If the equation x^{2}-cx+d=0 has roots equal | vedclass

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(D) Let $\alpha$ and $\beta$ be the roots of $x^{2}+ax+b=0$ and the roots of $x^{2}-cx+d=0$ be $\alpha^{4}$ and $\beta^{4}$.
From the relations betwe

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one positive and one negative

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(D) Let $\alpha$ and $\beta$ be the roots of $x^{2}+ax+b=0$ and the roots of $x^{2}-cx+d=0$ be $\alpha^{4}$ and $\beta^{4}$.
From the relations between roots and coefficients:
$\alpha+\beta=-a, \alpha\beta=b$ ...$(i)$
$\alpha^{4}+\beta^{4}=c, \alpha^{4}\beta^{4}=d$ ...$(ii)$
From $(ii),$ $c = \alpha^{4}+\beta^{4} = (\alpha^{2}+\beta^{2})^{2}-2(\alpha\beta)^{2} = ((\alpha+\beta)^{2}-2\alpha\beta)^{2}-2(\alpha\beta)^{2}$.
Substituting $(i)$: $c = (a^{2}-2b)^{2}-2b^{2} = a^{4}+4b^{2}-4a^{2}b-2b^{2} = a^{4}-4a^{2}b+2b^{2}$.
Thus, $2b^{2}-c = 2b^{2}-(a^{4}-4a^{2}b+2b^{2}) = 4a^{2}b-a^{4} = a^{2}(4b-a^{2})$.
Given $a^{2}>4b$, we have $4b-a^{2} < 0$, so $2b^{2}-c < 0$.
For the equation $x^{2}-4bx+(2b^{2}-c)=0$, the product of roots is $2b^{2}-c < 0$.
Since the product of the roots is negative, one root must be positive and the other must be negative.

If the equation $x^{2}-cx+d=0$ has roots equal to the fourth powers of the roots of $x^{2}+ax+b=0,$ where $a^{2}>4b,$ then the roots of $x^{2}-4bx+2b^{2}-c=0$ will be

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