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Find 3 consecutive positive integers such that the product of the first and the third is 29 more than the second?
Wiki User
∙ 13y ago
Let the three numbers be represented by, n, n+1 and n+2.
Then n(n+2) = (n +1) + 29
n2 + 2n = n + 30
n2 + n - 30 = 0
This can be factored (n + 6)(n - 5) = 0
When n + 6 = 0 then n = -6 : When n - 5 = 0 then n = 5.
As only the positive integer solution is required then the three numbers are 5, 6 and 7.
Wiki User
∙ 13y ago
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