SOLUTION: find three positive consecutive odd integers such that the square of the smallest is 9 more than the sum of the other two

SOLUTION: find three positive consecutive odd integers such that the square of the smallest is 9 more than the sum of the other two

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Question 727779: find three positive consecutive odd integers such that the square of the smallest is 9 more than the sum of the other two

Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
consecutive positive odd integers: (x-2), x, (x+2)

(x-2)^2 = 9 + x + (x+2)
x^2 - 4x + 4 = 2x + 11
x^2 - 6x - 7 = 0
(x - 7)(x + 1) = 0
x = 7, x = -1

integers are positive ----> integers are 5, 7, 9
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