SOLUTION: 5 consecutive positive integer are such that twice the product of the smallest and largest integers exceeds the square of the middle integer by 41. By letting x equal the smallest

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Question 757854: 5 consecutive positive integer are such that twice the product of the smallest and largest integers exceeds the square of the middle integer by 41. By letting x equal the smallest integer, set up and solve a quadratic equation to find the middle integer.
Answer by CubeyThePenguin(3113) About Me  (Show Source):
You can put this solution on YOUR website!
consecutive positive integers: x, (x+1), (x+2), (x+3), (x+4)

2(x)(x+4) = (x+2)^2 + 41
2(x^2 + 4x) = x^2 + 4x + 4 + 41
2x^2 + 8x = x^2 + 4x + 45
x^2 + 4x - 45 = 0
(x + 9)(x - 5) = 0
x = -9, x = 5

x has to be positive, so the middle integer is x + 2 = 7.