Question

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

• A. $\mid \frac{r}{p}-7\mid \ge 4\sqrt{3}$
• B. $\mid \frac{q}{p}-4\mid \ge 2\sqrt{3}$
• C. $\mid \frac{p}{q}-4\mid \ge 4\sqrt{3}$
• D. $\mid \frac{p}{q}-1\mid \ge \frac{\sqrt{3}}{2}$

Answer 1

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

As $p$,$q$,$r$ are in $A.P.,$ then $2q=p+r$

And as roots of $p{x}^2+qx+r=0$ are real, then

$q^2-4pr\ge 0$

$\Rightarrow {(p+r)}^2-16pr\ge 0\Rightarrow p^2-14pr+r^2\ge 0$

$\Rightarrow {(\frac{p}{r}-7)}^2\ge 48\Rightarrow |\frac{p}{r}-7|\ge 4\sqrt{3}$