If the expression `x^(2) - 11x + a and x^(2) - 14 x + 2a` have a common factor, then the values of 'a' are

Options

A. 0, 24
B. 0, -24
C. 1, -1
D. `-2, 1`

Answer

To solve the problem, we need to find the values of 'a' such that the expressions \( x^2 - 11x + a \) and \( x^2 - 14x + 2a \) have a common factor.

### Step-by-Step Solution:

1. **Identify the Common Factor**:
Let the common factor be \( \alpha \). Since \( \alpha \) is a root of both quadratic equations, it must satisfy both equations:
\[ \alpha^2 - 11\alpha + a = 0 \quad \text{(1)} \]
\[ \alpha^2 - 14\alpha + 2a = 0 \quad \text{(2)} \]

2. **Subtract the Two Equations**:
We can subtract equation (1) from equation (2) to eliminate \( \alpha^2 \):
\[ (\alpha^2 - 14\alpha + 2a) - (\alpha^2 - 11\alpha + a) = 0 \]
This simplifies to:
\[ -14\alpha + 2a + 11\alpha - a = 0 \]
Combining like terms gives:
\[ -3\alpha + a = 0 \]
Rearranging this, we find:
\[ a = 3\alpha \quad \text{(3)} \]

3. **Substitute \( a \) Back into One of the Equations**:
Now, substitute \( a \) from equation (3) back into equation (1):
\[ \alpha^2 - 11\alpha + 3\alpha = 0 \]
Simplifying this gives:
\[ \alpha^2 - 8\alpha = 0 \]
Factoring out \( \alpha \):
\[ \alpha(\alpha - 8) = 0 \]
This gives us two possible values for \( \alpha \):
\[ \alpha = 0 \quad \text{or} \quad \alpha = 8 \]

4. **Find Corresponding Values of \( a \)**:
Now, we will find the corresponding values of \( a \) using equation (3):
- If \( \alpha = 0 \):
\[ a = 3 \times 0 = 0 \]
- If \( \alpha = 8 \):
\[ a = 3 \times 8 = 24 \]

5. **Final Values of \( a \)**:
Therefore, the values of \( a \) are:
\[ a = 0 \quad \text{or} \quad a = 24 \]

### Conclusion:
The values of \( a \) for which the expressions \( x^2 - 11x + a \) and \( x^2 - 14x + 2a \) have a common factor are \( a = 0 \) and \( a = 24 \).

---