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

Mar 4, 2016

$- 15$ 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 ${n}^{2} > {\left(n + 1\right)}^{2}$ Thus, we then have the equation

${n}^{2} - {\left(n + 1\right)}^{2} = 29$

$\implies {n}^{2} - {n}^{2} - 2 n - 1 = 29$

$\implies - 2 n = 30$

$\implies n = - 15$

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