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Values of Coefficients in Equations.

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But the product (x - x)(-X") being arranged according
to
powers

of
I,
is

Io(' + 2")x+ xx"; which, being compared with its equal,

to +ar+b, gives

- (x+x)= a,

ac' x = that is, the coefficient of x is the negative of the sum of the roots of equation, and the term which does not contain x is the product of the roots.

276. Corollary. If the roots of the general equation of the third degree

23 + a +bx+c=0 are denoted by

X',x", x'", we have

zo+azo +6 1+c=(1-3)(cz)(zx)=0. But the product

(2 — 2) (2 — 2") (2 — qo'"') is, when arranged according to powers of x, 3(x+2+3)x2 + (x'x" + x'x' +'") <--- *' "; whence, by comparison with the given equation, we have

(sch + x + ''),
b = x 2 + x'x'' + 2" 2",

C=-ax' x'"; that is, the coefficient of zo is the negative of the sum of the roots, the coefficient of x is the sum of the products of the roots multiplied together two and two, and the term which does not contain a is the negative of the continued product of the roots.

a =

To find the Equal Roots.

277. Corollary. It may be shown in the same way that, in the equation

2" tar"-1+bx*-? + cx*—3 + &c. the coefficient of r*- is the negative of the sum of the roots; the coefficient of 2*-? is the sum of the products of the roots multiplied together two and two ; the coefficient of 2*-3 is the negative of the sum of the products of the roots multiplied together three and three ; and so on, the last term being the product of the roots when n is even, and the negative of this product when n is odd.

SECTION II.

Equal Roots.

278. Problem. To find the equal roots of an equation.

Solution. Let a' be one of the equal roots which occurs in times as a root of the given equation, the first member of which is therefore divisible by (x - 2)". If the quotient is a function of n denoted by X, the equation is, then,

(2 - 2')" X=0. The derivative of this first member is, as in art. 177,

n (3 — ')*-1 X+(2 — 2')» Y, provided that Y is the derivative of X. The factor 2 — occurs, then, (n − 1) times in this derivative of the first member, that is, once less than in the first member itself. The greatest common divisor of the first member and its derivative must, therefore, consist of the factors (x - 2) of

Examples of finding Equal Roots.

the first member, each being repeated once less than in the first member. No one of them is, then, a factor of the common divisor, unless it is more than once a factor of the first member, that is, unless it corresponds to one of the equal roots.

The equal roots of an equation are, therefore, obtained by finding the greatest common divisor of its first member and its derivative, and solving the equation obtained from putting this common divisor equal to zero.

279. Corollary. The common divisor must, itself, have equal roots, whenever a root is more than twice a root of the given equation.

280. EXAMPLES.

1. Find all the roots of the equation

203 — 7 + 16 x 12 = 0 which has equal roots. Solution. The derivative of this equation is

3 22

14 x + 16, the greatest common divisor of which and the given first member is

- 2. The equation

-2=0, gives

X = 2. Now since the given equation has two roots equal to 2, it must be divisible by

(x - 2)2 = 22 - 4x +4,

Examples of finding Equal Roots.

and we have

23-7x2 + 16x— 12 = (x - 2)2 (2-3) = 0; whence

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is the other root of the given equation. 2. Find all the roots of the equation

27 — 925 +624 + 1523 — 12 22 - 7x+6=0 which has equal roots. Solution. The derivative of this equation is

7 26 — 45 x4 + 24 x3 + 45 x2 — 24 2 — 7, the greatest common divisor of which and the given equation gives

23 - 2-1+1=0, which is an equation of the third degree, and we may consider it as a new equation, the equal roots of which are to be found, if it has any. Now its derivative is

3 - 2 - 1, and the common divisor of this derivative and the first mem

ber gives

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1= 0, or x = 1. Hence the first member of

23 - 2 - +1=0 must be divisible by

(x - 1), and we have indeed

23 — 2 — x+1=(1 — 1)2(x+1)= 0. The equal roots of the given equation are, therefore,

x = 1, and = l; and its first member is divisible by

(x - 1)3 (v + 1)?,

Examples of finding Equal Roots.

- 3.

and is found by division to be

(x - 1)3 (x + 1)2 (22+2-6). The remaining roots are, therefore, found from solving the quadratic equation

+ -6 = 0, which gives

x = 2, or = 3. Find all the roots of the equation

23 +3_-9x -- 27 = 0 which has equal roots.

Ans. = 3, or = 3. 4. Find all the roots of the equation

23 — 15.22 +752 — 125 = 0 which has equal roots.

Ans. x = 5. 5. Find all the roots of the equation

9.23 + 29 x — 39x + 18= 0 which has equal roots. Ans. x=1, or = 2, or = 3.

eq 6. Find all the roots of the equation

24 — 2 2:3 — 59.22 + 60x + 900 = 0 which has equal roots.

Ans. x=6, or =-5. 7. Find all the roots of the equation

24 - 6x2 8x —3=0 which has equal roots.

Ans. x=3, or =-1. 8. Find all the roots of the equation

24 + 12 23 +54 22 + 108 x +81 = 0 which has equal roots.

Ans. x = - 3.

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