Question

If the roots of the equations $a{x}^2+2bx+c=0$ and $b{x}^2-2\sqrt{ac}x+b=0$ are simultaneously real, then prove that $b^2=ac$.

Answer 1

Given: the roots of the equations $a{x}^2+2bx+c=0$ and $b{x}^2-2\sqrt{ac}x+b=0$ are simultaneously real.

Let $D_1$ and $D_2$ be the discriminants of the given equations respectively, then
$D_1\ge 0$ and $D_2\ge 0$

$\Rightarrow 4b^2-4ac\ge 0$ and $4ac-4b^2\ge 0$
$\Rightarrow b^2\ge ac$ and $ac\ge b^2$
$\Rightarrow b^2=ac$

Hence proved.