Find m so that the roots of the equation `(x^(2)-bx)/(ax-c)=(m-1)/(m+1)` may be equal in magnitude and opposite in sign.

Answer

To solve the problem, we need to find the value of \( m \) such that the roots of the equation

\[ \frac{x^2 - bx}{ax - c} = \frac{m - 1}{m + 1} \]

are equal in magnitude and opposite in sign. Let's go through the steps systematically.

### Step 1: Cross Multiply
We start by cross-multiplying the equation:

\[ (x^2 - bx)(m + 1) = (m - 1)(ax - c) \]

### Step 2: Expand Both Sides
Now, we expand both sides of the equation:

Left Side:
\[ x^2(m + 1) - bx(m + 1) \]

Right Side:
\[ (m - 1)(ax) - (m - 1)(c) \]

So we have:
\[ x^2(m + 1) - bx(m + 1) = (m - 1)ax - (m - 1)c \]

### Step 3: Rearrange the Equation
Rearranging gives us:
\[ x^2(m + 1) - bx(m + 1) - (m - 1)ax + (m - 1)c = 0 \]

### Step 4: Collect Like Terms
We can collect the terms involving \( x \):
\[ x^2(m + 1) + x[-b(m + 1) - a(m - 1)] + (m - 1)c = 0 \]

### Step 5: Identify Coefficients
This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where:
- \( A = m + 1 \)
- \( B = -b(m + 1) - a(m - 1) \)
- \( C = (m - 1)c \)

### Step 6: Roots Condition
Since the roots are equal in magnitude and opposite in sign, we can assume the roots are \( \alpha \) and \( -\alpha \). Therefore, the sum of the roots \( \alpha + (-\alpha) = 0 \).

Using the formula for the sum of the roots:
\[ \text{Sum of roots} = -\frac{B}{A} = 0 \]

This implies:
\[ -b(m + 1) - a(m - 1) = 0 \]

### Step 7: Solve for \( m \)
Now we can solve for \( m \):
\[ -b(m + 1) = a(m - 1) \]

Expanding both sides:
\[ -bm - b = am - a \]

Rearranging gives:
\[ bm + am = a - b \]

Factoring out \( m \):
\[ m(a + b) = a - b \]

Thus, we find:
\[ m = \frac{a - b}{a + b} \]

### Final Answer
The value of \( m \) such that the roots of the equation are equal in magnitude and opposite in sign is:
\[ m = \frac{a - b}{a + b} \]