Question

If the roots of the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ are equal in magnitude but opposite in sign, then their product is

• A. $\frac{1}{2}\left(a^{2}+b^{2}\right)$
• B. $\frac{-1}{2}\left(a^{2}+b^{2}\right)$
• C. $a b$
• D. $\frac{-1}{2} a b$

Answer 1

Answer: (B)

We have, $((x+b)+(x+a) c=(x+a)(x+b)$
$\Rightarrow x^{2}+b x+a x-2 c x+a b-b c-c a=0$
Now, let roots be $\alpha$ and $\beta$, then
$\alpha+\beta=0, $
$\alpha \beta=a b-b c-a c$
$\alpha+\beta=0 $
$\Rightarrow b+a=2 c$
and $\alpha \beta=a b-(b+a) c$
$\Rightarrow \alpha \beta=a b-\frac{(a+b)^{2}}{2}$
$\Rightarrow \alpha \beta=\frac{1}{2}\left(-a^{2}-b^{2}\right)$
$\therefore \alpha \beta=-\frac{1}{2}\left(a^{2}+b^{2}\right)$