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

Explanation

It is given that a and b are the roots of `x^2-3x+p=0`
`a+b=3 ` and `ab=p` ------(1)
Also c and d are the roots of `x^2-12x+q=0`
c+d=12 and `cd=q`-------(2)
It is given that a,b,c,d are in G.P.
Let `a=x` ,`b=xr` , `c=xr^2`, `d=xr^3` from (1) and (2)
we obtain `x+xr=3 implies x(1+r)=3`
`xr^2+xr^3=12 implies xr^2(1+r)=12`
on dividing
`(xr^2(1+r))/(x(1+r))=12/3`
`r=pm2` When `r=2` then `x=1` and `r=-2` then `x=-3`
Case 1:
When ` r=2` and `x=1` ,`ab=x^2r=2`, `cd=x^2r^5=32`
`(q+p)/(q-p)=(32+2)/(32-2)`=`17:15`
when`r=-2`, `x=-3`, `ab=x^r=-18 `, `cd=x^2r^5=-288`
`(q+p)/(q-p)`=`(-288-18)/(-288+18)`=`17:15`