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 is (a/x-a)+(b/x-b)=1 ⇒ a(x-b)+b(x-a)=(x-a)(x-b) ⇒ x2-x(a+b)+a b=a x-a b+b x-ab ⇒ x2-2 x(a+b)+3 a b=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 is $\frac{a}{x - a} + \frac{b}{x - b} = 1$
$\Rightarrow a (x - b) + b (x - a) = (x - a) (x - b)$
$\Rightarrow x^{2} - x (a + b) + ab = a x - ab + b x - ab$
$\Rightarrow x^{2} - 2 x (a + b) + 3 ab = 0$
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