Question

How do you find two consecutive negative integers such that the difference of their squares is 29?

Answer 1

`-15` and `-14`

Explanation:

If we let `n` be the lesser of the two integers, then our two integers are `n` and `n+1`. Because the integers are negative, we know that `n^2 > (n+1)^2` Thus, we then have the equation

`n^2-(n+1)^2 = 29`

`=>n^2-n^2-2n-1 = 29`

`=>-2n = 30`

`=> n = -15`

Therefore the integers are `-15` and `-14`.