How to solve the following radical equation?$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$

Question

How to solve the following radical equation? $$\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$$

I'm looking for an easy way to solve it.

Answer 1

Strong hint:

Notice your equation is the same as $$x=\sqrt{2+\sqrt{2-x}}$$

Answer 2

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

The difference between radicals and powers is that you can expand the right-hand side now and collect like terms. So far this seems pretty easy to me. Unfortunately, I don't think you can solve this one by hand easily once you've expanded and collected, because what you get is: $$x^{16} - 16 x^{14} + 104 x^{12} - 352 x^{10} + 660 x^{8} - 672 x^{6} + 336 x^{4} - 64 x^{2} + x + 2 = 0$$ According to Wolfram Alpha, however, there are closed forms for 8 out of 16 solutions, meaning you can use algebraic methods to solve this if you really want to, and the great thing about radicals is that the radicand cannot be negative, so it turns out that the only valid answer to your original equation is $$x=\frac{1+\sqrt5}{2}$$ which is actually the golden ratio. Neat!

Answer 3

As hinted

For real $x,\sqrt{2-x}\ge0\implies x\le2$

But $x=2$ doesn't satisfy the given equation

and $x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}>0$

So, $0<x,<2$

WLOG let $x=2\cos32t, 0<32t<\dfrac\pi2$

Use $\cos2x=2\cos^2x-1=2-2\sin^2x$

$$\sqrt{2-x}=+2\sin16t$$

Now $\sqrt{2+\sqrt{2-x}}=\sqrt{2(1+\sin16t)}=+\sqrt2(\sin8t+\cos8t)$ as $\sin8t+\cos8t>0$

$\implies\sqrt{2+\sqrt{2-x}}=2\cos\left(\dfrac\pi4-8t\right)$

$\sqrt{2-\sqrt{2+\sqrt{2-x}}}=\sqrt{2-2\cos\left(\dfrac\pi4-8t\right)}=2\sin\left(\dfrac\pi8-4t\right)$ as $\dfrac\pi8-4t>0$

$\implies\sqrt{2-\sqrt{2+\sqrt{2-x}}}=2\cos\left(4t+\dfrac{3\pi}8\right)$

Finally, $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}}=\sqrt{2+2\cos\left(4t+\dfrac{3\pi}8\right)}=2\cos\left(2t+\dfrac{3\pi}{16}\right)$

So, we have $$2\cos32t=2\cos\left(2t+\dfrac{3\pi}{16}\right)$$

$$\implies32t=2m\pi\pm\left(2t+\dfrac{3\pi}{16}\right)$$

Case $\#1:$ Taking the '+' sign, $$30t=2m\pi+\dfrac{3\pi}{16}=\dfrac{\pi(32m+3)}{16}$$

$$\implies32t=\dfrac{\pi(32m+3)}{16}\cdot\dfrac{32}{30}=\dfrac{\pi(32m+3)}{15}$$

We need $0<\dfrac{\pi(32m+3)}{15}<\dfrac\pi2$

$\iff0<2(32m+3)<15\implies m=0\implies x=2\cos\dfrac{\pi(32\cdot0+3)}{15}$

Case $\#2:$ Take the '-' sign

Answer 4

If $\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-x}}}} =x$

Then:

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

In general:

$x=\sqrt{2+\sqrt{2-\sqrt{2+\sqrt{2-\ldots}}}}$

Which means that:

$2-(x^2-2)^2=x$

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

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

$(x-1)(x^3+x^2-3x-2)=0$

$(x-1)(x^3+2x^2-x^2-x-2x-2)=0$

$(x-1)(x+2)(x^2-x-2)=0$

$(x-1)(x+2)\left(x - \dfrac{1 \pm \sqrt{5}}{2}\right)$

So the 'possible' answers for x would be :

$\left(1,-2 ,\phi,\dfrac{1}{\phi} \right)$

All are too small to be the answer of the original question except for $\phi$, the golden ratio. So that's the answer.