If (x+a) be a common factor of `x^2 + px + q` and `x^2 + p'x +q'`, then the value of a is:

Answer

To find the value of \( a \) given that \( (x + a) \) is a common factor of the quadratic equations \( x^2 + px + q \) and \( x^2 + p'x + q' \), we can follow these steps:

### Step 1: Understand the Factor Theorem
The Factor Theorem states that if \( (x - k) \) is a factor of a polynomial, then substituting \( x = k \) into the polynomial will yield zero. In this case, since \( (x + a) \) is a factor, we will substitute \( x = -a \).

### Step 2: Substitute \( x = -a \) into the first equation
We substitute \( x = -a \) into the first quadratic equation:
\[ (-a)^2 + p(-a) + q = 0 \]
This simplifies to:
\[ a^2 - pa + q = 0 \tag{1} \]

### Step 3: Substitute \( x = -a \) into the second equation
Now, we substitute \( x = -a \) into the second quadratic equation:
\[ (-a)^2 + p'(-a) + q' = 0 \]
This simplifies to:
\[ a^2 - p'a + q' = 0 \tag{2} \]

### Step 4: Set the two equations equal to each other
Since both equations (1) and (2) equal zero, we can set them equal to each other:
\[ a^2 - pa + q = a^2 - p'a + q' \]

### Step 5: Simplify the equation
We can cancel \( a^2 \) from both sides:
\[ -pa + q = -p'a + q' \]
Rearranging gives:
\[ -p a + p' a = q' - q \]
Factoring out \( a \) from the left side:
\[ a(p' - p) = q' - q \]

### Step 6: Solve for \( a \)
Now, we can solve for \( a \):
\[ a = \frac{q' - q}{p' - p} \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{\frac{q' - q}{p' - p}} \]