Question

find three consecutive positive even integers such that the product of the second and third integers is 20 more than 10 times the first integer.

Answer 1

Answer: 6, 8, 10

Explanation:

We can write the three numbers as: x, x+2, and x+4

(x+2)(x+4) = 10x + 20

x² + 6x + 8 = 10x + 20

x² - 4x - 12 = 0

(x-6)(x+2) = 0

x = -2,6

We will focus on x = 6 and ignore -2

To check our answer: (6+2)(6+4) = 10(6) + 20?

8*10 = 60 + 20?

Yes, 80 = 80