If m is chosen in the quadratic equation `(m^(2)+1)x^(2)-3x+(m^(2)+1)^(2)=0` such that the sum of its roots is greatest, then the absolute difference of the cubes of its roots is (A) `10sqrt5` (B) `8sqrt5` (C) `8sqrt3` (D) `4sqrt3`

Options

A. `10sqrt5`
B. `5sqrt5`
C. `8sqrt3`
D. `4sqrt3`

Answer

B

Explanation

Given quadratic equation is
`(m^(2)+1)x^(2)-3x(m^(2)+1)^(2)=0" "…(i)`
Let the roots of quadratic Eq. (1)mn are `alpha and beta.` so `alpha+beta(3)/(m^(2)+1)and alphabeta=m^(2)+1`
According to the question, the sum of roots is greatest and it is possible only when `(m^(2)+1)` is minimu and minimum value of `m^(2)+1=1,` when `m=0"`
`thereforealpha +beta=3` and `alpha beta=1, as m =0`
Now, the absolute difference of the cubes of roots
`=|alpha^(2)-beta^(3)|`
`=|alpha-beta||alpha^(2)+alphabeta|`
`=sqrt((alpha+beta)^(2)-4alphabeta)|(alpha+beta)^(2)-alpha beta|`
`=sqrt(9-4)|9-1|=8sqrt5`