Let `4x^(2)-4(alpha-2)x + alpha-2=0(alpha in R)` be a quadratic equation. Find the values of `alpha` for which
Atleast one root is positive

Answer

To find the values of \( \alpha \) for which at least one root of the quadratic equation \( 4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0 \) is positive, we will analyze the conditions under which the roots of the quadratic equation are positive.

### Step 1: Identify the coefficients
The given 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 at least one positive root
For a quadratic equation \( ax^2 + bx + c = 0 \) to have at least one positive root, we need to satisfy two conditions:
1. The discriminant \( D \) must be greater than or equal to zero (i.e., \( D \geq 0 \)) to ensure that the roots are real.
2. The quadratic function must take a negative value at \( x = 0 \) (i.e., \( f(0) < 0 \)) to ensure that at least one root is positive.

### Step 3: Calculate the discriminant
The discriminant \( D \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-4\alpha + 8)^2 - 4 \cdot 4 \cdot (\alpha - 2) \]
Calculating this:
\[ D = (16\alpha^2 - 64\alpha + 64) - 16(\alpha - 2) \]
\[ D = 16\alpha^2 - 64\alpha + 64 - 16\alpha + 32 \]
\[ D = 16\alpha^2 - 80\alpha + 96 \]

### Step 4: Set the discriminant greater than or equal to zero
To ensure the roots are real:
\[ 16\alpha^2 - 80\alpha + 96 \geq 0 \]
Dividing the entire inequality by 16:
\[ \alpha^2 - 5\alpha + 6 \geq 0 \]
Factoring the quadratic:
\[ (\alpha - 2)(\alpha - 3) \geq 0 \]

### Step 5: Determine the intervals
The roots of the equation are \( \alpha = 2 \) and \( \alpha = 3 \). We can test the intervals:
- For \( \alpha < 2 \): Choose \( \alpha = 1 \) → \( (1 - 2)(1 - 3) = (-1)(-2) = 2 \geq 0 \) (true)
- For \( 2 < \alpha < 3 \): Choose \( \alpha = 2.5 \) → \( (2.5 - 2)(2.5 - 3) = (0.5)(-0.5) = -0.25 < 0 \) (false)
- For \( \alpha > 3 \): Choose \( \alpha = 4 \) → \( (4 - 2)(4 - 3) = (2)(1) = 2 \geq 0 \) (true)

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

### Step 6: Condition for \( f(0) < 0 \)
Now we check the condition \( f(0) < 0 \):
\[ f(0) = c = \alpha - 2 < 0 \implies \alpha < 2 \]

### Step 7: Combine the conditions
From the discriminant, we have \( \alpha \leq 2 \) or \( \alpha \geq 3 \). From \( f(0) < 0 \), we have \( \alpha < 2 \). Therefore, the combined condition for at least one positive root is:
\[ \alpha < 2 \]

### Conclusion
The values of \( \alpha \) for which at least one root of the quadratic equation is positive are:
\[ \alpha < 2 \]