Question

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

• A.
• B.
• C. All p and r
• D. No p and r

Answer 1

The correct option is A

p,q,r are positive and are in A.P.
$\therefore q=\frac{p+r}{2}$ .....(i)
$\because$ The roots of $p{x}^2+qx+r=0$ are real
$\Rightarrow q^2\ge 4pr\Rightarrow {\left[\frac{p+r}{2}\right]}^2\ge 4pr$ [using(i)]
$\Rightarrow p^2+r^2-14pr\ge 0$
$\Rightarrow {\left(\frac{r}{p}\right)}^2-14\left(\frac{r}{p}\right)+1\ge 0$ $(\because p>0$ and $p\ne 0)$
$\Rightarrow {(\frac{r}{p}-7)}^2-48\ge 0\Rightarrow {(\frac{r}{p}-7)}^2-(4\sqrt{3}{)}^2\ge 0$
$\Rightarrow |\frac{r}{p}-7|\ge 4\sqrt{3}.$