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

The correct option is A $-\frac{1}{2}(a^2+b^2)$

$\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$
$c(x+a+x+b)=x^2+ax+bx+ab$
$x^2+x(a+b-2c)+ab-ac-bc=0$
Sum of roots is zero, hence $a+b=2c$
Product of roots $=ab-bc-ca$
$=$ $ab-\frac{(b+a)(a+b)}{2}$
$=$ $ab-\frac{a^2}{2}-\frac{b^2}{2}-ab$
$=$ $\frac{-a^2}{2}\frac{-b^2}{2}$

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