<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] Extraneous roots can be introduced by squaring.  We can avoid squaring by multiplying by the conjugate.
The equation
$$\sqrt{x + 1} - \sqrt{1 - x} = 1$$
imposes the restrictions that $x + 1 \ge 0 \implies x \geq -1$ and $1 - x \ge  0 \implies x \leq 1$.  Therefore, we know that any valid solution must satisfy $-1 \leq x \leq x$.
If we multiply both sides of the given equation by $\sqrt{x + 1} + \sqrt{1 - x}$, we obtain
$$\sqrt{x + 1} + \sqrt{1 - x} = 2x$$
This imposes the additional constraint that $x \geq 0$.  Hence, any valid solution must satisfy $0 \leq x \leq 1$.
We now have the system of equations
\begin{align*}
\sqrt{x + 1} - \sqrt{1 - x} & = 1\\
\sqrt{x + 1} + \sqrt{1 - x} & = 2x
\end{align*}
Adding the equations gives
$$2\sqrt{x + 1} = 1 + 2x$$
Squaring both sides of the equation yields
\begin{align*}
4(x + 1) & = 1 + 4x + 4x^2\\
4x + 4 & = 1 + 4x + 4x^2\\
3 & = 4x^2\\
\frac{3}{4} & = x^2\\
\frac{\sqrt{3}}{2} & = |x|\\
\pm \frac{\sqrt{3}}{2} & = x
\end{align*}
Since we require that $0 \leq x \leq 1$, we discard the solution $x = -\sqrt{3}/2$.
Direct calculation shows that $x = \sqrt{3}/2$ is a valid solution, so the solution set is $S = \{\sqrt{3}/2\}$.
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