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


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,
$$ x = \frac{1}{\sqrt{\alpha}} = i \, \sqrt{\beta} = \sqrt{\frac{\sqrt{5}-1}{2}} $$
