Let $\alpha, \beta$ be the roots of the equation $x^2-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots…
Let $\alpha, \beta$ be the roots of the equation $x^2-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^2-\mathrm{q} x+\mathrm{r}=0$. Then the value of r is
$\frac{2}{9}(p-q)(2 q-p)$
$\frac{2}{9}(q-p)(2 p-q)$
$\frac{2}{9}(q-2 p)(2 q-p)$
$\frac{2}{9}(2 p-q)(2 q-p)$
Solution
$\alpha$ and $\beta$ are the roots of $x^2-\mathrm{p} x+\mathrm{r}=0$
$\therefore \quad$ sum of roots $=\alpha+\beta$
$\Rightarrow \alpha+\beta=p...(i)$
$\frac{\alpha}{2}, 2 \beta$ are the roots of $x^2-\mathrm{q} x+\mathrm{r}=0$
$\therefore \quad$ sum of roots $=\mathrm{q}$
$\begin{aligned}
& \Rightarrow \frac{\alpha}{2}+2 \beta=q \\
& \Rightarrow \alpha+4 \beta=2 q...(ii)
\end{aligned}$ Subtracting (i) from (ii), we get
$\begin{aligned}
& 3 \beta=2 q-p \\
& \Rightarrow \beta=\frac{2 q-p}{3}
\end{aligned}$ From (i),
$\begin{aligned}
& \alpha+\frac{2 q-p}{3}=p \\
& \Rightarrow \alpha=\frac{2(2 p-q)}{3}
\end{aligned}$
Product of roots $=\alpha \beta$
$\begin{aligned} \Rightarrow r=\alpha \beta & =\frac{2(2 p-q)}{3} \times \frac{2 q-p}{3} \\ & =\frac{2}{9}(2 p-q)(2 q-p)\end{aligned}$