Question

The roots of $p{x}^2+2qx+r=0$ and $q{x}^2-2\sqrt{pr}x+q=0$ are simultaneously real, then

• A. $p=q,r\ne 0$
• B. $\frac{p}{q}=\frac{q}{r}$
• C. $2q=\pm \sqrt{pr}$
• D. none of these

Answer 1

The correct option is B $\frac{p}{q}=\frac{q}{r}$

The discriminant of the first equation is
$4q^2-4pr\ge 0$
$⟹q^2-pr\ge 0$ ------(1)
The discriminant of the second equation is
$4pr-4q^2\ge 0$
$⟹pr-q^2\ge 0$

$⟹q^2-pr\le 0$ --------(2)
Therefore, from (1) and (2)
$q^2-pr=0$
$⟹\frac{p}{q}=\frac{q}{r}$.