If the roots of the equations


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



It is given that the roots of the equation $a x^{2}+2 b x+c=0$ are real.

$\therefore D_{1}=(2 b)^{2}-4 \times a \times c \geq 0$

$\Rightarrow 4\left(b^{2}-a c\right) \geq 0$

$\Rightarrow b^{2}-a c \geq 0$    $\ldots \ldots(1)$

Also, the roots of the equation $b x^{2}-2 \sqrt{a c} x+b=0$ are real.

$\therefore D_{2}=(-2 \sqrt{a c})^{2}-4 \times b \times b \geq 0$

$\Rightarrow 4\left(a c-b^{2}\right) \geq 0$

$\Rightarrow-4\left(b^{2}-a c\right) \geq 0$

$\Rightarrow b^{2}-a c \leq 0$ $\ldots \ldots(2)$

The roots of the given equations are simultaneously real if (1) and (2) holds true together. This is possible if

$b^{2}-a c=0$

$\Rightarrow b^{2}=a c$

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