Question

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?

Answer 1

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

Explanation:

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

you get:
`(2n)(2n+2)=4(2n+4)-8`
`4n^2+4n=8n+16-8`
`4n^2-4n-8=0`
Using the Quadratic Formula you get:
`n_(1,2)=(4+-sqrt(16-4(4)(-8)))/8=`

`n_1=2`
`n_2=-1`

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

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