Q. Let $a,\ b,\ c$ be real, if $ax^{2} + bx + c = 0$ has two real roots $\alpha$ and $\beta$, where $\alpha < - 2$ and $\beta > 2$, then

A) $4 - \frac{2b}{a} + \frac{c}{a} < 0$
B) $4 + \frac{2b}{a} - \frac{c}{a} < 0$
C) $4 - \frac{2b}{a} + \frac{c}{a} = 0$
D) $4 + \frac{2b}{a} + \frac{c}{a} = 0$
Explanation

(a)
Given, $a,\ b,\ c$ are real, $ax^{2} + bx + c = 0$ has two real roots $\alpha$ and $\beta$, where $\alpha < - 2$ and $\beta > 2$
$\Rightarrow \ \ f( - 2) < 0$ and $f(2) > 0$
$\Rightarrow \ \ \ \ \ 4a - 2b + c < 0$ and $4a + 2b + c > 0$
$\Rightarrow \ \ \ 4 - \frac{2b}{a} + \frac{c}{a} < 0$ and $4 + \frac{2b}{a} + \frac{c}{a} > 0$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.