Q. If $p,q,r$ are positive and are in AP, then roots of the quadratic equation $px^{2} + qx + r = 0$ are complex for
A) $\left| \frac{r}{p} - 7 \right| \geq 4\sqrt{3}$
B) $\left| \frac{p}{r} - 7 \right| < 4\sqrt{3}$
C) All $p$ and $r$
D) No $p$ and $r$
Explanation
(b)
Since, $2q = p + r$
Given that, $px^{2} + qx + r = 0$ has complex roots
$\therefore\ \ \ \ \ D < 0$
$\Rightarrow q^{2} - 4pr < 0$
$\Rightarrow \ \left( \frac{p + r}{2} \right)^{2} - 4pr < 0$
$\Rightarrow \ \ p^{2} + r^{2} - 14pr < 0$
$\Rightarrow \ \ \frac{p^{2}}{r^{2}} + 1 - \frac{14p}{r} < 0$
$\Rightarrow \ \ \left( \frac{p^{2}}{r^{2}} - \frac{14\ p}{r} + 49 \right) - 48 < 0$
$\Rightarrow \ \left( \frac{p}{r} - 7 \right)^{2} < 48\ \ \ \Rightarrow \ \ \ \ \left| \frac{p}{r} - 7 \right| < 4\sqrt{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.