If two quadratic equations `x^(2) + ax + b = 0` and `x^(2) + bx + a = 0 (a ne 0)` hava a common root. Find the value of a + b.

Options

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

Answer

To solve the problem, we need to find the value of \( a + b \) given that the two quadratic 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.
- From the first equation:
\[ r^2 + ar + b = 0 \quad \text{(1)} \]
- From the second equation:
\[ r^2 + br + a = 0 \quad \text{(2)} \]

2. **Equate the two equations**:
Since both equations equal zero, we can set them equal to each other:
\[ r^2 + ar + b = r^2 + br + a \]

3. **Simplify the equation**:
By cancelling \( r^2 \) from both sides, we get:
\[ ar + b = br + a \]

4. **Rearrange the equation**:
Rearranging gives us:
\[ ar - br = a - b \]
Factoring out \( r \) from the left side:
\[ (a - b)r = a - b \]

5. **Consider two cases**:
- **Case 1**: If \( a - b \neq 0 \), we can divide both sides by \( a - b \):
\[ r = 1 \]
- **Case 2**: If \( a - b = 0 \), then \( a = b \). However, the problem states that \( a \neq b \), so we discard this case.

6. **Substitute \( r = 1 \)**:
Now we substitute \( r = 1 \) back into either of the original equations. Using equation (1):
\[ 1^2 + a(1) + b = 0 \]
Simplifying gives:
\[ 1 + a + b = 0 \]
Thus, we have:
\[ a + b = -1 \]

### Conclusion:
The value of \( a + b \) is \( -1 \).