Math question?

(x)(x + 2) = x + 4 + 8
x^2 + 2x - x - 12 = 0
x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x = 3
So the three consecutive odd integers are 3, 5, 7

Find three consecutive odd integers such that the product of the first and the second exceeds the third by 8.

Let n, n + 2 and n + 4 be the three consecutive odd integers.

n(n + 2) - (n + 4) = 8
n² + 2n - n - 4 = 8
n² + n - 12 = 0
(n + 4)(n - 3) = 0
n = -4 (rejected) or n = 3

The three consecutive odd integers are 3, 5 and 7.

3, 5, 7

Let the numbers be x, x + 2 and x + 4.

Product of the first and second numbers will be x(x + 2) = x^2 + 2x
Since this product exceed the third number by 8, we have:
(x^2 + 2x) - (x + 4) = 8
x^2 + 2x - x - 4 = 8
x^2 + x - 4 - 8 = 0
x^2 + x - 12 = 0
(x + 4)(x - 3) = 0
x + 4 = 0 or x - 3 = 0
x = -4 or x = 3

We will reject x = -4 since it is even instead of odd.

x = 3 means that:
x + 2 = 3 + 2 = 5
x + 4 = 3 + 4 = 7

Hence, the integers are 3, 5 and 7.