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(D) For the equation $x^2-px+r=0$,the roots are $\alpha$ and $\beta$. Thus,$\alpha+\beta=p$ and $\alpha\beta=r$.For the equation $x^2-qx+r=0$,the root

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

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(D) For the equation $x^2-px+r=0$,the roots are $\alpha$ and $\beta$. Thus,$\alpha+\beta=p$ and $\alpha\beta=r$.For the equation $x^2-qx+r=0$,the roots are $\frac{\alpha}{2}$ and $2\beta$. Thus,$\frac{\alpha}{2}+2\beta=q$ and $\left(\frac{\alpha}{2}\right)(2\beta)=r$.From the product of roots,we have $\alpha\beta=r$,which is consistent.We have the system of linear equations:$1) \alpha+\beta=p$$2) \frac{\alpha}{2}+2\beta=q \Rightarrow \alpha+4\beta=2q$Subtracting $(1)$ from $(2)$,we get $3\beta=2q-p \Rightarrow \beta=\frac{2q-p}{3}$.Substituting $\beta$ into $(1)$,$\alpha=p-\frac{2q-p}{3} = \frac{3p-2q+p}{3} = \frac{4p-2q}{3} = \frac{2(2p-q)}{3}$.Since $r=\alpha\beta$,we have $r = \left(\frac{2(2p-q)}{3}\right) \left(\frac{2q-p}{3}\right) = \frac{2}{9}(2p-q)(2q-p)$.

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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