If roots of the cubic equation $x^{3}$ + a $x^{2}$ + 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 $α^{2}$ + 6α + 2 = 146

$α^{2}$ + 2α - 48 = 0

$α^{2}$ + 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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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

If roots of the cubic equation $x^{3}$ + a $x^{2}$ + 146x + c = 0 are three consecutive positive integers. Find the sum of the roots.

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