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

Answer 1

dont just give the answer

Answer 2

Lets take our first integer to be x and second integer (x-1)
Thus the squares will be x² and (x-1)² respectively.
The sum of the squares will be x² + (x-1)²
Which equals to 41 and now we can set an equation
x² + (x-1)² = 41
Now simplify this equation

Answer 3

Let the two numbers be x and x+1
Their squares are x^2 and x^2 + 2x + 1
The sum of their squares is 2x^2 + 2x + 1 = 41
Subtract 41:
2x^2 + 2x - 40 = 0
Divide by 2:
x^2 + x - 20 = 0
Factor:
(x + 5) (x - 4) = 0 < Here is the answer.
Use the zero product principle twice:
x = -5, 4
x+1 = -4, 5
So the two consecutive negative numbers are -5 and -4.