Solution for $x$ where $\sqrt{x + 3} = −1 + \sqrt{x + 2}$ This equation has no solution in $\mathbb{R}$.  
$$\sqrt{x + 3} = −1 + \sqrt{x + 2}$$
The text I am reading is "Understanding Logic" by a small group of authors who do not offer a solution to the equation.  That is understandable given the subject of the book.  The equation is used as an example in a chapter about false hypothesis and bogus solutions. 
My simple algebra resolves the equation to $x=-2$.  Substituting for $x$ in the equation results in $1=-1$. That this can happen is new to me.
Does this equation have a solution that is imaginary?  Is there another set of numbers that contains a solution?
 A: Squaring cannot be reversed, so all you have calculated is a necessary condition, namely $x=-2$. This dosésn't mean it is sufficient, too. It is not, as back substitution showed.
By squaring the equation, you created $(-1)^2=1^2$, which is correct. The reverse is not.
A: This is known as an extraneous solution. Here's how it works:
$$
\begin{align*}
1 &= -1\quad &(\text{clearly false})\\
1^2 &= (-1)^2\quad &(\text{square both sides})\\
1 &= 1\quad &(\text{clearly true (!!)})\\
\end{align*}
$$
So you cannot assume that just because $x^2 = y^2$ that $x=y$. In your equation, you squared both sides of the equation, which possibly results in an extraneous solution appearing.
It also happens that this particular equation does not have complex solutions either, but sometimes you could see a complex solution in equations like these.
A: Presumably, your "simple algebra" starts with squaring, to get rid of one of the square roots. If you try to insert $x = -2$ in the new, squared version of the equation, everything fits. So that's what happened: Squaring made two unequal things (i.e. $-1$ and $1$) equal, and thus you gained a new solution that wasn't there before.
The equation doesn't have any real number solutions. We can see this if we plot the two sides as functions of $x$. The graphs are two half-parabolas of exactly the same shape, one is just a shifted-over version of the other. They will never intersect.
As for complex numbers, I personally don't like using square roots in that case, as they don't behave nicely. So in that case I don't think it's a well-formed equation in the first place.
