Let a,b,c and d are real numbers in GP. Suppose u,v,w satisfy the system of equations `u+2y+3w=6,4u+5y+6w=12` and `6u+9v=4`. Further consider the expressions
`f(x)=(1/u+1/v+1/w)x^(2)+[(b-c)^(2)+(c-a)^(2)+(x-b)^(2)]`
`x+u+v+w=0` and `g(x)=20x^(2)+10(a-d)^(2)x-9=0`
`(u+v+w)` is equal to

Options

A. `alpha, beta`
B. `-alpha,-beta`
C. `1/(alpha), 1/(beta)`
D. `-1/(alpha),-1/(beta)`

Explanation

Now `f(x)=(1/u+1/v+1/2)x^(2)x[(b-c)^(2)`
`+(c-a)^(2)+(d-b)^(2)]x+u+v+w=0`
`impliesf(x)=-9/10x^(2)+(a-d)^(2)x+2=0`
`impliesf(x)=-9x^(2)+10(a-d)^(2)x+2=0`
`impliesf(x)=-9x^(2)+10(a-d)^(2)x+20=0`…v
Given roots of `f(x)=0` are `alpha` and `beta`
Now replacing `x` by `1/x` in Eq v then
`(-9)/(x^(2))+(10(a-d)^(2))/x+20=0`
`implies20x^(2)+10(a-d)^(2)x-9=0`
`g(x)=0`
`:.` Roots of `g(x)=0` are `1/(alpha),1/(beta)`