<s>[INST] 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 
 [/INST] Well, let us start with something we all know and love, the quadratic formula:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
And I have a question for you.  Why is it that there is a $\pm$ in front of the square root?  Isn't it already the case that we have things like $\sqrt9=\pm3$?  And the answer is no.  By definition, we have it that $\sqrt x\ge0$, and thus, $\sqrt9=3$ since $-3<0$.
Also notice that $\sqrt x$ is undefined for $x<0$, so by looking at the problem, we can immediately deduce that
$$x-3=\sqrt{x+3}\ge0\implies x\ge3$$
$$\sqrt{x+3}\implies x+3\ge0\implies x\ge-3$$
Now, upon squaring the original equality, we have
$$x^2-6x+9=(x-3)^2=(\sqrt{x+3})^2=x+3$$
$$x^2-7x+6=0$$
$$(x-6)(x-1)=0$$
$$\implies x=1,6$$
Checking this, only the second value is $\ge3$ and checking them in the square roots, we find that
$$-2=1-3\ne\sqrt{1+3}=2$$
$$3=6-3=\sqrt{6+3}=3$$
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