Question

If the sum of the two roots of the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero, then the product of the two roots is

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

Answer 1

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

Simplifying we get
$\frac{2x+a+b}{x^2+(a+b)x+ab}=\frac{1}{c}$
$\Rightarrow x^2+x(a+b)+ab=2cx+bc+ac$
$\Rightarrow x^2+x(a+b-2c)+ab-bc-ac=0$
Hence it is given that the sum of roots is 0.
Now, we get
$a+b-2c=0$
$c=\frac{a+b}{2}$
Therefore, product of roots is
$ab-(a+b)c$
$=ab-\frac{(a+b{)}^2}{2}$
$=\frac{2ab-(a+b{)}^2}{2}$
$=\frac{-(a^2+b^2)}{2}$
Hence,answer is $\frac{-(a^2+b^2)}{2}$