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

解

t h e s u m o f t h e s q u a res o f tw o co n sec u t i v e in t e g ers i s 9 g re a t er t han 8 t im es t h e s ma ll er in t e g er . f in d t h e in t e g ers

The numbers : 4, 5 or - 1, 0

解答ステップ

Represent the numbers with

x

First number : x

Consecutive number : x + 1

A.T.Q.:

x^{2} + (x + 1)^{2} = 9 + 8 x

解く

x^{2} + (x + 1)^{2} = 9 + 8 x : x = 4, x = - 1

挿入

x = 4 : 5

挿入

x = - 1 : 0

The numbers : 4, 5 or - 1, 0