Question 190578: Find three consecutive integers such that the square of the first plus the product of the other two is 67. Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Find three consecutive integers such that the square of the first plus the product of the other two is 67.
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Smallest number = x
x^2 + (x+1)*(x+2) = 67
2x^2 + 3x + 2 = 67
2x^2 + 3x - 65 = 0
(x - 5)*(2x + 13) = 0
x = -6.5 Ignore this one, it's not an integer
x = 5
