Question

How do you find two consecutive even integers such that the square of the smaller is 10 more than the larger?

Answer 1

I found: `4 and 6`

Explanation:

Call the two numbers:
`2n`
and
`2n+2`
so we get:
`(2n)^2=(2n+2)+10`
`4n^2=2n+2+10`
`4n^2-2n-12=0`
solve for `n`:
`n_(1,2)=(2+-sqrt(4+192))/8=`
`n_(1,2)=(2+-14)/8`
`n_1=2`
`n_2=-12/8=-3/2`
so we have using `n_1`:

`4 and 6`