Question

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

Answer 1

Let the consecutive positive integers be x and x+1.
According to the question,
(x+1)^2=19+9x
x^2+2x+1=19+9x
x^2-7x-18=0
(x-9) and (x+2)=0
x=9 and x=-2(rejected)
Therefore, the consecutive positive integers are 9 and 10.