Question

The sum of the roots of $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero. The product of the roots is

• A. $-\frac{1}{2}(a^2+b^2)$
• B. 0
• C. $\frac{1}{2}(a+b)$
• D. $2(a^2+b^2)$

Answer 1

Answer: (A)

$-\frac{1}{2}(a^2+b^2)$

Consider the given equation.

$\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$

$\frac{x+a+x+b}{(x+a)(x+b)}=\frac{1}{c}$

$x^2+(a+b-2c)x+[ab-(a+b\left)c\right]=0$ $..........\left(1\right)$

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

Since,

$\alpha +\beta =0$

$a+b-2c=0$

$2c=a+b$

And $\alpha \beta =ab-(a+b)c$

$\alpha \beta =ab-(a+b)\frac{(a+b)}{2}$

$\alpha \beta =-\frac{(a^2+b^2)}{2}$

Hence, this is the answer.