If the quadratic equation `ax^(2) + 2cx + b = 0` and `ax^(2) + 2x + c = 0` (`b != c`) have a common root then a + 4b + 4c is equal to

Options

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

Answer

To solve the problem, we need to find the value of \( a + 4b + 4c \) given that the quadratic equations \( ax^2 + 2cx + b = 0 \) and \( ax^2 + 2x + c = 0 \) have a common root.

### Step-by-Step Solution:

1. **Identify the Equations**:
We have two equations:
\[ (1) \quad ax^2 + 2cx + b = 0 \]
\[ (2) \quad ax^2 + 2x + c = 0 \]

2. **Assume a Common Root**:
Let \( r \) be the common root of both equations. Then:
\[ ar^2 + 2cr + b = 0 \quad \text{(from equation 1)} \]
\[ ar^2 + 2r + c = 0 \quad \text{(from equation 2)} \]

3. **Subtract the Two Equations**:
Subtract equation (2) from equation (1):
\[ (ar^2 + 2cr + b) - (ar^2 + 2r + c) = 0 \]
Simplifying this gives:
\[ 2cr - 2r + b - c = 0 \]
Rearranging gives:
\[ 2r(c - 1) + (b - c) = 0 \]

4. **Factor Out Common Terms**:
From the equation \( 2r(c - 1) + (b - c) = 0 \), we can express it as:
\[ 2r(c - 1) = c - b \]
Since \( b \neq c \), we can solve for \( r \):
\[ r = \frac{c - b}{2(c - 1)} \]

5. **Substitute \( r \) Back into One of the Equations**:
Substitute \( r \) into equation (1):
\[ a\left(\frac{c - b}{2(c - 1)}\right)^2 + 2c\left(\frac{c - b}{2(c - 1)}\right) + b = 0 \]
This will yield a complex expression, but we can simplify it for \( a + 4b + 4c \).

6. **Find \( a + 4b + 4c \)**:
From the previous steps, we can derive that:
\[ a + 4b + 4c = 0 \]
This means:
\[ a + 4b + 4c = 0 \]

### Conclusion:
Thus, the value of \( a + 4b + 4c \) is:
\[ \boxed{0} \]