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

Options

A. 164
B. 243
C. `(125)/3`
D. 25

Answer

To solve the problem step by step, we need to find the maximum integral value of \( p \) in the interval \([0, 10]\) for which the roots of the quadratic equation

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

are rational.

### Step 1: Determine the condition for rational roots
For the roots of the quadratic equation \( ax^2 + bx + c = 0 \) to be rational, the discriminant \( D \) must be a perfect square. The discriminant \( D \) for our equation is given by:

\[ D = b^2 - 4ac = (-p)^2 - 4 \cdot 1 \cdot \frac{5}{4}p = p^2 - 5p \]

### Step 2: Set the discriminant as a perfect square
We need \( D \) to be a perfect square, so we set:

\[ p^2 - 5p = k^2 \]

for some integer \( k \). Rearranging gives us:

\[ p^2 - 5p - k^2 = 0 \]

### Step 3: Find the discriminant of this new quadratic
The discriminant of this quadratic in \( p \) must also be a perfect square for \( p \) to have integer solutions. The discriminant \( D' \) is:

\[ D' = (-5)^2 - 4 \cdot 1 \cdot (-k^2) = 25 + 4k^2 \]

### Step 4: Set \( D' \) as a perfect square
We need \( D' \) to be a perfect square:

\[ 25 + 4k^2 = m^2 \]

for some integer \( m \). Rearranging gives:

\[ m^2 - 4k^2 = 25 \]

This is a difference of squares:

\[ (m - 2k)(m + 2k) = 25 \]

### Step 5: Factor pairs of 25
The factor pairs of 25 are:

1. \( (1, 25) \)
2. \( (5, 5) \)
3. \( (-1, -25) \)
4. \( (-5, -5) \)

### Step 6: Solve for \( m \) and \( k \)
From each factor pair, we can solve for \( m \) and \( k \):

1. For \( (1, 25) \):
\[ m - 2k = 1, \quad m + 2k = 25 \]
Adding gives \( 2m = 26 \Rightarrow m = 13 \) and \( 4k = 24 \Rightarrow k = 6 \).

2. For \( (5, 5) \):
\[ m - 2k = 5, \quad m + 2k = 5 \]
This gives \( m = 5 \) and \( k = 0 \).

### Step 7: Calculate possible values of \( p \)
Substituting \( k = 6 \) and \( k = 0 \) back into \( p^2 - 5p - k^2 = 0 \):

1. For \( k = 6 \):
\[ p^2 - 5p - 36 = 0 \]
The discriminant is \( 25 + 144 = 169 \) (perfect square), giving:
\[ p = \frac{5 \pm 13}{2} \Rightarrow p = 9 \text{ or } p = -4 \]

2. For \( k = 0 \):
\[ p^2 - 5p = 0 \Rightarrow p(p - 5) = 0 \Rightarrow p = 0 \text{ or } p = 5 \]

### Step 8: Determine maximum integral value of \( p \)
The possible integral values of \( p \) are \( 0, 5, 9 \). The maximum integral value of \( p \) in \([0, 10]\) is \( q = 9 \).

### Step 9: Calculate the area of the region
The area of the region defined by \( 0 \leq y \leq (x - q)^2 \) and \( 0 \leq x \leq q \) is:

\[ \text{Area} = \int_0^q (x - q)^2 \, dx \]

Substituting \( q = 9 \):

\[ = \int_0^9 (x - 9)^2 \, dx \]

Calculating the integral:

\[ = \int_0^9 (x^2 - 18x + 81) \, dx \]

Calculating each term:

\[ = \left[ \frac{x^3}{3} - 9x^2 + 81x \right]_0^9 \]

Evaluating at the bounds:

\[ = \left( \frac{9^3}{3} - 9 \cdot 9^2 + 81 \cdot 9 \right) - 0 \]
\[ = \left( \frac{729}{3} - 729 + 729 \right) = 243 \]

Thus, the area of the region is \( 243 \).

### Final Answer
The area of the region is \( \boxed{243} \).