Alright, so I have √x−15=3−√x.
I first square both sides to get x−15=(3−√x)(3−√x) which simplifies to x−15=9−6√x+x.
I solved for x and got x=16, however, when I plug it in, the equation doesn’t work. Why does this happen?
You did good work in doing what many students forget to do – checking that your proposed solution actually is a solution.
Whenever you square both sides of an equation, you potentially introduce new solutions that may or may not be solutions of the original equation. Consider the simple example x=1. Obviously x=1 is the solution, but squaring both sides gives x2=1, which has solutions x=±1. Over R it is most definitely not true that −1=1, so x=−1 cannot be a solution here.
You should always be careful and check your proposed solutions by plugging them back into the original equation like you have done here.
Indeed, your equation actually has no solutions, which is what you should conclude when asked to solve it.