<s>[INST] Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the right solution, as $-\frac{\sqrt{3}}{2}$ returns $-1$, not $1$. My question is where did I miss a condition that excludes the negative "solution"? I expect somewhere along the line I squared where I wasn't allowed to square without an additional condition, hoping that I don't have to check these solutions every time.
 [/INST] Since $\sqrt{x+1} - \sqrt{1-x} = 1$, we have
$\sqrt{x+1} - \sqrt{1-x}>0$, which gives
$$\tag{1}\sqrt{x+1}>\sqrt{1-x}.$$ Therefore, if $x=-\frac{\sqrt{3}}{2}$,
inequality $(1)$ is not satisfied.
</s>