The sum of the squares of two consecutive integers is 9 greater than 8 times the smaller integer (Algebra 1)

Question

The sum of the squares of two consecutive integers is 9 greater than 8 times the smaller integer

Answer 1

Let x = the smaller integer and x + 1 = the larger one

 

x² + (x + 1)² = 9 + 8x

 

x² + x² + 2x + 1 = 9 + 8x

2x² + 2x  + 1 = 9 + 8x

2x² - 6x - 8 = 0

Divide both sides by 2:  x² - 3x - 4 = 0

                                   (x - 4)(x + 1) = 0

                                   x = 4 or x = -1

 

The consecutive integers are 4 and 5 or -1 and 0.