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 opposite in sign

Answer

To solve the problem, we need 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.

### Step-by-step Solution:

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

2. **Use the relationship between the roots:**
Let the roots of the equation be \( p \) and \( q \). According to Vieta's formulas:
- The product of the roots \( pq = \frac{c}{a} \)

Therefore, we have:
\[ pq = \frac{\alpha - 2}{4} \]

3. **Condition for opposite signs:**
For the roots \( p \) and \( q \) to be opposite in sign, their product must be negative:
\[ pq < 0 \]

Substituting the expression for \( pq \):
\[ \frac{\alpha - 2}{4} < 0 \]

4. **Solve the inequality:**
To solve the inequality \( \frac{\alpha - 2}{4} < 0 \), we can multiply both sides by 4 (since 4 is positive, the inequality sign remains unchanged):
\[ \alpha - 2 < 0 \]
Adding 2 to both sides gives:
\[ \alpha < 2 \]

5. **Conclusion:**
The values of \( \alpha \) for which both roots are opposite in sign are:
\[ \alpha \in (-\infty, 2) \]