# How do you find three consecutive even integers such that the product of the first and second is 8 less than the product of 4 and the third?

Nov 3, 2015

I found:
$4 , 6 , 8$
or
$- 2.0 , 2$

#### Explanation:

Call your consecutive even integers:
$2 n$
$2 n + 2$
$2 n + 4$

you get:
$\left(2 n\right) \left(2 n + 2\right) = 4 \left(2 n + 4\right) - 8$
$4 {n}^{2} + 4 n = 8 n + 16 - 8$
$4 {n}^{2} - 4 n - 8 = 0$
Using the Quadratic Formula you get:
${n}_{1 , 2} = \frac{4 \pm \sqrt{16 - 4 \left(4\right) \left(- 8\right)}}{8} =$

${n}_{1} = 2$
${n}_{2} = - 1$

So we have:
when $n = 2$
$2 n = 4$
$2 n + 2 = 6$
$2 n + 4 = 8$

when $n = - 1$
$2 n = - 2$
$2 n + 2 = 0$
$2 n + 4 = 2$