SOLUTION: Find three consecutive integers such that three times the sum of all three equals the product of the larger two.

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Question 663691: Find three consecutive integers such that three times the sum of all three equals the product of the larger two.
Answer by kevwill(135) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the first integer. Then the next two consecutive integers are x+1 and x+2.
Three times the sum of all three is 3*(x + (x+1) + (x+2))
The product of the larger two is (x+1) * (x+2)
So we have
(x+1) * (x+2) = 3*(x + (x+1) + (x+2))
x^2 + 3x + 2 = 3*(3x +3)
x^2 + 3x + 2 = 9x + 9
x^2 + 3x + 2 - 9x - 9 = 9x + 9 - 9x - 9
x^2 - 6x - 7 = 0
We can solve for x using the quadratic equation x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ with a=1, b=-6 and c=-7
x+=+%28-%28-6%29+%2B-+sqrt%28+%28-6%29%5E2-4%2A1%2A%28-7%29+%29%29%2F%282%2A1%29+
x+=+%286+%2B-+sqrt%28+36%2B28+%29%29%2F2+
x+=+%286+%2B-+sqrt%28+64+%29%29%2F2+
x+=+%286+%2B-+8%29%2F2+
x+=+14%2F2+=+7 or x+=+-2%2F2+=+-1
So there are two solutions to this problem:
7, 8, 9 and -1, 0, 1