If α and β are the roots of the equation x ^ 2 + px + 2 = 0 and...

$\alpha, \beta$ are roots of ${x}^{2} + px + 2 = 0$

$⇒ \alpha + \beta = - p, \alpha\beta = 2$

$\dfrac{1}{\alpha}, \dfrac{1}{\beta}$ are roots of $2{x}^{2} + 2qx + 1 = 0$

$∴ \alpha, \beta$ are roots of ${x}^{2} + 2qx + 2 = 0$

$⇒ \alpha + \beta = - 2q, \alpha\beta = 2$

$∴ p = 2q . . . ( i )$

$⇒ \left(\alpha - \dfrac{1}{\alpha}\right)\left(\beta - \dfrac{1}{\beta}\right)\left(\alpha + \dfrac{1}{\beta}\right)\left(\beta + \dfrac{1}{\alpha}\right) = \left\{\alpha\beta - \dfrac{\alpha}{\beta} - \dfrac{\beta}{\alpha} + \dfrac{1}{\alpha\beta}\right\} \cdot \left\{\alpha\beta + \dfrac{1}{\alpha\beta} + 2\right\}$

$⇒ \left\{2 - \left(\dfrac{{\left(\alpha + \beta\right)}^{2} - 2\alpha\beta}{\alpha\beta}\right) + \dfrac{1}{2}\right\} \cdot$\left\{2 + \dfrac{1}{2} + 2\right\}$</p><p>$⇒ $\left\{\dfrac{5}{2} - \left(\dfrac{{p}^{2} - 4}{2}\right)\right\} \cdot \left\{\dfrac{9}{2}\right\} = \dfrac{9}{4}\left(9 - {p}^{2}\right)$ or, $\dfrac{9}{4}\left(9 - 4{q}^{2}\right)$