Q. If $\alpha,\beta$ are the roots of the equation $ax^{2} + bx + c = 0$, then the value of $\frac{1}{a\alpha + b} + \frac{1}{a\beta + b}$ is equal to
A) $\frac{ac}{b}$
B) 1
C) $\frac{ab}{c}$
D) $\frac{b}{ac}$
Explanation
(d)
Here, $\alpha + \beta = - \frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$
$\therefore\ \frac{1}{a\alpha + b} + \frac{1}{a\beta + b} = \frac{a(\alpha + \beta) + 2b}{a^{2}\alpha\beta + ab(\alpha + \beta) + b^{2}}$
$= \frac{a\left( - \frac{b}{a} \right) + 2b}{a^{2}\left( \frac{c}{a} \right) + ab\left( - \frac{b}{a} \right) + b^{2}} = \frac{b}{ac}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.