Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

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 $(1)$

Answer 5

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 $b\ge0$

Answer 6

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: $$\boxed{x = \dfrac{\sqrt{3}}{2}}$$

Answer 7

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 $S = \{\sqrt{3}/2\}$.