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(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$. From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and

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$\frac{2}{9}(2p-q)(2q-p)$

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(D) Given that $\alpha$ and $\beta$ are the roots of $x^2-px+r=0$. From the relation between roots and coefficients,we have $\alpha+\beta=p$ $(i)$ and $\alpha\beta=r$. Given that $\frac{\alpha}{2}$ and $2\beta$ are the roots of $x^2-qx+r=0$. From the relation between roots and coefficients,we have $\frac{\alpha}{2}+2\beta=q$,which implies $\alpha+4\beta=2q$ $(ii)$. Also,the product of roots is $\frac{\alpha}{2} 	imes 2\beta = r$,so $\alpha\beta=r$. Subtracting $(i)$ from $(ii)$,we get $(\alpha+4\beta)-(\alpha+\beta)=2q-p$,which simplifies to $3\beta=2q-p$,so $\beta=\frac{2q-p}{3}$. Substituting $\beta$ into $(i)$,we get $\alpha=p-\frac{2q-p}{3} = \frac{3p-2q+p}{3} = \frac{4p-2q}{3} = \frac{2(2p-q)}{3}$. Since $r=\alpha\beta$,we have $r = \left(\frac{2(2p-q)}{3}ight) \left(\frac{2q-p}{3}ight) = \frac{2}{9}(2p-q)(2q-p)$.

Let $\alpha, \beta$ be the roots of the equation $x^2-px+r=0$ and $\frac{\alpha}{2}, 2\beta$ be the roots of the equation $x^2-qx+r=0$. Then the value of $r$ is

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