<s>[INST] Radical equation - can I square both sides with more than 1 radical on one side? I'm familiar with equations like:
$\sqrt{x+1} - \sqrt{x+2} = 0 $
Has no solutions, it's just an example off the top of my head
Just move the negative square root to the other side, square both sides and solve. 
$\sqrt{x+1} = \sqrt{x+2}$
$x+1 = x+2$
0 = 1
My question is, if there are two square roots on one side, then can I still square both sides in this way:
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$x+1 - (x+2) = x+3$
$x+1 - (x-2) = x+3$
$x = -4$
Or does squaring both sides cause something strange to happen on the left hand side?
 [/INST] Yes, you can square both sides.  But you do actually have to square both sides.
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$
$(\sqrt{x+1} - \sqrt{x+2})^2 = (\sqrt{x+3})^2$
$x+1 - 2\sqrt{x+1}\sqrt{x+2} + x + 2 = x + 3$.
$-2\sqrt{x+1}\sqrt{x+2} = -x$
$(-2\sqrt{x+1}\sqrt{x+2})^2 = (-x)^2$
$4(x+1)(x+2) = x^2$
$4x^2 +12x + 8 = x^2$
$3x^2 + 12x + 8 = 0$
$x = \frac {-12\pm{144-4*3*8}}{6} =4
$-2 \pm \frac{\sqrt{48}}6 = -2 \pm \frac 2{\sqrt 3}$.
BUt note:  Squaring both sides gives extraneous solutions because it ignores whether terms given are positive or negative.  These answers all don't actually work if we plug them back in.
.....
Obviously you can't just square parts of each side and expect the result to make any sense.
$\sqrt{x+1} - \sqrt{x+2} = \sqrt{x+3}$ obviously does not mean 
$(\sqrt{x+1})^2 - (\sqrt{x+2})^2 = (\sqrt{x+3})^2$
Because $(\sqrt{x+1} - \sqrt{x+2})^2 \ne (\sqrt{x+1})^2 - (\sqrt{x+2})^2$.
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