If `alpha` and `beta` are roots of `x^2+px+2=0` and `1/alpha,1/beta` are the roots of `2x^2+2qx+1=0`. Find the value of `(alpha-1/alpha)(beta-1/beta)(alpha+1/beta)(beta+1/alpha)1

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, we need to find the value of the expression \((\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha})\).

### Step 1: Identify the roots and their relationships
We know that:
- \(\alpha\) and \(\beta\) are the roots of the equation \(x^2 + px + 2 = 0\).
- The sum of the roots \(\alpha + \beta = -p\).
- The product of the roots \(\alpha \beta = 2\).

From the second equation \(2x^2 + 2qx + 1 = 0\), the roots are \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\).
- The sum of the roots \(\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{2q}{2} = -q\).
- The product of the roots \(\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{2}\).

### Step 2: Relate the sums and products
From the product of the roots:
\[ \frac{1}{\alpha \beta} = \frac{1}{2} \implies \alpha \beta = 2. \]
From the sum of the roots:
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = -q. \]
Substituting \(\alpha + \beta = -p\) and \(\alpha \beta = 2\):
\[ \frac{-p}{2} = -q \implies p = 2q. \]

### Step 3: Substitute values into the expression
Now we need to compute:
\[ (\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}). \]

1. **Calculate \(\alpha - \frac{1}{\alpha}\)**:
\[ \alpha - \frac{1}{\alpha} = \frac{\alpha^2 - 1}{\alpha}. \]

2. **Calculate \(\beta - \frac{1}{\beta}\)**:
\[ \beta - \frac{1}{\beta} = \frac{\beta^2 - 1}{\beta}. \]

3. **Calculate \(\alpha + \frac{1}{\beta}\)**:
\[ \alpha + \frac{1}{\beta} = \frac{\alpha \beta + 1}{\beta} = \frac{2 + 1}{\beta} = \frac{3}{\beta}. \]

4. **Calculate \(\beta + \frac{1}{\alpha}\)**:
\[ \beta + \frac{1}{\alpha} = \frac{\beta \alpha + 1}{\alpha} = \frac{2 + 1}{\alpha} = \frac{3}{\alpha}. \]

### Step 4: Combine all parts
Now substituting back into the expression:
\[ (\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \left(\frac{\alpha^2 - 1}{\alpha}\right)\left(\frac{\beta^2 - 1}{\beta}\right)\left(\frac{3}{\beta}\right)\left(\frac{3}{\alpha}\right). \]

### Step 5: Simplify the expression
This simplifies to:
\[ \frac{(\alpha^2 - 1)(\beta^2 - 1) \cdot 9}{\alpha^2 \beta^2}. \]
Using \(\alpha^2 \beta^2 = (\alpha \beta)^2 = 2^2 = 4\):
\[ = \frac{9(\alpha^2 - 1)(\beta^2 - 1)}{4}. \]

### Step 6: Final calculation
Now we need to find \((\alpha^2 - 1)(\beta^2 - 1)\):
\[ (\alpha^2 - 1)(\beta^2 - 1) = \alpha^2 \beta^2 - (\alpha^2 + \beta^2) + 1. \]
Using \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = p^2 - 4\):
\[ = 4 - (p^2 - 4) + 1 = 9 - p^2. \]

Thus, the final expression becomes:
\[ \frac{9(9 - p^2)}{4}. \]

### Conclusion
The value of \((\alpha - \frac{1}{\alpha})(\beta - \frac{1}{\beta})(\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha})\) is \(\frac{9(9 - p^2)}{4}\).