Find three consecutive integers such that that product of the first and third is one
1 answer:
Answer:
n = the first
n + 1 = the 2nd
n + 2 = the 3rd
n(n + 2) - (n + 1) = 10(n + 2) + 1
n2 + 2n - n - 1 = 10n + 20 + 1
n2 - 9n - 22 = 0
(n - 11)(n + 2) = 0
If two factors = 0, at least one must = 0.
n - 11 = 0
n = 11
11,12,13
n + 2 = 0
n = -2
-2,-1,0
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