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Question

If roots of the cubic equation x3 + a x2 + 146x + c = 0 are three consecutive positive integers. Find the sum of the roots.


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Solution

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


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