Question

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

• A. $-\frac{1}{2}(a^2+b^2)$
• B. $\frac{1}{2}(a^2+b^2)$
• C. $-\frac{3}{2}(a^2+b^2)$
• D. None

Answer 1

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

$\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$
$\Rightarrow \frac{x+b+x+a}{(x+a)(x+b)}=\frac{1}{c}$
$\Rightarrow 2cx+ac+bc=x^2+ax+bx+ab$
$\Rightarrow x^2(a+b-2c)x+ab-ac-bc=0$
Roots are equal in magnitude and opposite in sign
$\Rightarrow$ Coefficient of x is 0
$\Rightarrow a+b-2c=0$
$\Rightarrow 2c=a+b$
$\Rightarrow c=\frac{a+b}{2}$
Product of the roots $=ab-ac-bc=ab-c(a+b)$
$=ab-(a+b)\left(\frac{a+b}{2}\right)$
$=\frac{1}{2}[2ab-a^2-b^2-2ab]$
$=-\frac{1}{2}(a^2+b^2)$