Geometric Nightmare

First, let be $y=FD$ . As we can see, $\Delta FDE$ and $\Delta BAE$ are similar. Thus: $\frac{y}{x}=\frac{1}{1+x} \Rightarrow xy=x-y$ .

But, from triangle $FDE$ we can write: $x^2+y^2=1$ .

Subtracting $-2xy$ from both sides we obtain:

$(x-y)^2=1-2xy \Rightarrow (xy)^2=1-2xy$ .

Now, let's call $z=xy$

$z^2+2z-1=0$ .

As we know, $z=xy>0$ , then $z=\sqrt{2}-1$ .

Thus we have $\sqrt{2}-1=x-y \Rightarrow y=x-\sqrt{2}+1$

Substituting in the Pythagorean theorem: $x^2+(x-(\sqrt{2}-1))^2=1$

$2x^2-(2\sqrt{2}-2)x+2-2\sqrt{2}=0$

$x^2-(\sqrt{2}-1)x+1-\sqrt{2}=0$

$x=\frac{\sqrt{2}-1+\sqrt{2\sqrt{2}-1}}{2}$

backward solution: x / (x+1) = sqrt( 1-x^2 ) just substitute

First note that $\Delta FED$ and $\Delta FBC$ are similar. Thus

$\dfrac{|DF|}{|DE|} = \dfrac{|CF|}{|CB|} \Longrightarrow \dfrac{\sqrt{1 - x^{2}}}{x} = \dfrac{1 - \sqrt{1 - x^{2}}}{1}$

$\Longrightarrow \sqrt{1 - x^{2}} = \dfrac{x}{1 + x} \Longrightarrow 1 - x^{2} = \dfrac{x^{2}}{(1 + x)^{2}}$

$\Longrightarrow x^{4} + 2x^{3} + x^{2} - 2x - 1 = 0.$

Now there is a technique for factoring quartics, but it is quite a slog so I just used WolframAlpha to confirm that the posted answer is indeed a solution to the above quartic. If anybody has a more elegant way of algebraically finding this exact value I would be most interested.

How this equation is solve without using wolframalpha.....?

$x^4+2x^3+x^2-2x-1=0$