If the roots of the equation x^2+px+q=0 are a | vedclass

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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 $\alpha^4\beta^4 = s$. Consider the quadratic 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 $r = \alpha^4+\beta^4$,we get $D = 8(\alpha\beta)^2 + 4(\alpha^4+\beta^4) = 4(2\alpha^2\beta^2 + \alpha^4+\beta^4) = 4(\alpha^2+\beta^2)^2$. Since $(\alpha^2+\beta^2)^2 \geq 0$,it follows that $D \geq 0$. Therefore,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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