If the roots of the equation (a/x-a)+(b/x-b)=1 are equal in magnitude and opposite in sign, then

Question

If the roots of the equation (a/x-a)+(b/x-b)=1 are equal in magnitude and opposite in sign, then (A) a=b (B) a+b=1 (C) a-b=1 (D) a+b=0

Tardigrade Admin · Accepted Answer

Given equation i s (a/x-a)+(b/x-b)=1 ⇒ a(x-b)+b(x-a) =(x-a)(x-b) ⇒ x2-x(a+b)+ab =ax-ab+bx-ab ⇒ x2-2x(a+b)+3ab=0 So, sum of roots =α+(-α)=2(a+b) or a+b=0.

Q. If the roots of the equation $\frac{a}{x - a} + \frac{b}{x - b} = 1$ are equal in magnitude and opposite in sign, then

$a = b$

$a + b = 1$

$a - b = 1$

$a + b = 0$

Given equation i s $\frac{a}{x - a} + \frac{b}{x - b} = 1$
$\Rightarrow a (x - b) + b (x - a)$
$= (x - a) (x - b)$
$\Rightarrow x^{2} - x (a + b) + ab$
$= a x - ab + b x - ab$
$\Rightarrow x^{2} - 2 x (a + b) + 3 ab = 0$
So, sum of roots $= α + (- α) = 2 (a + b)$
or $a + b = 0$ .

Q. If the roots of the equation x−aa​+x−bb​=1 are equal in magnitude and opposite in sign, then

a=b

a+b=1

a−b=1

a+b=0