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(C) Given the roots of $x^2+px+q=0$ are $\alpha, \beta$,we have $\alpha+\beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2-rx+s=0$ are $\alpha^

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Both real

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(C) Given the roots of $x^2+px+q=0$ are $\alpha, \beta$,we have $\alpha+\beta = -p$ and $\alpha\beta = q$.Given the roots of $x^2-rx+s=0$ are $\alpha^4, \beta^4$,we have $\alpha^4+\beta^4 = r$ and $\alpha^4\beta^4 = s$.For the equation $x^2-4qx+2q^2-r=0$,the discriminant $D$ is given by:$D = (-4q)^2 - 4(1)(2q^2-r) = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$.Substituting $q = \alpha\beta$ and $r = \alpha^4+\beta^4$:$D = 8(\alpha\beta)^2 + 4(\alpha^4+\beta^4) = 4(2\alpha^2\beta^2 + \alpha^4+\beta^4) = 4(\alpha^2+\beta^2)^2$.Since $(\alpha^2+\beta^2)^2 \geq 0$,$D \geq 0$.Thus,the roots of the equation are always real.

If the roots of the equation $x^2+px+q=0$ are $\alpha$ and $\beta$ and the roots of the equation $x^2-rx+s=0$ are $\alpha^4$ and $\beta^4$,then the roots of the equation $x^2-4qx+2q^2-r=0$ will be

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