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to 21, the function X again changes its sign, and a second variation of sign is lost. Also, in passing from 24 to 3, the function X again changes its sign, and a third variation is lost; and there are no further changes of sign arising from the substitution of any number between 3 and + a.

Hence the given equation has 3 real roots; one situated between —2 and —3, one between 2 and 21, and a third between 27 and 3. The initial figures of the roots are therefore -2, +2, and +2.

There are three changes of sign of the primitive function, two of the first derived function, and one of the second derived function; but no variation is lost by the change of sign of either of the derived functions; while every change of sign of the primitive function occasions a loss of one variation.

2. Find the number and situation of the real roots of the equation

203-5x2 + 8x_1=0. Here we have

X=X3-5x2 + 8x-1,
X=3x2 - 10x+8,
R=2-31,

R,=-2295. When x=-oc, the signs are -+--, giving 2 variations, x=ta,

+++-,

1 Hence this equation has but one real root, and, consequently, must have two imaginary roots. Moreover, it is easily proved that the real root lies between 0 and +1.

3. Find the number and situation of the real roots of the equation 24 - 2003–72 +10x+10=0. Here we have

X=* -2.3-722 +10x+10,
X,=4x3 – 6x2 — 14x+10, or 2x3 -- 322—7x+5,

R=17c2-233-45,
R,=152x-305,

R= +524535. When x=- oc, the signs are +-+-+, giving 4 variations, x=

x=+ OC,

+++++, 0 Hence the four roots of this equation are real.

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Substituting different values for x, we find that when x=-3, the signs are +-+-+, giving 4 variations, x= -2,

-++-t,

3 x=-1,

-+

3
0,

++--+
-t,

2
x=+1,

+---t,

2 x=+2,

+---+

2 x=+21,

-0- ++,

1 x=+3,

+++++,

0 Ilence this equation has one negative root between-2 and -3, one negative root between 0 and -1, one positive root between 2 and 23, and another positive root between 27 and 3.

4. Find the number and situation of the real roots of the equation

203 — 7+7=0. Ans. Three; viz., one between -3 and -4, one be

tween 1 and 13, and the other between 13 and 2. 5. Find the number and situation of the real roots of the equation

2x4 -20x+19=0. Ans. Two; viz., one between 1 and 11, the other

between 13 and 2. 0. Find the number and situation of the real roots of the equation 25 +2.c* + 3x3+4x2+5x-20=0.

Ans. One, situated between 1 and 2. 7. Find the number and situation of the real roots of the equation

23 + 3x2 + 5x—178=0.

Ans. One, situated between 4 and 5. 8. Find the number and situation of the real roots of the equation

X4 - 12.2 + 12x—3=0.
Ans. Four; viz., one between 3 and -4, one be-

tween 0 and , one between 1 and 1, and the

other between 2 and 3. 9. Find the number and situation of the real roots of the equation

X4 —833 +14x2 + 4x—8=0.
Ans. Four; viz., one between - 1 and 0, one between

0 and +1, one between 2 and 3, and the other
between 5 and 6.

Solution of Simultaneous Equations of any Degree. 459. One of the most general methods for the elimination of unknown quantities from a system of equations, depends upon the principle of the greatest common divisor.

Suppose we have two equations involving x and y. We first transpose all the terms to one member, so that the equations will be of the form

A=0, B=0. We arrange the terms in the order of the powers of , and

X we will suppose that the polynomial B is not of a higher degree than A. We divide A by B, as in the method of finding the greatest common divisor, Art. 95, and continue the operation as far as possible without introducing fractional quotients having x in the denominator. Let Q represent the quotient, and R the remainder; we shall then have

A=BQ+R. But, since A and B are each equal to zero, it follows that R must be equal to zero. If, then, there are certain values of a and y which render A and B equal to zero, these values should be the roots of the equations

B=0, R=0. We now divide B by R, and continue the operation as far as possible without introducing fractional quotients having x in the denominator. Let R' denote the remainder after this divi. sion. For the same reason as before, R' must equal zero, and we thus obtain the two equations

R=0, R' =0, whose roots must satisfy the equations A=0, B=0. If we continue to divide each remainder by the succeeding, and suppose that each remainder is of a lower degree with respect to x than the divisor, we shall at last obtain a remainder which does not contain x. Let R" denote this remainder. The equation R"=0 will furnish the values of y, and the equation R'=0 will furnish the corresponding values of x.

If we have three equations involving three unknown quantities, we commence by reducing them to two equations with

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two unknown quantities, and subsequently to a single final equation by a process similar to that above explained. Ex. 1. Solve the two equations {a

x2 + y2—13=0,

x+y-5=0. Divide the first polynomial by the second, as follows:

22+ya-13 +y-5
22 +(y-5) 3-4+5
-(7-5)x+ y2-13
-(y-5).— 72 +10y-25

2y2-10y+12, the remainder. This remainder must be equal to zero; that is,

2y2--10y+12=0, whence

y=2 or 3. When

Y=2, x=3;
y=3, x=2.

x+xy-18=0,

xy+ay2-12=0. Multiply the first polynomial by y, to make its first term divisible, and proceed as follows:

xy(1+y3)—187 |xy(1+y)-12
xy(1+y3)—12(1-y+y2) 1-y+ya

12-30y+12y2, the remainder. Hence

12—30y+12y-=0; therefore

y=2 or 1. When

y=2, x=2,
y=1, x=16.

Q3y2 — 23 + xy +x-6=0,

Ex. 2. Solve the equations {

Ex. 3. Solve the equations (* = 3+2+3=0.

The first remainder is 3y—3, which, being placed equal to 0, gives y=1, whence x=3. Ex. 4. Solve the equations

23–3x2y +2(3y2–4+1)-43 +42—2y=0,

22--2xy + y2-y=0. The remainder after the first division is 2-2y, and after the second division ye-y. Hence we conclude

X—2y=0, and -y=0.

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Whence we have y=1 or 0,

w=2 or 0. Ex. 5. Solve the equations

oc2 + x(8y-13)+y2—7y+12=0,

22 — (4y+1) + y2 +5y=0. The remainder after the first division is

x (12y-12)-12y+12. Hence we have 12(7-1)(-1)=0, which equation may be satisfied by supposing y-1=0 or x-1=0. When

x=1, y=-1 or 0,
y=1, x= 2 or 3.

2038 +2xy(y2)=0.

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Ex. 6. Solve the equations { 2xy +299-54+2=0.

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The first division gives a remainder (y^2)+y2–4, whence we have

(y-2)(++2)=0; and we may have either

y-2=0, or x+y+2=0. If we divide the first member of the second equation by x+y+2, we obtain the remainder ya _5y+6, which also equals zero; whence y=2 or 3. When

y=2, x=-4 or 0, y=3, x=-5.

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