Official solution
\[\alpha \] and \[\beta \] are the roots of x2 - 3x + p = 0
\[ \therefore \] \[\alpha \] + \[\beta \] = 3 and \[\alpha \beta \] = p
\[\gamma \] and \[\delta \] are the roots of x2 - 6x + q = 0
\[ \therefore \] \[\gamma \] + \[\delta \] = 6 and \[\gamma \delta \] = q
Given that, \[\alpha \], \[\beta \], \[\gamma \], \[\delta \] are in G.P.
\[ \therefore \] Let \[\alpha \] = a, \[\beta \] = ar, \[\gamma \] = ar2 and \[\delta \] = ar3
As \[\alpha \] + \[\beta \] = 3
\[ \Rightarrow \] a + ar = 3
\[ \Rightarrow \] a(1 + r) = 3 ...(1)
Also \[\gamma \] + \[\delta \] = 6
\[ \Rightarrow \] ar2 + ar3 = 6
\[ \Rightarrow \] ar2 (1 + r) = 6 ... (2)
Dividing (2) by (1),
r2 = 2
\[ \Rightarrow r = \sqrt 2 \]
\[ \therefore \] \[a = {3 \over {1 + \sqrt 2 }}\]
So, \[\alpha = {3 \over {1 + \sqrt 2 }}\], \[\beta = {{3\sqrt 2 } \over {1 + \sqrt 2 }}\], \[\gamma = {{3 \times 2} \over {1 + \sqrt 2 }}\], \[\delta = {{3(2\sqrt {2)} } \over {1 + \sqrt 2 }}\]
\[ \therefore \] \[p = \alpha \beta = {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}\]
and \[q = \gamma \delta = {{36\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}\]
Now, \[{{2q + p} \over {2q - p}}\]
\[ = {{{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} + {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}} \over {{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} - {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}}}\]
\[ = {{72\sqrt 2 + 9\sqrt 2 } \over {72\sqrt 2 - 9\sqrt 2 }}\]
\[ = {{81} \over {63}}\]
\[ = {9 \over 7}\]