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 lie in `(0, (1)/(2))`

Answer

To find the values of \( \alpha \) for which both roots of the quadratic equation \( 4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0 \) lie in the interval \( (0, \frac{1}{2}) \), we will follow these steps:

### Step 1: Identify the coefficients
The quadratic equation can be expressed in the standard form \( ax^2 + bx + c = 0 \), where:
- \( a = 4 \)
- \( b = -4(\alpha - 2) = -4\alpha + 8 \)
- \( c = \alpha - 2 \)

### Step 2: Condition for real roots
For the roots to be real, the discriminant \( D \) must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-4\alpha + 8)^2 - 4 \cdot 4 \cdot (\alpha - 2) \geq 0 \]
Calculating \( D \):
\[ D = (16\alpha^2 - 64\alpha + 64) - 16(\alpha - 2) \geq 0 \]
\[ D = 16\alpha^2 - 64\alpha + 64 - 16\alpha + 32 \geq 0 \]
\[ D = 16\alpha^2 - 80\alpha + 96 \geq 0 \]
Dividing through by 16:
\[ \alpha^2 - 5\alpha + 6 \geq 0 \]

### Step 3: Factor the quadratic
Factoring the quadratic:
\[ (\alpha - 2)(\alpha - 3) \geq 0 \]
The roots are \( \alpha = 2 \) and \( \alpha = 3 \).

### Step 4: Analyze the intervals
Using a number line to analyze the sign of \( (\alpha - 2)(\alpha - 3) \):
- For \( \alpha < 2 \): both factors are negative, product is positive.
- For \( 2 \leq \alpha \leq 3 \): one factor is zero, product is non-negative.
- For \( \alpha > 3 \): both factors are positive, product is positive.

Thus, the solution for this condition is:
\[ \alpha \leq 2 \quad \text{or} \quad \alpha \geq 3 \]

### Step 5: Condition for roots to lie in \( (0, \frac{1}{2}) \)
Next, we need to ensure that both roots lie in the interval \( (0, \frac{1}{2}) \). This requires:
1. The vertex of the parabola (given by \( x = -\frac{b}{2a} \)) should lie in \( (0, \frac{1}{2}) \).
2. The function values at the endpoints \( f(0) \) and \( f\left(\frac{1}{2}\right) \) should be positive.

Calculating the vertex:
\[ x_v = -\frac{-4\alpha + 8}{2 \cdot 4} = \frac{4\alpha - 8}{8} = \frac{\alpha - 2}{2} \]
Setting the vertex condition:
\[ 0 < \frac{\alpha - 2}{2} < \frac{1}{2} \]
This simplifies to:
\[ 0 < \alpha - 2 < 1 \implies 2 < \alpha < 3 \]

### Step 6: Combine conditions
From the discriminant condition, we have \( \alpha \leq 2 \) or \( \alpha \geq 3 \). From the vertex condition, we have \( 2 < \alpha < 3 \).

However, there is no overlap between these two conditions. Therefore, there are no values of \( \alpha \) that satisfy both conditions.

### Conclusion
The final answer is:
\[ \text{There are no values of } \alpha \text{ such that both roots lie in } (0, \frac{1}{2}). \]