For the quadratic equation `(k-1)x^(2)=kx-1, k!=1` the roots are numerically equal but opposite sign, then k is greater than

Options

A. `-1`
B. `1`
C. 0
D. 2

Answer

To solve the problem, we need to analyze the quadratic equation given and find the conditions under which its roots are numerically equal but opposite in sign.

### Step-by-step Solution:

1. **Rewrite the Quadratic Equation**:
The given equation is \((k-1)x^2 = kx - 1\). We can rearrange it to standard form:
\[ (k-1)x^2 - kx + 1 = 0 \]

2. **Identify Coefficients**:
From the standard form \(ax^2 + bx + c = 0\), we identify:
- \(a = k - 1\)
- \(b = -k\)
- \(c = 1\)

3. **Condition for Roots**:
The roots of the quadratic equation are numerically equal but opposite in sign if they are of the form \(r\) and \(-r\). This occurs when the sum of the roots is zero. The sum of the roots is given by:
\[ \text{Sum of roots} = -\frac{b}{a} = -\frac{-k}{k-1} = \frac{k}{k-1} \]
Setting the sum of the roots to zero:
\[ \frac{k}{k-1} = 0 \]
This implies \(k = 0\).

4. **Condition for Real Roots**:
For the roots to be real, the discriminant \(D\) must be greater than or equal to zero:
\[ D = b^2 - 4ac \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (-k)^2 - 4(k-1)(1) = k^2 - 4(k - 1) = k^2 - 4k + 4 \]
We need \(D \geq 0\):
\[ k^2 - 4k + 4 \geq 0 \]
This can be factored as:
\[ (k - 2)^2 \geq 0 \]
The expression \((k - 2)^2\) is always non-negative and equals zero when \(k = 2\).

5. **Conclusion**:
Since we need \(D > 0\) for the roots to be distinct, we require:
\[ (k - 2)^2 > 0 \implies k \neq 2 \]
However, since we are looking for \(k\) such that the roots are numerically equal but opposite in sign, we need \(k\) to be greater than 2:
\[ k > 2 \]

### Final Answer:
Thus, the value of \(k\) must be greater than 2.