Question

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

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

Answer 1

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

$\because$ p,q,r are in AP.
$\therefore q=\frac{p+r}{2}$ $[\because p+r=2q]$
For the real roots $q^2-4pr\ge 0$
$\Rightarrow (\frac{p+r}{2}{)}^2-4pr\ge 0\Rightarrow p^2+r^2-14pr\ge 0\Rightarrow (\frac{p}{r}{)}^2-14\left(\frac{p}{r}\right)+1\ge 0\Rightarrow (\frac{p}{r}-7{)}^2\ge 48\Rightarrow |\frac{p}{r}-7|\ge 4\sqrt{3}$