Find 2 consecutive positive integers such that the square of the larger integer is 19 more than 9 times the smaller integer

Answer 1

This is a translation problem.

Two consecutive integers means that we have X and X+1.

We want the square of the larger (aka X+1) to be more than 9 times smaller by some amount (19).

So, (X+1)² > 9X. To make these equal, we add the 19 to the right side and obtain (X+1)² = 9X + 19.

From here, you are going to collect all the terms to one side and solve for X. There will be two solutions, and likely one of those solutions is negative. Discard the negative solution and you have your answer.