If `alpha, beta and gamma` are three consecutive terms of a non-constant G.P. such that the equations `ax^(2) + 2beta x + gamma = 0 and x^(2) + x - 1= 0` have a common root, then `alpha (beta + gamma)` is equal to

Options

A. 0
B. `alpha beta`
C. `alpha gamma`
D. `beta gamma`

Answer

D

Explanation

Given `alpha, beta and gamma` are three consectuve terms of a non-constant GP.
Let `alpha=alpha, beta=alphar, gamma=alphar^(2), {r ne0,1}`
and given quadratic equation is
`ax^(2)+2betax+gamma=0" "...(i)`
On putting the values of `alpha, beta, gamma` in Eq(i), we get
`alphax^(2)+2alphar^(2)=0`
`impliesx^(2)+2rx+r^(2)=0`
`implies(x+r)^(2)=0`
`impliesx=-r`
`because` The quadratic equationa `alphax^(2)+2betax+gamma=0` and `x^(2)+x-1=0` have a common root, so `x=-r` must be root of equation `x^(2)+x-1=0,so`
`r^(2)-r-1=0" "...(ii)`
Now, `alpha(beta+gamma)=alpha(alphar+alphar^(2))=alpha^(2)(r+r^(2))`
From the options,
`beta gamma=alphar. alphar^(2)=alpha^(2)r^(3)=alpha^(2)(r+r^(2))`
`" "[because r^(2)-r-1=0impliesr^(3)=r+r^(2)]`
`thereforealpha(beta+gamma)=betagamma`