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(D) Given that $\alpha, \beta$ are the roots of $x^2+p x+q=0$,we have $\alpha+\beta=-p$ and $\alpha \beta=q$. Since $\alpha^4, \beta^4$ are roots of $

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two real roots

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(D) Given that $\alpha, \beta$ are the roots of $x^2+p x+q=0$,we have $\alpha+\beta=-p$ and $\alpha \beta=q$. Since $\alpha^4, \beta^4$ are roots of $x^2-r x+s=0$,we have $\alpha^4+\beta^4=r$ and $\alpha^4 \beta^4=s$. Consider the quadratic equation $x^2-4 q x+2 q^2-r=0$. The discriminant $D$ is given by $D = (-4q)^2 - 4(1)(2q^2 - r)$. $D = 16q^2 - 8q^2 + 4r = 8q^2 + 4r$. Since $\alpha, \beta$ are real,$\alpha^4, \beta^4 \geq 0$,so $r = \alpha^4 + \beta^4 \geq 0$. Also,$8q^2 \geq 0$. Thus,$D = 8q^2 + 4r \geq 0$. Since the discriminant is non-negative,the equation $x^2-4 q x+2 q^2-r=0$ always has two real roots.

If $\alpha, \beta$ are the real roots of $x^2+p x+q=0$ and $\alpha^4, \beta^4$ are the roots of $x^2-r x+s=0$,then the equation $x^2-4 q x+2 q^2-r=0$ has always

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