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. Let's denote the common root as \( r \).

### Step-by-step Solution:

1. **Substituting the common root**:
Since \( r \) is a common root, it satisfies both equations:
\[ r^2 + ar + b = 0 \quad \text{(1)} \]
\[ r^2 + br + a = 0 \quad \text{(2)} \]

2. **Equating the two equations**:
From equations (1) and (2), we can set them equal to each other:
\[ r^2 + ar + b = r^2 + br + a \]
By simplifying this, we can cancel \( r^2 \) from both sides:
\[ ar + b = br + a \]

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

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

5. **Substituting \( r = 1 \)**:
Now, substitute \( r = 1 \) back into either equation (let's use equation (1)):
\[ 1^2 + a(1) + b = 0 \]
This simplifies to:
\[ 1 + a + b = 0 \]
Therefore, we can express \( a + b \) as:
\[ a + b = -1 \]

### Conclusion:
Thus, the value of \( a + b \) is \( -1 \).

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