`25p^2-55sqrt11p+198=0`
`25q^2+55sqrt11q+198=0`

Options

A. `pltq`
B. `pgtq`
C. `pleqq`
D. can't be determined

Answer

To solve the given quadratic equations and find the relationship between \( p \) and \( q \), we will follow these steps:

### Step 1: Write down the equations
We have two quadratic equations:
1. \( 25p^2 - 55\sqrt{11}p + 198 = 0 \)
2. \( 25q^2 + 55\sqrt{11}q + 198 = 0 \)

### Step 2: Identify coefficients
For the first equation \( 25p^2 - 55\sqrt{11}p + 198 = 0 \):
- \( a = 25 \)
- \( b = -55\sqrt{11} \)
- \( c = 198 \)

For the second equation \( 25q^2 + 55\sqrt{11}q + 198 = 0 \):
- \( a = 25 \)
- \( b = 55\sqrt{11} \)
- \( c = 198 \)

### Step 3: Apply the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

#### For \( p \):
Substituting the values into the formula:
\[ p = \frac{-(-55\sqrt{11}) \pm \sqrt{(-55\sqrt{11})^2 - 4 \cdot 25 \cdot 198}}{2 \cdot 25} \]

Calculating \( b^2 - 4ac \):
\[ (-55\sqrt{11})^2 = 3025 \cdot 11 = 33275 \]
\[ 4 \cdot 25 \cdot 198 = 19800 \]
\[ b^2 - 4ac = 33275 - 19800 = 13475 \]

Now substituting back into the formula:
\[ p = \frac{55\sqrt{11} \pm \sqrt{13475}}{50} \]

#### For \( q \):
Substituting the values into the formula:
\[ q = \frac{-55\sqrt{11} \pm \sqrt{(55\sqrt{11})^2 - 4 \cdot 25 \cdot 198}}{2 \cdot 25} \]

Calculating \( b^2 - 4ac \) (same as above):
\[ (55\sqrt{11})^2 = 3025 \cdot 11 = 33275 \]
\[ b^2 - 4ac = 33275 - 19800 = 13475 \]

Now substituting back into the formula:
\[ q = \frac{-55\sqrt{11} \pm \sqrt{13475}}{50} \]

### Step 4: Find the roots
1. For \( p \):
- \( p_1 = \frac{55\sqrt{11} + \sqrt{13475}}{50} \)
- \( p_2 = \frac{55\sqrt{11} - \sqrt{13475}}{50} \)

2. For \( q \):
- \( q_1 = \frac{-55\sqrt{11} + \sqrt{13475}}{50} \)
- \( q_2 = \frac{-55\sqrt{11} - \sqrt{13475}}{50} \)

### Step 5: Compare the roots
From the calculations, we observe:
- Both roots \( p_1 \) and \( p_2 \) are positive.
- Both roots \( q_1 \) and \( q_2 \) are negative.

### Conclusion
Since both roots of \( p \) are positive and both roots of \( q \) are negative, we can conclude that \( p \) is greater than \( q \).