Two positive numbers differ by 3, and the sum of their squares is 117. What is the larger number?

Explanation

1. 01

   Let them be $y$ and $y + 3$: $y^2 + (y + 3)^2 = 117$.

2. 02

   $2y^2 + 6y + 9 = 117$, so $y^2 + 3y - 54 = 0$.

3. 03

   $(y + 9)(y - 6) = 0$; positive root $y = 6$.

4. 04

   The numbers are 6 and 9 (check: $36 + 81 = 117$); the larger is 9.