Square root positive/negative What is the square root of $9$? Is it always $\pm 3$ or just positive $3$?
Trying to find the solution set of this equation : $x-3 = \sqrt{x+3}$
I want to understand the concept of square root to solve the problem.
Thanks 

There are two numbers $x$ such that $x^2=9$ : $x=3$ and $x=-3$, but with the symbol $\sqrt{9}$ we indicate only the positive one, so $\sqrt{9}=3$.
In other words, the symbol $\sqrt{x}$ indicates, by definition, the positive number $y$ such that $y^2=x$.
This implies that if we search the solution of the equation $x-3=\sqrt{x+3}$, than we must have $x-3\ge0$ and, if squaring we find a solution such that $x-3<0$ this is not a solution of the given equation.
We can say that the equation $x-3=\sqrt{x+3}$ is equivalent to the system:
$$
\begin{cases}
x-3\ge0\\(x-3)^2=x+3
\end{cases}
$$
