Let `'m'` be a real number, and suppose that two of the three solutions of the cubic equation `x^(3)+3x^(2)-34x=m` differ by `1`. Then possible value of `'m'` is/are

Options

A. `120`
B. `80`
C. `-48`
D. `-32`

Answer

A, C

Explanation

`(a,c)` Suppose that both `r` and `r+1` are solutions to the equation `x^(3)+3x^(2)-34x=m`
Then `r^(3)+3r^(2)-34r=m`, and also
`(r+1)^(3)+3(r+1)^(2)-34(r+1)=m`
Subtracting the first of these equations from the second gives.
`(3r^(2)+3r+1)+3(2r+1)-34=0`, or `3r^(2)+9r-30=0`
or `r^(2)+3r-10=0`
or `(r+5)(r-2)=0`
or `r=-5` or `r=2`. If `r=-5`
`:.m=(-5)^(3)+3(-5)^(2)-34(-5)`
`=-125+75+170=120`
For `r=2`, `m=-48`