SOLUTION: Find three consecutive positive odd integers such that the product of the first and the third is less than 7 times than the second.
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Question 862362: Find three consecutive positive odd integers such that the product of the first and the third is less than 7 times than the second.
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
consecutive odd integers: (x-2), x, (x+2)
(x-2)(x+2) < 7x
x^2 - 4 < 7x
x^2 - 7x - 4 < 0
integer solutions: x = 0, 1, 2, 3, 4 ---> odd: 1, 3
The integers have to be positive, so the integers are 1, 3, and 5.
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