Let α and β be the roots of x ^ 2 - 3x + p = 0 and γ and be the...

Let $\alpha = a, \beta = ar, \gamma = a{r}^{2}, \delta = a{r}^{3}$

$\alpha + \beta = 3 ⇒ a + ar = 3$ ...........\left(1\right)$</p><p>$\gamma + \delta = 6 ⇒ a{r}^{2} + a{r}^{3} = 6$.........\left(2\right)$

By \left(1\right)$and \left(2\right) ⇒ \dfrac{a{r}^{2}\left(1 + r\right)}{a\left(1 + r\right)} = \dfrac{6}{3}$

$⇒ {r}^{2} = 2$

$∴ \dfrac{2q + p}{2q - p} = \dfrac{2\gamma\delta + \alpha\beta}{2\gamma\delta - \alpha\beta} = \dfrac{2{a}^{2}{r}^{5} + {a}^{2}r}{2{a}^{2}{r}^{5} - {a}^{2}r} = \dfrac{2{r}^{4} + 1}{2{r}^{4} - 1} = \dfrac{8 + 1}{8 - 1} = \dfrac{9}{7}$