If the equation `ax^(2)+bx+c=0 and cx^(2)+bx+a=0 a ne c`, have negative common root then the value of a-b+c is

Options

A. 0
B. 1
C. 2
D. none

Answer

To solve the problem, we need to find the value of \( a - b + c \) given that the equations \( ax^2 + bx + c = 0 \) and \( cx^2 + bx + a = 0 \) have a negative common root. Let's denote the common root as \( \alpha \).

### Step-by-step Solution:

1. **Set up the equations with the common root**:
Since \( \alpha \) is a common root, it must satisfy both equations:
\[ a\alpha^2 + b\alpha + c = 0 \quad \text{(1)} \]
\[ c\alpha^2 + b\alpha + a = 0 \quad \text{(2)} \]

2. **Subtract the two equations**:
Subtract equation (2) from equation (1):
\[ (a\alpha^2 + b\alpha + c) - (c\alpha^2 + b\alpha + a) = 0 \]
Simplifying this gives:
\[ (a - c)\alpha^2 + (c - a) = 0 \]
Rearranging gives:
\[ (a - c)(\alpha^2 - 1) = 0 \]

3. **Analyze the factors**:
Since \( a \neq c \) (given), we have:
\[ \alpha^2 - 1 = 0 \]
Thus, \( \alpha^2 = 1 \), which means \( \alpha = \pm 1 \).

4. **Determine the negative root**:
Since we are told that the common root is negative, we take:
\[ \alpha = -1 \]

5. **Substituting the value of \( \alpha \)**:
Substitute \( \alpha = -1 \) into either of the original equations. We can use equation (1):
\[ a(-1)^2 + b(-1) + c = 0 \]
This simplifies to:
\[ a - b + c = 0 \]

6. **Final result**:
Rearranging gives:
\[ a - b + c = 0 \]

Thus, the value of \( a - b + c \) is \( 0 \).