If the quadratic equations, `a x^2+2c x+b=0 and a x^2+2b x+c=0(b!=c)` have a common root, then `a+4b+4c` is equal to: a. -2 b. 2 c. 0 d. 1

Options

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

Explanation

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 \( ax^2 + 2bx + c = 0 \) have a common root. Let's denote the common root as \( \alpha \).

### Step-by-Step Solution:

1. **Set Up the Equations**:
We have two equations:
\[ ax^2 + 2cx + b = 0 \quad \text{(1)} \]
\[ ax^2 + 2bx + c = 0 \quad \text{(2)} \]
Since \( \alpha \) is a common root, we can substitute \( \alpha \) into both equations.

2. **Substituting the Common Root**:
For equation (1):
\[ a\alpha^2 + 2c\alpha + b = 0 \quad \text{(3)} \]
For equation (2):
\[ a\alpha^2 + 2b\alpha + c = 0 \quad \text{(4)} \]

3. **Subtract the Two Equations**:
Now, we subtract equation (4) from equation (3):
\[ (a\alpha^2 + 2c\alpha + b) - (a\alpha^2 + 2b\alpha + c) = 0 \]
This simplifies to:
\[ 2c\alpha + b - 2b\alpha - c = 0 \]
Rearranging gives:
\[ 2c\alpha - 2b\alpha + b - c = 0 \]
Factoring out \( \alpha \):
\[ 2\alpha(c - b) + (b - c) = 0 \]

4. **Factor and Solve for \( \alpha \)**:
We can rewrite this as:
\[ (2\alpha - 1)(c - b) = 0 \]
Since \( b \neq c \), we have:
\[ 2\alpha - 1 = 0 \implies \alpha = \frac{1}{2} \]

5. **Substituting \( \alpha \) Back**:
Now, we substitute \( \alpha = \frac{1}{2} \) back into equation (3):
\[ a\left(\frac{1}{2}\right)^2 + 2c\left(\frac{1}{2}\right) + b = 0 \]
Simplifying gives:
\[ \frac{a}{4} + c + b = 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ a + 4c + 4b = 0 \]

6. **Rearranging for the Required Expression**:
We need to find \( a + 4b + 4c \):
\[ a + 4b + 4c = 0 \]

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