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. $|\frac{r}{p}-7|<4\sqrt{3}$
• C. All $p$ and $r$
• D. No $p$ and $r$

Answer 1

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

Given, $p,q$ and $r$ are in $AP$
$\therefore 2q=p+r$
The roots of $p{x}^2+qx+r=0$ are real, if
$q^2-4pr\ge 0$
$\Rightarrow {\left(\frac{p+r}{2}\right)}^2-4pr\ge 0$
$\therefore {\left(\frac{r}{p}\right)}^2-14\left(\frac{r}{p}\right)+1\ge 0$
$\Rightarrow |\frac{r}{p}-7|\ge 4\sqrt{3}$.