Find two consecutive positive integers such that the square of the larger integer is nineteen more than nine times the smaller integer.

解

F in d tw o co n sec u t i v e p os i t i v e in t e g ers s u c h t ha t t h e s q u a re o f t h e l a r g er in t e g er i s nin e t ee n m ore t han nin e t im es t h e s ma ll er in t e g er .

The numbers : 9, 10

解答ステップ

Represent the numbers with

x

First number : x

Consecutive number : x + 1

A.T.Q.:

(x + 1)^{2} = 19 + 9 x

解く

(x + 1)^{2} = 19 + 9 x : x = 9, x = - 2

Select positive solutions:

9

x + 1 = 9 + 1 = 10

The numbers : 9, 10