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.

Yet another way :
Let $\sqrt{1+x}=a, \sqrt{1-x}=b$
For real $x, a\ge0, b\ge0 $
By the given condition,
$$a-b=1$$  and $$a^2+b^2=2\implies2=(b+1)^2+b^2\iff2b^2+2b-1=0$$
$\implies(2b+1)^2=3\implies2b+1=+\sqrt3$ as $b\ge0$
