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Step-by-Step Solution
Step 1: Express the given quadratic equation in a standard form
The given equation is
$x^2 - p x + \frac{5}{4} \, p = 0.$
Multiply through by 4 to clear the fraction:
$$
4x^2 - 4px + 5p = 0.
$$
Step 2: Determine the condition for the roots to be rational
For the roots to be rational, the discriminant $D$ must be a perfect square. The discriminant of
$4x^2 - 4px + 5p = 0$
is given by
$$
D = (-4p)^2 - 4 \cdot (4) \cdot (5p)
= 16p^2 - 80p
= 16p(p - 5).
$$
For $D$ to be a perfect square, $16p(p - 5)$ must be a perfect square. We also have the restriction $p \in [0,10].$
Step 3: Find the maximum integral $p$ within $[0,10]$ for which $D$ is a perfect square
We check integer values of $p$ to ensure $16p(p - 5)$ is a perfect square. It turns out that $p = 9$ is the largest integer in $[0,10]$ that makes $16 \cdot 9 \cdot (9 - 5) = 16 \cdot 9 \cdot 4 = 576,$ which is $24^2,$ a perfect square. Hence,
$$
q = 9.
$$
Step 4: Set up the integral for the area
The region described is
$$
\{(x,y) : 0 \leq y \leq (x - q)^2,\; 0 \leq x \leq q\}.
$$
Substituting $q=9,$ the upper boundary for $y$ is $(x - 9)^2,$ and $x$ ranges from $0$ to $9.$ Therefore, the area $A$ is computed by the definite integral:
$$
A
= \int_{x=0}^{9} (x - 9)^2 \, dx.
$$
Step 5: Evaluate the integral
Expand and integrate:
$$
(x - 9)^2 = x^2 - 18x + 81.
$$
So,
$$
\int_{0}^{9} (x - 9)^2 \, dx
= \int_{0}^{9} (x^2 - 18x + 81) \, dx.
$$
Integrate term by term:
$$
\int x^2 \, dx = \frac{x^3}{3},
\quad
\int (-18x) \, dx = -9x^2,
\quad
\int 81 \, dx = 81x.
$$
Therefore,
$$
\int_{0}^{9} (x^2 - 18x + 81) \, dx
= \left[\frac{x^3}{3} - 9x^2 + 81x\right]_{0}^{9}.
$$
Step 6: Substitute the limits
Substitute $x = 9$ and then $x = 0$:
$$
\left(\frac{9^3}{3} - 9 \cdot 9^2 + 81 \cdot 9\right)
- \left(\frac{0^3}{3} - 9 \cdot 0^2 + 81 \cdot 0\right).
$$
Simplify:
$$
\frac{729}{3} - 9(81) + 81(9)
= 243 - 729 + 729
= 243.
$$
Hence, the area is
$$
A = 243.
$$
Final Answer
The area of the given region is $\boxed{243}$ square units.
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