(A) For the equation $ax^2 + 2bx + c = 0$,the roots are real,so the discriminant $D_1 \ge 0$.$D_1 = (2b)^2 - 4ac = 4b^2 - 4ac \ge 0 \implies b^2 \ge a

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(A) For the equation $ax^2 + 2bx + c = 0$,the roots are real,so the discriminant $D_1 \ge 0$.$D_1 = (2b)^2 - 4ac = 4b^2 - 4ac \ge 0 \implies b^2 \ge ac$.For the equation $bx^2 - 2\sqrt{ac}x + b = 0$,the roots are real,so the discriminant $D_2 \ge 0$.$D_2 = (-2\sqrt{ac})^2 - 4(b)(b) = 4ac - 4b^2 \ge 0 \implies ac \ge b^2$.Since $b^2 \ge ac$ and $ac \ge b^2$,it must be that $b^2 = ac$.

Class 11 Mathematics 4-2.Quadratic Equations and Inequations Solution of quadratic equations and Nature of roots

Easy

If the roots of the equations $ax^2 + 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real,then:

A
$ac = b^2$
B
$4b^2 - ac = 0$
C
$a = b, c = 0$
D
$a = b = c$

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4-2.Quadratic Equations and Inequations

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Solution of quadratic equations and Nature of roots

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If the roots of the equations $ax^2 + 2bx + c = 0$ and $bx^2 - 2\sqrt{ac}x + b = 0$ are real,then:

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Find the complete set of values of $k$ for which the equation $4^x - (k + 2)2^x + 2k = 0$ has exactly one positive root.

The number of real solutions of the equation $3(x^2 + \frac{1}{x^2}) - 2(x + \frac{1}{x}) + 5 = 0$ is:

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