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 $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is :

Accepted Answer

Given equation : $4{x^2} - 4px + 5p = 0$

for rational roots, D must be perfect square

$$D = 16{p^2} - 4 \times 4 \times 5p = 16p(p - 5)$$

So, max. Integral value of $p = 9$ for making D is perfect square

$$\therefore$$ $q = 9$

Area of shared region

$$ = \int\limits_0^9 {{{(x - 9)}^2}dx} $$

$$ = \left. {{{{{(x - 9)}^3}} \over 3}} \right|_0^9 = {{{9^3}} \over 3} = 243$$ sq. units

Solutions · 1