If a and b are the roots of x^2-3x+p=0 and c,d are roots of x^2-12x+q=0,where a,b,c,d form a G.P. Prove that (q+p):(q-p)=17:15

Since x^2-3x+p=0 a+b=3 and ab=pAnd x^2-12x+q =0c+d=12 and cd=qAnd a,b,c,d are in GP Let a =a ,b=ar,c=ar^2 and d=ar^3a+b/c+d= 3/12a+ar/ar^2+ar^3=1/4a(1+r)/ar^2(1+r)=1/41/r^2=1/4r=2Now (q+p):(q-p)= (cd+ab):(cd-ab) =(a^2r^5+a^2r):(a^2r^5-a^2r) =a^2r(r^4+1):a^2r(r^4-1) =(r^4+1):(r^4-1) =(2^4+1):(2^4-1) =(16+1):(16-1) =17:15 proved