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

Let α, β be the roots of the equation x² - px + r = 0 and α/2, 2β be the roots of the equation x² - qx + r = 0. Then the value of r is

$\alpha, \beta:\ \text{roots of } x^2 - px + r = 0$

$\dfrac{\alpha}{2},\ 2\beta:\ \text{roots of } x^2 - qx + r = 0$

$\text{To find: the value of } r$

$\text{Both equations have the same constant term } r$

$\text{Use sum and product of roots (Vieta)}$

$\alpha + \beta = p$

$\alpha\beta = r$

$\dfrac{\alpha}{2} + 2\beta = q$

$\dfrac{\alpha}{2} \cdot 2\beta = r$

$\dfrac{\alpha}{2}\cdot 2\beta = \alpha\beta \Rightarrow \text{same } r$

$\text{Two equations in } \alpha, \beta:$

$\alpha + \beta = p$

$\dfrac{\alpha}{2} + 2\beta = q$

$\alpha = p - \beta$

$\dfrac{p - \beta}{2} + 2\beta = q$

$\dfrac{p}{2} - \dfrac{\beta}{2} + 2\beta = q$

$\dfrac{p}{2} + \dfrac{3\beta}{2} = q$

$\dfrac{3\beta}{2} = q - \dfrac{p}{2}$

$\beta = \dfrac{2q - p}{3}$

$\alpha = p - \beta = p - \dfrac{2q-p}{3}$

$\alpha = \dfrac{2(2p - q)}{3}$

$r = \alpha\beta = \dfrac{2(2p-q)}{3}\cdot\dfrac{2q-p}{3}$

$r = \dfrac{2}{9}(2p-q)(2q-p)$

$\alpha + \beta = \dfrac{2(2p-q) + (2q-p)}{3} = p\ \checkmark$

$\dfrac{\alpha}{2} + 2\beta = \dfrac{2p-q + 2(2q-p)}{3} = q\ \checkmark$

$\boxed{r = \dfrac{2}{9}(2p-q)(2q-p)}$

$\text{Option } (4)$