Question

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

• A. $0$
• B. $\frac{2ab}{b+c}$
• C. $\frac{2bc}{b+c}$
• D. $-\frac{2bc}{b+c}$

Answer 1

The correct option is D $-\frac{2bc}{b+c}$

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

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

$(b-a)(x+c)=x^2+ax+bx+ab$

$bx+bc-ax-ac=x^2+ax+bx+ab$

$x^2+2ax+ab+ac-bc=0$

Given that product of roots $=0$

$a=0$

$ab+ac=bc$

$a(b+c)=bc$

$a=\frac{bc}{b+c}$

Sum of roots $=-2a$

$=\frac{-2bc}{b+c}$