Question

The sum of the squares of 2 consecutive negative integers is 41. What are the numbers?
Which of the following equations is the result of using the factoring method to solve the problem?
(n - 5)(n - 4) = 0
(n - 5)(n + 4) = 0
(n + 5)(n - 4) = 0
(n + 5)(n + 4) = 0

Answer 1

Let the greater number be n.
n^2 + (n - 1)^2 = 41.
n^2 + n^2 + 1 - 2n = 41.
2n^2 - 2n - 40 = 0.
n^2 - n - 20 = 0.
Splitting the middle term,

n^2 - 5n + 4n - 20 = 0
n(n - 5) + 4(n - 5) = 0
(n + 4)(n - 5) = 0. This is the answer to the second part of the question.

Moving on, n + 4 = 0 or n - 5 = 0.
So, n = - 4, 5.
But, it is given that n is negative. So, n = -4 and n - 1 = -5.

So, the numbers are - 5 and - 4.

Answer 2

This assumes a well ordering for the integers.....