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

Options

A. 0, 7
B. 5, 20
C. 0, 24
D. 1, 3

Answer

To solve the problem, we need to determine the values of \( a \) for which the polynomials \( x^2 - 11x + a \) and \( x^2 - 14x + 2a \) have a common factor.

### Step-by-Step Solution:

1. **Identify the Polynomials**:
We have two polynomials:
\[ P(x) = x^2 - 11x + a \]
\[ Q(x) = x^2 - 14x + 2a \]

2. **Set Up the Condition for Common Factors**:
If these two polynomials have a common factor, then there exists a value of \( x \) such that both polynomials equal zero. Let’s denote the common root by \( r \). Therefore, we have:
\[ P(r) = 0 \quad \text{and} \quad Q(r) = 0 \]

3. **Express the Conditions**:
From \( P(r) = 0 \):
\[ r^2 - 11r + a = 0 \quad \text{(1)} \]
From \( Q(r) = 0 \):
\[ r^2 - 14r + 2a = 0 \quad \text{(2)} \]

4. **Subtract the Two Equations**:
Subtract equation (1) from equation (2):
\[ (r^2 - 14r + 2a) - (r^2 - 11r + a) = 0 \]
Simplifying this gives:
\[ -14r + 2a + 11r - a = 0 \]
\[ -3r + a = 0 \]
Thus, we find:
\[ a = 3r \quad \text{(3)} \]

5. **Substituting \( a \) Back**:
Now substitute \( a = 3r \) back into equation (1):
\[ r^2 - 11r + 3r = 0 \]
Simplifying gives:
\[ r^2 - 8r = 0 \]
Factoring out \( r \):
\[ r(r - 8) = 0 \]
This gives us two possible values for \( r \):
\[ r = 0 \quad \text{or} \quad r = 8 \]

6. **Finding Corresponding Values of \( a \)**:
- If \( r = 0 \):
\[ a = 3 \cdot 0 = 0 \]
- If \( r = 8 \):
\[ a = 3 \cdot 8 = 24 \]

7. **Conclusion**:
The values of \( a \) for which the polynomials have a common factor are:
\[ a = 0 \quad \text{and} \quad a = 24 \]

### Final Answer:
The values of \( a \) are \( 0 \) and \( 24 \).