If roots of the cubic equation x3 + a x2 + 146x + c = 0 are three consecutive positive integers. Find the sum of the roots.
Solution: Let the roots be α, α + 1, α + 2
Sum of the root taken 2 at a time
α(α + 1) + (α + 1)(α + 2) + (α + 2)α = 146
3 α2 + 6α + 2 = 146
α2 + 2α - 48 = 0
α2 + 8α - 6α - 48 = 0
(α + 8)(α - 6) = 0
α = 6, α = -8
One root = 6
Other roots are 7, 8
Sum of the root = 6 + 7 + 8 = 21