Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
Both the roots are equal

Answer

To find the values of \( \alpha \) for which both roots of the quadratic equation \( 4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0 \) are equal, we need to set the discriminant of the quadratic equation to zero.

### Step-by-step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the standard form \( ax^2 + bx + c = 0 \). Here, we have:
- \( a = 4 \)
- \( b = -4(\alpha - 2) \)
- \( c = \alpha - 2 \)

2. **Write the discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = \left(-4(\alpha - 2)\right)^2 - 4 \cdot 4 \cdot (\alpha - 2) \]

3. **Simplify the discriminant**:
Calculate \( D \):
\[ D = 16(\alpha - 2)^2 - 16(\alpha - 2) \]
Factor out the common term \( 16 \):
\[ D = 16\left((\alpha - 2)^2 - (\alpha - 2)\right) \]

4. **Set the discriminant to zero**:
For the roots to be equal, we set \( D = 0 \):
\[ 16\left((\alpha - 2)^2 - (\alpha - 2)\right) = 0 \]
This implies:
\[ (\alpha - 2)^2 - (\alpha - 2) = 0 \]

5. **Let \( y = \alpha - 2 \)**:
Substitute \( y \) into the equation:
\[ y^2 - y = 0 \]

6. **Factor the equation**:
Factor out \( y \):
\[ y(y - 1) = 0 \]
This gives us two solutions:
\[ y = 0 \quad \text{or} \quad y = 1 \]

7. **Substitute back for \( \alpha \)**:
Recall that \( y = \alpha - 2 \):
- If \( y = 0 \), then \( \alpha - 2 = 0 \) which gives \( \alpha = 2 \).
- If \( y = 1 \), then \( \alpha - 2 = 1 \) which gives \( \alpha = 3 \).

### Conclusion:
The values of \( \alpha \) for which both roots of the quadratic equation are equal are:
\[ \alpha = 2 \quad \text{and} \quad \alpha = 3 \]