GRAPHIC ALGEBRA s •Tl ^^y&fe- THE MACMILLAN COMPANY NEW YORK ■ BOSTON • CHICAGO ATLANTA • SAN FRANCISCO MACMILLAN & CO., Limited LONDON • BOMBAY • CALCUTTA MELBOURNE THE MACMILLAN CO. OF CANADA, Ltd. TORONTO GRAPHIC ALGEBRA BY ARTHUR SCHULTZE, Ph.D. ASSISTANT PROFESSOR OF MATHEMATICS, NEW YORK UNIVERSITY HEAD OF THE DEPARTMENT OF MATHEMATICS, HIGH SCHOOL OF COMMERCE, NEW YORK THE MACMILLAN GOMPANY 1909 All rights reserved S3 Copyright, 190S, By THE MACMILLAN COMPANY. Set up and electrotyped. Published February, 1908. Reprinted January, 1909. • • ■ ... • • » • * ■ * • • • • • • .• • • • • • . • - * > * • • , *•. • NorfoooD $wsa J. 8. Cushing Co. — Berwick & Smith Co. Norwood, Mass., U.S.A. PREFACE It is now generally conceded that graphic methods are not only of great importance for practical work and scientific investigation, but also that their educational value for sec- ondary instruction is very considerable. Consequently, an increasing number of schools have introduced graphic algebra into their courses, and several elementary text-books on graphs have been published. This book gives an elementary presentation of all the funda- mental principles included in such courses, and contains in addition a number of methods which are shorter and require less numerical work than those usually given. Thus, for the solution of a cubic or biquadratic by the customary method a great deal of calculation is necessary to determine the co- ordinates of a number of points. To avoid these calculations and to make the work truly graphic, the author has devised a series of methods for solving quadratics, cubics, and biquad- ratics by means of a standard curve and straight lines or circles. Two of these methods — the solution of quadratics by a parabola (§ 30) and of incomplete cubics by a cubic parabola (§ 49) — are but slight modifications of methods previously known ; the others are original with the author, who first pub- lished them in a paper read before the American Mathematical Society in April, 1905. The author desires to acknowledge his indebtedness to Mr. Charles E. Deppermann for the careful reading of the proofs and for verifying the results of the examples. ARTHUR SCHULTZE. New York, December, 1907. 260013 CONTENTS PAET I General Graphic Methods CHAPTER I PAGE Definitions 1 CHAPTER II Graphic Representation of a Function of One Variable . 4 CHAPTER III Graphic Solution of Equations involving One Unknown Quantity ........... 13 CHAPTER IV Graphic Solution of Equations involving Two Unknown Quantities 17 PAET II Solution of Equations by Means of Standard Curves CHAPTER V Quadratic Equations 26 vii Vlll CONTENTS CHAPTER VI PAGE Cubic Equations 42 CHAPTER VH Biquadratic Equations 69 APPENDIX I. Graphic Solution of Problems 73 II. Statistical Data Suitable for Graphic Representation 78 III. Tables 84 Answers to Exercises 89 GRAPHIC ALGEBRA it i > PAET I GENERAL GRAPHIC METHODS CHAPTER I DEFINITIONS A N A" 1. Location of a point. If two fixed straight lines XX' and YY meet in at right angles, and PM _L XX', and PN± YY, then the position of the point P is determined if the lengths of PM and PN are given. 2. Coordinates. The lines PN and PM are called the coordinates of point P; PN, or its equal OM, is the abscissa; and PM, or its equal ON, is the ordinate of point P. The abscissa is usually- denoted by x, the ordinate by y- The line XX' is called the jr-axis or the axis of abscissas, YY' the y-zxis or the axis of ordinates. The point is the origin, and M and N are the projections of P upon the axes. Abscissas measured to the right of the ori- gin, and ordinates above the x-axis, are considered positive; hence coordinates lying in opposite directions are negative. -(P & 3 M4 D 2 GRAPHIC ALGEBRA 3. The point whose abscissa is x, and whose ordinate is y, is usually denoted by (x, y). Thus the points A, B, C, and D are respectively represented by (3,4), (-2, 3), (-3, - 2), and (2, - 3). The process of locating a point whose coordinates are given is called plotting the point. 4. Since there are other methods of determining the location of a point, the coordinates used here are sometimes, for the sake of distinction, called rectangular coordinates. Note 1. While usually the same length is taken to represent the unit of the ahscissas and the unit of the ordinates, it is sometimes convenient to draw the x and the y on different scales. Note 2. Graphical constructions are greatly facilitated by the use of cross-section paper, i.e. paper ruled with two sets of equidistant and par- allel lines intersecting at right angles. (See diagram on page 29.) , \Y A ! ! ! i t __^ T . p 3 2 1 ■u X' i X x '-[3 -2 "T" l ) 1 2 3 J/4' i i i 1 -1 C -& =3- i D EXERCISE 1 1. Plot the points: (3, 2), (4, -1), (-3, 2), (-3, -3). 2. Plot the points: (-2, 3), (-5, 0), (4, -3), (0, 4). 3. Plot the points : (0, - 4), (4, 0), (0, 0). 4. Draw the triangle whose vertices are respectively (4, 1), (-1,3), and (1, -2). 5. Plot the points (—2, 1) and (2, —3), and measure their distance. 6. What is the distance of the point (3, 4) from the origin? 7. Where do all points lie whose ordinates equal 4? DEFINITIONS 3 8. Wliere do all points lie whose abscissas equal zero ? 9. "Where do all points lie whose ordinate equals zero? 10. What is the locus of (x, y) if y = 3? 11. If a point lies in the o^axis, which of its coordinates is known? 12. What are the coordinates of the origin? CHAPTER II GRAPHIC REPRESENTATION OF A FUNCTION OF ONE VARIABLE 5. Definitions. An expression involving one or several let- ters is called a function of these letters. x 2 — x + 7 is a function of x. q Vy y 2 is a function of y. y 2 xy — y 2 + 3 y 3 is a function of x and y. If the value of a quantity changes, the value of a function of this quantity will change, e.g. if x assumes successively the values 1, 2, 3, 4, x 2 — x + 1 will respectively assume the values 7, 9, 13, 19. If x increases gradually from 1 to 2, x 2 — x + 7 will change gradually from 7 to 9. A variable is a quantity whose value changes in the same discussion. A constant is a quantity whose value does not change in the same discussion. In the example of the preceding article, x is supposed to change, hence it is a variable, while 7 is a constant. 6. Temperature graph. A convenient method for the repre- sentation of the various values of a function of a letter, when this letter changes, is the method of representing these values graphically ; that is, by a diagram. This method is frequently used to represent in a concise manner a great many data refer- ring to facts taken from physics, chemistry, technology, eco- nomics, etc. To give first an example of one of these applications, let us suppose that we have measured the temperatures at all 4 FUNCTION OF ONE VARIABLE hours, from 12 m. to 11 p.m., on a certain day, and that we have found : At 12 m. 3°C. At 6 p.m. 5° a At 1 P.M. 5°C. At 7 p.m. 3£°C. At 2 p.m. 6i° C. At 8 p.m. 2° a At 3 p.m. 7° C. At 9 p.m. 2 v * At 4 p.m. 6f°C. At 10 p.m. -1°C. At 5 p.m. 6° a At 11 P.M. -2i°C >• 7° b fl° V i ( 1 r° n 1° 17 [ - rt° 4 1 0° 1° i X > ) ; i i I ; ) f r i > ) 10 1 1 P M. i ..V To represent graphically one of these facts, e.g. the tem- perature at 6 p.m. was 5°, construct a point G, whose abscissa is 6 and whose ordinate is 5, taking any convenient lengths as uuits. Eepre- senting in a similar way the temperatures at all hours, we obtain the points (0, 3), (1, 5), (2, 61), (3, 7), etc., i.e. A, B, G, D, . . . M. The diagram thus con- structed contains all the information given in the table, but it gives it in a clear and concise form, that at once impresses upon the eye the relative values of the temperatures and their changes. In a similar manner we may plot the temperatures at any time between 12 m. and 11 p.m. Thus, to represent the fact that the temperature at 1.30 p.m. was 6°, construct the point (Xh 6)- 7. If we represented the temperatures of every moment between 12 m. and 11 p.m., we would obtain an uninterrupted sequence of points, or a curved line, as shown in the next dia- gram. This curve is said to be a graphical representation or a graph of the temperatures from 12 m. to 11 p.m. It is, of course, not possible to construct an infinite number of points, GRAPHIC ALGEBRA hence every graph constructed in the above manner is only an approximation, whose accuracy depends upon the number of points constructed. To find from the diagram the temperature at any time, e.g. at 2.30, measure the ordinate which corresponds to the abscissa 21; to find when the temperature was 4°, measure the abscissa that corresponds to the ordinate 4, etc. Y a 1 1 1 X i 2 1 • \ 3 1 ' i y\ 1 1 p. d. EXERCISE 2 1. From the diagram find approximate answers to the follow- ing questions : a. Determine the temper- ature at : 5 p.m., 1.30 p.m., 5.45 p.m., 11.45 a.m. b. At what hour or hours was the temperature 6°, 5°, 1°, - 1°, 0° ? c. At what hour was the temperature highest ? d. What was the highest temperature ? e. During what hours was the temperature above 5° ? /. During what hours was the temperature between 3° and 4°? g. During what hours was the temperature above 0° ? h. During what hours was the temperature below 0° ? i. How much higher was the temperature at 4 than at 8 p.m. ? k. At what hour was the temperature the same as at 1 p.m. ? I. During what hours did the temperature increase ? m. During what hours did the temperature decrease ? n. Between which two successive hours did the temperature change least ? o. Between which two successive hours did the temperature increase most rapidly ? FUNCTION OF ONE VARIABLE 7 2. Construct a diagram containing the graphs of the mean temperatures of the following four cities : ^H i-t T-H rH rH T— 1 i-H th H i—i tH a •4 £ a w PS < 3 1* < P o p O 5 d a 1-5 fc* S «f y l-B i-s .:: 12.S 17.1 ■23.0 50.6 a. Represent graphically by a point air of 30° C. which holds 15 grams of water vapor per cubic meter. b. If such air would cool, represent the change graphically by a line. c. At what temperature would such air become saturated, i.e. contain all the moisture it can hold?* * This temperature is called the dew-point. 8 GRAPHIC ALGEBRA d. If the same air was cooled to 5°, how many grams of moisture would be condensed per cubic meter ? e. How much more moisture per cubic meter can air of the kind mentioned in Ex. a hold ? * [For more statistical data suitable for graphic representation see Appendix II.] 8. Graph of a function. The values of a function for the va- rious values of x may be given in the form of a numerical table. Thus the table on page 84 gives the values of the functions r- 1 x 2 , x 3 , V as, - for x = 1, 2, 3, • • • up to 100. The values of functions may, however, be also represented by a graph. E.g. to construct the graph of x 2 construct a series of points whose abscissas represent x, and whose ordinates are x% i.e. construct the point (-3,9), (-2,4), (-1,1), (0, 0) • • • (3, 9), and join the points in order. If a more exact diagram is required, plot points which lie between those drawn above, as (^, \), C4, 21), etc. Since the squares of the numbers increase very rapidly, it is convenient to make the scale unit of the x 2 smaller than that of the x. The graph on page 29 was constructed in this manner. To find from the graph the square of — 2.5, measure the ordinate corresponding to the abscissa —2.5, i.e. 6.25. To * Many meteorological facts can be explained by the graph of Ex. 3, e.g. the meaning of "dew-point," relative and absolute humidity, the fact that the mixing of two masses of saturated air of different temperatures produces precipitation, etc. L_ i\ p- \ / \ i / \ / \ / \ \ \ s V y X f - 2 + fl L 2 T l . 11 i i \ -1 1 V r: B ^4- 10 GRAPHIC ALGEBRA 10. The graph of an equation of the form of ax? + bx + c is called a parabola. Thus the graph of \ x 2 — \ x — 3 is a parabola. EXERCISE 3 Draw the gr; aphs of the following functions : # 1. x + 2. 9. x 2 -l. 17. X 2 — X + 1. 2. 3 z + 5. 10. X 2 + 05. 18. 6 +05 — aA 3. 2 as -7. 11. x 2 -2x. 19. 2-as-aj 2 . 4. fa;. 12. 4 a; — ar 8 . 20. 10 - 3 x - x 2 . 5. 1 — re. 13. a? 2 — 4*4-4. 21. 2 x 2 + 5 x - 20 6. 2-3a,\ 14. x 2 — x — 5. 22. ar 3 . 7. — 3 as. 15. as* -3 a; -8. 23. a,* 3 — 2 a5. 8. \tf. 16. a?* + a; — 2. 24. ar 3 — a; + 1. 25. Draw the graph of x 2 from 05 = — 4 to a; = 4, and from the diagram find : a. (3.5) 2 ; 6. (-1.6)*; c. (-2.8) 2 ; d. (1.9) 2 ; e. V6\25; /. Vl2^25; gr. V5 ; ft. V^3. 26. Draw the graph of x 2 — 4 05 + 2 from 05 = — 1 to a; = 4, and from the diagram determine : (a) The values of the function if as = — ^, 1^, 2^. (6) The values of as, if x 2 — 4 a; + 2 equals — 2, 1, 1£. (c) The smallest value of the function. (d) The value of x that produces the smallest value of the function : (e) The values of a; that make x 2 — 4 x + 2 = 0. (/) The roots of the equation 05* — 4 x + 2 = 0. (p*) The roots of the equation as 2 — 4a? + 2=— 1. (ft) The roots of the equation ar — 4 as + 2 = 2. * If necessary, use for the ordinates a smaller auit than for the ab- scissas. FUNCTION OF ONE VARIABLE 11 27. Draw the graph oiy = 2 -\-2x-x 2 , from x = — 2 to x = 4, and from the diagram determine : (a) The values of y, i.e. the function, if x—%, — 1£, 2\. (b) The values of x if y = — 2. (c) The greatest value of the function. (d) The value of a; that produces the greatest value of y. (e) The values of x if the function equals zero. (/) The roots of the equation 2 + 2 x - x 2 = 0. (#) The values of x if y = 1. (/i) The roots of the equation 2 + 2cc— a^ = l. 28. The formula for the distance traveled by a falling body- is S = ±gt 2 . (a) Eepresent i gt 2 graphically from t = to t = 5. (Assume g = 10 meters, and make the scale unit of the t equal to 10 times the scale unit of the ^ gt 2 .) (b) How far does a body fall in 2\ seconds ? (c) In how many seconds does a body fall 25 meters ? 11. A function of the first degree is an integral rational func- tion involving only the first power of the variable. Thus, 4x + 7orax + 6 + c are functions of the first degree. 12. It can be proved that the graph of a function of the first degree is a straight line, hence two points are sufficient for the construction of these graphs. (This is true if the abscissas and ordinates are drawn on different scales or on the same scale.) It can easily he shown that the preceding proposition is true for any particular example, e.g. 3 x + 2. If a; =-3, -2, -1, 0, 1, 2, 3, then 3x + 2=-7, — 4, — 1, 2, 5, 8, 11; i.e. if x increases hy 1, 3 x + 2 increases hy 3. Hence if a straight line be drawn through (—3,-7) and ( — 2, — 4), this line will ascend 3 units from x = — 3 to x = — 2. Obviously the prolongation of this line will ascend at the same rate throughout, and it will pass through (— 1, — 1), (0, 2), etc. 12 GRAPHIC ALGEBRA Instead of plotting (—3, —7) and (—2, —4), any other two points may be taken. It is advisable not to select two points which lie very closely together. EXERCISE 4 Draw the graph of 1. 3 a? -10. 3. 2 a;— 7. 5. 6+ a. 2.5^ + 2. 4. 2 — 3 x. 6. | x — 5. 7. Degrees of the Fahrenheit scale are expressed in degrees of the Centigrade scale by the formula C. = -| (F. — 32). (a) Draw the graph of f (F. - 32), from F. = - 5, to F. = 40. (6) From the diagram find the number of degrees of Centi- grade equal to - 1° F., 9° F., 14° F., 32° F. (c) Change to Fahrenheit readings: - 10° C, 0° C, 1° C. 8. Show that the graphs of 3 x + 2 and 3 x — 1 are parallel lines. CHAPTER III GRAPHIC SOLUTION OF EQUATIONS INVOLVING ONE UNKNOWN QUANTITY 13. Degree of an equation. A rational integral equation which contains the nth. power of the unknown quantity, but no higher power, is called an equation of the 7ith degree. x 5 — 5 x 3 + 2 x 2 — 7 = is an equation of the fifth degree. X s — 2 x 2 — 5 x + 1 = is an equation of the third degree. A quadratic equation is an equation of the second degree. A cubic equation is an equation of the third degree. A biquadratic equation is an equation of the fourth degree. x 3 + 2x + 3=0isa cubic equation. x 4 + 3 x 3 + 2 x — 7 = is a biquadratic equation. 14. Solution of equations. Since we can graphically deter- mine the values of x that make a function of x equal to zero, it is evidently possible to find graphically the real roots of an equation. Ex. Find graphically the real roots of the equation !B 8 4-aj 8 -9a!-7 = 0. (In computing the values of y use table on page 84.) X X 2 X s — 9a; x 3 + x 2 — 9x x 3 +x 2 — 9x— 7 or y -4 16 -64 36 -12 -19 -3 9 -27 27 9 2 -2 4 - 8 18 14 7 -1 1 - 1 9 9 2 - 7 1 1 1 - 9 — 7 -14 2 4 8 -18 - 6 -13 3 9 27 -27 9 2 4 16 64 -36 44 37 13 14 GRAPHIC ALGEBRA Obviously the values of the f uuction for x > 4 will increase rapidly, and for values of x < — 4 will be less than — 19. Locating the points (-4, -19), (-3, 2) (-2, 7) ... (4, 37) and joining produces the graph ABC. Since ABC intersects the avaxis at three points, P, P', and P", three values of x make the function zero. Hence there are three roots which, by measuring OP", OP', and OP, are found to be approximately —3.1, -.8, and 2.9. To find a more exact answer for one of these roots, e.g. OP, we draw the portion of the diagram which contains P on a larger scale. If x = 2.9, the function equals —.301, i.e. it is negative. Hence it appears from the diagram that the roots must be larger. Substituting x = 3 produces x 3 + x' 2 — 9 x — 7 = 2, a positive quantity. The root there- fore must lie between 2.9 and 3. Making the unit of length ten times as large as before, we locate the points (2.9, — .301) and (3, 2), i.e. B' and C , in diagram II. Since in nearly all cases small portions of the curve are almost straight lines, we join the two points by a straight line B'C, which intersects the x-axis in P. The measurement of P gives the root x = 2.915. If a greater degree of accuracy is required, a third drawing on a still larger scale must be constructed. 15. The diagram of the last exercise may also be used to find the real roots of an equation of the form a^-f x 2 — 9x— 7 = ra, when m represents a real number. To solve, e.g., the equation a^ + ar — 9 x — 7 = 2, determine the points where the function is 2. If cross-section paper is used, the points may be found by inspection, otherwise draw EQUATIONS INVOLVING ONE UNKNOWN QUANTITY 15 through (0, 2) a line parallel to the £-axis, and determine the abscissas of the points of intersection with the graph, viz. -3,-1,3. 16. It can be proved that every equation of the nth degree has n roots ; hence if the number of the points of intersection is less than n, the remaining roots are imaginary. Thus, x* + x 2 — 9x — 7 = 13 has only one real root, viz. 3.4; hence two roots are imaginary. If, however, the line parallel to the x-axis is tangent to the curve, the point of tangency represents at least two roots, and hence the preceding paragraph cannot be applied. EXERCISE 5 Solve graphically the following equations : 1. 4a; — 7 = 0. 14. 2 x 2 - 4 x -15=0. 2. 2 z + 5 = 0. 15. 2 x 2 + 10 a;-7 = 0. 3. 6-x = 0. 16. 3ar 2 -6cc-13 = 0. 4. 8-3 a; = 0. 17. X s - 3^-1 = 0. 5. x 2_ a; _ 6 = 18< ar?- 12 £ + 18 = 0. 6. ar-£-5 = 0. 19. £ 3 -4« + l=0. 7. a? — 2 x — 7 = 0. 20. x 3 + £-3 = 0. 8. £ 2 -6£ + 9 = 0. 21. £ 3 + 3£-ll = 0. 9. x 2 + 5 a j_4 = o. 22. 2£ 3 -6£ + 3=0. 10. £ 2 -5£-3 = 0. 23. £ 3 -5a; 2 -9£ + 50 = 0. 11. x 2 -Sx-6 = 0. 24. x 3 - 13 £- + 38 £ + 17=0. 12. £ 2 -2£-9 = 0. 25. X 4 -10x 2 +8 = 0. 13. 3£ 2 -3£-17 = 0. 26. £ 4 -4£ 2 + 4x-4 = 0. 27. x 4 -6x 3 + 7x 2 + 6x-7 = 0. 28. x 5 - x 4 - 11 x 5 + 9 x 2 + 18 x - 4 = 0. 29. 2 x + x — 4 = 0. 16 GRAPHIC ALGEBRA 30. If y = x s + 5 x 2 - 10, (a) Solve y = 0. (c) Solve y = — 5. (6) Solve y = 5. (d) Solve?/ = -15. (e) Determine the number of real roots of the equation y=-2. (/) Determine the limits between which m must lie, iiy = m has three real roots. (g) Find the value of m that will make two roots equal if y = m. (h) Find the greatest value which y may assume for a negative x. (i) Which negative value of x produces the greatest value ofy? 31. Ify=za? — 7x + 3, (a) Solve y = 0. (d) Solve y = 10. (6) Solve y = 3. . (e) Solve y = -15. (c) Solve y = — 3. (/) Determine the number of real roots if ?/ equals 15, 10, 5, or - 7. (#) Determine the number of imaginary roots if y = — 10, if ?/ =12, if y = 2. CHAPTER IV GRAPHIC SOLUTION OF EQUATIONS INVOLVING TWO UNKNOWN QUANTITIES 17. Graphs of functions of two unknown quantities. In § 8 the graph of the function \ x 2 — \x — 3 was discussed. If £ ajS _ £ x — 3 is denoted by y, then the ordinate represents the various values of y, and the annexed diagram represents the equation y = hx*-\x-3. (1) The coordinates of every point of the curve satisfy equation (1), and every set of real values of x and y satisfying the equation (1) is represented by the coordinates of a point in the curve. Similarly, to represent — ^— = 2 graphically solve for y, i.e. y-5 x 2 + x + 10 \ y> v V l \ \ \ X' \ 1 X 4 - 3 \ 2 - i c . ll i i \ -I K r; B y = - and construct the graph of ■ • 2 18. The curve representing an equation is called the graph or locus of the equation. 19. If an equation containing two unknown quantities can be reduced to the form y = f(x), when f(x) represents a func- tion of x, then the equation can be represented graphically. 17 18 GRAPHIC ALGEBRA Ex. 1. Eepresent graphically 3 x — 2 y = 2. 3z-2 Solving for y, y = ■2 v 3 .' .*2 - S ^ -^ ^> % 1r U __„2I _ ^ £- ,<"2 3 <^s ~ 2 ,*' ^ 3 —1 ^^ ^sf -<" t^ : ZfZ L. _ _x 20 GRAPHIC ALGEBRA By measuring the coordinate of P, we obtain the roots, x = 3.15, y = .57. 23. The roots of two simultaneous equations are represented by the coordinates of the point (or points) at which their graphs intersect. 24. Since two straight lines which are not coincident nor parallel have only one point of intersection, simultaneous linear equations have only one pair of roots. If two equations are inconsistent, as2x + ?/ — 2 = and 2x + y — 4 = 0, their lines are parallel lines (§21). If two equations are dependent, their graphs are identical, as * + 1 = l &n&Sx + 2y = 6. 2 3 Obviously inconsistent and dependent equations cannot he used to deter- mine the roots of a system of equations. 25. Equations of higher degree can have several points of intersection, and hence several pairs of roots. Ex. 1. Solve graphically the following system : x> + f = 25, (1) [3x-2y=-6. (2) Solving (1) for y, y = V25 - x 2 . Therefore, if x equals -5,-4,-3,-2,-1, 0, 1,2, 3, 4, 5, y equals respectively 0, ± 3, ± 4, ± 4.5, ± 4.9, ± 5, ± 4.9, ± 4.5, ± 4, ± 3, 0. Locating the points (-5, 0), (-4, + 3), (_4, -3), etc., and joining, we obtain the graph (a circle) ABC of the equation x 2 + y 2 = 25. Locating two points of equation (2), e.g. (- 2, 0) and (0, 3), and joining by a straight line, we obtain DE, the graph of 3x-2y=-Q. Since the two graphs meet in two points P and Q, there are two pairs of roots, which we find by measurement, x s: 1J, y = 4$, or x =- 4, y =- 3. Y E • ,?P- X / / \ B / . X' O \x - ■ / 1 i i ! ' r 5i' -1 /r -2 [ 3/ \t -4 ' C J V 1 SOLUTION OF SIMULTANEOUS EQUATIONS 21 Ex. 2. Solve graphically the following system: xy 12, (1) (2) From (1) y = 12 x —y = 2. Hence, by substituting for x the values — 12, — 11, • •• to + 12, we obtain the following points : (— 12, — 1), (— 11, — 1^), (- 10, - 1|), (- 9, - H), (-8, -li), (- 7, - If), (- 6, - 2), (- 5, -2|), (-4, -3), (-3, -4), (-2, -6), (-1. -12), (0, ±ao), (1, 12), (2, 6), etc., to (12, 1). Locating these points and joining them produces the graph of (1), which consists of two separate branches, CD and EF. Locating two points of equation (2) and joining by a straight line, we have the graph AB of the equation (2). The coordinates of the two points of inter- section P and P' are the required roots. By act- ual measurement we find x = 4.5+, y =2.5+, or x = — 2.5, y = — 4.5. To obtain a greater degree of accuracy, the portion of the diagram near P is represented on a larger scale in the small diagram. Since the small part of CD which is represented is almost a straight line, it is sufficient to locate two or three points of this line. By actual measurement we find : x - 4.606, y = 2.606. Evidently the second pair is as = —2.606, y = -4.606. By increasing the scale further, any degree of accuracy may be obtained. . ,Y-' C ■ ■- ■ 1 ^R 1 z -,£ I z X 7 > /_ \ / \ ~? ^t^ ~X- P*^ v' "X" F ~ — _ Z X ^ /- PjV JT )/^ 3- P4- -B'Z q- \- sr -k ~2_ \ ^ T^z ~y A ^J^ 7 ~2_ i 6 J2,_ ~? ~7 7 Nj tz / ^ JZ i" jk ._:&. z£j_ jL 2.5 a: 4.5 4.6 4.1 22 GRAPHIC ALGEBRA EXERCISE 7 4 Ux + 3y=12, \ x + 5y=6. 5. 3x + 5y = 7. 5 x — y = 7. 3. Solve graphically the following simultaneous equations ■3x + 4y = 8, 2x-3y = 6. 3x + 4y = 10, 4 x + y = 9. 2a;-32/ = 7, 3z + 2?/=-8. 6. Show graphically that the following system cannot have finite roots : \2x-y=2, [2x-y = 6. 7. Show graphically that the following system is satisfied by an infinite number of roots : 4 + 3 ' 3x + 4y=12. 8. Without constructing the graphs, determine the relative positions of the loci of 14 x — 7 y -f 2 = and 14 x — 7 y + 5 = 0. Solve graphically : 10. 11. 12 r^ + 2/ 2 =i6, # + 2/ = 5, ./•// = 6. «-y = l, x 2 + 2/ 2 = 25. * - y = 2 > xy = 8. 13. 14. 15. \ 16. (4x-5y = 10, \xy = 6. x 2 — y 2 = 4, 25. = 12, = 8. a- = 2 y. f.r//= 6, x 2 + xy-- x 2 - y 2 : 26. The equation of the circle. 77ie locus of an equation of the form x 2 -\- y 2 = r 2 (1) is a CiYcZe whose center is the origin and whose radius is r. SOLUTION OF SIMULTANEOUS EQUATIONS 23 For the distance from the origin of a point P in the locus, OP=V.?+? (1) = Vr 2 = r. But if the distance of every point in the locus from is equal to r, then the locus is a circle whose center is and whose radius is r. Thus, x 2 + y 2 — 16 is a circle whose center is and whose radius equals 4, x 2 + y 2 ~ 10 is a circle whose center is and whose radius is VlO. Note. The square root of a number can often be represented by the hypotenuse of a right triangle whose arms are rational numbers. Thus, VlO = V3 2 + l 2 , hence VlO, equals a line joining (0, 0) and (3, 1). i ,r P u X' X X • \y' 27. The locus of expressions of the form (x-ay+(y-by = r* is a circle whose center is (a, b) and ivhose radius eqxials r. Let P beany point in the locus, and C= (a, h). (2) Draw CD || OX ; ±£^k CP = CD" +DP\ But CD = x — a, and DP =y — b. .:CP 2 =(x-a) 2 +(y-by. Hence, from (2), CP 2 = o 2 , or CP=r. I.e. the distance of any point iu the locus from C equals r, or the locus is a circle whose center is (a, b) and whose radius is r. Thus, (x — 2) 2 +{x + 4) 2 ^8 represents_a circ le whos e center is (2, — 4) and whose radius equals V8. Since V8 = V2 2 + 2 2 , it is easily con- structed. Note. The equation (x — a) 2 + (y — b) 2 = r 2 , however, represents a circle only if the scale units of the abscissas and ordinates are equal. If the two scales are unequal, the locus is an ellipse. 24 GRAPHIC ALGEBRA Ex. 1. Construct the locus of x 2 + 2x+y 2 -4y-5 = 0. Transpose and complete the squares of the expressions involving x and y, (x2+2x + l) + ( 2 /2-4 2 / + 4) = 5 + 6, (x +1) 2 + 0-2)2 = 10. I.e. the required locus is a circle *vhose center is ( — 1, +2) and whose radius is vTo. 0. 3 l 9 l 9 2 + TS T" ?J Ex. 2. Construct the locus of 2a; 2 + 2?/ 2 -3a; + 6?/ + 3 Dividing by 2, and transposing, X 2_3 x + 2/2 + 3 2/= _3 < Completing the squares, X 2_3 x+( 3 )2 + 2/ 2 +3 y +( | )2= . (^-f) 2 +(2/ + l) 2 = fi- I.e. the locus is a circle whose center is (f , — |) and whose radius IV21. 28. The preceding examples show that the locus of a quad- ratic function involving two variables is a circle, if the func- tion does not contain xy and if the coefficients of x 2 and y 2 are equal. EXERCISE 8 is Solve graphically : * 2 + 2/ 2 = 4, x + y = 3. x 2 + y 2 = 16, x — y = 4. a; 2 + 2 / 2 = 50, x — y = — 6. fx 2 + 2 / 2 = 9, [x — 2y = 2. x? + y 2 =16, 2y-3a = 6. 6. 7. 4. 5. 9. 10. [x 2 -2x + y 2 -± y = 0, [y = 2x. x 2 -4 : x + y 2 +2y+3=0, x — y = 3. x 2 -10x + y 2 = 0, x 2 + 6 a; + y 2 = 16. (x- r -l) 2 -( 2/ -l) 2 = 2, (o;-l) 2 +(2/ + l) 2 = 8. 'aj 2 + 2/ 2 = l, (a:-l) 2 + y 2 = 2. PART II SOLUTION OF EQUATIONS BY MEANS OF STANDARD CURVES 29. A disadvantage of the preceding graphic methods is the fact that they often require a great deal of numerical calcula- tion, and that the necessary curves are difficult to draw. In the following chapters, methods will be given for the solu- tion of quadratics, cubics, and biquadratics by means of one standard curve, and straight lines or circles ; i.e. one curve may be used to solve all quadratics or all cubics, etc. The construction of these curves requires very little calculation, and once constructed, each curve may be used for the solution of many problems. Three curves are used in the following chapters, viz. a parabola y = x 2 , a cubic parabola y = oc?, and an equilateral hyperbola y = -. y = x 2 was drawn and discussed in § 8. A locus of the form y = - was given in § 25, Ex. 2, and the graph of x y-=x z will be given in § 49. Any one of these three curves may be used to solve with rules and compasses either quadratics or cubics, but only the parabola and equilateral hyperbola yield simple solutions for biquadratics. 25 CHAPTER V QUADRATIC EQUATIONS 30. To solve the quadratic ax 2 + bx + c = (1) by means of a standard curve, we split the equation (1) into two simultaneous equations, one of which is the standard curve, while the other is a straight line or circle. Thus, if ax 2 + bx + c = 0, (1) Let y = x*. 1 (2) Substituting in (1), ay + bx + c = 0. J (3) The solution of the system (2), (3) for x produces the required roots of (1). x But the graph of (3) is a straight line, while the graph of (2) is identical for all quadratic equations. Hence, after the graph y = x 2 (see annexed diagram) has been constructed, any quadratic equation may be solved by the Construction of a straight line, provided the roots lie within the limits of the represented abscissas (—6 and + 6). Ex. 1. Solve 11 x 2 + 30 x - 165 = 0. (1) Let y = x 2 . (2) Then 11 y + 30 x — 165 = 0. (3) In (3), if x = 0, then y = 15 ; if y = 0, then x = 5 J. Tbe straight line joining the points (0, 15) and (5|, 0) is the graph of (3), which intersects the graph of (2) in P and P'. By measuring the abscissas of P and P', we have x = 2.7, or x = — 5.5. Ex.2. Solve 5 x? - 14 x- 65 = 0. (1) Let y = a; 2 . (2) Then by- 14* -65 =0. (3) 26 QUADRATIC EQUATIONS 27 Locating two points of the equation (3), e.g. (0, 13) and (5, 27), and joining by a straight line produces the graph of (3), which intersects the graph of (2) in Q and Q'. Measuring the abscissas of Q and Q', we obtain a = 5.3, or x = — 2.5. :"::::::::::::""::x"::^- T-T"" : "-"-"""" : " F: 35t — p-i — | 1 — L -r J -d —i- \ i Yd' -Of _ 30—1 p LJ yZ- >^ ^ ^ *yQ S. v *. 7^ * ^L/u — - — MVi- ' =P- S4--"' v \ ^§0. - T-2--' ' £ ' V 3>7 ffly / ... _ ^ >-~ >w ■<& -7 V" S>7T - - — -N - *■?*>- t s v s -^%"" 7t "is ■ST ^P> 5> *£_ US. 2"i £ -* d.5. *C -j , > ■' ■ - __,v\ ____ >>? 7 ^ ^ z \ ^ ^ / ^h V \ "fe p 12. "^ 7 _ ^ ^ > / p i \ ^ >v h : :::: ::: :::^-i: * $N " "- ,,'fe I"! .5 _ * * -- - «= ^4 i S ^ s? %. ' T =* - *■ ^ s >» X- "X <*' ' ""^dr t_s^ri __ -iX^, -£,■' -i4.: =3 -.? -1 *] ' J -2_ -3 "• -5 -2k Jy* I ' -M^ x : zm 3Z ZL. 31. In the equation ay + bx + c = 0, if x = 0, then y = , and if y = 0, then *= — ?• Hence, by laying off on the a>axis the distance — - and on the w-axis the distance , and apply- b a ing a straight edge, the roots of the equation ax 2 + bx + c = can frequently be determined by inspection. If the two points constructed on the axes lie very closely together, the drawing is likely to be inaccurate, and it is better to locate one or both points outside the axes. 28 GRAPHIC ALGEBRA EXERCISE 9 Solve the following equations by the graphical method : 1. x t -x-6 = 0. 8. 4x 2 -25« + 20 = 0. 2. ^_|_a;_2 = 0. 9. 3ar + 20o; + 12 = 0. 3. a 2 - 3 a -18 = 0. 10. a^ + aj — 5 = 0. 4. a? 2 + 3aj-10 = 0. 11. x 3 — 2x — 9 = 0. 5. a 2 -2a;-8 = 0. 12. 3x* + 7x-42 = 0. 6. ^ + 2^-4 = 0. 13. 2^ + 5^-20 = 0. 7 . ar ! _5a;-15 = 0. 14. 5r-4a;-5 = 0. 32. Solution for large roots. By changing the unit of the abscissas and the unit of the ordinates, the same diagram may be used to represent y = v? for various values of x. For in the dia- gram we may assign any values to the abscissas, provided the corresponding ordinates are made equal to the squares of the abscissas. Thus after the graph of y = x 2 has been drawn from x = — 10 to x = 10, we may multiply the numbers on the «-axis by any number, e.g. 3, and thereby extend the diagram from x = — 30 to x = 30, provided we multiply the numbers on the ?/-axis by 3 2 , or 9. This change of scale units does not affect the character of the locus ay + bx + c = 0, for this equation is a straight line whether the abscissas and ordinates are drawn on the same or different scales. The annexed diagram can be used directly for roots between — 10 and + 10. If the roots are larger, but lie between — 100 and +100, multiply the units by 10 and 10 2 respectively; i.e. omit the decimal points in the diagram. For roots still larger, add another cipher to the values of the abscissas and two ciphers to the values of the ordinates. Ex. 1. Solve graphically x 2 — 16 x - 4400 = 0. Let 2/ = x 2 ; then y- 16 x- 4400 = 0. QUADRATIC EQUATIONS 29 Obviously the regular diagram does not contain the required roots. Hence multiply the values of the abscissas by 10 and the values of the ordinates by 10 2 ; i.e. disregard the decimal points. 30 GRAPHIC ALGEBRA C b Since ^ is very large, locate two points as follows If x = 0, y = 4400. If x = 100, y = 6000. The line joining (0, 4400) and (100, 6000) intersects y = a?m >^3 1 S-pvgH — 5--r--S— - S-- ' . ?. - ± - - i r 3a L"aL _L =>V "5 J 9 =2 A! - s e>" "o" ©' *e s» e>" 3:? c '| ■ ' 1 ,' ■■ f I , iv = ^ j k o H 7 c 3T - «; - Jr^ - — => i - - X X___ 2 -- . -X __x ._ _ __/ _ -s. t - si - : . : x x -7 = > V \h — ;- -- :--; v 5i i iSL 30 / X \, v>t t X'v $#■ X " X "" " "X " 2 z X ~~ s\ W t : x : : : ^::_ . _.- 3 _ ^^ 1 " . S * _ x .=! . _^ ; ~ I V * o ~ / ^ r^ . : :: :: ::: ::: 1:2::: : 1 s Y^ r V \ "V : x x" :: ' ' J x . x - - ::_: ■ > n - ' - 3 ° i S ,N - / " t *o. XX--' X — S S- : _ -X : x_ : i : :.5S : J~ S . d / e " L O / E :::::::.:::::: s V -^ / X -IV-r \ / - x ritt W-- . 1 i =_ _ : __ _ _: / s £ : ._ 3jt — : ;: ; — v -at - — » r , _ '•? -4 v< :z _ - X ^ r . ± x^: xz : : : _. - ts xp _ .. . £:_:: _- :: ::t: . : .- _ _ 3 :>=> - - . -_ = t. rT x _: : : :: i _ - _ <o 1 -- - - - // = _ : _ __ tijz - i_ _ : _ ±3. _x ;: :_ ::.x::.:3 =----- ^ y i \ u + _,__ xx ::: :: :: x : : : ::x :: ::: :: : x I - - X ^■'3 : . _ _ _=:-_ - -. . -- -P. id >^ ji x : :: - x x s :_ : --± 1 _ x *k : X _ _ _ x 3) ^ ** 3> CB' - ^ > ^, """x ^ 3: : 3" : 3: ~ ■- £ 1 m c ' US ^ ' ° Ja 3 ::ji -, 4, x^ .,x_-.jix.---i.._.t QUADRATIC EQUATIONS 31 P and P. By measuring the abscissas of Panel P, we obtain x = 74+ and x = - 59 + . 33. Small roots. Tor small roots multiply the values of the abscissas by a fraction, most conveniently by .1, and the values of the ordinates by .01 ; i.e. place the decimal point in front of each number given in the diagram (except x = 10 and y = 100, which become 1.0 and 1.00 respectively). Thus, 1.0, 2.0, 3.0, etc., become .10, .20, .30, etc. As this shifting of the decimal point is a simple operation, it may be done mentally, without any actual alterations of the numbers in the diagrams. Ex. 2. Solve 10 ar> + 5a? = l. Let y = x 2 . Then 10 y + bx = 1. If x = 0, y = A. If x=l, y -- A. Since in the original diagram such small fractions of y cannot be well represented, multiply the numbers on the x-axis by .1 and the numbers on the ?/-axis by .01; i.e. imagine the decimal point to be placed in front of each number. Then the straight line that joins (0, .1) and (1, —.4) intersects the parabola in Q and Q'. The measurement of the abscissas of Q and Q' gives the roots x = .15, or x = — .65. Note. The student should draw a diagram similar to the one used in the text, but on a larger scale. The cross-section paper employed should have each unit divided into 10 parts. EXERCISE 10 Solve graphically : 1. x 2 - 15 x -4500 = 0. 6. x 2 + 80 x = - 700. 2. x 2 - 10 x - 3000 = 0. 7. x 2 - 10 x- 600 = 0. 3. or + 80 x + 1200 = 0. 8. .^ + 8^-128 = 0. 4. x 2 + 40a;=1200. 9. x 2 - 30 x- 1800 = 0. 5. x 2 + 30 x = 4000. 10. x 2 + 33 x - 1210 = 0. 32 GRAPHIC ALGEBRA 11. 2 or 2 4- 3 a -1500 = 0. • 12. 3 or 2 + 10 a; = 3000. 13. ar + 29 a =210. 14. 3x £ + 200 a- 1200 = 0. 15. 50^-15^-6 = 0. 16. 10a^-6a;-l=0. 17. 4af ! + 5a;-l = 0. 18. 20 x 2 + 3 x -1 = 0. 19. 50ar-5.r-3 = 0. 20. 25 a 2 + 10 a- 3 = 0. 21. 50.c 2 + 5a;-l = 0. 22. 8 x 2 -2 x — 1 = 0. (1) (2) (3) 34. Graphic representation of a quadratic function. Consider the equation x 2 +px + q = 0. Let y = x 2 . Then y +px + g = 0, or 2/ = — pa — q. In the annexed diagram, let COD represent the parabola y — x 2 , and BR the straight line y + px + q = Q,ov y = — px — q. Let CL4 or x' be any particular value of x, then (L4 = x' 2 , and 2L1 = — px' — g. Hence CB=OA-BA = x' 2 + px' + g. I.e. the value of the function x 2 + px + q for any particular value x' is represented by that part of the corresponding ordinate which is intercepted between the straight line y + px + q = 0, and the parabola y = x 2 . The distance is measured from the straight line, and is taken positive if it extends upward, negative if it extends downward. Thus, in the annexed diagram, y = x 2 — \x — 1 , and we have : \ ■y f \ O _ _ - : s r jl zlizii _. _ -. .- ::=i - :: zizz _- = RK s. V 9 * \ i *■< V \* T' i c.;;:z.s3: s '/ <= 4- - - I - ^v v?b -? x X \ % v.P. - ± : :: j \ i^'- : : x x : i :: : i =_ _ ::. : . _ : :; : ^s r i . _:: . : ::=> :.: Izz - c ^r- t- ^V*. ■■€■ 7 - -f X t- SX .. ... Z.. ' . Z. ; X -- i 55,. - - 2 l ^i / C- -- -x_ - -- -x: -X __ :v :: : :?_ :: ,*__ _ .: ::= a k >? ' -# -4* V- - j! . : :::::: j\ . ±z: : :*> : : : r :..:: ::: : - jJj X - ^r^W- -1 oi \SC ^ => r- cj| | V " ' . 4 , X ^S ► £ : : : ; js 1 :±: : : : v -,__ _ _ 3-xx s tzz z i .: .:: 2l i :: . _ & 2 j.. - -. -- _ - s . : . -£4x -1 z : -r-V- wztz z» ^ix & x ri \S / Zp. , V i ... -i^A'X ._ _ _ -V -- --u = yj--a, --_--_-------. -- H X 'L±X ^ :r __ _ _ _. __2_^<= s- _. . --* ■-■; --* \ r ----- X h - r. -X \.;,.X. .. X 1 t J :±: 5c. x --- X i :.-.. j ± X x = .. - _ _:± i -_ :i2 - zl° - x X ~ "X k - X-jX- - ::~:± x -------- --.± t / / ±: - j :i \ N/ / ?i .:. . : .- j - v?x ~ x x -- i ----- -&/ 5 x v>^ -X t _ _ : : 1 IP' v ° : 1 " :i' :: " * X -- - t 3$v -■*- ' -I X - ^S>X - z x . i^x : / - : - -- => _ ___ _ "f rf~t_ --t -- ; ;_ ' a 4 \ • ' y 1 u ' - - IX X-,' X .- 12, z : x -- i " z = / y => b ' ^ ■-- ^ /■\ * \ 1 / M ; ' >f ^ 1 - -x»"J'- J -p- - - -, -- , zf sj \ 1 = :: _: sj _ _ r_ x = g ^ \ =• .__.. - ,^- _ _ X-- -- c - - - =i * J s*\ * *- ^ jS* => /^ ' - z, ^ '.-.'■ i *r ?i fc * '" ^ - 1 Ej , '■' ' 'L 1 38. If the roots are equal or nearly equal, the graphic method is, however, liable to be inaccurate, since a slight inaccuracy in the construction of ay + &.« + c = usually produces a considerable error in the value of x. 36 GRAPHIC ALGEBRA 39. Complex roots. If the line ay + bx + c = does not intersect the parabola y = x 2 , the roots are complex, and their values may be found by means of the following theorems. 40. Let x 2 + px + q = represent an equation whose roots are equal; then these roots are, by the general formula: Hence Assuming that d is a positive' quantity, it can easily be shown that x 2 -\-px-\-q — d = has the roots — - ± y^~. n\ x 2 +px + q = has the equal roots — -P, — ^; (2) — — x 2 +px + q + d = has the roots — -2 ± A /_ c i /$\ The roots of (3) are complex and cannot be found directly by the graphic method, but if we solve (1) instead, we only have to multiply the irrational parts of the answer by V — 1 to obtain the roots of (3). The straight line which serves to solve (1) can be obtained from the one which solves (3) by means of the following proposition. 41. If x 2 -\-px + q = has equal roots, the three straight lines y+px + q-d = 0, (la) y+px + q =0, (2 a) y+px + q + d = 0, (3 a) are parallel, and the second one is equidistant from the other two. These lines (AB, CD, and EF in annexed diagram) are parallel by §21. By making x = 0, we obtain : OA = -q + d, OC = -q, * Schultze's Algebra, p. 269. QUADRATIC EQUATIONS 37 OE- -q-d. Hence AC=CE = d; I. e. CD is equidistant from AB and EF. 42. Hence if EF is known, draw the tangent CD || EF,make CA=EC, and construct AB \\ EF; then AB is the required line whicli produces the roots of (1). 43. The construction, however, is simplified by the following theorems: 1. TJie abscissa of the point of con- tact (Gr) is equal to the rational part of the roots of (1) and (3) . For this abscissa = — " • 2 2. A parallel to YY' through the point of contact (G) bisects the chord KB, and hence any chord parallel to the tangent. For the abscissas of K, H, and B are respectively OL OM-. -P-Vd; 2 _£• ' 2' 2 Hence LM = MN = Vd. Therefore, according to a geometric theorem,* KH= HB. To solve the equation (1) 44. Graphic solution for complex roots. x 2 + bx + c = 0, which has imaginary roots. Construct the locus EF of y -f- bx -f- c = 0, and draw any chord PQ \\ EF. (See diagram, page 38.) Through M, the midpoint of PQ, draw RI || YY' intersecting * Schultze and Sevenoak's Geometry, § 144. 38 GRAPHIC ALGEBRA the parabola in G, and EF in J. Make GH = IG, and through .H" draw i£B || -Ei^ 7 . Then the abscissa of H is the real part, and the difference of the abscissas of B and H mul- tiplied by V — 1 is the imagi- nary part of the required roots ; i.e. x = 031 ± MN x V — 1. Note. The line RI may also be constructed by drawing an ordinate b |, or if through one point [0, V 2 a l o = l, through (o, --Y 45. If the coefficient of x 2 is a, the solution is the same as if this coefficient was unity; for by dividing ax 2 + bx + c = by a, we obtain aj 2 +^aj + -=0. (2) a a The straight line which serves to solve (2) is therefore a a or a?/ + 6x + c = ; t'.e. we may substitute y = x 2 directly into the given equation. Ex. l. Solves 8 -4 a; + 13 = 0. Let Then If If y = %'. y- 4x + 13 = 0. x = 0, y = - 13. x = 5,y= 7. Join (0, - 13) and (5, 7) by line EF, and through the midpoint (R) of any parallel chord draw RIWYY'. Make GH = IG and draw KB II EF. By measuring we obtain : The abscissa of H — 2. \ Y '35- /l \ / 'b- \ • ' / \ • \\/ / T **.„ . M 111' F, ;. ■x • / A ' K/ ■>• U; V / A > - i N I ' 1 : ! > r - 1 P ( V / 3 -If, K QUADRATIC EQUATIONS 39 The difference of the abscissas of B and H= 3. Hence the required roots are x = 2 ± 3 V~^l or 2 ± 3 j. Ex.2. Solve 2 a; 2 +5 a; + 15 = 0. (1) Let y — X 2. (2) Then 2 y + 5 a; + 15 = 0. (3) Construct the line (3), i.e. LN, and through the midpoint (M) of any parallel chord draw MP\] YY'. Make QS = PQ and draw SU\\ LN. The roots produced by TUare - 1.3 ± 2.4. Hence the required roots are - 1.3 ±2.4V- 1, or - 1.3 ± 2.4 i. EXERCISE 12 Solve the following equations graphically : 1. x 2 - 10a; + 25 = 0. 8. a; 2 - 5a; + 15 = 0. 2. x 2 - 6 x + 13 = 0. 9. x 2 + 3 x + 27 = 0. 3. a; 2 + 4 a; + 8 = 0. 10. a; 2 4- 9 a; + 36 = 0. 4. 0^ + 8 a; + 20 = 0. 11. x 2 + x + 1 = 0. 5. ar 9 - 8 a; + 25 = 0. 12. a; 2 + 2 a; + 1 = 0. 6. x 2 - 10a; + 29 = 0. 13. 2 2 a; + 2^ + 3 = 0. 7. ar + 7a; + 21 =0. 14. 4 a^ - 12 a; + 25 = 0. 46.* Solution of quadratic equations by means of the standard curve y = -. x As stated in § 29,"the parabola y = a; 2 is not the only curve that may be used for the graphic solution of quadratic equations by means of straight lines. A curve that gives a very convenient solution is the equilateral hyperbola y = -, which is plotted in the annexed diagram. It consists of two disconnected branches which approach the axes indefinitely. Note. To plot this curve exactly, it is^necessary to locate several points between x = and x = 1. Thus, if y'= 2, x = J ; If y = 3, x = I ; if y = 4, x = \ ; etc. (See table on page 84.) * Paragraphs marked by asterisk * may be omitted. 40 GRAPHIC ALGEBRA 47.* To solve the equation ax 2 + bx + c = 0. Let 1 1 y = -, or x = ~. x y (1) (2) Partly substituting this value for x in equation (1), Or y y ax + b + cy=0. 1 (3) (2) The solution of the system (2), (3) for x produces the re- quired roots of (1).* Ex. 1. Solve x 2 + 2 x- 8 = 0. Y' \\ O-VfJ -TS pStf= J! A ' "NZ-ii 4—= n l) ) \ > 3 •: 2 \ -1 \ \\ > V ■V Let y = Ex. 2. If Then x + 2-8y = 0. liy = 0,x=—2;iiy = l,x = 6. The straight line that joins (-2, 0) and (6, 1) intersects (2) ^ in P and P'. Measuring the ab- scissas of P and P', we obtain x = - 4 or + 2. If the line ax + b + cy = is tangent to y = - . equal. Solve 4ar*-4a + l = 0. 1 the roots are then y 4x — 4 + y x 0. (2) (3) The line (3) touches (2) at Q. Hence there are two equal roots, \ and \. * This method may be used for all equations of the form ax f(x) + bf(x) + c = 0. Let Then y — — —' ax + b + cy = 0. QUADRATIC EQUATION'S 41 48.* Complex roots can be found by a method similar to tlie one given in § 44. Students who wish to derive this method may be guided by the follow- ing suggestions : 1. Consider the same equations as in § 40. 2. These equations are repre- sented by the lines x+p + (q-d)y -0, (la) x+p + qy = 0, (2 a) x+p + (q + d)y = 0. (3 a) 3. Instead of being parallel (as in § 41) the lines (la), (2 a), and (3 a) meet in a point (B) on the x-axis whose abscissa is —p. 4. The lines (1 a), (2 a), (3 a) intercept equal parts on any line parallel to the x-axis. 5. A parallel to the y-axis through the midpoint of OB intersects y = - at C, the point of contact of (2 a), x The annexed diagram solves the equation x 2 — x + 2 =0. The line x — 1 + 2y — 0, or BA, does not intersect the curve, but the correspond- ing line BB produces the roots .5 ± 1.3. Hence the required roots are .5 ±1.3 xV^l. EXERCISE 13 Solve by means of the equilateral hyperbola the following equations : 1. a? — 2x — W — 0. 4. 35*+ 5a; + 4 = 0. 2. x 2 -x-6 = 0. 5. a 2 — 2» + 10 = 0. 3. ar-6a + 5 = 0. 6. ar + 6a; + 6 = 0. [For more examples see Exs. 9 and 12.] Note. The solution of quadratics by means of the cubic parabola y — x 3 is given in § 60. CHAPTER VI CUBIC EQUATIONS 49. Solution of incomplete cubics. To solve an incomplete cubic of the form ax 3 + bx + c = 0, the method that was used for quadratics (§ 30) may be employed.* Thus, to solve ax 2 + bx + c = 0, (1) let y = rf.) (2) Then ay + bx + c = 0. J (3) The solution of the system (2), (3) for x produces the re- quired roots. But the graph of (3) is a straight line, while the graph of (2) is a cubic parabola which is identical for all cubic equa- tions. Hence after the graph of the cubic parabola (AOP in the diagram) has been constructed, any cubic may be solved by the construction of a straight line. Ex.1. Solve 4 ^-39 x + 35 = 0. (1) Let y = x 3 . (2) Then 4 y - 39 x + 35 = 0. (3) In (3), if x = 0, then y - — 8|, and if x = 4, then y = 30 J. The line joining (0, — 8J) and (4, 30£) intersects the graph of (2) in P, P', and P". By measuring the abscissas of P, P', and P", we find x = — 3£, or + 1, or 2\. 50. In the equation ay -f bx + c, if x = 0, then y = — ; if c a y = 0, then x = — r. Hence, by taking on the a>axis the point G C —r, on the ?/-axis the point — , and applying a straight edge, the roots of the equation ax 3 + bx + c = can frequently be * This method may be used for any equation of the form af(x) + bx + c = ; e.g. ax 5 — bx + c = 0, or x — e sin x — 0, etc. 42 CUBIC EQUATIONS 43 determined by inspection. If the two points thus constructed on the axes lie very closely together, the drawing is liable to be inaccurate, and it is better to locate one or both points out- side the axes. Ex. 2. Solve z 3 + 6 x- 15 = 0. (1) Let y = % % . (2) Then y + 0x-15 = 0. (3) Hence, the distances cut off by (3) on the x- and y-axes are respec- tively 2\ and 15, and the line (3) is easily constructed. As there is only one point of intersection, Q, the equation has only one real root, viz. 1.7+. 45 Y A *>> / "/ ^ H '^o P < ^/G -4 -3 -2 -1 P n*^ /'SS 2\ 3 4 X X V II „ «y V <& p /J p -45 Y 44 GRAPHIC ALGEBRA 51. Solution for large roots. One diagram may be used for the solution of large and small roots. For in the diagram we may assign any values to the abscissas, provided the corre- sponding ordinates are the cubes of the abscissas. Thus, after the cubic parabola y = a,* 3 has been drawn, we may multiply the numbers on the x-axis by any convenient number, e.g. 3, provided we multiply the values of the ordi- nates by the cube of the number, i.e. 27. Similarly, to find small roots, mnltiply the values of the abscissas by a r~ fraction e.g. \, and the values of the corre- sponding ordi- nates by the cube of this fraction, i.e. i. Ex. 3. Solve graphically x*+ 2^-320=0.(1) Let y = x 3 . (2) Then y + 2 x - 320 = 0. (3) If x = 0, y = 320, and if x = 8, y = 304. Obviously the preceding diagram cannot contain the roots, and the position of (3) shows that there cannot be a negative root. Hence, multiply the values of the abscissas in the diagram by 2. Then the values of the ordinates must be multiplied by 8. (The resulting values are given in parenthesis.) Joining the points (0, 320) and (8, 304), we obtain the real root 6.8", ■while the other roots are imaginary. Note. The student should draw the graph of y = x 3 from x = — 3| to x = + 31 (or from - 4 to + 4) on a large scale, and use one curve for the solution of a number of equations. The table on page 84 will be found useful for the construction. CUBIC EQUATIONS 45 EXERCISE 14 Find graphically the real roots of the following equations : 1. a 8 + 4 a; — 16 = 0. 13. aj 3 -10a?-48 = 0. 2. ar* - 5 a; - 12 = 0. 14. £ 3 -9£ + 54 = 0. 3. a? - 2 x + 4 = 0. 15. £ 3 -14£ + 24 = 0. 4. 2 or 5 - 9 £ + 27 = 0. 16. tf _ 30 x - 18 = 0. 5. or 3 - 7 £ + 6 = 0. 17. or 3 + 10 £-13 = 0. 6. 4. r 3 - 39 ^-35 = 0. 18. x 3 - 45 x - 152 = 0. 7. of _ 5 a + 20 = 0. 19. £ 3 - 60 x + 180 = 0. 8. a? _ 5 a _15 = 0. 20. x? - 90 x + 340 = 0. 9. cc 3 — 5 x — 5 = 0. 21. £ 3 -75£-250 = 0. 10. a* -32 a; -80 = 0. 22. £ 3 - 100 £ + 500 = 0. 11. 2£ 3 -5£ + 20 = 0. 23. £ 3 + 120 £-560 = 0. 12. ar 5 + 8 x - 64 = 0. 24. a? -200 £ + 1200 = 0. 52. Graphic representation of a cubic function. Consider the equation Let Then or £ 3 +j5£ + g = 0. y = X s . y +px + q = 0, y = -px-q. / k i \ /3 7 L 5 K^/ &x >- >-X- > 1 O '^f^ L S/ I Sy-- z% % * 'A G { H S^A & E /^i s^\ & 7U> 1 (1) (2) (3) 46 GRAPHIC ALGEBRA In the annexed diagram, let COD represent the cubic pa- rabola y = x 3 , and BE the straight line y -{- px + q = 0, or y = —px—q. Let OA or x' be any particular value of x. Then CA = x' 3 , and BA = —px' — q. Hence CB = CA- BA = x' s +px' + q. I.e. the value of the function x 3 -{-px + q for any particular value x' is represented by that part of the corresponding ordinate which is intercepted between the straight line y+px + q = 0, and i v (if 'B • 3 tpx-hg 7 L '13 c K^ ^X >- P-XJr-x — > i L ^i_ 7l z V, X 'A G ( H ^ji j & E /^ y^ *y D | tfie cm&i'c parabola y = a 3 . The distance is measured from the straight line, and is taken positive if it extends upward, negative if it extends downward. Thus in the annexed diagram y = x 3 — Qx + %, and we have if x = -2,y = FG = 5, x = -l\,y = HI=l, x = l%,y= KL = — 2, etc. Ex. 1. Find the greatest value of the function a; 3 — 7x + 6, for a negative x. CUBIC EQUATIONS 47 Construct AB, the locus of y — 7 x+ 6 =0. Draw CD parallel to AB, touching the cubic parabola in E ; then FE, or 14, is the required value. Ex. 2. Which values of x will make the function X s — 7 x + 6 equal to 4, ie. 3^—735 + 6=4? On any ordinate, from the straight line AB, lay off 4 units upward, as JFT^. Through G draw HI parallel to AB, intersecting the cubic parabola in P, P', and P". By measuring the abscissas of P, P', and P", we find x = — 2f , or J, or 2J. Note. If we consider the distances cut off from SB by the ordinates, as abscissas, e.g. ST=1%, then the cubic parabola represents the function x 3 +px + q in so-called " oblique coordinates." 53. To construct the graph of x?+px + q in the usual manner (rectangular coordinates), make K'L'=KL, M'N' = MN, 0'B'=OR, etc. The curve L'N'B' is the required graph of X s +px + q. 48 GRAPHIC ALGEBRA 54. Tlie value of the function ax 3 + bx + c is equal to a times the part of the corresponding ordinate which is intercepted by the straight line ay + bx + c — 0, and the cubic parabola y = x 3 . The proof is similar to that of § 36. EXERCISE 15 Find graphically : 1. The value of a 3 + 4 a; — 16, if x equals —3, —2.5, -2.1, 3.5. 2. The value of x 3 + 4 x — 8, if x equals — 1.6, — 1.5, 2, 1.5. 3. The value of x s - 6 x - 15, if x = - 3, - 2, 1.5, 3.5. 4. The value of x 3 - 5 x + 18, if a? = - 8, - 5, +3, +1. 5. The value of x, if x 3 — 5 x — 12 = 5. 6. The value of a-, if x 3 - 5 a; - 12 = - 10. 7. The value of a?, if x 3 - 5 a; -12 = -40. 8. The value of x, if a,- 3 - 5 x — 12 = 10. 9. The smallest value of x 3 — 5 a; — 12 for a positive a?. 10. The greatest value of ar 3 — 5 x + 10 for a negative a\ 11. Construct the graph of ar 3 — 12 x — 30 = 0. 12. Construct the graph of x 3 — 8 = 0. Find : 13. The value of 2 x 3 + 9 a; + 20 = 0, if x equals 3, 2.5, - 1.5. 14. The value of 3 x 3 + 9 x — 25 = 0, if x equals —3,-5,-2. 15. The smallest value of 3 x 3 — 9 a* —25, for a positive a,\ 55. The preceding paragraphs may be used to locate the line ay -\-bx-{-c = by determining two values of the function ax 3 -f bx + c. In applying this method it is advisable to reduce the coefficient of a; 3 to unity by dividing by a. E.g., let 2ar i -12a,- + 3 = 0. Dividing by 2, a; 3 — 6 x + f = 0. If a; = 3, a 3 -6a; + ! = 10 J. If ic = -3, flJ»_6a; + 4 = — 74. CUBIC EQUATIONS 49 Through the point A (whose abscissa = 3) draw an ordinate meeting the cubic parabola in B, and on BA lay off downward jBC=101 Similarly, through the point E (whose abscissa = —3) draw a perpendicular EF upward equal to 1\\ join FC, which is the required line. 56. Equal roots. If the line ay + bx + c = is tan- gent to the cubic parabola, two points of intersection coincide, and two roots of the equation ax 3 + bx + c = are equal. The straight line must intersect the parabola at least once ; hence every cubic equation has at least one real root. 57. It can be proved that the sum of the roots of an in- complete equation of the form ax s + bx + c = is equal to zero. Hence if one root is m, and the other two are equal, then these equal roots are each — ^; i.e. if the abscissa of P=m, the abscissa of C, the point of contact, equals — — . Since it is difficult to locate graphically a point of contact with accuracy, it is advisable to deter- mine equal roots by the preceding relation. 58. Complex roots of incomplete cubics; If the line ay + &.r + c = meets the cubic parabola in only one point, then two roots are complex. To find complex roots of the form n ± V^i, we employ the same method as for 50 GRAPHIC ALGEBRA quadratic equations ; viz. we determine the line that produces the roots n ± V £. If the equation ax 3 + bx + c = has one root equal to m, the left member is divisible by x — m, and the equation may be represented in the form a(x — m)(x 2 +px-{-q) = 0. Supposing that x 2 + px + q = has equal roots, and that d is a positive quantity, we consider the equations : a(x — m)(x 2 +px+q— d) = 0, (1) a( x — m)(x 2 -{- px + q) =0, (2) a(x — m^x 2 + px + q + d) =0. . (3) In the same manner as in § 40 it follows that the roots of the equations (1), (2), and (3) are respectively m, - 1 ± Vd ; P V m, —-, - 1 - ; ' 2' 2' m, V ± V^d. Hence the roots of (3) can be found by solving (1). But the three straight lines (l a ), (2°), and (3 a ) which serve to solve (1), (2), and (3) re- spectively are connected by the following geometric rela- tions : 1. TJie three lines (1°), (2°), and (3°) meet in a point (m, m 3 ), i.e. P. For m is a root of the three equa- tions (1), (2), and (3). 2. Tlie three lines intercept equal parts on an ordinate drawn through the point of contact C, or DC— CE. CUBIC EQUATIONS 51 2 For according to § 52, CD is equal to the value of (3) if x = mrt Similarly, it follows from (1) that CE = -^; i.e. CD and OE are equal and lie on opposite sides of C. 3. 77ie line (2 a ) is tangent to the cubic parabola at C. This follows from § 56. 4. 77*e abscissas of D, C, and E are equal to — —, hence EF = |FP(§57). 59. Construction of complex roots. Let ax 3 + bx + c = have two com- plex roots. Substitute y = x 3 . Then ay + bx + c = 0. (3) Construct PF, the locus of (3), and let it meet the parabola in one point, P, and the ?/-axis in F. Produce PF by one half its length to E, and through E draw an ordinate, meeting the cubic parabola in C. Produce EC by its own length to D and draw PD, intersecting the curve in Q and R. Then the abscissa of D is the real part, and the difference of the abscissas of Q and D is the imaginary part of the required roots ; i.e. x=OG±GSV^l. Ex. 1. Solve a? + x - 10 = 0. (1) Let y = x 3 . (2) Then y+x- 10 = 0. (3) Construct the locus of (3), i.e. FF, which intersects the cubic parabola in one point, viz. P. 52 GRAPHIC ALGEBRA Hence the equation has one real root, which equals 2, and two imaginary roots. Produce PF by one half its length to E. Through E draw an ordinate which meets the curve in C. Produce EC by its own length to Z>, and draw PD, intersecting the cubic parabola in Q and R. The abscissa of Z>= — 1, the difference of the ab- scissas of Q and D = 2. Hence the complex roots are — 1 ± 2 V— 1. 1 F Y 10 1 X I 5 % X 3 - 9 c . !! < -o / } -10 ^lS ^JO -1 I R • -25 T EXERCISE 16 Find the real and complex roots of the following equations 1. x z ~ 3 x -2 = 0. 2. a*-3o; + 2 = 0. 3. tf + x _|_ io _ 0. 4. 4x- 3 -lla-10 = 0. 5. 4^-3^-26 = 0. 6. x" + 9 a; + 26 = 0. 7. x 3 -9x +28 = 0. 8. a? — 9x+ 280 = 0. 9. 8ar 3 -12a + 9 = 0. 10. 4^-9^-14 = 0. 11. a!* + 4aj — 5 = 0. 12. a: 3 + 2a,-+6 = 0. 60.* Solution of quadratics by means of cubic parabolas. To solve the quadratic x 2 +px + q=0 (1) by means of a cubic parabola, mul- tiply by x— p, i.e. introduce the new root p, x 3 + (q — p-) x — pq — 0. (2) Or if y = x*, (3) y+(q-p 2 )x-pq=0. (4) The line (4) passes through (p, p s ) and (0, pq). Thus, to solve a; 2 + 4 a; + 3 = 0, Y' P *, uu 1U X =4, and on OT layoff OB =pq = 12. The line PB determines the roots (P' and P"). x = — 1 or — 3. Note. For examples see Exercise 9. 61. Complete cubic equations. To determine a method for the graphic solution of complete cubic equations, consider first a concrete example. To solve or 5 + 9 a? + 20 x + 12 = 0. (1) Substitute x = z — (i X second coefficient). Or x = z - 3. (2) Then 0-3) 3 + 9 (2 -3) 2 + 20 (2- 3) +12 = 0. (3) If (3) were simplified, it would not contain the second power of z, for the first term produces — 9 z 2 , the second term + 9 z 2 , and the other terms do not contain z 2 . Hence equation (3) can be solved by one of the methods for incomplete cubic equations, but the one given in § 55 is the more convenient, since it does not require the simplification of the equation. If z = 3, z - 3 = 0, and ( 2 _3) 3 + 9(z-3) 2 + 20(2-3) + 12 = 12. If 2 = -l,z-3==-4, and (z_3) 3 + 9(2-3) 2 + 20(z-3) +12 = 12. Consider z as abscissa, y as ordinate, and construct the cubic parabola y = z s . Through (3, 0) and (-1, 0) draw ordinates and let them meet the curve in A and C. On the ordinates lay off downward AB = 12, and CD = 12, and draw BD. By measuring the abscissas of the points of intersection, we obtain the roots : z= 2, 1, Hence x = — 1, —2, and and 3. 6. 54 GRAPHIC ALGEBRA 62. In the preceding diagram z represents the abscissas, but by changing the location of the ?/-axis we can obtain abscissas which equal x. On OX lay off 00' = 3. Consider 0' as the new origin, and the ordinate Y Y ', drawn through 0', as the new ?/-axis. Then the abscissa of any point is smaller by 3 than the old ab- scissa z, or the new abscissa is 2 — 3, i.e. x. By thus introduc- ing a new axis, the entire work of the preceding paragraph can be done without introducing z at all. Thus, instead of saying : If z - 3 = - 4, (z - 3) 3 + 9 (2 - 3) 2 + 20 (z- 3) + 12 = 12, we have briefly : If *=-4, ar 3 + 9a; 2 +20a + 12 = 12. Similarly, instead of measuring the z, we may directly meas- ure the x, and thus obtain the roots of (1). 63. A change in the position of the axes is called a trans- formation of coordinates. To solve x 3 -+- bx 2 + ex + cZ = 0, (4) we locate 0', the new origin, at the point ( ^, V and consider the ordinate through 0' as the new y-axis. If z is the old abscissa, then the new abscissa x = z— ^, and this value sub- stituted in (4) produces an equation without z 2 . Similarly, to solve aa? + bx 2 + ex + d = 0, _6 3a* make 00' 64. The method for solving complete cubics, which was derived in the preceding paragraphs, may be summarized as follows : CUBIC EQUATIONS 55 To solve the complete cubic, ace 3 + bx 2 + ex + d = 0, divide by a : or + -or -j — x+ - = 0. a a a Construct the standard cubic parabola, and a/Yer ££ *'s cow- structed change the origin to the point ( — , ]. \3a ) Locate two points by the method of § 55. The line which joins these points intersects the cubic parabola in one or more points whose abscissas are the required roots. Note. In finding real roots, all work except the construction of the cubic parabola refers to the new y-axis, and the old axis may be omitted. Ex.1. Solve 2 x 3 - 15 ar + 31 x - 12 = 0. Dividing by 2, and denotiug the left member by y, we have 2x 3 - 15x 2 + 31x-12 y = 2 = 0. After drawing the standard cubic parabola {i.e. y = z 3 ), lay off on the x-axis 00' = | (— -V 5 -), i- e - — 2\, and consider 0' as the new origin. If a5 = 0, y = -6. If x = 2,y= 3. Let the new y-axis (i.e. Yo To') meet the cubic parab- ola in A, and the ordinate through (2, 0) meet the curve in C. On AO' lay off upward AB = 6, and on tffe ordinate through C lay off downward CD = 3. Draw BD and measure the abscissas of the points of inter- section P, P', and P". Thus we obtain : x = \, 3, and 4. 65. Complex roots of complete cubics are determined by applying §§59 and 64. In using § 59 we find the ordinate 56 GBAPHIC ALGEBRA through the point of contact by producing the line PF from P to the ?/-axis by one half its own length. The student should bear in mind that this refers to the old y-axis, or that F lies in TY'. Ex.2. Solve 4 « 3 + 18 a 2 + 24 a -17 = 0. 4x 3 + 18x 2 + 24x-17 Dividing by 4, y = :0. 2/=-8f. Construct the cubic parabola, lay off on the x-axis 00' = % of *£-, i.e. 1J, and consider 0' the new origin. If x = 0, If K = -3,. Locate the points .4 and ^4' in the usual manner (AB = — 4J, A'B'=-8%), and draw ^1^1', which meets the cubic parabola in P and the old y-axis in F. Produce PF by one half its own length to E, and let the ordinate through E meet the curve in C. Produce EC by its own length to Z>, and draw PD meeting the cubic parabola in P' and P". The abscissa of D is — f , and the difference of the abscissas of P' and D is Hence the required roots are _ | + 3 v^i, _ | -| V^T, and EXERCISE 17 Find graphically the real roots of the following equations : * 1. x 3 -3x 2 -x + S = 0. 3. z 3 -6x 2 + 3a; + 10 = 0. 2. x 3 -9 z 2 + 23x- 15 = 0. XT 8 a 2 4- 17 a -10 = 0. *For most of the following examples, a graph of the cubic parabola from x = — 3 J to x = 3\ is sufficient. In other cases, apply the method of § 51. CUBIC EQUATIONS 57 5. ar* +7 a; 2 + 14 a + 8 = 0. 12 . 5ar 5 -3a; 2 -20a:+12=0. 6 . a 3 -2a: 2 -5a; + 6 = 0. 13. 2ar 5 -4a; 2 - 10x + 9=0. 7. x 3 -2x 2 -4 : x + 2 = 0. 14. 2ar 3 -5ar 2 -4a; + 3 = 0. 8 . a *-3x i + x+7 = 0. 15. 4x 3 -12» 2 -19x+12=0. 9. ar 5 + 4ar J -2x-5 = 0. 16. 4ar 5 -12a; 2 -31a;+18=0. 10. a? + x 2 + x + 5 = 0. 17. ^+6 a; 2 -24 x + 60=0. 11. 2a; 3 + 8a; 2 + 2a;-3 = 0. Find the real and complex roots of the following equations : 18. a?-3x i + x + 5=0. 22. a; 3 -6 x 2 + 11 a; -12 = 0. 19. a^ + 6a; 2 + 10a: + 8 = 0. 23. x 3 + x 2 - 2x + 12 = 0. 20. a 3 -3ar 7 + 2a; + 6 = 0. 24. ar 3 + ar-7z + 15 = 0. 21. a; 3 + 6a; 2 + 13a: + 20 = 0. 25. x 3 -9ar + 28x-20 = 0. 66. Values of a complete cubic function. The method for finding the values of a function for various values of x, as given in § 52, is true also for the complete cubic equation. Thus, in the example of § 65 : If x = -3, 4a; 3 + 18a; 2 + 24a-17 = 4(yl'B') = -35. If x = 1, 4 ar 5 + 18 a; 2 + 24a; - 17 = 4 (IK) = 29, etc. Note. In order to make the new y-axis coincide with one of the lines of the cross-section paper, it is sometimes advisable to take the unit of the abscissas equal to the length of three squares of the paper. 67. Construction of the graph of a complete cubic function. f Ex. 3. Construct the graph of y = x s + 4:X 2 — x — 4. 00' = |-4 = f. Take the unit of abscissas equal to the length of three squares (Note, §66). 58 GRAPHIC ALGEBRA Construct the cubic parabola, and place the new origin at the point OK If x = 0, y = - 4. If x=-2,y = 6. Locate the points A and B in the usual manner, and draw AB. Draw a new x-axis X X ', and make CD' = CD, E'F> = EF, G'W = GH, etc. By joining the points Z>', F', N', etc., in succession the required graph IF'KL is obtained. 68. If the coefficient of x 3 is a, the student should keep in rnind that in applying the above method every ordinate has to be multiplied by a. EXERCISE 18 1. If y = x 3 — 4 ar + 2 x + 5, determine graphically the value of y if (a) x = i (6) x = If, (c) x = 2. Construct, by means of the standard curve, the graphs of the following functions in rectangular coordinates : 2. y = x i + 2a?-5x-l. 6. y = x 3 + 6a; 2 - x - 30. 3. y = x 3 + 5 x 2 — 3. 7. y = x 3 + x 2 — x + 15. 4. ?/ = aj*+ 4 x 2 + x + 2. 8. ?/ = ar 3 — 3 x* 2 + 7 a; + 5. 5. y = 4 ar 3 - 12 x 2 - 19 x + 12. 9. ?/ = x* + 6 x 2 + 2 a- - 9. Note. The solution of a cubic equation by means of a parabola or a rectangular hyperbola is given on §§ 75 and 84. CHAPTER VII BIQUADRATIC EQUATIONS 69. Solution of biquadratics in which the second term is want- ing. To solve an incomplete biquadratic of the form x* + bx 2 + ex + d = 0, (1) write this equation as follows : x A + (6 - 1) x 2 + x 2 + ex + d = 0. Let y = ^\ (2) Then f + (6 - ±)y + sc 2 + ex + cZ = 0. f (3) The solution of the system (2), (3) for x produces the re- quired roots. But the graph of (2) is a parabola which is identical for all biquadratic equa- tions, while the graph of (3) is a circle (§ 27). Ex. 1. Solve x* - 15 x 2 - 10 x + 24 = 0. (1) Separate — 15 x 2 into two parts, one of which is x 2 : x* - 16 x 2 + x 2 - lOx + 24 = 0. Let 2/=x 2 . W2) Then ?/ 2 -16y+x 2 -10x+24=0. J (3) To construct (3) transpose 24 and com- plete the squares, y 2 - 16 y + 64 + x 2 - 10 x + 25 = - 24 +64 + 25. Or ( 2/ _8) 2 + (x-5) 2 =(V65) 2 . (3) I.e. (3) is a circle whose center C is the point (5, 8) and whose radius equals V65.* 1 - ? P ' 1 iO' \ it \ \ (p It) J \ o c / i / / 1 P' t V (P XK r f i i ri < LA 2 3 4 5 £ALj 1 1 1 L J * V65 = V8 2 + l 2 , hence the line joining C and (4, 0) is the radius. In other cases, use table of square roots, Appendix III. 59 60 GRAPHIC ALGEBRA Equation (2) is the standard parabola, which is intersected by the circle in four points, P, P, P", and P". The abscissas of P, P, P", and P'" are the required roots. .-. x = — 3, — 2, 1, or 4. Note. The student should remember that in problems involving cir- cles, the same scale unit must be used for abscissas and ordinates. 70. Formulae for radius and origin. According to the preced- ing paragraph, the equation a 4 + bx 2 + ex + d = (1) is solved by the system y = «?,\ (2) v? + cx + f+(b-l)y + d = 0. J (3) Transposing and completing the squares in (3), 2 ^ + «+(5Y+^+(&-i) y +^Y«(|Y+feiY-d Or r+ i' 2 &-1V + 2 -d. IT > iT" P < JT ' 1 r 14^ TO" i* \ [p 1U" ' / c / < / o / 1 /' N ,Q V' c //' ■ z $ -■ji -rrr^i < LA 2 3 4 5 If we denote the coordinates of the center of the circle by x and y , and the radius by r, we have c 9> (4) ?/o = : Hp> (-5) r 2 = a- 2 4-7 /o 2 -d. (6) Ex. 2. Solve by means of the formulae : aj» 3aj 2 + 4a? + 3 = 0. a; — — 2, r 2 = 5. /.e. the center C" of the circle is ( — 2, 2), and its radius is V5, or the line that joins C and (— 1, 0). BIQUADRATIC EQUATIONS 61 There are only two points of intersection, and hence two roots are real, and two complex. The real roots are - .6 and— 2.1. 71.* The expression Vx 2 + y 2 — d can be con- structed geometrically. If C is the center of the circle, lay off on the x-axis OD = 0(7, and draw the ordinate DE, which equals x 2 + y 2 . On ED lay off EF = d and draw FH || XO. The segment HG in- tercepted on this parallel by the y-axis and the "* parabola, equals r. EXERCISE 19 Find the real roots of the following equations : * 1. z 4 + 5a 2 + 4;c-28 = 0. 2. a; 4 -15ar + 10a; + 24 = 0. 3. x A — x 2 + 4 x — 4 = 0. 4. 3^-19^ + 2 x + 56 = 0. 5. x 4 -5r + 4=0. 6. a 4 -7 a; 2 -12 a + 18=0. 7. a; 4 — 4a? + 12ar + 9=0. 8. a 4 -7ar-6a; + 12 = 0. 9. z 4 - 9 « 2 -2^ + 6 = 0. 10. a; 4 + 4 x 2 — 5 x — 55 = 0. 11. x- 4 - 6^ + 30; + 2 = 0. 12. x 4 - 15a 2 - 10x + 24 = 0. 72. Solution for large roots. To use the same diagram of the standard curve for the finding of large and small roots of the equation x * + bx 2 + ex + d = 0, (1) multiply the values of the abscissas and ordinates in the dia- gram by any number, as p. Then the equation of the parabola becomes PV = *?• ( 2 ) , Equation (1) may be written in the form x* + (b -p 2 ) x 2 +p 2 x 2 + ex + d = 0. Partly substituting py for x 2 , p 2 y 2 + (b — p^py +p 2 x 2 + ex + d = 0. * For the following exercises a graph from x — — 4 to x = + 4 is sufficient. 62 GRAPHIC ALGEBRA The last equation is easily transformed into the following one (§ 27) : x + + [y + 2p 0\ 9 - jr y _ 2^ 2 + 5 — p' 2p 2\ 2 -|. (3) Xn Equation (3) represents a circle whose center and radius are determined by the formulae (4) (5) (6) 2p 2 ' _p 2 — 6 2p r 2 = x 2 + ?/ 2 p 2 The abscissas of the points of in- tersection of the circle (3) and the parabola (2) are the required roots. Ex. Solve x i _ 37 x 2 - 24 x 4- 180 = 0. Since obviously x and y are very large, multiply the values on the two axes by 2 ; i.e. make p = 2. (The new values are given in parentheses.) Applying formulae (4), (5), and (6), we have : x =3, 2/o = 10J, r = 8.3+.* Construct the circle and measure the abscissas of the points of inter- section. Hence x = -5, -3, 2, 6. EXERCISE 20 Solve graphically: 1. x 4 - 45 jc 2 - 40 a; + 84 = 0. 3. x A - 23 a 2 - 18 x + 40 = 0. 2. ic* - 42 x 2 - 64^ + 105 = 0. 4. z 4 -37ar-24x + 180 = 0. * To compute »•, use table of squares and square roots in Appendix III. BIQUADRATIC EQUATIONS 63 5. a**- 75 x 2 - 70 x + 144 = 0. 8. z 4 -58a 2 + 441 =0. 6. a 4 - 63 cc 2 + 50 x + 336=0. 9. a?* -49 a? + 36 a? + 252 = 0. 7. z 4 -55a; 2 -30£ + 504 = 0. 10. a,- 4 -49ar -36x- + 252 = 0. 73. Complex roots. If an equation has two real and two complex roots, the roots may be found by a method similar to the one employed for quadratics and cubics (§§41 and 59). Let the equation x 2 +px-\-q = have equal roots, and d be a positive quantity. Consider the following three equations which are supposed not to contain x 3 when simplified : (x-a)(x-b)(x 2 +px + q + d) = 0, (1) (x-a)(x-b)(x 2 +px + q)=0, (2) (a: -a){x- b) (x 2 +px + q-d)=Q. (3) Then the roots are (§ 58) respectively : P a,b,- T} ±V-d a, b, a P P. V ,b, —r>± Vd. I.e. the roots of equation (1) are complex, but they may be found by solving (3) instead. The circles C, C, and C", which represent respectively equa- tions (1), (2), and (3), are connected by simple geometric relations, which make it possible to construct the third circle (C") when the first one (O) is given. 1. The three circles C, O, and C", pass through two points, P and P', in the parabola, the abscissas of P and P' being a and b. Obviously a and b are roots of the equations (1), (2), and (3). 64 GRAPHIC ALGEBRA 2. TJie three centers C, C, and C", lie in the perpendicular bisector of PP. 3. The second center, C, bisects the line joining the other two, Cand C",orCC'=C'C". If the ordinate of C" is m, then the ordinate of C is m — - , for the coefficient of x 2 in (1) is greater by d than the coefficient of x 2 in (2) (§ 70). Similarly, the ordinate of C n equals m + -. 74. Hence if circle C is given and it intersects the parabola in P and P, construct AB, the perpendicular bisector of PP, and in AB determine C, the center of the circle that passes through P and P, and touches the parabola in another point, E. Produce CC by its own length to C", and from C", with a radius equal to C"P, draw a circle. This circle intersects the parabola in two other points, Q and Q'. If the abscissas of Q and Q' are m + n and m — n, the required roots are respectively m + n V — 1 and m — n V — 1. The abscissa of E is always equal to m, and the difference of the abscissas of Q and E (or E and Q') is equal to n. A convenient method for constructing the circle C, which touches the parabola and passes through P and P, is the fol- lowing : Let the perpendicular bisector of PP' meet PP in A, and the ?/-axis in B. Produce AB by its own length to D, and let the ordinate through D meet the parabola in E. Then E is the point of contact, and the perpendicular bisector of DE meets CD in the required point C". Ex. Solve the equation x* — x 2 — 4 x — 4 0. BIQUADRATIC EQUATIONS 65 According to § 70, x = 2,y = l,r = 3. The circle drawn from (2, 1), or C, as center with a radius equal to 3 intersects the parabola in only two points, P and P'. Hence there are only two real roots, viz. — 1 and 2. Draw AB, the perpendicular bisector of PP', and let it meet the y-axis in B. Produce AB by its own length to D, and draw the ordinate DE, E being a point in the parabola. The perpen- dicular bisector of DE meets AB in C, the center of the second circle. Produce CC by its own length to C" and from C" as a center draw a circle through P and P'. This circle, C", meets the parabola in two other points, Q and Q'. The abscissa of E, i.e. — |, is the real part, and the difference of the abscissas of E and Q (or E and Q'), i.e. 1.3, is the imaginary part of the required roots. Hence the roots are : \ /\ A /p m D a_„// \" ry (/ / n P ^ X * -e *V i * 'i ■J±1.3aA 1,2, and — 1. EXERCISE 21 Find the real and complex roots of the following equations : 6. x 4 + 3 x 2 + 6 x - 10 =0. 7. «*- 11a 2 - 14a? + 24=0. 1. ^-8^ + 8^ + 15 = 0. 2. a 4 - 7 a 2 + 12.x + 18 = 0. 3. x 4 -4x 2 -12x-9 = 0. 4. x 4 -5ar-10x-6 = 0. 5. x 4 - 5 x 2 - 4 a? + 12 = 0. 8. x*- 10.x 2 + 20 a -16 = 0. 9. x 4 -5x-' + 10.x -6 = 0. 10. x 4 -8x 2 + 8x + 15=0. 75.* Solution of cubic equations by means of the standard parabola. To solve the cubic x s + bx* + ex + d = (1) by means of a parabola, multiply by (x — b), i.e. introduce the new root b. x* + (c - 6 2 )x 2 + (d - bc)x -bd = 0. 66 GRAPHIC ALGEBRA Hence, applying § 70, we have d \P / , JO 10 lo 1- 11 10- f-1 M \ / \ / \ / \ U // \ 5 1 V \ 4 4 V \ ./ / A -i , / X' V V X ^-5-4-^-2-1^123' o 2 - c + 1 (2) (3) The formula for the radius is not neces- sary, since the circumference must pass through the point of the parabola whose abscissa is b, i.e. (b, ft 2 ). Thus, to solve a? + 3 x 2 - 6 x - 8 = 0, (4) either multiply by x — 3, obtaining x i — 15 x 2 + 10 x + 24 = 0, or apply directly formulae (2) and (3). Hence x = — 5, 2/o = 8. From (—5, 8) as a center construct a circle passing through A, i.e. the point in the parabola whose abscissa is 3. Hence the roots of (4) are — 4, — 1, 2. [For examples see Exercise 16.] 76. Power of a point with respect to a circle. If r is the radius of a circle C, and d the distance of a point P from its center, d 2 — r 2 is called the power of the point P icith respect to circle G. If P lies without the circle the power is j^ositive and equal to the square of the tangent drawn from P to the circle.* If the point lies within the circle, as P', the power is negative. If a chord DE is drawn perpendicular to CP\ the power of P' is equal to - (P'D) 2 . 77. Values of a biquadratic function. To solve x A + bx> + cx + d = 0, (1) we substitute y = x 2 , (2) * Schultze and Sevenoak's Geometry, § 311. BIQUADRATIC EQUATIONS 67 and obtain the equation of a circle (§ 70), (x-x ) 2 + (y-y ) 2 -r 2 = 0. If any point P, whose coordi- nates are x' and y\ is joined to (x , y ) i.e. C, we have (Geometry, § 310) 2/o) 2 - (3) pc- = (x^-x Q y + - x o y + (y' - y ) 2 - r 2 . I.e. if we substitute the coor- dinates of any point P in the left member of (3), this member becomes equal to the power of P with reference to circle (3). If the point P is located in the parabola, then y' — x' 2 and the left member of (3) becomes equal to the left member of (1). Hence, the value of the function (1) for any particular value x' is equal to the poiver of point (x', x' 2 ) ivith respect to circle (3). 78. Thus, to find the various values of the function y = x* — 11 x 2 — 4 x + 6, construct a circle so that 3^ = 2, 2fo = 6,r = V34. (§70) To find y if x = — 3|, locate in the parabola a point R, whose abscissa is — 31, and draw the tangent EH to circle C. The required value equals (RHy = (6-) 2 = 36-. Similarly, if x = — 1-i, locate in the parabola a point S whose abscissa equals — 1\, and draw ST1. CS, then y = _ (ST) 2 . To find (ST) 2 graphi- cally, make OA = ST, then the ordi- nate AB = (ST) 2 , or the function equals — AB = — 7.7. 68 GRAPHIC ALGEBRA Many other problems relating to the value of the functions may be solved by such a diagram. Thus, to find which value of x be- tween x = — 4 and x = produces the smallest value of y, determine in the parabola the point nearest to C by drawing an arc EFD from C, touch- ing the parabola in F. The abscissa of F, i.e. — 2.2, is the required value. Similarly, we can determine the greatest value of the function, the value of x, if the function is given, etc. 79. The graph of a biquadratic func- tion in rectangular coordinates can be constructed by means of § 77. Since y = PC 2 —^, construct first the curve whose ordinates are PC 2 ; i.e. make AP' = CP 2 , EB' = CB 2 , etc. (Use table in Appendix III, and draw new ordinates on smaller scale.) Locate in this manner a sufficient number of points, P', B', F, D, etc., and draw the curve P'B'FD. Make O'O" = r 2 and through 0" draw XX' ±0'0". Then the curve P'B'CD re- ferred to " as origin is the required graph. EXERCISE 22 If y = x*- 3 x 2 + 4 x + 3, find graphi- cally the value of y, 1. If a: = 2. 3. Ifz = 4. 2. Ifa = 3. 4. Ifa; = 3.2. 5. Ifx = 2.7. A E O' 6. Construct the graph of y in rectangular coordinates. BIQ UA BRA TIC EQ UA TIONS 69 80. Complete biquadratic equations. The complete biquad- ratic equation x* + ax* -f bx 2 + cx + d = (1) is transformed into another equation without the cubic term by the substitution x — z a 4" a The resulting equation can be solved, and by subtracting ~ from the answers the roots of (1) are obtained. 4 w Ex. Solve aj* + 4sc 3 -5£c 2 -22a;-8=:0. (1) Substituting x=z — \=z — 1, (2) (3_l)4 + 4(2-l) 3 -6(s-l) 2 -22(S-l) -8 = 0. Simplifying, 2 4_ii 2 2_ 4 .~ + g = o.* •'• 2o = 2,y = 6 : r = V§4. Drawing tbe circle and measuring tbe abscissas of the points of intersection, we obtain s=-3, -1, .6,3.4. Hence, from (2), x=-4, -2, -.4, 2.4. 81. In most cases, the method of the preceding exercise is the best. It is possible, however, to derive general formulae. 4A— / 1 / i f? P' A" S? A' •'• 1 ( W i i i If x A + aa 3 + bx 2 + ex + d = 0, let # = z+p, where p = — a <1) (2) * Students who are familiar with the general method for removing the second term should of course use this method. See Schultze's Advanced Algebra, § 566. 70 GRAPHIC ALGEBRA Then equation (1) becomes y A + (6 p 2 + 3 ap + b)z 2 + (4p 3 + 3 ap 2 + 2bp + c)z +p 4 + op 3 + &p 2 + op + d = 0. (3) a Considering that p= — j, we can easily obtain the following values : a —2 ap 2 — 2 ftp — c *° 2 ' 2/o : 2-3ap-2 6 r 2 = # ? + 2/o 2 - (j? 4 + ap 3 + ftp 2 + <%> + d)* Constructing the circle (z , y , r) and measuring the abscissas of the points of intersection, produces the roots of (3), and hence those of (1). Thus, in the preceding equation, a!* + 4^-5^ -22 x — 8 = 0, we obtain p = -l. -8-10 + 22 i k 1 Q t t / H I / Id" T ' i« 1 / 11 n //< W \ / n \ / V I \ / \\ / - \ ■) I \ V 1 \ !'■ > \ X' V St X - -3 -2 -1 ( UMM =2. 2/o : 2 + 12 + 10 = 6. r 2 =4+36-(l-4-5+22-8)=34. Ex. Solve 4 x* + 16 x 3 - 31 x 2 - 139 a? -60 = 0. Dividing by 4, x 4 + 4 x 3 - Y a; 2 - ^ x -15 = 0. * Students familiar with Calculus can, by means of Taylor's series, obtain - f'(P) „ - 2-/"(P) — o > ^o — ,. r 2 = *o 2 +yo 2 -/(P)- BIQUADRATIC EQUATIONS 71 Hence p = — 1, z = 5 f> 2/o = 7f, r = 8.8. The construction of the circle produces the values z = — 3, — 1£, ^, 4. Hence a; = z +p = s — 1. Or x = - 4, - 2£, - i, 3. EXERCISE 23 Solve the equations : 1. x 4 + 4ar 5 -9ar 2 -16x + 20 = 0. 2. z 4 -4ar 5 -17aj 2 + 24a: + 36 = 0. 3. z 4 - 8 af 3 + z 2 + 78 .r- 72 = 0. 4. x i + 8x* + Ux i -8x-15 = 0. 5. a; 4 + 4a; 3 -4ar 2 -16a; = 0. 6. o; 4 -2a; 3 -16ar 9 + 2a; + 15 = 0. 7. a 4 + 4x 3 -21ar-64a + 80 = 0. 8. x 4 + Sx 3 -3x 2 -62x + 56 = 0. 9. a; 4 -10 a 3 +35 a,- -50 a; + 24 = 0. 10. a; 4 + 3 a 3 - 8 x 2 - 12 x + 16 = 0. 11. a: 4 -4ar s -5ar + 22a,--8 = 0. 82.* Solution of biquadratics by means of the hyperbola y = -. x Let us first consider the equation x 4 + ax 3 + bx 2 + ex +1 = 0. (1) Partly replacing x by-> y x_.ax.J>.c.j _ o, y-2 y2 yl y Or x 2 + ax + b + q/ + y 2 = 0. Applying § 27, 7.e. the required roots are determined by the points of intersection of the standard curve y = - (2) and the circle (3). 72 GRAPHIC ALGEBRA The circle is determined by the formula x o — ~ n ' ^o — _^.r 2 r 2 = x 2 + 2/ 2 - 6. 83.* The equation x 4 + ax 3 + 6x 2 + ex + d = may be solved by the circle : a •2 dP 2/o= --^,^ 2 = V + 2/o 2 --i 2d* d 2 i The abscissas of the points of intersection multiplied by d ¥ are the required roots. 84.* Solution of a cubic by the hyperbola y = - To solve x s + 6x 2 + ex + d = 0, (1) + 1 = 0. multiply by x + - , i.e. introduce the new root d d Hence, by § 82, we have to construct the circle that is determined by the f ormulge : Vo The formula for the radius is not necessary, since the circle must pass through the point f — , — d ] • Ex. Solve a; 3 - ar- 4 a + 4 = 0. ^0 = — 2 ( — " 1 + ?) = 8» 2/ =-K 4-1) = -i. From (x , y ) as center draw a circle through (— J, —4), r.e. A. By meas- uring the abscissas of the other points of intersection, we obtain x = - 2, 1, 2. Note. The preceding construction can be used advantageously for large roots, since the ordinates do not become as large as in the case of the cubic parabola. APPENDIX I. GRAPHIC SOLUTION OF PROBLEMS 85. Problems are usually solved in algebra by expressing the conditions of the problems in the form of equations. By using the graphic method, however, many problems can be solved directly, without obtaining equations. The fact that the graph of two proportional variables is a straight line is often useful. Thus, if x and y are the coordi- nates of a point, the following variables are represented by straight lines : x — time, y = distance covered by body moving uniformly ; x = time, y = work done by a person ; x = volume, y = weight of a body ; x = time, y = quantity of water flowing through a pipe at a uniform rate, etc. 86. Uniform motion. To represent graphically the motion of a person traveling three miles per hour, it is only necessary to locate one point, e.g. (1, 3) or A, and to connect this point to the origin. The increase of the or- dinate per hour equals the rate of travel, i.e. 3 miles per hour. Similarly, CD repre- sents the motion of another person who started two hours later and traveled 1\ miles per hour. EFOH represents graphically that a third person had a start of 4 miles, traveled for 2 hours at the rate of 4 miles per 73 F (0 HI £ < F T \ IU / •J ? o/ n — & I 4 A H c HC URS X < ) i M^ L 5 < r 6" l 74 APPENDIX hour, then rested 2 hours, and finally returned to the starting point at the rate of 2 miles per hour. IK represents graphically the motion of a fourth person who started 3 hours after the first and traveled in the opposite direction at the rate of 1 mile per hour. Ex. 1. A and B start walking from two towns 15 miles apart, and walk toward each other. A walks at the rate of 3 miles per hour, but rests 1 hour on the way; B travels at the rate of 4 miles per hour and rests 3 hours. In how many hours do they meet ? Construct the graphs OA'A"A'" and BB'B"B'". The abscissa of C, the point of intersection, is the required time. Hence A and B meet in 4^+ hours. Ex. 2. A stone is dropped into a well, and the sound of its impact upon the water is heard at the top of the well 5 seconds later. If the velocity of sound is assumed as 360 meters per second, and g = 10 meters, how deep is the well ? (A body falls in t seconds f- 1 2 meters.) B \ \ A/ 12 \ B' B"J >B" 4 V 1 /* \ 7 1 . 2 3 4 hou 1 1 RS Construct the graph ODA of the falling body, making the distances negative, to indicate the downward mo- tion. Since the motion of the sound is an upward motion, its graph CB is obtained by joining (4,-360) and (5,0). The ordinate of the point of intersection D is the required number. Hence depth of well = 110 meters. 87. Problems relating to work done, and to quantity of water flowing through a pipe, are quite similar to those of the preceding paragraph. APPENDIX 75 -;2 — n * s /) • \ i. A B o z 1- o 1- ^ c DAYS 2 k 4 & 6 " Ex. 3. A can do a piece of work in 3 days, and B in 6 days. In how many days can both do it, working together ? Make the hour equal to the unit of abscissas, and the work to be done equal to the unit of the ordinates. Then OA and OB represent the work done by A and B respectively. To obtain the graph of the work done by both to- gether, we add the ordinates correspond- ing to any particular time, e.g. 3 hours ; i.e. produce CA to D so that AD = CE. Then OD is the graph of the work done by both, and the required time is equal to abscissa of S, or two days. Hence both working together will do the work in two days. Ex. 4. At what time between 5 and 6 o'clock are the hands of a clock at right angles ? Let the abscissa represent the time from 5 to 6, and the ordinate the hour spaces. It can easily be seen that AB repre- sents the motion of the hour hand. A point 90° distant from the hour hand moves in the same time from 2 to 3 or from 8 to 9. Hence the motion of such a point is represented by CD or EF. But the graph of the motion of the minute hand is OG-. Therefore the abscissas of the points H and I represent the required time. Or the hands are at right angles at 5 : 11 and 5 : 43|. EXERCISE 24 1. A sets out walking at the rate of 3 miles per hour, and 3 hours later B follows on horseback, traveling at the rate of 6 miles per hour. After how many hours will B over- take A, and how far will each then have traveled ? 76 APPENDIX 2. A and B set out walking at the same time in the same direction, but A has a start of 3 miles. If A walks at the rate of 2\ miles per hour, and B at the rate of 3 miles per hour, how far must B walk before he overtakes A ? , , h 3. A train traveling 30 miles per hour starts \] of an hour before a second train that travels >'?■ miles an hour. In how many hours will the first train be overtaken by the second ? 4. A sets out walking at the rate of 3 miles per hour, and one hour later B starts from the same point, traveling by coach in the opposite direction at the rate of 6 miles per hour. After how many hours will they be 27 miles apart ? 5. A and B start walking at the same hour from two towns 17^ miles apart, and walk toward each other. If A walks at the rate of 3 miles per hour and B at the rate of 4 miles per hour, after how many hours do they meet, and how many miles does A walk ? 6. An accommodation train runs according to the following schedule : ' Station Distance feom A Arrives Leaves A 2 B 10 2:20 2:24 C 15 2 :32 2:35 D 25 2:50 2:55 E 40 3:20 3 :21 F 50 3:40 An express train leaves A at 2:15 and reaches Fat 3:25. Where does it overtake the accommodation train, if we assume that both trains move uniformly ? (a) A in (P) A in (<0 A in (d) A in in 6. in 4. in 6 in 5. APPENDIX 77 7. In how many days can A and B working together do a piece of work if each alone can do it in the following number of days ? 6, B 12, B 12, B 20, B 8. A can do a piece of work in 18 days, B in 9, and C in 12 days. In how many days can all three do it working to- gether ? 9. At what time between 2 and 3 o'clock are the hands of the clock together ? 10. At what time between 3 and 4 o'clock are the hands of a clock in a straight line and opposite ? 11. At what time between 6 and 7 o'clock are the hands at right angles ? 12. A cistern can be filled by two pipes in 3 and 6 hours respectively. In how many hours can it be filled by the two running together ? 13. A cistern can be filled by pipes in 3, 4, and 5 hours re- spectively. In how many hours can it be filled by all three together ? 14. A stone is dropped into a well and the sound of its impact upon the water is heard at the top of the well (a) 4 (6) 6 seconds later. If the velocity of sound is assumed as 360 meters per second, and g = 10 meters, how deep is the well ? (A body falls in t seconds 1 1 2 meters.) 78 APPENDIX II. STATISTICAL DATA SUITABLE EOR GRAPHIC REPRESENTATION 1. Table of Population (in Millions) of United States, France, Germany, and British Isles Tear U.S. France Germany British Isles 1800 5.3 27.2 22.0 16.0 1810 . 7.2 28.8 23.4 17.6 1820 . 9.6 30.5 26.2 20.5 1830 . 12.9 32.4 29.7 24.0 1840 . 17.0 34.0 32.4 26.4 1850 . 23.2 35.6 35.2 27.2 1860 . 31.4 37.3 38.1 28.7 1870 . 38.6 36.1 40.5 31.2 1880 . 50.2 37.6 45.2 34.5 1890 . 62.6 38.6 49.4 37.5 1900 . 76.3 38.9 56.4 41.2 2. Arrival of Immigrants (in Ten Thousands), 1891-1905 From '91 11 '92 13 '93 10 '94 6 '95 4 '96 3 '97 2 '98 2 '99 2 '00 '01 '02 3 '03 '04 '05 Germany 2 2 4 5 4 Italy 8 6 7 4 4 7 6 6 8 10 14 18 23 19 22 Russia 5 8 4 4 3 5 3 3 6 9 9 10 14 15 18 3. Population of New York City 1653 1661 1673 1696 1731 1750 10,000 1,120 1756 . . . . . . . 10,530 1,743 1771 . . . . . . . 21,865 2,500 1774 . . . . . . . 22,861 4,455 1786 . . . . . . . 23,688 8,256 1790 . . . . . . . 33,131 0,000 1800 . . . . . . . 60,489 APPENDIX 79 3. Population of New York City — Continued 1805 75,587 1810 96,373 1816 100,619 1820 123,706 1825 166,136 1830 202,589 1835 253,028 1840 312,710 1845 358,310 1850 515,547 1855 629,904 1860 813,669 1865 726,836 1870 942,292 1875 1,041,886 1880 1,206,299 1890 1,515,301 1893 1,891,306 1898 (all Boro's) . . 3,350,000 1899 (all Boro's) . . 3,549,558 1900 (all Boro's) . . 3,595,936 1901 (all Boro's) . . 3,437,202 1902 (all Boro's) . . 3,582,930 1903 (all Boro's) . . 3,632,501 1904 (all Boro's) . . 3,750,000 1905 (all Boro's) . . 3,850,000 1906 (all Boro's) . . 4,014,304 Population (in Hundred Thousands) of Illinois, Massachu- setts, New York, and Virginia State 1800 1810 1820 18.30 1840 1850 8.5 1860 17.1 1870 1880 30.8 1890 1900 Illinois .5 1.6 4.8 25.3 38.3 48.2 Mass. 3.4 3.8 4.0 4.5 4.7 5.8 6.9 7.8 9.3 10.4 11.9 N. York 6.9 9.6 13.7 19.2 24.3 31.0 38.8 43.8 50.8 60.0 72.7 Virginia 8.8 9.7 10.7 12.1 12.4 14.2 16.0 12.3 15.1 16.6 18.5 5. Table of Mortality Com- Number Deaths Number of Number Dying an- nually out of Each 1000 pleted Age Surviving at Each Age in Each Year Years Expectation 10 100,000 749 48.7 7.49 11 99,251 746 48.1 7.52 12 98,505 743 47.4 7.54 13 97,762 740 46.8 7.57 14 97,022 737 46.2 7.60 15 96,285 735 45.5 7.63 16 95,550 732 44.9 7.66 17 94,818 729 44.2 7.69 18 94,089 727 43.5 7.73 19 93,362 725 42.9 7.77 20 92,637 723 42.2 7.81 80 APPENDIX 5. Table of Mortality — Continued Com- Number Deaths Number of Number Dying an- nually out of Each 1000 pleted Age Surviving at Each Age in Each Tear Tears Ex- pectation 21 91,914 722 41.5 7.86 22 91,192 721 40.9 7.91 23 90,471 720 40.2 7.96 24 89,751 719 39.5 8.01 25 89,032 718 38.8 8.07 26 88,314 718 38.1 8.13 27 87,596 718 37.4 8.20 28 86,878 718 36.7 8.26 29 86,160 719 36.0 8.35 30 85,441 720 35.3 8.43 31 84,721 721 34.6 8.51 32 84,000 723 33.9 8.61 33 83,277 726 33.2 8.72 34 82,551 729 32.5 8.83 35 81,822 732 31.8 8.95 36 81,090 737 31.1 9.09 37 80,353 747 30.4 9.23 38 79,611 749 29.6 9.41 39 78,862 756 28.9 9.59 40 78,106 765 28.2 9.79 41 77,341 774 27.5 10.01 42 76,567 785 26.7 10.25 43 75,782 797 26.0 10.52 44 74,985 812 25.3 10.83 45 74,173 828 24.5 11.16 46 73,345 848 23.8 11.56 47 72,497 870 23.1 12.00 48 71,627 896 22.4 12.51 49 70,731 927 21.6 13.11 50 69,804 962 20.9 13.78 51 68,842 1,001 20.2 14.54 52 67,841 1,044 19.5 15.39 53 66,797 1,091 18.8 16.33 54 65,706 1,143 18.1 17.40 55 64,563 1,199 17.4 18.57 56 63,364 1,260 16.7 19.89 57 62,104 1,325 16.1 21.34 58 60,779 1,394 15.4 22.94 59 59,385 1,468 14.7 24.72 60 57,917 1,546 14.1 26.69 61 56,371 1,628 13.5 28.88 62 54,743 1,713 12.9 31.29 63 53,030 1,800 12.3 33.94 64 51,230 1,889 11.7 36.87 APPENDIX 81 6. Table of Mortality — Continued Com- Number Deaths Number of Number Dying an- nually out of Each 1000 pleted Surviving at in Each Years Ex- Age Each Age Year pectation 65 49,341 1,980 11.1 40.13 66 47,361 2,070 10-5 43.71 67 45,291 2,158 10.0 47.65 68 43,133 2,243 9.5 52.00 69 40,890 2,321 9.0 56.76 70 38,569 2,391 8.5 61.99 71 36,178 2,448 8.0 67.67 72 33,730 2,487 7.6 73.73 73 31,243 2,505 7.1 80.18 74 28,738 2,501 6.7 87.03 75 26,237 2,476 6.3 94.37 76 23,761 2,431 5.9 102.31 77 21,330 2,369 5.5 111.06 78 18,961 2,291 5.1 120.83 79 16,670 2,196 4.8 131.73 80 14,474 2,091 4.4 144.47 81 12,383 1,964 4.1 158.61 82 10,419 1,816 3.7 174.30 83 8,603 1,648 3.4 191.56 84 6,955 1,470 3.1 211.36 85 5,485 1,202 2.8 235.55 86 4,193 1,114 2.5 265.68 87 3,079 933 2.2 303.02 88 2,146 744 1.9 346.69 89 1,402 555 1.7 395.86 90 847 385 1.4 454.55 91 462 246 1.2 532.47 92 216 137 1.0 634.26 93 79 58 .8 734.18 94 21 18 .6 857.14 95 3 3 .5 1,000.00 6. Railway Accidents in the United States Total Killed Injured 1897 6,437 36,731 1898 6,859 40,882 1899 7,123 44,620 1900 7,865 50,320 1901 8,455 53,339 1902 8,588 64.662 1903 9,840 76,553 1904 10,046 84, 155 82 APPENDIX 7. Amount of $1 at Compound Interest from One to Thirtt Years Tears 3J Per Cent 4 Per Cent 5 Per Cent 6 Per Cent 1 1.035 1.040 1.050 1.060 2 1.071 1.081 1.102 1.123 3 1.108 1.124 1.157 1.191 4 1.147 1.169 1.215 1.262 5 1.187 1.216 1.276 1.338 6 1.229 1.265 1.340 1.418 7 1.272 1.315 1.407 1.503 8 1.316 1.368 1.477 1.593 9 1.362 1.423 1.551 1.689 10 1.410 1.480 1.628 1.790 11 1.460 1.539 1.710 1.898 12 1.511 1.601 1.795 2.012 13 1.564 1.665 1.885 2.132 14 1.618 1.731 1.979 2.260 15 1.675 1.800 2.078 2.396 16 1.734 1.873 2.182 2.540 17 1.794 1.947 2.292 2.692 18 1.857 2.025 2.406 2.854 19 1.922 2.106 2.527 3.025 20 1.989 2.191 2.653 3.207 21 2.059 2.278 2.786 3.399 22 2.131 2.369 2.925 3.603 23 2.206 2.464 3.071 3.819 24 2.283 2.563 3.225 4.048 25 2.363 2.665 3.386 4.291 26 2.446 2.772 3.555 4.549 27 2.531 2.883 3.733 4.822 28 2.620 2.998 3.920 5.111 29 2.711 3.118 4.116 5.418 30 2.806 3.243 4.321 5.743 APPENDIX 83 8. Amount of $1 Annually Deposited at Compound Interest Years 3 J Pbk Cent 4 Per Cent 5 Per Cent 6 Per Cent 1 1.000 1.000 1.000 1.000 2 2.035 2.040 2.050 2.060 3 3.106 3.121 3.152 3.183 4 4.215 4.246 4.310 4.374 5 5.363 5.416 5.525 5.637 6 6.550 6.633 6.801 6.975 7 7.779 7.898 8.142 8.393 8 9.052 9.214 9.549 9.897 9 10.368 10.582 11.026 11.491 10 11.731 12.006 12.577 13.180 11 13.142 13.486 14.206 14.971 12 14.602 15.025 15.917 16.869 13 16.113 16.626 17.713 18.882 14 17.677 18.291 19.598 21.015 15 19.296 20.023 21.578 23.276 16 20.971 21.824 23.657 25.672 17 22.705 23.697 25.840 28.212 18 24.500 25.645 28.132 30.905 19 26.357 27.671 30.539 33.760 20 28.280 29.778 33.066 36.785 21 30.270 31.969 35.719 39.992 22 32.328 34.248 38.505 43.392 23 34.460 36.617 41.430 46.995 24 36.666 39.082 44.502 50.815 25 38.949 41.645 47.727 54.864 26 41.313 44.311 51.113 59.156 27 43.759 47.084 54.669 63.705 28 46.290 49.967 58.402 68.528 29 48.910 52.966 62.322 73.639 84 APPENDIX III. TABLES TABLE 1 Squares, Cubes, Square Roots and Reciprocals of Numbers from 1 to 100 The squares, cubes, and reciprocals of decimal frac- tions can be obtained by shifting the decimal point. Thus 4.2 2 = 17.64, 4.2 3 = 74.088, — = .24. For square roots, how- ever, this method fails, and Table 2 has to be used. ■ X* as 3 •£ X x r I I I I. OOO I. OOO I 2 4 8 I.4I4 .500 2 3 9 27 1-732 •333 3 4 16 64 2.000 .250 4 5 2 5 125 2.236 .200 5 6 36 216 2.449 .167 6 7 49 343 2.646 • 143 7 8 64 512 2.828 .125 8 9 81 729 3.OOO .111 9 IO 1 00 1 000 3.162 .100 10 ii 1 21 I33 1 3-3I7 .091 11 12 144 1 728 3-464 .0S3 12 13 1 69 2197 3.606 .077 13 14 1 96 2744 3-742 .071 14 15 2 25 3 375 3-873 .067 15 16 256 4096 4.000 .063 16 17 289 49i3 4-123 •059 17 18 324 5832 4-243 .056 18 19 361 6859 4-359 •053 19 20 400 8 000 4.472 .050 20 X St* X3 V* 1 X x APPENDIX 85 X X* X s v'x; 1 x X 21 441 9 261 4-583 .048 21 22 484 10648 4.690 •045 22 23 529 12 167 4.796 .043 23 24 576 I3824 4.899 .042 24 25 625 I5 6 25 5.000 .040 25 26 676 I7576 5-°99 .039 26 27 729 19 683 5- I 96 •037 27 28 784 21 952 5.292 .036 28 29 841 24 389 5-385 •034 29 30 900 27 OOO 5-477 •033 SO 31 9 61 2979I 5.568 .032 31 32 10 24 32768 5-657 .031 32 33 1089 35 937 5-745 .030 33 34 n 56 39 304 5-83I .029 34 35 1225 42875 5.916 .029 35 36 12 96 46656 6.000 .028 36 37 1369 50653 6.083 .027 37 38 1444 54872 6.164 .026 38 39 15 21 59 319 6.245 .026 39 40 1600 64000 6.325 .025 40 4i 1681 68921 6.403 .024 4i 42 1764 74088 6.481 .024 42 43 1849 79 507 6-557 .023 43 44 1936 85 184 6.633 .023 44 45 2025 91 125 6.708 .022 45 46 21 16 "• 97 336 6.782 .022 46 47 22 09 103 823 6.856 .021 47 48 2304 1 10 592 6.928 .021 48 49 24 01 117 649 7.000 .020 49 5° 2500 125 000 7.071 .020 50 5i 2601 132 651 7. 141 .020 51 52 2704 140 608 7. 211 .019 52 53 2809 148877 7.280 .019 53 54 29 16 157464 7-348 .019 54 55 3025 166375 7.416 .018 55 56 3136 175 616 7-483 .018 56 57 3 2 49 185 193 7-55° .018 57 58 33 6 4 195 112 7.616 .017 58 59 34 8i 205 379 7.681 .017 59 60 3600 216 000 7.746 .017 60 X as2 SC3 ^x 1 X X 86 APPENDIX X X* X 3 *X 1 X X 61 37 21 226981 7.810 .016 61 62 38 44 238 328 7.874 .016 62 63 39 69 250 047 7-937 .016 63 64 4096 262 144 8.000 .016 64 65 4225 274 625 8.062 .015 65 66 43 56 287 496 8.124 .015 66 67 4489 300 763 8.185 .015 67 68 46 24 3H43 2 8.246 .015 68 69 4761 338 5°9 8.307 .014 69 70 4900 343 000 8.367 .014 70 7i 50 4I 3579" 8.426 .014 7i 72 5184 373 248 8.485 .014 72 73 53 29 389017 8-544 .014 73 74 54 76 405 224 S.602- .014 74 75 5625 421 875 8.660 .013 75 76 57 76 438 976 8.718 .013 76 77 59 29 456 533 8-775 .013 77 78 6084 474 552 8.832 .013 78 79 6241 493 039 8.888 .013 79 80 6400 5 1 2 000 8-944 .013 80 81 6561 53i 441 9.000 .012 81 82 6724 551368 9-055 .012 82 83 6889 57i 787 9.110 .012 83 84 7056 592 704 9.165 .012 84 85 7225 614 125 9.219 .012 85 86 73 96 636 056 9.274 .012 86 87 7569 658 5°3 9-327 .Oil 87 88 77 44 681 472 9.381 .Oil 88 89 79 21 704 969 9-434 .Oil 89 90 81 00 729000 9.487 .Oil 90 9i 8281 753 571 9-539 .Oil 9i 92 8464 778 688 9-592 .Oil 92 93 8649 804 357 9.644 .Oil 93 94 8836 830 584 9.695 •Oil 94 95 9025 857 375 9-747 .Oil 95 96 92 16 884 736 9.798 .010 96 97 9409 912673 9.849 .010 97 98 9604 941 192 9.899 .010 98 99 ^801 970 299 9-95° .010 99 100 100 00 1 000 000 10.000 .010 100 * x2 X* Vx I X X APPENDIX 87 TABLE 2 Square Roots of Numbers from 1 to 9.9 o.o O.I 0.2 0.3 0.4 0-5 •0.6 0.7 0.8 0.9 o o.ooo 0.316 0.447 0.548 0.632 0.707 °-775 0.837 0.894 0.949 I I.OOO 1.049 1.095 1. 140 1. 183 1.225 1.265 1.304 1.342 i-37* 2 1.414 1.449 1.483 1-517 1-549 1.581 1.612 1.643 I.673 1-703 3 !-732 1. 761 1.789 1.817 1.844 1.871 1.897 1.924 1.949 J-975 4 2.000 2.025 2.049 2.074 2.098 2.121 2.145 2.168 2.191 2.214 5 2.236 2.258 2.280 2.302 2.324 2-345 2.366 2.387 2.408 2.429 6 2.449 2.470 2.490 2.510 2.530 2 -55° 2.569 2.588 2.608 2.627 7 2.646 2.665 2.683 2.702 2.720 2-739 2-757 2.775 2-793 2.81 1 8 2.828 2.846 2.864 2.881 2.898 2.915 2-933 2.950 2.966 2.983 9 3.000 3017 3-033 3.050 3.066 3.082 3.098 3-"4 3-130 3.146 ANSWERS TO EXERCISES Exercise 1. Pages 2, 3 5. 5.66. 6. 5. 7. In a line II XX' passing through (0, 4). 8. In the ?/-axis. 9. In the x-axis. 10. The line II XX' passing through (0, 3). 11. The ordinate. 12. (0, 0). Exercise 2. Pages 6, 7, 8 1. (a) 6°, 5.9°, 5.25°, 2.5°. (6) 1 : 40 p.m. and 5 p.m; 1 p.m. and 6 p.m. ; 11 a.m. and 8.40 p.m. ; 10 p.m. ; 9.20 p.m. (c) 3.15 p.m. (d) 7+°. (e) 1 p.m. to 6 p.m. (/) 12 m. to 12.30 p.m. ; 6.40 p.m. to 7.20 p.m. (g) 11 a.m. to 9.20 p.m. (h) 9.20 p.m. on. (£) 4.75°. (&) 6 p.m. (Z) 11 a.m. to 3.15 p.m. (m) 3.15 p.m. on. (n) Between 3 p.m. and 4 p.m. (o) Between 12 m. and 1 p.m. ; and between 11 a.m. and 12 m. 2. (a) San Francisco. (&) Bismarck. (c) April 20 and Sept. 20. (d) During April. (e) 25°. 3. (c) 18° C. (d) 8.1 grams. (e) 15 grams. Exercise 3. Pages 10, 11 (rt) 12.25. (6) 2.25. (c) 7.84. (d) 3.61, (e) 2.5. (/) 3.5. (g) 2.24. (ft) .55. (a) 4.25, - 1.75, - 1.75. (6) 2 ; 3.73 and .27 ; 3.87 and .13. (c) -2. (