Solution

Given:

$ x^2 - 3x + p = 0 \Rightarrow \alpha + \beta = 3,\ \alpha \beta = p $

$ x^2 - 6x + q = 0 \Rightarrow \gamma + \delta = 6,\ \gamma \delta = q $

Given $ \alpha, \beta, \gamma, \delta $ are in GP.

Let GP be:
$ \alpha = a,\ \beta = ar,\ \gamma = ar^2,\ \delta = ar^3 $

From sum:
$ \alpha + \beta = a(1 + r) = 3 \quad ...(1)$

$ \gamma + \delta = ar^2(1 + r) = 6 \quad ...(2)$

Divide (2) by (1):
$ r^2 = \frac{6}{3} = 2 \Rightarrow r = \sqrt{2} $

From (1):
$ a(1 + \sqrt{2}) = 3 \Rightarrow a = \frac{3}{1 + \sqrt{2}} $

Now,

$ p = \alpha \beta = a^2 r = a^2 \sqrt{2} $

$ q = \gamma \delta = a^2 r^5 = a^2 ( \sqrt{2} )^5 = 4\sqrt{2}, a^2 $

So,
$ 2q = 8\sqrt{2}, a^2 $

Now ratio:
$ (2q + p) : (2q - p) $

$ = (8\sqrt{2}a^2 + \sqrt{2}a^2) : (8\sqrt{2}a^2 - \sqrt{2}a^2) $

$ = 9\sqrt{2}a^2 : 7\sqrt{2}a^2 $

$ = 9 : 7 $