If a and b are the roots of

Solution:

We have,

a +b = 3, ab = p, c + d =12 and cd = q

a, b, c and d form a G.P.

∴ First term = a,  b = ar, c = ar2 and d = ar3

Then, we have

a + b = 3  and c + d = 12

$\Rightarrow a+a r=3$

$\Rightarrow a(1+r)=3$    ...(1)

Similarly, $a r^{2}(1+r)=12$    ...(2)

$\Rightarrow \frac{a r^{2}(1+r)}{a(1+r)}=\frac{12}{3}$

$\Rightarrow r^{2}=4$

$\Rightarrow r=2$

$\therefore a(1+r)=3$

$\Rightarrow a=1$

Now, $p=a b$

$\Rightarrow p=a \times a r=2$ And,

$q=c d$

$\Rightarrow q=a r^{2} \times a r^{3}=2^{5}=32$

$\therefore \frac{q+p}{q-p}=\frac{32+2}{32-2}=\frac{34}{30}=\frac{17}{15}$