Question

The condition that the equation $\frac{1}{x}+\frac{1}{x+b}=\frac{1}{m}+\frac{1}{m+b}$ has real roots that are equal in magnitude but opposite in sign is

• A. $b^2=m^2$
• B. $b^2=2m^2$
• C. $2b^2=m^2$
• D. $b^2=3m^2$

Answer 1

The correct option is B $b^2=2m^2$

It can be observed that one root is $m$.
Therefore, other root will be $-m$
$\Rightarrow \frac{1}{-m}+\frac{1}{-m+b}=\frac{1}{m}+\frac{1}{m+b}$
$\Rightarrow \frac{1}{b-m}-\frac{1}{b+m}=\frac{2}{m}$
$\Rightarrow \frac{b+m-b+m}{b^2-m^2}=\frac{2}{m}\Rightarrow 2m^2=b^2$