subtract 32 from each Fahrenheit tempera- ture before we can conveniently compare it with Centigrade temperature. Hence, F— 32 is the quantity to be compared with C, if F and C stand for the readings of the two thermometers at the same time and place. Since these compare as 9 compares to 5, we have 7^—32 9 C~ = 5' and we say that F— 32 is proportional to C, the constant quo- tient (or ratio) being I. Multiplying both sides by 5, also by C, we have 5CF-32) =9C, or, 5 F - 160 = 9 C. To draw the corresponding picture, we may, as on p. 23, take this equation (which is given in the Tables without argument) and plot several points. We need only plot two of them, though three are better. If C = 0, F=32, as above (this is the freezing point). If C = 20, F = 68 (this is the ordinary temperature). If C = 100, F = 212 (this is the boiling point of water). Plot these points ; they are L, M, N, respectively, in the figure ; the straight line through these is the graph. Notice that it is here not F but rather F — 32 which is pro- portional to C. Hence, the dotted line which is just 32 points below Fig. 22. 144 FRACTIONS [Ch. V the line LMN is the line which represents the proportion of F — 32 ami C. The heavy line really represents the corresponding values of F and C. Given values of F, we can find the corresponding C either from the figure (approximately) or from the equation 5 F - 160 = 9 C, and conversely F can be found if C is given. Thus, if a Centigrade thermometer reads 28°, find the temperature in the ordinary (Fahrenheit) scale. In the figure we move out along the line marked C to 28, then directly upwards till we meet the straight line LMN. The height is the value of F; it is seen to be 82, approximately. From the equation, 5 F - 160 = 9 C = 9 x 28 = 252. Add 160 to each side : 5 F = 252 + 160 = 412. Divide each side by 5 : F = 82.4. The exact answer is therefore 82°.4; notice that the answer, 82°, obtained from the figure, is not exactly correct. In any case it is well to get a result from the figure and a result from the equation ; the result from the figure is only approximately correct; that from the equation is exact. The result from the figure serves as a check on the exact answer from the equation. In many cases an exact answer is not necessary. In the tempera- ture example above, it is often desirable to know approximately the temperature; in fact, very few thermometers, except those especially made for experiments in physics, etc., are really accurate. The error made above ( T 4 „ of a degree) is smaller than the error made by most thermometers in ordinary use. The student should in no case express an answer more exactly than the circumstances warrant ; thus, no ordinary thermometer, even of those used in physical laboratories, will read correctly nearer than tenths of a degree. Hence, in such an expression as 35°.74, the last figure does not really mean anything and should be omitted.* * It is suggested in the text that to plot the graph of y = kx, the two determining points may be chosen as (0, 0) and (1, k). This, as well as the ordinary practice of plotting ax + by + c — 0, by locating the points (0, — c/6), ( — e/a, 0) is unwise as a general rule. The disadvantages are two in the latter case, and at least one applies to the former also : 1. The numbers that fix the points thus located may be fractions with large denominators which cannot be located accurately ; proper 80] PROPORTION 14f> EXERCISES X: CHAPTER V By the principles of § 78, solve the following equations (Exs. 1-6): 1. -2— = ** • [Use II or V or VI (2).] 4. -^- = 8 . p-8 5 L WJ x+2 3 2. a !±i! = n . [Use II or V or VI (2).l 5. 4 = ^i-. _ a^-3 2 a- -2 2 c + ^ then « = £. a c 6 a" 25. If a2 + 52 = ac + ^,theng = ^ ac + 6d tr' + d 2 ' & d 26. Express the simple interest, i, at 6 % on $ 500 for t years. Draw the graph. In how many years will the inter- est be $90? $60? What quantities are here proportional ? 27. Express the amount in the above problem. Draw the graph on the same diagram as that for example 26. What quantities are proportional ? State and explain the relative position of the two figures. 28. Compare by an equation the readings of a Centigrade and a Reaumur thermometer as given in the table at the back of the book. What quantities are proportional ? Draw the graph. Compare the readings of a Fahrenheit and a Reaumur ther- mometer. What quantities are proportional ? Draw the graph. 80] PROPORTION 147 29. John can run 20 feet a second ; Henry, 15. Compare the distances, x, y, from the same starting point of John and Henry at any time. What quantities are proportional ? Draw the graph. 30. If John gives Henry a start of 5 feet, compare the distances from John's starting point, in Ex. 29. What quanti- ties are proportional ? Draw the graph. Explain the relation to the figure for Ex. 29. 31. If John and Henry, in Ex. 30, meet, show that y = x for the point of meeting. Draw the graph giving all values for which this relation holds. Where will John meet Henry in Ex. 30 ? Solve this problem also by means of the equations. 32. Try to find graphically, as in Ex. 31, the distance at which John and Henry meet in Ex. 29. Explain your failure. 33. State the general formula for the interest, /, in terms of the principal, p, the rate per cent, r, and the time, t, for simple interest. If the principal and time are constant, to what is the interest proportional ? To what other quantities can we make the interest proportional by keeping certain quantities constant ? 34. What is the number of minute spaces, m, traveled by the minute hand of a clock in 60 minutes ? the number, h, traveled by the hour hand ? What is the relation between m and h ? Draw the graph. At 12 o'clock what are the posi- tions of the two hands? Thus, m and h may represent the positions of the two hands m minutes past twelve. 35. What are the positions of the two hands at 3 o'clock ? If m and h are to represent the number of minute spaces dis- tance from the figure "XII" of the two hands at m minutes past three, what equation connects m and h ? Draw the graph. Find graphically when m = h ; that is, when the hour and minute hands will be together. Solve also from the equation. 148 FRACTIONS 36. Find graphically and from the equation, when, after 6 o'clock, the hour and minute hands of the clock will be to- gether. 37. Find graphically and from the equation, when, after three o'clock, the hour and minute hands will be exactly oppo- site one another. Find also when before three o'clock this will happen. 38. What is the total surface area, A, of a closed cylinder whose height is h and the radius of whose base is r ? Factor the result. If r is constant, to what variable quantity is A proportional ? Assume some convenient value for r, say r = 7 or | (take tt = 3}), and plot the relation between A and h. Plot on the same diagram the surface area, B, of a corre- sponding cylinder, open at the top. Plot on the same diagram the surface area, C, of a cylinder open at both ends. Compare the three figures. 39. Each of two boxes has a square bottom and rectangular sides. The heights are the same. The first is 2 feet wide and has a closed bottom and top. The second has a bottom, but no top; and it is 3 feet wide. Find the common height of the boxes if the amounts of lumber used in making them are in the ratio of 8 to 13. PART IV. FRACTIONAL EQUATIONS 81. Fractional Equations. A fractional equation is one which contains fractional expressions. Equations with simple fractional coefficients are not usually in- cluded, because they can be so easily solved; but they are none the less fractional equations. Some such equations were solved on p. 58. Ex. 1. Thus, f x + 5 = 7 is an easy fractional equation. Subtract 5 from each side : f x = 2. Divide each side by 2 : ^ x = 1. Multiply each side by 3 : x = 3. 2 x 1 6 &• Ex. 2. A more typical example is x + 2 3 x + 10 Multiply each side by x + 2 and then by 3 x + 10 : (3 x + 10) (2 x - 1) = 6 x (x + 2), or, x 2 + 17 x - 10 = 6 z a + 12 x. Subtract 6 x 2 from each side : 17 a:- 10 = 12*. Subtract 12 x from each side, and add 10 to each side : ox = + 10, or, x = 2. This we verify by trying x = 2 in the given equation. Tims, 2-2-1 6-2 3 12 = — = , or, - = — , 2 + 2 3-2 + 10 4 16 which is true ; hence we conclude that x = 2 is a correct answer. As in this example, it is usually best to dear of fractions (§ 78) immediately. To do so, multiply both sides by the L. C. M. of the denominators, if it can be found, or at least by as loiv a multiple as is practicable. Or we may simply cross-multiply, as in § 78, No. V, p. 138. 149 150 FRACTIONS [Ch. V Ex.3. _^- + ^i = 2x-6 5 x — 5 a; — 1 2 a: The L. C. M. of the denominators is 10 x(x — 1) ; multiplying both sides by this gives, 2 x • 3 + 10 x • (x — 4) = 5 (x — 1)(2 x — 6). (This is also found hy cross-multiplying, § 70, No. V.) Or, 6 x + 10 x 2 - 40 x = 10 a; 2 - 40 x + 30. Subtract 10 x 2 — 40 x from each side : 6 x = 30, or, x = 5. ™ , 3 ,5-4 2-5-6 nr 3,1 4 Check : = > or, — + - = — - . 5-5-55-1 2-5 20 4 10 In case the answer found makes any denominator of the original equation zero, that answer must be discarded, since division by zero is impossible. (See § 46, p. 75 ; § 65, p. 106; § 67, p. 109.) EXERCISES XI : CHAPTER V Solve the following equations for the unknown quantities : i. y~ 3 =~jl a. 2. 2y-5 Vy + 2 3*-4 Gt -11 t + 2 2* + l x-1 x + 3 2x + l 2 x + 12 M 2t + a 2^ + 6a 4. = • t — a t + a [Find «.] 10. 1] 3z-5 ._ 6z-25 12 ' 2z-15 4z-33' 6 m 8 + 4 = m + l . 2m 2 7 a 2 + a + 3 == a- r -l i 14 2a + l 2 10 & 4- 35 5 fe + 5 2 fc + 8 k-1 x 2 + 2 x -3 x — 1 3 a; 2 + 4 a: + 1 sb + 1 2 n + 5 _ n-2 6 n + 20 3h + 4 vi -2 ' »-4 a + 16 # w + 2 a; + 3 _ j a; + 5 c + 6^ c + 9* 10 a; -5 4«-l 4 — 5a; l-2o; fc + 18 _ A; + 14 A; + 3 fc + 2 81-82] FRACTIONAL EQUATIONS 151 15 3c-20 _3c-12 18 x + 2 ,_2_ =1 c _2 c + 3 a? + 3 a; 4- 5 16 a ^±l + Ezi^ = 2. 19. 2iH^ + i^5 = 5 . a; — 3 a; jp-o yi-10 _2a^ x + 6 =3 2a 3j^_2z ± U = 2> a: + 1 a; 4- 3 » + 5 2 z + 1 82. Other Cases. If the terms containing x 2 and other higher powers do not cancel as they do above, the resulting equations often may be solved nevertheless. (See § Gti, p. 106.) Ex. 1. 2x ~ S 5x x + 2 3 x + 10 Clear of fractious : 6x 2 + 11 x - 30 = 5x 2 + 10 x. Subtract 5 x 2 + 10 x from each side : x 2 + x - 30 = 0, or, (x+ 6)(x -5)=0. Hence, either x + 6 = or x - 5 = 0. See § 66. Whence, x = — 6 or x = 5. Check for x = 5 : o . 5 _ 3 5.5 7 25 , . -v = - — , or, -= — (check). 5 + 2 3-5 + 10 7 25 v Check for x = — 6 : g-(~ 6 )- 3 = 5 (~ 6 >- , or, -li5 = -30 -6+2 3(- 6) + 10 -4 -8 v Thus there are two answers, either of which is correct. EXERCISES XII: CHAPTER V Solve for the unknown quantity : 2ai + 3 _ 5a; 23 + 5 _ z + 5 _ x + 2 ~2a: + l" ' z + 1 " ' z + l' t + 3 = 3t + 5 n-1 = 2n-18 t-3 2t-5 ' 2n-5 3w-20 152 FRACTIONS [Ch. V 5 2a — 4 = 3a — 6 [ 1-x 43-2 , a - 3 a — 2 ' ' » + 33aj — l" 6 6 + 1 ■ = 26 - 1 , 9 3a? + l = 2 a? + 3 "3 6-2 36 + 2* 5x 3x + 4* o x + 3 _3* + 2 1Q fc + 1 , ^-4 = 1> 2 a; -5 a; + 2 A: + 3 ft-1 ii. p+I + Zp- 1 ^. p-1 2p-3 p* - 5 p + 3 = 2 p 2 -10 p + 15 p-9 2p 12. 83. Linear Equations ; Other Equations. The equations in §§ 81 and 82 are distinguished by having only the first power of x in their reduced form after clearing of fractions and simplifying. Such an equation is called a simple equa- tion, a linear equation, or an equation of the first degree. (See pp. 25, 58.) If x 2 is the highest power of x in the reduced form, the equation is called a quadratic equation or an equation of the second degree. If x 3 is the highest power of x in the reduced form, the equation is called a cubic equation, or an equation of the third degree. If ic 4 is the highest power of x in the reduced form, the equation is called an equation of the fourth degree, and so on. In general, if the reduced form contains x n , but no higher power of x, the equation is called an equation of the nth degree. In these statements it is understood that nothing but simple powers of x multiplied by constant coefficients remain as terms in the reduced form. 82-85] FRACTIONAL EQUATIONS 153 84. Operations. In this chapter we shall deal princi- pally with equations of the first degree or linear equations. The operations of importance are .- 1. Clearing of fractions : explained in § 81, p. 149. (See also § 78, V.) 2. Multiplying, dividing, except by zero, adding, or sub- tracting by the same number on each side : explained in §§ 85, 65, pp. 55, 106. 3. Transposing a term. To transpose a term is to carry it from one side of the equation to the other and cha7ige its sign ; this amounts to subtracting the term from each side. (Compare § 38, pp. 58-59.) Thus, if 2 x + 4 = 8, subtracting 4 from each side, we have 2 x = 8 — 4, which amounts to carrying the term 4 to the other side an4- N \l -S V ^ss :r_ N \7 > hv Vr " io O r?\ V X- \-X S s \ \ -t \ 5 s Fig. 24. ifa; = 0,y = 4, (/)) if y = 0, X = 5. (f) Drawing these points, we find the lines Afi and CD to represent equations (1) and (2), respectively. The only common point P gives x = 1.7, y = 2.8, about. Let us try to solve and find the answers exactly. Method I. By Addition or Subtraction. j3a;+2y = 12, (1) 1 4 x + 5 y = 20. (2) In order to obtain an equation that contains only the letter y, multiply both sides of (1) by 4, (3) 12x+8// = 48, and multiply both sides of (2) by 3, (4) 12 a: + 15 y = 60. iiedrick's el. alg. 11 1&2 SIMULTANEOUS LINEAR EQUATIONS Ch. VI Subtract (3) from (4) : 1 s if. In actual work we write this as follows : ly = 12, 3 x + 2 y = 12 4z + 5?/ = 20 7 2/ = 12 y=lf The numbers — 4 and 3 indicate the multipliers; notice that — 4 is written in place of + 4, in order that we may add instead of subtract. Likewise, we find an equation that contains only x : 5 _ o 3 x + 2 y = 12 4x + 5# = 20 whence, Hence, Check : 7 x =20 x = 2f. ™ 06 y — ]5 3(2f ) + 2(l f )- 60 + 24 - 8 7 4 = 4(2 . ) + B(1O= B0 4-60 = 140 12. 20. Notice that the answers obtained from the figure are slightly incorrect : x = 2f = 2.8555 ••• (we found 2.8 in the figure). y — 14 = 1.714 ••• (we found 1.7 in the figure). We might also find x after finding y by putting the value of y found (y — 14) in place of x in either of " ■ 1 , t X rJWwtv- \~ * t W- (0 °'<)_»Jl / 1) \- , 5 TO~ •v J abuut ' . _ _ J 3 ± "«»" '.i „ ; - -1. 7 1 z\ i-/ A i£x \ Z A - rr - „-, -W_ sh-(? 5:XT; it it i A d= _ _ the given equations; thus, putting y = \$ in (1), we find 3 x + 2 (If) = 12, whence x = 2f. But it is better to do the work as shown, because a mistake in finding y does not then cause an error in the value of x also. Ex. 3. y = 3 X — 8, i 4aj + 2y-5 = 0. Transpose the terms to get the equation in the form 3x — y = 8, 4 x + 2 // = 5. The figure is as shown. The results are therefore about x = 2, y= — If. Fig. 86-87] SIMULTANEOUS LINEAR EQUATIONS Ki3 The solution (Method I, by addition and subtraction) is 3 a:- y = 8 -4 3x-y = 8 4 x + 2 n = 5 3 | 4 a; + 2 y = 5 10 »■ 21 10y = - 17 a; 2.1 y = - 1.7 Check •• // = = 3x- 8: - 1.7 = 3(2.1) - 8 = 6.3 — 8 (correct) 4x4- 2y- 5 = 4(2 1) + 2( - 1-7) - 5 = o, or i 8.4 -3.4- •5 = (correct). Notice that the answers from the figure are not precisely correct ; they agree with the precise result as nearly as we could expect. 87. Formal Rule. Elimination. The method of solu- tion just given may be summed up in the following rule: Multiply loth sides of each equation by the coefficient of y [or of x] in the other equation; subtract the two resulting equations. The new equation has no term in x [or y~\; solve it for y [or x\. Proceed similarly to find x [or y~\. Any process which, as above, results in getting rid of one of the unknown quantities, is called elimination. We say that y (for example) has been eliminated when we find a new equation which is free from y. The purpose of eliminating one unknown quantity is to find an equa- tion in one letter alone. Such an equation can be solved by previous rules. In equations that contain fractions we first clear of fractions, then proceed as above. EXERCISES I: CHAPTER VI [In each of the following, draw the graph, and estimate solutions; then solve by addition and subtraction.] 1. The sum of two numbers is 19; their difference is 6. What are the numbers ? 2. The sum of two numbers is — 10, their difference is 2. What are the numbers? 164 SIMULTANEOUS LINEAR EQUATIONS [Ch.VI 3. Divide 20 into two parts, one of which shall be four times the other. 4. Divide 35 into two parts, which shall be in the ratio of 3 to 2. 5. What fraction becomes equal to f if each term is de- creased by 1, and to f if each term is increased by 1 ? 6. Twice the difference of two numbers is 12; twice the greater number exceeds the lesser number by 20. What are the numbers ? 7. Half one number exceeds one third another number by unity ; the first number is less than the second by unity. What are the numbers ? 8. What number of two digits exceeds 7 times the sum of its digits by 3, and exceeds 16 times the difference between the tens' and units' digit by 4 ? Solve the following pairs of equations for the letters appearing in them : 9. 10. 11. 12. 13. 14. 21. 6p + 5? = 2, p + 3 q = 9. x + t = 7, 3 x - 10 1 = 8. 2m — n = 1, 6 n - 11 m = 7. Ux + 3y = 12, [9 x-Uy = 27. y = 5 x + 3, x=2y 24. 4fc-3r = l, 6 r -2k= 1. 15. 16. 17. 18. 19. 20. (8s + 5k + l = 0, [4:8.= 10 k + 7. f / — 5 n = 4 / — 6 n, {/_(5„_/) + (£ + 4) = 0. (-3u + 8v = 5, (_ u + v = 2. |(6- a; )-( ?/ + 3.r) = -l, \(x + y) + (4x-3y) = 12. la + f # = 3, 9 , X y = \. x y _ 3 4 ' 5 s -3z/ = 3. < = 1, 22. 7 6 3 p ir 2 V 23. z 3~ t _ 4~ 1 12 3 2~ 3« 16 : 1 2 87-88] SIMULTANEOUS LINEAR EQUATIONS li;,", Solve the following for the letters indicated; draw the figure for any convenient choice of the other letters: 24. x + y = a, x-y = b. 0, y.) 25. ^S 7T ^^^ ffni f m v H '\_; < 5 0) ^<-P>^ S^ S^S S^ sr Fig. 27. Likewise, (2) is represented by the upper straight line. Now these lines never cross ; tor if they did, their common point would be on both, i.e. there would be a pair of numbers whose sum is 5 and whose sum is also 7, which is absurd. Note 1. The kind of argu- ment just used is often called reductio ad absurdum, or reduction to an absurdity ; we prove that the statement made (in this case the statement that (1) and (2) do not cross) is true by showing that otherwise an absurd (incorrect) conclusion would follow. Note 2. Straight lines in the same plane that never cross are called parallel lines. Two simultaneous equa- tions in two unknown quan- tities that correspond to par- allel lines in the figure have no pair of solutions. A pair of numbers that are the solutions of a pair of simul- taneous equations corresponds to a point on both lines. ■2y = 6x-5, (1) $x-3y = -10. (2) Here it is not easy to see by mere inspection that the equa- tions are not solvable. But if we solve for y in each one, we find fy = 3x-f, (4), from (1) \y = 3 x + S°, (5), from (2) FlG . 28 . These are represented by parallel lines, for each one is parallel to the line. («) y = 3 x. Ex. 2. 88] SIMULTANEOUS LINEAR EQUATIONS 107 Iu fact, (5) is formed by raising (6) vertically by J 3 > while (4) is formed by lowering (ti) vertically by |. (See p. 25.) It is clear that the three lines are parallel, i.e. that they never meet. Notice that if one does not suspect that a pair of equations has no pair of solutions, attention will be called to the suspicious character by a figure. If a figure shows the lines fairly parallel, we work as above. The line y — k.r + c is parallel to the line y = Tex; hence two lines are parallel if the coefficient of x is the same in each after solving for y in each. An attempt to solve two such equations will call attention to the peculiar nature of the example. (2y = 6x-5, (1) EX ' 3, Ux-3y = ~lQ. (2) We should transpose terms to get them in the form below ; the solu- tion would then be -3 Gx-2y = 5 (3), from (1) 2 9 x - 3 y = - 10 (4), the same as (2) + = - 35 This is absurd ; — 35 cannot be equal to zero. If we did not suspect that there was no solution before, such an absurd i-esult should make us carefully investigate, as above. Notice that elimination of either letter also eliminates the other. EXERCISES II: CHAPTER VI In each of the following plot the equations ; solve each equation for one of the unknown letters and compare results : 3y = 6* + 2, 4a;-2y = 3. 2x-y=7, 3x-y = ll-^¥. p-2q=7, 5p-q = 3(5 + 3q). r 3 n + 4 r = 6, { (2 r - n) - 2 (n + 3 /•) +12 = 0. ltJ8 SIMULTANEOUS LINEAR EQUATIONS [Ch. VI 5. Find two integers whose sum is 12, such that the sum of the integers next following them is 18. 6. What fraction becomes equal to f if its terms are either each increased by 5 or each decreased by 2 ? 89. Equivalent Equations. A problem may lead to two equations which are essentially the same one. Ex. 1. 2y = 6a-4, 9x-3y-6 = 0. Solving for y in each, we get y = 3 x - 2, y = 3 x - 2. (1) (2) (3), from (1) (4), from (2) In such a case any pair of numbers that satisfies one equation also satisfies the other. The equations are called equivalent, for one may be reduced to the other by opera- tions which are correct. The figure in this case is the same for both equations ; it is a straight line, since each equation is of the first degree. There are, of course, an indefi- nitely large number of pairs of numbers that satisfy both equa- tions, for any point on the straight line gives such a pair. To test for this kind of problem, we solve for y in each. If the result is precisely the same in each case, we X know the equations are equivalent. Attention will be called to such a case (just as in § 88) by drawing a figure, or also by an attempt to solve by the Fig. 29. ordinary method. It is well to notice that if two straight lines have two points in common, they are the same line; i.e. if two 88-80] SIMULTANEOUS L1NEAK EQUATIONS 169 equations of the first degree have two pairs of solutions, they are equivalent equations, and therefore have an in- definitely large number of pairs of solutions. Two simultaneous equations of the first degree have intersecting lines. one solution no solution many solutions . if the two lines are . parallel lines the same line EXERCISES III: CHAPTER VI In each of the following, plot the equations ; solve each equation for one of the unknowns and compare the results: 1. Wy = 6x-2. 2. J 3 [4i>-6v = 60. f9Z + 2 = 2(5-m), 3 - l 4 i+»-M- f ax — by = a 2 + b 2 , x-a ■_ y_±b = Q (Solve for x, y.) y b a x = 10 y + 3, (3x-25y)-5(y + 2) = -l. 6. Twice the sum of two numbers is 12; the difference be- tween the numbers lacks twice the smaller number of being equal to 6. What are the numbers ? 7. A number of two digits is equal to four times the sum of its digits; if the digits are reversed, the new number is equal to seven times the sum of the digits. What is the number ? 8. What fraction becomes equal to § if its terms are each diminished by 1 ; or also if its denominator is diminished by 7 and its numerator by 5 ? 170 SIMULTANEOUS LINEAR EQUATIONS [Ch. VI 90. Solution by Substitution. The problems given above may be solved by other methods ; one of these is the method called solution by substitution. In this method, also, we first eliminate one of the unknown let- ters by a somewhat different process illustrated in the following examples, and then solve for the one which remains, as before. rx + y = 45, (1) Ex.1. (Ex.l )P .160.){^ = 18 (2 j Solve (1) for y : (3) y = 45 - x. Substitute the value found (45 - x) in the place of y in (2) : x — (15 — x) = 18, or, 2 x - 45 = 18. Solving for x, we have x = 31|. Put this value of x in place of x in (3) : y = 45 - 31| = 131, These are the answers found above (p. 161). Check: x + y = 31| + 13| = 45 (correct). x - y = 31 ! - 131 = 18 (correct). 3as + 2y = 12, (1) Ex.2. (Ex.2,p.l61.) l4ic + 5y = 2a Solve (1) for y : (3) Substitute for y in (2) 12 - 3 x '12-3^ Clear of fractions : 8 x + 60 - 15 x = 40. Simplify : 7 x = 20. Solve for x : x = 2f . Substitute 2f for x in (3) : 12-3(2f)^-y _24_12_ 1 , y 2 2 14 7 These are the answers found above (p. 162). 90] SIMULTANEOUS LINEAR EQUATIONS 171 Check : Sx + 2 y = 3(^) + 2( V) = 60 + 24 = 1- (correct) . 4ar + 5y = 4(^) + 5(Y): 7 80 + 60 = 20 (correct). Ex. 3. (Ex. 3, p. 162.) y = 3 x - 8, (1) (2) 4 a- + 2 ;y - 5 = 0. Substitute (3x — 8) for // in (2): (3) 4x + 2(3x-8)-5 = 0, or, 10x-16-5=0, or, 10a; = 21, or, x = 2.1. Substitute 2.1 for x in (1): y = 3 (2.1) - 8 = 6.3 - 8 = - 1.7. These are the answers found above (p. 163). Check : y = 3x-8; - 1.7 = 3 (2.1) - 8 (correct). 4x4-2^-5 = 0; 4 (2.1) + 2 ( - 1.7) - 5 = 8.4 - 3.4 - 5 = (correct). This last example is easier by this method than by the previous method. EXERCISES IV: CHAPTER VI Solve the following by substitution ; in each case draw the figure : 4. x + y = 17, 3a — 5y = 11. 6 x + 9 * = 54, 5 x - 3 1 = 3. ' 2 a - 5 b = 9, .a + 3 6 = -l. 3. 7 y + 9 w = 8, 5 ?t — 3 y — 1 . f an -f- &?>i = 0, I com —bu = l (m 2 + ?i 2 ). (Solve for a, 6.) 6 _ (Sx + 4:y = 5, iSx + 9y = 10. 8. f* 3j, O 1 ^4-JU=7. 5 14 6 4 2' x y _ 1 4~6~12 172 SIMULTANEOUS LINEAR EQUATIONS [Ch. VI 91. Solution by Comparison. A third method of solu- tion, called solution by comparison, is illustrated by the fol- lowing examples. This is also a process of elimination. (x + y = 45, (1) \x-y = 18. (2) Solve each equation for y : y = 45 — x, from (1), (3) y. = x - 18, from (2). (4) Since 45 — x and x — 18 are each equal to y, they are themselve? equal, for they are the same number as y ; hence, 45-x = x- 18. Solve for x: x — 31^. Substitute 31$ for x in (3) : y = 45 - 31 J = 13|. These are the answers found above (p. 161). Chech: x + y = 31£ + 13| = 45 (correct). X - y = 31£ - 13$ = 18 (correct). r3a; + 2y = 12, (11 Fy 2 [4:X + 5y = 20. (2) Solve for x in each : * = S=**, (3) 20- 5w x = £. 4 (4) Hence, 12 - 2 y _ 20 - 5 y 3 4 Solving this fractional equation for y, we find 1, — i 2 — 15 y — r — 1 t Substituting l f a for y in (3), we find _ _ 1 2 - 2 Q 7 *) _ *■* - Y _ 60 _ 20 _ „ , 3 3 ~21~7~~ 7 Hence, ar =2£, 3/ = If, which are the answers found above (p. 162). 6 01-92] SIMULTANEOUS LINEAR EQUATION'S 173 EXERCISES V: CHAPTER VI Solve the following by comparison. Draw the graphs: 3. 92. x - y = ' s > H r6r-s + l=0, 3x+5y= 25. 10r— 3s+ll=0 2 m - 3 n = 2, 5 I -4,- + y = l, 5 m + n = 56. 1 .2x + 3y = 2. P - aq = k, x , y « 3 + 5 = 4 ' ap + q = 1. (For Pi 1 Fig. 30. If two weights, w and H 7 , balance when placed on opposite sides of a lever at distances d and D, respectively, from the single intermediate point (known as the fulcrum) on which the lever rests, then it is known that dw = DW. 93] SIMULTANEOUS LINEAR EQUATIONS 177 12. Weights of 12 and 18 pounds are fastened to the ends of a ten-foot pole. Where must the pole be supported in order that the weights may balance ? 13. An unknown weight 3 inches from the fulcrum of a lever is balanced by another unknown weight 9 inches from the fulcrum ; an addition of 2 pounds to the first weight neces- sitates the removal of the second weight 3 inches farther from the fulcrum in order to preserve the balance. What are the two weights ? 14. Two unknown weights balance when placed 6 and 9 inches from the fulcrum of a lever ; if their positions are re- versed, 2 pounds must be added to the lesser weight to restore the balance. What are the weights ? 15. An 8-pound weight is placed at an unknown distance from the fulcrum of a lever ; on the opposite side is placed an unknown weight. In order to make the lever balance we may either increase the unknown weight by 1 pound and station it 3 inches from the fulcrum, or else station it 4 inches from the fulcrum and add 2 pounds to the other weight. Find the unknown weight and the distance from the fulcrum to the other weight. 16. The sum of the three angles of a triangle is 180°. What are the acute angles of a right-angled triangle if one is twice the other ? 17. Determine the three angles of a triangle if the sum of the first and second is twice the third, and the sum of the first and third is the second. 18. The equation or + if = ax + by + c is to be satisfied by x = 4, y = 2 ; x = 4, y = — 4 ; and x = 5, y = 1 . Determine a, b, and c. 19. A boatman in 2 hours rows a certain distance up a stream where the rate of the current is known to be 2 miles an hour ; he then rows back to a place 1 mile beyond the starting point in 1 hour. Find the distance he rows each way, and the rate of the boat in still water. hedrick's el. alg. — 12 178 SIMULTANEOUS LINEAR EQUATIONS [Ch. VI 2. 3. REVIEW EXERCISES Solve the following sets of quantities : 7 x + 5 y = 3, 2 x + 3 y = 4. 5 a - 7 6 - c = 16, 3 a = 2 6 + 2 c = 10, l2a + 6 + 3c = 6. a 2 + f = 13, I. or — y- = 5. [Suggestion. First solve for x 2 and #' 2 ; then find x and ?/.] f 5& 2 -r-3r 2 = 32, i 2 j- 2 - 3 k 2 = 15. 7(p+9)-8(i>-g)=23, 4(p + g)+Q>-g) = 41. (c + a) + (a + &) = 1, (a + &) + (& + c) = 0, . (6 + c) + (c + a) = - 1. VIII: CHAPTER VI equations for the unknown 9. 8 x y + 1 = 3, 5 + ^=3. 5 — x* y + 1 [Suggestion. First solve for 4. :6, M x y 1-1=4. x y [Suggestion. First solve for and -; then find x and y.~\ ' — a' V IV o 1,1 7 ^ n i> 4 [Suggestion. First solve for _, -, — ; then find ?/, r, ?o.] u o jo 5 -x y + 1 J ' a . b - + - =.P> 10. [Solve for a, 6 ; also for x, y.~\ f 1 + 1 -8, 2 + x x + y 11. ■ i + 1 = 7 > x- + 2/ 2/ + ^ 1 + 1 =7. y + z z + x [SUG( jestion. First solve for , and — ; then x + y x + z H find x + y, x + z, and y + z; then solve for x, y, z.] 3 + -*--«> 12. ^ x - y x+y 15 2 . x — y a? + ?/ n 5 = 3. 13. x 2 y- 5, A + 3 - 22. 5ar y 2 93] REVIEW 179 14. x'+y ; + -5 + 3x 2 -2y 2 6 x* + y- Sx*-2tf 9 : 50' 2 5* 15. m — n = ", m -f-n 1 m — ?i m + »i [Solve for m, ».] 16. A and B can do a piece of work in 2 days ; but an equal piece of work when A puts in only half his time and B only one third his time requires 4^ days. How long would the work take A and B each, working alone ? 17. Two pipes empty a reservoir in 1 hour and 12 minutes. When one pipe is used to fill and one to empty the reservoir, it becomes full in 3 hours. How long would it take each pipe separately to empty the reservoir? 18. A boat can be rowed downstream 4 miles in the same time as it can be rowed upstream 3 miles. A trip 6 miles downstream and back requires 3^ hours. Find the rate of the boat in still water, and the rate of the current. 19. An investment yielding simple interest amounts to $ 2080 in 4 years, and to $2210 in 6 years. Find the amount of the investment and the rate of interest. 20. A, B, and C are at various times engaged to do certain equal pieces of work. B and C together complete the work in 2 hours; C and A in H hours; A and B in 1| hours. In what tinie can each do the work alone, and in what time can all three do the work ? 180 SIMULTANEOUS LINEAR EQUATIONS SUMMARY OF CHAPTER VI: SIMULTANEOUS LINEAR EQUATIONS, pp. 159-180 Figure : two intersecting straight lines. A nswers : pair of values at point of intersection. Algebraic Solution, Method I : remove one letter by addition or sub- traction. Elimination : any process for removing one letter. § 86, pp. 159-163. Formal Rule, Method I, by Addition or Subtraction: essentially, multiplication of both sides of each equation by the coefficient of one letter in the other; Exercises I. § 87, pp. 163-165. Impossible Case : figure, parallel lines ; no answers ; contradictory conditions, revealed by attempt to solve ; Exercises II. § 88, pp. 165-168. Equivalent Equations : figure, coincident lines ; answers, one pair of solutions given by any point on the line. Classification of Equations : Correspondence of one solution to intersecting] parallel \ lines coincident J Exercises III. § 89, pp. 168-169. no solution many solutions. Solution by Substitution {Method II) : essentially, solve one equa- tion for one letter and substitute this value in the other ; Exercises IV. § 90, pp. 170-171. Solution by Comparison {Method III) : essentially, solve each equa- tion for one of the letters and equate the values found ; Exercises V. § 91, pp. 172-173. Three Unknowns : elimination of one letter, reduction of problem to two equations in two unknowns ; Exercises VI. § 92, pp. 173-175. English Problems : solution of typical examples ; Exercises VII. § 93, pp. 175-177. Review Exercises for Chapter VI: Exercises VIII. pp. 178-17!*. CHAPTER VII. SIMPLE POWERS AND ROOTS PART I. POWERS AND ROOTS OF NUMBERS 94. Introduction. The student is already familiar with ordinary powers and roots (§§ 9-10, pp. 8-9). By definition, x n =x-x-x--x (n times) is called the nth power of a. (See p. 8.) The process of raising a number to a power is often called involution. By definition, if b — a ■ a • a ■■• a (n times), then a = \/£>, i.e. the nth root of b. (See p. 9.) The process of extracting a root is often called evolution. [Let the student again state the definitions given in § 9, pp. 8-9.] There are often two answers for a root of a number. Thus, 4 = 2-2; hence, 2 is one answer for the square root of 4. But 4 = (— 2) • (—2) also ; hence, — 2 is also an answer for the square root of 4. When no sign is written be/ore a ^/ we shall understand that the positive answer is intended, and we shall agree to distinguish between the two roots by writing, for example : + 2 = V4, - 2 = - V4, the negative answer in any case being said to be the nega- tive square root. Similarly, Va 2 = + «, although —a is also one square root of a 2 . There is only one cube root of any number among numbers we know at present. Thus, 8 = 2-2-2, hence 2 = ^8, but -2 does not equal \^8, for (-2)(-2)(-2) = -8. In fact, ->2 = -s/- 8. 181 182 SIMPLE POWERS AND ROOTS [Ch. VII A negative number has no square root among numbers we now know ; for the square of any number we know at present is positive, except zero, whose square is zero. In fact, since an even number of either -f or — factors gives + in the product, an even root of a negative quantity is meaningless, at present, to the student. Such roots, even roots of negative quantities, are called imaginary. Later we shall discuss them (Appendix, § 31), and we find it possi- ble to use them sometimes. Just now we shall have nothing to do with them. In this chapter, in even roots, we shall assume that the letters used have positive values. If any even root (i.e. square root, fourth root, etc.) of any number is known, then the negative of the known answer is also an answer, as above. On the other hand, the negative of a known answer for an odd root (i.e. cube root, fifth root, etc.) is not an answer, for the odd number of negative signs produce — , not +, in the answer. There is only one answer (among num- bers we know at present) for any odd root of a number, since an increase or decrease in the answer would cor- respond to a larger or a smaller number than the given number. 95. Figure. We have seen how to indicate the squares of numbers in a figure. (See p. 30.) Thus, if x = a given number and y = the square of the given number, we have y = x 2 . Let us give x various values, as on p. 30, and find values of y. This gives a table as follows: X 1 1 -1 1 2 4 -2 4 ±3 9 ±4 16 ±5 ±6 &c. + h ±1 ±1 ±1 1 5 4 5 ±1 9 4~ ±i ±1 &c. 25 i i 1 5 & ¥ [Let the student fill in the blank spaces and extend the table, espe- cially by putting in more fractional values.] 94-95] POWERS AND ROOTS OF NUMBERS 183 Drawing these values in a figure, as on p. 30, we get a graph like Fig. 31. Since this is very narrow for use, we take a longer unit space (one heavy- marked space) horizontally to produce Fig. 32, which represents the same figure on an enlarged scale horizontally (enlarged one way only). To find the square root of a number, we merely look for the given number in the table marked y ; its square root is the corresponding x, for we squared each x to get the corresponding y. Or, in the figure, we count off the given number vertically (on the y line), find the corresponding point on the curve, and then count horizontally to find x. Thus, if y = 9, we rise 9 spaces on the y line ; the corresponding point of the curve is C; the horizon- tal space out to is 3. The point C is also at a height 9 ; its x is — 3. Hence, the square root of 9 is either 3 or — 3. Fig. 31. Fig. 32. 184 SIMPLE POWERS AND ROOTS [Ch. VII 96. Inexact Roots. There are numbers that are not in the y table. For example, 15 is not to be found in the y table, no matter how complete the table be made. We may still carry out the process in the figure. Thus, rising 15 points, we find a point marked M on the curve; counting horizontally out to it, we find x = 3.8 + . In fact, we have: (3.8) 2 = 14.44, and (3.9) 2 = 15.21. Hence, (3.9) 2 is too large, while (3.8) 2 is too small; that is, x = 3.8 gives a point (x = 3.8, y = 14.44) below M, while x = 3.9 gives a point (x= 3.9, y = 15.21) above M. With a more precise figure we could find values of x still closer together — differing in any place of decimals we please — which give points above and below M. Thus, while many numbers do not give exact square roots, we can interpolate the number between two which differ in a small amount. This process gives a result to any number of decimals desired, but it is not exact. The negative of the answer thus found is also an answer, as above. Any inexact root may be left merely indicated by means of the radical sign ; thus Vl5. Any expressed root, whether exact or inexact, is called a radical. Any ex- pression which contains a radical is called a radical expres- sion. (See also pp. 186, 192, 284.) 97. Formal Process. A process for finding the decimal places is based on the rule (x + y~) 2 = x 2 + 2 x y + y 2 , p. 93. We found VT5 = 3.8+. Suppose the digit in the next place of decimals is d, then squaring 3.8 + <7(.01), we find {3.8 + rf(.Ol)} 2 = (3.8) 2 + 2d (3.8) (.01) + rf 2 (.0001) = (3.8) 2 + d(.076)± (very small number). Since we wish to get near to 15, we write (3.8) 2 + tf (.076) = 15 (nearly), or, d (.076) = 15 - (3.8)2 (nearly), , 15 -(3.8)2. ' or, d = *■ — *- (nearly). .076 v 3J We can tell by this very closely what d must be ; since d is an integer, we can guess it exactly. !Mi-!>7] TOWERS AND ROOTS OF NUMBERS 185 The work is written as follows (3.8)2 15 14.44 13.817 2 x 3.8 x .01 = (.01)- x 7 = .0700 .0007 .0767 We continue upon the same plan: 2 x 3.87 x .001 = .007740 (.001) 2 x 2 = .000002 .0(17742 .5600 .5369 2 x 3.872 x .0001 (.0001)- x 9 .00077440 .00000009 .00077449 .0231 .023100 13.871219 .015484 .00761600 .00697041 .00064559 and so on. We really omit many of the decimal points and zeros, these are useless if ice keep the places right ; notice that in order to do so two zeros are to be placed after the difference after subtraction, and one zero after the " trial divisor." The trial divisor is twice the amount previously found with the decimal point shifted as above. The whole work in the above example would be written as follows : 15.06060606 13.8729, etc. 9 2 x 3 = 60 600 8 68 544 2 x 38 = 760 5600 i 767 5369 2 x 387 = 7740 23100 7742 15484 2 x 3872 = 77440 9 761600 77449 697041 64559, etc. (The small dots placed over every other place beginning with the units' place in the original number are helpful in bringing down the figures, especially if the given number contains decimal digits.) 186 SIMPLE POWERS AND ROOTS [Ch. VII The result obtained is a decimal fraction, whose square is less than 15 ; but the square of the decimal just higher in the last place is greater than 15, at each stage : (3.8) 2 <15<(3.9) 2 , (3.87) 2 <15<(3.88) 2 , (3.872) 2 <15<(3.873) 2 . We say that Vl5 is determined by this sequence of numbers, i.e. by the decimal fractions found as above. (This is really the first definition of VT5; the idea of drawing a smooth curve through the points we can locate in Fig. 31 contains the essentials of a strict definition. See Appendix, § 29.) In carrying out any operation upon Vlo we shall really operate upon these decimal fractions; the result is said to be the number defined by these decimal results. (See also Appendix, §§ 29-30.) The same statements apply, of course, to all inexact square roots as well as to V15. Any inexact root, i.e. a root that cannot be expressed exactly in fractions or integers, is called a surd. Any expression containing a surd is called a surd expression. The quotient formed by dividing any integer by another integer is called a rational fraction. A number that is an integer or a rational fraction is often called a rational number. Any number that is neither an integer nor a rational fraction is called irrational. (See pp. 192, 284, and Appendix, § 29.) Thus, every surd is an irrational number. EXERCISES I : CHAPTER VII Find the square roots of the following numbers: • i. 441. 4. 1253.16. 7. .0625. 2. 324. 5. 24.1081. 8. .1369. 3. 55,225. 6. 27.6676. 9. .056644. 97-98] POWERS AND ROOTS OF NUMBERS 187 From the figure on p. 183 (y = af), read oft' the values of the following radicals : 10. V2. 11. V6. 12. VlO. 13. Vl8. 14. V2J5. Calculate the values of the following radical expressions to three places of decimals: 15. V2. 17. Vl2. i6. Vu. is. V33. 23. \2-V3. 19. V19. 20. V71. 24. Vi - V Vrr^J 21. V109. 22. V2 + V3. 98. Powers of Radicals. We are justified in saving that the square of V2 is 2, or, in general,* if a is positive, (Vay a. If we do so, we can easily square such expressions as 3 4- 2 V5. For by p. 93, Chapter IV, we have (x + y)- = x 2 + 2xy + y 2 , hence, (3 + 2 V5) 2 = (3) 2 + 2(3)(2V5) + (2 V5)* = 9 + 12 a/5 + 4(V5~) 2 = 9+ 12V5 + 20 = 29 + 12 VS. Check: V5 = 2.236 (nearly). Now 3 + 2V5 = 7.472 (nearly), and 29 + 12 Vs = 55.832 (nearly), (7 .472) 2 = 55.832 (nearly). We shall need this process in the following chapter. Other operations of this kind will require some further explanation and are deferred until Chapter XI, p. 284. * See, however, Chapter XI, p. 280, and Appendix, § 31. , 188 SIMPLE POWERS AND ROOTS [Ch. VII EXERCISES II: CHAPTER VII Square the radical expressions : 1. 2 + V3. 6. 2V7-5. n. Vfr-4. 2. 2-V3. 7. iV5-l. 12. 2V11-7. 3. 1+3V2. 8. x + Vy. 13. «V6-c. 4. 3V5 — 7. 9. x+Va-r. 14. mx+mVy 2 —x 2 > 5. V8-3. 10. 1-Vo^l. 15. fcaj-ZV^+1. 16. A-x + lyly 2 ' tfx 2 I 2 99. Simple Operations. We shall also need to notice that Vl2 =_VT^3 = V3V3 = 2V3 for example; or, in general, Va 2 b = Va 2 Vb = aVb, if a and 6 are positive. This is justified by observing that (2V3) 2 = 4(V3) 2 = 12, whence 2V3=Vl2; or, in general, if a and b are positive (a Vby = a 2 (V6) 2 = a 2 b, whence a VI = V a 2 b. Likewise, \- = — - = — - ; M V4 2 , J~b Vb VI or, in general, V— = — — = v a 2 Va 2 a These operations are more fully studied in Chapter XI, p. 284. EXERCISES III: CHAPTER VII Remove from under the radical sign all square factors in the following, assuming that the letters used have positive values: 1. 2. Vl8. V50. 5. 6. V98. V24. 9. 10. V 12 JKJ 2 . vs. 13. J5 3. 4. Va 3 b 2 . V75. 7. 8. V99. Vttrxy 3 . 11. 12. V|. v/ T . 14. J«*V. 1 98^100] POWERS AND ROOTS OF NUMBERS 180 15 is. V T % (= V/A) 17. V T \ (= y/ffr). is. vi. Express the following quantities in a form entirely under the radical sign : mn~vnp. 19. 5^3. 20. 3Vo. 21. 6V3. 22. 23. 1V3. 24. -VV8. 25. 26. 27. Vfc a 6 m I n af-r 100. Cube Roots. analogous process. We may find cube roots by an First let us draw a figure, where x means any number and y means the cube of that number i.e. y = $. A table is + 1 + 1 -1 -1 + 2 + 8 -2 -8 ±3 ±27 ±4 ±64 ±5 ±6 etc. ±1 ±1 ±57 ^ 3 4- 8 ±57 4- 3 ± 8 ±1 ±5 etc. [ Let the student fill in the blank spaces and expand the table.] From this table we may draw a figure that will appear as shown, the scale being taken for convenience : 1 vertically = 1 small space ; 1 horizontally = 5 small spaces. Fi'om this figure we can read off, approximately, the cube of any number, or also the cube root of any number, as for square roots above. Thus, the cube of 1.8 is seen to be about 5f (really 5.832); the cube root of 11 is seen to be a little over 2.2. More accurate results can be found by drawing the figure on a larger scale. Fig. 33. 190 SIMPLE POAVERS AND ROOTS [Ch. VII A method similar to that of § 97 can be devised for find- ing further decimal places. (See Appendix, § 20.) But it is really seldom used because there is a much quicker process. (See Chapter XIV on Logarithms, p. 339.) 101. Higher Roots. A graphical process like that above applies to finding any root of any number approximately. The figure shows the pictures for (1) y-«S and (2) y = a?, Y . P J _ _ „t f - . I__ _ 7 -,- f . " 1 X i / 2, , / ^ra J #17 // / /; > / a ■-- — f — 6-r-- I - J$.Z ._/__ 4,- -, t Js-l ■4 -i- % + i i -j t t i _, J j! t t -7 \ - t-i J . \ -. J / \£. ~t Is -£ ' ' S V v i +-2?7 -£&' TnT J& - + ■ ^_ <<^ rTTi - 1 1 i—i - -3r t r C ^? ,_T ± ii 'c 1, 'y J**f y' ±: ± ^ i \S ' ' r X ± S / i_d;_ / J* T . ,* ij i lit 1 ' ' 1 ■Jl/ 1 III i X -+--1(1 ill - " t if ± III _!___ Fig. 34. from which fourth and fifth roots may be found approxi- mately. These roots may be found more accurately either (1) By drawing a larger figure. (2) By a method similar to that of § 97. (3) By logarithms. (See Chapter XIV, p. 339.) 100-101] POWERS AND ROUTS OF NUMBERS 191 The last method is by far the easiest; further explana- tion of it will be found in the Chapter on Logarithms (Chapter XIV). EXERCISES IV: CHAPTER VII Draw accurately figures for y = x", y = x\ y = x", as indi- cated on p. 190, and estimate as accurately as possible the values of the following radicals; check by raising the result to the proper power: 1. Vvf. 3. a/17. 5. ^50. 7. ^50. 9. \/76. 2. ^17. 4. a/25. 6. a/34. 8. = x n+m . In multiplying, add exponents. hedrick's el. alg. — 13 194 SIMPLE POWERS AND ROOTS [Ch. VI] We also have III. (xyy i = x- y - x-y • x - y ••• x> y(n times) = [x • x • x---x(ji times)] x[y • y • y ••• y(n times)] = x" xy n . Any simple power of a product is equal to the product of the same powers of the separate factors. III a. Similarly, — = (-) • This rule follows from in, also: y n K y J Any simple power of a quotient is equal to the quotient of the same powers of numerator and denominator. These rules assist in reducing the work of involution : To raise a monomial to a simple power, raise the coefficient to that power, multiply the exponent of each letter by the degree of the power, and make sure that the sign follows the Rule of Signs. The Rule of Signs (p. 71) shows that (1) an even number of negative factoi-s gives + in the product; (2) an odd number of negative factors gives — in the product; (3) any number of positive factors gives + in the product. This rule is simply the written expression of what we already did in § 103 ; if this rule is forgotten, the examples can all be solved by § 103. 105. Advantage of Fractional Notation. Practice with the fractional notation of S 102 shows its convenience. Ex.1. V16 o 4 6 2 = 4 a 2 b, for (4 a 2 b) (4 a 2 b) =16 a 4 b 2 , or, (16a 4 6 2 )* = 4a 2 &. Note that this follows the Rules of § 104 ; for the work (10 aW)* = (16)^ 4x ^ 2x ^ = 4 aVi is correct, although we had no reason to expect it. 104-105] POWERS AND ROOTS OF .MONOMIALS 196 Ex. 2. Trying the same rules of § 103 for V— 27 aW, Ave find : (-27 8 x *= - 3 ax 2 , ?'-/^7,7t ?'s correct, for (—3 «.r)( — 3 ckb 2 ) (—3 ax ! ) = — 27 aV : . It is necessary for the student to check his work as above, to avoid errors, at least until he becomes expert. Rule I of § 104 holds when the power is a fraction, that is, for roots, if the root can be found otherwise (i.e. for exact roots. See § 94, p. 182, and § 98, p. 187): for or • a K • ■ (a k -"y< = a !; - n -n = a \ a k (n times) = a k ' n . Later (§ 136, p. 285), we shall see that the rules of § 104 always hold when we properly extend our notation. Likewise, Rule III holds for exact roots : 1 i 1 O")" x (b n y = (a n b n y, for each side is equal to ab. EXERCISES VI: CHAPTER VII Perform the following involutions by the aid of Rule I : 2afyV V a m z p ~ q J / — 5 m?n s p \ 2 V 3 r°V 5 J ' io. f^^rr. V af-'y 2 J Perform the following evolutions by the aid of Rule I; check by actual multiplication, when the index is numerical : 11. V64 a*b 6 . 13. y/2187 p 7 q 21 . 1. -2frbcy. 2. ( {—xyz') 2 . 3. I [atfc 5 )*. 4. I [— 5 mny*) 4 . 5. | [— c(x r i/' 2, z 3t y 2 . 6. ( [atfc?)™. 7. (x m -"y n - p z p - q ) a . 8. { -**&* )*. 11 crz"-* J 12. V^MafyV 2 . 14. ^625 rV 6 . 196 SIMPLE POWERS AND ROOTS [Cn. VII 15. V-27z 6 . 18. VoV. 16 - X ~> '7 °» "5 ' 19 - V.Tf'y^z^jr-q^-' 1 * 24o if ?/" 1? « 256 A 12 20 . A^H^L. 106. Longer Expressions. Binomials and longer ex- pressions can be raised to powers by simply applying the rules for multiplication ; or also by using the rules of Part III of Chapter IV, p. 91. (a±by~d?±2ab + b\ (Seep. 93.) (a ± by = a 3 ± 3 a 2 b + 3 ab 2 ± b*. (See p. 99.) (a ± by = a 4 ± 4 a% + 6 a 2 P ± 4 ob* + b\ (See p. 99.) Ex. 1. (4 tff _ 3 xff = (4 xhff - 3(4 ccy-)\3 xtf) + 3(4 a; 3 i/ 2 )(3 xff - (3 xff = 64 x 9 y e - 144 a; 7 ?/ 7 + 108 aft/ - 27 aft/ 9 . Test this by direct multiplication of three factors, each 4 a- 3 ^ 2 — 3 a'^ 3 . Roots of longer expressions may be found by inspec- tion, comparing with the formula written above. Ex. 2. Find the square root of 4 m 2 — 12 mn + 9 n 2 . Comparing with (a — b)' 2 we see a — 2 m, J = 3n will fit to give the given expression : (2 m - 3 n) 2 = 4 m 2 - 12 mn + 9 ?* 2 ; hence, (I ?n 2 — 12 mn + 9 n 2 )^ = (2 m — 3 n). Another answer is — (2 m — 3 n), by the general rule that the negative of one answer for a square root is also an answer ; for, [_ (2 m - 3 n)Y = (- 2 m + 3 n) 2 = 4 m 2 - 12 mn + 9 n 2 . It is not certain which of these answers is the negative one until we know whether 2m — 3n is positive or negative. If m = 5 and r» = 1, 2 m — 3 n = 7; (— 2 m + 3 n) = — 7 ; the given expression is of course the square of this, i.e. 49. 105-107] POWERS AND ROOTS OF MONOMIALS 197 Check: If m = 5 and n — 1, 4 m 2 -12 n*n + 9n 2 = 4(5) 2 -12(5)(l) + 9(1) 2 = 100-60 + 9 = 49. A numerical check of this kind is most valuable in avoiding errors. Ex. 3. Find the square root of 4 or 2 + 9 //- + 10 z 2 - 12 xy + 16 xz - 24 yz. Comparing with a 2 + b- + c- + 2 ab + 2 ac + 2 be = (a + b + c) 2 , from p. 92, we see that a = 2 x, b = — :3 y, c = 4 z will fit ; for, (2 x - 3 y + 4 z)' 2 = 4 .r 2 + 9 y 2 + 16 2 2 - 12 a# + 16 as - 24 yz. Hence, (4 x 2 + 9 y 2 + 16 2 2 - 12 a:y + 16 xz - 24 yz) ^ = 2ar-3y + 4z. CAec£ : Put a: = 1, y = 2, z = 3. [4(l) 2 +9(2) 2 +16(3) 2 -12(l)(2) + 16(1)(3) - 24(2)(3)]* = 64* = 8. 2 x - By + 4tz = 2(1) - 3(2) + 4(3) = 8. Another answer is - (2 x - 3 # + 4 2) = — 2 a: + 3 y — 4 2. Check it. EXERCISES VII: CHAPTER VII Perform the indicated operations : 1. (ax + by) 2 . 3. (2 p -7q) 3 . 5. (m + 7n) 4 . 2. (ab — xy)*. 4. (3a-2 6) 5 . 6. (c-2d + e) 2 . 7. (6c + ca + «6) 2 . 12. (a 2m + 12a" , & n + 36& 2 ") 2 \ 8. (2x-3y + 4:zy. / o 3 13. r +- - 9. (ap-6g) 2 + (a6+p 9 ) 2 . V 3 *. 10. (*<-8.i- 3 +24a- 2 -32.r + 16)*. 14 (& + vt y. 11. (z 3 -12z 2 + 48z-64)l ^" ' 15. (4 a; 2 + y 2 + 9 z 2 - 6 yz + 4 .17/ - 12 zx)i. 107. Process for Square Roots. Roots of longer expres- sions may also be found by the following process, which is occasionally convenient. In simple examples, however, roots can be best found by inspection when there is any exact answer. 198 SIMPLE POWERS AND ROOTS We know (a + bf = a 2 + 2 aJ 4- 5 2 = a 2 + (2 a + 5)6. This we write as follows : a 2 + 2a6 + 6 2 l« + & [Ch. VII a 2 2 a + b 2a+0 2ab + b 2 2ab + b 2 If there were more terms, we should proceed similarly ; the important step is to form the " trial divisor " (2 a, above), by multiplying the part already found by 2 ; to this we add the second term (ft, above) and multiply the sum (2 a +b, above) by this second term ; this gives the total original amount as shown above. Ex. 1. Find (4 ra 2 - 12 mn + 9 »")* (see Ex. 2, § 106, p. 196). The work may be written. 4 m 2 - 12 mn + 9 n 2 \ 2 m - 3w 4m 2 2 x 2 m = 4 m 3n 4ib-3 n - 12 »m + 9 n 2 - 12»nw + 9 /i 2 Ex. 2. Find (4 x 2 + 9 f + 16 z 2 - 12 xy + 16 a» - 24 yz) K 4 x 1 + 9 y 2 + 16 z 2 - 12 xy + 16 xz - 24 yz |2x-3y + 4 « 4x 2 it// 2 + 16 Z 2 - 12 xy + 16xz - 24 yz 2 x 2x = 4x (arranged) -3y - 12 xy + 9 y 2 + 16 « 2 + 16 xz - 24 yz 4x — 3 y - 12 xy + 9 // 2 2(2 a: - 3 y) = 4 a: - 6 y 16z 2 + 16 xz- 24 yz + 4z (arranged) 4x — Qy + 4z 16 xz- 24 yz+ 16z 2 16 xz- 24 yz + 16z 2 107] POWERS AND ROOTS OF MONOMIALS 199 la this example the terms cannot be arranged until we see which term will afford an exact quotient with the trial divisor. Practice will indicate the best arrangements. Only exact square roots should be attempted by this process. The process is valuable also in discovering whether or not a given expression is a perfect square. Inexact roots are treated (for numbers only) in Part I, Chapter VII, § 97, pp. 184-186. EXERCISES VIII: CHAPTER VII Extract the square root of the following expressions: 1. 16 I r + 2rjq°--A0pq. 2. 9 c- + 400 dW + 120 cdx. 3. a 2 + 25 V + 9 r 2 - 30 be + 6 ca - 10 ab. 4. x 4 + 2 x'y + 3 xhf + 2xf + y\ 5. 4x- 4 -12^ + 13ic 2 -6x- + l. 6. 4 - 4 a - 7 a 2 + 4 a 3 + 4 a 4 . 7. 9x 4 - 12 x* - 26 a; 2 + 20 x + 25. 8. x- 4 — 2 a?y — x-if + 2 xif + y\ 9. a 6 - 2 a 5 + 3 a* - 4 a 3 + 3cr - 2 a + 1. 10. l-«(66+4)+a- , (9 6 2 + 126 + 4). 11. 9 a G - 12 a 5 6 + 10 a'b' 2 - 4 a?b s + a 2 6 4 . 12. 49 - 42 * + 37 f 2 - 12 J 3 + 4 £ 4 . 13. 4 6 2 c 2 - 16 abc 1 + 16 a 2 c~ - 12 a6 2 c + 24 d 2 bc + 9 a 2 6 2 . 14. a 4 + 4 a 8 6 + 6 a 2 6 2 + 4 ab 3 + b\ 15. r 6 - 6 r 5 + 15 r*s 2 - 20 rV + 15 ?~Y - 6 rs 5 + s 6 . 200 SIMPLE POWERS AND ROOTS [Ch. VII REVIEW EXERCISES IX : CHAPTER VII By use of the figures ou preceding pages, estimate the fol- lowing roots : l. V60. 2. V60. 3. V60. 4. V60. 5. MVtiO. By rules for use of fractional exponents, so far as they have been given (§ 105, pp. 194-195), show that 6. VVoTUa/60. 8. \'^=t/c = -\[V^ Plot the following curves, making all necessary calculations by use of the figures (Figs. 32, 34, pp. 183, 190). 10. y = x + x 2 . 12. y = x + -\/x. 14. y = V* 2 . 11. y = x — -y/x. 13. y = Vx 3 . 15. y = x 2 + Va;. Perform the operations indicated : 16. (5-2V6) 2 . 20. (5V2 + 7)(5V2-7). 17. (5 + 2V6) 2 . 21. (3Vx-Jy) 2 . 18. (5 + 2V6)(5-2V6). 22. (6x-y+5z) 2 . 19. (5V2-7) 2 . 23. ( x + y + z) s . 24. (27 a 3 - 27 a 2 + 9 a - 1)1 25. (16 a; 4 - 96 x 3 y + 216 a; 2 ?/ 2 - 216 xtf + 81 ?/ 4 ) 4 . Find the square root of each of the following expressions : 26. 9x 10 + 6a; 8 -5x 6 -2.*r 3 + l. 27. x 4 - 6 x*y + 13 a?tf - 12 xtf + 4 y\ 28. a?-2ab-3b 2 + — + — • a a 2 29. « 4 + 6 2 + c 2 + 3* + 2 a 2 6 + 2 « 2 c + 2 bd 2 + 2 cd 2 + 2 av 7 s + 2 &c. 30. ./' + 2 a- 5 + 3 x 4 + 4 X s + 3 x 2 + 2 .»• + 1 . 31. a 6 - 4 a 5 + 1 a 4 - 20 a 3 + 25 a? - 24 a + 1 6. 107] SUMMARY 201 SUMMARY OF CHAPTER VII: SIMPLE POWERS AND ROOTS, pp. 181-200 Part I. Powers and Roots of Numbers. pp. 181-192. Definitions: elementary definitious recalled ; involution and evolu- tion. Number of Roots: odd roots, one answer; even roots of positives, two answers; even roots of negatives ("imaginary numbers"), no answers among numbers now known to student. Notation : V~ sign for even roots denotes the positive answer; the negative answer indicated by — V~. § 91, pp. 181-182. Figure for Square Roots : graph of y = x 2 drawn ; square roots from figure. § 95, pp. 182-183. Inexact Square Roots : approximate square roots of inexact squares found from figure. § 96, p. 184. Forma! Process for Square Roots: arithmetic process for closer ap- proximation ; formal process, using " trial divisor " ; root actu- ally defined. Definitions: rational fraction — quotient of two integers ; rational number — integer or rational fraction ; irrational number — not rational ; surd — irrational radical. Exercises T. § 97, pp. 184-187. Squares of Radicals: (Va) 2 = a; longer forms by previous rules. Exercises II. § 98, pp. 187-188. Simple Operations on Quadratic Radicals : insertion and removal of simple factors ; simple multiplications and divisions by in- tegers. Exercises III. § 99, pp. 188-189. Cube Roots : figure, y = x s ; approximate cube roots ; other methods suggested. § 100, pp. 189-190. Higher Roots: figures for x\ x 5 ; other methods. Exercises IV. § 101, pp. 191-192. Part II. Simple Powers and Roots of Polynomials. p. 192. i Fractional Notation for Roots : equivalence of x n to Vx ; positive answer intended if n is even. § 102, p. 192. Monomials: powers — direct extension of multiplication; roots — direct reversing of multiplication. Exercises V. § 103, pp. 192-193. 202 SIMPLE POWERS AND ROOTS Formal Rules ; Powers of Monomials : essentially, raise coefficient to power, multiply exponents. § 104, pp. 193-194. Roots by Fractional Notation : correctness of rules of § 104 for exact roots in fractional notation. Exercises VI. § 105, pp. 194-196. Longer Expressions : powers by previous rules ; roots by inspection. Exercises VII. § 106, pp. 196-197. Formal Process; Square Roots of Polynomials : illustrative prob- lems ; key is " trial divisor " ; restricted usefulness. Exercises VIII. § 107, pp. 197-199. Review Exercises for Chapter VII : Exercises VIII. p. 200. CHAPTER VIII. QUADRATIC EQUATIONS PART I. METHODS OF SOLUTION; CHARACTER OF THE ROOTS 108. Quadratic Equations. If an equation when cleared of fractions and radicals, and simplified, contains the square, but no higher power, of the unknown quantity, it is called a quadratic equation, or an equation of the second degree. We have solved some such equations (see §§ 64-07, pp. 103-115, and § 82, p. 151) ; we shall now show how to solve any such equation. Ex. 1. Given the equation 2 x 2 — 9 x ■+- 4 = 0. We notice the factors (2/-1) and (x - 4) on the left ; hence we (2x- l)(x -4) =0. - 1 = 0, or x - 4 = 0, x = \, or x = 4. If these factors were not noticed, or if the example were so difficult that the factors could not be seen by inspec- tion, we could proceed as follows : Trying various numbers for x, we should naturally try the numbers a: = 0, x = i, x = 2, etc., to see if we could find a correct answer by trial. Letting x = 1, for example, the left side is not zero; it is 2-9 + 4=- 3. Trying various numbers, as suggested, we should find a table like this : write Whence, and 'ix X I 4 1 2 3 4 5 6 22 7 etc. - 1 -2 - 3 - 4 etc. - 3 -6 -5 9 etc. 15 etc. 203 204 QUADRATIC EQUATIONS [Ch. VIII where / means the value of the left side of the given equation, that is, l = 2x 2 -9x + ±. [Let the student fill in the blank spaces.] These values may be plotted in a figure, as on p. 183. The figure shows the value of the left side for any value of x ; we wish to have the left side, I, equal to zero, in order to satisfy the given equation. One value of x for which / is zero was discovered l>y trial on p. 203, namely, x = 4. It is clear from the figure that there is another value some- where between and 1, for the curve crosses the horizontal line, i.e. the left side is zero, somewhere between and 1, about .5. These answers are correct, as we found by a different method. This process will not always give an accurate answer, of course, because the figure is not entirely accurate. — / 3 I I t t t L ts XX- 1 I J 1 . 1 i t i t t „r r "t --,- _ t r 4 4 J- t It t -Jv t I = k jTX"--U X T 4 V .Y Fig. 35 EXERCISES I : CHAPTER VIII [The first of these exercises are easy examples by the factoring method ; the student should solve them by factoring as on pp. 107 and 151. Then he should draw the figure as illustrated above and compare the results.] f _^ 1. x 2 - 7 x + 6 = 0. 2. r 2 -r-12 = 0. 3. * 2 =3* + 10. 4. z 2 = 9z-20. 5. 2 t 2 - lit + 12 = 0. 6. 3 f- 7 £-6 = 0. 7. 13i? = 2p 2 + 15. 8. 2m 2 + m = 21. 9. 10< 2 + <-3 = 0. 10. 3v 2 = 34w-40. [The second half of these exercises includes examples that the student will find too difficult by the factoring method. In these he should only draw the figure and estimate the answers.] 108-100] METHODS OF SOLUTION 205 11. n 2 - 4 n + 2 = 0. 15. 2aj 2 -3aj-6 = 0. 12. m 2 — 6 m + 6 = 0. 16. 3 2- - 7 2 + 3 = 0. 13. t 2 - 5 t + 3 = 0. 17. 2r 8 + 3r — 1 = 0. 14. a* - 7 .r - 4 = 0. 18. X 2 - a; - 1 = 0. 109. General Solution. If an absolutely precise answer is needed, it can be found by the process illustrated below, no matter how difficult the example may be by factoring. Ex. 1. x 2 - Ax -5 = 0. Transpose 5 and write, x 2 — 4 x — 5. Add -1 to each side : x- — 1 x + 4 = 9. The purpose of this is to make the expression on the left side a perfect square. Evidently it is (a; — 2) 2 , and we have : (x - 2)2 = 9. Whence, x — '2 = ±3, and x = 2 ± '■> — 5 or — 1. Check : For x = 5, 5 2 — 1 • 5 — 5 = (correct). For x = — 1, (- l) 2 - 1 (- 1) - 5 = 1 + 4 - 5 = (correct). Problems should always be carefully checked. An iiiiswer is a value of x (or whatever letter is used) that satisfies the original equation when x is replaced by it. Such an answer is often called a solution, or a root of the equation. Ex. 2. 2 x 2 - 9 x + 4 = 0. (See Ex. 1, p. 203.) Divide by 2 on both sides, and transpose 4 : x 2 _ | x = _ o. Desiring to make a perfect square on the left side we add /<> each side the square of\ of the coefficient of x, that is (^ x — |) 2 = f^, (* - !) 2 = «■ Whence, a? — f = ± J, or, x=|±£ — 4or£. 206 QUADRATIC EQUATIONS [Ch. VIII Check : For x = 4 ; 2 • 4 2 - 9 • 4 + 4 = (correct). For x = i ; 2 ■ Q) 2 - 9 • Q) + 4 = (correct). These are the same answers as those found above for this problem. 110. Completing a Square. Remembering the rule (x + a) 2 — x 2 + 2 ax + a 2 , we notice that the last term (a 2 ) in a perfect square is the square of ^ the coefficient of x, if the first term is x 2 . The preceding work depends on this, and we may definitely state the rule : To solve a quadratic equation : (1) Notice what letter denotes the unknown quantity ; we shall here call it x. (2) Transpose the terms in x 2 and in x to the left side of the equation, the other term or terms to the right. (3) Divide both sides by the coefficient of x 2 . (4) Add to each side the square of half the coefficient of x, so that the left side is a perfect square. (5) Extract the square root of both sides, taking care to put both signs ± on the right. (6) Solve the resulting equations by transposing to one side all but the term in x. (7) Check each answer by substitution in the original equation. In step (5) the sign ± might be put on both sides, since a square root is extracted on each side. But the effect is the same as that given in (5), since any equation ± A = ± B with ± on both sides is really the same as the equation A = ±B with ± on one side only. [The student should now carefully examine the solutions of the problems solved in § 109 and notice that this rule is followed there.] The solution may not come out in rational form, as illustrated by the following example: 109-110] METHODS OF SOLUTION 207 Ex. 1. x 2 -4 x+\ = 0. Transpose 4- 1 and add (\ ■ 4)' 2 or 4 to each side : x 2 - 4 x + 4 = 3. Take the square root of each side : x - 2 = ± V3, or, x = 2±v^ = 2+V3, or, 2 - V3. Check: For 2 + V3: (2 + V3)2 _ 4 (2 + V3) + 1 = 4 + 4V3 + (V3> - S - 4 V3 + 1 = 0. For 2- v'3: (Correct.) (2 - \/3) 2 - 4 (2 - V3) + 1 = 0. (Correct ; student finish the work.) The figure is drawn as in § 108. X I 1 1 2 3 _ 2 4 1 5 6 6 7 8 etc. -1 _ 2 -3 -4 etc. _ 2 -3 6 From the figure the answers are seen to be x = 0.3 (about), and x = 3.7 (about). This compares well with the results above, for V-3 = 1.73+ ; hence the answers found above are x = 2 - 1.73+ and x = 2 + 1.73+, or. x = .27+ and x = 3.73+. It is seen that the figure shows approxi- mately these square root values. EXERCISES II: CHAPTER VIII [In solving these exercises, the student should draw a figure for each one as in § 108. The answers found from the figure will serve as a check on the correctness of the work of solution.] 1. x 2 -lo.r + 44 = 0. 4. .t 2 2. p* — 15p + 54 = 0. 5 Fig. 36. 15* = 76. d 2 - 15 d = 154. 3. a 2 -15 a = 100. 6. 2 r*-r = 36. 208 QUADRATIC EQUATIONS [Ch. VIII 7. 10 w 2 - 31% + 24 = 0. 17. m 2 + 6m = 3. 8. 3a 2 -lla + 10 = 0. 18. p 2 -10p = 5. 9. 4 # 2 + 4 .« = 15. 19. x 2 + x = 7. 10. b 2 — 7a + 3 = 0. 20. a 2 + 3 a; = 10. 11. a- 2 -5.r + 3 = 0. 21. 2x 2 + 5x = A. 12. >r-3rc-3 = 0. 22. 3i> 2 + 4^ = 6. 13. x 2 -x-3 = 0. 23. a¥ + 2 ate = « 2 - b 2 . 14. r/ + -6p + 3 = 0. 5. z 2 + 7z + 13 = 0. 15. 2.r>-4a; + 3=0. 6. m 2 -9m + 21=0. 16. 9s 2 -30s + 25 = 0. 7. .r 2 -8.« + 16 = 0. 17. £ 2 -12* + 36 = 0. 8. v 2 -'Sv + 3 = 0. 18. a? -7 a; + 13 = 0. 9. z 2 -4z + 8 = 0. 19. 2?/ 2 -5?/ + 4 = 0. 10. r 2 -8r + 20 = 0. 20. 3m 2 -4m + 3 = 0. 114. General Solution by Formula. We shall now ex- press the solution of any quadratic equation. The stu- dent has seen above that we could always do so, though the expressed solution may not have a meaning. Any quadratic equation is of the form : ax 2 + bx + c = 0, where a, b, c are fixed, known numbers in any one example, and a is not zero. Divide both sides by «, as on p. 206. a? + -x+-=0. a a c f b \ 2 Transpose - and add — to both sides : a \2 a) a \2aJ \2aJ a The left side is found to be the square of (x A ) : b \ 2 b 2 — 4 ac x + 2 a 4 a 2 113-114] METHODS OF SOLUTION 213 Taking the square root of each side, we find z + A =± J^!zii^ =± V52 - 4 ^ . (See §99, p. 188.) •la y Aa* 2 a b Transposing — — , we find — ' f b V A 1 ' — 4 ac X ~~2a ± 2a ' — b ± V6 2 — 4 ac * = 2a ' Check : These answers are correct, for we may reverse each step, starting with the answers and going backwards through the work to the original equation. Let the teacher guide the students in doing this, accounting for each step as it is taken. The answers may also be checked by direct substitution in the given equation, ax 2 + bx + c = 0. This work is not written here be- cause it is long. The teacher should guide the students in actually doing it. Ex. 1. x> - 4 x - 5 = 0. (See Example 1, § 109.) Comparing with ax 2 + bx • -f- c = 0, we find a = 1, 6 = — 4, c = — 5. Putting these numbers in the general result just found, we get _ ( _4)±V(-4)2-4(l)(-5) 2(1) = + 4±V36 = 4 ± 6 = _ l0 5 2 2 which are the answers found before. Ex. 2. 2x 2 -9x + 4 = 0. (See example 2, § 109.) a = 2, 6 = - 9, c = 4. .-. // 2 - 4 ac = (- 9) 2 - 4 (2)(4) = 49, = _ ( -9) : fcV49 = 9_±jr = 4OT 2-2 4 * It is often best, as here, first to calculate the value of b 2 — 4 ac, the quantity under the radical sign. One reason for this is seen in the next article. 214 QUADRATIC EQUATIONS [Ch. VIII The other examples solved in the text above are given in the following table, solved by the preceding formula. No 3 4 Example 1 b -4 c 1 b*-i 0, fAere are two answers ("unequal real roots "). (In a figure, drawn as in § 108, the curve cuts the main horizontal in two points.) A. 2. If b 2 — 4 ac = 0, there is only one answer (" equal roots"}. (In a figure, drawn as in § 112, the curve just touches the main horizontal line.) A. 3. If b 2 — 4 ac < there are no ansivers (" imayinary ^ roots''''). (In a figure, drawn as in § 113, the curve does not cut the main horizontal.) For b 2 — Aac is the quantity under the radical sign ; if it is negative we have the square root of a negative quantity, which is meaningless to the student at present. B. If b 2 — 4 ac is an exact perfect square, the ansivers are rational; otherwise, the ansivers are irrational, provided a, o, c are rational. 114-115] METHODS OF SOLUTION 215 On account of tlu-se facts the quantity //- - lac is often called the discriminant; the knowledge of its value enables us to discriminate among the cases mentioned. Notice that if these rules are forgotten, all this information can be tumid in any one example by attempting to solve as in §§ 109-113, by completing the square. EXERCISES V: CHAPTER VIII Draw the figure, solve by completing the square, and also solve by the formula. Compare the three answers. 1. or' - 10 x + 1(3 = 0. 9. 6r 2 + r-2 = 0. 2. 3 a; 2 + 5 x - 12 = 0. 10. 5 z 1 - 3 z -36 = 0. 3. 2z 2 -9z + 4 = 0. ll. 3 n 2 -14 n — 5 = 0. 4 . sf — Sy + 5 = 0. 12. 5J9 2 — 17j> + 6 = 0. 5. 2 1* + t- 15 = 0. 13. 7 x 2 - 10 x-\- 3 = 0. 6. 6 u 2 - 13 u + 6 = 0. 14. 9 L a - 30 L + 25 = 0. 7. 3 ) = 85120, P p 2 + ip - 6080 = 0, or, on solving, p = — 80 or 76. The solution of the problem is p = 76 centimeters ; also, we see easily that v = 1120 /. 26. For a certain mass of gas in a collapsible reservoir, K = 277,500. A decrease of 1 cm. in pressure causes an increase of 50 /. in volume. Find the original volume and pressure. 116] PRACTICAL APPLICATIONS; PROBLEMS 219 Solve the following similar problems: 27. K = 162,060. Increase in p, 1 cm. Decrease in v, 30 h 28. K = 275,625. Inciease in p, 1.5 cm. Decrease in v, 75 I. 29. K= 112,480. Decrease in p, 2 cm. Increase in v, 40 1. 30. Two tanks of equal capacity are emptied by unequal pipes; it is observed that the tank having the larger pipe is empty two hours sooner than the other. Both pipes attached to a single tank till it in 2| hours. How long do the pipes separately require to fill or empty the tanks? 31. Two equal tanks are filled by pipes, one requiring 3 hours longer than the other. Both pipes together fill one tank in 2 hours. How long does each pipe require to fill one tank ? 32. Two pipes are used, the larger to fill a tank, the smaller to empty it. When they are both open, the tank is filled in 15 hours. The pipes are then detached and used separately to empty two tanks each equal to the first. It is observed that this work is done 4 hours sooner by the large than by the small pipe. Find the number of hours required by each pipe. 33. A tank is filled by a certain pipe in an hour less than is required for a larger pipe to fill a tank of twice the capacity of the first. Both pipes together fill the large tank in 3 hr. 45 min. In what time could each pipe fill the small tank ? 34. Two tanks whose capacities are as 2 to 3 are emptied by pipes, the larger tank requiring an hour longer than the other. Both pipes together fill the larger tank in 11 hr. How long would each separately require to fill the small tank ? 35. Two men are on streets at right angles to each other, distant 7 and 8 feet from the crossing. If they approach the corner at the same rate, how far must each walk so that their distance apart shall be 5 feet ? Comment on the two answers. 36. Two men standing 1 foot and 8 feet from the crossing of two streets at right angles to each other walk away from the crossing at the same rate. When will they be 13 feet apart ? Interpret the two answers. 220 QUADRATIC EQUATIONS [Ch. VIII ,Mirror 37. Two men stand on streets meeting at right angles, in positions 3 and 5 feet from the crossing. How far must each walk toward the crossing at the same rate in order that they may be 10 feet apart? Interpret the two answers. Here neither really satisfies the implied conditions of the problem. 38. Two electric cars are on tracks meeting at right angles. One starts from the crossing at the rate of 20 feet a second, the other starts 100 feet away from the crossing at the rate of 10 feet a second. When will they be 200 feet apart ? A concave mirror is formed from a part of a spherical shell of radius r. If the distance of an object (O) from the mirror is denoted by u, and the distance of the image (/) of the object from the mirror is denoted by v, it is known that approximately M-i. u v r (The proof of this formula need not be attempted ; being an ap- proximation, its derivation is not a direct proof.) Note that image and object are situated on oppo- site sides of the center (C) of the mirror, also that placing the ob- ject where the image was, throws the image where the object was. 39. How far is an object from a concave mirror of radius 20 cm. if its reflection is 15 cm. farther from the mirror ? 40. Where are the object and its reflection in the mirror of Ex. 39, if their distance apart is 48 cm. ? 41. What radius must be chosen for a concave mirror in order that an object 1\ cm. from the mirror may be reflected 5 cm. farther from the mirror than the center ? 42. An object is placed beyond the center of a concave mirror so that the distance from the object to the center of the mirror shall be twice the distance from the center to the reflection. Where is the object placed ? (State the result in terms of the radius of the mirror.) Comment on the two results. 116] PRACTICAL APPLICATIONS; PROBLEMS 221 43. I desire to place an object before a concave mirror so that the center of the minor may lie halfway between the object and its image. Can I do so ? 44. A cylindrical box 6 inches high, open at the top, has a surface of 64 it sq. in. What radius must be chosen ? 45. If the box of Ex. 44 is 6 in. high and if the radius of its base is 4 in., show that an error of e (in in.) in measuring the radius causes an error E (in sq. in.) in the total surface, such that E = 7T (20 e + e 2 ). Find e if E = 21 tt ; if E = 5 tt. 46. About how much error e would be caused if we took 7r = 3 in place of tt = 3| in the value 64 n mentioned in Ex. 44? 47. A solid cylinder 4i inches high is entirely covered with 45 tt square inches of paper. Find the radius of the base. 48. Show that the error E in the computed total area of the cylinder of Ex. 47, caused by an error e in measuring the radius, is E = (2 e 2 + 21 e)ir. Find e if E = 50 it ; if E = - 45 tt. Solve for e in terms of E in general. 49. Find the radius of the base of a cone whose slant height is 7 cm. and whose total surface area is 800 -n- sq. cm. 50. A box which is open at the top is to be constructed 20 inches high, with a square base. What are the dimensions of the base, if the surface of the box is 1536 square inches ? 51. If an error e, in inches, is made in measuring the side of the base in Ex. 50, show that the error E in the computed area of the box, in square inches, is E = e 2 + 192 e. Find E if e = 1 ; 2 ; 3 ; - 1 ; etc. Plot the graph. Find e if E = 985. 52. How nearly must the measurement side of the base in Ex. 51 be made in order that the computed area may be correct to within 1 sq. ft. ? Is it practicable to do this with a foot rule ? 53. A solid block, 10 in. high, is to be twice as long as it is wide. Find its dimensions, if the surface area is 1008 sq. in. Tf a body is dropped from a point near the earth's surface, the number of feet it falls in / seconds is given by the relation .s = 16 P. Tf instead of being merely dropped, the body is thrown downward at a 222 QUADRATIC EQUATIONS speed of v feet a second, the relation is s = 16 f 2 + vt. If the body is thrown upward, the relation is s = 16 t' 1 — vt. Notice that in the last instance, since s represents distance downward, and the body starts upward, .s will be negative at first, until t = v — 16. 54. A body is thrown downward with a speed of 50 feet a second from a distance of 225 feet from the earth. When will it strike the ground ? 55. When will a body thrown upward with the same speed from the same place as in Ex. 54, strike the ground ? 56. When will a body, thrown as in Ex. 55, descend to the level from which it was thrown ? Comment on the two answers. 57. When will a stone, thrown downward at a speed of 74 feet a second from a height of one mile, reach the earth ? 58. If the error in measuring the time of fall of a body which is dropped from a height is e, in seconds, show that the error in the computed value of the distance fallen through is E = 32 te + 16 e 2 , where t is the real time of fall. If t = 10, E = 320 e + 16 e\ Find E if e = 1, 2, \, \ , - 1, - 2, etc. Plot the graph. Find e if E = 516. How carefully must the time be measured in order that the error in the computed distance may be less than 50 ft. ? How nearly can the distance be com- puted with a stop watch which measures to fifths of a second ? 59. A body is thrown upward from a height of 1728 feet at a speed of 48 feet a second. When will it reach the earth ? When does it reach the level from which it is thrown ? How much time elapses between its reaching the level from which it is thrown and its reaching the earth ? With what speed should it be thrown doumward in order to reach the earth in this time ? Comment on the result. 60. Show that the error E in the computed value of the distance fallen through in Ex. 54, caused by an error e in measuring the time of fall, is E = 130 e -f 16 e 2 . Plot the graph. Find e if Z£ = 69; if E = — 159. How nearly can the distance be measured by means of a stop watch that reads to fifths of a second? PART III. PROPERTIES OF QUADRATIC EQUATIONS 117. Given Roots. Factor Theorem. We can now manufacture equations which shall have any roots we wish. Ex. 1. If 2 x 2 - 9 x + 4 = 0, (,_4)(2,:-l)=0. Hence, x — 4 = 0, or, 2x — 1 = 0, that is, x = 1, or, x = J. Suppose we wish to make an equation with roots 4 and 1, then we write , As , , % n and tind the product of these factors. We get x 2 - | x + 2 = 0, or, multiplying' both sides by 2 to clear of fractions, we get 2 x°- - 9 j; + 4 = 0. In yeneraljhe roots will be r and s if the equation is (x — r) (.r — s) = 0, or, x 2 — (r + s)x + rs = 0. (See § 60.) 7^ fo«7f? aw equation with two given roots write the prod- vet of x less the first root and x less the second root, and set this product equal to zero. Ex. 2. Given the roots 2 -+- V3 and 2 — V3, find the equation. The equation is |> - (2 + v 7 :!)] [x - (2 - V3)] = 0. Multiplying, we get ^ __ 4 j; + , = 0> which is the desired equation. (See Example 1, p. 207.) Ex. 3. Given the roots -f- 3 and — V2, find the equation. The equation is (x — 3) (x + V2) = 0, or, x 2 (3-V2)x-3V2 = 0. The coefficients in this equation are surds ; this will always happen if there are surd roots, unless the example is especially selected with roots of the form a + vb and a — Vb. 223 224 QUADRATIC EQUATIONS [Cn. VIII The principle of this article may also be stated in a form called the factor theorem (See also Appendix, § 5) : If s is a solution of the equation ax 2 + hx + c = 0, then (x — 8 ) {g a factor of the expression ax 2 + bx + c. 118. Relation of Roots to Coefficients. If r and s are the roots of a quadratic equation, the equation is x 2 — (r + 8)x + rs = 0, as seen above. Hence, (1) The sum of the roots with the sign changed is the coefficient of x in the equation. (2) The product of the roots is the last {or constant) term. Notice that the following problem arises from this : Having di- vided both sides by the coefficient of x 2 , any quadratic equation has the form above; to find its roots we must find two numbers whose sum and product we know. Problems of this kind lead to quadratic equations. Thus, if we know that the sum of two numbers is 6 and their prod- uct is 8, these numbers must be the two solutions of the equation x 2 —6x+ 8 = 0, i.e.'2 and 4. Another method will be found later (See Chapter X, p. 253.) EXERCISES VII: CHAPTER VIII Find quadratic equations whose roots are : 1. 2,3. 5. -3, -5. 9. -2,2*. 2. 3, 5. 6. 7, - 9. 10. 6, - 6. 3. 3, -5. 7. 3,*. 11. 2|, -If 4. -3,5. 8. 4,-lf 12. -3.4,7.1. 13. 3+Vo, 3-V5. 18. 2-V3, 3-v3. 14. _ 4 _ Vl7, - 4 + Vi7. 19. a + b, a - 6. 15. 3 — V7, 3 + V7. 20. a, a. 16. — 6-r-Vo, 2 — Vo. 21. a,— a. 17. 3 + V5, 2 - VS. 22 - « + V&, a - V6. 117-1 lit] PROPERTIES OF QUADRATIC EQUATIONS 225 State the sum and product of the roots in each of the follow- ing equations ; then check by solving and actually finding the sum and product : 23. .>•- - 8 x + 12 = 0. 24. x*-3 x = 28. 25. .r + 7x = 30. 26. x~ + 4 x + 3 = 0. 27. 2 x 2 - x -10=0. 28. 6 a 2 + a — 40 = 0. 29. ./•- - 2y;x + jr — v ft 2 — 4 we The solutions of ax' 2 + bx + c = are x = — — ± - -; 2 a 2 a hence, , , r f-h Vl> 2 -iac\lf f-b Vfe 2 - 4 ac \ 1 where the factor a on the outside is chosen so that the term in x 2 will be ax-. The remit is meaningless if b 2 — 4 ac < 0, for in that case the expressed radical V&' 2 — 4 ac has no meaning at pres- ent. It is helpful to notice the following scheme of results : 110-f2O] PROPERTIES OF QUADRATIC EQUATIONS 227 Discriminant (b- -4ac) Solutions of ,i.r- t- bx + c = Factors or = or < 0. When b 2 — 4 ac ^ 0, find the roots of the corresponding equations, and thus factor the expressions : 1. x *_$ x + u. 8 . 3 v 2 - 30?; + 75. 15. 7r-9r + 4. 2. tf-lp + 13. 9. 4 x 2 - 20 a; + 25. 16. 2 s 2 - 10s + 121 3. 6m-m 2 -9. 10. 7?r-12?t + 4. 17. llz 2 + 2z-40. 4. 3t 2 + t-2. 11. 5u 2 -3u + 2. 18. 9f 2 + 42*+49. 5. z i + z + \. i2. 5z 2 -3.c-2. 19. 13« 2 + 5*. 6. 2 -'_z + l. 13. 6c 2 -lie + 3. 20. 3< 2 -2< + 105. 7. 2 2 + 2 _ 1> 14 t 2 -l2t. 21. £ 2 + .3.c+.02. .F*Vs£ factor each of the following expressions, then find the roots of the corresponding equation : 22. a; 2 -4. 26. .r 2 + 8a;-33. 30. 6z 2 +z-35. 23. 4 a 2 -9. 27. r' + 5^. 31. 4 m 2 — 2m- 12. 24. 25m 2 -l. 28. 10 y 2 -y. 32. 15 r 2 -34 r + 15. 25. ^-17^ + 30. 29. 4.T 2 . 33. x J -.3.c+.02. 120-121] PROPERTIES OF QUADRATIC EQUATIONS 229 121. Literal Coefficients. The student has solved some exercises in which letters occurred in the coefficients. We shall now solve such equations systematically. Ex. 1. Given or - 2 mx + (»r - 1) = 0. Transpose m' 1 - 1 and add m- to both sides : x' 2 — 2 mx + m? = — (ni 2 — 1) + m 2 , or, (x — ?h)- = 1, or, x — m=± 1, or, x = m ± 1. This solution holds for any value of the letter ?n. (Check it.) Ex. 2. Given .r - 12 xy -f 4 */- = 0. Solving for x, we find x' 2 — 12 xy + 36 y 2 = 32 y 2 , or, x — 6 y = ± V'32 y' 2 , or, x = 6 y ± V32^ = y (6 ± 4 V'2) . (Check it. ) Solving for y, we find 4 y 2 — 12 xy + 9 a; 2 = 8 x 2 , or, 2y -Sx=± VWx 2 , or, j / =§-£±^ V87 2 = ?(6±4V2). (Check it.) Thus, we may solve the same equation for any letter in it. * Note ox the Discriminant. The given equation has equal roots if the discriminant is zero. Let d stand for discriminant ; then the roots of the given equa- tion are real and unequal if d > 0; imaginary if d < 0; equal if d = 0. Ex. 3. Given a- 2 + kx + (3 + fc) = 0. Comparing with ax' 2 + bx + c = 0, we find a = 1, h = k, c — 3 -f k; hence, the discriminant b' 2 - 4 ac = IP - 4 (3 + fc) = fc 2 - 4 fc - 12. Let us try several values of fc. If k = 0, d = - 12, and the roots are imaginary; in fact, the equation is x' 2 + • x + (3 + 0) = 0, or, x 2 + 3 = 0, or, x — ± V — 3, which is imaginary- * This work may be omitted unless especially desired ; in any case it should not be attempted until the student is quite proficient. 230 QUADRATIC EQUATIONS LCh. VIII Draw the figure for k = 0, and show that it does not cut the main horizontal line. If k = 1, d — 1 — 4 — 12 =— 15; roots imaginary. What is the original equation in this case (k = 1)? Solve it. Are the solutions imaginary ? Draw the figure. If k = 6, d = 36 - 4 • 6 - 12 = 0; roots equal. What is the original equation in this case (k = 6) ? Solve it. Are the roots equal (i.e. only one solution). Draw. If k = 10, d = 100 — 4 • 10 — 12 = 48; roots real and unequal. What is the original equation in this case (Jk = 10) ? Solve it. Are the roots real and unequal? Draw. Trying several other values of k, we make this table : k -12 1 -15 2 ;; etc. 5 7 9 etc. - 1 -7 -2 - 3 etc. d — 7 9 roots imag. imag. imag. equal real imag. equal imag. [Let the student fill in the blank spaces and extend this table to k = + 10 and backward to k = — 10.] We may draw these values of k and d as shown in Fig. 40. From this it is clear that: d = only when k = — 2 and when k — 6 (i.e. the original equation has equal roots only when k = j- 6 or- 2). d < ivhen k has any ralue between — 2 and -f (i.e. the original equation has imaginary roots when and only when k is a number between — 2 and + (J). d>() when k is less than —2, also when k is greater than + G (/.e. the original equation has real and unequal roots when &< — 2, and when k > + 6). These results may also be found by factoring the discriminant: Fig. 40. d = fc 3 - 4* - 12 = (k + 2) (*• - 0). 1n «_ "J! u £ t -j£ IT 4 + i - 4 1 1 n it -h- T ~~1 ? 1 K 1 L dU -k~ lik =43- l-~ £ rt t H J \ t V 4 X- t \-10,_ / -y- u f fc T X t ±-\ J- 35 32 121] PROPERTIES OF QUADRATIC EQUATIONS 231 EXERCISES IX: CHAPTER VIII Solve the following equations for the letters indicated : 1. x- — 5 ax = — 6 a 2 . (First for x ; then for a.) 2. x 2 — 6 ax 2 — — 5 a 2 , (x, a.) 3. m 2 — 6 utx = r — 9 x 2 . (m, x.) 4. wt r 2 — ( m + » ) »*s + » 2 -2pm-(2m + l)=0. ^.) 8. ^—6 ay -6a — 1 = 0. (?/.) 9. t 2 — mt + m = l. (t.) 10. or 2 + 6 ?/- + 6 z 2 + 12 yz — 5 za — 5 xy = 0. (x, y, z.) Find for what values of A; the roots of the following equa- tions will be real and unequal, real and equal, or imaginary ; in the first two cases, solve the equations. [Omit these until very proficient.] 11. x*-kx + k = 0. 12. x*-(k + 2)x + 2(k + 2)=0. 13. x 2 + 2 (k - 3) x + (k + 3) = 0. 14. kx 2 + (k + 5) x + (k + 5) = 0. 15. _^- + ® + ^- = 0. k+3 k k + 3 16. The area of a square field measured in square rods is equal to the length of its whole perimeter measured in rods less twice a given number k. For what values of k is the problem possible ? impossible ? For what values are there two answers for the size of the field ? When is there just one answer? 232 QUADRATIC EQUATIONS [Cii. VIII REVIEW EXERCISES X: CHAPTER VIII Solve the following equations : 1. z 2 -5z-300 = 0. 11. 12^-^-20 = 0. 2. ^-3 1- 108 = 0. 12. 18 a 2 -79 a- 3100 = 0. 3. ^ + 34 x -800 = 0. 13. 4A 2 + 7fc-147 = 0. 4. a,- 2 -29 a; + 168 = 0. 14. 13 k 2 + k- 120 = 0. 5. p2_ 6^-247 = 0: 15. 28 x 2 - 3 a- -135 = 0. 6. 6 a- 2 - 7 a-- 20 = 0. 16. 2 y 2 - oy - 88 = 0. 7. 15r 2 -22r-91 = 0. 17. 18/ -31?/ + 6 = 0. 8. 6 q 2 + 5 9 - 781 = 0. 18. 6 a 2 + 7 a; - 49 = 0. 9. 14^ — 3^-270 = 0. 19. 12 x 2 + 5 x -72 = 0. 10. 8 s 2 - 18 s -425 = 0. 20. 35A: 2 -A-6 = 0. Factor the following expressions by first solving the corre- sponding equations : 21. 6 a- 2 -x -77. 26. 21r 2 + 13r + 20. 22. 24 p 2 - Up -63. 27. 6z 2 -7z-24. 23. 6 k 2 -7 A; -33. 28. 50 r 2 - 5 r - 36. 24. ?/ 2 - 66 y - 675. 29. 21 m 2 - 20 m - 96. 25. 12 A 2 + A - 130. 30. 24 a- 2 + 19 x - 35. Form equations whose roots shall be : 2—3 —1—2 „ — 2 31. 6,-8. 33. I -A. 35. _,_._ 37. 7, — 32. 1,1 34. 5, =£. 36. 3,1 38. -5,1*. 39. a-36,2a-6. 40. f, -• 121] REVIEW 233 Find the discriminant of each of the following equations; determine the character of the roots, and solve if the roots are real : 41. 14 a 2 + 29 a; -15 = 0. 45. $x t -llx+ 11 = 0. 42. 81 g 2 - 198 z + 121 = 0. 46. 6 .r-- 11 .c + 4 = 0. 43. 8 P- 13 1 + 6 = 0. 47. x 2 -(k + 3)x + k 2 = 0. 44. 5/+lG.y + ll = 0. 48. i 2 -(fc+3)*--fc= : 0. 49. (k-l)r + 2kr+(k+l)=0. 50. (A- + 1) .r - (3 k + 1) x + (k + 1 ) = 0. 51. The sum of two numbers is 16; the sum of their squares is 130. What are the numbers ? 52. The sum of two numbers is s ; the sum of their squares is s. Show that the quadratic equation found in order to determine one number has for its other root the other number. Sketch ok Solution. Let n r and n., denote the two numbers. Then n l + n 2 = s or n t = s - n., ; since n 2 + » 2 2 = s, the equation for either n is of the form n- + (s - n)' 2 = s; hence the two n'a are precisely the two roots of this equation. 53. The sum of two numbers is s ; if each of the numbers is divided into A, the sum of the quotients is s. Show that the same quadratic equation has both numbers for its roots. 54. Four consecutive integers have as the sum of their squares 54. What are the integers ? 55. A and B together can fold 1000 circulars in an hour. It is observed that when each works separately at 1000 circu- lars, A finishes 50 minutes before B. How many circulars can each man fold in an hour ? 56. Two tanks have capacities in the ratio 3 to 4. When unequal pipes are attached, the small tank is filled in 2 hours less time than the large tank. The two tanks are then connected, and both pipes used to empty them. This process requires 2 hours 48 minutes. How long would be required for each pipe alone to empty each tank? 234 QUADRATIC EQUATIONS [Ch. VIII [Suggestion. The student may introduce the idea of a " unit tank " whose capacity is one third that of the smaller tank ; use as the principal unknown the time, t, required to fill the small tank with its pipe, and express the capacity of each pipe (r x and c q ), i.e. the amount each can carry in one hour, in terms of f.] 57. An open box 8 cm. high, whose base is a rectangle with sides in the ratio 3 to 4, is to have a surface area of 640 sq. cm. What are the dimensions of the base ? 58. A regular pyramid on a square base is to be made by folding a figure like that shown. The altitude of the triangular sides is to be 6 in. The sur- face area of the pyramid is not to exceed 100 sq. in. What must be the side of the base to obtain precisely that surface area ? (Solve first graphically and then directly from the equation.) Fig. 41. 59. What three consecutive integers can measure the sides of a right triangle ? 60. Show that if one perpendicular side and the hypotenuse of a right triangle are measured by consecutive integers, the square of the other perpendicular side must be measured by an odd number. Choose for the square of this side successively 9, 25, 49. In each case what are the other two sides ? 61. A body is thrown upward at a height of 500 feet with a speed of 30 feet a second. When will it reach the earth ? (Solve first graphically, then directly from the equation.) See p. 221. A body thrown horizontally into the air at a height of h feet with a speed v feet a second, follows a path thus described : in t seconds the horizontal flight, x feet, and the distance from the earth, y feet, of the body are given by x = vt, y = h — If) t 2 . A body thrown into the air from the earth's surface at an angle of 15° with a speed of v follows the path x = — , y = — - 16 t 2 . V2 V2 121] REVIEW 235 62. What is the relation between x and y, i.e. the equation of the path of the body, in each of the above cases? 63. A stone is thrown horizontally from a cliff 400 feet high at a speed of o0 feet a second. When will the stone strike the earth, and how far from the foot of the cliff ".' 64. When will the horizontal distance traveled by the stone be equal to its height? What will this distance be? (Solve graphically first.) 65. A body is thrown into the air at an angle of 45° with a speed of 37 ft. a second. At what horizontal distance will it be 5 ft. from the ground ? How long has it then traveled ? 37 2 66. At what horizontal distance will the body be feet 128 from the ground ? Consider the graph carefully in connection with Exs. 65, (5G. [The following exercises are intended only for use upon a review of the whole book and are not to be solved until the student has completed the study of Chapter XII.] Solve the equations : 67. aJ*+6ar* = 40. _. /—T* , 1 qh T 71. V.c + i H = = o\\. 68. Vx + 5^/x = U. VaT+7 69. (.r , + l; 2 + 8(.r 2 + l) = 180. 72. .r 2 -3.e-3Va; 2 -3a;+7 = 3 70. V2x + 3 + 0a- = 71. 73 - VA- + \/.r = 2. 74. 2 .r + 7 x + 6 - 3 V8 x- + 28 x - 1 1 = 0. 75. 2 (x + -Y- 9 fx + -V 10 = 0. 23G QUADRATIC EQUATIONS SUMMARY OF CHAPTER VIII: QUADRATIC EQUATIONS, pp. 203-235 Part I. Methods of Solution, Character of Roots. pp. 203-215. Definition of Quadratic Equations : contains square of unknown. First Methods of Solution : factoring, as in Chapter IV, if easy; otherwise, drawing figure. Exercises I. § 108, pp. 203-205. General Solution, Completion of Square: typical example; definition of root; insistence on verification. § 109, pp. 205-206. Formal Rule for Completion of Square : making left-hand side perfect square by adding a number. Exercises II. § 110, pp. 206-208. Second Method: multiplication by 4 x 1st coefficient; previous method with kx in place of x. Exercises IIT. §111, pp. 208-209. Equal Roots : single answer; perfect square; curve "tangent " to main horizontal line. § 112, pp. 209-210. Imaginary Roots : no answer; curve misses main horizontal; defi- nition of imaginaries. Exercises IV. § 113, pp. 210-212. Formula : general solution ax' 2 + bx 4- c = 0. § 114, pp. 212-214. Discriminant: correspondence of b' 2 — 4«e>0 to 2 ("real and unequal ") roots, = to 1 (" equal ") root, < to (" imagi- nary ") root ; irrational roots. Exercises V. § 115, pp. 214-215. Part II. Practical Applications; Problems, pp. 216-222. Practical Examples : solution ; suggestions. Exercises VI. § 116, pp. 216-222. Part III. Properties of Quadratic Equations, pp. 223-235. Given Roots: correspondence of roots r and s to equation ( x _ r ) (x - s) = 0. Factor Theorem : discovery of factor x — r for any root r. § 117, pp. 223-224. Relation of Roots to Coefficients: sum of roots = — coefficient of x; product = constant term. Exercises VII. § 118, pp. 224-225. Factoring Quadratic Expressions: use of factor theorem; nature of factors from discriminant. § 119, pp. 225-227. Solution by Factoring: factors of ax' 2 + bx + c ; method advisable ouly in simple cases. Exercises VIII. § 120, pp. 227-228. Literal Coefficients: typical problems; discriminant; figure for discriminant. Exercises IX. § 121, pp. 229-231. Review Exercises for Chapter VIII : Exercises X. pp. 232-235. CHAPTER IX VARIATION: INDETERMINATE EQUATIONS 122. Simple Variation. We have discussed before quan- tities that vary. Thus, the cost of an amount of butter varies with the number of pounds bought. (See p. 23.) Whenever the quotient of two varying quantities y and x is a constant k, <- = k, or, y = kx, x J the variables y and x are said to be in proportion (see pp. 25, 140, etc.), by which we mean that any pair of values of y and x form a proportion with any other pair. We also say in this case that/ varies directly as x. We have seen, p. 141, that the corresponding figure is a straight line through the starting point. The following are therefore synonymous : (1) y is proportional to x. (2) y varies directly as x. (3) The quotient y -f- x is constant ; or, y = kx. (4) The graph for y and x is a straight line through the starting point. Instead of the quotient y -s- x, we may speak of the ratio of y to x and write it y : x ; and we may say that the ratio y : x is a constant. This constant, which is called k above, is called the ratio of proportionality. Thus, if batter costs 30 per pound (see p. 28), the ratio c :p or -, where c denotes the cost in cents and p denotes the number of P 237 238 VARIATION: INDETERMINATE EQUATIONS [Ch. IX pounds, is always 30. The ratio of proportionality is 30. Any pair of values of c and p, say c x and p v give the same ratio as any other pair, c 2 and p 2 : £i = £z (=30), or, ?i=Pi. (See VI, (1), p. 138.) Pi Pi c 2 lh If one varying quantity, 2, varies directly as the prod- uct, x x «/, of two other quantities x and y, then z evidently varies directly as x when y is constant, and as y when x is constant. For if z = k • x ■ y, where k is a constant, then z = (k • x) - y, whence z is a constant times y if x is a con- stant. Likewise z =(k ■ y~) • x, whence z is a constant times x if y is a constant. Thus, the area, A, of a rectangle is given by the formula A = b x h, where b is the base and h is the height of the rectangle, measured in feet or in any other unit of length. If b is constant, A varies as h, i.e. the areas of rectangles with equal bases are to each other as the heights. Likewise A varies as b if h is constant, i.e. the areas of rec- tangles of the same height are to each other as the bases. The reverse statement is also true : if z varies as x when y is constant, and as y when x is constant, then z varies as the product x x y. Thus, if we know that the area, A, of a rectangle varies as h (the height) when b (the base) is a constant, and as b when h is a constant, we may conclude that A varies as b x h. A formal proof of this last form of statement is deferred. 123. Linear Variation. We have also seen that a linear equation (or equation of the first degree) of the form (1) y = ax + b, where a and b are constants, is represented by a straight- line graph. (See pp. 2fi, 143.) This kind of relation is often called linear variation, and we say that y is a linear function of x. Examples of this occur on pp. 25, 143. 122-124] VARIATION: INDETERMINATE EQUATIONS 239 Ex. 1. As another example consider the equation 2 x — 3 y = 6. Solve for y : y = | * - 2, which is of this same type, i.e. the curve is a straight line. To draw it we need only plot two points (see footnote, p. 144). For example, (x = 0, y = - 2) and (y = 0, X = 3) r as can be seen also from the origi- nal equation 2 .»■ — 3 y = 6. The graph is as shown in Fig. 42. Any equation of the form (2) Ax + By + C = can be reduced to the type (1) pro- ' vided B =£ 0, as in the preceding example. If 11 = 0, the equation -C , — — — 1 jS ■zy . j>&' _j_ n *UB y 1 ^" " T + s jS i* / yf I Ax + C = 0, or, j- = yl Fig. 42. r/of.< no* contain y. Hence, a: = ^e some ^m? for every possible y, and the curve is a vertical straight line. In all eases, without exception, an equation of the first degree is represented in a figure by a straight line. 124. Inverse Variation. It may happen that one vary- ing quantity increases as another decreases, in such a way that their product is constant. Thus, if a train goes 20 miles per hour on a trip 600 miles long, the time taken is 30 hours ; if it goes 25 miles per hour, the time taken is 24 hours; if it goes 30 miles per hour, the time taken is 20 hours. Notice the product speed x time = constant = total distance : s ■ t = d, where s stands for speed, t for time, d for distance. A table follows : s 1 2 5 10 15 20 30 40 50 60 etc. -10 -30 etc. etc. * 600 300 120 (in 40 30 20 15 12 10 etc. -60 -20 d 600 600 600 600 600 600 000 600 600 600 000 000 240 VARIATION: INDETERMINATE EQUATIONS [Ch. IX Here (/ is a constant, 600 (in miles). But s and t vary. The graph of the relation between s and I is shown in Fig. 13. Negative values of s and t correspond to backward motion. A relation between two varying quantities — say y and x — such that , , ,n x y = k ( or x = -, or y = - ), \ y xj where k is a constant, is called inverse variation, and we say that/ varies inversely as x. — I — 1 — 1 1 1 — — 1 — t I 4 1 1 1 it 1\ |\ + 4^- -nio ■ 10 ""j ■— — t^-_,^_ -~. o in lui . S \ \ «fi=600 + " \ n - t- I -oo_ i Fig. 43. Curves shaped like the above, i.e. curves corresponding to an equa- tion of the form x>/ = k, we shall call inverse variation curves. Simi- larly, we notice that direct variation curves are straight lines. EXERCISES I: CHAPTER IX Draw graphs to represent the following relations : 1. y = 16x. 5. 3 a; + 7 # = -18. 2. y = — 12x + 7. 6. y = 5(x — 2). 3. x — 3 y — 5. 4. 20Q. 17. Suppose y varies directly as x and the ratio of propor- tionality is 2 ; write the equation ; draw a graph ; find y when x = 1, when x = 2. 18. Suppose y varies d.irectly as x and x=2 when y = 10; write the equation; draw a graph; find y when as =.3. 19. Suppose y varies inversely as x and x = l when y = 2; write the equation ; draw the graph ; find y when x = 2. 20. Suppose y varies as the square of x and x = 1 when ?/ = 4 ; write the equation ; draw the graph ; find y when x = 2, when x = 3. 21. The volume, V, of a box whose height is 4 ft. is V= 4 wl, where w is the width and I is the length (in feet), and where V is the volume (in cubic feet). Show that V varies as iv when I is constant, and as I when iv is constant. 22. Show that the area of an open circular cylinder varies as the height (h) when the radius (r) of the base is a constant, and as r when h is constant. (See Tables.) 23. Find from the Tables all geometrical figures whose areas, or volumes, vary directly as certain of their dimensions, and express each of these both by formulas and in words. 24. The cost of any number of pounds of butter varies as the number of pounds. If one pound costs 30 cents, express the cost of any number of pounds; draw the graph; what is the cost of 5 pounds ? 25. For a certain mass of gas, it is observed that pressure times volume is equal to 120,000. Plot the graph. (See p. 218.) 26. Indicate by a picture the relation between base and alti- tude of a triangle of constant area. it; 242 VARIATION : INDETERMINATE EQUATIONS [Ch. IX 27. If a certain stretched wire is caused to vibrate, then the number n of vibrations a second, the radius r of the wire in centimeters, and its length / in centimeters are observed to satisfy the relation rirl = 16000. For a wire 50 cm. long, plot the relation between n and r. 28. Express by a figure the radius and length of all wires sounding F oi the middle register (n = 320 per second). 29. The surface area of a cylindrical ring is 330 sq. in. Express by an equation, and plot a figure for all possible values of the radius and height of the ring. (Take 7r= ? f 2 -.) 30. The relation between the mass, volume, and density of a body is m = vd. Choose a suitable constant value for the density, and plot the relation between volume and mass. 31. Choose a suitable constant value for the mass, and plot the relation between volume and density. 32. Express graphically the relation between the total sur- face area and the slant height of a right circular cone whose base has a radius of 7 inches. 125. Indeterminate Equations. The examples above are all examples of indeterminate equations, i.e. equations in which each of the unknown letters has not one fixed value, but rather an indefinite number of possible values. Sometimes a problem is of such a nature that only a few possibilities remain ; we may then fix definite values for the letters. Ex. 1. A carpenter has boards 8 in. wide and others 6 in. wide. How many of each may he take to make a walk 50 in. wide, the boards being laid lengthwise ? Let x = number of 8-inch boards, and y = number of 6-inch boards. Then 8 x + 6 y = width, or, 8 x + 6 y = 50. Let us now try various numbers. Tf x = 0, y — ^ 5 = 8^. (A in Fig.) [f y = 0, x = *£ = (i\. (B in Fig.) 124-125] VARIATION: INDETERMINATE EQUATIONS 243 Y 1 V — III .1 l i B\ - Fig U. The figure is a straight line, for the equation is of the first degree. Hence, the possible values of x and y correspond to the points on the straight line joining .4 and B Now the carpenter must choose an integral number of each sort of boards. The values of x and y must then be integers, i.e. the corre- sponding points lie at the intersections of the square paper. We notice three points that may serve : x=lA ( X = M \ x = ^ y = 7,J h/ = 6,j U = 3, for each of these lies at least very close to the straight line. Trying m. these, we find x = 2, y = 6 gives a total width 2 • 8 + 6 • 6 = 52 This is wrong. The point (x = 1, y = 7) gives 1 • 8 + 7 • 6 = 50. This is correct. Likewise (x = 4, y = 3) is correct. The carpenter may, therefore, choose 1 board 8 in. wide and 7 boards 6 in. wi :e, or 4 boards 8 in. wide and 3 boards 6 in. wide. If he has more 8-inch boards than 6-inch boards, he should clearly choose the latter scheme. 2U VARIATION: INDETERMINATE EQUATIONS [Ch. IX Ex. 2. A man agrees to exchange young hogs at $ 3.00 each for sheep at $ 5.00 each. How many sheep and hogs may be traded practically ? — •S II \ o R 1/ 1 ) 20 h A Fig. 45. Let h = the number of hogs, and s = the number of sheep. Then 3A = 5s, or, 3 h — 5 s — 0. If h = 0, s = 0. If h = 10, s = 6. The graph is surely a straight line since the equation Bh — 5s= 0, is linear. Drawing a figure, the possibilities are seen to be only these : Hogs .... 5 10 15 20 etc. Sheep .... 3 6 9 12 etc. Ex. 3. A frame (box without a bottom or top) 2 ft. high is to be constructed out of a board 1 ft. wide and 24 ft. long. Find how to cut the board to make the largest box. 125] VARIATION: IN DETERMINATE EQUATIONS 245 .9 Let s = length of one side, then 6 — s = length of other side. Then area of bottom = s(6 — s). 0-s The largest box is that which has the largest bottom. We wish to find when h = 8 (t! — s) is largest, where b means the area of the bottom in square feet. If s = 0, b = ; if s = 1, t = 5 ; if s = '2, b = 8 ; etc. Proceeding in this way, we find a table as follows : Fig. 46. s 1 2 3 4 5 G etc. b 5 8 9 8 5 etc. From the figure, the highest value of b is seen to be where s = 3, at least approximately. Hence, the largest box will result by taking the side approximately 3 ft. The shape of the bottom will then be a square ; and, as a matter of fact, this is the shape that will give the greatest volume to the box. = / \ / \ / \ / I / \ / \ / \ 1 J 1 si U l\ 1 \ 1 / 6 = !?( I-.S ) 1 1 1 1 \i ! / 1 1 1 -10 1 Fig. 47. 246 VARIATION: INDETERMINATE EQUATIONS [Ch. IX EXERCISES II: CHAPTER IX 1. Find all positive integral solutions of the equation 2x + 7y = 50. 2. Find some integral solutions of the equation 3x—5y=7. 3. Find all integral solutions of the equation -+ ^ = 3. 3 4 4. Find all even positive integral solutions of the equation 5 x + 7 y = 96. 5. Find all positive integral solutions of the equation 5 x + 7 y = 99, such that x is odd and y even. 6. Show that the equation 5x-\-7y = 99 is satisfied by all values of x and y given by the formulas x = 17 — 7 s, y = 5 s + 2. For what values of s are the results of Ex. 5 obtained ? 7. Show that the equation 5 x — 6 y = 7 is satisfied by the formulas x=6s — 1, y — 5s — 2. Give several values to s, and determine values of x and y. Also solve graphically. Find the least positive integral solution of the following equations : 8. 7x — 8y = 17. 10. 3x-Uy = l. 9. 12 x + 5y = 49. 11. 49 x- 53 y = 17. 12. How can I pay a debt of 60 cents in quarters and dimes ? Give all selections. 13. How can I pay $57 in two- and five-dollar bills ? Which method uses the fewest bills ? 14. A farmer pays a debt of $4.50 by giving chickens valued at 30 cents each and turkeys valued at 80 cents each in trade. How many of each sort of fowl are needed ? 15. I desire to weigh 100 pounds by means of 3- and 1 1-pound weights placed in the same pan. How can this be done ? 125-426] VARIATION : INDETERMINATE EQUATIONS 247 16. How can this be done if the two kinds of weights are to be placed in opposite pans, the 3-pound weights being in excess ? 17. How can this be done, the 11-pound weights being in excess? 18. A Avalks 3 hours to reach a neighboring town ; B, start- ing from the same place, walks 5 hours and is still a mile from the town. If each man walks in an hour an integral number of miles, how fast do the men walk, and how far distant is the town ? Note that one condition of this problem is the limitation of possible speed of human walking. 19. Find a number that leaves a remainder of 11 on division by 16, and of 4 on division by 11. Show also that every number of the form 176 t + 59, where t is a positive integer, has this property. 20. What is the perimeter y of a rectangle having one side x and area A ? Plot the relation between y and x for several values of A. On each curve note the value of x, which makes y a minimum. What rectangle of given area has the minimum perimeter ? 21. A window having the shape of a rectangle surmounted by a semicircle has a total perimeter of 200 inches. What dimen- sions should it have to admit the greatest amount of light ? (Take ■*■ = ^.) What is then the area ? 126. Other Cases. Many other forms of indeterminate equations arise. Some figures have been drawn on pp. 30, 182, 204, to illustrate these, particularly in connection with quadratic equations. In general, given an equation containing tivo varying quan- tities, we give several values to one of them, find the corre- 248 VARIATION: INDETERMINATE EQUATIONS [Ch. IX spondiny values of the other, and plot in a figure the points that correspond to each pair of values. Finally we connect these points by a smooth curve. In case there is any doubt as to how the curve should be drawn, we simply plot more points, as above, until the curve is fairly out- lined by them. Among interesting curves are the following : J£x. 1. y = kx 2 , where k is constant. This arises often practically. Thus, the price of painting a square surface varies as the square of one side. p = k • s 2 where k is the price (in dollars) for painting 1 sq. yd., and where p means the total price (in dollars), and s the length of one side in yards. The length / (in feet) of a pendulum varies as the square of the time t of vibration (in seconds): ?^ = 3.26 (nearly). _ X X 1 1 1 1 x x:__ _ T T to) wn '..-a ;^lj 1 fcpiL .. -ju l-Ak-1 5£ L C4 tl 1 1 .11) 1 4 +- --20 hf /- 1 1 1 1 1 ' : "Yir x 7 tx xt-, I ~ ---^X 1 \ \ 1J l! / fe M l 717 n-V IE II iR'JcAL \-\\ tl-3 IJJ -iit.I 4! ft Jjj ^ILl lL\ 177 ir 77 >\\ T],r> \i // XL--*/ IT *K^ Fig. 49. Z=ifc • t 3 where k- The distance d (in feet) traversed by a body falling from rest varies as the square of the time t (in seconds) it falls : d = k-t 2 where k = 16.08. The equations of the form I = ax 2 + bx + c considered in the previous chapter are of the same type, but are somewhat more complicated than this example, (a) Let us draw y = x' 2 , i.e. the case Making in which k a table as before, we find : = 1. X 1 2 3 4 16 5 6 7 etc. - 1 - 2 -3 -4 etc. y 1 4 9 1 4 etc. Xet the StlK ent fill in t lie bla nks.] 126] VARIATION: INDETERMINATE EQUATIONS 249 The graph is as drawn, in Fig. 49, the curve marked (a). (A) If y = 2 x% we find X •1 2 3 4 •'. 6 etc. -1 -2 -3 - 4 j etc. y 2 8 18 32 2 8 The graph is marked (6) in Fig. 49. It should be noticed that (b) (i.e. the curve for fc=2) is just twice the height of (a) (i.e. twice the height of the curve for k — 1) at each point. (c) Tf y = \ x 2 , the curve is just half as high at each point as is (a). [Let student make a table, and actually draw the graph.] ('/) [Let the student draw other curves to represent y = 3 x 2 , y = 4: A y = ox% y = 10x 2 , y = i* 2 , y = ^^ y = - x\ y = -2 x% y = _i x 2] Different colors of ink or pencil may be used to advantage for the different curves. All these should be drawn on the same sheet of paper and marked so that they can be recognized. Ex. 2. y = lcx\ This equation is readily plotted as above. Only the figure is here given ; and that only for k = 1, k = 2, k =£. [Let the student make a table for each of these.] Y 1 Fig. 00. 250 VARIATION: INDETERMINATE EQUATIONS [Ch. IX Ex.3. x 2 + y 2 = k 2 . (a) Uk = l, x 2 + y 2 = 1. Take a point P and mark the values of x and y as in the figures : x = OM, y = MP. Now OM 2 + MP 1 - OP 2 since OMP is a right-angled triangle. [The student should know that the square of the diagonal side of a right-angled triangle is equal to the sum of the squares of the other two sides. See Tables.] Hence, x 2 + y 2 = OP 2 ; but x 2 + y 2 = 1 ; hence, OP 2 = 1 en 1 OP = 1. Consequently the point P is at a distance 1 from ; the points for which this is true lie on a circle of radius 1 whose center is at 0. Fig. 51. The equation x 2 + y 2 =l is represented by a circle of unit radius about the origin as center. (Ii) If k = 2, the reasoning is the same except that OP = 2 ; hence, x 2 + y 2 = 4 is represented by a circle of radius 2 about the origin. (c) In any case x 2 + y 2 = t 2 is represented by a circle of radius k about the origin, for OP = k, whatever k may be. EXERCISES III: CHAPTER IX Plot the following curves : 100 3. y=x 2 -7x + 10. i. y = X 2 . 20 2. y = x-\ « X 4. y = 50 10-z' 126] VARIATION 1 : INDETERMINATE EQUATIONS 251 50 , 50 7. y = x 2 + x. 5. y = 1 • a x 10 — x 6. y = x + Va;. 8. y = 2x 2 —9x-\-S. For various values of k draw the curves: 9. y = |. ii. (aj-fc) 2 + ^*l. ar 10. y = » 2 + 2aj + A;. 12. » 2 +(y — fc) a =l. For various values of k and ?, draw the curves : 13. y = kx + 1 -- 15. (x-k) 2 +(y-l) 2 = l. x 14. ?/ = x 2 + A'.t- + /. 16. y = kx — lVx. 17. Represent graphically the relation between the ratios of each perpendicular side of a right triangle to the hypotenuse. Note. These ratios are called the " sine " and the " cosine " of one of the angles of the triangle ; it is suggested that they be denoted here by the letters s and c ; and that the base, altitude, and hypote- nuse be denoted by b, a, and h, respectively. Show that s = — , c =-; h h then since a' 2 + b 2 — A 2 , show that s 2 + c 2 = 1, by dividing both sides by h\ 18. Represent by a figure the relation between the ratio of the hypotenuse of a right triangle to a perpendicular side, and the ratio of the other perpendicular side to the former. Note. Denote these ratios by x ( = - ] and by tl = - ] (called " tangent "). Prove first that x 2 = 1 + t 2 . 19. Represent by a figure the relation between the ratio (c) of one side of a right triangle to the hypotenuse and the ratio (oj) of the hypotenuse to that same side. Show first that x x c= 1. 20. Find the maximum rectangle of given perimeter k. (Solve for various special values of k ; then note the results, and state generally.) 252 VARIATION: INDETERMINATE EQUATIONS SUMMARY OF CHAPTER IX : VARIATION ; INDETERMINATE EQUATIONS, pp. 237-251 Simple Variation : equivalence to proportion between variables ; also to equation y = kx ; figure, straight line. § 122, pp. 237-238. LinearVariatioh : equivalence to equation y = ax + b; figure, straight line. General Linear Equation: figure always straight line. § 123, pp. 238-239. Inverse Variation: constancy of product of two variables; typical problem ; Fig. 43, inverse variation curve. Exercises I. § 124, pp. 239-242. Indeterminate Equations: definition; typical problems ; figures. Ex- ercises II (for graphical solution). § 125, pp. 242-247. General Indeterminate Equations : general case of variation ; plotting curves by assignment of values to one letter. Special Types: type y = kx 2 , i.e. variation as the square; y = kx 3 ', circle x 2 + y 2 = r 2 , center at origin, radius r. Exercises III. § 126, pp. 247-251. CHAPTER X. SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS PART I. ONE LINEAR AND ONE QUADRATIC 127. Introduction. When two simultaneous equations are given, one of which is linear, the other quadratic, it is usually best to use the method of substitution, similar to that of § 90, p. 170. Ex. 1. A wagon bed is to be made to hold 2 cubic yards. It must be 4 feet wide and six times as long as it is high. Find the dimensions of the wagon bed. Let x be the height of the bed in feet and y the length in feet. Since the length is to be six times the height, we have \^ s s. N \^ X (1) y = Qx. Fig. 52. The volume is 4 xy (in cubic feet), for the volume of a rectangular box is the product of its three dimensions. The volume is to be 2 cu. yd. = 54 cu. ft. (2) .\ 4x#=54, or 2xy = 27. (This is called a quadratic equation in x and y, for the sum of the ex- ponents of x and y is 2.) Let us now solve the equations we have found : y = 6x, (1) 2 xy = 27. (2) Substitute 6 x for y from (1) in (2) : 2 x(6 x) = 27, or, 12 x 2 = 27, whence, x- = f , or, x - ± § . 253 254 If SIMULTANEOUS EQUATIONS = + !, [Cm. X x y=6x:=9. Hence, the result is x — §, y = 9. Check- : y = Qx; 9 = 6x| (correct). 2xy = 27; 2 • f • 9 = 27 (correct). — 3 ~ 2> If X y = 6 x = — 9. Hence, the result is x = — |, y = — 9. Check: y-%x; — 9 = 6 ( — |) (correct). 2x^ = 27; o ( _ I) ( _ 9) = 27 (correct) . In this example only the first set of answers has a real meaning. In other applications of these equations both sets of answers may be useful. Check : The dimensions of the wagon bed are : width = 4 ft; height =1| ft; length = 9 ft. The volume is then 4 x 1| x 9 cu. ft. = 54 cu. ft. = 2 cu. yd., as required. Y Fig. We may also draw a figure, Fig. 53, as on p. 239. Thua,(l) (»/ = 6 x) is a straight line, as drawn. (Compare pp. 141, 160.) To draw (2) we give x various values and find the corresponding values of y. (See p. 239 and § 126, p. 247) 127-128] ONE LINEAR AND ONE QUADRATIC 255 X 1 2 3 V 4.5 4 5 6 7 S 910 &c. -1 —2 -3 -4 &c. y ¥ ¥ ¥ 3.37 27 10 ■i - 1 2 -¥ -V y (reduced) 13.5 6.76 2.7 2.41 -13.5 -6.75 [Let the student fill in the blank spaces and continue the table.] The picture for (2) drawn from this table, as on p. 254, is as shown ; it has two parts and is surely not straight. In fact, comparing with § 124, p. 239, we see that (2) is a curve of inverse variation. Every point on the straight line (1) gives a pair of numbers that satisfy y — 6x. Every point on the inverse variation curve (2) gives a pair of numbers that satisfy 2 xy = 27. The points that lie on both (1) and (2) satisfy both equations, i.e. each point of intersection gives a pair of numbers that are a pair of simultaneous solutions of (1) and (2). Compare § 86, pp. 159-160. These points in the figure are P and Q. P gives (x = 1\, y — 9) Q gives (x = - 1^, y = - 9), which are in fact the answers found above. This serves as a check on the work. We shall later find this graphical process invaluable in solving approximately, in difficult examples. Compare p. 276. 128. To find the degree of a term, add together the exponents of each of the important (i.e. unknown) letters. (See § 83, p. 152.) The degree of an equation is the degree of its term of highest degree after it is freed of fractions and radical signs, and is simplified as far as possible (See p. 152.) An equation of the first degree in the important letters is called a linear or simple equation. (See p. 152.) An equation of the second degree in the important let- ters is called a quadratic equation. (See p. 152 and com- pare p. 208.) In Part I of this chapter we consider pairs of equations, one of which is linear, the other quadratic. 256 SIMULTANEOUS EQUATIONS [Ch. X 129. Rule. To solve for two unknown letters in a pair of equations, one of which is linear, the other quadratic : (1) Solve the linear equations for y (or x) in terms of x (or y). (2) Substitute in the quadratic equation for y (or x) the value just found. (3) The new equation will he a quadratic in x (or ?/), or else a linear equation (in rare cases). (4) Solve this equation for x (or y), and substitute both values found in the linear equation to find the values of y (or x). (5) Draw the fiyure as a check on the ivork. (6) Substitute each pair of answers in the original equa- tions as a complete check. Notice that there will usually be two pairs of solutions, for we solve a quadratic equation during the process. Care should be taken to pair off the values of x and the values of y that belong together. In doing this the figure will be of help in avoiding errors. (2x + y = 5, (1) \x 2 + f = 25. (2) Solve (1) for y : y-5-2x. Substitute (5 - 2 x) for y in (2) : x 2 + (5 - 2 x) 2 = 25, or, x 2 + 25 - 20 x + 4 x 2 = 25, whence, 5 x 2 — 20 x = 0, or 5 x(x — 4) = 0. Hence, 5 x = 0, or x - 4 = 0, that is, x = 0, or x = 4. If x = 0, If x = 4, y = 5 [from (1)]. 8 + y = 5, or y = - 3 [from (1)]. The pairs of solutions are therefore (x = 0, y = 5) and (x = 4, y = -3). 129] ONE LINEAB AND ONE QUADRATIC 257 Chech: x = I r 2 x + # = 5 111 \ 1 ** + y* = 25 J y = 5j x = 4 ) f 2x + y = 5 ' 1 x* + y 2 = 25 give give 2.0 + 5 = 5 2 + 5 2 = 25 ' 2- 4 + (-3) = 5 42 + (-3) 2 = 25 (correct). (correct). The graph is as shown in Fig. 54: (1) is a straight line, drawn as in § 80, p. 140, etc. (2) is a circle with radius 5, and center 0, as shown; see p. 250, where the equation x- + y' 2 = 25 is studied. The values (x = 0, y = 5) and (x = 4, y = — :{) as we have paired them correspond to the points of intersection. It will be instructive for the student to see what happens if he pairs off the values incorrectly : thus,(x = 0, y — — 3) and (x = 4, y = 5); do these give points on the curves? Do these pairs satisfy the given equations? }' / *^ (2) / \ / / (1) o v \ Q\ { \ \ V ^ -^ S \ \ \ \ \ > Fig. 54. Ex. 2. Solve (1) for y : Substitute in (2) fcc + 2/=10, tar 2 + tf = 25. (1) (2) + (10 y = 10 — x. x) 2 = 25, or, or. x 2 + ioo - 20 x + x 2 = 25, 2 x 2 - 20 x + 75 = 0. 258 SIMULTANEOUS EQUATIONS [Cii. X The solutions of this equation are imaginary, for a = 2, b = — 20, c = 75 in the notation of § 114, p. 212. 6 2 - 4ac = 400 - 4 x 2 x 75 = - 200. Hence, there is no value of x among numbers we know at present that satisfies the equation. (See § 113, p. 211 ; also Appendix, § 31.) The original example therefore has no solutions. This is clearly brought out by the graph (Fig. 55) : (2) is a circle of radius 5 and center O, as before; (1) is a straight line, as shown. It is clear that these two curves have no point of intersection ; hence, there is no pair of numbers that satisfy both equations. (See p. 211.) Later (see Appendix, § 37), it is desirable to solve the quadratic. We can then proceed to get answers, but these would be meaningless to the student at present. ] T \ \ \ .1 (1) cars <> /; \ \ \ \ s \ Ex. 3. Solve for y (1) : Substitute in (2) or, Divide by 25 : Fig. 55. lx> 2 4-/ = 25. 25 - 4 x y — — : x! + p-4*y = , 5> 25 x 2 - 200 x + 625 = 225. x 2 - 8 x + 25 = 9, (1) (2) 129] ONE LINEAR AND ONE QUADRATIC J.V.t Transpose 9: x- -8x + 16 = 0, or, (./• - 4) a = 0. Hence, x = 4, there being only one solution. (See § 112, p. 209; this is the case of "equal roots.") Since x = 4, y = '■'> from (1). The only pair of solutions is therefore (x = 4, y = 3). Check: fx = 41 . f4a: + 3y = 25] . f 4 • 4 + 3 • 3 = 25 U = 3j m i ,••-' + //- = 25 J glV6S [42 + 32 = 25 (correct), \ ■> V \i 11 1 *) <■' ^ V i\ 3] ,B Fig. 56. Figure 56 makes clear why there is but one answer. The equation (2) gives the same circle as before, i.e. radius 5, center 0. The straight line (1) goes through the point A : (x '= 0, y = 8|) and £> : (x = 6^, y = 0). It also goes through (x = 4, ^ = 3). If carefully drawn as shown here, it is seen that the straight line and the circle have only one point P in common. The straight line A B is said to be tangent to the circle. In drawing a graph, if the equation is not one we have already studied, it must be plotted out by points as in § 126, pp. 247-248. 260 SIMULTANEOUS EQUATIONS [Ch. X EXERCISES I: CHAPTER X Solve the following pairs of equations when a solution exists. Note in each case whether the quadratic in one vari- able obtained in the course of the solution has equal or distinct roots. Always draw the corresponding graphs, check the numerical results by substitution in the given equations, and note the interpretation of two, one, or no roots : *1. •3.1 ar + ?/ 2 = 34, x + y = 2. x 2 + y 2 = 25, x + y + 7 = 0. x 2 + y 2 = 169, . 5 x + 12 y = 169. #4 ff + ^=17, { x + y = 5. f5 jay = 12, x + y = 8. t6. ' xy = 9, _x + y = 6. f7 fay = 16, \x-y = 2. ta 2:c + 3y = 9, „2 t9- • §10 1^-^ = 8. ' a;?/ = 6, 3z-2y=16. ?/ 2 = 4 x, 'y = S. * See § 126, p. 248. t See § 124, p. 240. til. 1 12. x 2 -y* = $, 2x + y=7. 2 x — y = 6. 13 [9^ + 16^ = 25, 'l.T + 7/=2. 14. '9a; 2 + 16 y 2 = 25, . as + y = 0. 15 f.x 2 + ^=15, " 1 2 a + y = 8. 16. ( ^ + ^ = 21 ' I » — 2/ = — 1. 17 j*7/ + y 2 = -2, "U + 2y=l. .r y 4 as — 3 y = 1. a; + ?/ + .»# = 11, »-y = l. y 2 - 10 or = 15, y - 3 as = 2. J See § 131, p. 269. § See § 126, p. 248. 18. 19. 20. 129-180] ONE LINEAR AND ONE QUADRATIC 261 130. Ordinary Quadratic. The following special ex- ample leads to a new figure for graphical solution of quadratic equations in one letter. Ex. 1. f y = 2 x + 8, (1) t y = rf (2) Substitute for y from (1) in (2): 2 x + 8 = x' 2 , or, x 2 - 2 x - 8 = 0. Solving, we get x — — 2, or, x = + 4. If x = - 2, If x = + 4, y = 2 x + 8 = 2 • (- 2) + 8 = 4. y = 2 • 4 + 8 = 16. CAeci . (correct). 4 = 2(-2)+8 k2 fx = -2|. fy = 2* + 81 . f4 = 2(-2 { in \ give ^ fx = 4 ) fv = 2x + 8] fl6 = 2.4 + 8] i in { > give -I ^ (correct). b = 16j U=x 2 J & [16 = 4 2 J V The figure is as shown on p. 264 : (1) is a straight line (draw it) ; (2) is as shown (see § 95, p. 183). From the figure there are two points of intersection, P : (x = — 2, ;/ = 4) and Q : (x = 4, y = 16) ; these pairs of numbers are therefore solutions, as we found above. From this example it is clear that the solution of the quadratic equation H x 2 -2x-8 = gives the same values of x as are given for x by the simultaneous Pair y = **! y = 2x+ 8. . Likewise, the values of x found in x 2 +px + q = are also found from r .> y = x\ y = -p X -q, for, substituting for y, the last pair give the previous equation. The advantage of this graphical picture over 262 SIMULTANEOUS EQUATIONS [Ch. X those in Chapter VIII is that the curve y = x 2 is the same for all quadratics by this method. This curve being drawn once for all, nothing remains but to draw the straight line y = — px — q. Y 1 1 1 w J + r 1 m. 1 X- L- \ I J \ J ( I \1 \~Z Of i_ "t\ r i/j_ r ^t X i \ ~% ~£_ X_ ~~\ ~i_ r -X- f ~j K / B H - i L i \j i p\/ l "X 4 t\ -2 ± 7 1 J- A-t ±2 7 ° T J t :u X Fig. 57. Several examples may be drawn in the same figure, as follows : Solve the following list of examples graphically : (1 ) 2 x 1 - 9 x + 4 = 0. See p. 208. (4) x' 1 -4 x+ 5 = 0. (2) ar'-4a:+l = 0. Seep. 207. (5) .^+2 =0. See p. 211. (3) »»-4a>+4=0. See p. 209. (6) x*+2x+5=0. Seep. 211. 130] ONE LINEAR AND ONE QUADRATIC 263 la (1) it is desirable first to reduce the coefficient of x 2 to unity by dividing both sides by 2 ; this gives (1) x 1 -\x + 2 = Q. J (8) ~H A \-$t 1 - tltt H$- 4^ it-W- \- - -,-lut t 4UY t ttt-t i_ i: izx i fe J/DL It - Jo - (z| l " _ H&rf V(ti') * / \ L jf w v r jf F \ \ /// V \ L /// \ ' Wj \ I ill j C tj' \. \ / / \ \ h l\ll \ \ I \ ^ r w V \ Hill \ \ not Mo~ -; ^ ^ 'Jpf $ no j fiL (5) \ l\ 1 (5) ^7 Tf t n ri^ 5j "?// ^_ J#A / ( 7 V ' / - G > (pr /» 2 1 2 m - — mn - n = = 3, 4. 4. I 2 2 - t2w- o ° - o ?r = 38, ^. - 3 u = = 4. 5 fft + Z + fcZ: = 5, •-'. ' . 1 5 * - -21 = = 1. 10. The diagonal of a rectangle is 13 inches long. What are its dimensions, if one side is 7 inches longer than the other ? 11. It takes 2 hours longer for one pipe to empty a tank than for a second pipe to empty an equal tank. Both pipes together can fill either tank in 1 hour 20 minutes. How long would it take each separately to do so ? 12. The fence around the outside of a walk 5 feet wide sur- rounding a park lot is 340 feet long ; the area of the lot itself is 5000 square feet. What are the dimensions of the lot ? 13. What integer can be taken such that the sum of all integers from 10 Up to and including the chosen one shall be 1230? Solution. Let n be the chosen number. Then the sum is 10 + 11 + 12 + .» + (n - 1) + n = 1230. Simply reversing this, we get n 4 ( n - 1) + O - 2) ••• + 11 +10 = 1230. Adding, (10 + n) + (10 + n) + ••• +(10 + n) + (10 + n) = 2400. 266 SIMULTANEOUS EQUATIONS If there are t terms, we have t (10 + n) = 2160. But it is easy to see that 10 + t - 1 = n. Solving these two equations, as above, we find, n = 50, t = 41. 14. If we add together the numbers obtained on starting with — | and increasing by unity successively, where must we stop in order to have the sum 8 ? 15. Solve Ex. 9 of Chapter VIII, Exs. VI, p. 216, by the use of two unknowns. 16. Solve Ex. 18 of Chapter VIII, Exs. VI, p. 217, by the use of two unknowns. 17. Solve Ex. 30 of Chapter VIII, Exs. VI, p. 219, by the use of two unknowns. 18. Solve Ex. 54 of Chapter VIII, Exs. VI, p. 221, by the use of two unknowns. 19. Solve Ex. 51 of Chapter VIII, Exs. X, p. 233, by the use of two unknowns. 20. Solve Ex. 56 of Chapter VIII, Exs. X, p. 233, by the use of two unknowns. As suggested in Exs. 15-20, many problems solved in pre- vious chapters by means of one unknown may now be solved, in some cases more expeditiously, by the use of two or more unknown numbers ; it is recommended that many problems from previous lists be now solved in this manner under the guidance of the teacher. PART II. SIMULTANEOUS QUADRATICS 131. Simultaneous Quadratics. Two quadratic equations which are both to hold true are called a pair of simultane- ous quadratics. We can often solve such pairs by methods similar to those of § 129, p. 256, and §§ 86-89, pp. 159-172. The following examples will illustrate these methods : f.r + ?/ 2 =16, (1) Ex. 1. Given r * , )> [y=\ v- ( 2 ) Solve (2) for x 2 : x 2 = 6 y. Substitute y for x 2 from (2) in (1) : (3) y a + 6y = 16. Complete the square : y' 2 + 6 y + 9 = 25, or, y + 3 = ± 5. Hence, either, y — 2, or y = — 8. Uy = 2, Uy=-8, x 2 = 12, x 2 = - 48, and x = ± Vl2. x = ± v — 48 (imaginary). The real solutions are (x = + VT2, y — 2) and (x = — Vl2, ^ = 2). The other expressed solutions are meaningless to the student at present. They are (x = V — 48, y = — 8) and (x = — V— 48, y = — 8) ; we shall not regard them as solutions here (see Appendix, §§ 31-38). Check for real solutions: x = ± Vl2 1 . f * 2 + y* = 16 1 . f 12 + 4 = 16 1. fs2 + 2 ,3 = i ^t -/ \ ^t / \ A \ 7 \ T x ~ 7 \ ° T \ Qs^ — ^^u " J \& /|V •l\ • ^5_ 'S,,o _„> ' | \ 7 V ->/ V y ■j Fig. 59. The whole figure is as drawn. From it we note the two points of intersection, P and Q: P is (x = 3.5 (about), y = 2) and (2 is (x = — 3.5 (about), y = 2). These agree with the results above as closely as we could expect; in fact Vl2 = 3.464 ••• as will be found by the ordinary process for square root in arithmetic. The solution of this example illustrates the method of substitution. (Compare p. 170 and p. 256.) Ex.2. * + f = 25, tf _ f - 7. (1) (2) We proceed either as before, by substitution, or as follows: 131] SIMULTANEOUS Ql'ADUATICS 269 1 x 2 + f- = 25 1 x*-y* = 7 2 x 2 =32 x 2 =16 a; =±4 1 x 2 + y a = 25 -1 3?-y* = 7 2y» = 18 # 2 = 9 y =±3 The possible combinations are (compare example 1, pp. 256-257): f x = + 4 f x = + 4 f x = - 4 (x = -4 yl: ; J5 : ; C: J ; Z>: J „■ The figure shows these points clearly: equation (l)is the circle drawn before (radius 5, center 0) ; equation (2) gives the following tal >le : y ±0 ±1 ±V8 ±2.8. ±2 ±3 ±4 ±5 ±6 ±7 ±8 ±9 &c. X ±V7 ±VTT ±Vl6 ±V23 ±V32 x (reduced) ±2.66 ±3.3 ±4 ±4.8 ±5.7 and the figure is as shown in [(2) in figure]. The points A, B, C, D of intersection corre- spond to the pairs of solutions just found. Check for Jx = ±4) _ I y = ± 3 J ' i 42 + 32 _ 25, j 4 2 - 3 2 = 7. The solution of this illustrates the method of addition or subtraction, (Compare p. !(>:>.) Y \ a) t / (2J\ /m 1 3 \ j > \ 1 1 \ / 1/ 1 ; \ / \ / *- \ Fig. 60. It is always successful for pairs of equations of the type a.i- + by- = c. 270 SIMULTANEOUS EQUATIONS [Cii. X In solving exercises, ingenuity is sometimes required. Often, however, some simple process, usually the method of substitution or the method of addition or subtraction will suffice in the following exercises. (See also Chapter XII, p. 318, and Appendix, §§ 39-42.) EXERCISES IV : CHAPTER X Solve the following equations; always draw the figure: 1. 3. y = 5 X 2 + 3, x 2 + 2y 2 = 129. a 2 + b 2 = 85, a 2 - b 2 = 77. x 2 + 2 xy = 55, 2 x 2 — xy = 35. 8. [Sr<;<;F.sTiON. Solve first for x 2 and xy.~\ ( 3 t 2 - 5 at = 42, 4 ' {ot 2 -4 at = 161. 10. p 2 + ^ with the first equation.] 12. \ { 2 ^ + 3 aw + 4 - m 2 + n 2 = 29, mn = 10. + 3 aw + 4^ ==9. [Suggestion. Eliminate the terms in xy.~] 132. Raising the Graph Vertically. It will often hap- pen that there will be no solution. Examples of the various possibilities follow in § 133. 131-432] SIMULTANEOUS QUADRATICS 271 Let us first notice the effect of writing (1) 2/=J.'---r£ in place of 1 ->-2 (-') y in example 1, § 131, where k is some fixed number. If we wnute (3) y = \x* + % for example, the curve in the figure will be just 2 units vertically above the curve for equation (2), since each value of x makes y just l! units greater. The figure is drawn for the equations (2), (3), and also for (1) y = \x n - + b, (5) y = ±x 2 -5. There is a curious optical illusion about such figures, which may deceive the student; although they may not seem to be so, these curves are the some shape and size, and any one is formed from any other by simply raising it or lowering it. The student will convince himself of this by cutting out a piece of paper to fit in one of them and then raising or lowering it. Y Fig. 61. 272 SIMULTANEOUS EQUATIONS [Ch. X 133. Various Possibilities Illustrated. Consider now the following examples : * 2 +?/ 2 =i6, (l) r^+?/ = i6, (i) y=K- (2) ' U=i^+io. (2) n {•- + r = i6, (i) [rf+j^ie, (i) ly=ia- 2 + 2. (2) ' l y= i a .2_4. (2 ) x 2 -f-/ = 16, (1) r* 2 + .r = 16, (1) 2/ = io; 2 + 4. (2) L * ly^rf-S. (2) Iy (* + »■« 16, (1) ;ar + 2/ 2 =16, (1) [y = l* + B. (2) ly^aF-lO. (2) These eight examples are all very similar to example 1, p. 267 ; and they may be solved in the same way. But a glance at the figure shows that Exs. I and II have two solutions each (A and B, and C and D, in Fig. 62) ; Ex. Ill, only one solution (E in Fig. 62) ; Exs. IV and V, no solutions; Ex. VI, three solutions (F, G, H, in Fig. 62) ; Exs. VII and VIII, no solutions. For the equation (1) is the same in all of them, while equation (2) differs; in the figure, equation (1) is a circle of radius (I) about 0, while equation (2) is the curve of § 131 moved upward or downward. The student will see just when there are solutions by cutting out a piece of paper the shape of the curve and moving it up or down. Let us solve Ex. VI by the algebraic methods to check these results: In Ex. VI put x 2 = 6 y + 24 from (2) in (1) ; then f + 6?/ + 24 = 16. Solving this equation, we find y = — 4 or j = - 2. If y = - 4, If y = - 2, _ 4 = i x 2 _ 4 f r o m (2), - 2 = \ x 2 - 4, or x 2 = 0, or x = 0. or x 2 = 12, or x — ± VT2. _,,.. f x = f x = + VT2 f x = - Vl2 1 he three answers are \ ; ; i ; b=~4' U=-2 U=-2 which are the points marked F, G, H, respectively, in Fig. 62. Each of the other examples may be solved in a similar manner; the results will he found to agree with the graph. SIMULTANEOUS QUADRATICS 273 . .v.. 1 i.u IV. \ \ ^p \\ ' 1 \ \ 1 f //(in. \\\ \ ' //" ml) 1 fill \ tin 1 " 1 ////i/i (IX!) , //// 7/ v i- V »' // // JL y I/) III II A 1 il j! ) \ _\ /j ill ;> ii .) L_V V_V\ \\ tt V \ \ \ \\ V 7) / \ \ \ \\ V / / A i // / J h \) j V^ 7/7/ III \ ////- 1 V F v w_ / // / / / ftxjio HH \ \ Vv / \\ vo# /!/'/ \\ \ , / •// / ' '" hill .L^ r b. ' i \\\ \ "-o^. /I/ / A xn > V \ \ > \ \ V^ ■rx \ J J1 J \ V\i \ u }rfK // 7 V! // 1 III \ \\ ' V t2, III \ "AY* \ ^SvL -xT/f / \ / -\ /. \\ \ \^ "e sS^ ssHz 1 AI / i\ v? sU /•> 1 \ \^j > / y \ / 1 / i Y \\ ?_ ■} y / \i \ -£ 1 ', / -/^/ IP t^ _L ** V / V ^ y \^ // / i w ■ i (VIII \^ L X V v L \ ^^^ \/ \ \ V s XI. ._ v., 1 y^sxi. i i i 1 1 l Fig. 62. 274 SIMULTANEOUS EQUATIONS [Ch. X In the same figure we have drawn the new circle x 2 + f - 100, which is a circle of radius 10 about 0. This shows some of the details lost because of small size in the other figure. The figure using the larger circle shows graphically the solutions for the follow- ing examples : v (a? + y* = 100, vn , f*» + y» = 100, y = %x>. ly = \x i —£. n , f x- + f = 100, vnr l .r + f = 100, {y = \x> + 2. U = i^-10. ni» {^ 2 + 2/ 2 =ioo, IX , j^+2/^ioo, IV , 1^+^=100, x , r^+r = ioo, y=\x 2 + 5. \y = $a?-15. x 2 + y- = 100, XI , f x 2 + / = 100, y = ^ + 10. ' l^i^-lSi VI r fa- 2 +/=ioo, xir r^+2/ 2 =ioo, These are seen by the figure to have the following number of solu- tions (which can be judged approximately from the figure): 1', two (at A' and B') ; II', two (where in Fig. ?) ; III', two (where in Fig. ?) ; IV, two ; V, one (at /) ; IX', none ; VI', two at G' and H' ; VII', two ; VIII', three, atL,M, N; X',four, at P, Q, R, S; XI', two, at T, V; XII', none. Of these the most remarkable are X' with four solutions, and XI' with its curious two solutions. There are similar ones with the small circle ; but they cannot be seen readily in the figure because of the small size. The position in XI' is determined as follows, so that there shall be just two roots at T and U. We try y - J a: 2 + k and solve : (1) (x* + y* = 100, (2) {y = lx* + k. From (2), x* = 6 y - 6 k. Substitute in (1) : y 2 + 6 y - 6 k = 100, or, y 2 + 6 y + ( - 100 - 6 k) = 0. 133] SIMULTANEOUS (.U'ADRATICS 275 The condition that this equation should have equal roots is that b- — 4 ac = ; we have a = 1, h — 0. c = — 100 — 6 k ; hence, b 2 - 4 ac = 36 - 4 ( - 100 - 6 k) = 0, or k = - 18£, which was the value taken in XI'. It is to be noticed that the value of ij may be real although x is imaginary, and via V( rsa, and as in the first case solved (p. 267). The student will find other possibilities by cutting out the figures shown above and moving them around across each other. Some of these possible positions correspond to fairly complicated pairs of equations. Other curves may be tried, as suggested in the exercises below. A complete study of the possibilities — at least a full understand- ing of them — cannot be hoped for here. It is only after a study of Analytic Geometry, in which such questions are discussed at length, that the student will really appreciate all that is involved. A few special rules are sometimes given ; these are really not particularly useful. See Appendix, §§ 39-42. EXERCISES V: CHAPTER X Draw figures for Exs. 1-5, note the number of solutions, esti- mate their numerical values, and solve algebraically: 1. x~ -y- = 16, tf- = 2x-16. (No real solution.) f .r — >/- = 16, 3. \ , ' I ir = 2 x. $ -f- = 16, y°- = 2x-S. (One set of solutions.) 4. \ I f = 2 x + 8. x 2 — y 2 = 16, [y 2 = 2x + 16. (How many sets of solutions in each case?) Similarly the following exercises (6-8) 6 - 1 la f x 2 - if =16, + tf = 9. 7. .r-r = 16, x 2 + y 2 = 16. Similarly the following exercises (9-16) 9. x 2 + y 2 = i9. 10. 8. x- — y= 16, x 2 + y 2 = 25. ?/- = !, (x-o) 2 + y 2 = 19 276 SIMULTANEOUS EQUATIONS [Ch. X (a - 6) 2 + f = 49. I (x — 12. 13. x 2 -y 2 = l, (x-iy+if = 4Q. yf 2/ 2 = l, (x-8)* + y 2 = 49. 14. 15. 16. x 2 -y 2 = 1, (x _ 10) 2 + f = 49. ' a* 2 — y 2 = 1, . ( x _ 15) 2 + 2/ 2 = 49. ar - ?/ 2 = 1, ( >T _20) 2 + ?/ 2 =49. Treat the following similarly, except that an algebraic solu- tion need not be obtained ; instead, check the estimated results by substituting in the equations : 17. xy = 6, ,y = 3 x 2 . 18. .y = 'Sx 2 -9. 19. xy = 6, y = 3 x 2 - 18. 134. Graphical Solution. Graphical methods of solution have been mentioned above in practically all cases as a con- venient check and as a method for finding approximate answers. There are many problems so difficult that no method except the graphical one is really convenient, or even possible, with the student's present knowledge. Such is the example below. (x 2 + y = 7, (1) \x + y 2 = ll. (2) Ex.1. This pair of equations defies all the methods of elementary algebra except the graphical method. Equation (1) may be written in the form (1) y = - x 2 + 7. A table of values of x and y is : X ±0 ±1 ±2 ±3 ±4 ±5 1 ±7 1 » ±9 ±10 etc. V 7 6 3 _ o -9 -18 133-131] SIMULTANEOUS QUADRATICS 277 and the figure is as shown in Fig. 63. It is the same as the curve of Fig. 31, p. 183, turned upside down and then raised vertically 7 points. Equation (2) may be written x = - >f + 11. A table of values of x and y is : X 11 10 ± 1 7 ±3 — 5 ±4 - 14 ±G ± 8 etc. !l ± ±2 ±5 ±7 and the figure is as shown. It is the same as the curve of Fig. 31, p. 1S3, turned on its side (i.e. turned through 90°) and then moved horizontally 11 points to the right. Y 1 r \ v Bl r L 1 (2 > i \ A' / \ y / ' ID ! / -5 U)a--+ 2/=7 \ (2) a • 1- .'/ -=n . Fig. 63. These curves are seen by the figure to intersect in four points, which, if the figure is carefully drawn, are seen to be about f x = 2 f x = - 1.8 ( x = ?>.l f x = - 3.3 r : U = 3 ; B: b = 3.6 ; C: l,/=-2.8 ; ^ : U=-3.8 The results for /?, C, D are not, of course, precisely accurate; the answers can be found to any number of decimal places, however, by drawing the figure on a sufficiently large scale on a large sheet of paper. The exact results to four places are : x= -1.8481 f ar = 3.1313 (x= -3.2832 7: J y=3 B: C +3.5844' ' [y= -2.8051 D, y = -3.7794 278 Ex. 2. SIMULTANEOUS EQUATIONS 0^+^ = 25, y=x 3 . [Ch. X (1) (2) La ^ / / 1 ri) / J r I \ \ \ 'p N l D 1 Fig. (i4. The figure is as shown (Fig. 64); (1) is a circle of radius 5, (2) is the cubic curve drawn on p. 189. A table of values for (2) is : X IJ 1 1 2 3 4 64 etc. -3 5 7 etc. - 1 _ o etc. 8 27 -27 125 343 - 1 - 8 Ex. 3. The solutions are seen to be about (a: = 1.7, y = 4.7) and s x _ _ i7 j y __ 4j). They can be found more accurately from a larger figure. y = *, (1) y = 5 X - 2. (2) Equation (1) is the cubic curve drawn above. Equation (2) is a straight line through (.<• = 0, y = -2) and (x = 5, .'/ = 23). The. figure is as shown (Fig. 65), with ty small spaces as unit ver- tically, and one large space (5 small spaces) as unit horizontally; this is convenient, as will be seen by trying to draw the figure. The solu- tions are (approximately) : '•{;::}» •■(;:!)■ <"•-" 134] SIMULTANEOUS QUADRATICS 279 It is interesting to notice that we have really solved the equation (3) x 3 - 5 x + 2 = 0. For, if we subtract the sides of (2) from those of (1) respec- tively, we get equation (3). Hence, the values of x just found are the solutions of (3). (Coin- pare p. 261.) Another method of solving (3) will also be given. We note by trial that x = 2 is a solution. Hence, x — 2 is a factor of the left side (see § 117, p. 223, and Appendix, § 5). Actually divid- ing x 3 — 5x + 2 by x — 2 we find, x 3 -5x + 2=(x-2)(x* + 2x-\), hence, (3) is the same as (x-2)(a: 2 + 2x-l)=0, whence, either x - 2 = 0, or x 2 + 2 x - 1 = 0. That is, either x '., or x = - 1 ± V2. Since V2 = 1.1142 • •., the answers are J J 1 / 1 J f t J / / / / '. 1 1 If ¥ 3 i \ . 1 ft , f j i I j j 5 .' j j I 1 j ^~F t f ■* i / / 7t / / i I < I l\ i / T /;// ,, 3 -3" -2 ~k~J~ i MA / SCALE- . j J i i t t- jt H -t 3 t 1 r t 3 - -Jt --& - -6 .-- T -i 1 4 t t ■ - - zi2 M-A / -,- 1/ =8 — 1- z 3 / -r + J -,' n9 r i ,. t__c 3.1ft it /- M- ± z 7T i" r W- / / _L 7 t = X!i IT ±_i_ MU ± rr _1 & in x - ^ :I :r__ 3/T -Pf _ 2° j£ ^i3 -JA 23 _L IZI _ - _M X Fig. 05. x >, or x = + .4142 ..., or x = - 2.4142 -, which are more accurate values than those found by our figure. Subtracting these values in (2), we find y=S, or y = + 0.0710 ..., or y = - 14.0710-..; hence, the more precise answers for the simultaneous equations (1) and (2) are : .4 r x = 2 U = 8 B x = .4142 1 y = .0710j C: x= -2.4112 y = -- 14.0710 280 SIMULTANEOUS EQUATIONS [Ch. X EXERCISES VI: CHAPTER X Solve graphically : 3. 4. x 2 + 3y = 7, f + 3 x = 1. { ah = 21, \b = a 2 -2. y + z = 2, z 2 -y 2 = 16. ^ + 2^ = 25, ay =12. \p 2 +pg = 30, p — 5 (f. 8. x 2 + 2/ 2 = 100, 2/ = ^. m« = 10, ?/ = m s — 3. I z = * + 6. 9. .r 3 — 7 a; — 5 = 0. 10. £/ 3 -'°>y + 2 = 0. 11. a^-a 2 — 2 = 0. REVIEW EXERCISES VII : CHAPTER X Solve the following, graphically and analytically : 1. 3. z = k 2 + 7, z 2 - 25 fc 2 = 31. | a; + ?/ + xy = 13, I a* + y2 + ajy = 43. I 2 + 3 Zm = 10, Im — m 2 = 1. w 2 + uv + v 2 = 7, 2 v 2 — 3 ?r = 5. [ r 2 + s 2 = 45, 1 + 1 = 1 r+s 2' J/ 2 = 5x, x 2 + 7y 2 = 36. 8. 10. 11. 12. 13. J/ 8 = _ 5 a?, a 2 + 7 2/* = 36. .6- + 2/ + 3J 2 = 4, a? + 4a!2/ + 2y 2 = 17. x*-y* = 7, xy = 12. mn — w 2 = 6, m 2 — mw = 15. a 2 -ft 2 = 32, (a-3) 2 +6 2 = 13. a 2 + /r = 25, (a-5) 2 +(&-3) 2 = 5. x 2 + y 2 + x + y = 98, x 2 -y 2 + x-y=U. 134] REVIEW 281 Solve graphically : ar + ?/ = 54, 16. x" -39 x- 70 = 0. x + y- = 32. 17. x* - 9 x 2 - 4 x + 12 = ( i. ar - y 2 = 15, a^ = 56. 18. s 3 -s 2 _4s + 5 = 0. 15. 19. A 24-foot rope is exactly long enough to surround a right triangle, whose hypotenuse is 10 feet long. How long are the other sides ? 20. The diagonal of a rectangle whose area is 60 square inches is 13 inches long. What are the length and breadth of the rectangle ? 21. I find that I can walk half a mile farther in an hour than John ; also it takes me half an hour less to walk 14 miles ; how fast do we each walk ? 22. The radius of curvature of a concave mirror is 15 cm. How far from the mirror must an object be placed in order that its image be distant 20 cm. farther ? (See p. 220.) 23. A loop of twine 30 inches long is to be stretched over four pegs so as to form a rectangle whose area shall be 44 square inches. What must be the sides of the rectangle ? 24. The radius of curvature of a concave mirror is r. Where must an object be placed in order that the center of the mirror shall lie halfway between the object and image? (In solving graphically, choose any convenient value for r.) 25. Find two numbers whose sum is 27, and whose product is 126. (See also p. 224.) 26. Find two numbers whose difference is 19, and whose product is 216. 27. In order that an object 20 cm. farther from a concave mirror than its center may produce an image 30 cm. from the mirror, what radius of curvature must be chosen, and where must the object be placed ? 282 SIMULTANEOUS EQUATIONS [Ch. X 28. A can fold 3000 advertising circulars in three hours less time than B. The two working together can fold 7500 of them in five hours. How many can each fold in one hour? If there are n numbers, the first of which is a and the last /, such that the differences between consecutive pairs are all equal to the same number d, then it is known (see Chapter XIII, p. 324, § 157) that I = a + (n — l)d, and that, if s represent the sum of the n numbers, «=jj(a+0. 29. A row of numbers is written down, beginning with 3, such that each number is obtained from the preceding by add- ing 4. If the sum of the numbers is 55, find all the numbers. Here a, d, and s are known ; we therefore have one linear and one quadratic equation to determine n and I. When n is known, all the numbers can at once be written down. 30. The sum of all the integers from 100 up to a certain integer is 2800. What is this integer ? 31. The sum of all the odd integers from 1 to a certain num- ber is 324. What is the last odd number of the list ? 134] SUMMARY 283 SUMMARY OF CHAPTER X: SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS, pp. 253-282 Pakt I. One Linear and Onk Quadratic, pp. 253-266. Figure for one Linear and one Quadratic: straight line and curve as in Chapter IX. Answers: pairs of values at points of intersection; approximate answers from figure; check on algebraic solution. § 127, pp. 253-255. Definitions : degree found by adding exponents of unknowns. § 128, p. 255. Formal Rule for Algebraic Solution : essentially, solve the linear equation for one letter and substitute this value in the quad- ratic. Exercises I (§§ 127-129). § 129, pp. 255-260. Second Graphical Method, Ordinary Quadratic : equivalence of roots of x' 2 + px + q = to roots of J ^ ~ I ; list of exam- [ y = - px - q J pies in one figure. Exercises IT. § 130, pp. 261-265. Review Exercises for Part 1, Chapter X: Exercises III. pp. 265-206. Part II. Simultaneous Quadratics, pp. 267-282. Simultaneous Quadratics : Figure — two curves; answers — pairs of values at points of intersection; algebraic solution illustrated — methods of Chapter VI. Exercises IV. § 131, pp. 267-270. Possibilities : possibility of no answers ; effect of raising a curve vertically ; all cases shown — no answers up to four sets of an- swers. Exercises V. §§ 132-133, pp. 270-276. Necessity of Graphical Solution : failure of algebraic methods ; re- course to approximate graphical solution. Exercises VI. § 134, pp. 276-280. Review Exercises for Part II, Chapter X : Exercises VII. pp. 281-2*2. CHAPTER XL RADICALS; FRACTIONAL AND NEGATIVE EXPONENTS PART I. OPERATIONS; FRACTIONAL AND NEGATIVE EXPONENTS 135. Essential Rules. We have seen (§ 105) that roots may be operated upon in the fractional exponent notation (§ 102) whenever the roots can be found otherwise. The student should read again and state the definitions of § 94, p. 181, and § 97, p. 186; and he should review §§ 98, 99, 102, 103, 11)4, 105, with great care. A few important statements are now repeated in some cases in greater detail. A rational fraction is the quotient formed by dividing one integer by another. A rational number is an integer or a rational fraction. Any number that is not rational is called irrational. (§ 97.) All rational and irrational numbers are called real numbers. A radical is any expressed root, whether rational or irrational, and any expression that contains a radical is called a radical expression. The index of the expressed root is the degree of the radical. Thus, V'J or 2*, Vi or 4^, Vb 2 - 4 ac or (b 2 - 4 ac) * are radicals of the second degree, or quadratic radicals; Vo or 5», Vx 2 or (x 2 ) 3 , are radicals of the third degree, or cubic radicals; 3 — V2, 2 Vo — 1, x + vx 2 are radical expressions. A surd is a radical that is irrational. A surd expression is one that contains surds; such an expression is said to be surd. Any radical expression is either surd or rational. 284 FRACTIONAL AND NEGATIVE EXPONENTS 285 Thus, of the radicals mentioned above, V'2 and v.i are surds; VI is not a surd : \ .;- and Vb' 2 — lac may or may not be surds according to the values of the letters: if x — 2, 1/x' 2 = y/i (surd) but if x = 8, ^2=\/6T=4 (not surd); in Chapter VIII, § 114, the radical expression — — — is used extensively; in some examples 2 it it is a surd expression, in others it is not. (See § 114, p. 212.) In this Chapter, we shall not deal with even roots of negative numbers, which are often called imaginary num- bers. (See § 94 and § 113.) The definitions given above exclude such roots. (See Appendix, § 31.) We shall frequently use the rules of § 10-1 : I. (x m y = z'"- n . II. 3? " X X" = X m+n . III. x n xy n = (x>yy- i III a. ^ = f-X. These rules were proved for simple powers only, i.e. for positive integral values of m and n. (See, however, § 105.) Let us now try to follow out the work already done by extending our notation as in § 102 ; and in doing so let us observe the rules I, II, and III, as well as the rules of p. 35, if possible. 136. Meaning of Fractional Exponents. The student already knows the notation of § 102 : (1) x^=^/x; X 3/~~ (2) X s = Vx, etc.; or, in general, i (3) x n = m -Z/3 a'b* ' -V{y-z)(z-x){x- y) 15 27 ays . 18 3& 2 -26c + < t V3^z (36 + c)V6-c lSnwi-^ 19 (1+V2) 2 . 16. ■v/5 m?n 2 pr 20. (5-3V6) 3 . 137-138] FRACTIONAL AND NEGATIVE EXPONENTS 289 21. (7-Vo) 2 . 25. (x + yVz)(x\/z — y). 22. (V7-V2) 2 . 26. (2 + ^4)(^2-3). 23. (3 + 2 V5)(2 - 3 V3). 27. (6 + V5)(l+V2). 24. (3 -a/2) 3 . 28. (2-V2) 2 + (3 + V2) 2 . 29. (2 + V7)(l-3V7)(5V7-19). 30. (6-V3 + V2) 2 -(l + 2V3-3V2) 2 . 138. Addition; Similar Radicals. Radicals can be added together (or subtracted) only when they involve the same root of the same number. Thus, 2 V3 + 5 V;> = 7 V3; but 2 y/S + 3 Vo must remain as it is. We can in any case, of course, find approximately the value of the radicals, and add these approximate values. Thus, V3== 1.732 (approximately) and Vo = 2.236 (approximately); hence,2 Vd + 3 Vo = 10.172 (approximately). Radical monomials whose radical factors are the same roots of the same quantities are called similar, otherwise they are dissimilar. By § 137 many forms which appear dissimilar may be made similar. They may then be added. Ex. 1. Add 7 V3 + Vl2 - (75)1 7V3 + Vl2 - (75)4 = 7 V3 + V4 x 3 - V2o x 3 = 7 V3 + 2 V'3 - 5 V3 = 4V3. EXERCISES II: CHAPTER XI Simplify the following expressions: 1. 3V98-V50 + 2V8. 4. 4^54 -^250 -^16. 1 rt /H rt OvJ 2. V300 _ 3 V27 + 5 V12. 5. (256)3 - + 2(4)4- 2(108) ' 3. 2(54)*+4(2.1G)*-(150)*. 6. 7 Vio- 2(20)* + V2ooo. 290 RADICALS [Ch. XI 7. VaW-VW + V6. 9- ^54^- ^16^+ ^250^. 8. 6V8¥-V32l-+(162)U-. io. V^ + VWtf-itey 2 ^. 11. VO- 2 -7x + 6) (x - 1) + V4 a; - 24. 12. Var 3 + 3 a; 2 + 3 x + 1 - Var + a; 2 . 13. Var 5 — 6 a; 2 + 9 x — -Jx' + V9 x. * 139. Reduction to Different Degree. Consider the ex- ample V25. We have or, by Rule I, 121 a/25 = (5 2 ) ? = 5 1 = 5* = V5, which is correct if Rule I is to hold. p*n In general, (\/>) = z« = x*' n = *V^% which enables us to reduce radicals to different degrees. This rule will appear more reasonable if we notice that, on raising each side of xi = x qn to the power qn, we find \.c*) x= Kqn or a^" = xP" by Rule I ; but the last result is surely correct. If we agree to take the positive root when there are two possible values, we may re- turn from the last result to the preceding rule, which is then justified. By this new rule we can reduce radicals of different degrees to radicals of the same degree. Ex. 1. Multiply V3 by -y/2. V3 = 3* = 8$, and $2 = 2* = 2* ; hence, VSx\ / 2 = 3* x2«= (3 3 )' x (2 2 )^ = (27 x 4)* = (108)*, or v^108. 13&-139] FRACTIONAL AND NEGATIVE EXPONENTS 291 Ex. 2. Multiply ^/3aF 2 by Vtfb. 4 ) *, and Va*& = (a 8 6) * = (a 8 />) ' = («»6») *. Hence, \/3~aP x Va% = (9 a 2 6 4 ) ■" x (a 9 6 8 ) * = (9a u & 7 )* = (a 8 & 6 X 9a 6 />)* = (a% 6 )*x (9a 5 6)« = a6(9« 5 6)^ or aft\^9a«6. Ex. 3. Simplify V2a /2 = \/\Al+ = 1.12H Ex 5 ^^ -= W* . = («&)* = («&)* x ah* = (cfb^xjtfb 2 ,* _ (a 7 6 5 )^ _a(a6 g )i_(a6 5 )^ nr 1 6 a*6* CAec*; Let a = 6 = 2; ^L = |= 1, and^ V^=l. 2 = 1. Vab* 2 6 2 B-l n~\ In general, ,4 ix(?" AC n AVC' 1 ' 1 ~D-!}//~i i. "— 1 n £C A radical expression is said to be in its simplest form when there are no radicals in any denominator and when each radical in the numerator contains no factor that might be taken out. 294 RADICALS [Ch. XI EXERCISES IV: CHAPTER XI Rationalize the denominators of the following; check the numerical ones by calculating the value to two or three deci- mal places before and after rationalization : ! J_. , V6kz 5 A. 7 6a& 3- r • J - 3/- '• V2 ' V7^- V3 Vl8a 2 1_ 4> S^5. 6 xyz m Q _*-£ i/2 Vtx 2 wjjw VW^y) cr + 2ab + b\ 2m 2 + 3mn + n 2 -\/3(a + &) 2 Vm + nV2m + n xy — xz 10. ,_ „ , 12. -y/x -yy — z 13. ^ a 3 -l J/ a 2 + a + 1 • V «' 2 - 2 a + 1 142. Rationalization of Binomial Quadratic Denominators. When the denominator is not a monomial it may be very difficult to rationalize it. We shall treat only the case of simple binomial denominators, involving square roots. Ex.1. _j_ = , 1X(2-V3 L = 2-V3 = 2 _ V 3 J 2 + V3 (2 + V3) (2 - V3) 4-3 Check: V3 = 1.732+ ; hence, — r— r^ - — 0.28 - , 2+V3 3.732+ and 2 - V3 = 2 - 1.732+ = 0.28- These examples result surprisingly; it is very clear that the riual result is much simpler than the given form. The general principle upon which such examples are done is that Q~Va — V7> )(Va + Vj) = a — b. The radical binomials V« + Vo and va — V7; are called conjugate if a and b are rational. Their product, a — b, is rational. If the denominator is the difference of two quantities, one of which is a radical, multiply both numerator and 141-Hi'] FRACTIONAL AND NEGATIVE EXPONENTS 295 denominator by the sum of the same two quantities ; and vice versa. In general, the proper multiplier is the con- jugate of the given denominator. Ex 2 V« + V& = (Va+V&)x(Vct + V6) = a+b+2Va b Va-Vb (Va-V&)x(Va+V6)~ a- 6 Check : Let a — 9 and & = 4; then Va+V^ = 3 ± 2 = 5) Va - Vfi 3 - - and q + ft + 2v^ft _ 9 + 4 + 2-6 = 25 _ 5 a- b 9-4 5 " Ex 3 x ~ "^ = ( x ~ v '^ x ( A ' ~ ^^ — x2 + y — 2 x Vy . x + Vy (» + Vy) X (a; — Vy) ** ~~ # Check : Put ar = 3, y = 4 ; student carry out. Take care not to alter the value of the given expression. EXERCISES V: CHAPTER XI Rationalize the denominators of the following; check the first seven by calculating numerical values to two decimal places before and after rationalization : __2 „ V2+V3 V3-1 2 + V3 2- V3 V3+2V2 V3-V2 13. 14. 15. 1. V2-1 1 4. 3 + 2 V2 1 + V2 3 - 2 V2 10. 1 5. 6. Vic + V X - -1 ii. Vz-VT= m Vr+-VT= m V* + jf + x — y 7. 2- V3 8. 2+ V5 2V5-3 9. 1 + 2V5 1- V2 my/ a — ?iV& m Va + wV& a^ r 1-Vl -xy m x + y — Vx* + y 2 y — V# 2 — 2 wias 296 RADICALS [Ch. XI 143. Exponent Negative or Zero. It is convenient to use negative numbers as exponents in the following manner. Since, by Rule II, . x m x x n = x m+n , it follows that x n x r or, on putting m + n = r, n = s, — = x r *• (See p. 76.) Us This rule works perfectly when r is greater than s. Thus, x 5 -=- X s = a,- 5-3 = x 2 , (see p. 76) ; and x z -r- x s = x 2 3 = a; 6 , which is correct, 112 3 4 2 1 n X^ __ X* • X 3 _ X 6 - X s _ X 6 _ X • X 6 __ 1 X 3 X 3 • X 3 XXX Even in the last example the new rule is the easier. When we come to examples in which s > r, we are led to write down negative exponents thus, by the rule above : ■ /y*u O . — . /v» — • JU but ^ = 1- X 5 X 2 hence, we say that x~ 2 means — , in order that the rule above may remain true. In general, x 2 x n X 2 " by the rule above ; but ryn— 2rt __ iy~n x n Jin ~w ' V X 71 hence, we say that x n means — . of 1 Any letter with a negative exponent is equal to the recip- rocal of the expression formed by changing the sign of the exponent. 143-144] FRACTIONAL AND NEGATIVE EXPONENTS 297 Likewise, if r = s, we find ^ = x r ~ r = x°; x r but - = 1; x r hence, we say that x°= 1. Any number with the exponent is equal to 1. In this statement 0° must be excepted, for - ■ has no meaning. The expression 0° therefore has no meaning. 144. With this understanding we can work certain problems more quickly. It is to be remembered that x 1 =x. Ex. 1. ^3 (Ex. 4, p. 293). V2 i,- i -\/4 4 s 2 1 2 _1 2_I 1 I! = t = 2 s x- = 2 5 x2 2 = 2* * = 2*. (See p. 293.) V2 l\ 2* Ex.2. 4 « W a 3 /> 2 4 a 2 i 3 c° 4 a 2 fr 3 x a-%- 2 = 4 a 2 - 3 & 8 - 2 , since c° = 1, 4a" 1 & a 8 6 2 46 a Ex. 3. ^H ( S ee Ex. 5, p. 293). 12«^6 a c*p- 1 g- a > , Perform the operations indicated : 16. 5 afy-V x 3 x~ 3 yz- 4 . 21 (x-hfz~%) 5 . 17- _ 3 V 22. (o-^-'ct-*) *. 18. 2ajV»~ i x3a5~V 2 » 2 - 23 - ^^s^rl 19. (30x 2 //-^)^-(5x- 2 //-^). 24. (x*-y-*)\ 20. 7a 8 6" i c*x5a*6 2 c- 1 x2(a6c)"i 25. (a* + a~^) 2 . Change from the sign V to fractional exponents ; then simplify : 26 v^yvi . 28 . ***** Vx-yz J/r-\s~ l ty 27. — 29. 3 , / • 144] REVIEW: PART I; OPERATIONS 29t> REVIEW EXERCISES VII : CHAPTER XI Simplify : 1. . 3. Vz-2=14-a\ 11. 2V2z-3=z-4. 4. \Ac - 2 = x - 14. 12. Vfc 2 -3A7-6 + 8 = 2£. 5. x + V3 x - 2 = 4. 13. V2 x 2 + 5 x + 2 + 2 = 2. 6. 2 ? / + V*/ 2 + 7=-2. 14. V3 ? / 2 + y + 2 = //+2. 7. a + V2 a — 1 = 8. 15. V» + V» — 7 = 7. 8. 3v-Vv 2 + 3'u-l = 3. 16. Vj) + V2/>-2 = 7. * In most of the exercises below, the radicals that occur are of the second degree. In exercises in which radicals of higher degree occur, raise both sides to the corresponding power. 146-147] RADICAL EQUATIONS 305 17 Vp + 3 + Vi>-2=5. 19. y/a?+x + 6 = V» 8 -9. 18. V3s + 1 -V2s-l = l. 20. V«+l + Va;-4 = Vaj + 17. 21. V3z-2 + Vz^l-V4z-r-l=0. 22. y/a? + 7 = x + l. 24. Va? — 3 + v » + 3 = V2 a*. 25 V2 a; + 5 + V.c — 1 _ ^ 23. VaJ 2 + *-V^=0. • V 2^T5-vaT=n: 147. Square Root of Surds. The preceding exercises were especially selected so that the answers should not be surd. Otherwise we should have to find the square root of a surd in checking the problem. Ex. 1. (1 ) V2 x + 4 - -y/x - 1 = 2. (Compare Ex. 2, p. 303.) Solving as above, the student will find x = 7± V32 = 7 + 4 V2 or 7 - 4a/2. Cftecifc: Vl8 + 8V2 - V6 + 4V2 = 2 (for x = 7 + 4V2), Vl8 - 8V2 - Ve - 4V2 = 2 (for x = 7 - 4V2). We must then find, for example, V 6 + 4 V2. The answer is 2 + V2, for (2 + V2) 2 = 2 2 + 4 V2 + (V2) 2 = 6 + 4 V2. But such an answer cannot be found by inspection. To find it directly suppose (2) V6 + 4 V2 = Vr + V.7, where r and s are two unknown integers or fractions (i.e. rational numbers, hence not themselves surd, p. 284). Squaring both sides, we get (3) 6 + 4V2 = r + 2\/n + s. 300 RADICALS [Ch. XI If we can find values of r and s so that (4) r + s = Q, and (5) 2 Vrs~ = 4 V2, these values will satisfy (3), since (3) holds if (4) and (5) both hold. Square both sides of (4) and both sides of (5) ; then subtract ; we find (6) r 2 - 2 rs + s 2 = 4, or, (7) r - s = ± 2. Solving (4) and (7) as linear simultaneous equations, we find (r = 4, s = 2) or (r = 2, s = 4) ; these satisfy (4) and (5), hence they satisfy (2), as will be found on trial. Substituting in (2), we have V 6 + 4 y/2 = n/4 + y/2 = 2 + V2, which is the value used above. 148. Equality of Surd Expressions. In § 147 we re- placed equation (3) by equations (4) and (5). Let us show that this is surely correct. Transpose (/• + s) in equation (3) : [6- (r + s)] +4 V2=.2V7i. Square both sides : [6 - (r + s)] 2 + 2 [6 - (r + s)] • 4 V2 + 32 = 4rs. We can now solve this equation for V2 unless its coefficient [6 — (?' + *')] is zero ; if we could solve we should get V2 = 4rs-r6-(r + s) ]g-32 8 [6 - (r + *)] where the right side contains only rational expressions. This is absurd, for \/2 is not rational. (See p. 1S4.) The only escape from i his absurdity is that the coefficient [(5— (r + *)] mentioned above should be zero : 6 _ (r + ,) = , or r + s = 6, which is (4) of § 147. Subtracting this new equation from (3), we get 2Vrs = i\ / 'J, which is (4) of § 147. This argument is another instance of reduction to an absurdity (reductio ad absurdum). See § 88, p. 166. This justifies completely the work of § 147. 147-148] RADICAL EQUATIONS 307 In general, by a similar argument, we may say that if a -f- b Vc = u + v y/w where a, b, c\ u, v, iv are rational (i.e. rational fractions or integers) and v'V is irrational, then u = a and iWw = b Vc, i.e. the rational terms are equal, and the irrational terms are equal. It will be found that some surds cannot have a square root of the form Vr + v« where r and s are rational. For example, the surd expression \9 + 2V2 will give no rational values of r and s. EXERCISES IX: CHAPTER XI Extract the following square roots ; check by squaring the results : 1. V3 + 2V2. 2. V7-4V3. 3. V12 + 4V8. 4. Vl3-2V42. 5. ^19 +6V10. 6. V27-6V20. 8. Vl4 + 2 Vl3. 9. V40 + 10 VI5. 10. Vl8 - 2 V72. 11. V7 + 2 Vio. 12. V18-4V20. 13. Vl + 2 Va - a 2 . 14. A'2^-2vy-a 2 . 7. V16 + 0V7. 15. V2 (m 2 + n 2 + Vm 4 + mV + w*)- Solve the following equations, and check each result : 16. V2 x - 6 + V27 - 4 x = 3. 17. V6« + 2 + Vl7-4x = 5, 18. V±x — 6 + V2 a; - 3 = 1. 308 RADICALS [Ch. XI 19. V33-10- Vz-4 = 2. 20. V3Z-10 + Vz-4 = 2. 21. V2 x 2 + 2 a; + 20 - V2 a 2 + 13 = 1. 22. V4 ar 8 + 2 + V12 - a? = 5. 23. Vl4 + 4aj-ar 2 + Vl4cc-£c 2 -21=5. 24. V»*-2a; + 28- Var* - 6 a; + 39 = 1. 25. V90 - 18 x - 5 x 2 - V34 - 2 a: - 5 x 2 = 4. 149. Radical Coefficients. Equations may involve radi- cals in their coefficients only; in that case we may solve directly. Ex.1. (2+V3)a-4=V3. Transpose 4 : (2 + V3) z = 4 + V3. Divide by 2 + V3 : x = i±J^§ 2+ V3 _ (4 + V3) x (2 - V3) = 8- 2 V-3 - 3 (2 + V3) x (2 - V3) 4-3 = 5 - 2 V3. CAecifc: (2 + V3) (5 - 2 V3) -4 = V3, or, (10 + V3 - 2 x 3) - 4 = V3 (correct). Ex.2, x 2 + Ax -2V3 = 0. Transpose 2 V3 : x 1 + 4 x = 2 V3. Complete the square : z 2 + 4x + 4 = 4 + 2 V3. Take square roots : x + 2 = ±V4 + 2 V3. or, x = - 2 ± V4 + 2V3. i > V4 + 2 V3 = Vr + V». Then, r + s = 4, r - s = ± 2. (See § 148.) 148-149] RADICAL EQUATIONS 309 Whence, r = 3 or 1 and s = 1 or 3. V4 + 2 V3~ = V3 + VI = 1 + V3 ; hence, x = - 2 ± (1 + V3), or, x = — 1 + V3 or — 3 — V3. C%rc£ /or x = - 1 + V3 : (- 1 + VS)« + 4(- 1 + V3) - 2 V3 = 0, or, 4-2V3-4 + 4V3-2V3=0 (correct). Check for x = - 3 - V3~ : ( - 3 - V3> + 4(- 3 - V3) - 2 V3 = 0, or, 12 + 6 V3 - 12 -4 V3 -2 V3 = (correct) . EXERCISES X: CHAPTER XI Solve the following equations : 1. (3 - V2> - 4 = V2. 5. « + l = V2(a!-3). 2. (V3 + V2> = V6. 6 2 + x> = 3 2-Z 2 3. (5 + V3> + 7 = 3V3. , 7. ~ ^ = -4: • 4. 4a; + l = V3(a;+2). 3 + a 2 V7 8. z 2 + (1 - V3) a;- 2 (1 -I- V3) = 0. 9. ar'-5ar+2(V2-l)=0. 10. x 2 - 9 x + (13 - 3 V5) = 0. 11. af'-(5- V3)*b- V3 = 0. 12. ar-3V3-x + 4+ V6 = 0. 13. « 2 -5V2aj + 5-Vl4 = 0. 14. ar°-3(V5- V3)x + 16-5V15 = 0. 15. (V2 - 1) k 2 + (3 V2 - 6)aj + (2 V2 - 1) = 0. 16. r V3" • x + V2 . ?/ = l, I V2-a>-V3-y^=l. 310 RADICALS [Ch. XI : + (3-V% = 4-V£, 17. * 2 a; + (1 + 2 Vo) */ = 2 + 3 V5. ar* + ?/ 2 = 20, a^ + x _ y _ _ 3. (Suggestion. Subtract twice the second equation from the first.) x + y +xy = 5, 1 X 2_y2 + Sy=1 Q_ (Suggestion. Solve the second equation for x or y, and substitute in the first.) J xy + 7 x = 46, 2 °" I 7 a,- 2 + 10 xy = 299. REVIEW EXERCISES XI : CHAPTER XI Solve the equations : 1. Va?-7 + V2aj-7 = 8. 2. V7x + 2-V3o;-2 = 2. 3. vV-3a;-l-Var + 3a + 9 = -4. 4. a-V4a*-25a; + G = 6. 5. Var° _ 2 a; - 10 + VV - 6 a: - 6 = 6. 6. Var* - 14 x + 55 - Va,~ + 2 a; - 17 = 4. 7. V2a^-2a; + ll-V2a/ 2 -4a,- + 8 = l. 8. V2 a.- 2 - 4 a; - 17 - V2 a; 2 - 12 a; + 15 = 2, 9. V4a- + 74-ar 2 -V24a;-66-x 2 = 2. 10. V.''- + 2 + V4 a 2 + 2 = 3. ll. V6x--2V8-2a; = V48-10a-. 149J REVIEW: PART II; RADICAL EQUATIONS 311 12. Vx + 2 - V2 x - 10 = V3 x - XO. 13. V22- HI a- x n ~-y n = ( - ] ; basis for extension; even roots of negative numbers — called imaginary — excluded. § 135, pp. 284-285. p Meaning of Fractional Exponents : extension under Rule I ; x q = (v / x)p; take positive answer if two exist. § 136, pp. 285-286. Multiplication and Division, Radicals of Same Degree : equivalence of any root of a product to product of roots ; reverses ; divi- sion ; removal of factors. Exercises I. § 137, pp. 286-289. Addition, Similar Radicals: essentially, principle a (b-\-c) = ab + ac. reduction to similar radicals. Exercises II. § 138, pp. 289-290. Reduction to Different Degree : essentially, fractional exponents obey laws of fractions. Exercises III. § 139, pp. 290-292. Rationalization of Parts: monomial denominator — multipy by radi- cal in denominator; quadratic binomial denominator — multipy by conjugate. Exercises IV and V. §§ 140-142, pp. 292-295. Negative or Zero Exponent: x° = 1'; x~ n = — ; illustrative prob- lems. Exercises VI. §§ 143-144, pp. 296-298. Review Exercises, Part I, Chapter XI : Exercises VII. p. 299. Part II. Radical Equations. pp. 300-311. Generalities: possibility of elusive answers; graph; reversibility of steps necessary ; check necessary. § 145, pp. 300-302. Methods: clearing of radicals by throwing radicals to one side and raising to power. Exercises VIII. § 146, pp. 303-305. Square Roots of Quadratic Surds : examples. § 147, pp. 305-306. Equalities of Quadratic Surds : rational parts equal ; surd parts equal ; justification of § 147. Exercises IX. § 148, pp. 306-308. Radical Coefficients : examples. Exercises X. § 149, pp. 308-310. Review Exercises, Part II, Chapter XI : Exercises VII. pp. 310-311. CHAPTER XII EQUATIONS SOLVED BY SUBSTITUTION 150. Substitution. In the preceding Chapters we have solved a few examples (see pp. 178, 270) by substituting new letters in the place of inconveniently long expressions. Thus, in squares of binomials we used (x+ y) 2 = x 2 + 2xy + y 2 as a standard (p. A3). Given a more complicated example, say (4 m 2 n — 3 niri) 2 , we substituted 4 m 2 n for x and — 3 mn for y : (4 m 2 n - 3 mn) 2 = (4 m 2 n) 2 + 2 (4 m 2 n) ( - 3 nro) + (- 3 nm) 2 = 16 m 4 n - 24 m s n 2 + 9 nhn 2 by comparison with the formula above. This process, which we have already used so often, is called substitution. Apparent answers are sometimes found that are not really roots (or solutions) of the given equations. The best possible safeguard is to check the final ansivers by direct substitution in the given equations. 151. Equations solved as Linear. Many equations may be solved by substitution more easily than by direct processes. Ex. 1. —i— + 3 = 5. x 2 - 2 Put s = ; then 4 s + 3 = 5 ; whence, s — \, or, 1 1 x 2 - 2 2 Clear of fractions : x 2 — 2 = 2. Transpose : x 2 = 4, or, x = ± 2. 313 314 EQUATIONS SOLVED BY SUBSTITUTION [Ch. XII 4 Check x = ± 2 gives 4-2 + 3 = 5 (correct). In this example the advantage of substituting a new letter is not very great ; however, the principle illustrated by this simple example will be found useful below. Ex. 2. Put then, whence, 10 x-1 2y+3 = 3, 6 + La; — 1 2 y + 3 1 = 4. = u, x-l~ 2y + 'd i + 10t> = 3, + 5 v = 4, = »; (2u (6u v = |. — i ~ 2' Replacing the letters w and « by their meaning as above, 1 = 1 1 = 1 x-1 2' 2y + S o or, x -1 = 2, 2>/ + 3 = 5; a; =3, y = 1. hence, Check Setting x = 3, # = 1 in the given equation gives 2 + -J/ = 3 (correct), |+|=4 (correct). [M-ft Ex. 3. * */ LUo. L* j. Hence, the given equations have no solutions. 151] EQUATIONS SOLVED BY SUBSTITUTION 315 EXERCISES I: CHAPTER XII 8 + 5 = 7. af' - 21 5-20T 5 \ 5-2ar 3 1 x + y x — y 10 3 Vx + y x-y 10 4 = 1, = -2. = 3, 4. <^ r — s 2 r — 5 s 5 +-«-=3. r — s 2 r — 5 9. 2 w + 3 v 5u-2v 5, Ctt+9r+ o u — 2v = 26. ^ + ^ + -^-5 = 7, 10. x 2 — y 2 11. 3ar° + 3*/ 2 ^— = 11. x- - y- 2« 2 -3* 2 + — ^ =9, 6 n 2 -9f- 13_ 7f 2 -3n 2 : = 0. 5. ' 1 +^U=i. 2p + 3^ p — 3q 7 2 .2/>-|-3(/ p — 'Sq _ 9 12. r i a; + 2/ + ^ + 2/ 2 = 62, 2x 2 + 2?/ 2 - J50_ = 72. 6. ; 15 7 = 8 ^ + 82 r2_ S 2 5' 50 21 j .T^ + S 2 l^ — S? 13. < 15 10 a-2_|_y2_5 a?+y— 5 45 2 = 11, la^+i, 2 — 5 x+y-5 7. < 7 1 8 2m 2 -n 2 m 2 -2?r 2' 14 14. 3 + 2 ?/t 2 — « 2 ??i 2 — 2 )i' 2 = 4. y-4a? 25 y — 4x x l + tf = 25 } 2x- 2 + 2y 2 = 73. 8. x + 2/ + — — = 10, x — y 3a; + 3i/--^L=17. a; — 2/ 15. < 3* 2 + 3^+3y 2 -i™=37, •'• + . V -11 9 P 2 P\ 2 4 f * + 3 = 10, 3 11 (x - yf (x + yf = 4. - = 80, y a- + 1 = 24. y _2 = n+3 (m — 3) 2 to — 3 x t-^ = 9, 12. ^ 3 X-4: 13. { (to - 3) 2 2(w+3) 2 =-28. y _^ 2^ = 63. (y-sy («-4)> 2 r^+j/ 2 + x _ y _ 7 ^ 4^±l 2 + (x-2/) 2 =14. a;?/ rr 2 14. ! +-A-=3, 4 r — s r — 1 (r 2 -* 2 ) 2 s 2 =5 (4 r - s) 2 (r - l) 2 322 EQUATIONS SOLVED BY SUBSTITUTION SUMMARY OF CHAPTER XII: EQUATIONS SOLVED BY SUB- STITUTION, pp. 313-322 Substitution : new letters in place of longer combinations. § 150, p. 313. Equations solved as Linear: illustrations. Exercises I. § 151, pp. 313-315. Equations solved as Quadratic: illustrations. Exercises II. § 152, pp. 316-317. Radical Equations : illustrations. Exercises III. § 153, pp. 318-319. Simultaneous Equations : illustrations. Exercises IV. § 154, pp. 319-322. CHAPTER XIII. PROGRESSIONS OR SEQUENCES PART I. ARITHMETIC SEQUENCES 155. Sequences or Progressions. A set of numbers that follow one another in a definite order is called a sequence or a progression. Any one of the numbers is called a term or an element of the sequence. In this Chapter only those sequences are treated in which there are a definite number of terms, so that the sequence finally stops. Such a sequence is called finite. Other sequences exist ; for example, the integers 1, 2, 3, 4, etc., ••■ form an unending or infinite sequence. (See Appendix, §§ 26-28.) 156. Arithmetic Progressions or Sequences. A sequence is called an arithmetic progression (or arithmetic sequence) if the difference between any term and the term which precedes it is always constant. This difference is called the common difference. We shall use the word sequence by preference from now on, instead of progression. For example, the sequence of integers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 is an arithmetic sequence, for the difference between consecutive terms is one in all cases. Similarly, the sequences (a) 3, 5, 7, 9, 11, 13 (common difference 2) (b) 5, 10, 15, 20 (common difference 5) (c) —4, —1, +2, +5, +8 (common difference 3) (d) 7, 3, — 1, — 5 (common difference — 4) (e) — \, +1, +|, +4, + y (common difference V) 323 324 PROGRESSIONS OR SEQUENCES [Ch. XIII and, in general, (/) a, a + d, a + 2 d, etc., •••, a + nd are all arithmetic sequences. The first term will usually be denoted by a, the com- mon difference by e?, as in example (/)• If d is positive, the terms increase ; in that case the se- quence is called increasing. If d is negative, the terms decrease ; the sequence is then called decreasing. 157. Any Term. Any term may be found by adding to the first term the common difference as many times as there are intervening steps ; thus, the second term is a+d, the third is a + 2 c?, etc. ; in general, the number of steps is one less than the number of the term : hence the nth term is a + (w — 1)<#, or, t n = a+ (n — V)d where t n means the wth term. If there are only n terms, the With is the last one, and we write (I) l = a+(n-l)d, where I now denotes the last term. Similarly, we may find the first term if we know the fourth (say) and the common difference : in this case, a = t i — Sd; or, in general : a = t n — (n— V)d where t n means the wth term. Ex. 1. If an arithmetic sequence starts with the number 4 and has a common difference 5, find the 10th term. Here a = 4, d = 5, n - 10 ; hence, t l0 = 4 + 9 • 5 = 49. Ex. 2. a = - 71, d = 2, to find t u . Here n = 12 ; hence, * 12 = - 7$ + 12 . 2 = 17J. Ex. 3. a = + 6, d = — 4, to find t 3 . * 8 = 6 + 3(-4)=-6. 156-158] ARITHMETIC SEQUENCES 325 158. The Sum. To find the sum of all the terras of an arithmetic sequence s=a+ (a + d) + (a + 2tf) -\ h (I - d) + I, where I denotes the last term, write this in the form s = I + Q — d) + (I - 2 d) H h (a + d) +a and add these two equations; we get 2 8 = (a + + (a + + (a + + ••• + (a + 1) + (a -fr Q, the term (a + £) occurring w times if there are w terms ; hence, n , 7N 2 8 = ?i(a + £), or, (II) » = g («+«)■ From I and II, if we know the values of any three of the quantities, a, w, df, Z, s, the other two may be found. Thus, given «, 8, d, to find w and Z, we substitute the value of I from I into II : s = -(a + a+ (n — V)d), or, (III) s = ?(2a + (n-l)d), and so on. When necessary, to avoid ambiguity, we denote the ft sum of n terms by s n instead of s and write s„ = - (a + I), etc. Ex. 1. Given a = 2, / = 14, w = 6, to find s and d : I becomes 14 = 2 + (6 — 1)^/, whence d = ^ 2 . II becomes s = f (2 + 14), whence s = 48. Check : The sequence is 2, 4|, 6|, 9£, llf, 14 ; its sum is 2 + 4f + 6| + 9$ + llf + 14 = 45 (correct). 326 PROGRESSIONS OR SEQUENCES [Ch. XIII Ex. 2. Given d = 3, n — 4, s = — 1. I becomes I =a + (4 — 1)3 or / = a + 9. II becomes - 1 = £ (a + /) or 2 (a + I) = — 1 ; whence, solving these simultaneous linear equations for a and /, we find a — — l 3 1 — XX Check: The sequence is -19-7+5+17. 4 ' 4 ' then s = { -~- \ + \ —^ \ +i'i,\ + \-r\+-~r = -h as given. The general method is illustrated by the examples : first insert the given values in formulae I and II ; then solve the two resulting equations for the two unknown letters. Since n must be an integer, only integral values of n are given in problems. If, in a problem in which n is to be found, the value is not an integer, the problem is impossible, i.e. no arithmetic sequence exists for that problem. EXERCISES I : CHAPTER XIII (The first examples should be checked as above by actually writing down the sequence.) Find the quantities not given : a d n I 1. 5 3 8 2. 5 -3 8 3. i 2 17 10 4. 16 -3 -2 5. -3 1 6. i 5 7. -1 7 -4 8. i i 6 3 -9 158] ARITHMETIC SEQUENCES 327 a a 9. 15 10. 7 11 12. -24 13. 5 14. 6 -1 15. -1 16. i 1 17. 100 18. 19. -9 20. o w n j s 19 69 5 G5 5 1 -5 3 1G 13 45 -10 7 56 1G GO 300 40,200 6 1 -24 7 -3 7 21 21. Show that t m — t n = (m — n)d. 22. Show that the sum of all the terms of the sequence from the with to the »th inclusive is (n — m + 1) — '-^—^ — — '■ — *— • 23. Given a = — 12, d =2, s = - 40, find n. Write out the sequence, and indicate why there are two values of n. Check the result of example 22, by finding the sum of all terms of this sequence from the sixth to the eighth inclusive. 24. Given 1 = 11, d = %, s = 116; find ?* and a. 25. Given a = 2, (7 = 3, s = 77; find n and 1. How many true solutions of the problem are there in this case ? PART II. GEOMETRIC SEQUENCES 159. Geometric Sequences. A sequence is a geometric sequence if the ratio of any term to the preceding one is al- ways the same ; this common ratio is an important number. Thus, 1, 2, 4, 8, 16, 32, 64 is a geometric sequence, the common ratio being 2. Also, 3, 12, 48, 192 (common ratio, 4), 5, -10, +20, -40, +80 (common ratio, -2), 2, 2, |, 2 2 T , (common ratio, 0, and, in general, a, ar, ar 2 , ar 3 , ••■, ar n (common ratio, r). 160. Formulae. As in the case of arithmetic sequences, we may find two formulae : one for the nth term, the other for the sum of n terms. If the first term is called a and the common ratio r, the second term is ar, the third is ar 2 , etc.; since we multiply by r at each step, it is clear that the nth term is t H = ar n ~\ where t n means the nth term. If the last term is called I, we may write (I) l = ar"-* if there are just n terms. Again, the sum is (1) s =a + ar + ar 2 + ar 3 +- ••• + ar n ~ 2 + ar n ~ l . Multiplying by — r, we find (2) — rs = — ar — ar 2 — at 3, — ••• — ar"~ 2 — ar 71 ' 1 — ar", whence, adding (1) and (2), s — rs = a — ar n = a(\ — r"), 328 (II) s = a GEOMETRIC SEQUENCES 329 l-rl After this formula has been obtained, it is easy to see that it is correct, for, by actual division, whence, on multiplying by a, /i _ >.«\ a I = a + ar + ar 2 + ••• + ar"- 1 = s. These formulae I and II may be used as were the formulas for arithmetic sequences. "Ex. 1. Given a = 3, r = 2, n = 7, we have I = 3 • 26 = 192, s = 3(y^| ) = 3 { II ^y) = 3 • 127 = 381. Check: The sequence is 3, 6, 12, 24, 48, 96, 192; the last term is 192 (as found) ; the sum of the terms is 3 + 6 + 12+ ••• + 192 = 381. Ex. 2. Given r = 3, n = 4, s = 120. II becomes 120= n(|— |*) = «(^"!r) = 40a > whence, a = W = 3- I becomes Z = 3 - 3 3 = 81. Check : The sequence is 3, 9, 27, 81 ; it is easy to see that all the conditions of the problem, and the answers, hold good. The same process is used in all cases: the known num- bers are substituted for the corresponding letters in I and II; the resulting equations are then solved for the two unknown quantities. Problems in which n is not given, but is to be found, are not given because the solution requires logarithms (see Chapter XIV) ; these problems are also not particularly valuable since n is by its nature an integer; finally, it is always easy to solve such a problem hy trial, since only integral values of n need be tried. 330 PROGRESSIONS OR SEQUENCES [Cii. XIII EXERCISES II: CHAPTER XIII Find the quantities not given : a r n I s 1. 2 3 5 2. 16 -| 6 3. 2 5 31 4. 4 3 19 5- $ 5 1 6-314 7. 2 3 98 8. 3 6 .00003 9. V3 5 1 10- T V 5 111.11 11. 2 4 16 12. -2 4 16 13. 16 5 81 14. -1 13 -1 15. 3 2 3 16- -i 5 51 17. 1 4 27 18. 3 80 105 Suggestion. Here ± = n ( ? ' 2+r + 1 > =*±1±1. I ar 2 r 2 19. Show that t m — t n = ar'- 1 (r m ~" — 1) . 20. Show that the sum of all the terms of the sequence from the nth to the mth inclusive is a r-1 160-161] GEOMETRIC SEQUENCES 331 161. Theorems in Factoring. The formula II is also useful in factoring; dividing both sides of it by a : (1) Ll^ = l +/ . + r -2 + ... +r / l -l, 1 — r which may be used as a type form for factoring expressions of the form 1 — r", as in Chapter IV, pp. 91-10± Again, setting r = ^ and clearing of fractions, we get x (2) ^1/1 = x n-l + x n-2 y + x n Sy2 ^ h/ 1 " 1 , x-y which is a type form for factoring expressions of the form x n - y". If in (1) we put r = — s, we g sret 1 — s n (3) — - = 1 — s + s 2 — s 3 + • • • — s" \ if n is an even in teger, 1 + s or, 1 + s" (4) — — - = 1 — s + s 2 — s 2 + • • • -f- s* , if n is an odd integer. Setting s= - in these, we find the following type forms: wH yll (5) -J- =x"- 1 —x n -y + x" 3 / 2 — • • • — /" - 1 , if n is even, x -hy or, (6) *l + yl _ x n 1 _ x n i y + x n-'iy2 y y n - 1 j if „ j s odd. x -\-y In English, from (2) jr— / is a factor of x n —y", if n is any positive integer ; from (5) x+y is a factor of x"—y n , if n is even ; from (6) x +y is a factor of x n +/", if n is odd. Comparing with the type forms of p. 101, we see that x 2 — y 2 falls under (2) and under (5); in fact a? — y 2 is divisible by x — y and by x +y, as we already know. The 332 PROGRESSIONS OR SEQUENCES [Ch. XIII form x 3 — y s falls under (2) only ; it is divisible by x — y, and the quotient is x 2 + xy + y 2 . Likewise, X s + y 3 = O + #)0 2 -xy+ y 2 ) by (6) (see p. 101); x*- y*= (x- y)(x s + a% + xy 2 + # 3 ) by (2), or also, = (x + y){x^ - x 2 y + xy 2 + y 8 ) by (5), or also, ={x- y)(x + y) (x 2 + y 2 ) by (2) and (5) , x 5 — y b = (# — y) (a; 4 + a 3 y 4- x 2 y 2 + a**/ 3 + ?/ 4 ) ; x b + y b = (x + #) (a; 4 - a% + ^ 2 i/ 2 - ## 3 + J/ 4 ), and so on. Compare also Appendix, § 6. Ex. 1. Factor 32 a 5 - b 5 . This compares to (2) if x = 2 a, y = b, n = 5 ; hence, 32 a 5 - b 5 = (2 a - b) [(2 «) 4 + (2 a)% + (2 a)% 2 + (2 u)b 3 + &«] = (2 a - 6)(16 a 4 + 8 a?b + 4 a 2 £- + 2 «6 3 + ?/ 4 ). Ex. 2. Factor 81 * 4 - s s . This corresponds to (2) if x — 3 /, ?/ = s 2 , n = 4 ; hence, ar — y =3 < — s 2 is a factor. By (5) x + y = 3 1 + s 2 is also a factor ; hence, (3 1 — s 2 ) (3 1 + s 2 ) is also a factor, i.e. 9t 2 — s 4 is a factor. This results also by taking x = 9 1 2 , y — s*, n = 2, so that x 2 — y 2 = 81 f* - s s . From either we find 81 1* - s s = (9 t 2 - s 4 )(9 / 2 + s 4 ) = (3 « - s 2 )(3 t + s 2 )(9 ? 2 + * 4 ). EXERCISES III: CHAPTER XIII Factor the following expressions : 1. x a -y 6 . 5. x n ±y u . 9. 1 + 1024 x 5 . 2. 1 +y 7 . 6. 8m 8 + 27w 8 . 10. 1 - 64afy 12 - 3. a: 7 -?/ 7 . 7. 81 a 4 - 10 & 4 . 11. 32?>i 5 -f-243// ,i . 4. I-*/ 12 . 8. 125a 6 6 9 -l. 12. 625 a 4 / -A 161] SUMMARY 333 SUMMARY OF CHAPTER XIII: SEQUENCES OR PROGRES- SIONS, pp. 323-332 Sequence (or Progression) : set of numbers in definite order. § 155, p. 323. Arithmetic Sequence: common difference between terms. § 156, pp. 323-324. Formula : t„ = a + (n - l)r/; s = (-)(a + I). Exercises I. V *- 7 § 157-158, pp. 321-327. Geometric Sequence: constant common ratio. § 159, p. 328. Formulae : t n = ar n ~ l ; I = at*- 1 ; s = ^ — ^—* • Exercises II. C 1 _ r ) § 160, pp. 328-331. Factoring x m ± y n : x — y a factor of x n —y n ; x + y a factor of x n — y n when n is even ; of x" + y n when n is odd. Exercises III. § 161, pp. 331-332. CHAPTER XIV LOGARITHMS 162. Introduction. We shall now study a method by which long numerical computations are greatly simpli- fied. Before proceeding with this chapter, the student should review the entire subject of exponents. (See §§ 135-144.) The following examples may serve as a basis of this review. EXERCISES I: CHAPTER XIV 1. Simplify the following as much as possible without per- forming any long numerical computations : a. x' x x 2 . b. 127 5 x 127 9 . c. 74 7 x 74. d. 3 27 x 2 27 ; 7 8 x 5 9 . e. 3 17 X 3 17 . Can this be done in two ways ? /. g 7 + cf. g. (a 8 ) 5 , h. vV-. i. ■"= d. of n -x n = f. (x m ) n = h. ar n = 4. 136, State in words the laws expressed in Ex. 3. (See §§ 135, 143.) 334 LOGARITHMS 335 5. Complete the following table by giving all the powers of 16 at intervals of ', from — 4 to + '• 16° =1 lo n =1 16* = 2 16"* = £ 16* = 4 16""* = \ 16* = 8 16"* = i 16* = 65536 16-* = ^ T ^ 6. By means of the table in example 5 find the value, wherever it can be done without any long computation, of the following expressions : a. 512 x 128 = what ? Solution 512 x 128 = 16* x 16* = 16* + * = 16* =65536. 6. 16384-64. d. ^/128 x 16384 3 - c 512x32768 , , ^^(S), / 16384 2048 X8192 263 3 The preceding method alone is not, completely successful in this case, because we cannot perforin the subtraction by means of expo- nents. g. Can the following operations be performed by means of your table ? Why not ? ^1024, a/572. 163. A Table of Exponents. Our table is not suffi- ciently complete for practical purposes, as was seen when we attempted to solve 6 g, above, by means of it. In order to make a table that will be of practical use in examples involving any number, it will be more convenient to use 10 as a basis instead of 16, as 10 is better suited to the 336 LOGARITHMS [Ch. XIV decimal system. Let the student complete the following table and make himself familiar with it: 10° =1 10!= 10 10 2 =100 10° =1 io-i=. 1 10- 2 =.01 10 6 = 1000000 io- 6 =.000001 We may use this table as we did the other in solving certain simple problems, but it has the same difficulty as the other, for such numbers as 273, 1772, etc., are not found in it; but it has an advantage over the other table in that it is easier to remember. Let us now make a similar table containing many more numbers. To do this, let us make a graph of the table, in other words, a graph of the equation — ~\(\l The preceding table gives values of n and of L as follows : ' Point in Fig. 08 : M L -2 K -1 A B 1 c 2 L: -6 — 5 -4 o — o 3 4 5 6 n : .0000001 .1 1 10 100 1000000 [Let the student complete the table.] Since n can never become negative, we take the starting point near the left edge of the paper. '1 1 K fhi f J in '" '• ■ 1 A a s ■ > ; 53 = 100 x 3.39 = 339, nearly; that is, 2.53 = log 339. 340 LOGARITHMS [Ch. XIV Ex. 4. Find L if 10 z = 7250. We find from the table that 7.25 = 10- 86 , nearly. But 1000 = 10 3 . Hence, 7250 = 1000 x 7.25= 10 3 x 10- 86 = 10 3 - 86 . Hence, Z = 3.86. That is, log 7250 = 3.86. Ex. 5. Find log (.0725); that is, find L if 10 L = .0725. As above, 7.25 = 10- 86 , .01 = 10" 2 . Hence, .0725 = .01 x 7.25 = 10-2 x io-86 _ 10-2+ .86, Hence, log .0725 = - 2 + .86. It is usually more convenient, when the integral part of the logarithm is negative, to leave the decimal part always positive, instead of adding ; thus, we write — 2 + .86 and not — 1.14. 164. Simple Computation by Exponents. We can now perform any of the simple operations of multiplication, division, raising to powers, or extraction of roots upon any number whatever very easily by logarithms, though our results will not be precisely accurate. Ex. 1. Multiply 2 by 3. We can do this easily without logarithms: 2x3 = 6. Notice that 2 = 10 30 , 3 = 10- 48 . Hence, 2x3 = 10- 30 x 10- 48 = 10 78 = 6. Let the student make the measurements on his figure. The student's measurements may not agree precisely with these, owing to inaccuracy in the figure used. Ex. 2. Multiply 3f by 7.3. Let the student use his own figure. 3f = 10- 56 , 7.3 = 10- 86 . Hence, 3f x 7.3 = 10 142 = 10 1 x 10- 42 = 10 x 2.6 = 26. Multiplying by the ordinary method, we see that this is not precisely accurate, owing to the inaccuracy of our table, but this degree of accuracy would be sufficient in many practical problems. Ex. 3. Divide 13 by 7. 13 = 10x1.3 = 101 x 10 U = 10 1U ; 7 = 10- 85 . Hence, 13 -*■ 7 = 10 1U +-10- 85 = 10- 26 = 1.8, nearly. Ex. 4. Eind (4.3) 17 . 4.3 = 10 63 . Hence, 4.3 17 = 10 17 *- 63 = 10 in71 = 10 10 x 10 n = 10000000000 x 5.1 = 51000000000. Of course, this is far from accu- 163-165] LOGARITHMS .",11 rate; we are sure only of the first one or two figures, but the student should notice the great saving in labor, and that frequently the accuracy here attained would he sufficient. With the more extended table given below, greater accuracy is obtainable. Ex. 5. Extract the seventh root of 7825. The work would be extremely long by methods previously used. We have here 7825 = 1000 x 7.825 = 10 3 x 10- 89 = 10 389 , nearly. Hence, v*7825 = VlO 3 -* = lO- 5 * 5 = 3.6, nearly. In finding the loga- rithm of 7.825 by the table we took the nearest Logarithm found iu the table. A more accurate table will be found on p. 3-48. EXERCISES II: CHAPTER XIV Simplify the following by the aid of the preceding table ; compare each result with the figure : i 3 /T^J 1. 4.3 x 23. 4. V1730. 7. 45 2 + V(l.o) 2 . 2. 230^17. 5. 2.3'. a („ ;u ) • 3. 2.7 12 . 6. ^3. 9 . .00o 4 x 5200 4 . (•Ml) 5 U (8 6)l 165. Definitions and Principles. A fundamental num- ber must always be chosen as the base; this is usually 10. The logarithm of a number to a given base is the exponent with which the base must be affected to produce the number. In other words \§ L = n and log n = Z, have precisely the same meaning. We have taken 10 as our base, as usual. Henceforth, when the base is not specified, it will be understood that the base 10 is used. When the base 10 342 LOGARITHMS [Cn. XIV is used the logarithms are called common logarithms or Briggs's logarithms. Note 1. The base 10 is most common, but others may be used. When necessary, we write log 10 n to avoid ambiguity. In general, if the base is b, the logarithms are denoted by log/, n. The base may be any number except or 1. Note 2. We have not explicitly defined the meaning of an irra- tional exponent, and the student should not be burdened at this stage with the idea. It is sufficient to call attention to the fact that the use of the smooth curve (§ 163) through certain points that can be located involves the essential idea of a rigorous treatment. (See Appendix, § 30.) From the principles of exponents (§§ 104, 135, pp. 193, 285, and p. 334), it is evident that I. TJie logarithm of a product is equal to the sum of the logarithms of the factors, for 10™ x 10 n = 10 m+M . II. The logarithm of a quotient is equal to the logarithm of the dividend less that of the divisor, for 10'" -=- 10" = 10'"~". III. The logarithm of a power of a number is equal to the logarithm of the number multiplied by the exponent of the power, for (10'")" = 10 m ". This principle includes the extraction of roots by using fractional exponents, for, by § 139, 1 1 m — tnx — — (10™)" = 10 " = 10". These may be stated as follows : I. Log (m x n ) = log m + log n. II. Log -- = log m — log n. n III. Logn K = K log n. (K may be an integer or a fraction.) The logarithm of \ to any base is zero. Since b° = 1. The logarithm of the base is 1. Since b 1 = 6. The common logarithm of a number between one and ten lies between zero and one. See the table or the figure. 165-166] LOGARITHMS 343 166. Characteristic and Mantissa. Let us now study the curve 10 X = », or log n = I for a greater range of values. From the table already constructed (§ 168, p. 33U) we find, for example, 10 53 = 3.4. Hence, 10 L68 = 34. 10- 1+53 =.34 10 253 =340. 10- 2+53 =.034 10 353 = 3 400# IO-3+.53 = m 0034 10 4 - 63 = 34000. 10-4+53 = .00034 The student should notice that increasing the logarithm by 1 corresponds to multiplying the number by ten (i.e. the decimal point is changed one place). Likewise, decreasing the logarithm by 1 corresponds to dividing the number by ten. If lo; m -lyo • Joa ■-; no: ;• ,- n- — -- OhunUti-iistiVi ! • ----I- ■ -t -- h -H-B=fO t-H ' ±T - ■ [ - m,l 1 - r.i — 1 I' ■ ; ' : . ' , ' ' ' 1 [i 1 I 1 :.il_I_ — ajll rti.-illy" norizontally-l- M' 1 MM TUT 1 i 'I i hi 1 1 1 1. 1 1 1 MM 1 1 1 1 1 1 i hi 1 1 1 1 166-167] LOGARITHMS 345 To find the Logarithm of a Given Number from the Table. Ex. 1. Find the logarithm of 3. In the columns marked N we look for 30 (the decimal point after the 3 is omitted). On a line with this in the column headed we find 4771. Hence, log 3 = .4771 Ex. 2. Find the logarithm of 4.6 In the column marked N we find 46. On a line with it in the column marked we find 6628. Hence, log 4.6 = .6628 Ex. 3. Find the logarithm of 3.76 In the column marked N we find -)7 . On a line with this in the column marked 6 we find 5752. Hence, log 3.76 = .5752 Ex. 4. Find the logarithm of 3760. By example 3, 10- 5752 = 3.76. 1 Hence, 10 3 - 5752 = 3760 But, 10 3 =1000. or, log 3760 = 3.5752 Ex. 5- Find the logarithm of .0376. By example 3, 10-5752 _ 3.7^5 But, 10- 2 = .01 This is sometimes written log .0376 = 8.5752 Ex. 6. Find log 3764. From the table log 3.76= .5752 (.4) log 3.77 = .5763 (B) We will plot these points on a very large scale, drawing only a small part of the curve. A is the point where n = 3.76, L = .5752 B is the point where n = 3.77, L = .5763 E is the point where n := 3.764 We wish to find L. Hence, 10" 2 +- 5752 = .0376 or, log .0376 = - 2 + -5752 10. r = lc K n I •1 * .5- in f-y? O 3* E fr w 1 3 1 .5755 f\ u 1 1 1 .! D \ < f - -.0 II - • " :S. 75 a 755 3 7i ■> Ti ". ::: 7 t cale 111 b< ni so it. 11 ■ ' 1 .mil '1 1 ir- :! l.v 1 Fig. 71. 346 LOGARITHMS [Ch. XIV If we draw the straight line AB, cutting DE at F, then F will lie very near to E and it will be sufficiently accurate to find the height of F instead of E. D is just as high as A, i.e. .5752; CB = .0011 ; DF — A of .0011 = .0004 (dropping decimal places after the fourth). Hence, the height of F is .5752 + .0004 = .5756 That is, 10- 5756 = 3.764 10 3 = 1000. Hence, 10 3 - 5756 = 3764. Hence, log 3764 = 3.5756 This process of finding a result not given explicitly in the table is called interpolation. Ex. 7. Find the logarithm of .02756 Consider 2.756 log 2.75 = .4393 log 2.76 = .4409 .0016 log 2.756 = .4393 + (.6 x .0016)= .4393 + .0010 (nearly) = .4403 10 .4403 _ 2.756 Hence, 10 " 2 +- 4403 = .02756 Hence, log .02756 = - 2 + .4403 = 8.4403 - 10. The student should draw a figure illustrating this work. To find the Number corresponding to a given Logarithm. When the decimal part of the logarithm can be found in the table, the corresponding number can be written down at once. Ex. 1. Find n if log n = 4.9175 Looking in the table, we find 9175 on a line with 82 (in column N) and in the column marked 7. Hence, 10- 9175 = 8.27 Hence, 10 4 - 9175 = 82,700. Hence, log 82,700 = 4.9175 When the decimal part of the logarithm cannot be found in the table, we follow a process similar to that in examples 6 and 7 above. Ex. 2. Find n if log n = 2.4574 From the table we find log 2.87 = .4579 log n = 2.4574 log 2.86 = .4564 log 286 = 2.4564 Difference = .0015 = .0010 167-168] LOGARITHMS 347 The difference between log 286 and log n = .0010 This is {§ of the difference between log 286 and log '2*7. Hence, .4574 = log (2.86 + J§ of .01). Hence, 2.4574 = log 286.7 The student should construct a figure like that above for this case. Notice that the differences found above are given in the column marked D in the table. These differences are not always exact, since the real differences often vary in the same row ; but they are always sufficiently accurate for use with this table. EXERCISES III: CHAPTER XIV 1. Find the logarithm of : (a) 7 (c) 600 (e) .032 (g) .0467 (J) 2.473 (b) 40 (d) 4.7 (/) 2.43 (h) 57,200 (j) .04257 2. Find the number whose logarithm is : (a) .3010 (c) 2.3802 (e) 5.4533 (g) 2.6290 (7) 2.0027 (6) 3.4771 (d) 3.3075 (/) 1.5732 (/<) 1.4563 0') -0317 168. Computation by Logarithms. Ex. 1. Simplify j]¥LKIM. 1 J V (16.3) 2 3 1 247 x 3.42 8 _ »/ l0*«>" x 10- 5350 3 /10 2 - 9277 "* (16. 3) 2 * (101-2122)2 : \l()2.4244 _ v'lO-sosa = 10 .i677 = i.47i This work may be tabulated as follows : 047 v ;{ 4-->x 1 [iQ^yT = r d Og 247 + log 3.428 - 2 log 16.3]. log 247 = 2.3927 log 3. 428 = .5350 log yj- Divide by 3 log numerator = 2.9277 log 16.3 = 1.2122 log (1 6.3) 2 = 2.4244 log quotient = .5033 log answer = .1678 hence, answer = 1.471 348 LOGARITHMS [Ch. XIV N 1 2 3 4 5 6 7 8 9 D 10 0000 0043 0086 0128 0170 0212 0253 0294 0334 0374 42 11 0414 0453 0492 0531 0569 0607 0645 0682 0719 0755 38 12 0792 0828 OS64 0899 0934 0969 1004 1038 1072 1106 35 13 1139 1173 1206 1239 1271 1303 1335 1367 1399 1430 32 14 1401 1492 1523 1553 1584 1614 1644 1673 1703 1732 30 15 1761 1790 1818 1847 1875 1903 1931 1959 1987 2014 28 16 2041 2068 2(1! 15 2122 2148 2175 2201 2227 2253 2279 26 17 2304 2330 2355 2380 2405 2430 2455 2480 2504 2529 25 18 2553 2577 2601 2625 2648 2672 2695 2718 2742 2765 24 19 2788 2810 2833 2856 2878 2900 292;; 2945 2967 2989 22 20 3010 3032 3054 3075 3096 3118 3139 3160 3181 3201 21 21 3222 3243 3263 3284 3304 3324 3345 3365 3385 3404 20 22 3424 3444 3464 3483 3502 3522 3541 3560 3579 3598 19 23 3617 3636 3655 3674 3692 3711 3729 3747 3766 3784 18 24 3802 3820 3838 3856 3874 3892 3909 3927 3945 3962 18 25 3979 3997 4014 4031 4048 4065 4082 4099 4116 4133 17 26 4150 4166 4183 4200 4216 4232 4249 4265 42S1 4298 16 27 4314 4330 4346 4362 4378 4393 4409 4425 4440 4456 16 28 4472 4487 4502 4518 45:;:; 4548 4564 4579 4594 4609 15 29 4(524 4639 4654 4669 4683 4698 4713 4728 4742 4757 15 30 4771 4786 4800 4814 4829 4843 4857 4871 4886 4900 14 31 4914 4928 4942 4955 4969 4983 4997 5011 5024 5038 14 32 5051 5065 5079 5092 5105 5119 5132 5145 5159 5172 13 33 5185 5198 5211 5224 5237 5250 5263 5276 5289 5302 13 34 5315 5328 5340 5353 5366 5378 5391 5403 5416 5428 13 35 5441 5453 5465 5478 5490 5502 5514 5527 5539 5551 12 36 5563 5575 5587 5599 5611 562:; 5635 5647 5658 5670 12 37 5682 5694 5705 5717 5729 5740 5752 5763 5775 5 7. SO 12 38 5798 5809 5821 5832 584:; 5855 5866 5877 5888 5899 11 39 5911 5922 5933 5944 5955 5966 5977 5988 5999 6010 11 40 6021 6031 6042 6053 6064 6075 6085 6096 6107 6117 11 41 6128 6138 6149 6160 6170 61SO 6191 6201 6212 6222 10 42 6232 6243 (5253 6263 6274 6284 6294 6304 6314 6325 10 43 6335 6345 6355 6:365 6375 6385 6395 6405 6415 6425 10 44 6435 6444 6454 6464 6474 6484 6493 6503 6513 (5522 10 45 6532 6542 6551 6561 6571 65S0 6590 6599 6609 6618 10 46 6628 6637 6646 (5656 6665 6675 6684 6693 6702 6712 9 47 6721 6730 6739 6749 6758 6767 6776 6785 6794 6803 9 48 6812 (J821 6830 6839 6848 6857 6866 6875 6884 689.". 9 49 6902 6911 6920 6928 6937 6946 6955 6964 6972 6981 9 50 6990 6998 7007 7016 7024 703:5 7042 7050 7059 7067 9 51 7676 7084 7093 7101 7110 7118 7126 7135 7143 7152 8 52 716(1 7168 7177 7185 7193 7202 7210 7218 7226 7235 8 53 7243 7251 7259 7267 7275 72S4 7292 7300 7308 7316 8 54 = 7324 7332 7340 7348 7356 7364 7.472 7380 7388 7396 8 168] FOUK PLACE TABLE 349 N O 1 2 3 4 5 <> 7 8 9 D 55 7404 7412 7419 7427 7435 7443 7151 7459 7466 7171 8 56 7482 7490 7197 7505 7513 7520 7528 7536 754:; 7551 8 57 7559 7566 7574 75S2 7589 75! 17 7604 7612 7619 7627 8 58 7634 7642 7649 7657 7664 7672 7679 7686 7694 7701 7 59 770'.) 77 It; 7723 7731 7738 7745 7752 7760 7767 7774 7 60 7782 7789 7796 7803 7810 7818 7825 7832 7839 7846 7 61 7853 7 still 7868 7875 7882 7889 7896 7903 7910 7917 7 62 7924 7931 7938 7945 7952 7! 159 7966 797:; 7980 7987 7 63 7993 8000 8007 8014 8021 8028 8035 80 1 1 8048 si 155 7 64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 7 65 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 7 66 8195 si'i (2 8209 S215 8222 8228 8235 8241 8248 8254 7 67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 6 68 8325 8331 8338 8344 8351 8357 8363 8370 8376 8382 6 69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 6 70 8451 8457 8463 8470 8476 84S2 84SS 8494 8500 8506 6 71 8513 8519 8525 8531 8537 8543 8549 S555 8561 8567 6 72 8573 8579 8585 8591 8597 8603 si 109 8615 SI 121 8627 6 73 8633 8639 8645 8651 Si 157 8663 8669 SI 175 8681 8686 6 74 8692 8698 8704 8710 8716 8722 8727 8733 8739 8745 6 75 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 6 76 SSI IS ■SSI 4 8820 8825 8831 8837 8842 8848 SS54 8859 6 77 8865 8871 8876 SSS2 8887 8893 8899 8904 8910 S915 6 78 8921 8927 8932 8938 89-13 8949 S954 8960 8965 S971 6 79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 5 80 0031 9036 9042 9047 9053 9058 9063 9069 9074 9079 5 81 9085 9090 9096 9101 9106 9112 9117 9122 9128 9133 5 82 9138 9143 9149 9154 9159 9165 9170 9175 9180 9186 5 83 9191 9196 9201 9206 9212 9217 92' '2 9227 9232 9238 5 84 9243 9248 9253 9258 9263 9269 9274 9279 9284 9289 5 85 9294 9299 9304 9309 9315 9320 9325 9330 9335 9340 5 86 9345 9350 9355 9360 9365 9370 9375 9380 9385 9390 5 87 93115 9400 9405 9410 9415 9420 9425 9430 9435 9140 5 88 9445 9450 9455 9460 9465 9469 9474 9479 9484 948! 1 5 89 9494 9499 9504 9509 9513 95 IS 9523 9528 9533 95:38 5 90 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 5 91 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 5 92 9038 9643 9647 9652 in 157 9661 9666 9671 9675 9680 5 93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 5 94 9731 9736 9741 9745 9750 9754 9759 9763 9768 9773 5 95 9777 9782 9786 9791 9795 9800 9S05 9 si 19 9814 9818 5 96 9823 9827 9832 9836 9841 9845 9S50 9S54 9859 9863 5 97 9868 9872 9877 9SS1 9SS6 9890 9S94 9S99 9903 9! His 4 98 9912 9917 9921 9926 9930 9934 99: 19 9943 994S 9952 4 99 995C, 9961 9965 51969 9974 ! m 17s 99s:; 9987 9991 999f. 4 350 LOGARITHMS [Ch. XIV Ex. 2. Simplify ^43 + 5^278 The operations of addition and subtraction cannot be performed by logarithms. We may find 5 V278 by logarithms, add this to 43, and then perform the remaining operations by logarithms. We handle the problem as if all the numbers were positive and then put the proper sign before the answer, which in this case is negative. Ex. 3. Find S/.0247 Log >^0247 = \ log .0247 =\ (- 2 + .3927) = (- .6667 + . 1309) =- .5358 We cannot find this in the table, since it is negative. Hence, we write in another form : log VT0247 = \ (- 2 + .3927) = \ (8.3927 - 10) = (2.7976 - 3.3333)= 2.4643 - 3 =1.4643 We may do this in another and simpler way. By subtracting 1 from the — 2, and adding 1 to the mantissa, we obtain \ ( — 3 + 1.3927) = 1 + .4642 Hence, log ^0247 = L4642, whence ^.0247 = .2912 Ex.4. Solve 3* = 17. We might put L = 3 1 and try various values of x and plot a curve. X 1 2 3 5 1 1 21 H L 3 9 27 15.59 20.51 Thus, we would find an approximate value without logarithms, but the computation is very long. It can be done much more simply by logarithms. 8 3' = 17 log 3* = log 17 x log 3 = log 17, by III, § 165 x = loo- 17 1.0304 log 3 .4771 = 2.579 168] LOGARITHMS 351 EXERCISES IV: CHAPTER XIV Compute by the use of logarithms : 1. 43 x 75. 20. (99. 43) 5 . 2. 437 x 9.63. 21. (1.01) 9. .0024-1.034. 5 ^ ^ V S" 25 1800 3. 439 x. 0372, 22. (1.04) 4. 243.7 x .179.2. 23 VMM 5. .8752 x .01529. 6. 1.002x3.075. 7. 20 -=- 685. 8. .257 -=-1.73. 24. V. 005726. 25. ^/12a3. 26. ^/L072. 27. V200. 10. 10.07-4.617. 28 V - m 11. 25,680-152,980. 29. ^f. 12. 3 - .002463. 30. \/4. o 13. 47x6.3x250. ^ 4 83 x .035 - 2.461 14. 246 x .0072 x 102. 5x4.6x3547 15. 76 x 50.04 x .06004. 32 642 x 3729 x .0007 1705 x 2.004 x .4 546 x .0001 x .4040 16. 1205 x 6 1 x 54f 17. 241x174x2464. 33 2.02 x .003 x .2623 18. (1.47) 8 . 19. (3.057/. 34. A /4fi^3^ 352 LOGARITHMS [Ch. XIV 35 ° P47 X 23 91. (4.QQ7)* x (.003*) ^14.62x24 43 ' ^/4^T"" 36. 5 X 24 X 3.1 44. (4)3^2J_3. V'2 x 18 x 640 ' 7 = 45. V-38V95. 37. v ¥ 2 T x 1 ^x240. 38. yjl9Ax^x' 46. 17 / == 3 A/- 2 °- -v/475 39. 40. 41. 976 3.95 47. \16-v92. \ 89.32x34.95 48 - TT "' V16-V92 5020V.00437 4^97~3 49 ^/ 247 x V44 ___ 392-V44 16S/17 x 43 ^ 247 6- 1 54. 17 2 = §. 57. 2* 2 =47. 59. 256=4.7*-'. 60. Find the compound amount of $ 100 for 15 years at 4 %. 61. Find the compound amount of $ 1 for 100 years at 6 %. 62. What amount put at compound interest at 4 % at a child's birth will amount to $ 1000 when he is 21 years of age ? 63. At what rate at compound interest will $ 1 amount to $ 10 in 40 years ? 64. In how many years will $ 1 amount to $ 2.00 at 4 % compound interest ? 168] SUMMARY 353 SUMMARY OF CHAPTER XIV: LOGARITHMS, pp. 334-352 Review of Exponents : restatements. Exercises I . § 162, pp. 334-335. Temporary Table : logarithms as exponents of a chosen base ; fig- ure for n = 10 L ; short table from figure and computation, — base 10. § 163, pp. 335-340. Simple Computation by Exponents : illustrations of use of exponent laws. Exercises II. § 164, pp. 340-341. Formal Definitions: base, — the number whose exponents are used; logarithm, — the exponents that produce given num- bers; equivalence of n = lO- 6 and log n = L. Formal Rules: logarithm of product, — sum of logarithms; loga- rithm of quotient, — difference of logarithms; logarithm of power, — power times logarithm. § 165, pp. 341-342. Characteristic : integral part; increase of 1 for every digit place; judgment of characteristic without table. Mantissa : fractional part ; independent of decimal point ; tables necessary. § 166, pp. 343-344. Four Place Table : logarithm of given number ; number for given logarithm ; interpolation. Exercises III. § 167, 344-347. ■ Computations: illustrative examples; tables given (pp. 348-349). Exercises IV. § 168, pp. 347-352. APPENDIX NOTE I. DETACHED COEFFICIENTS 1. Detached Coefficients. We may considerably shorten the labor of many operations in polynomials by writing only the coefficients, taken in their natural order after the polynomials are arranged in ascending or descending powers of some one letter. 2. Multiplication. Thus, to multiply 3 a- 3 — 7 x 2 + 5 x — 6 by 2x* — 4 x + 3, we merely write 3- 7 + 5-6 2-4+3 6-14 + 10-12 - 12 + 28 -20 + 24 9-21 + 15-18 6 - 26 + 47 - 53 + 39 - 18 and write the product 6 x 5 - 26 x 4 + 47 x s - 53 x 2 + 39 x - 18, the power of x which belongs in eacli term being clearly indicated by the position of the term. 3. Division. Likewise, if we are to divide G x 5 - 26 x* + 47 aj 8 - 53 x 2 + 39 x- 18 by 2 x 2 - 4 x + 3, 354 DETACHED COEFFICIENTS 355 we write onlv the coefficients, as follows: m Dividend : 6 - 26 + 47 - 53 + 39 - 18 6 -12+9 - 14 + 38 -53 -14 + 28 -21 10 - 32 + 39 10 -20 + 15 - 12 + 24 - 18 - 12 + 24 - 18 4 + 3 Divisor 3 — 7 + 5 — 6 Quotient Remainder Whence, the quotient is 3 x 3 - 7 x- + 5 x - 6. Compare with the multiplication performed above. The student must be extremely careful not to omit terms even when the coefficient is zero, for the position of the term indicates the degree. Thus, x 3 + 2 x - 5 should be written 1 + + 2-5, not 1 + 2-5, for 1 + 2 — 5 would represent x 2 + 2 x — 5. 4. Division by x — a. Division by a simple binomial of the form (x — a) is especially easy. Let us divide sc 3 4- 2 x — 5 by x — 3. We write Dividend : 1+0 + 2-5 1-3 + 3 + 2 + 3-9 1—3 Divisor 1 + 3 + 11 Quotient 11-5 11 - 33 28 Remainder Whence, ** + 2 * ~ ~° = x* + Bx + 11 + ~^-- x - 3 x-3 Note that this becomes even more simple if we write none of our numbers twice. Thus, we may write Dividend : 1 + + 2 - 5 |1 - 3 Divisor _ 3 _ 9 _ 33 Quotient: 1 + 3 + 11|+ 28 Remainder which gives all the information in shorter time. This shorter form may be used only when the divisor is of the form x ± a. 356 APPENDIX EXERCISES I: NOTE I— DETACHED COEFFICIENTS Perform the indicated operations : 1. (ar ? + 2a,- 2 -5.r + 3)x(a; + 2). 2. (5x i -3x 2 + 7)x(3x 2 + x-5). 3. (x + 2x 2 + 3x 3 + x i )x(x-2x* + 5x i ). 4. (a,' + 2 + 2ar 2 -3ar 2 )x(z + 3-ar 1 ). 5. (lx i + 2x-l)x(3x 2 + 2x). 6. (3x 7 + 2x s -5x + 4)x(3x 2 -7). 7. (2 x z -3x + 7)x(x 2 -2x + 1). 8. (x* + 6x 2 + 12x + $) + (x + 2). 9. (6^ + 8x + 16)-f-(a; + 4). 10. (8 re 3 -36 a; 2 + 54 a; -27) ^-(2 a; -3). 11. (12x 2 -Ux + 16) + (6x + 5). 12. (14 z 5 - 13 x* + 27 a; 2 - 5) -- (2 a; 2 + 3 x - 7). 13. (25 or 4 + 5 x- 3 - 2 ar 2 + ar 1 + 5) -- (5 ar 2 + 3 ar 1 + 2). 14. (3ar 5 + 2ar'-5a; + 7)-f-(3a;-l). 15. (ar 3 + a; 2 -4 x + 6) -s-(aj + 3). 16. (re 4 - 4 jb 2 - 5 x + 10) -r- (a: - 2). 17. [(a^ + a? — a-l)-s-(a;-l)]-s-(a; + l). 18. (a; 3 + 7 a; 2 + 15 x + 25) -5- (a; + 5). 19. ( x *-x s —6x i + x — 3)+(x — 3). 20. [(a; 2 + 2 a; + 1) x (x + 3)] - (a; + 1 ). 21. [(2 x 3 + 5 x 2 + 12 a: + 5)-s-(2 a; + l)]-s-(3 » + 5). 22. [(2 a? + 4 ar* + 3) (8 ar 2 + 16 x + 16)] h- [(4 aj + 4) (as + 3)]. 23. [(3 ar 5 + 4 a; + 7) (3 x 2 - 5)] -*- [(a; + 5) (9 x 2 + 2x+ 3)]. 24. [x- & + 4 a;- 4 + 3 ar 3 - 2 ar 2 + ar 1 + 5 + 3 x] X (ar 1 + 3). NOTE II. REMAINDER THEOREM; FACTORING 5. Factor Theorem. If a polynomial P = Ax n + Bx nl + • • • + Lx + N has a factor of the form z — a, we have P = (x — a) • Q. Hence, the equation P=0 is equivalent to the equation (x — a) Q = ; since x = a is a solution of this equation (as is seen by actually substituting a for x), we may say : I. If P has a factor (x — a), then a is a root of the equa- tion P=Q. Conversely, suppose a is a root of the equation P = 0. Dividing P by x — a, we should get some quotient Q and some remainder R, where R is a number independent of x : P = Q( x - a) + R. Now set x = a. Since x = a is a solution of the equation P = 0, we have first P = when x = a. Next, (x — a) is zero when x = a. Hence, the above equation reduces to = + i2, that is, R = 0. In other words, the remainder obtained by dividing P by x — a is zero, or P is exactly divisible by x — a. II. If x = a is a root of P = 0, then P is divisible by x — a. 6. Factors of x n ±y". These facts enable us to factor in many cases. For example, x n — 1 = is satisfied by setting x = 1. Likewise, x 3 + 1, z 5 + 1, etc., or, in general, of 1 + 1 where n is odd, is divisible by £ + 1, for the equation x n + 1 = (n odd) is satisfied by setting x = — 1. Finally, 3f l — l is divisible by x + 1, if w is even, f or x n — 1 = is satisfied by x = — 1. 367 358 APPENDIX In like manner, (1 ) x n — y n is always divisible by x — y. (2) x n — y n is divisible by x + y if n is even. (3) x n + y n is divisible by x + y if n is odd. Many forms may be factored upon this basis. Thus, ar 3 — if is divisible by x — y. By actual division, we find x a - y s = (x - y)(x 2 + xy + y' 2 ). Check : 1 + 1 + 1 i-e- x 2 + xy + y 2 1 — 1 i.e. x — y 1 + 1 + 1 — 1 — 1 — 1 multiplied 1 + 0+0-1 x 3 - y* We find also,x 8 + y 3 = (x + y)(x 2 - xy + «/ 2 ) ; * 4 - y* = O 2 ) 2 - Of 2 ) 2 ! = (* 2 - ? 2 )0 2 + V 2 ) = ( x - 2/)(- l ' + y)(** + y"); x 5 - ?/ 6 = (a; - //) (x 4 + x 3 # + x 2 y 2 + x,>/ 3 + y A ) ; a;6 + ^ _ (x + y)(x 4 - x 3 y + x 2 // 2 — x// 3 + y 4 ); x e - / = [(x 3 ) 2 - O 3 ) 2 ] = (x 3 - ,/ 3 )(x 3 + y*) = (x - y)(x' 2 + xy + if 2 )(x + y)(x' 2 - xy + # 2 ); or,also, x*-y* = [(x 2 ) 3 - - (.y 2 ) 3 ] = (x 2 - y 2 ) (x 4 + x 2 ,/ 2 + y 4 ) = (x - y)(x + //)(x* + x'V 2 + /) ; whence, comparing with the preceding, we find : x* + xY + # 4 = (x 2 + *y + y 2 )(x 2 - xy + .v 2 ), x 7 ± y 1 = ( x ± y) (x 6 T x 6 . The work is even shorter in detached coefficients, see p. 355. EXERCISES III : NOTE II — REMAINDER THEOREM Find by the remainder theorem the value as shown in each of the following examples. Verify the first three by actually substituting the indicated value for the unknown. 1. P=: or 5 -5 a; 2 + 15 a; -75, for a; = 5. 2. P = x 5 - 27 x* + 15 x 2 - 35 x + 125, for x = - 3. 3. P = a;- - 35 x + 17, for x = 2. 4. P = or 5 - 15 x 4 + 25 ar J -125» B + 50, for a,- = -4. 5. P = x 3 - 3 a,- 2 + 15 x - 20, for x = 3. 6. P=a; 4 + 4ar J -12ic 2 + l=0, for a-- = -2. NOTE III. CHOICE AND CHANCE; PERMUTA- TIONS AND COMBINATIONS 9. Choice. The arrangements of objects and the possi- bilities of choice form the basis of this note. As a typical example, suppose that I wish to select a route from Chicago to Liverpool, via New York, from among four railroads from Chicago to New York which I would consider, and three lines of steamers from New York to Liverpool. Having taken any railroad to New York, I may go to Liverpool on three different lines. Since I have four railroads from which to choose, the total number of ar- rangements is 4 x 3, or 12. In general, if one choice is made in n ways and then another independent choice made in m ways, the total num- ber of arrangements for the two choices is n x m, or nm. Likewise, for any series of choices, the total arrange- ments of all choices is the product of the separate numbers for the separate choices. 10. Chance. The chance of selecting a given object from among a number of objects decreases as the number of objects increases. If among five balls in a bag there is one white one, the chance of selecting the white one at a random choice is 1 to 5, or |. In general, the chance of selecting one special object among n objects is - • The chance increases if there are more favorable possibilities. If there are ttco white balls among five, the chance of drawing a white ball is twice as great as it is if there is only one, i.e. the chance is f . In general, the chance of selecting one of m objects out m of a total of n objects is J n 361 362 APPENDIX EXERCISES I: NOTE III — CHOICE AND CHANCE 1. Two doors in one room and three doors in another (not adjoining) room open on to"" a common court. In how many ways may one go from one room to the other ? 2. A man has three coats, two vests, and five pairs of trousers. In how many ways can he dress ? 3. A boy can go to school by five different roads. In how many ways may he go and return ? In how many ways can he arrange his trips on six different days ? 4. There are eleven horses in a pasture and nine saddles in a barn. How many choices of saddle and horse may be made ?• 5. What is the chance of selecting a black ball out of a bag containing five black balls and one white ball ? 6. A bag contains six grains of white corn and five grains of yellow corn. What is the chance of selecting a white grain the first time a grain is taken ? If a white grain is drawn the first time and kept out, what is the chance of getting another white grain on the second drawing ? What is the total chance that a white grain will be drawn both times ? 7. What is the chance of throwing "heads" in tossing a coin? 8. If three coins are thrown up together, what is the chance that at least one will fall " heads " ? What is the chance that two will fall "heads"? 9. Dice are usually cubes marked on the six faces with the numbers from 1 to G. What is the chance of throwing a "2" with one die? What is the chance of throwing a "2" with two dice? What is the chance of throwing a double "2" (i.e. a "2" on each die) with two dice? What is the chance of throwing two numbers whose sum is ten with two dice ? CHOICE AND CHANCE M\:\ 11. Permutations. The number of arrangements of a given set of objects in order is called the number of permu- tations of them. Thus, given five pictures to be hung upon a wall, any one may be placed at the extreme right, then any of the four remaining ones next, then any one of the three remaining ones next, then any one of the two remaining ones next, and finally the last one must be hung at the extreme left. The number of possibilities is given by § 9. There are five distinct choices; the first with 5 objects from which to choose, the second with 4, and so on ; hence, the total number of possible arrangements isox4x3x2xl = 120. In general, if n objects are to be arranged in an order, any one of the n objects may be put first, then any of the remaining n — 1 next, then any of the remaining n — 2 next, and so on to the last. The total number of possible arrangements is n(n— l)(w— 2) -"4 • 3 • 2 • 1. 12. Permutations among a Limited Number. If less than the whole number of objects are to be arranged in order, the number of possible arrangements is evidently reduced. Thus, if we desire to select four out of ten candidates for an office and arrange them in order of merit, the number of possible arrange- ments is smaller than if all ten were to be arranged in order of merit. Any of the ten may take first rank, any of the nine remain- ing second rank, any of the eight remaining third rank, any of the seven remaining fourth rank; but here we must stop. The total number of arrangements (or permutations) is 10x9x8x7 = 5040 ; whereas, if all ten were to be arranged in order, the total number of arrangements (or permutations) would be 10x9x8x7x6x5x4x3x2x1= 3,628,800. In general, if from among n objects we are to make an arrangement of m objects in an order, we ma) 7 choose any one first, then any of the remaining n — 1 second, and so on for m steps. The last choice will be n — m + 1 objects, 364 APPENDIX since n — m will still remain after the last choice. The number of arrangements is, therefore, n(n — l)(w — 2) ••• (n — m 4- 2)(n — m+ 1). 13. Factorial Notation. A convenient notation for the kind of expressions just found consists in writing 2 ! for 2 • 1 ; 3 ! for 3-2-1; 4 ! for 4 • 3 • 2 • 1, etc. ; in general, w! = w(>-l)0-2)...4.3.2.1. The sign n ! is read " factorial /*." The permutations of n objects, m at a time, is often de- noted by P B) m . If m = w, all the objects are to be arranged ; we then write P n , n for the number of permutations. Using this notation, the results found above are P =n\ and P B] m = — — - • in — m) ! EXERCISES II: NOTE III — FACTORIALS ; PERMUTATIONS Compute the value of: 1.5! 2.6! 3.12! 4. (8!) -s-(4!). 5. (5 ! x 4!)-(6 !). ^- P 3, 1 5 -* 5, 2 ) Pi, 4 5 -* 8, 8 ' •'15, 5 5 " { 9, 4 5 -^ n+1, n-1 5 *n+k, k 'l -*12, 10- Write in abbreviated form and compute the number of per- mutations of: 7. 8 objects taken 3 at a time. 8. 15 objects taken 7 at atime. 9. k objects taken 1 at a time. 10. n + 1 objects taken n — 1 at a time. 14. Combinations. If we merely wish to select objects from among a given set of objects without arranging them in order, many of the permutations become equivalent. Thus, if among the ten candidates of the problem in § 12 we wish to select four without placing them in order of merit, the same four men would be arranged in 4 ! = 4 • 3 • 2 . 1 = 24 different arrangements in § 12 ; these arrangements are all equivalent if we do not specify the PERMUTATIONS AND COMBINATIONS 365 order. Hence, the number of combinations {i.e. not counting differ- ent arrangements of the same ones) is ^ ? of the previous number, i.e. in. 9- 8-7 210. 4 • 3 • 2 • 1 In general, if C„ fTn represents the number of combina- tions {i.e. selections not counting different arrangements of the same objects) among n objects chosen m at a time, EXERCISES III: NOTE III — COMBINATIONS 1. Find the value of (7 !)-s-(4 ! X 3 !). 2. Find the value of C 7i 3 ; C 5>4 ; C 5iS . 3. Find the values of C 4|2 ; A, 2 5 nn( l C^j-f-P^a- 4. Find P 5>5 ; P g5 ; hence, find (7 8|5 . 5. If a farmer has twelve horses, how many different teams of two horses each may he use ? 6. How many different and distinct committees of five may be selected from a group of 15 men ? 7. In how many ways may 3 books be selected from 12 books ? 8. In how many ways may the sum seven be thrown with two dice marked on the faces with the numbers from 1 to 6 ? With three dice ? 9. In how many ways may three debaters be chosen from a squad of 18 ? How many if six must be chosen ? 10. In a plane are ten points, no three of which are in a straight line. How many triangles may be formed with three of the points as vertices ? 11. In how many ways can four men and three women be selected from eight men and seven women? NOTE IV. INEQUALITIES 15. Operations on Inequalities. We have used a few inequalities, but we have not worked with them systemati- cally. The signs < (read " less than ") and > (read " greater than ") are already known. A few statements that will be understood at once are now given : (1) \i a~>b, then b b and b >c. (3) If a > b and b > c, then a>c. (^ is read " greater than or equal to," ami < is read "less than or equal to/') (4) If a ^b and b < c, then a 5, then ka> kb if k > 0. (6) If a>b, then ka b and c> d, then a + c>b -f d. (8) If a > b, then a ± x>b ± x where x is any number. These rules, together with a clear understanding of what is intended, enable us to work with all ordinary inequalities. Ex. 1. For what values of a; is 2x — 3>6 — x? Subtracting 6 — x from each side, we get (2a:-3)-(6-a;)>0, or, 3 x - 9 > 0. Add 9 to each side : 3 x > 9. Divide both sides by 3 : x > 3. Hence, if x > 3, then 2 x — '3 > 6 — x. 366 INEQUALITIES 367 16. Graphical Solutions. The problem just solved may be done graphically. Tims, let I = 'lx — 3, r = 6 — x where r and I denote the right and left sides of the above inequality. Plot- ting each of these on squared paper, we have two straight lines, as shown. It is clear from such a figure that / > r whenever the line l = 2x — 3 is above the line r= 6 — x. This happens evidently for all values of x greater than x = 3, for the lines cross at x=3, and they surely do not cross again. We may use this fact to advantage in harder examples. Let us find the values of x for which x 2 + 2 x - 8 > 0. Call I = x 2 + 2 x - 8 the left side and draw the figure ; it is as shown (compare pages 183, 204). We see that I > when the curve is above the main horizontal ; this hap- pens twice, once to the left of x = — 4, once to the right of x = + 2. These points are to be found by solving the equation £ = 0, i.e. Tori J% s& iV V J/ 7 N>/ / / Fig. 72. i J J if>- V I I l:o r -6- A t 4 *- r A t 1 l>0 \ ^ t nr o J \ 1 + t - 7 I 7 \ Kb 7 \ n=i 2 + 2.c-8 \ / x 7 A r "52 X 2 + 0, Fig. 73. of which the solutions are x = + 2, x = — 4. Likewise, / < 0, i.e. x 2 -(- 2 x - 8 < for all values of x between x = - x = + 2. 4 and 368 APPENDIX Similarly, in any case we may represent the left and right sides graphically and see when one exceeds the other. It is evident from a figure that we should first find when the two sides are equal, for the points for which this is true are points of intersection of the graphs. EXERCISES I: NOTE IV — INEQUALITIES Write down in the following examples the conclusions you conld draw from what is given, as directed: 1. 10 > 6 ; multiply both sides by 5 ; by — 5 ; by ^ ; by — ^ ; divide by 2 ; divide by i ; add 4 to each side ; add — 4 ; add — 20 ; subtract 4 ; subtract 6. (Perform each operation on the given inequality only.) 2. - 10 < - 6 ; multiply by + 5 ; by - 5 ; by \ ; by - \ ; add 10 to each side; subtract 4. 3. 3>2; multiply by x (if x is positive); divide by x (positive) ; add x to each side ; subtract x. 4. 3 > 2; multiply by x (if x < 0); add x; subtract x. 5. Given Sx — 2 > a; + 4, subtract x from each side ; then add 2 to each side of the resulting inequality; then divide both sides by 2. Draw the figures and find the values of x for which : 6. 3 x — 2 > x + 4. 9. x 2 > 6 — x. 12. x 2 + 4 x — 5 > 0. 7. * - 2 > 3 — x. 10. x 2 > 5 x - 4. 13. x 2 - 2 x - 3 > 0. 8. 4.x-2>2x + 3. 11. x"0. NOTE V. THE BINOMIAL THEOREM 17. Formula. We have learned how to write down certain powers of binomials by inspection. (See §§ 57, 62, pp. 93, 99.) Thus, we had (1) (a + by = a*+2ab + b°- ( V . 93), (2) (a + bf = a 3 + 3 a% + 3 aP + & 3 (p. 99), (3) (a + by= a* + 4 a s b + 6 aW + 4 a6 3 + 6 4 (p. 99), and so on. We shall now prove the following formula, called the binomial theorem, which gives the result for any positive integral value of the exponent n : (1st term) {2d term) (3d *erm) (4) (a + 6)" = a n + na n ^b + n(n ~ *) a n ~W (Itth term) wQ-l)Q-2) 3&3 frfwkrrf'writel 3 f 1 5th term J (WA term) . w(w-l)(n-2) ••• (w-r + 2) a „-r+ijr-i (r-1)! ((/' + j)*a (wm) | ft(w-l)Q-2) ••• (w - r + 1) aW _ rftr + _ r ! (»itfA terwi) (/as< term) + ab n ~ l + 5 n , where r ! means 1 • 2 • 3 ••• (r — 1) • r, as on p. 364. To prove this formula, we shall show that if it holds for any positive integral value of n, it holds also for the next higher value of n. For this purpose, assume for a 369 370 APPENDIX moment that (4) is correct and multiply both sides by (a + 6) ; this gives (a x 1st term) (b x 1st term) (a x 2d term) (5) (a + 6)» +1 = [a n+1 ] + [a n b + na n b] (fix 2d term) (a x 3d term) (h x 3d term) + (a x hth term) r ,,~ nfn — 1 ^ ,,oH [ student write \ (6 x r-l) s . 3. (x-y) 5 . 6. (4m 2 + v 8 )*- 9- (rs-O 4 - Write out the first three terms of: 10. (x-yy°: 12. (a + 6) 20 . 14. (1 - x) 10 . 11. (4x 2 + 3?/ 3 ) 5 . 13. (m; + l) 9 . 15. (2 + 4 a) 6 . 16. Write out the 6th term of (a + ?;) 10 . 17. Write out the 5th term of (1 + x) u . 18. Write out the 12th term of (2 a - 3 6) 20 . NOTE VI. EUCLIDIAN METHOD H. C. F. AND L.C.M. 19. Euclidian Method. In Chapter V, p. 118, we defined H.C.F. and L. CM., and showed how to find them if the given expressions could be easily factored. The following method applies to all polynomials, no matter how difficult the factoring may be. Ex. 1. Find the H.C.F. of ^ = 3ar-llx + G and B = 12x 3 +x 2 -12a; + 4. Divide the expression of higher degree (B) by the other (J ) : 3x 2 - llx + 6 = ^4 4x + 15 = Quotient B=12x s + x 2 - 12 x + 4 12 x 3 - 44 x 2 + 24 x 45 x 2 - 36 x + 4 45 x 2 - 165 x + 90 129 x - 86 or 43 (3 x - 2) = Remainder Calling the quotient Q and the remainder R, we have, as always, B= Q- A + R i.e. 12 x 3 + x 2 - 12 x + 4 = (4 x + 15)(3 x 2 - 11 x + 6) + 43 (3 x - 2). Any factor of both A and B must also be a factor of the remainder R, for R = B — QA, so that a factor of both A and B is a factor of B — QA, which is nothing but R. For example, a factor of 12 x ? + x 2 — 12 x + 4 and 3 x 2 — 11 x + 6 must also be a factor of 43 (3 x — 2). All common factors of B and A are also common factors of A ami R, and vice versa. Hence, we may take instead of the given problem the simpler one : to find the common factors of A and R; i.e. of 3 x l - 11 x + G and 43 (3 x - 2). This new problem may be done by inspection, or if this is too diffi- cult in any case, by repeating the process. In our example it is clear that 3 x — 2 is a common factor; it is also the only one, by the argu- ment just used. Hence, the required H. C. F. is 3 x — 2. 373 374 APPENDIX Ex.2. Find the H.C.F. of A = x* + ±x 3 + 2ar*-f-a-2 and B = 2x 5 + 5x i -7x 3 -2x i -5x + 7. Dividing as above, we find B = 2 x 5 + 5 x 4 - 7 x 3 - 2 x 2 - 5 x + 7 2 x 6 + 8 x 4 + 4 x 3 + 2 x 2 - 4 a _ - 3 x 4 - 11 x 8 - 4 x 2 - x + 7 - 3 x 4 - 12 x 3 - 6 x 2 - 3 x + 6 x 3 + 2x 2 + 2x + l = Remainder As above, we may now take, instead of the given problem, the new problem of finding the H. C. F. of the divisor and the remainder : x* + 4 x 3 + 2 x 2 + x -2 = ^1 3 = Quotient x 4 + 4 x 8 + 2 x 2 + x - 2 x 4 + 2 x 3 + 2 x 2 + x 2x 3 -2 2x 8 +4x 2 + 4x + 2 x 3 + 2 x 2 + 2 x + 1 x + 2 — 4x 2 — 4x — 4 = — 4 (x 2 -f x + 1) = Remainder Again, take the divisor and the remainder and discard the factor — 4, which clearly cannot be a common factor: x 2 + x + 1 X + 1 x 8 + 2x 2 + 2x + l X 8 + X 2 + X X 2 + X + 1 X 2 + x + 1 = Remainder Since this remainder is zero, x 2 + x + 1 is a factor of x 8 + 2 x 2 + 2 x + 1 ; hence, it is the H. C. F. of the given expressions. At any stage in the process we may find the H. C. F. by inspection, as in example 1, if it is easy to do so. Other- wise we repeat the same process as often as is necessary until we find a remainder zero ; the last divisor is then the required H. C. F. Numerical factors may be inserted or taken out at any, time, for it is easy to see by inspection whether or not such factors are common factors of the yiven expressions. Usu- ally no account is taken of these purely numerical factors. Any common factors that can be found by inspection EUCLIDIAN METHOD: H.C.E. AND L.C.M. 375 should be removed at once. Sometimes the H. C. F. can be found in this manner, as in Chapter V. The L. C. M. of two polynomials can be found by § 74, p. 129, after finding their H. C. F. A precisely similar process holds for numbers ; but care should be taken not to insert or discard numerical factors in dealing with problems in numbers. EXERCISES: NOTE VI — H.CF. AND L.C.M. [Solve by inspection as in Chapter V wherever possible; otherwise use the process explained above.] \ Find the H. C. F. and the L. C. M. of : 1. 35, 75, 25, and 65. 2. 42, 105, 147, and 63. 3. 1884 and 2079. 4. 3718, 5269, and 12,168. 5. 35 a 4 b\ 84 a%\ and 63 a 3 6 3 . 6. or 9 - 4 x — 21 and x 2 - 5 x - 14. 7. a 8 - 1, x 2 - 1, and (x - 1) 2 . 8. x A + 3 x 2 + 3 x + 2 and x 3 - 2 x 2 - x - 6. Find the H. C. F. of : 9. & + tf _|_ 3 x + io and a? + x 2 - 5 x - 6. 10. ar 5 — x A — x + 1 and 5 aJ 4 — 4 x 3 — 1 . 11. x 2 + 6 x + 9 and ar + ar + x + 21. 12. z 4 - 2 0? + 3 ai 2 - 4 a; + 2 and a? - 2 a; 4 + 3 X s + a- 2 - 8 x + 5. 13. 4 a,- 1 - 18 a- 2 - 3 + 19 aj and 19 a- 2 - 12 ar 5 + 2 a; 4 + 9 - 6 x. 14. x* — x 3 + x — 1 and a; 4 + ar 3 — a; — 1. 15. 3 a; 4 - 4 x 2 - 4 and 3 x 4 - 8 x 2 + 4. 376 APPENDIX 16. X s — 1 and ar 4 + 2 ar 3 + 3 ar' + 2 a; + 1. 17. x 3 + 3x* + 7 x + 5, x* - 2 ar 3 + 4 x 2 + 2 x — 5, and x 5 + 5 ar 3 f a^ + 5. 18. ar 5 - 5 x + 4, a; 4 - 2 ar 5 + 1, and x 6 + 4 ar 3 - 3 x - 2. f ar* + 3 a; 4 + 4 a; 3 - 9 x 2 - 5 a? + 6, 19 ' { aJ> + 3 x * + 7 ^ + 10 x _ 12. f 4 + 7 a; + ar 2 + a,* 5 - x 4 , 20. < [7x + 4-2ar 5 -3ar 3 -3a: 4 . J x*-3x 3 + a.- 4 + 3 x -2, 21 ' |2a,'-9ar 2 + 4ar 3 + 3. 22. ar 3 -6ar 2 + lla;-6, as 8 - 9 a; 2 + 26a:- 24, and x'-Sx 2 f 19 x -12. Find the L. C. M. of the following : 23. 9 xy 2 , 6 x 2 y 3 , and 15 y 2 z 2 . 24. ar 5 — 1 and x 3 + 1. 25. 4 a; 2 - 9 ?/ 2 and 4 x 2 - 12 ar?/ + 9 y 2 . 26. (a + 3), (a 2 - 9), 3 a + 15, and 15 a - 45. 27. 2 ar 3 - ar 2 + 2 - 3 x, 6 ar 2 + 4 x 3 - 4 - 2 x, and 4 ar 3 - 5 aj +2. 28. ar 2 — 11 x + 24, x 2 — 6 a; — 16, and ar 2 — a: — 6. 29. c 2 - (a + &) 2 , b 2 - (a + c) 2 , and a 2 - (6 + c) 2 . 30. (a,- 2 -2x + 1), (.« 2 - l) 2 , ar 3 - 1 . 31. a; 4 — ar 3 + x — 1 and a; 4 + ar 3 — a? — 1 . 32. ar 3 - 9 x 2 + 26 x - 24 and ar 3 - 10 x 2 + 31 a? - 30. 33. x 2 + 6 x + 9 and a; 3 + .r + a? + 21. 34. (a + &) 2 - (c + d) 2 and (a + 6) 3 - (c + d) 3 . 35. a; 4 + 11 x 2 + 25 a; and x- 4 + 2 a- 3 + 11 x 2 + 10 a: + 25. 36. ar 3 + 3 x 2 + 7 a; + 5 and e 8 + 5 ar 3 + x 2 + 5. 37. ar 3 + y 3 , x 3 — y 3 , (x — y) 3 , and ar 3 + 3 x 2 // -f 3 ar?/ 2 + 1/ 3 . 38. ar 2 + 2 xy + ?/ 2 , ar 2 — 2 xy + ?/ 2 , a: 2 — y 2 , and a; 2 + y 2 . NOTE VII. CUBE ROOT AND HIGHER ROOTS 20. Introduction. When cube roots and higher roots of numbers are needed in practical work, they can be found approximately most easily by the use of loga- rithms (see § 163), which computers almost always use. Any root of any number can be found approximately by trial, as in § 97. This process may be very long. An old method, analogous to that used in § 97 for square root, is given here chiefly for its historic interest. 21. Cube Roots of Numbers. Ex. 1. Find the cube root of 91609.86. Notice that l 3 = 1, 10 3 = 1000, 100 3 = 1,000,000, and so on. Also, .l 3 = .001, .01 s = .000,001, etc. Mark off the number into periods of three figures each, in each direction from the decimal point, thus : 9i, 009.86. This assists in estimating the first figure of the root. The process is based on the formula (a + b) 3 = a 3 + 3 a 2 b + 3 ab 2 + b 3 = « 3 f 6(3 o 2 + 3 ab + ft 2 ), and consists in choosing convenient values of a and of h and veri- fying; and then in repeating the process, choosing as a new value for a the whole root already found, after the manner of § 97. 40 s = 3 x 40 2 = 4800 91,009.86 40 64,000 5 27,609.86 45 91,609.86 45 91,125 .07 484.86 45.07 91,609.86 45.07 91,550.911843 58.948157 45 3 = 3 x 45 2 = 6075 45.07 3 = We take for a the largest convenient number whose cube is con- tained in the given number, in this case a = 10. If l> is the remain- ing part of the root, then 3 a-b + 3 ab' 2 + b s is the remainder of the cube. Since b is small compared with a, the principal term of this is '-'xi-b. 377 378 APPENDIX Hence, 3 a% is nearly equal to 27,609.86. That is, 3 x 40 2 x b = 27,609.86 nearly. Hence, b = (27,609.86) -=- (3 x 40 2 ) nearly = 5 nearly ; this is called the trial divisor. Hence, the root is a little over 45. Start again : a = 45, cube a, subtract a 3 , determine b — .07, etc., repeating this process, each time taking as a the part of the root already found. Each time b is found by dividing the remainder by 3 a 2 . The last two figures of the root are found by simple division. This work can be tabulated in slightly different form, as follows : 91,609.86 40 40 3 = 64,000 3 x 40 2 = 4800 27,609.86 5 3 x 40 x 5 = 5 2 = 600 45 5 x 5425 = 27,125 3 x 452 6075 484.86 .07 3 x 45 x .07 = .072 = 9.45 .0049 .07 x 6084.4549 = = 425.911843 45.07 3 x 45.07 2 = 6093.9147 58.948157 .0097 45.0797 Any Answer to four places. EXERCISES I: NOTE VII — CUBE ROOTS OF NUMBERS Find the cube root of : 1. 262,144. 3. .001906624. 5. 12.812904. 2. 69,426,531. 4. 259,694.072. 6. 64,144.108027. Find the cube root of the following to four figures : 7. 25,473. 8. 46.32. 9. 3.4674. 10. 65,463,257.0423. 22. Cube Roots of Polynomials. The cube root of a polynomial that is a perfect cube may be found in a similar way. The polynomial should always be arranged according to the ascending or descending powers of the same letter. The most important points to remember are (1) the trial divisor is 3 times the square of the part of the root already found ; (2) the complete divisor is 3« 2 + .3 ^ + p where a is the part of the root previously found and b is the new term. CI BE ROUT AND HIGHER ROOTS 379 Ex. 1. Find the cube root of : 114 x 4 - 171 x 2 - 135 x - 27 + 8 a 6 + 55 x* - 60 a? 5 . 8x 6 -60 x 5 + 114x 4 + 55 x 3 - 171 x 2 - 135x- 27 2x 2 8x 6 _ -5x 12 x* -30x 3 + 25x 2 -60x 5 + 114x 4 + 55x 3 -171x 2 -135x-27 - 60 x 5 + 150 x 4 - 125 x 3 12 x* ■ -30x 3 + 25x 2 12 x* 12 x 4 - -60x 8 + 75x 2 - 18 x 2 -60x 3 + 57x 2 -t- 45 x + !> + 45 x + _ 36 x 4 + 180 x 3 - 171 x 2 - 135 x - 27 - 36 x 4 + 180 x 3 - 171 x 2 - 135 x - 27 Answer : | 2 x 2 — 5x — 3 The first term of the root is evidently 2x 2 ; hence, the first trial di- visor is 3(2x 2 )= 12 x 4 . Dividing the first term of the first remainder by this trial divisor, we find the next term of the root : — 5 x. The first divisor completed is, therefore, 12 x 4 + {3(2 x 2 )( — 5 x) + ( — 5 x) 2 }= 12 x 4 — 30 x 3 + 25 x 2 . The remaining steps are repetitions of these. 23. Higher Roots. Since the fourth power is the square of the square, the fourth root is the square root of the square root. Likewise, the sixth root is the cube root of the square root. Other roots may be found in a similar manner, or by a method similar to that used for cube root. Any root of any number can be found approximately by a figure (p. 190), or by logarithms (§ 165, p. 342). EXERCISES II: NOTE VII — CUBE ROOTS OF POLYNOMIALS Find the cube root of: 1. 343a 3 -441a 2 6 + 189o6 2 -27& 3 . 2. 8 m s - 12 m s n + 30 m*n* - 25 m s n* + 30 mhi* - 12 mn 5 + 8 n G . 3. k« + 12 k" + 63 k* + 184 k 3 + 315 k 2 + 300 k + 125. 4. a 3 -12a 2 + 54a-112 + ^- 4 ?+4- a or a 6 5. 6.^ + - 4 + ^ + 20 + ^ + ^+15ar°. x A as" x 2 6. 184 c* + 315 c 2 + 63 c 4 + 300 c + 125 + 12 & + c . NOTE VIII. LIMITS AND INFINITE SERIES; IRRATIONAL NUMBERS 24. Introduction. In this article we shall state briefly the fundamental notions connected with limits. These propositions are of a far more intricate character than are most of the topics treated in this book, and it is recom- mended that these articles be studied only by mature students. 25. Limits. In case the difference between a variable quantity v and a constant quaiiUty k can be made to become and remain as small numerically as we please, the variable v is said to approach the_ constant k as its limit, and we write Lim v = k. Usually we shall have some definite process for deter- mininsr whether or not the difference between the variable and the constant can be made to become and remain as small as we please. Ex. 1. Imagine any porous body (e.g. a brick) soaked in water till its weight is increased. Let k be the weight of the body when dry, and let v be the weight when wet. If the body is heated, the water passes off in steam, and the weight v is variable. It is evident that by continued heating we can drive off as much of the water as we please. Hence, we can make the difference v — k become and remain as small as Ave please ; we say : Lim v = k. Ex. 2. It is found by trial that a rubber balloon will burst when it is so filled with gas as to have a diameter of six feet. If v means the volume of the balloon, it is evident that we may 380 LIMITS AND INFINITE SERIES 381 expand balloon so that its volume is as near as we please to ~ • (6) 3 cu. ft. = 288 7T cu. ft. ; hence, o Lim v = 288 w. In this example the variable v cannot be made equal to its limit, for the balloon would burst if expanded to exactly six feet diameter. In example 1, on the contrary, the variable may equal its limit, for we may drive off all the water and leave v = k. Ex. 3. Consider the expression : 1 - Q)» V = ±J-L- . 1-1 We may. make this expression become and remain as near to ^ as we please, by making n sufficiently large. For let ) Then, v -I or ' 2"- 1 2 «2 ••• ((«-l)times)"-2 Since the denominator contains as many factors 2 as we please, this fraction can evidently be made to become and remain as small as we please, numerically, by taking n large enough. Hence, * where we understand that the limit is to be taken by making n as large as we please. Ex. 4. Similarly, 1 — f- x 1 _o 1-i 1-0)" i _ __ffi. = -(I) 1-i 1-i 1-i 1 1 1-- X —' where x is any number numerically greater than 1, approaches as its limit i x or 1_I x ~ l 382 appp:ndix For let then, or, (v-k) v — k -©' 1 x 1-i 1-1 r i X 1-1 a; (x — 1) x n_1 (a; — 1) • a; • x ••• ((«.— 1) times) If x is numerically greater than 1, the denominator can be made as large numerically as we please ; hence, v — k can be made to be- come and remain as small numerically as we please by taking n suffi- ciently large. Hence, x x^l' Lim 1 1-1 x 1-1 X if x is numerically greater than 1. 26. Infinite Series. The last examples above have a direct application in the question of infinite series. By an infinite series we mean an unending sequence of terms connected by + signs: a + «x+ « 2 + a 3+ '*' + a n + •"• (It is improper to say that an infinite series is the sum of an unending sequence of terms, for the word " sum " is defined only for a finite number of terms.) We cannot find the sum of such an infinite series directly, but we can add together as many of the terms as we please ; thus, we can find the sum of the first n terms : S n = a Q + a l + a 2 + ••• + a„_ 1 by direct addition. We say that the sum of the series is the limit of S n as n increases indefinitely, if there is a limit : S= Lim S„. if Lim S n exists. F »i LIMITS AND LNTINITE SERIES 383 27. Infinite Geometric Series. As an example, consider an infinite geometric series (see § 159, p. 328): a -+- ar + ar 2 + ar 8 + •• • + ar n + •••, where r is numerically less than 1. The sum of the first n terms (see p. 329) is \ r n S = a + ar + «r 3 + • • • + ar" -1 = a ■ 1 — r The sum of the infinite series is therefore / 1 _ r "\ iS= Lim (*S'„) = Lim ( a ■ J • Since r is numerically less than 1, let us put r = - ; then, x is numerically greater than 1, and we have x S = Lim ( a • ] = Lim i-i'- lv ' r. a x = a i-l x or, S=a a -, by example 4, § 25. 1-r 1-r Hence, the sum of an infinite geometric series in which r is numerically less than 1, is a 1-r Ex. 1. In the infinite series a = 1, r = ^ ; hence, # Ex. 2. The series a _ o 1-r 1-i gives 1 111 i_u i i o-i ... a = 1, r = - 1; hence, # = -^— = - — — — - = -. 1-r l-(-i) 3 384 APPENDIX Ex. 3. The series 3 + * + A + tK + A + - gives a = 3, r = i; hence, £ = ^- = ^ = ^ = 3f. Ex. 4. The series 3 _|_ _3_ _l_ _ 3 i 3 _ i . . . ° ' 10 i 1 00 I T000 ^ gives a = 3, r = ^; hence, # = -^— = —?— = ^ = 12. 10 1-r 1- t l 9 3 Note: this series may be written 3.3333 ••• in the form of a repeat- ing decimal, and we may write 3£- = 3.3338 ••• Ex. 5. The repeating decimal .27272727 •••, which is often indicated by .27, is equivalent to the geometric series : 2 7 _1_ 2 7 |_ 2 7. _|_ . . . . itto • i oooo i i oooTro~o ' > here a = JUL, r = * ; hence, S = — — = tVo =?I = A. 1 0U' ioo> 1 _ r 1 _ l QQ 11 ' consequently, J\ = . 272727 •••. T »o Jy 1X Ex. 6. The decimal 5.743216216216216 ••• (often written 5.743216) repeats the figures 216 forever; it is equivalent to the SerieS *16 216 5.743 + £±2 + ^ + .... 10 6 10 9 T Taking a = , r = — -, we find s 10 6 10 3 216 216 216 10 s 216 24 10 6 10 9 = 1 _ J_ ~ 999000 ~ 111000' 10 3 hence, 5.743216216216 = 5 + — + 24 = 5 + 8249T . 1000 111000 111000 In general, any repeating decimal may be regarded as a terminating decimal + a certain geometric series ; such a deci- mal may therefore be evaluated by the preceding formula for the sum of an infinite geometric series ; the result is always a rational fraction. LIMITS AND INFINITE SERIES 385 EXERCISES I: NOTE VIII — INFINITE GEOMETRIC SERIES Find the values of the following infinite geometric series: ° '>■ ° + 3. 3 + 2 + | + f + ^H 5 ' ,r ' 5 3 9!j.fxf±ij.... * "T^TfTf 1. 2+=+-+—+ 4 5 l IS I 4 5 _i_ 186 l . . , 9 O 9 o 9 ^ _j_ w w _i_ A w ?"•"« K3' ' 5 _*>_ 3_ 9._27 8.1 O.)"0 **■ w 2 8 32 128 . Find the values of the following repeating decimals ; check each answer by long division : 6. 2.222 •-.. 9. 5.133333--. 12. 10.1010101010—. 7. .7777—. 10. 42.716161616.-.. 13. 26.308308308 •••. 8. .232323-... 11. .0454545-... 14. 83.83838383 •••. 28. Other Infinite Series. We have seen how to find the sum of an infinite geometric series if r is numerically less than 1. In general, as in § 26, we say that the sum of any series (1) tfo+fli + '^-l \-a n +-- is S= himS n = Lim (a + a x + a 2 + ••• + a n _ 1 ) if this limit exists. If Lim S' n exists, the series (1) is called convergent ; otherwise (1) is called divergent. Ex.1. Thus, 1 + 2 + 4 + 8+ ••• is an infinite geometric series which diverges, for #i = lj #2 = 3, £3 = 7, £ 4 = 15, — , £ n =(l + 2 + 4+...+2«- 1 ), and S n becomes greater than any quantity one may name; hence, Lim S n does not exist, and the series diverges. 386 APPENDIX Ex. 2. Another divergent geometric series is 1 — 1-f-l — 1-f-l — 1 ••-, where a = 1 and r = — 1 ; here S n = + 1 if n is even, and S n = if n is odd. There is therefore no constant which S n approaches ; hence, Lim (M n ) does not exist, and the series diverges. Ex. 3. A series which is not geometric, and which diverges, is i_ J _i + i + i : + i + i + i+ i+ i + .... This series diverges, for if Ave arrange it in the form inclosing the next eight terms ending with T \, then 16 terms ending with gV, etc., we see that each parenthesis exceeds £ ; hence, S n > 1 + \ + \ + • • • (as many times as we please) + \. It follows that S n approaches no limit; hence this series diverges. EXERCISES II: NOTE VIII — OTHER INFINITE SERIES 1. Show that the series 1+3+5+7+9H diverges. 2. Show that the series 2— 1+3— 1+4— lH diverges. 3. Show that the series 1 + \ - 1 + \ + 1 + 1- - 1 + T V + • •• diverges. 4. Show that any series A + A { -{- A 2 -\ will diverge if A n does not approach zero as n increases indefinitely. 5. Show that if a + a i + a 2+ '•' is a convergent series of positive terms, and if 6 + b x + b.,-\ is another series of positive terms snch that b u < a n , then the second series converges also. 29. Irrational Numbers. We have already considered many irrational numbers ; thus, V2 is irrational. (See pp. 184, 186.) In fact, if the positive integer a is not the perfect square of an integer, Va is irrational, for the sup- position Va = — , where m and n are each integers without n o a common factor, leads to the absurdity a = — -, i.e. an integer is equal to a fraction. LIMITS AND INFINITE SERIES 387 A rational number is the quotient of two integers ; an irrational number is any real number that is not rational. (See pp. 180, 284.) In case of V2 we saw that we could get a rational number whose square is as close to 2 as we please, and we wrote (1.4)2 <2< (1.5)2, (1.41)2 < 2 < (1.42)2, (1.414)2 < 2 < (1.415)2. In fact, the square of any rational number is either greater than 2 or less than 2. Thus, the rational numbers are divided into two classes according as their squares are greater than or less than 2. This division into two classes practically defines V2, as above. Whenever, in such a fashion, all rational numbers are divided into two classes such that any number in one of the classes is greater than any number in the other class, we say that this division defines a certain irrational number, which is called the cut number. Thus, V2 is the cut num- ber which separates the rational numbers whose squares are less than 2 from those whose squares are greater than 2. We may also think of V2, for example, as the limit of a sequence of rational numbers a , a v a v •••, a n , •••, where (2 — a 2 „) approaches zero as n increases indefinitely. Thus, V2 is the limit of the sequence of the numbers 1.4, 1.41, 1.414, 1.4142, •••, which are obtained in the ordinary square root process. 30. Operations on Irrationals. Any irrational can be expressed by means of rationals to any degree of exactness 388 APPENDIX required. Thus, a/2 may be expressed, by means of one of the numbers, 1.4, 1.41, 1.414, ••• to any desired number of decimal places. Similarly, for any irrational, we may decide upon two integers between which it lies, then tenths between these, and so on ; eventually the irrational is expressed to any number of decimal places desired. If two irrationals are given, say V2 and V% we may express each of them to any number of decimal places. Having done so, we may add, subtract, multiply, or divide these approximate values of the two to get the sum, dif- ference, product, or quotient of the approximate values. We now define the sum of two irrationals, or of one rational and one irrational, to be the limit approached by the sum of the approximations as the number of true decimal places in each approximation increases indefi- nitely. Likewise, the difference, product, quotient, etc., are defined as the limits of the corresponding approximate values. It can be shown that by these definitions the axioms of § 24, p. 35, remain true. As an example consider V2 x V3. The rationals that define V2 are all those whose squares are less than 2 ; the rationals that define V3 are all those whose squares are less than 3. If we multiply these approximate rational values together, it is clear that the approximate products are all those rational numbers whose squares are less than 6. It follows that V2 • V3 = V6, a result previously ob- tained by analogy, without strictly logical proof. Simi- larly, Va • V£= Va6, and, in general, Va ■ Vb = -\/ab. This is the rule of p. 286, which may now be strictly proved. The operations on radical quantities and on other irra- tionals used above should finally be revised in the manner just shown in order to justify them completely ; but it will not be advisable to do this in detail in this book. It should be noticed that the definition given on p. 183, by LIMITS AND INFINITE SERIES 389 means of the graph y = :c 2 , really includes the notion of approximation just used. In general, for elementary pur- poses, a geometrical definition of irrational numbers, by the graph, mentioned on p. 186, will be found most suitable, since it is easy to follow and is also quite in accordance with the most logical definitions. EXERCISES III : NOTE VIII — OPERATIONS ON IRRATIONALS 1. Write out a definition of V2 as a cut number. 2. Give five terms of a sequence of increasing numbers whose limit is Vo. 3. Give five terms of a sequence of increasing numbers whose limit is V 5 ; also a sequence of decreasing numbers whose limit is v'3. 4. Multiply Vo by V5; show the result is a cut number which expresses the product of the cut numbers which give V3 and Vo, respectively. 5. Add the increasing sequences of numbers which ap- proach V3 and V5, respectively, term by term ; define the sum Vo + V5 by this means as a cut number. 6. Prove V« X V& = Va&. 7. Prove Va-s- V& = \/-- ' b 8. Show that the graphs of y = x' 2 and y = 2 serve to define the cut number V2. 9. Show that the graphs y = x- and x- + y=l serve to define the cut numbers x= ± V^. 10. Show that the graphs of y = x 3 , y = x*,y = x 5 , serve to define all cube, fourth, fifth roots. (See p. 190.) 11. Show that the curve x = 10" (see p. 336) defines log x as a cut number for every value of x. Hint. Show that lO can be really found for every rational y; hence log x can be found for values of x as near any given value of x, as we please. NOTE IX. IMAGINARY AND COMPLEX NUMBERS 31. Introduction. In the work of this book we have used the following classes of numbers : (1) Positive integers, defined by counting (see p. 1). (2) Positive rational fractions, the quotients of pairs of integers (see pp. 1, 118, 186). (3) Zero, defined by the equation + a = a (see p. 15). (4) Negative integers and fractions, defined by the equation a +(— a) = (see p. 14). . (5) Irrational numbers, defined by a cut (p. 387). These five classes of numbers are called real numbers. Just as mankind has found it desirable to invent and use these various kinds of numbers, it is convenient to invent and use a sixth kind of number. With real numbers alone we may perform any addi- tions, subtractions, multiplications, divisions, except divi- sion by zero. We may also raise any number to any integral power, and we may extract integral roots of posi- tive numbers. Indeed, we may extract any odd root of a negative number ; but an even root of a negative number cannot be expressed by the numbers we know at present. For this reason, the new kind of numbers arise in solv- ing even very simple quadratic equations. (See pp. 182, 212.) Thus, x 2 + 1 = gives z? = - 1 or x = ± V^T, but V— 1 is an even root of a negative number, and cannot be expressed in terms of the numbers of the five classes above. 32. Definitions. We shall denote the new number V — 1 by the letter i : (1) V-l=i, i* = (V-T) a =»-l; 390 IMAGINARY AND COMPLEX NUMBERS 391 and we shall work with this new number, as far as pos- sible, according to the rules for real numbers, observing always the axioms of § 24, p. 35. (See note, § 30.) The product bi of any real number b and the new num- ber i is called a pure imaginary number. The square of such a number is (2) (In) 2 = (bi) (hi) = b -i-b ■ i = bbii = bH 2 = b% - 1) = - b 2 by IV, p. 35. Likewise, (— ln") 2 = — b 2 . Hence, we say (3) V^2 = ± bi, and V^ = V-(±Vi) 2 = ± i/x • i if x > 0, so that the square root of any negative number — x may be written as the pure imaginary ± Vx • i. We shall always do this at once in any problem.* The sum (a + bi), where a and b are real numbers, is called a complex number, or simply an imaginary number. 33. Direct Operations. Addition, subtraction, and mul- tiplication are performed by working as if i were an un- known letter, and replacing i 2 by — 1 wherever it occurs. Ex. 1. (2 + 3 i) + (4 + 5 1) = 6 + 8 i, and, in general, (a + bi) + (c + di) = (o + c) + (b + d)i. Ex. 2. (2 + 3 i)(4 + 5 i) =8 + 22 i + 15 r = 8 + 22 /-15=-7 + 22t, and, in general, (a + bi)(c + di) = ac + (be + ad)i + bdi 2 = ac + (be + ad) i — bd = (ac — bd) + (be + ad)*\ These operations employ only the direct results of the axioms (p. 35), and the statement r = —1. * There is not the same advantage in dealing with square roots of nega- tive quantities and with square roots of imaginary numbers, in deciding which of the two answers shall be denoted by the sign V , as there was in dealing with square roots of positive quantities. In examples, however, we shall take V— x = + Vas i, to avoid undesirable complication. 392 APPENDIX Ex. 3. (2 + 3 i)(2 - 3 = 2 2 - (3 t) s = 4 - (- 9) = 13, and, in general, (a _|_ bi)(a - U) = ar- (bif = or - (- b 2 ) = a 2 + b 2 . The result in this problem does not contain the letter i\ hence, it is a real number. 34. Conjugate Complex Numbers. The importance of Ex. 3 above leads us to emphasize it by calling a— hi the conjugate to a + hi. The 'product of two conjugate complex numbers a + hi and a— hi is the real positive number a 2 + h 2 . This is often stated by saying that the sum of two squares is the product of the imaginary factors a + hi and a — hi : (4) a 2 + b 2 = (a + hi) (a - hi) . 35. Division. Division of complex numbers is based on this fact, for the divisor (or denominator) may be made real by multiplying both divisor and dividend (or both numerator and denominator) by the conjugate to the divisor (or denominator). _ 7 + 22 i = - 7 + 22 i 2-3t = -14 + 65 i- 66 i 2 X ' * 2 + 3* 2 + 3 1 '2-3* 4-9 i 2 -14 + 65i + 66 = 52 + 65i ^ 4 5 . 4 + 9 13 " Check i (2 + 3 i) (4 + 5 i) = - 7 + 22 i. (See Ex. 2, § 33.) Ex.2. l = l.=i = = *==±=-i. i i —i —%- +1 (We might have multiplied numerator and denominator by + t instead of — i; but — iwas taken in order to observe the general rule.) Ex.3. -1_=-A_.lzii = 2^2 1 = 1 _. Ex. 4. 1 + ?: 1 + * l — i 2 4-5* 4-5i 3 + 2 i 12-7i-10i* 3 -2 1 3-2/ 3 + 2i 9-4i 8 _ 12-7t + 10 __ 22 -7 i _ 23 _ ^ 9 + 4 13 13 13 i. IMAGINARY AND COMPLEX NUMBERS 393 36. Further Examples. If square roots of other nega- tive numbers than — 1 occur, we reduce them at once to the form above by the relation (3), i.e. V— x = ±^/xi. In the following examples the sign + is chosen in each case : Ex.1. (2 + V-4) + (3 + 2V-9) = (2 + 2 V- 1) + (3 + 2 • 3 V^I) = 5 + 8 i. Ex.2. (4-V^9)x(2 + V^l6) = (4-30(2 + 4i) = 20 + 10 i. -c, ol+V-4 1+2 i 1+2* -3 + 4 i 3,4. 1_V^4 1-2* l + 2i 5 5 5 Ex. 4. (2 + V^3) (1 + V^3) = (2 + VS 0(1 + V3 i) = 2 + 3 V3 i + 3 i 2 = - 1 + 3 V3 i. Ex.5. 4+V-5 = 4 + V5i 2-V3i 2 + V^3 2+V3i 2-V3J = 8 +(2 V5 - 4 V3) i - Vl5 i* 4-3 i 2 = (8 + V15) + (2 V5 - 4 V3) % 7 _8 + Vl5 , 2V5 -4V3 . 7 7 Failure to follow the directions given above may lead to error, for although some of the rules of ordinary algebra hold, there are other rules which do not hold (under the agreements we have made) for imaginary numbers. For example, V-l V^l = i • i = i 2 = -l; but if we attempted to use the rule of ordinary algebra: Va s/b = y/ab } we should obtain the incorrect result V^T V^l = + V+T = l. 894 APPENDIX for we have agreed in this book that V+ 1= +V+ 1 = + 1. Avoid the rule Va V&= Va6 for imaginary numbers, until after very much more thorough study of the subject, by pro- ceeding as directed above. EXERCISES I: NOTE IX— OPERATIONS ON IMAGINARIES Perform the operations indicated, where i means V— 1; when other square roots of negative quantities occur, take V— x = + -y/x i in these exercises : 1. (V^) 2 - 13. 2V^9 + 5^-3V^8. 2 (~ 3 *) 2 - 14. (2+ V^9) + (3-2V^9). 3. (2 0(3 0- 4. (3+2i) + (2+5»). 5 . (6 _4i)-(8 + 0. 16> ( 3 + V- 2 )( 4 -V-5). 6. (1-0 2 . 17. 15. (4 + V-5)(4-V : -5). (3 + V 5-V- 7. (4 + 5 0(3-20- 2 + V-4 8 . *=« is. 3 + v z?. 2 + 3* 2-V-2 j3 2i 6-i 9 - k 19. i±^=?. 3 + V-l 10. 2-5i 20. (_^+iV-3)(-i-iV-3). 3 11. 2 — Ai-\ o-i /_ i i iV— .W T 5 _ 8 i 21 - y 2 + 2 V a; • 12. SV^ + T?:. 22. (-|-W^3) 3 - Write the real and imaginary factors of the following : 23. x 2 + y 2 . 25. 9x 2 + 16y 2 . 27. x 4 -y\ 24. m 2 + w 2 . 26. 4m 2 + w 2 . 28. 16# 4 + ?/ 4 . 37. Quadratic Equations. In solving quadratic equa- tions we found just such numbers as those above. We may now use these answers intelligently ; hence, we may regard them as answers. Occasionally, such answers may have some actual meaning in a problem. IMAGINARY AND COMPLEX .NUMBERS 395 Ex.1. Given af' + 2 = 0, we found (p. 210) x=±V^2. These answers actually satisfy the equation, for (±V-2r + 2 = -2 + 2 = 0. Ex.2. Given a? — 4a; + 5 = 0, we found x=+2 iV^l, or x = 2 ± i. Substituting these answers for x, we find : (1) for x = 2 + i: (2 + 2 -4(2 + + 5=(3 + 4f)-(8 + 4 + 5 = O; (2) for x = 2-i: (2 - if- 4(2 - i) + 5 = (3-4i)-(8 - 4 i) +5 = 0; hence, these answers really satisfy the given equation. Ex. 3. Given any quadratic ax*+ bx+c=0, we found (p. 213) b L V& 2 - 4 ac x = ± . 2a 2 a The answers are real, equal, or imaginary, according as b 2 — iac is >0, = 0, <0. See p. 21E The only difference in our present rules from those of p. 214 is that we now understand how to work with imaginaries, whereas such numbers were then meaningless. EXERCISES II: NOTE IX — QUADRATIC EQUATIONS Solve the following quadratic equations; check each result: 1. x 2 -2x + 2 = 0. 5. 2 a; 2 - 3 a + 5 = 0. 2. a?-4a> + 8 = 0. 6. ^-6 x + 11 = 0. 3. x* + 4:X + 8 = 0. 7. x--3x + 3 = 0. 4. 4^-431 + 10 = 0. 8. x- + x+l = 0. 38. Other Operations. Many other operations in imagi- naries are possible, but we shall content ourselves with the preceding after giving a few examples. Ex.1, -y/— 1 = VV— l = ^/i = x + yi. Squaring both sides, we have i = (x 2 — y 2 ) + 2 xyi, whence, r- — //'- = and 2 xy = 1, or, (x — y)(x + y) = and 2 xy = 1. 396 APPENDIX Solving for x and y, we find x= ±^\= ±|V2, or, x = ±^l=± l -V2, but the last result may be discarded since we wish to have x real. The corresponding values of y are y — ± jV2. It follows that x + yi = ±|V2(1 — /). If we consider also the other possibility V— 1 = V — i, we find two other answers: x + yi= ±JV§(1 + 0- These results may all be checked by raising any one of the results to the fourth power. Ex. 2. Solve the equation x 3 — 1 = 0. Noting that we can factor the left side, (x - l)(x 2 + x+ 1) =0, we see that x = 1, or else x 2 + x + 1 = 0, an equation whose roots are x = — \ ±^V— 3. There are therefore three answers. (Verify each of them.) Ex. 3. Solve x- 4 + x 2 + 1 = 0. Factoring, we find x* + x 2 + 1 = (x 4 + 2 x*+ 1) - x 2 = (x 2 + l) 2 - x 2 = [(x 2 + 1) - x] [(x 2 + 1) + x] = (x 2 -X + l)(x 2 + X + 1). Hence, x 4 + x 2 + 1 = gives either x 2 - x + 1 = 0, or, x 2 + x + 1 = 0, hence, x = \ ± \ V— 3, or, x = — f ± |V '— 3. Check : Substitute x = \ + §V— 3 in x 4 4 x 2 + 1 ; we find (I + |V3s)4 + q + j vri)^ + i = (-$- jvrs) + (_.] +^vC3) + i=o. The answers x = i — £ V— 3, x = — £ ± £V— 3 may be verified in an exactly similar way. No other problems will be solved because the best method — known as De Moivre's theorem — is beyond the scope of this book. Logarithms also lead to imagi- nary numbers ; the discussion of these and other more intricate matters is left to more advanced courses. NOTE X. SIMULTANEOUS QUADRATICS 39. One Equation Factorable. We have studied in Chap- ter X pairs of simultaneous equations of which at least one is a quadratic equation. We shall add, in this note, several additional methods. We saw that any such pair, one of which is linear, may be solved by substitution (p. 255). This method applies also whenever one of the equations can be factored as in the following examples. Ex j [2x>-?>xy + tf = 0, (1) la5 a + 2/ 2 -10a;=75. (2) (1) may be written in the form (2x-y)(x-i / )=0, (1) which is equivalent to the two equations (In) 2x-y = 0; (lb) x-y = 0. Solving each of these in combination with (2) by the method of p. 256, we find from (1 a) : from (1 b) : x = f + 5 V7, 1 J" x = \ - 5 V7, y = f + 5V7,J ' ly = f-5V7. These answers may all be verified by the student. These solutions are shown in Fig. 74 by the points marked A, B, C, D, respectively. Equation (2) gives the circle of center (5, 0) and radius 10; equation (1) gives the two straight lines (1 a) and (1 b). This process may be used whenever one of the equations is of the form (I) Ax* + Bxy + Cy 2 = 0, for this equation is always factorable by the methods of §§ 61, 119, pp. 98-225. Such an equation as (I), i.e. an equation in which the degree of each term is the same, is called homogeneous. 397 ■x= 5,1 or f x = - 3, .y = io,J U = -6. 398 APPENDIX Y : Jr t na Ab A J 1 )^ \>^ /^* L. ySC** ~? ~1 2 ~£_ / \K2) ~7 i / —t f 7 I / / 1— f k^ "7^/ <>// v ~T_ T XDf a /\Jb / / f\\ 2 f ^' _^^-^^ J_ Fig. 74. A case of another type is illustrated by the example following: : Ex.2. f--?/ 2 + 2.r-f2 2/== 0, l*»-y«0. Factoring (1), gives (x — y)(x + .'/) + 2(x + y) = 0, or, (x + y)(x-y + 2) = 0, which is equivalent to the two equations : (la) x + y = 0; (\ l>) x-y + 2 = 0. (1) (2) SIMULTANEOUS QUADRATICS 399 Solving each of these in combination with (2) by the methods of p. 2oo, gives from (1 a) from (lb): x = 0, ] f x = - 1, } or y = o. j u = + 1- a: = -1,1 (x = + 2, t or y = + h) U = + 4. These may be verified by the student. The figure for (1) in Fig. 75 is a pair of lines (1 a) and (1 b) ; that for (2) is the curve y = x 2 , 1 vt J *v / I r -t ' i i ^t 4 \ / 4^ y j ^> \ / t y 4 X I ^Z 4 ZT 7 \ . f / T~t + 4- n ^ x y ^z i 4 t^v t \ // •4- 4 -S X- U- -v^- \ (U) A \ \ / I \ \ / j \ \ / j \ \ / (2)/ \ i / 1 y (la>\ \| /" |/ \| \ ■> / / \ \ / I 5 A Z /L \ \ • / \\ 7 / (-l. + l^/ j / Z x's / 7 Z "fs^ ^ 1 zK s ^>dj)p -4- t -3 ^2 - "\ £ *! s \^ \ ^ ■ •i \ ^ ^ ^ ^ 42 \ Fig. 75. 400 APPENDIX which we have drawn before. The point (-1), (+1) appears twice as a solution ; the figure makes clear why this is true. In general, any quadratic equation containing y 1 may be factored as in Example 2 if an attempted solution for y involves no radicals in x. Ex. 3. The equation (1) tf — 3a? — 2xy + 8x — 4 y + 3 = may be written f _ 2(a: + 2)y = 3 x 2 - 8 x - 3. Solving for y by the usual method (see p. 206 ), we find y = (* + 2)±(2*-l), i.e. (la) y = 3 x 4- 1 ; or, (16) y = - x + 3. Equation (1) therefore represents a pair of straight lines (1 a) and (1 b) (Fig. 76) ; hence, if (1) occurs as one of a pair of simultane- ous quadratics, the pair may be solved as above. Y N s \ / la i \:ifi \ \ \ 0j 1 ) i ) -5 Fio. 76. SIMULTANEOUS QUADRATICS 401 40. Type Ax 2 + Bxy + Cy 2 = D. If neither of the given pair of quadratics is factorable in the sense of § 39, it is often possible to form a new equation from them which is factorable. Ex.1. {* + ** = *> (1) 1^ = 1. (2) Multiplying both sides of (1) by — 1, both sides of (2) by 5, and adding, we find (3) x 2 -5xy + iy 2 = 0, which is factorable (see § 39) and is equivalent to (3 a) x-iy = 0, (36) x - y = 0. Solving each of these with (2), we find from (3 a) : from (3 b) : { X = 2 ' 1 or / X = ~ 2 ' 1 { X = h 1 or f * = ~ 1j These answers may be verified by the student. This process may be used whenever both equations are of the form (II) Ax 2 + Bxy + Cy 2 = D ; the rule is to eliminate the constant term, as above, so as to get an equation of the type (I).* * This type of equations also yields to the following method : Let y = vx in both equations ; equate the values of x' 2 in the resulting equations, and solve for v. Thus, in the preceding example, r. (1) becomes x 2 + 4 v 2 x 2 = 5, or x 2 = 1 + 4 »a (2) becomes vx 2 = 1, x 2 = • v 5 1 Equating the values of x 2 : = -, or 4 v 2 — 5 v + 1 =0. 1 + 4 v 2 v Hence, v = } or 1, whence x = ± 2 or ± 1, and y = vx = ± J or ±1, which, when properly paired, give the pairs of solutions found above. 402 APPENDIX 41. Symmetrical Type. Other equations yield to the method of § 40. Ex (2xy-x-y + 5 = 0, (1) \x i +6xy + y 2 —4:X-4:y-5 = 0. (2) Multiply both sides of (1) by 2 ; multiply both sides of (2) by - 1, then add : (3) x 2 + 2 xy + y 2 -2x-2y- 15 = 0, or, (x + y yi-2(x + y) - 15 = 0, which is the same as \_(x + y) — 5] \_(x + y) + 3] = 0, and is therefore equivalent to the two equations (3rt) x + y-5 = 0, (3 6) x +y + 3 = 0. Solving each of these with (1), we find from (3 «) : from (3 //) : far = 5,1 r* = 0,j (x = l, ) fx=-4,l This method is successful whenever both equations are symmetrical ; i.e. of the form (III) Ax 2 + Bxy + Ay* + Dx + Dy + F=0, so that an interchange of x and y does not affect the equa- tion ; the rule is to make the quadratic part of the new equation a perfect square, by multiplying each equation by the value of B — 2 A in the other equation, and then sub- tr acting ; this is practically what we did above.* 42. General Method of Inspection. It can be shown that any pair of simultaneous quadratic equations can be * This type (both equations symmetrical) can be solved also by mak- ing the substitution x = u -{- v, y = u — v ; in the example above this gives from (1) 2 «2 _ 2 v 2 - 2 u + 5 = 0, from (2) 8 rfi - 4 v* - 8 u - 5 = 0. Adding these, after multiplying the first by — 2, we get 4 U 2_4h- 15 = 0, which gives u — § or — \ ; hence, v 2 = \ (8 u' 2 — Su — 5)= %£- and v = ± f , whence, x = u + v = 5 or 0, or 1 or — 4, and y = u — v = or 5, or — 4 or 1, which give the preceding solutions when properly paired. SIMULTANEOUS QUADRATICS 403 solved in a similar manner if the proper multipliers can be found. Thus, on multiplying both sides of the first by X and adding it to the second, there are always three values of X which make the new equation factorable in the sense of § 39. In many examples, however, it may be impossible for the student at present to find these values of\. Ex 1 {^-^/ + ?/ + 2.r-y = 3, (1) • [a? + tf + 4:X-2y = 5. (2) Choose X — — 2 ; i.e. multiply (1) by — 2 and then add equation (2), (•'») - x 2 + 2 xy — y 2 = — l, or, changing signs and transposing, we have (x - y y -i = o, whence, (3) is equivalent to the two equations (:5 a) x - y + 1 = 0, (:} h) x -y-l = 0. J It j _J__L -j- y l2 > s. -^ v^p /' v : u& st t~ / '3 IS. / ^ / ^\~^ / / / V / / / \ ! f 1 ^~~~~ ' ~*^^\ // •***" ^^ L / 1) S ' ** y^V\ v^\* / S / 1 \\ $f I jT / f 1 \\ s ( ■■■* f / i 1 y s 1 ' / ./ V / / 1 /_ /L^&b i / / / /// l? j\ / ZZ W - \lA^ l\ / x . °' 4P - - 5 k / -'I -? y^ -1 -W ^^Ti - -^-3-4- \ / ^y / ^'//i\ \ / y' ^1 / // 1 \ / »/ -^/ — ^ f/J ' X y ^y^ / / y^x " ~" \ / ^ t\tS/m J S \ ^ W' 1 (56) l-J«r " "* 'J-9^^1 ' ' "_^_ ^ t^ -+■ x *>«; *s^ / ^"""" — ' — "" SJ ^s' / // ■ \ I \ '■■*' / j/ / V7 ^ / / —3 / / £> v^/ / i: :js£: ^^^fet ztzp ± i r 3i Fio. 77. 404 APPENDIX Combining each of these with (2), we find the solutions (Fig. 77) from (3 a) : from (3 b) : ly=2,j b=-2,j ly = o,J ly = -2j which the student may verify directly. The figure shows the picture for each of the equations used, and also those which follow. The choice = — 1 is equally successful (Fig. 77, 5 a and 5 b). The choice = — f is equally successful (Fig. 77, 4 a and 4 b). As a general method this is undoubtedly superior ; its details are too lengthy for discussion here. Notice all the methods given above are special cases of this one. EXERCISES I: NOTE X — SIMULTANEOUS QUADRATICS Solve each of the following pairs of equations and draw a figure showing the curves and their points of intersection : (x 2 -y 2 = 0, ix>-5xy + 4:y 2 = 0, \x 2 -2y = 0. ' [xy = 36. lx 2 -5 xy + 6 y 2 = 0, f x 2 - 2 xy + f - 3(x -y) = 0, 2 ' \x 2 + y 2 =100. *' [y = x 2 . (y 2 -3 x 1 -2xy + 8x-4:y-t3 = (see p. 400), | x 2 - y 2 = 33, J x 2 + xy + f = 21 , 6 \ X y + y' = U. 8 ' [x 2 -xy + y 2 -2x-2y = 23. lx 2 +3xy+y 2 +2x+2y = 17, {x*-xy + y 2 = 3, 7 ' \x i + xy + y 2 = 7. 9 ' \x 2 + xy + y 2 = 7. | x 2 + 4 xy + y 2 + 4 x + 4 y = 89, 10 " \x 2 -2xy + y 2 -2x-2y = ll. \x 2 -2 xy + y 2 + 4 x + 4 y = 16, [2x 2 + xy + 2y 2 -2x-2y = 7. SUMMARY 40.J SUMMARY OF APPENDIX, pp. 354-404 Note I. Detached Coefficients. pp. 354-355. 1. Detached Coefficients. p. 354. 2. Multiplication ; illustrative examples. p. 354. 3. Division : illustrative example. pp. 354-355. 4. Division by (x — a) : illustrative example. Exercises I. pp. 355-356. Note II. Remainder Theorem; Factoring. pp. 357-360. 5. Factur Theorem: proof of theorem ; converse. p. 357. 6. Factors of x n ±y n : application of § 5 ; type-forms. Exer- cises I. pp. 357-359. 7. Factors of Polynomials : factors by trial. Exercises II. p. 359. 8. Remainder Theorem : extension of § 5 ; calculation of polyno- mial values. Exercises III. p. 360. Note III. Choice and Chance; Permutations and Combina- tions, pp. 361-365. 9. Choice: general formula. p. 361. 10. Chance : general formula. Exercises I. pp. 361-362. 11. Permutations: definition; general formula. p. 363. 12. Permutations among a Limited Number : discussion; formula. pp. 363-364. 13. Factorial Notation: definition of n! P„ iB ; P, u m ; formulas. Exercises II. p. 364. 14. Combinations: definition; formula. Exercises III. pp. 364-365. Note IV. Inequalities. pp. 366-3tis. 15. Operations on inequalities. p. 366. 16. Graphical solutions. Exercises I. pp. 367-368. Note V. Binomial Theorem. pp. 369-372. 17. Formula: proof for positive integral exponents; mathemati- cal induction. pp. 369-371. 18. Notes and Examples. Exercises I. pp. 371-372. Note VI. Euclidian Method; H.C.F. and L.C.M. pp. 378-376. 19. Euclidian method : examples; statement of principles. Exer- cises I. pp. 373-376. 400 APPENDIX Note VII. Cube Root and Higher Roots. pp. 377-379. 20. Introduction. p. 377. 21. Cube Roots of Numbers. Exercises I. pp. 377-378. 22. Cube Roots of Polynomials : example. pp. 378-379. 23. Higher Roots. Exercises II. p. 379. Note VIII. Limits and Infinite Series; Irrational Numbers. pp. 380-389. 24. Introduction. p. 380. 25. Limits: definition ; illustrations; examples. pp. 380-382. 26. Infinite Series : definition; definition of sum. p. 382. 27. Infinite Geometric Progressions : general formula for sum ; ex- amples ; repeating decimals. Exercises I. pp. 383-385. 28. Other Infinite Series: definitions of convergence, divergence; examples. Exercises IT. pp. 385-386. 29. Irrational Numbers : definitions ; illustrations ; direct defini- tions by " cuts." pp. 386-387. 30. Operations on Irrationals . approximations ; definition of " sum," etc. Exercises III. pp. 387-389. Note IX. Imaginary and Complex Numbers. pp. 390-396. 31. Introduction. p. 390. 32. Definitions. pp. 390-391. 33. Direct Operations. pp. 391-392. 34. Conjugate Complex Numbers. p. 392. 35. Division. p. 392. 36. Further Examples. Exercises I. pp. .393-394. 37. Quadratic Equations: imaginary solutions. Exercises II. pp. 394-395. 38. Other Operations : examples. pp. 395-396. Note X. Simultaneous Quadratics. pp. 397-404. 39. One Equation Factorable : homogeneous type, A x 2 + Bxy + Cy 2 . pp. 397-400. 40. Type Ax 2 + Bxy + Cy 2 = D: elimination of constant, p. 401. 41. Symmetrical Type : Ax 2 + Bxy + Ay 2 + Dx + Dy + F = 0. p. 402. 42. General Method of Inspection: three possible multipliers. Ex- ercises I. pp. 402-404. TABLES [The student doubtless knows many of these ; some others should he learned, stress heing laid ou the tables in the metric system.] I. TABLE OF SIGNS + , read plus. — , read minus. X , or •, read times, multiplied by, multiplied into, or into (axb = a • h — ah). -=-,:, or /, read divided by or over. = , read equals, or is equal to. =£, read is not equal to. >, read is greater than. >, read is greater than or equal to. <, read is less than. <, read is less than or equal to. [The signs just written may be easily remembered by noting that the opening faces the greater quantity.] a 2 , read a square = a x a. a 3 , read a cube = a x a x a. a n , read a to the nth poiver, a to the power n, or a with an exponent n = a x a ■■• to n factors. ( )> I ]' { 1' ~ , called signs of aggregation; a general term used for them is parentheses, and quantities inclosed by them are read the quantity ■■■ or the expression •••. To distinguish them, we call ( ) parentheses ; [ | brackets, { } braces ; ~ the vinculum. Va, read the square root of a ; this is also written a- and read a to the poiver \, see pp. 192, 285. v'a, read the cube root of a; this is also written a* and read a to the poiver \, see pp. 192, 285. n , , 1 v a, read the nth root of a : this is also written a», and read a to the power i, see pp. 192, 285. 407 408 TABLES II. TABLES OF WEIGHTS AND MEASURES [The customary abbreviation of each measure is indicated, portaut units are printed iu black-faced type.] A. Measures of Quantity The most im- 1. Table of Measures of Common Objects 12 units = 1 dozen (doz.). 20 units = 1 score. 12 dozen = 1 gross (gr.). 12 gross = 1 great gross (G. gr.). !. Table of Measures of Paper 24 sheets = 1 quire. 20 quires = 1 ream. 2 reams = 1 bundle. 5 bundles = 1 bale. B. Measures of Value 1. Table of American Money 10 cents (?) or (ct.) = 1 dime. 10 dimes = 1 dollar ($) 10 dollars = 1 eaqie. 2. Table of English Money 4 farthings (far.) = 1 penny (d.). 12 pence = 1 shilling (s.). 20 shillings = 1 pound (£). Other measures are the crown (= 5 the half-dollar (= 50 shillings) and the guinea (= 21 shil- lings). 4. Table of German Money 100 pfennigs (pf.) = 1 mark (mk.). 5. Table of Italian Money 100 centisimos = 1 lira (h\). Other measures are the quarter (= 25 cents) cents), and the mill (=i 1 5 cent). 3. Table of French Money 10 millimes = 1 centime. 10 centimes = 1 decime. 10 decimes = 1 franc (fr.) In the following tables of equivalents, three signijicant figures are given, except in a few important instances. The student should realize that it is not the number of decimal places, but rather the num- ber of significant figures, which determines the degree of accuracy. 6. Table of Eqtiivalents hi. = $ 0.0203. 1*. = $ 0.243. 1 £ =$4.8665. 1 f r. =$0,193. 1 mk. = $ 0.238. 1 lr. = $0,193. 0. Measures of Length Table of English Measure of Length 12 inches (in.)= 1 foot (ft.). 3 feet = 1 yard (yd.). 5£ yards = 1 rod (rd.). 320 rods = 1 mile (mi.). WEIGHTS AND MEASURES 409 2. Table for Metric Measure of Length 10 millimeters (mm.)= 1 centimeter (cm.). 10 centimeters = 1 decimeter (dm.). 10 decimeters = 1 meter (m.). 10 meters = 1 dekameter (Dm.). 10 dekameters = 1 hektometer (Hm.). 10 hektometers = 1 kilometer (Km.). 10 kilometers = 1 myriameter (Mm.). The meter is approximately one four-millionth of the circumference of the earth, measured along a meridian through Dunkirk, France. 3. Table of Equivalents ENGLISH TO METRIC METRIC TO ENGLISH 1 in. =2.54 cm. 1cm. =0.3937 in. 1 ft. = 30.5 cm. 1 m. = 39.4 in. = 3.28 ft. 1 yd. = 91.4 cm. = .914 m. 1 Km. = 0.621 mi. 1 mi. = 1.61 Km. D. Measures of Area 1. Table of English Measure of Area 144 square inches (sq. in.) = l square foot (sq. ft.). 9 square feet = 1 square yard (sq. yd.). 30} square yards = 1 square rod (sq. rd.). 160 square rods = 1 acre (A.). 640 acres = 1 square mile (sq. mi.). 2. Table of Metric Measures of Area 100 square millimeters (sq. mi.) = 1 square centimeter (sq. cm.). 100 square centimeters =1 square decimeter (sq. din.). 100 square decimeters = 1 square meter. 100 square meters = 1 square dekameter (sq. Dm.) or are (a.). 100 square dekameters = 1 square hektometer (sq. Hm.) or hektare. 100 square hektometers =1 square kilometer (sq. Km.). 410 TABLES 3. Table of Equivalents ENGLISH TO METRIC METRIC TO ENGLISH 1 sq. in. = 6.45 sq. cm. 1 sq. cm. — .155 sq. in. 1 sq. yd. = 0.835 sq. m. 1 sq. m. = 1.20 sq. yd. 1 sq. rd. = .253 a. la. 3.95 sq. rd. E. Measures of Volume 1. Table of English Measure of Volume 1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.). 27 cubic feet = 1 cubic yard (cu. yd.). 128 cubic feet = 1 cord (of wood) (cd.). 2. Table of Metric Measurements of Volume 1000 cubic millimeters (cu. mm.) = 1 cubic centimeter (cu. cm.). 1000 cubic centimeters = 1 cubic decimeter (cu. dm.). 1000 cubic decimeters = 1 cubic meter (cu. m.) or 1 stere (of wood) (st.). 3. Table of Equivalents ENGLISH TO METRIC METRIC TO ENGLISH 1 cu. in. = 16.4 cu. cm. 1 cu. cm. = 0.0612 cu. in. 1 cu. yd. = 0.76 en. m. 1 cu. m. = 1.31 cu. yd. 1 cd. = 3.6 st. 1 st. = 0.28 cd. F. Measures of Time 60 seconds (sec.) = 1 minute (min.). 60 minutes = 1 hour (hr.). 24 hours = 1 day (da.). 7 days = 1 week (wk.). 14 days = 1 fortnight. 12 months = 1 year (yr.). 365 days = 1 year. 366 days = 1 leap year. 10 years = 1 decade. 100 years = 1 century. The number of days in the different months vary. February has 28 days, except for leap year when the extra day is added to it, giving it 29. Septem- ber, April, June, and November each have 30 days. The other months each have 31 days. Every year divisible by four, except eQn.t;ennial. years not di- visible by 400, is a leap year. WEIGHTS AND MEASURES 411 G. Measures of Capacity 1. Tables of English Measurement of Capacity DRY MEASURE LIQUID MEASURE 2 pints (pt.) = 1 quart (qt.). 4 gills (gi.) = 1 pint (pt.). 8 quarts = 1 peck (pk.). 2 pints = 1 quart (qt.). 4 pecks = 1 bushel (bu.). 4 quarts = 1 gallon (gal.). Other measures are the barrel (bbl. = 31 \ gal.) and the hogshead (hhd. = 2 bbl.). The gallon contains 231 cu. in. 2. Tables of Metric Measurement of Capacity 10 milliliters (ml.) = 1 centiliter (cl.). 10 centiliters = 1 deciliter (dl.). 10 deciliters = 1 liter (1.). 10 liters = 1 dekaliter (Dl.). 10 dekaliters = 1 hectoliter (III.). 10 hectoliters = 1 kiloliter (Kl.). The liter contains 1 cu. dm. or 61.02 cu. in. 3. Table of Equivalents ENGLISH TO METRIC METRIC TO ENGLISH 1 dry qt. = 1.10 1. 11.= 0.908 dry qt. 1 liquid qt. = 0.947 1. 11.= 1.0567 liquid qt. 1 bu. = 35.2 1. 1 III. = 2.84 bu. 1 gal. = 3.79 1. 11.= 0.264 gal. H. Measures of Weight 1. Table of Avoirdupois Weight 16 drams (dr.). = 1 ounce (oz.). 16 ounces = 1 pound (lb.). 100 pounds = 1 hundredweight (cwt.). 112 pounds = 1 long (or English) hundredweight. 20 hundredweight = 1 ton (T). 20 long hundredweight = 1 long (or English) ton. 7000 grains = 1 pound. 412 TABLES 2. Table of Troy Weight 3. Table of Apothecaries' Weight 24 grains (gr.) = 1 pennyweight 20 grains (gr.) = 1 scruple (3). (pwt.). 3 scruples * = 1 dram (3). 20 pennyweights = 1 ounce (oz.). 8 drams = 1 ounce (§)• 12 ounces = 1 pound (lb). 12 ounces = 1 pound (lb.)- 4. Table of Metric Measure of Weight 10 milligrams (mg.) = 1 centigram (eg.). 10 centigrams = 1 decigram (dg.). 10 decigrams = 1 gram (g.). 10 grams = 1 dekagram (Dg.). 10 dekagrams = 1 hektogram (Hg.). 10 hektograms = 1 kilogram (Kg.). 1000 kilograms = 1 metric ton (T.). Table of Equivalents English to Metric Metric to English 1 oz. Troy = 31.1 g. 1 g. = 0.0322 oz. Troy. 1 oz. Av. = 28.4 g. 1 g. = 0.0353 oz. Av. 1 lb. Av. =0.454 kg. 1 Kg. = 2.2046 lb. Av. 1 ton = 0.907 metric ton. 1 metric ton = 1.10 tons. 7. Measurement of Angles 60 seconds (") = 1 minute ('). * = 3.14159265 .... 60 minutes = 1 degree (°). = 3.1416- (nearly). 90 degrees = 1 right angle. = 3i (roughly). 360 degrees = 1 perigon (circumference). 1 radian = **°! = , ™° = 57° 17' 45" (nearly). it 3.141b— 1° = 3.U16-. radians _ 0.0175- •• radians. 180 WEIGHTS AND MEASURES 413 J. Measurement of Temperature The unit of temperature is the degree. The Centigrade thermometer is graduated so that the temperature at which water freezes is 0°, and the tem- perature at which it boils is 100°. The Fahrenheit thermometer is graduated so that the temperature at which •water freezes is 32°, and the temperature at which it boils is 212°. The Reaumur thermometer is graduated so that the temperature at which water freezes is 0°, and the temperature at which it boils is 80°. If C denotes the Centigrade record of temperature and F the Fahren- heit, then, C = $ (F - 32°), F= f C + 32°. (See p. 143.) If R denotes the Reaumur temperature, R = 4C = *(F-32°). A comparison of these three scales of temperature may be readily seen from the following table : Centigrade Fahrenheit Keacmi k From boiling to freezing 100 100 32 212 180 80 80 III. TABLES OF SIMPLE FORMULAS FROM PHYSICS 1. Falling Bodies. "With the notation s = space passed over (in ft. or cm.), t = time (in sec), v = velocity (in ft. per sec. or cm. per sec), g = gravitational acceleration = 32.16 ft. per sec. per sec = 981 cm. per sec. per sec If dropped from rest : v = gt, s = \ gf 2 , v* = 2 gs. In general, v — v = gt, s — s = \ gt 2 + v t, if v = initial velocity, s = initial distance. 2. Lever, or Balance. dW = Dw. See Fig. 30, p. 176. 3. Boyle's Law : Pressure and Volume. pv = constant. See p. 218. Other formulas given as needed in the text. 414 TABLES IV. TABLES OF GEOMETRIC MENSURATION FORMULAS q Triangle Area = \ (base x altitude) = \ AB ■ CX = \b ■ a, where ft = base = A B, and a = altitude = CX. Sum of angles = ZA + ZB + ZC = 2 rt. A = 180°. Bight Triangle ZB = rt.Z = 90°, Z A + Z C = 90°. (Hypotenuse) 2 = sum of squares of the perpendicular sides : -^ B AC 2 — AB 2 + BC 2 . Right Triangle If A C = h(= hypotenuse), AB -b(= base), CB = a(= altitude), the formula becomes, n 2 = a 2 + b 2 . If and D a h Ratios in Bight Triangle s ( = sine of Z A), - = c (= cosine of ZA), h -= t (= tangent of ZA), then 6 a * = < Square 1 1 +±= t — /2 s 2 +c 2 = l Area = (length of side) 2 = AB 2 — s 2 , where s = AB = length of side. Rectangle Area = base x altitude = AB x. BC = b ■ a, where b = base = A B, and a = altitude —BC. D X C D C Rectangle B A V B Parallelogram Parallelogram Area = base x altitude = AB x XY = b ■ a, where b = base = AB, and where a — altitude — XY. MENSl RATION FORMULAS 415 Circumference Area where and, see p. 112, Circle — 2 it x radius = 2 -n-r ; = 7r x (radius) 2 = 7r/- 2 , r = radius of circle, 7T = :j.Ul(j... = 3} (nearly). Cube Circle Volume = (length of side) 8 = s 8 , where s = length of side. Pyramid Cure Rectangular Parallelopiped Volume = (area of base) x altitude. Pyramid - any Base Rkctanqular Parallelopiped Volume = \ (area of base) x altitude. Regular Pyramid (Base a regular polygon ; perpen- dicular from vertex meets base at center.) Lateral area = \ (perimeter of base) x (altitude of face). The altitude of face is the altitude of Regular any one of the triangular faces, often called Pyramid slant height. Right Cylinder Lateral area = (circumference of base) x altitude. Cylinder Volume = (area of base) x altitude. Right Cone Lateral area = \ (circumference of base) x slant height (i.e. length of ele- ment) . Volume = J (area of base) x altitude. Sphere Area = 4 n x (radius) 2 = 4 irr 2 . Sphere Volume = f it x (radius) 8 = $ ttt 8 . Right Cone INDEX (The numbers refer to pages, v. = vide.) Abbreviations (v. Substitution and Signs), 2. Addition, 1, 34. axioms of, 35, 56. of expressions, 49. • of fractions, 125, 129. of monomials, 44. of negatives, 15, 36, 39. of radicals, 289, 388. Algebra, 1. Alternation (in proportion), 139. Answers (v. Solutions and Roots), 19, 104, 109, 205. checking of, v. Check. false, 108, 300, 313. irrational, 206, 214. negative, 109. Approximations (v. Errors and Graphs), 19. 30, 144. Arithmetic (v. special headings), 1. Arithmetic Sequences, v. Sequences. Associative Law, of addition, 35. of multiplication, 35. Averages, 41. Axioms, of addition and multiplication, 35. of exponents, 285. "of operations on equations, 56. Base (of logarithms), 341. Binomial, 8. Binomial Theorem, 369. Boyle's Law, 218. Braces, v. Parentheses. Brackets, v. Parentheses. Cancellation, in equations, 153. in fractions, 132. Centigrade, v. Thermometer. Chance, 361. Changes, Permissible, in equations (v. Equations), 55, 106, 137, 153. in fractions, 119, 124. in proportion, 137. in rationalization,. 292. Characters, v. Signs. Characteristic (of logarithms), 343. Checks (v. Graphs), 5. complete, 5. in addition, etc., 45, 78. in English problems, 5, 61. in equations, 5, 57, etc. in radical equations, 302. iu radicals, 287. in substitution methods, 313. Choice, 361. Circle, 250. Clearing of Fractions, 138, 149, 153. Clearing of Radicals, 292, 300, 304. Coefficients, 8. choice of, 45. detached, 354. radical, 308. relation to roots, 224. Combinations, 364. Common Factors, v. Factors. Common Multiples, v. Multiples. Commutative Laws, 35. Comparison, solution by, 172. Completing a Square, 20(5. Complex Fractions, 134. Complex Numbers, 391. Composition (in proportion), 137. Computation, 340, 347. Conjugate Complex Numbers, 392. Conjugate Radicals. 294. Convergent Series, 385. Cosine, 251, 414. Cross-multiplication (in proportion), 138. hedkiok's el. alu. — 27 417 418 INDEX Cube Root, v. Root. Curve, v. Graph. Degree, of equation, 152, 255. of radicals, 284. of terras, 152, 255. reduction of, 290. Denominator (v. Fraction), 118. Detached Coefficients, 354. Diagram, v. Graph. Discriminant, 214, 227, 229. Divergent Series, 385. Division (v. Fractions), 1, 74. by zero, 56, 75, 79, 106, 119, 150. exact, 84. in proportion, 78. of fractions, 74, 134. of longer expressions, 79, 84. of monomials, 77. of powers, 76. of radicals, 286. Element (of a sequence), 323. Elimination, 163. English Statements (v. Problems), 103. Equal, 2. Equal Roots, v. Roots. Equality, v. Equations. Equations, 2. answers to, v. Answers. cancellation in, 153. cubic, 152. degree of, 152, 255. equivalent, 168. ' formation of, 223. fractional (v. Proportion), 149. graph of (v. Graph), 23. homogeneous, 397. indeterminate, 159, 242. of degree 1, v. Equations, Linear. of degree 2, v. Equations, Quad- ratic. of degree n, 152. simple, v. Equations, Linear. solution of, v. Ausivers, and Roots, and Problems. Symmetrical, 317, 402. Equations, Linear (v. Equations, Simultaneous Linear, and Varia- tion, Linear), 25, 140, 152, 239. graph of, v. Graph. [Equations, Linear], operations on, 55, 106, 137, 153. solution of, 57. Equations, Quadratic, 109, 152, 203, 253, 255, 396, 316. character of roots, 214. discriminant, 214, 227, 229. given roots, 223. graph, 204. literal, 229. relation of roots to coefficients, 224. simultaneous, 267. solution of, 109 4 151, 205, 206, 209, 212, 227, 261, 316, 396. Equations, Radical, 146, 318. Equations, Simultaneous, Linear, 159, 160, 163, 170, 172, 173, 319. Linear and Quadratic, 253, 319. Quadratic, 267, 276, 397. Errors (v. Approximations) , 108. Euclidian Method, H. C.F. aud L.C.M., 373. Evolution, 181. Exponents (v. Logarithms) , 8. fractional, 192, 194, 285. negative, 29(5. rules for, 194, 285, 334. table of, 335, 339. zero, 296. Expression, 7. quadratic, 225. radical, 184, 192, 284. surd, 186, 284, 306. Factorial, 364. Factors, 91. by grouping, 100. common, 119. difference of powers, 99, 331, 358. difference of squares, 94, 99. highest common, 120, 373. monomial, 79. of perfect square, 93. polynomial, 91, 121, 128, 359. quadratic expressions, 98, 107, 225. sum of cubes, 95). sum of powers, 99, 331, 358. theorem, 223, 357. trinomial, 96, 98, 105, 107, 225. type forms, 91, 331, .",58. Fahrenheit, v. Thermometer. INDEX 419 Falling bodies, 221,234. Fractions (v. L'), 1, 118. addition of, 125, 131. clearing of, v. Clearing. complex, 134. division of, 74, 134. multiplication of, 07, 131. rational, 186, 284. reduction of, 118. Fractional Equations, v. Equations. Fractional Exponents, v. Exponents. Function, linear (v. Variation, Linear), 238. Geometric Sequences, 328. Geometry, formulas of, 414. Graphs (v. Curves, and Straight Line), lit, 23. characteristic of logarithm, 343. cube roots, 189s higher roots, 100. inequalities, 366. linear variation, 25, 140, 238. logarithms, 330, 338. of prices, 20. plotting of, 247. quadratics, 204. raising vertically, 25, 271. simultaneous linear equations, 150.^. square root, 182.^t various, 248. Grouping, v. Parentheses. Highest Common Factor, v. Factor. Hindu Method (in quadratics), 200. Homogeneous Equations, 397. Imaginary Numbers, 182, 210, 285, 300. Indeterminate Equations, v. Equations. Index of root, 9. Inequalities, 366. Infinite Series, v. Series. Integers, 1. Intersection, point of, 160, 255. Inversion (in proportion), 139. Involution, 181. Irrational Numbers (v. Sunl), 186, 284, 342, 380. Lever, Law of, 176. Limits, 380. Line, v. Graph, Curve, Straight Line. Linear Equations, v. Equations. Logarithms (v. Exponents), 337. Briggs', 342. common, .'i42. rules, 342. table, 330, 348. Lowest Common Multiple, v. Multiple. Lowest Terms, v. Terms. Mantissa (of logarithms), 343. Marks, v. Signs. Mathematical Induction, 371. Measurement, errors in, 108. Measures, 408. Mensuration Formulas, 414. Metric System, 408. Mirror, Concave, 220. Monomials, 8. addition of, 44. division of, 77. multiplication of, 71, 73. powers of, 102. roots of, 192. subtraction of, 45. Multiples, 127. common, 127. lowest common, 128, 140, 375. Multiplication (v. Product), 1, 68. by zero, 68, 106, 109. of fractions, 67, 131. of longer expressions, 78. of negatives, 70. powers, 72. radicals, 286. Negative Exponents, 295. Negative Numbers, 14. addition of, 15, 36, 39. as answers, 109. averages of, 41. multiplication of, 70. Numbers, 1. complex, 391. conjugate complex, 392. imaginary, 182, 210, 285, 390. irrational, 180, 284, 342, 386. rational, 186, 284, 387. real, 284, 390. Numerator, 118. Operations, in equations, etc., v. Changes, ' 420 INDEX Parallel, 166. Parentheses, 7, 10, 47, 53. Permutations, 3(53. Physics, formulas of, 413. Polynomials, 8, 9, 88, 91, 121. Powers, 8, 181. divisiou of, 76. fractional, 192, 194, 285. multiplication of, 72. of longer expressions, 196. of monomials, 192. of radicals, 187. positive integral, 8, 193. rules for, 194, 285. simple, 8, 193. Prices, graph of, 20. equation of, 24. Problems in English, directions, 5, 59, 61, 103, 109. structure of, 103. Product (v. Multiplication) , 1, 8, 68. equal to zero, 107, 153. sum and difference, 94. Progressions, v. Sequences. Proportion (v. Equations, Fractional and Variation, Linear), 2, 25, 137. between variables, 139, 237. operations in, 137. Quadratic Equations, v. Equations. Quotient, v. Fraction aud Division. Radicals (v. Surd and Irrationals), 184, 192, 284. addition of, 289. conjugate, 294. degree of, 284. division of, 286. equations, v. Equations. expressions, 184, 192, 284. multiplication of, 286. operations on, 188. powers of, 187. rationalization of, 292. reduction of, 290. sign, 2, 192, 285. similar, 289. simplest form, 'J'.».">. Ratio (v. Fraction), 137, 237. common (in sequence), 328. [Ratio], in right triangle, 251, 414. of proportionality, 237. Rational, fraction, 186, 284. number, 186, 284, 387. Rationalization (of radicals), 292. of denominator, 292, 294. Real Numbers, 284, 390. Reaumur, v. Thermometer. Reciprocal, 134, 296. Reductio ad absurdum, 166, 306. Remainder, in division, 85. Remainder Theorem, 360. Reversible Process, 302. Roots, 8, 182. cube, 9, 189, 377. even, 182. higher, 190, 379. imaginary, 182, 210,285, 390. index of, 9. longer expressions, 19(3. monomials, 192. odd, 182. square, 2, 104, 181, 184, 197, 305. Roots of an Equation (v. Answers), 205. equal, 209, 214. given, 223. imaginary, 210, 214. unequal, 214. Sequences (v. Series), 323. arithmetic, 282, 323. geometric, 328. Series (v. Seqitences) (convergent, di- vergent, etc.), 382. Signs, 1. rules of, 71, 75, 124. tables of, 407. Similar, v. Terms and Radicals. Simple, v. Equations and Powers. Simultaneous, v. Equations. Sine, 251, 414. Solutions, v. Ausivcrs and Equations. Square. 4. 8, 30. completing a, 206. of difference, 93. of sum, 93. Square Root, v. Root. INDEX 421 Straight Lines, 25, 140, 239. parallel, 166. plotting, 141', 144. Structure (of problems), 103. Substitution, 45, 78, 91. method of, 313. solution by, 170, 250. Subtraction (v. Addition), 1, 15, 39, 45, 51, 125. Surds (v. Radicals and Irrationals), 186, 214, 284. equality of, 306. expressions, 18G, 284. square roots of, 305. Symbols, v. Signs. Tables, 407 et seq. Tangent (to a curve), 259. Tangent (ratio in right triangle), 251, 414. Temperature (v. Thermometer) (curves), 17. Terms (of an expression), 7. like, or similar, 8, 45. Terms (of a fraction), 118. lowest, 120, 122. Terms (of a sequence), 323. Thermometer, 413. Centigrade, 13, 142. Fahrenheit, 13, 142. Reaumur, 146. Transposition, 59, 153. Trial Divisor, 185, 198, 378. Trinomial, 8. factors of, 96, 98, 105, 107, 225. Variables, 25, 139, 237. Variation, 2-'i7. as cube, 249. as square, 248. inverse, 239. linear, 125, 140, 238. simple, 237. simultaneous, 238. various, 248 et seq. Vinculum, v. Parentheses. Weights, 408. Zero, 15. division by, 56, 75, 79, 106, 119, 150 exponent, 296. multiplication by, 68, 106, 109. s , UNIVERSITY OF CALIFORNIA LIBRARY Los Angeles This book is DUE on the last date stamped below. QL.'APR 0