If `x^(2)+ax+b=0` and `x^(2)+bx+a=0,(a ne b)` have a common root, then `a+b` is equal to

Options

A. 0
B. 1
C. `-1`
D. none of these

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.

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

3. **Substituting into the second equation**:
\[ r^2 + br + a = 0 \quad \text{(2)} \]

4. **Setting the two equations equal**:
From equations (1) and (2), we can set them equal to each other since both equal zero:
\[ r^2 + ar + b = r^2 + br + a \]

5. **Simplifying the equation**:
Cancel \( r^2 \) from both sides:
\[ ar + b = br + a \]

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

7. **Considering the case \( a \neq b \)**:
Since \( a \neq b \), we can divide both sides by \( a - b \):
\[ r = 1 \]

8. **Substituting \( r = 1 \) back into one of the original equations**:
We can substitute \( r = 1 \) into equation (1):
\[ 1^2 + a(1) + b = 0 \]
Simplifying gives:
\[ 1 + a + b = 0 \]
Thus, we find:
\[ a + b = -1 \]

### Final Answer:
\[ a + b = -1 \]