Question 871313: Find 3 consecutive odd integers such that 3 times the product of the first and third exceeds the product of the first and second by 28
Answer by CubeyThePenguin(3113) (Show Source):
You can put this solution on YOUR website! consecutive odd integers: (x-2), x, (x+2)
3(x-2)(x+2) = (x-2)(x) + 28
3(x^2 - 4) = x^2 - 2x + 28
3x^2 - 12 = x^2 - 2x + 28
2x^2 + 2x - 40 = 0
x^2 + x - 20 = 0
(x + 5)(x - 4) = 0
The integers must be odd, so the integers are -7, -5, and -3.
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