For what value of k , are the roots of the quadratic equation (k + 1) `x^(2) - 2(k - 1) x + 1 = 0` real and equal?

Options

A. k = 0 only
B. k = - 3 only
C. k = 0 or k = 3
D. k = 0 or k = - 3

Answer

To find the value of \( k \) for which the roots of the quadratic equation \( (k + 1)x^2 - 2(k - 1)x + 1 = 0 \) are real and equal, we need to use the condition for real and equal roots. This condition states that the discriminant \( D \) of the quadratic equation must be equal to zero.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \).
Here,
- \( a = k + 1 \)
- \( b = -2(k - 1) = -2k + 2 \)
- \( c = 1 \)

2. **Set up the discriminant**:
The discriminant \( D \) is given by the formula:
\[ D = b^2 - 4ac \]
For the roots to be real and equal, we need:
\[ D = 0 \]

3. **Substitute the coefficients into the discriminant**:
\[ D = (-2k + 2)^2 - 4(k + 1)(1) \]

4. **Expand the discriminant**:
\[ D = (4k^2 - 8k + 4) - 4(k + 1) \]
\[ D = 4k^2 - 8k + 4 - 4k - 4 \]
\[ D = 4k^2 - 12k + 0 \]

5. **Set the discriminant equal to zero**:
\[ 4k^2 - 12k = 0 \]

6. **Factor out the common term**:
\[ 4k(k - 3) = 0 \]

7. **Solve for \( k \)**:
This gives us two solutions:
\[ 4k = 0 \implies k = 0 \]
\[ k - 3 = 0 \implies k = 3 \]

### Conclusion:
The values of \( k \) for which the roots of the quadratic equation are real and equal are \( k = 0 \) and \( k = 3 \).

### Final Answer:
The correct answer is \( k = 0 \) or \( k = 3 \).

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