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. $b^2=2m^2$
• B. $b^2=m^2$
• C. $2b^2=m^2$
• D. $b^2=4m^2$

Answer 1

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

Clearly, $x=m$ is a root of the equation.
Therefore, the other root must be $-m$.
Putting the value of $x=-m$ in the given equation, we get
$\frac{1}{-m}+\frac{1}{-m+b}=\frac{1}{m}+\frac{1}{m+b}\Rightarrow \frac{2}{m}=\frac{1}{b-m}-\frac{1}{b+m}\Rightarrow \frac{2}{m}=\frac{2m}{b^2-m^2}\Rightarrow m^2=b^2-m^2\Rightarrow b^2=2m^2$