If the roots of `(x^(2)-bx)/(ax-c)=(k-1)/(k+1)` are numerically equal but opposite in sign, then k =

Options

A. c
B. `(1)/(c)`
C. `(a+b)/(a-b)`
D. `(a-b)/(a+b)`

Answer

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

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

are numerically equal but opposite in sign.

### Step-by-Step Solution:

1. **Rewrite the Equation**:
Start with the given equation:
\[ \frac{x^2 - bx}{ax - c} = \frac{k - 1}{k + 1} \]
Cross-multiply to eliminate the fraction:
\[ (x^2 - bx)(k + 1) = (k - 1)(ax - c) \]

2. **Expand Both Sides**:
Expand both sides:
\[ (k + 1)x^2 - b(k + 1)x = (k - 1)ax - (k - 1)c \]
Rearranging gives:
\[ (k + 1)x^2 - (b(k + 1) + (k - 1)a)x + (k - 1)c = 0 \]

3. **Identify Coefficients**:
From the standard form \( Ax^2 + Bx + C = 0 \), we identify:
- \( A = k + 1 \)
- \( B = -[b(k + 1) + (k - 1)a] \)
- \( C = (k - 1)c \)

4. **Condition for Roots**:
Since the roots are numerically equal but opposite in sign, their sum must be zero. The sum of the roots is given by:
\[ -\frac{B}{A} = 0 \]
Thus, we have:
\[ B = 0 \implies -[b(k + 1) + (k - 1)a] = 0 \]
This simplifies to:
\[ b(k + 1) + (k - 1)a = 0 \]

5. **Solve for \( k \)**:
Rearranging gives:
\[ b(k + 1) = -a(k - 1) \]
Expanding both sides:
\[ bk + b = -ak + a \]
Bringing all terms involving \( k \) to one side:
\[ bk + ak = a - b \]
Factoring out \( k \):
\[ k(b + a) = a - b \]
Thus, we find:
\[ k = \frac{a - b}{a + b} \]

6. **Final Answer**:
Therefore, the value of \( k \) is:
\[ k = \frac{a - b}{a + b} \]