If `alpha,beta` are the roots of the quadratic equation `x^2-2(1-sin2theta)x-2 cos^2(2theta) = 0` , then the minimum value of `(alpha^2+beta^2)` is equal to

Options

A. `-4`
B. 8
C. 0
D. 2

Answer

To find the minimum value of \( \alpha^2 + \beta^2 \) given the quadratic equation \( x^2 - 2(1 - \sin 2\theta)x - 2\cos^2(2\theta) = 0 \), we can follow these steps:

### Step 1: Identify coefficients
The given quadratic equation can be compared with the standard form \( Ax^2 + Bx + C = 0 \). Here:
- \( A = 1 \)
- \( B = -2(1 - \sin 2\theta) \)
- \( C = -2\cos^2(2\theta) \)

### Step 2: Calculate \( \alpha + \beta \) and \( \alpha \beta \)
Using the relationships from Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -\frac{B}{A} = 2(1 - \sin 2\theta) \)
- The product of the roots \( \alpha \beta = \frac{C}{A} = -2\cos^2(2\theta) \)

### Step 3: Use the formula for \( \alpha^2 + \beta^2 \)
We can express \( \alpha^2 + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha \beta \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]

### Step 4: Substitute values into the formula
Substituting the values we found:
\[ \alpha^2 + \beta^2 = (2(1 - \sin 2\theta))^2 - 2(-2\cos^2(2\theta)) \]

### Step 5: Simplify the expression
Calculating \( (2(1 - \sin 2\theta))^2 \):
\[ = 4(1 - \sin 2\theta)^2 = 4(1 - 2\sin 2\theta + \sin^2 2\theta) \]
Now substituting back:
\[ \alpha^2 + \beta^2 = 4(1 - 2\sin 2\theta + \sin^2 2\theta) + 4\cos^2(2\theta) \]
Using the identity \( \sin^2 2\theta + \cos^2 2\theta = 1 \):
\[ = 4(1 - 2\sin 2\theta + 1) = 8 - 8\sin 2\theta \]

### Step 6: Find the minimum value
To find the minimum value of \( \alpha^2 + \beta^2 = 8(1 - \sin 2\theta) \), we need to maximize \( \sin 2\theta \). The maximum value of \( \sin 2\theta \) is 1, which occurs when \( 2\theta = \frac{\pi}{2} \) or \( \theta = \frac{\pi}{4} \).

Substituting \( \sin 2\theta = 1 \):
\[ \alpha^2 + \beta^2 = 8(1 - 1) = 0 \]

### Conclusion
Thus, the minimum value of \( \alpha^2 + \beta^2 \) is \( \boxed{0} \).