Answer to Question #92168 in Algebra for Raven

Question

The sum of the squares of two consecutive integers is 85. What are the integers?

Answer 1

Let these integers be n and (n+1). We have:

$$n^2+(n+1)^2=85$$

$$n^2+n^2+2n+1=85$$

$$2n^2+2n-84=0$$

$$n^2+n-42=0$$

We need to solve this quadratic equation:

$$D=1^2-4(1)(-42)=169=13^2$$

$$n_1=\frac{-1-13}{2(1)}=-7$$

$$n_2=\frac{-1+13}{2(1)}=6$$

So, we hawe two pairs of consecutive integers:

$$(-7,-6)\ and \ (6,7)$$