Why aren't these negative numbers solutions for radical equations? I was working on radical equations and I came across a few problems where I got answers that worked when I checked, but were not listed as solutions. My teacher's only explanation was, "just because." Here is one problem where the only solution is $1$.
$x=\sqrt{2-x}$
How I solved it
$x^{2}=2-x$
$x^{2}+x-2=0$
$(x+2)(x-1)=0$
$x= \{-2, 1\}$
Then plugging both numbers back in, I get
$1 = \sqrt{2-1}$
$1 = \sqrt{1}$
$1 = 1$
and 
$-2 = \sqrt{2--2}$
$-2 = \sqrt{4}$
The square root of $4$ can be both $-2$ because $-2 \times -2 = 4$ and $2$. $1$ is the only solution listed and my teacher says that it's right.
What is the explanation for this? Why isn't $-2$ a solution for the problem?

Omar already gave a good answer, but I wanted to expand on this part:

You just have to remember that when squaring, you can always introduce
  incorrect 'solutions' simply because A = B is not logically equivalent
  to A^2 = B^2.

If you have an equation, and you multiply both sides by another equation, you don't always get the same equation.
For example, $(x-2)=0$.  If you multiply both sides by $(x+5)$, you'll get $(x-2)(x+5)=0$.  The second equation has solutions $x=2,-5$, and the first has $x=2$ only.
In Omar's example, since $A=B$, he is multiplying both sides by $A$.  In general, you can only multiply both sides by a non-zero constant to always preserve logical consistency.  In your example, you're multiplying both sides by $(2-x)^{1/2}$.  Since $(2-x)^{1/2}$ is not a non-zero constant, it does not always preserve logical consistency.
For example, you can multiply both sides by $4$, and it doesn't change the solutions to the equation.  However, it will sometimes preserve the equation in something like the following example: $(x-4)^4=0$, if you multiply both sides by $(x-4)^2$, the result, $(x-4)^6=0$, will have the same set of solutions.
