If `alpha and beta` are roots of the equation `x^(2)+px+2=0` and `(1)/(alpha)and (1)/(beta)` are the roots of the equation `2x^(2)+2qx+1=0`, then `(alpha-(1)/(alpha))(beta-(1)/(beta))(alpha+(1)/(beta))(beta+(1)/(alpha))` is equal to :

Options

A. `(9)/(4)(9+p^(2))`
B. `(9)/(4)(9+q^(2))`
C. `(9)/(4)(9-p^(2))`
D. `(9)/(4)(9-q^(2))`

Answer

To solve the problem step by step, we will analyze the given equations and their roots.

### Step 1: Identify the roots and their relationships
We have two equations:
1. \( x^2 + px + 2 = 0 \) with roots \( \alpha \) and \( \beta \).
2. \( 2x^2 + qx + 1 = 0 \) with roots \( \frac{1}{\alpha} \) and \( \frac{1}{\beta} \).

From the first equation, we can use Vieta's formulas:
- Sum of roots: \( \alpha + \beta = -p \)
- Product of roots: \( \alpha \beta = 2 \)

### Step 2: Analyze the second equation
For the second equation, again using Vieta's formulas:
- Sum of roots: \( \frac{1}{\alpha} + \frac{1}{\beta} = -\frac{q}{2} \)
- Product of roots: \( \frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{2} \)

### Step 3: Find expressions for the roots
From the product of the roots of the second equation:
\[ \frac{1}{\alpha \beta} = \frac{1}{2} \implies \alpha \beta = 2 \quad \text{(which we already know)} \]

From the sum of the roots of the second equation:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{-p}{2} \implies -\frac{q}{2} = \frac{-p}{2} \implies q = p \]

### Step 4: Calculate the expression
We need to find:
\[ (\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) \]

#### Step 4.1: Simplify each term
1. \( \alpha - \frac{1}{\alpha} = \frac{\alpha^2 - 1}{\alpha} \)
2. \( \beta - \frac{1}{\beta} = \frac{\beta^2 - 1}{\beta} \)
3. \( \alpha + \frac{1}{\beta} = \frac{\alpha \beta + 1}{\beta} = \frac{2 + 1}{\beta} = \frac{3}{\beta} \)
4. \( \beta + \frac{1}{\alpha} = \frac{\beta \alpha + 1}{\alpha} = \frac{2 + 1}{\alpha} = \frac{3}{\alpha} \)

#### Step 4.2: Combine the terms
Now substituting these into the expression:
\[ (\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \left(\frac{\alpha^2 - 1}{\alpha}\right)\left(\frac{\beta^2 - 1}{\beta}\right)\left(\frac{3}{\beta}\right)\left(\frac{3}{\alpha}\right) \]

This simplifies to:
\[ \frac{(\alpha^2 - 1)(\beta^2 - 1) \cdot 9}{\alpha^2 \beta^2} \]

Using \( \alpha \beta = 2 \):
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = 4 \]

Thus, we need to calculate \( (\alpha^2 - 1)(\beta^2 - 1) \):
\[ (\alpha^2 - 1)(\beta^2 - 1) = \alpha^2 \beta^2 - (\alpha^2 + \beta^2) + 1 \]

Using \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = p^2 - 4 \):
\[ = 4 - (p^2 - 4) + 1 = 9 - p^2 \]

### Final Expression
So, we can substitute back:
\[ \frac{(9 - p^2) \cdot 9}{4} = \frac{9(9 - p^2)}{4} \]

### Conclusion
The final answer is:
\[ \frac{9(9 - p^2)}{4} \]