GRAPHIC ALGEBRA
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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.
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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°
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r°
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1°
17
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0°
1°
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1 P
M.
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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
<f
CD
O
to
43
R
34
H
New York City
30
32
37
48
60
69
74
72
66
55
52
San Francisco
50
52
54
55
57
58
58
59
60
59
56
51
56
Tampa
59
66
66
72
76
80
82
81
80
73
65
63
72
Bismarck
4
10
23
42
54
64
70
68
57
44
26
15
40
a. Which of these cities has the most uniform temperature ?
b. Which one has the greatest extremes of temperature ?
c. When is the mean temperature in San Francisco the
same as in New York ?
d. When does the mean temperature of New York rise most
rapidly ?
e. What is the difference between the mean temperatures of
New York and Bismarck on Jan. 15 ?
3. By using the annexed table represent graphically the
greatest amount of water vapor which a cubic meter of air can
hold at various temperatures.
Degrees of Centigrade
— 25
— 20
— 15
-10
— 5
5
10
15
20
25
3n
30
35
39.3
40
Grams of Vapor
.8
1.0
1.5
2.3
3.4
4.8
6.9
:>.::
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
<d
FUNCTION OF ONE VARIABLE
find VT, measure the abscissa whose ordinate is 7, i.e. +2.6
or - 2.6.
Ex. Draw the graph of i x 2 — A x — 3.
To obtain the values of the functions for the various values
of x, the following arrangement may be found convenient :
(Compute each column before commencing the next, and see table on
page 84.)
X
X 2
t* 2
-i*
■^x- — -g X — O
- 4
16
8
.8
5.8
-3
9
4.5
.6
2.1
-2
4
2
.4
- .6
- 1
1
.5
.2
-2.3
-3
1
1
.5
-.2
-2.7
2
4
2
-.4
-1.4
3
9
4.-3
-.6
.9
4
16
8
-.8
4.2
Locating the points (-4,5.8), (-3,2.1), (- 2, - .6), ••-, (4,4.2),
and joining in order produces the graph ABC.
9. For brevity, the function is frequently represented by a
single letter, as y. Thus, in the above
example,
i
y = i x ~-i
x-
3;
if # = i, we fiud from the graph
1 x 2 — i x — 3 or y = — 3, if x — 1\,
y = — .4, etc.
For values of x greater than 4, the func-
tion will obviously be always positive and
increase when x increases. Hence the curve
will continue to go upward beyond C, and
similarly above A.
Graphs should always be drawn until they reach their ultimate direc-
tion at both ends.
\ A
Y'
,
\
1
\
\
\
X'
\
t
X
i •
3 \
>
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
.'
<V/
9
1
0/
X'-
2 -
1
/
:
i ;
i X
/
/
r
/
Y'
Hence, if » equals — 2, — 1, 0, 1, 2, 3 ;
then y equals — 4, — 2£, — 1,^,2,3$.
Locating the points (— 2, — 4),(— 1,— 2J),
etc., and drawing a line through them, we ob-
tain the graph of the equation, which is a
straight line.
20. The graph of an equation of the
first degree involving two unknown quan-
tities is always a straight v
line, and hence it can be AY
constructed if tioo points
are located (§ 12).
Ex. 2. Draw the locus of 4 x + 3 y = 12.
Hai = 0,y = 4; if y - 0, x = 3.
Hence, locate points (0, 4) and (3, 0), and join them
by a straight line AB. AB is the required graph.
Note. Equations of the first degree are called linear
equations, because their graphs are straight lines.
21. If two linear equations differ only in their
absolute terms (i.e terms not containing x or y)
as 2 x + y = 4 and 2 x + y = 2, their graphs are
parallel lines.
EXERCISE 6
Draw the loci of the following equations :
1. x + y = 4:. 9. 12 x + 15 y = 48.
2. x — 2?/ = 4. 10. x 2 — \/ + 2 = 0.
3. 2a? — 3y = 12. ll. 2x 2 -y- x = 0.
4. a — ?/ = 0. 12. a 3 + ?/ = 0.
5. x+y=— 10. 13. y- — x = 2.
6. 2/ = - 4.
7. .r + 2/ = 0.
8. y = 2x.
y—x+2+l
X X
14
15. a 2 -2/ 2 =16
0.
SOLUTION OF SIMULTANEOUS EQUATIONS
19
16. A body moving with a uniform velocity of 3 yds. per
second moves in t seconds a distance cl = 3 1.
Kepresent this formula graphically.
17. If two variables x and y are directly proportional, then
y = ex, where c is a constant.
Show that the graph of two variables that are directly pro-
portional is a straight line passing through the origin (assume
for c any convenient number).
18. If two variables x and y are inversely proportional, then
y — -, where c is a constant.
x
Draw the locus of this equation if c = 12.
19. The temperature remaining the same, the volume v of a
gas is inversely proportional to the pressure p. For a certain
body of gas, v = 2 cubic feet, if p = 15 lbs. per square inch.
Represent the changes of p and v graphically.
22. Graphical solution of a linear system
To find the roots of
the system :
2x + 3y = 8, (1)
x-2y = 2. (2)
By the method of
the preceding article
construct the graphs
AB and CD of (1)
and (2) respectively.
The coordinates of
every point in AB
satisfy the equation
(1), but only one point
in AB also satisfies
equation (2), viz. P,
the point of intersection of AB and CD.
\
__J_ —
,r v
A *v
-ft- — s
^ '-'
*S ^r
S S,^-E> .*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
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A ^J^ 7
~2_
i 6 J2,_
~?
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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.
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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
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XZ X - __ - -
: : :j-/::
,Z_ .1 -------
£ /
.1 ^
r A ^
" : :: _i " ::::: : i
X - X
" ::: ::: : : jtw z
. - - _ - x
• M jfv-,
l A* : cW/' 'Sh
-^ T^ >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
<?'
\
4^
V
y*
3
&
^
\
F 8
A
1
\^-
\,
T
2
i
O
r'
4
If
If
If
x = -l,y = HF=\,
x= l,y = KI = ~l
x — 2, y = 2, etc.
QUADRATIC EQUATIONS
33
Note. If we consider the distances cut off from SB by the ordinates,
as abscissas, e.g. SK= x = 1, then the parabola represents the function
x 2 + px + q in so-called " oblique coordinates."
Ex. 1. Find the values of x which will make the function
x 2 —\x — \ equal to 2, i.e.
x 2
\x-l = 2.
On YY' lay off SL = 2, and through L draw PQ II BE, meeting the
parabola in P and Q. By measuring the abscissas of P and Q we find
x —— |, or 2.
Ex. 2. Find the smallest value of the function x 2 — 2 x — 1.
Construct AB, the locus of y — 2x — 1=0. (See next diagram.)
Draw a tangent parallel to AB, touching the parabola in C ; then
DC = — 2 is the required value.
35. To construct the graph of
x 2 + px + q in the usual manner
(rectangular coordinates), make
H'G' = HG, E'F'=EF, I'K'= IK,
etc. The curve G'F'C'L' is the
required locus of x 2 +px + q (i.e.
of x 2 — 2a: — 1).
Note. § 35 makes it possible to con-
struct the locus of a quadratic function
without any computation.
36. The value of the function ax 2
+ bx + c is equal to a times the
part of the corresponding ordinate
ivhich is intercepted by the straight
line ay + bx + c = 0, and the pa-
rabola y = x 2 .
For ax 2 +bx+c =a(x 2 -\- ■
b c
But the function x 2 H — sc + - is represented by that part of
a a ,
the ordinate that lies between y — x 2 and y + -x +- = or
a a
ay + bx + c = 0. Hence ax 2 + bx + c is equal to a times this
intercept.
-x+ c -
a a
34 GRAPHIC ALGEBRA
EXERCISE 11
Find graphically :
1. The value of x 2 - 2 x - 2, if x equals - 3, - 2, .5, 1±.
2. The value of x?-\- x — 3, if a; equals — 1.5, — 1, 0, 2.
3. The value of x 2 — 5 x — 12, if x equals — 2.1, — 1.5, 3.5„
4. The value of x 2 + 4 x + 5, if sc equals — 7.5, 9.4, — 8.8.
5. The values of x if x 2 — 2 x — 10 = 4.
6. The values of x ii x 2 + 10 x + 10 = 5.
7. The smallest value of x 2 — 4 a; + 3.
8. The smallest value of x 2 + 10 x — 5.
9. The smallest value of x 2 -f 7 a; — 3.
10. The smallest value of cc 2 — 5 x + 2.
11. The value of 2 x 2 + 6 a + 7, if a; equals - 3, 6.5, 7.
Without calculating the various values of the function, con-
struct the loci of : ,
12. x 2 + Qx + 10. 15. tf + hx—'S.
13. a^ + 4a;-5. 16. a 2 — 3 a; + 7.
14. a; 2 — 4 a; + 7. 17. a' 2 + .c + l.
18. a; 2 + 4 a; — 7.
37. Equal roots. If the line ay + bx + c = is a tangent
to the parabola y = x 2 , the two points of intersection coincide,
and the two roots of ax 2 + bx + c — are equal.
Ex. 1. Solve jb 2 - 8 x + 16 = 0.
Let 2/ = x' 2 .
Then ?/-8x+l6 = 0.
If x = 0, ?/ = -16.
If * = 10, y = G4.
The line RR', which joins (0, —16) and (10, 64), is tangent to the
parabola at the point R.
Measuring the abscissa of R, we obtain two equal roots, 4 and 4.
QUADRATIC EQUATIONS
35
=: __ *s" „ K __:__ _ __ __^» _:___: : _2__ =
1 * s 1 \ J' '3
s^ \ /
=> _ _ - : 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<Z-
>-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
<S <
J
X
'
3
¥^~
1
:
j :
i ■
I I
1
r.
-20
^40-
CUBIC EQUATIONS
53
take in the cubic parabola a point P whose abscissa =$> =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 + <y
Hence PC 2 - r 2
= (x> - 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. (<f) 2. (e) 3.41 and .59.
(/) 3.41 and .59.
(g) 3 and 1. (h) and 4.
(a) 2.75, -3.25, 1.5. (6) 3.24, -1.24.
(c) 3. (d) 1
(e) 2.73, - .73. (/) 2.73, - .73. (g) 2.4, -
.4. (ft) 2.4, -A
25.
26.
27.
28. (6) 31.25 meters. (c) 2.24 seconds.
Exercise 4. Page 12
7. (&) -18i°C, -12|°C, -10°C, 0°C. (c) 14°F., 32°F., 33|°F.
89
Exercise 5. Pages 15, 16
1.
1.75.
5. 3, - 2.
9. .7, -5.7.
13.
-1.93, 2.93.
2.
-2.5.
6. 2.79, -
■1.79. 10. 5.54, -.54.
14.
-1.92,3.92.
3.
6.
7. 3.83, -
-1.83. 11. 4.37, -1.37.
15.
-5.62, .62.
4.
2.67.
8. 3, 3.
12. -2.16, 4.16.
16.
-1.31, 3.31.
17.
-1.53,
- .35, 1.88.
20. 1.21, 2 imag. 23.
-3.1,
3.5, 4.6.
18.
-4.05,
2 imag.
21. 1.78, 2 imag. 24.
-.39,
5.44, 7.95.
19.
-2.11,
.25, 1.86.
22. - 1.94, .55, 1.39. 25.
±.94,
±3.02.
26. -2.5, 1.73, 2 imag. 28. - 2.99, - 1.15, .21, 1.9, 3.05.
27. -.97, .85, 2.15, 3.97. 29. 1.38.
30. (a) -4.51,-1.75,1.26. (6) -4.12,-2.4,1.52.
(c) - 4.78, - 1.14, .92. (d) - 5.19, 2 imag.
( e ) 3. (/) -10 to 8.5+. (g) - 10 or 8.5+. (h) 8.5. (i) -3.33.
31. (a) - 2.84, .44, 2.4. (b) - 2.65, 0, 2.65.
(c) - 3, 1, 2. (d) - 1.68, - 1.38, 3.05.
(e) - 3.49, 2 imag. (/) 1, 3, 3, 1. (g) 2, 2, 0.
Exercise 7. Page 22
1. x = 2.8, y = — .1. 8. Parallel. 11. x = 4, y = 3.
2. x = 2, y = 1. 9. x - 3.6, y = — 1.6. x = — 3, y = - 4.
3. x = - .8, y = — 2.8. x = - 1.6, 2/ = 3.6. 12. x = 4, j/ = 2.
4. x = 2.6, y = .7. 10. x = 3, ?/ = 2. x = — 2, y = — 4.
5. x = 1.5, y = .5. x = 2, y = 3.
13. x = 4.3, J/ = 1.4. 14. x = 2. 3, y = 1.15.
x = - 1.8, y = - 3.4. x = -2.3, y= -1.15.
15. x = ± 4.8, ?/ = ± 1.3. 16. x= ± 3, y = ± 1.
x = ± 1.3, y = ± 4.8. x=± 00 ,2/= : Foo-
Exercise 8. Page 24
1. x = 1.82, y = - .82. 6. x = 2, 2/ = 4.
x = - .82, 2/ = 1.82. x = 0, 2/ = 0.
2. x = 4, y = 0. T. x = 3, y = 0.
x = 0, 2/ = — 4. x = 1, 2/ = — 2.
8. as = — 7, 2/ = — 1. 8. x = 1, 2/ = ± 3.
x = 1, 2/ = 7. 9. x = .22, y = 1.72.
4. x = 2.96, 2/ = .48. x = -1.72, 2/ = - .22.
X = - 2.16, 2/ = - 2.08. 10. x = 0, 2/ = ± 1.
5. x = - 3.4, 2/ = - 2.1.
x = .03, y = 3.95.
ANSWERS
91
Exercise 9. Page 28
.i-"5
1.
3, - 2. 5,
4,-2.
(
3. - 6, - .67.
12. 2.7, -5.1.
2.
1,-2. 6.
1.24, - 3.24.
10. 1.8, - 2.8.
13. 2.1, -4.6.
3.
6, - 3. 7.
7.1, -2.1.
11. 4.2, -2.2.
14. 1.5, - .7.
4.
2, - 5. 8.
.95, 6.3.
Exercise 10.
Pages 31, 32
1.
- 60, 75.
7. - 20, 30.
13. -35,6.
19. -.2, .3.
2.
- 50, 60.
8. - 16, 8.
14. - 72.2, 5.5.
20. - .6, .2.
3.
- 60, - 20.
9. - 30, 60.
15. -.225, .525. 21. - .2, .1.
4.
- 60, 20.
10. - 55, 22.
16. - .136, .736. 22. - .25, .6.
5.
- 80, 60.
11. - 28.3, 26.8
17. -1.425, .17
5.
6.
- 70, - 10.
12. - 33.33, 30.
18. - .311, .161.
Exercise 11.
Pages 34, 35
1.
13,6, -2.75,
- 2.75.
7.
— 1 when x =
2.
2.
- 2.25, - 3, -
-3,3.
8.
— 30 when x =
= -5.
3.
2.91, -2.25,
- 17.25.
9
— 15.25 when
x = -3.5.
4.
31.25, 130.96,
47.24.
10
, — 4.25 when x
; = 2.5.
5.
4.87, -2.87.
11.
7, 130.5, 147.
6.
- 9.47, - .53.
Exercise 12.
Page 39
1. 5, 5.
5. 4±3z.
9.
-1.5 ±4.97 i.
13. -.5±1.12i.
2. 5 ± 2 i.
6. 5 ± 2 i.
10.
- 4.5 ± 3.97 tf.
14. 1.5 ± 2 i.
3. - 2 ± 2 i.
7. _3.5±2.96i.
11.
-.5 ±.87 i.
4. - 4 ± 2 i.
8. 2.5 ± 2.96 i.
12.
-1,-1.
Exercise 13.
Page 41
1. 5, -3.
3. 5, 1.
5. 1 ± 3 i.
2. 3, -2.
4. -4, -1.
6. -1.27, -4.73.
Exercise 14.
Page 45
1. 2.
9. 2.7.
17.
1.1.
2. 3.
10. 6.6.
18.
8.
3. -2.
11. -2.6.
19.
4.5,4.5, -9.
4. -3.
12. 3.3.
20.
-11.
5. 1,2, -3.
13. 4.5
21.
- 5, - 5, 10.
6. -21, -1,
3£. 14. -4.6.
22.
-11.9.
7. -3.3.
15. -4.5
23.
4.1.
8. 3.2.
16. -.6,
6.7,
- 5.2. 24.
- 16.5.
92
l.
2.
3.
4.
5.
6.
7.
1.
2.
3.
4.
5.
6.
^sio^s
Exercise 15.
Page 48
-55, -41.6, -33.7, 40.9.
8. 3.4.
-18.5, -17.4,8, 1.4.
9. -16.
-24, -11, -20.6, 6.9.
10. 14.
- 454, - 82, 30, 326.
13. 101, 73.7, -.2.
3.2.
14. -133, -445, -
-67.
2.4, .4, -2.1.
15. -31.
-3.6.
Exercise 16. Page 52
-1,-1, 2.
-2,1,1.
- 2, 1 ± 2 i.
2, -1 ± .5 i.
2, -1-1-1.5*.
- 2, 1 ± 3.46 i.
7. -4, 2 ±1.73*.
8. _ 7, 3.5 ± 5.27 i.
9. - 1.5, .75 ± .43 i.
10. 2, - 1 ± .87 i.
11. 1, - .5 ±2.18 t.
12. - 1.46, .73 ± 1.9 i.
Exercise 17. Pages 56, 57
1. -1,1,3.
2. 1, 3, 5.
3. - 1, 2, 5.
4. 1, 2, 5.
5. - 1, - 2, - 4.
6. -2, 1, 3.
7. - 1.52, .43, 3.09.
8. -1.18, 2 imag.
9. -4.19, -1, 1.19.
10. - 1.88, 2 imag.
11. -3.61, -.87, .48.
12. - 2, .6, 2.
13. - 1.9, .76, 3.14.
14. - 1, .5, 3.
15. - 1.5, .5, 4.
16. - 2, .5, 4.5.
17. -9.28, 2 imag.
18. - 1, 2 ± i.
19. - 4, - 1 ± i.
20. - 1, 2 ±1.41*.
21. - 4, - 1 ± 2 i.
22. 4, 1 ± 1.41 i.
23. -3, 1 ±1.78*.
24. -3.84, 1.42 ±1.37 i.
25. l,4±2i.
1. (a) 5.26.
Exercise 18. Page 58
(6) 1.85. (c) 1.
Exercise 19. Page 61
-2, 1.66, 2 imag.
-4, -1,2,3.
— 2, 1, 2 imag.
-4, -1.83, 2, 3.83.
-2, -1,1,2.
1.
2.
3.
4.
5.
6. 1, 3, 2 imag.
7. - 2.69, - .63, 2 imag.
8. 1, 2.76, 2 imag.
9. - 2.73, - 1, .73, 3.
10. -2.22, 2.54, 2 imag.
11. -2.62, -.38, 1, 2.
12. -3,-2, 1,4.
ANSWERS
Exercise 20.
Pages 62, 63
1.
2.
3.
4.
5.
-6,-2, 1, 7.
-5,-3, 1, 7.
-4, -2,1,5.
- 5, - 3, 2, 6.
-8,-2,1,9.
6. -8,-2, 3, 7.
7. -6,-4, 3, 7.
8. -7,-3, 3, 7.
9. - 7, - 2, 3, 6.
10. -6, -3, 2,7.
Exercise 21. Page 65
1.
2.
3.
4.
5.
- 3, - 1, 2 ± i.
-3, -1,2 ±1.41 i.
-1,3, -1± 1.411
- 1, 3, - 1 ± i.
1.47, 2, - 1.73 ± 1.04 i.
6. -1.82, 1, .41 ±2.31 i.
7. 1, 3.61, - 2.31 ± 1.15 i
8. - 4, 2, 1 ± i.
9. - 3, 1, 1 ± i.
10. - 3, — 1, 2 ± i.
93
Exercise 22. Page 68
1. 15. 3. 227. 5. 45.07.
2. 69. 4. 89.94.
Exercise 23. Page 71
1.
-5,-2, 1,2. 7. -5, -3, 1, 4.
2.
-3, -1, 2,6. 8. -7, -4, 1, 2.
3.
-3, 1,4,6. 9. 1,2, 3, 4.
4.
-5,-3,-1,1. 10. -4,-2,1,2.
5.
-4,-2, 0, 2. 11. - 2.41, .41, 2, 4.
6.
-3, -1,1,5.
Exercise 24. Pages 75, 76, 77
1.
6 hrs. ; 18 miles. 5. 2J hrs. ; 7£ miles. 9. 2 : 10.9.
2.
18 miles. 6. 2 : 50. 10. 3 : 49.1.
3.
4J hrs. 7. (a) 3. (6) 3. (c) 4. (d) 4. 11. 6 : 16.4, 6 : 49.1
4.
3f hrs. 8. 4. 12. 2 hrs.
13. 1.28 hrs. 14. (a) 72.2 meters.
(6) 155 meters.
ELEMENTARY ALGEBRA
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