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 imaginary

Answer

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

### Step 1: Identify the coefficients
The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). Here, we can identify:
- \( a = 4 \)
- \( b = -4(\alpha - 2) \)
- \( c = \alpha - 2 \)

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

### Step 3: Simplify the discriminant
Calculating \( D \):
\[ D = 16(\alpha - 2)^2 - 16(\alpha - 2) \]
Factoring out \( 16 \):
\[ D = 16\left((\alpha - 2)^2 - (\alpha - 2)\right) \]
Let \( y = \alpha - 2 \):
\[ D = 16(y^2 - y) \]

### Step 4: Set the discriminant less than zero
For the roots to be imaginary, the discriminant must be less than zero:
\[ 16(y^2 - y) < 0 \]
This simplifies to:
\[ y^2 - y < 0 \]

### Step 5: Factor the inequality
Factoring the quadratic:
\[ y(y - 1) < 0 \]

### Step 6: Determine the intervals
The roots of the equation \( y(y - 1) = 0 \) are \( y = 0 \) and \( y = 1 \). We analyze the sign of the product \( y(y - 1) \) in the intervals:
- \( y < 0 \) (both factors negative, product positive)
- \( 0 < y < 1 \) (one factor positive and one negative, product negative)
- \( y > 1 \) (both factors positive, product positive)

Thus, the inequality \( y(y - 1) < 0 \) holds for:
\[ 0 < y < 1 \]

### Step 7: Substitute back for \( \alpha \)
Recalling that \( y = \alpha - 2 \):
\[ 0 < \alpha - 2 < 1 \]
Adding 2 to all parts of the inequality:
\[ 2 < \alpha < 3 \]

### Conclusion
The values of \( \alpha \) for which both roots of the quadratic equation are imaginary are:
\[ \alpha \in (2, 3) \]