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 opposite in sign.

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 opposite in sign, we can follow these steps:

### Step 1: Identify the coefficients
The given quadratic equation can be rewritten in the standard form \( Ax^2 + Bx + C = 0 \):
- \( A = 4 \)
- \( B = -4(\alpha - 2) \)
- \( C = \alpha - 2 \)

### Step 2: Condition for roots to be opposite in sign
For the roots of the quadratic equation to be opposite in sign, the product of the roots must be negative. The product of the roots \( r_1 \) and \( r_2 \) is given by:
\[ r_1 \cdot r_2 = \frac{C}{A} \]
Thus, we have:
\[ r_1 \cdot r_2 = \frac{\alpha - 2}{4} \]
To ensure that the roots are opposite in sign, we need:
\[ \frac{\alpha - 2}{4} < 0 \]

### Step 3: Solve the inequality
To solve the inequality \( \frac{\alpha - 2}{4} < 0 \), we can multiply both sides by 4 (since 4 is positive, the direction of the inequality remains unchanged):
\[ \alpha - 2 < 0 \]
This simplifies to:
\[ \alpha < 2 \]

### Step 4: Conclusion
The values of \( \alpha \) for which both roots of the quadratic equation are opposite in sign are:
\[ \alpha < 2 \]