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Find two consecutive even integers whose product is 1,520

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2n = 1st even integer
2n + 2 = 2nd even integer

Product = 2n(2n + 2) = 4n^2 + 4n = 4(n^2 + n)
1520 = 4(n^2 + n)
1520/4 = 4(n^2 + n)/4
380 = n^2 + n
0 = n^2 + n - 380
0 = (n + 20)(n - 19)
n = -20, +19

Answer can not be negative so n = +19

2n = 2(19) = 38
2n + 2 = 2(19) + 2 = 40

ANSWER: the integers are 38 and 40


Posted on Dec 02, 2009

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