Question

If $p,q$ and $r$ are positive and are in AP, the roots of the quadratic equation $p{x}^2+qx+r=0$ are real for

• A. $|\frac{r}{p}-7|\ge 4\sqrt{3}$
• B. $|\frac{p}{r}-7|<4\sqrt{3}$
• C. all $p$ and $r$
• D. no $p$ and $r$

Answer 1

The correct option is A $|\frac{r}{p}-7|\ge 4\sqrt{3}$

$p,q,r$ are in AP
$\therefore 2q=p+r$
Since, $p{x}^2+qx+r=0$ has real roots
So, $D\ge 0$
$\therefore q^2-4pr\ge 0$
$\Rightarrow {\left(\frac{p+r}{2}\right)}^2-4pr\ge 0$
$\Rightarrow p^2+r^2-14pr\ge 0$
$\Rightarrow \frac{p^2}{r^2}+1-14\frac{p}{r}\ge 0$
$\Rightarrow {(\frac{p}{r}-7)}^2\ge 48$
$\Rightarrow |\frac{p}{r}-7|\ge 4\sqrt{3}$