If the roots of the equation ax2-2bx-c-0 and bx2-2-acx-b-0 are simultaneously real then prove that b2-ac-

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Question

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

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Solution

Given, the roots of both the equations are real.

First equation:

$a x^{2} + 2 b x + c = 0$

Its discriminant, $D \geq 0$

⇒ $(2 b)^{2} - 4 (a c) \geq 0$

⇒ $4 b^{2} - 4 a c \geq 0$

⇒ $4 b^{2} \geq 4 a c$

⇒ $b^{2} \geq a c$ ... (1)

Second equation:

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

Its discriminant, $D \geq 0$

⇒ $(2 \sqrt{a c})^{2} - 4 (b^{2}) \geq 0$

⇒ $4 a c - 4 b^{2} \geq 0$

⇒ $4 a c \geq 4 b^{2}$

⇒ $a c \geq b^{2}$ ... (2)

The results of equation (1) and (2) are simultaneously possible in only one case when $b^{2} = a c .$

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If the roots of the equation ax2+2bx+c=0 and bx2−2√acx+b=0 are simultaneously real then prove that b2=ac.

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