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.
 A: The two numbers $\pm\frac{\sqrt{3}}{2}$ are solutions to $$\sqrt{1+x}-\sqrt{1-x} = \pm 1.$$
Notice that after squaring this original equality, you arrive at your equations, independent of the sign of the right hand side.
A: 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.
A: Notice that the initial equation can be rewritten as
$$2x = \sqrt{1+x}+\sqrt{1-x}$$
This immediately enforces the condition that $x>0$ and negative solutions can be thrown out.
A: Another way avoiding squaring which often introduces extraneous roots?
As $-1\le x\le1,$ WLOG $x=\cos2t,$
Using principal values, $ 0\le t\le\dfrac\pi2\ \ \ \  (1)$
$$1=\sqrt2(\cos t-\sin t)\iff\dfrac12=\cos\left(t+\dfrac\pi4\right)$$
$$\implies t+\dfrac\pi4=2m\pi\pm\dfrac\pi3$$
$+\implies t=2m\pi+\dfrac\pi{12},2t=?\cos2t=?$
$-\implies t=2m\pi-?$ which is untenable by $(1)$
A: 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 $b\ge0$
A: Using a trigonometric substituition:
As $-1 \le x \le 1$, then $x = \cos \theta$:
$$\sqrt{1+x} = \sqrt{1-\cos \theta} = \sqrt{2\sin^2 \frac{\theta}{2}} = \sqrt{2}\left|\sin \frac{\theta}{2}\right|$$
$$\sqrt{1+x} = \sqrt{1+\cos \theta} = \sqrt{2\cos^2 \frac{\theta}{2}} = \sqrt{2}\left|\cos \frac{\theta}{2}\right|$$
Then the equation becomes
$$\left|\sin \frac{\theta}{2}\right| + \left|\cos \frac{\theta}{2}\right| = \frac{1}{\sqrt{2}}$$
As we care only about the value of $x$, we can restrict the value of $\theta$ to be at the interval $\left[0, \ \pi\right]$:

*

*If $\theta \in \left[0, \ \dfrac{\pi}{2}\right]$:


$$\sin \frac{\theta}{2} + \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sqrt{2} \sin \underbrace{\left(\dfrac{\theta}{2} + \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{\sqrt{2}}$$
$$\sin \alpha = \dfrac{1}{2} \Rightarrow \alpha = \dfrac{\pi}{3} \ \ \text{or} \ \ \alpha = \dfrac{2\pi}{3}$$
Then $\theta = 2\left(\alpha - \frac{\pi}{4}\right)$ can assume two values: $\theta = \frac{\pi}{6}$ or $\theta = \frac{5\pi}{6}$.
Only $\theta = \frac{\pi}{6}$ is on the interval $\left[0, \ \frac{\pi}{2}\right]$, then we exclude the second value. Therefore:
$$\boxed{x = \cos \frac{\pi}{6} = \dfrac{\sqrt{3}}{2}}$$


*

*If $\theta \in \left[\dfrac{\pi}{2}, \ \pi\right]$

$$\sin \frac{\theta}{2} - \cos \dfrac{\theta}{2} = \frac{1}{\sqrt{2}}$$
$$\sin \underbrace{\left(\dfrac{\theta}{2} - \dfrac{\pi}{4}\right) }_{\alpha}= \dfrac{1}{2}$$
$$\alpha = \frac{\pi}{3} \ \ \text{or}  \ \  \alpha =  \frac{2 \pi}{3}$$
$$\theta = \frac{7\pi}{6} \ \ \text{or}  \ \  \theta =  \frac{11 \pi}{6} $$
As both $\theta$ are outside the interval $\left[\dfrac{\pi}{2}, \ \pi\right]$, there's no solution for this interval.

Then we got only one solution:
$$\boxed{x = \dfrac{\sqrt{3}}{2}}$$
A: 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 $S = \{\sqrt{3}/2\}$.
