Let `a, b, c, d` be real numbers in `G.P.` If `u, v, w` satisfy the system of equations `u + 2y +3w = 6,4u + 5v + 6w =12 and 6u + 9v = 4` then show that the roots of the equation `(1/u+1/v+/w)x^2+[(b-c)^2+(c-a)^2+(d-b)^2]x+u+v+w=0` and 20x^2+10(a-d)^2 x-9=0` are reciprocals of each other.

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)`