Q. If $\alpha,\ \beta$ be the roots of the quadratic equation $ax^{2} + bx + c = 0\ and\ k$ be a real number, then the condition so that $\alpha < k < \beta$ is given by
A) $ac > 0$
B) $ak^{2} + bk + c = 0$
C) $ac < 0$
D) $a^{2}k^{2} + abk + ac < 0$
Explanation
(d)
Since, $\alpha,\ \beta$ are the roots of the equation $ax^{2} + bx + c = 0.$
$\therefore\ ax^{2} + bx + c = a(x - \alpha)(x - \beta)$
$\Rightarrow \ \alpha,\ \beta$ be the roots of $ax^{2} + bx + c = 0$. Also $\alpha < k < \beta$
So, $a(k - \alpha)(k - \beta) < 0$
Also, $a^{2}k^{2} + abk + ac = a\left( ak^{2} + bk + c \right) = a^{2}(k - \alpha)(k - \beta) < 0$
$\Rightarrow \ a^{2}k^{2} + abk + ac < 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.