Question

The value of $m$ for which the equation $\frac{a}{x+a+m}+\frac{b}{x+b+m}=1$, has roots equal in magnitude but opposite in sign is

• A. $\frac{a+b}{a-b}$
• B. $0$
• C. $\frac{a-b}{a+b}$
• D. $\frac{2(a-b)}{a+b}$

Answer 1

Answer: (B)

$0$

From the given equation we get,

$a(x+b+m)+b(x+a+m)=(x+a+m)(x+b+m)$

$\therefore x^2+2mx+m-ab=0$

Let $\alpha$ and $\beta$ be the roots of the equation

Therefore, $\alpha +\beta =-2m$
But it is given that, $\alpha =-\beta$
Hence, $\alpha +\beta =-2m=0$
$\therefore m=0$