Question

Find three positive consecutive integers such that the product of the first and second is two more than three times the third.

Answer 1

Let the three consecutive positive integers be the following:
x, x + 1, and x + 2
The product of the first two integers is x * (x + 1)
Three times the third is 3(x+2)
Two more than three times the third is 2 + 3(x+2)
Putting these expressions together according to the dictates of the problem:
x(x+1) = 2 + 3(x+2)
s^2 + x = 2 + 3x + 6
x^2 + x = 8 + 3x
x^2 - 2x - 8 = 0
Use the Zero Product Property
(x-4)*(x+2) = 0
x = 4 or x = -2
The integers are positive so -2 is excluded.
x = 4
x + 1 = 5
x + 2 + 6