Q. If the roots of the equation $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$ be equal in magnitude but opposite in sign, then $\alpha + \beta$ is equal to
A) 0
B) 1
C) 2
D) None of these
Explanation
(a)
Given equation $\frac{\alpha}{x - \alpha} + \frac{\beta}{x - \beta} = 1$ can be rewritten as
$x^{2} - 2(\alpha + \beta)x + 3\alpha\beta = 0$
Let its roots be $\alpha'$ and $- \alpha'$.
$\Rightarrow \ \alpha^{'} + \left( - \alpha^{'} \right) = 2(\alpha + \beta)$
$\Rightarrow 0 = 2(\alpha + \beta)$
$\Rightarrow \ \alpha + \beta = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.