Question

How do you find two consecutive odd integers such that the square of the first, added to 3 times the second is 24?

Answer 1

I found: `3 and 5`

Explanation:

Let us call our odd integers:
`2n+1` and `2n+3` so we have:
`(2n+1)^2+3(2n+3)=24`
`4n^2+4n+1+6n+9-24=0`
`4n^2+10n-14=0`
Using the Quadratic Formula to solve for `n` we have:
`n_1=1` so that the two integers will be:
`2+1=3` and `2+3=5`
or:
`n_2=-7/2` so that the two integers will be:
`2(-7/2)+1=-6` and `2(-7/2)+3=-4` but these are even!