Question

If the roots of the equation $p{x}^2+2qx+r=0$ and $q{x}^2-2\sqrt{pr}x+q=0$ be real, then

• A. $p=q$
• B. $q^2=pr$
• C. $p^2=qr$
• D. $r^2=pq$

Answer 1

The correct option is B $q^2=pr$

Equation $p{x}^2+2qx+r=0$ and $q{x}^2-2\sqrt{pr}x+q=0$ have real roots, then

$b^2-4ac\ge 0$ for both equations

From first

$4q^2-4pr\ge 0\Rightarrow q^2\ge pr$ ...............(1)
and from second

$4\left(pr\right)-4q^2\ge 0$ (for real root)
$\Rightarrow pr\ge q^2$ ...............(2)

From (1) and (2), we get result $q^2=pr$