If `(x^(2)-x+p)(11y^(2)-4y+2)=(9)/(2)` have exactly one ordered pair of `(x, y)` then find `p`.

Explanation

To solve the problem, we need to find the value of \( p \) such that the equation

\[ (x^2 - x + p)(11y^2 - 4y + 2) = \frac{9}{2} \]

has exactly one ordered pair of \( (x, y) \).

### Step 1: Identify the Quadratic Functions
We have two quadratic functions:
1. \( f(x) = x^2 - x + p \)
2. \( g(y) = 11y^2 - 4y + 2 \)

### Step 2: Find the Minimum Values of Each Quadratic
To have exactly one ordered pair \( (x, y) \), the product of the minimum values of these two functions must equal \( \frac{9}{2} \).

#### Minimum of \( f(x) \):
The minimum value of a quadratic function \( ax^2 + bx + c \) is given by

\[ \text{Minimum} = \frac{4ac - b^2}{4a} \]

For \( f(x) = x^2 - x + p \):
- \( a = 1 \)
- \( b = -1 \)
- \( c = p \)

Calculating the minimum value:

\[ \text{Minimum of } f(x) = \frac{4(1)(p) - (-1)^2}{4(1)} = \frac{4p - 1}{4} \]

#### Minimum of \( g(y) \):
For \( g(y) = 11y^2 - 4y + 2 \):
- \( a = 11 \)
- \( b = -4 \)
- \( c = 2 \)

Calculating the minimum value:

\[ \text{Minimum of } g(y) = \frac{4(11)(2) - (-4)^2}{4(11)} = \frac{88 - 16}{44} = \frac{72}{44} = \frac{18}{11} \]

### Step 3: Set Up the Equation
We need the product of the minimum values to equal \( \frac{9}{2} \):

\[ \left( \frac{4p - 1}{4} \right) \left( \frac{18}{11} \right) = \frac{9}{2} \]

### Step 4: Solve for \( p \)
Multiply both sides by \( 4 \) to eliminate the fraction:

\[ (4p - 1) \left( \frac{18}{11} \right) = 18 \]

Now, multiply both sides by \( 11 \):

\[ 18(4p - 1) = 198 \]

Distributing \( 18 \):

\[ 72p - 18 = 198 \]

Adding \( 18 \) to both sides:

\[ 72p = 216 \]

Dividing by \( 72 \):

\[ p = 3 \]

### Conclusion
The value of \( p \) is

\[ \boxed{3} \]