Let `alphaa n dbeta` be the roots of `x^2-x+p=0a n dgammaa n ddelta` be the root of `x^2-4x+q=0.` If `alpha,beta,a n dgamma,delta` are in G.P., then the integral values of `pa n dq` , respectively, are `-2,-32` b. `-2,3` c. `-6,3` d. `-6,-32`

Options

A. `-2,-32`
B. `-2,3`
C. `-6,3`
D. `-6,-32`

Explanation

Let `r` be the common ratio of the GP then
`beta=alpha r, gamma = alpha r^(2)` and `delta=alpha r^(3)`
`:.alpha+beta=1-impliesalpha +alphar=1`
Or `alpha(1+r)=1`…I
and `alpha beta=p=alpha(alpha r)=p`
or `alpha^(2)r=p`…….ii
and `gamma +delta=4impliesalphar^(2)+alhar^(3)=4`
or `alphar^(2)(1+r)=4`
`implies(alpha r^(2))(alphar^(3))=q`
or `alpha^(2)r^(5)=q`......iv
On dividing Eq iii by Eq i we get
`r^(2)=4impliesr=-2,2`
If we take `r=2` then `alpha` is not integer so we taker `r=-2` on substituting `r=-2` in Eq. i we get `alpha=-1`
Now from eqs ii and iv we get
`p=alpha^(2)r=(-1)^(2)(-2)=-2`
and `q=alpha^(2)r^(5)=(-1)^(2)(-2)^(5)=-32`
Hence `(p,q)=(-2,-32)`