Question

The product of the first and third of three consecutive integers is 3 more than 3 times the second integer. Find the integers.

Answer 1

Let:

x = the middle integer

Then the 3 consecutive integers are:

x - 1, x, x + 1

The product of the first and the third integer:

(x - 1)*(x + 1) = x^2 - 1

is equal to:

3 more than 3 times the second integer:

3x + 3

So:

x^2 - 1 = 3x + 3

Solve:

x^2 - 3x - 4 = 0

(x-4)(x+1) = 0

x = -1 and 4

Since x is the middle integer, we have two sets of solutions:

3,4,5 and -2,-1,0