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

Options

A. 2
B. `2 sqrt(2)`
C. `2 + sqrt(2)`
D. none of these

Answer

To find the minimum possible value of \( \alpha^2 + \beta^2 \) where \( \alpha \) and \( \beta \) are the roots of the equation

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

we can follow these steps:

### Step 1: Identify the coefficients
For the quadratic equation \( ax^2 + bx + c = 0 \), we have:
- \( a = 1 \)
- \( b = p \)
- \( c = -\frac{1}{2p^2} \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas:
- 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 3: Express \( \alpha^2 + \beta^2 \)
We can express \( \alpha^2 + \beta^2 \) in terms of \( \alpha + \beta \) and \( \alpha \beta \):
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \]
Substituting the values from Vieta's formulas:
\[ \alpha^2 + \beta^2 = (-p)^2 - 2\left(-\frac{1}{2p^2}\right) \]
This simplifies to:
\[ \alpha^2 + \beta^2 = p^2 + \frac{1}{p^2} \]

### Step 4: Find the minimum value of \( p^2 + \frac{1}{p^2} \)
To find the minimum value of \( p^2 + \frac{1}{p^2} \), we can use the AM-GM inequality:
\[ \frac{p^2 + \frac{1}{p^2}}{2} \geq \sqrt{p^2 \cdot \frac{1}{p^2}} = 1 \]
Thus,
\[ p^2 + \frac{1}{p^2} \geq 2 \]

### Step 5: Conclusion
The minimum value of \( \alpha^2 + \beta^2 \) is therefore \( 2 \).

### Final Answer
The minimum possible value of \( \alpha^2 + \beta^2 \) is \( \boxed{2} \).