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 943384: find three positive consecutive odd integers such that the square of the smallest is 9 more than the sum of the other two
Found 2 solutions by CubeyThePenguin, MathTherapy:
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
consecutive odd integers: (x-2), x, (x+2)

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

The integers could be {5, 7, 9} or {-1, 1, 3}.

Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

find three positive consecutive odd integers such that the square of the smallest is 9 more than the sum of the other two
With the fist being F, that's SIMPLY F2 = F + 2 + F + 4 + 9 , which forms the following quadratic equation: F2 - 2F - 15 = 0, and the first ODD INTEGER, 5. 

Now, the others are easy to decipher.