Let q be the maximum integral value of p in [0,10] for which the roots of the equation x^2…

Question

Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-px+\dfrac{5p}{4}=0$ are rational. Then the area of the region $\{(x,y): 0\le y\le(x-q)^2,\ 0\le x\le q\}$ is:

Accepted Answer

Discriminant $D=p^2-5p=p(p-5)$ must be a non-negative perfect square. Testing integers in $[0,10]$: $p=9$ gives $D=36=6^2$ (and $p=10$ gives $50$, not a perfect square), so $q=9$.

$$\text{Area}=\int_0^9(x-9)^2\,dx=\left[\frac{(x-9)^3}{3}\right]_0^9=\frac{729}{3}=243.$$

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