If `alpha and beta` be the roots of the equation `x^(2) + px - 1//(2p^(2)) = 0`,
where `p in R`. Then the minimum value of `alpha^(4) + beta^(4)` is

Options

A. `2sqrt2`
B. `2 - sqrt2`
C. `2`
D. `2 + sqrt2`

Explanation

To find the minimum value of \( \alpha^4 + \beta^4 \) where \( \alpha \) and \( \beta \) are the roots of the equation

\[ x^2 + px - \frac{1}{2p^2} = 0, \]

we can use the properties of the roots of a quadratic equation.

### Step 1: Identify the sum and product of the roots
From Vieta's formulas, for the quadratic equation \( ax^2 + bx + c = 0 \):

- The sum of the roots \( \alpha + \beta = -\frac{b}{a} = -\frac{p}{1} = -p \).
- The product of the roots \( \alpha \beta = \frac{c}{a} = -\frac{1}{2p^2} \).

### Step 2: Express \( \alpha^4 + \beta^4 \) in terms of \( \alpha + \beta \) and \( \alpha \beta \)
We can express \( \alpha^4 + \beta^4 \) using the identity:

\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2\beta^2. \]

### Step 3: Find \( \alpha^2 + \beta^2 \)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):

\[ \alpha^2 + \beta^2 = (-p)^2 - 2\left(-\frac{1}{2p^2}\right) = p^2 + \frac{1}{p^2}. \]

### Step 4: Find \( \alpha^2 \beta^2 \)
We have:

\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(-\frac{1}{2p^2}\right)^2 = \frac{1}{4p^4}. \]

### Step 5: Substitute back into \( \alpha^4 + \beta^4 \)
Now substituting \( \alpha^2 + \beta^2 \) and \( \alpha^2 \beta^2 \) into the expression for \( \alpha^4 + \beta^4 \):

\[ \alpha^4 + \beta^4 = \left(p^2 + \frac{1}{p^2}\right)^2 - 2\left(\frac{1}{4p^4}\right). \]

### Step 6: Simplify the expression
Calculating \( \left(p^2 + \frac{1}{p^2}\right)^2 \):

\[ \left(p^2 + \frac{1}{p^2}\right)^2 = p^4 + 2 + \frac{1}{p^4}. \]

Thus,

\[ \alpha^4 + \beta^4 = p^4 + 2 + \frac{1}{p^4} - \frac{1}{2p^4} = p^4 + 2 + \frac{1}{2p^4}. \]

### Step 7: Find the minimum value
Let \( y = p^4 + \frac{1}{2p^4} \). To find the minimum value of \( y \), we can use the AM-GM inequality:

\[ p^4 + \frac{1}{2p^4} \geq 2\sqrt{p^4 \cdot \frac{1}{2p^4}} = 2\sqrt{\frac{1}{2}} = \sqrt{2}. \]

Thus, the minimum value of \( y \) is \( \sqrt{2} \) when \( p^4 = \frac{1}{2p^4} \) or \( p^8 = \frac{1}{2} \).

So the minimum value of \( \alpha^4 + \beta^4 \) is:

\[ \sqrt{2} + 2. \]

### Final Answer
The minimum value of \( \alpha^4 + \beta^4 \) is:

\[ 2 + \sqrt{2}. \]