If the roots of the equations ax2 + 2bx + c = 0 and bx2−2ac−−√x+b=0

The given equations are

$a x^{2}+2 b x+c=0 \ldots \ldots$ (1)

$b x^{2}-2 \sqrt{a c} x+b=0 \ldots \ldots$ (2)

Roots are simultaneously real

Then prove that $b^{2}=a c$.

Let $D_{1}$ and $D_{2}$ be the discriminants of equation (1) and (2) respectively,

Then,

$D_{1}=(2 b)^{2}-4 a c$

$=4 b^{2}-4 a c$

And

$D_{2}=(-2 \sqrt{a c})^{2}-4 \times b \times b$

$=4 a c-4 b^{2}$

Both the given equation will have real roots, if $D_{1} \geq 0$ and $D_{2} \geq 0$

$4 b^{2}-4 a c \geq 0$

$4 b^{2} \geq 4 a c$

$b^{2} \geq a c$......(3)

$4 a c-4 b^{2} \geq 0$

$4 a c \geq 4 b^{2}$

$a c \geq b^{2} \ldots \ldots(4)$

From equations (3) and (4) we get

$b^{2}=a c$

Hence, $b^{2}=a c$