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to 23, the function X again changes its sign, and a second variation of sign is lost. Also, in passing from 21 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 24 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

23 — 5x2 +8—1=0. Here we have

X=X3-522 +82-1,
X,=3x2-10x+8,
R=2-31,

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

+++-,

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

+ --2003-742 +10x+10=0. Here we have

X=x*-2.03-702 +10x+10,
X,=4x36x2 – 14x+10, or 2203 — 322—7x+5,

R=17c2-23-45,
R,=152.–305,

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

+++++,

0 " Hence the four roots of this equation are real.

t,

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

-++-t,

3
x=-1,

-++-t, 3
0,
++--+,

2
x=+1,

+-

2 x=+2,

+---+

2 x=+27,

-0-++,

1 x=+3,

+++++

0 IIence 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

2C3 — 7x+7=0. Ans. Three; viz., one between 3 and -4, one be

tween 1 and 13, and the other between 14 and 2. 5. Find the number and situation of the real roots of the equation

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

between 13 and 2. 6. Find the number and situation of the real roots of the equation 23 +224 + 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

203 + 3x2 +5x-178=0.

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

x-12.02 +12.-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

* -8x3 +14x2 + 4x-8=0. Ans. Four; viz., one between - 1 and 0, one between

O 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 x, and 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 ~ 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 ac 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 ~ in the denominator. Let R' denote the remainder after this division. 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 2. 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 two unknown quantities, and subsequently to a single final equation by a process similar to that above explained. Ex. 1. Solve the two equations (22+42–1320,

+y5=. Divide the first polynomial by the second, as follows:

22 + y2-13 x+y-5
202 +(7-5) 2-4+5
-(-5)x+ y2-13
-(7-5) – y +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. Ex. 2. Solve the equations

x+xys-18=0,

1 xy +xy-12=0. Multiply the first polynomial by y, to make its first term divisible, and proceed as follows: wy(1+y3)— 18y

xy(1+y)-12 xy(1+y3)—12(1-y+y?) 1-y+ya

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

12-30y+12yo=0; therefore

y=2 or 1. When

y=2, x=2,

y=t, x=16. Ex. 3. Solve the equations ( 2*y*—x+xy+~—6=0,

c°g –co+z-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-y+1)-48 + y2—2y=0,

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

x-2y=0, and y-y=0.

,

Whence we have y=1 or 0,

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

x2 + (8y-13)+y2—7y+12=0,

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

x(12y12)–12y+12. Hence we have 12 (4-1)(x-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.

203 +2x2y + 2xy(y-2)+42—4=0.

22 The first division gives a remainder a(y-2)+yo—4, whence we have

(y-2)(x+y+2)=0; and we may bave 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 y2-5y+6, which also equals zero; whence y=2 or 3. When

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

Ex. 6. Solve the equations ** 224 +2y –5y+2=0.

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