If the equations `ax^2+bx+c=0` and `cx^2+bx+a=0, a!=c` have a negative common root then the value of `a-b+c=`

Options

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

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 this common root as \( -\alpha \), where \( \alpha \) is a positive integer.

### Step 1: Substitute the common root into both equations
Since \( -\alpha \) is a root of the first equation, we can substitute it into the equation:

\[ a(-\alpha)^2 + b(-\alpha) + c = 0 \]

This simplifies to:

\[ a\alpha^2 - b\alpha + c = 0 \quad \text{(Equation 1)} \]

Similarly, substituting \( -\alpha \) into the second equation gives:

\[ c(-\alpha)^2 + b(-\alpha) + a = 0 \]

This simplifies to:

\[ c\alpha^2 - b\alpha + a = 0 \quad \text{(Equation 2)} \]

### Step 2: Set up the equations
Now we have two equations:

1. \( a\alpha^2 - b\alpha + c = 0 \)
2. \( c\alpha^2 - b\alpha + a = 0 \)

### Step 3: Subtract Equation 2 from Equation 1
Subtracting Equation 2 from Equation 1, we get:

\[ (a\alpha^2 - b\alpha + c) - (c\alpha^2 - b\alpha + a) = 0 \]

This simplifies to:

\[ (a\alpha^2 - c\alpha^2) + (c - a) = 0 \]

Factoring out \( \alpha^2 \):

\[ \alpha^2(a - c) + (c - a) = 0 \]

### Step 4: Factor out \( (a - c) \)
We can factor this equation:

\[ (a - c)(\alpha^2 - 1) = 0 \]

### Step 5: Analyze the factors
Since \( a \neq c \) (given in the problem), we must have:

\[ \alpha^2 - 1 = 0 \]

This implies:

\[ \alpha^2 = 1 \implies \alpha = 1 \quad (\text{since } \alpha \text{ is positive}) \]

### Step 6: Substitute \( \alpha = 1 \) back into the equations
Now, substituting \( \alpha = 1 \) into either of the original equations, we can use Equation 1:

\[ a(1)^2 - b(1) + c = 0 \]

This simplifies to:

\[ a - b + c = 0 \]

### Final Result
Thus, we find that:

\[ a - b + c = 0 \]

### Conclusion
The value of \( a - b + c \) is \( 0 \).

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