Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$

Question

Question:

Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$

My approach:

$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^2-2\right)^2\\4x^4(1-x^2)=(x^4-x^2+1)^2$$

I don't know how can I proceed from here.

Is there a way so that not using the complicated expansion of polynomials?

I'm looking for methods that doesn't use $4$ or higher degree polynomial expansions.

Answer 1

Firstly $x>0$

Let $\theta=\sin^{-1}{x} \Rightarrow x=\sin \theta$

We get $$\sin \theta(1+\cos \theta)=\sqrt{1+\sin ^2 \theta} $$ Squaring both sides: $$\begin{aligned} & \left(1-\cos ^2 \theta\right)(1+\cos \theta)^2=2-\cos ^2 \theta \\ \Rightarrow \quad & \cos ^4 \theta+2 \cos ^3 \theta-\cos ^2 \theta-2 \cos \theta+1=0 \\ \Rightarrow \quad & \left(\cos ^2 \theta+\frac{1}{\cos ^2 \theta}\right)+2\left(\cos \theta-\frac{1}{\cos \theta}\right)-1=0 \end{aligned}$$ $$\begin{aligned} & \cos \theta-\frac{1}{\cos \theta}=u \\ \Rightarrow & u^2+2+2 u-1=0 \\ \Rightarrow \quad & u=-1 \\ \Rightarrow \quad & \cos ^2 \theta+\cos \theta-1=0 \\ \Rightarrow \quad & \cos \theta=\frac{-1+\sqrt{5}}{2} \end{aligned}$$ $$\Rightarrow \quad x=\sin \theta=\sqrt{\frac{\sqrt{5}-1}{2}}$$

Answer 2

We quickly observe that $x>0$.

Letting $1-x^2=u^2, \,u≥0$ we have:

$$x^2(1+2u+u^2)=1+x^2$$

This implies that,

$$\begin{align}&(1-u^2)(u^2+2u)=1\\ \implies &u(u+1)(u+2)(u-1)=-1\\ \implies &(u^2+u)(u^2+u-2)=-1\end{align}$$

Finally letting $t=u^2+u$, we have:

$$(t-1)^2=0\implies t=1$$

This leads to:

$$\begin{align}&u^2+u-1=0,\,u>0\\ \implies &u=1-u^2\\ \implies &x=\sqrt u=\sqrt {\frac{-1+\sqrt 5}{2}}~~~.\end{align}$$

Answer 3

Starting from: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ then $$ x^2 \, (2 - x^2 + 2 \, \sqrt{1-x^2}) = 1 + x^2 \\ 2 \, x^2 \, \sqrt{1-x^2} = 1 - x^2 + x^4 \\ 4 \, x^4 \, (1-x^2) = 1 - 2 x^2 + 3 x^4 - 2 x^6 + x^8 \\ x^8 + 2 x^6 - x^4 - 2 x^2 + 1 = 0 \\ (x^4 + x^2 - 1)^2 = 0 \\ x^4 + x^2 - 1 = 0 \\ (x^2 + \alpha)(x^2 + \beta) = 0 \\ (x + i \sqrt{\alpha})(x - i \sqrt{\alpha})(x + i \sqrt{\beta})(x - i \sqrt{\beta}) = 0.$$ This gives the solutions as $$ x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha}, i \sqrt{\beta}, - i \sqrt{\beta} \}, $$ or $$ x \in \left\{ \frac{1}{\sqrt{\beta}}, - \frac{1}{\sqrt{\beta}}, \frac{1}{\sqrt{\alpha}}, - \frac{1}{\sqrt{\alpha}} \right\} $$ where $2 \, \alpha = 1 + \sqrt{5}$ (the golden ratio), and $2 \, \beta = 1 - \sqrt{5}$ (inverse golden ratio). Now that there are values to consider it can be shown that $x \in \{ i \sqrt{\alpha}, - i \sqrt{\alpha} \}$ do not satisfy the original equation. Of the remaining values only one satisfies the equation, namely, $$ x = \frac{1}{\sqrt{\alpha}} = i \, \sqrt{\beta} = \sqrt{\frac{\sqrt{5}-1}{2}} $$