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 negative

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 negative, we will follow these steps:

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

### Step 2: Ensure the roots are real
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 \( c \):
\[ D = [-4(\alpha - 2)]^2 - 4(4)(\alpha - 2) \geq 0 \]
\[ D = 16(\alpha - 2)^2 - 16(\alpha - 2) \geq 0 \]
Factoring out \( 16(\alpha - 2) \):
\[ 16(\alpha - 2)[\alpha - 3] \geq 0 \]

### Step 3: Solve the inequality
The critical points are \( \alpha = 2 \) and \( \alpha = 3 \). We can test intervals around these points:
- For \( \alpha < 2 \): \( 16(\alpha - 2)(\alpha - 3) > 0 \) (both factors negative)
- For \( 2 < \alpha < 3 \): \( 16(\alpha - 2)(\alpha - 3) < 0 \) (one factor positive, one negative)
- For \( \alpha > 3 \): \( 16(\alpha - 2)(\alpha - 3) > 0 \) (both factors positive)

Thus, the solution to the inequality is:
\[ \alpha \leq 2 \quad \text{or} \quad \alpha \geq 3 \]

### Step 4: Ensure both roots are negative
For both roots to be negative, we need the following conditions:
1. The sum of the roots \( \frac{-b}{a} \) must be positive.
2. The product of the roots \( \frac{c}{a} \) must be positive.

**Condition 1: Sum of the roots**
\[ \frac{-b}{a} = \frac{4(\alpha - 2)}{4} = \alpha - 2 > 0 \implies \alpha > 2 \]

**Condition 2: Product of the roots**
\[ \frac{c}{a} = \frac{\alpha - 2}{4} > 0 \implies \alpha - 2 > 0 \implies \alpha > 2 \]

### Step 5: Combine conditions
From the conditions derived:
- From the discriminant: \( \alpha \leq 2 \) or \( \alpha \geq 3 \)
- From the negativity of roots: \( \alpha > 2 \)

The only overlapping solution is:
\[ \alpha \geq 3 \]

### Final Answer
The values of \( \alpha \) for which both roots of the quadratic equation are negative are:
\[ \alpha \in [3, \infty) \]