If `x^(2) + ax + b = 0, x^(2) + bx + a = 0` (`a != 0`) have a common root, then a + b =

Options

A. 3
B. 2
C. 1
D. -1

Answer

To solve the problem, we need to find the value of \( a + b \) given that the equations \( x^2 + ax + b = 0 \) and \( x^2 + bx + a = 0 \) have a common root.

### Step-by-Step Solution:

1. **Let the common root be \( r \)**:
Since both equations have a common root, we can substitute \( r \) into both equations.

For the first equation:
\[ r^2 + ar + b = 0 \quad \text{(1)} \]

For the second equation:
\[ r^2 + br + a = 0 \quad \text{(2)} \]

2. **Subtract the two equations**:
We can subtract equation (2) from equation (1):
\[ (r^2 + ar + b) - (r^2 + br + a) = 0 \]
This simplifies to:
\[ ar + b - br - a = 0 \]
Rearranging gives:
\[ (a - b)r + (b - a) = 0 \]

3. **Factor out the common terms**:
We can factor out \( (a - b) \):
\[ (a - b)(r - 1) = 0 \]

4. **Analyze the factors**:
Since \( a \neq b \) (as per the problem statement), we must have:
\[ r - 1 = 0 \quad \Rightarrow \quad r = 1 \]

5. **Substitute \( r = 1 \) back into one of the original equations**:
We can substitute \( r = 1 \) into equation (1):
\[ 1^2 + a(1) + b = 0 \]
This simplifies to:
\[ 1 + a + b = 0 \]

6. **Solve for \( a + b \)**:
Rearranging gives:
\[ a + b = -1 \]

### Final Answer:
Thus, the value of \( a + b \) is:
\[ \boxed{-1} \]