If the roots of the quadratic equation ax^2+b | vedclass

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(C) Given that the quadratic equation $ax^2+bx+c=0$ has imaginary roots,the discriminant $D = b^2 - 4ac < 0$,which implies $b^2 < 4ac$.Let $f(x) = 3a^

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$> -4ac$

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(C) Given that the quadratic equation $ax^2+bx+c=0$ has imaginary roots,the discriminant $D = b^2 - 4ac Let $f(x) = 3a^2x^2 + 6abx + 2b^2$.Since the coefficient of $x^2$ is $3a^2 > 0$,the expression has a minimum value.The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $\frac{4AC - B^2}{4A}$.Here,$A = 3a^2$,$B = 6ab$,and $C = 2b^2$.Minimum value $= \frac{4(3a^2)(2b^2) - (6ab)^2}{4(3a^2)} = \frac{24a^2b^2 - 36a^2b^2}{12a^2} = \frac{-12a^2b^2}{12a^2} = -b^2$.Since $b^2  -4ac$.Thus,the minimum value is greater than $-4ac$.

If the roots of the quadratic equation $ax^2+bx+c=0$ are imaginary,then for all real values of $x$,the minimum value of the expression $3a^2x^2+6abx+2b^2$ is

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