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 equal in magnitude but opposite in sign

Answer

To solve the problem step by step, we need to find the values of \( \alpha \) for which the quadratic equation \( 4x^2 - 4(\alpha - 2)x + \alpha - 2 = 0 \) has roots that are equal in magnitude but opposite in sign.

### 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: Set the condition for equal magnitude but opposite sign roots
For the roots of a quadratic equation to be equal in magnitude but opposite in sign, the coefficient \( b \) must be zero. This is because if \( b = 0 \), the equation simplifies to \( ax^2 + c = 0 \), leading to roots of the form \( \pm \sqrt{-\frac{c}{a}} \).

### Step 3: Set \( b = 0 \)
We set the coefficient \( b \) to zero:
\[ -4(\alpha - 2) = 0 \]
Solving for \( \alpha \):
\[ \alpha - 2 = 0 \implies \alpha = 2 \]

### Step 4: Check the condition for \( c/a > 0 \)
Next, we need to ensure that the value of \( c/a \) is positive for the roots to be real. We calculate \( c/a \):
\[ c = \alpha - 2 \quad \text{and} \quad a = 4 \]
Thus,
\[ \frac{c}{a} = \frac{\alpha - 2}{4} \]
For the roots to be real, we need:
\[ \frac{\alpha - 2}{4} > 0 \]
This implies:
\[ \alpha - 2 > 0 \implies \alpha > 2 \]

### Step 5: Conclusion
We have found that \( \alpha = 2 \) from the condition \( b = 0 \), but this value does not satisfy the condition \( \alpha > 2 \). Therefore, there are no values of \( \alpha \) that satisfy both conditions simultaneously.

### Final Answer
There are no values of \( \alpha \) for which both roots are equal in magnitude but opposite in sign.

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