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

It is given that the roots of the equation $a{x}^2+2bx+c=0$ are real.
$$\begin{gathered} \therefore D_1={\left(2b\right)}^2-4\times a\times c\ge 0 \\ \Rightarrow 4\left(b^2-ac\right)\ge 0 \\ \Rightarrow b^2-ac\ge 0.....\left(1\right) \end{gathered}$$

Also, the roots of the equation $b{x}^2-2\sqrt{ac}x+b=0$ are real.
$$\begin{gathered} \therefore D_2={\left(-2\sqrt{ac}\right)}^2-4\times b\times b\ge 0 \\ \Rightarrow 4\left(ac-b^2\right)\ge 0 \\ \Rightarrow -4\left(b^2-ac\right)\ge 0 \end{gathered}$$
$\Rightarrow b^2-ac\le 0.....\left(2\right)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if
$$\begin{gathered} b^2-ac=0 \\ \Rightarrow b^2=ac \end{gathered}$$