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

\(\alpha\) and \(\beta\) are the roots of \(x^2-\mathrm{p} x+\mathrm{r}=0\)
\(\therefore\) sum of roots \(=\alpha+\beta\)
\(\Rightarrow \alpha+\beta=p...(i)\)
\(\frac{\alpha}{2}, 2 \beta\) are the roots of \(x^2-\mathrm{q} x+\mathrm{r}=0\)
\(\therefore\) sum of roots \(=\mathrm{q}\)
\(\Rightarrow \frac{\alpha}{2}+2 \beta=q \)
\( \Rightarrow \alpha+4 \beta=2 q...(ii)\)
Subtracting (i) from (ii), we get
\(3 \beta=2 q-p\)
\(\Rightarrow \beta=\frac{2 q-p}{3}\)
From (i),
\(\alpha+\frac{2 q-p}{3}=p \)
\(\Rightarrow \alpha=\frac{2(2 p-q)}{3}\)
Product of roots \(=\alpha \beta\)
\(\Rightarrow r=\alpha \beta =\frac{2(2 p-q)}{3} \times \frac{2 q-p}{3} \)
\( =\frac{2}{9}(2 p-q)(2 q-p)\)