Let `4x^2 - 4(alpha - 2)x + alpha - 2 = 0 (alpha in R)` be a quadratic equation. Find the values of 'a' for which
(i) Both roots are real and distinct.
(ii) Both roots are equal.
(iii) Both roots are imaginary
(iv) Both roots are opposite in sign.
(v) Both roots are equal in magnitude but opposite in sign

Answer

To solve the given quadratic equation \(4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0\) for different conditions regarding its roots, we will analyze the discriminant and the nature of the roots.

### Step-by-Step Solution:

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

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

3. **Factor the discriminant**:
To find the conditions for different types of roots, we will factor the discriminant:
\[ D = 16(\alpha^2 - 5\alpha + 6) \]
Factoring the quadratic:
\[ D = 16(\alpha - 2)(\alpha - 3) \]

4. **Analyze the conditions**:
- **(i) Both roots are real and distinct**:
For the roots to be real and distinct, the discriminant must be greater than zero:
\[ D > 0 \implies (\alpha - 2)(\alpha - 3) > 0 \]
This inequality holds for:
\[ \alpha < 2 \quad \text{or} \quad \alpha > 3 \]

- **(ii) Both roots are equal**:
For the roots to be equal, the discriminant must be zero:
\[ D = 0 \implies (\alpha - 2)(\alpha - 3) = 0 \]
This gives:
\[ \alpha = 2 \quad \text{or} \quad \alpha = 3 \]

- **(iii) Both roots are imaginary**:
For the roots to be imaginary, the discriminant must be less than zero:
\[ D < 0 \implies (\alpha - 2)(\alpha - 3) < 0 \]
This inequality holds for:
\[ 2 < \alpha < 3 \]

- **(iv) Both roots are opposite in sign**:
The product of the roots (given by \(\frac{c}{a}\)) must be negative:
\[ \frac{\alpha - 2}{4} < 0 \implies \alpha - 2 < 0 \implies \alpha < 2 \]
Additionally, the discriminant must be greater than zero:
\[ D > 0 \implies \alpha < 2 \quad \text{or} \quad \alpha > 3 \]
The intersection gives:
\[ \alpha < 2 \]

- **(v) Both roots are equal in magnitude but opposite in sign**:
For this condition, we need:
\[ \alpha + \beta = 0 \quad \text{and} \quad \alpha \beta < 0 \]
This means:
\[ -\frac{b}{a} = 0 \implies -\frac{-4(\alpha - 2)}{4} = 0 \implies \alpha - 2 = 0 \implies \alpha = 2 \]
And since \(\alpha < 2\) must also hold, this condition does not yield any valid \(\alpha\).

### Summary of Results:
- (i) Real and distinct roots: \(\alpha < 2\) or \(\alpha > 3\)
- (ii) Equal roots: \(\alpha = 2\) or \(\alpha = 3\)
- (iii) Imaginary roots: \(2 < \alpha < 3\)
- (iv) Opposite sign roots: \(\alpha < 2\)
- (v) Equal in magnitude but opposite in sign: No valid \(\alpha\)