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(D) For the equation $x^2 - px + r = 0$,we have $\alpha + \beta = p$ and $\alpha \beta = r$. For the equation $x^2 - qx + r = 0$,we have $\frac{\alpha

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For the equation $x^2 - px + r = 0$,we have $\alpha + \beta = p$ and $\alpha \beta = r$. For the equation $x^2 - qx + r = 0$,we have $\frac{\alpha}{2} + 2\beta = q$ and $(\frac{\alpha}{2})(2\beta) = \alpha \beta = r$. From the first equation,$\alpha + \beta = p \implies 2\alpha + 2\beta = 2p$. Subtracting $\frac{\alpha}{2} + 2\beta = q$ from $2\alpha + 2\beta = 2p$,we get: $(2\alpha - \frac{\alpha}{2}) = 2p - q \implies \frac{3\alpha}{2} = 2p - q \implies \alpha = \frac{2(2p - q)}{3}$. Now,$\beta = p - \alpha = p - \frac{2(2p - q)}{3} = \frac{3p - 4p + 2q}{3} = \frac{2q - p}{3}$. Since $r = \alpha \beta$,we have: $r = \left(\frac{2(2p - q)}{3}\right) \left(\frac{2q - p}{3}\right) = \frac{2}{9}(2p - q)(2q - p)$.

If $\alpha, \beta$ are the roots of the equation $x^2 - px + r = 0$ and $\alpha/2, 2\beta$ are the roots of the equation $x^2 - qx + r = 0$,then what is the value of $r$?

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