If $$\alpha$$ and $$\beta$$ are the roots of the equation $$x^2 + px + 2 = 0$$ and $$\frac{1}{\alpha}$$ and $$\frac{1}{\beta}$$ are the roots of the equation $$2x^2 + 2qx + 1 = 0$$, then $$\left(\alpha - \frac{1}{\alpha}\right)\left(\beta - \frac{1}{\beta}\right)\left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right)$$ is equal to:

Let us denote the two unknown roots of the first quadratic by $$\alpha$$ and $$\beta$$. They satisfy the equation

$$x^{2}+px+2=0.$$

For any quadratic $$x^{2}+Sx+P=0$$ the sum and product of its roots are given by Vieta’s formulae:

$$\alpha+\beta=-S,\qquad \alpha\beta=P.$$

Applying these to the first equation we have

$$\alpha+\beta=-p\qquad\text{and}\qquad\alpha\beta=2.$$

Now we are told that $$\dfrac1\alpha$$ and $$\dfrac1\beta$$ are the roots of the second quadratic

$$2x^{2}+2qx+1=0.$$

Again invoking Vieta’s formulae for this quadratic, we get

$$\frac1\alpha+\frac1\beta=-\frac{2q}{2}=-q,\qquad \frac1\alpha\cdot\frac1\beta=\frac1{\alpha\beta}=\frac12.$$

Because $$\dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta},$$ we equate the two expressions for this sum:

$$\frac{\alpha+\beta}{\alpha\beta}=-q.$$

Substituting $$\alpha+\beta=-p$$ and $$\alpha\beta=2$$ from the first quadratic, we get

$$\frac{-p}{2}=-q\quad\Longrightarrow\quad p=2q.$$

This relation between $$p$$ and $$q$$ will not be directly needed in the numerical value we are about to compute, but it is reassuring to see the two equations are consistent.

Our aim is to evaluate

$$E=\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right)\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right).$$

We treat the two bracketed pairs separately.

1. Calculating $$\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right)$$

First expand:

$$\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right) =\alpha\beta-\frac{\alpha}{\beta}-\frac{\beta}{\alpha} +\frac1{\alpha\beta}.$$

The quantities inside can be rewritten with the symmetric sums we already know.

• We already have $$\alpha\beta=2.$$

• Next, $$\frac{\alpha}{\beta}+\frac{\beta}{\alpha} =\frac{\alpha^{2}+\beta^{2}}{\alpha\beta}.$$

We still need $$\alpha^{2}+\beta^{2}$$. Use the identity

$$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2\alpha\beta.$$

Substituting $$\alpha+\beta=-p$$ and $$\alpha\beta=2$$ gives

$$\alpha^{2}+\beta^{2}=(-p)^{2}-2\cdot2=p^{2}-4.$$

Hence

$$\frac{\alpha}{\beta}+\frac{\beta}{\alpha} =\frac{p^{2}-4}{\alpha\beta} =\frac{p^{2}-4}{2}.$$

• Finally, $$\dfrac1{\alpha\beta}=\dfrac1{2}.$$

Substituting all these results back into the expansion we obtain

$$\left(\alpha-\frac1\alpha\right)\left(\beta-\frac1\beta\right) =2-\frac{p^{2}-4}{2}+\frac12 =\frac{4}{2}-\frac{p^{2}-4}{2}+\frac12 =\frac{4-(p^{2}-4)+1}{2} =\frac{9-p^{2}}{2}.$$

2. Calculating $$\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right)$$

Expand directly:

$$\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right) =\alpha\beta+\alpha\cdot\frac1\alpha +\frac1\beta\cdot\beta+\frac1\beta\cdot\frac1\alpha.$$

Simplify each term:

$$\alpha\beta=2,\qquad \alpha\cdot\frac1\alpha=1,\qquad \frac1\beta\cdot\beta=1,\qquad \frac1\beta\cdot\frac1\alpha=\frac1{\alpha\beta}=\frac12.$$

So

$$\left(\alpha+\frac1\beta\right)\left(\beta+\frac1\alpha\right) =2+1+1+\frac12 =\frac92.$$

3. Combining the two factors

Multiply the results from steps 1 and 2:

$$E =\left(\frac{9-p^{2}}{2}\right)\left(\frac92\right) =\frac{(9-p^{2})\cdot9}{4} =\frac94\,(9-p^{2}).$$

This matches Option D.

Hence, the correct answer is Option D.