Question

If the roots of the equation $\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ are equal in magnitude but opposite in sign, then find their product

Answer 1

We have,

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

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

$\Rightarrow \frac{2x+b+a}{x^2+ax+bx+ab}=\frac{1}{c}$

$\Rightarrow 2xc+bc+ca=x^2+ax+bx+ab$

$\Rightarrow x^2+ax+bx+ab=2xc+bc+ca$

$\Rightarrow x^2+(a+b-2c)x+(ab-bc-ca)=0$

Sum of roots $=-\frac{coeff.ofx}{coeff.of{x}^2}=-\frac{(a+b-2c)}{1}=2c-a-b$

According to given,

Sum of roots $=0$

$2c-a-b=0$

$a+b=2c$

Product of roots $=\frac{\text{constant}term}{coeff.of{x}^2}=\frac{ab-bc-ca}{1}=ab-bc-ca$

Product of roots $=$

$ab-bc-ca$

$=ab-c(b+a)$

$=ab-c\left(2c\right)$

$=ab-2c^2$

Hence, this is the answer.