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

Question

ETA: since commenters noted I missed a sign I am correcting that.

This is one of those Mathsolutionzz instagram problems. I was curious if I did this correctly, because I looked at the comments, and the solutions others came up with are rather different (and not integers).

The problem:

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

So my first step was simple: square both sides and get rid of one of the radicals. We then get

$$2+\sqrt{2-\sqrt{2+x}} = x^2 \tag{1}$$ which I can rearrange to: $$\sqrt{2-\sqrt{2+x}} = x^2-2 \tag{2}$$

Then we repeat the process, square both sides again: $${2-\sqrt{2+x}} = (x^2-2)^2 \tag{3}$$

rearrange: $$-\sqrt{2+x} = (x^2-2)^2-2 \tag{4}$$

rinse, repeat: $$2+x = ((x^2-2)^2-2)^2 \tag{5}$$

and where I originally had a solution at $x=2$, because I messed up a sign, this time we get a polynomial: $$x^8-8x^6+20x^4-16x^2-x+2 \tag{6}$$

Which is the really ugly polynomial that @Lubin noted (thanks!) and has several roots.

All that said, were I trying to find the roots by hand, I know the roots can't be negative (since this problem doesn't seem to allow for imaginary solutions and we are sticking to real numbers).

So my next go-to would be factoring out the polynomial. When I originally did this I would try factoring one side of the equation with the opposite-sign binomial, like this:

$$2+x = ((x^2-2)^2-2)^2$$ $$2+x = (x^2-2)^4-4(x^2-2)^2+4$$ $$x-2 = (x^2-2)^4-4(x^2-2)^2$$ $$x-2 = [(x^2-2)^2-2(x^2-2)][(x^2-2)^2+2(x^2-2)]$$ $$x-2 = (x^2-2)[(x^2-2)-2][(x^2-2)+2]$$ $$x-2 = (x^2-2)(x^2-4)(x^2)$$ $$x-2 = (x^2-2)(x-2)(x+2)(x^2)$$

Which at least tells me that +1, +2, -1, -2 don't work, really except insofar as x=2 makes the whole thing zero. But I am curious if this is a bad method, or simply erroneous. (I do realize that ultimately I'd probably want to move the x-2 to the RHS and expand it leaving me with the degree 8 polynomial anyway, but I thought this worth a shot).

Answer 1

Substituting $x$ with $2\cos\theta$ will suffice to get the result to your question $$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x$$ will be $\sqrt{2+\sqrt{2-\sqrt{2+2\cos\theta}}} = 2\cos\theta$

On LHS by Half angle cosine formula

$\sqrt{2+\sqrt{2-2\cos\frac{\theta}{2}}}$ = $\sqrt{2+2\sin\frac{\theta}{4}}$ = $\sqrt{2+2\cos(\frac{\pi}{2}-\frac{\theta}{4}})$ = $2\cos(\frac{\pi}{4}-\frac{\theta}{8})$

Now $\theta = \frac{\pi}{4}-\frac{\theta}{8}$ on further simplification we can get $\theta$ = $\frac{2\pi}{9}$ which is $40°$

Therefore $$x = 2\cos\frac{2\pi}{9} \qquad \text{or}\qquad 2\cos40°.$$

Answer 2

By squaring$$\sqrt{2+\sqrt{2-\sqrt{2+x}}} = x,$$ you arrive at $${2+\sqrt{2\color{red}-\sqrt{2+x}}} = x^2,$$ then $$ 2\color{red}-\sqrt{2+x}=(x^2-2)^2$$ and $$ 2+x=((x^2-2)^2-2)^2.$$ While $x=2$ and $x=-1$ are guessable solutions to this, they are not solutions to the original equation. The problem is that each squaring could introduce spurious solutions, i.e., $a=b$ implies $a^2=b^2$, but not vice versa. In fact, using the correct minus sign (you somehow replaced it with a $+$) leads to the same final polynomial even though the solutions cannot be the same.

Answer 3

Well, thanks to Hagen and to you, the polynomial that would have the solution(s) among its roots is the octic $$ x^8-8x^6+20x^4-16x^2-x-2\,. $$ Rational Root Theorem tells you where to look, $\pm1$ and $\pm2$, if you want rational numbers. Of these, only $+2$ and $-1$ are roots, but neither of these is a solution to your problem. Looking at the Desmos graph, I see that there are eight, count’em eight, real roots: four positive, four negative. The four negative ones are excluded by the statement of the problem, and $2$ doesn’t work. I guess it remains to check the other three.