Let q be the maximum integral value of p in 0 , 10 for which the roots of the equation x 2 - p x + 5 4 p = 0…

$x^2-px+\frac{5p}{4}=0$

For rational roots, $D$ must be a perfect square.

$D=p^2-5p=p\left(p-5\right)$

Now put values in reverse order as $q$ is the maximum value of $p$

At $p=10$,

$D=50\to \text{not a perfect square}$

At $p=9$,

$D=36\to \text{perfect square}$

So, the maximum integral value of $p$ is $9$

$\therefore q=9$

$0\le y\le {\left(x-9\right)}^2$

Area of shaded region $={\int }_0^9{\left(x-9\right)}^{2}dx$

$={\left[\frac{{\left(x-9\right)}^3}{3}\right]}_0^9=243 {\text{unit}}^2$