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. $ab$
• D. $\frac{-1}{2}ab$

Answer 1

Answer: (B)

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