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To determine the situation of these roots we substitute different numbers, commencing at 0 and working in each direction, thus:

f(x) fi(x) $2(2) 13 (20) X-00


+ 3 variations. =-5

3 variations. X=-4

2 variations. X=-3

2 variations. x= -2

2 variations. X=-1

2 variations. X=0

1 variation. X=1

1 variation. X=2

1 variation. x=3

+ no variation. 30 =0

+ + + + no variation. Thus one root lies between —4 and -5; a second lies between O and – 1, and the third lies between 2 and 3.

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EXAMPLES L. g. Determine the number and situation of the real roots of: 1. X3 — 4x2 - 6x +8=0.

2. 2x4 – 11x2 + 8x — 16=0. 3. 23 - 73+7=0.

4. 24 — 4x3 +6x2 — 12x+2=0. 5. 24 — 4x3 +2 +6x+2=0. 6. 24 - 23+*—1=0. 7. 23 – 9x2+23x - 16=0.

8. 25 + x3 — 2x2 +23-1=0.



Fig. 1.

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579. Two lines drawn at right angles to each other as in Fig. 1

form a simple system of lines of Y

reference. Their intersection,

0, is called the origin. Distances P

from 0 along XX' are called abscissas; distances from XX'

on a line parallel to YY' are 7

called ordinates.

580. Abscissas measured to

b. the right of the origin are con-8

sidered positive, and to the left, negative. Ordinates measured above XX' are considered posi

tive, and when taken below XX' Y

are negative.

- a




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581. The abscissa and ordinate of a point are called the co-ordinates of that point, and the lines XX' and YY' are called Co-ordinate Axes, Axis of X, and Axis of Y, or Axis, of Abscissas, and Axis of Ordinates.

582. Any point in the plane can now be given by means of these co-ordinates: thus the point P of the figure of Art. 579 is located by measuring off the distance a to the right of the origin on the axis of abscissas and then taking a distance b vertically upwards.

Since a and b can be either positive or negative, a point P' is found by taking a positive and b negative, p' is found by taking a negative and b negative, and Pl is found by taking a negative and b positive.

Abscissas and ordinates are generally represented by x and y respectively. Thus for the point P, x=a, and y=b; for P', X=a, and y=-b, etc.

583. Instead of writing “ the point whose co-ordinates are 5 and 3,” a more concise form is used : thus the point (5, 3) means that the point will be found by taking an abscissa of 5 units and an ordinate of 3. Locate the points

(3,-2); (5,8); (-4, 4); (-8, -3).

FIG. 2.


584. Let f(x) be any rational integral function of x, and let us place it equal to y. If we give a series of numerical values to x we can obtain corresponding values for y. Now laying off the values of x as abscissas and the

X corresponding values of y as ordinates, we have a series of points which lie upon a line called the graph of the given function.

Example 1. Construct the graph of 2x – 1.

Let 2x – 1=y. Giving to x successive values, we


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obtain the corresponding values of y as follows:

x= -2, y=-5.

2=-1, y=-3.

x=1, y=1.
x=2, y=3.

y=5. Locating these points as explained in Art. 582 and

drawing a line through ix them, we have in this case

the straight line AB, Fig. 2, as the graph required.

Example 2. Plot the equation x2–2x—4.

Putting y=22—23 — 4, we obtain the following values:

x= -2, y=4.
x= -1, y=-1.
x=1, y=-5.


y=4. The points lie on the line ABC, Fig. 3, which is the graph of the given equation.

Example 3. Construct the graph of 23 — 2x.

Assuming y = 23 — 2x, we have

x= -2, y=-4.

=-1, y=1. X=0,

y=-1. x=2,

y=4. The line ABCD, Fig. 4, is the required graph.

By taking other values between those assumed, we may locate the curve with greater precision.


FIG. 4.







585. In giving values to x it is evident that the substitution of a root gives y=0, that is the ordinate, or distance from the Axis of Abscissas, is 0; hence where the graph cuts the Axis of X we have the location of a real root, and the graph will cross the Axis of X as many times as the equation has real and unequal roots. If the roots of the equation be imaginary, the curve will not touch the Axis of X.


Construct the graphs of the following functions : 1. 2x-3. 2. 22+2x+1. 3. 22-5. 4. x3+1. 5. 23—2. 6. 28–5x+3. 7. 24 — 322 +3. 8. 23 + 3x2 + 5x – 12.


COMMENSURABLE Roots. 586. A root which is either an integer or a fraction is said to be commensurable.

By Art. 553 we can transform an equation with fractional coefficients into another which has all of its coefficients integers, that of the first term being unity: hence we need consider only equations of this form. Such equations cannot have for a root a rational fraction in its lowest terms [Art. 546], therefore we have only to find the integral roots. By Art. 541 the last term of f(x) is divisible by every integral root, therefore to find the commensurable roots of f(x) it is only necessary to find the integral divisors of the last term and determine by trial which of them are roots.

587. If the divisors are small numbers we may readily ascertain by actual substitution whether they are roots. In other cases we may use the method of Arts. 533 and 537 or the Method of Divisors, sometimes called Newton's Method. Suppose a to be an integral root of the equation 2"+P.xn-1+p, 1*-2+

+Pn-1X+Pn=0. By substitution we have

an+P, an-1+p2an-2+ ...... +Pn-1a+Pr=0. Transposing and dividing throughout by a, we obtain


-Pn-1 - P2an-3_P, an-2-an-1,





in which it is evident that Pn must be an integer. Denoting

Pn by Q and transposing - Pn-1, Q+Pn-1=

- P2an-3P, an-2-an-1. Dividing again by a gives Q+Pn-1

-P,*-4-Pan-3--013-2 Again, as before, the first member of the equation must be an integer. Denoting it by Q, and proceeding as before, we must after n divisions obtain a result Qn-1+P1




Hence if a represents one of the integral divisors of the last term we have the following rule:

Divide the last term by a and add the coefficient of x to the quotient.

Divide this sum by a, and if the quotient is an integer add to it the coefficient of x2

Proceed in this manner, and if a is a root of the equation each quotient will be an integer and the last quotient will be 1.

The advantage of Newton's Method is that the obtaining of a fractional quotient at any point of the division shows at once that the divisor is not a root of the equation.

Example.. Find the integral roots of 24 +4x3 — 22 — 16x — 12=0. By Descartes' Rule the equation cannot have more than one positive root, nor more than three negative roots.

The integral divisors of — 12 are +1, +2, +3, +4, +6. Substitution shows that – 1 is a root, and that +1 is not a root. To ascertain if 2 is a root, arrange the work as follows:

1+4- 1–16 – 12 12 --1-6-11- 6

-2-12-22 Hence 2 is a root.

[Explanation. The first line contains the coefficients of the original equation, and the divisor 2. Dividing the last term, – 12, by 2 gives a quotient –6; adding – 16, the coefficient of x, gives

- 22. Dividing – 22 by 2 gives -11; adding – 1, the coefficient of x, gives - 12. Dividing – 12 by 2 gives - 6; adding +4, the coefficient of x', gives - 2, which divided by 2 gives a final quotient of – 1, hence 2 is a root.]

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