If the equation `(x^(2)-bx)/(ax-c) =(m-1)/(m+1)` has roots equal in magnitude but opposite in sign, then m is equal to

Options

A. `(a+b)/(a-b)`
B. `(a-b)/(a+b)`
C. `(b-a)/(b+a)`
D. none of these

Answer

To solve the equation \(\frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1}\) under the condition that the roots are equal in magnitude but opposite in sign, we can follow these steps:

### Step 1: Rearranging the Equation
We start by cross-multiplying to eliminate the fraction:
\[ (x^2 - bx)(m + 1) = (ax - c)(m - 1) \]

### Step 2: Expanding Both Sides
Next, we expand both sides of the equation:
\[ (m + 1)x^2 - b(m + 1)x = (m - 1)ax - (m - 1)c \]
This simplifies to:
\[ (m + 1)x^2 - b(m + 1)x = (m - 1)ax - (m - 1)c \]

### Step 3: Rearranging into Standard Quadratic Form
Now, we rearrange this into standard quadratic form \(Ax^2 + Bx + C = 0\):
\[ (m + 1)x^2 - \left[b(m + 1) + (m - 1)a\right]x + (m - 1)c = 0 \]

### Step 4: Identifying Coefficients
From the standard form, we identify:
- \(A = m + 1\)
- \(B = -\left[b(m + 1) + (m - 1)a\right]\)
- \(C = (m - 1)c\)

### Step 5: Condition for Roots
Since the roots are equal in magnitude but opposite in sign, their sum must be zero. The sum of the roots for a quadratic equation is given by \(-\frac{B}{A}\). Therefore, we set:
\[ -\frac{-\left[b(m + 1) + (m - 1)a\right]}{m + 1} = 0 \]
This implies:
\[ b(m + 1) + (m - 1)a = 0 \]

### Step 6: Solving for \(m\)
Rearranging the equation gives:
\[ b(m + 1) = -(m - 1)a \]
Expanding and simplifying:
\[ bm + b = -am + a \]
Combining like terms:
\[ bm + am = a - b \]
Factoring out \(m\):
\[ m(b + a) = a - b \]
Thus, we can solve for \(m\):
\[ m = \frac{a - b}{a + b} \]

### Final Result
Therefore, the value of \(m\) is:
\[ \boxed{\frac{a - b}{a + b}} \]