solving $\sqrt{3-\sqrt{3+x}}=x$.

Can we solve the following equation in $\mathbb{R}$ without expanding it into a fourth degree equation :

$$ \sqrt{3-\sqrt{3+x}} = x.$$

squaring both sides and squaring again is the only thing I could done, If you have any other idea just post hints.

Here's an approach that is also algebraic, but involves only quadratic equations (the straightforward approach produces a 4th degree equation).

Denote $y = \sqrt{3+x}$. Then we have a system: $$ \begin{array}{rcl} y & = & \sqrt{3+x} \\ x & = & \sqrt{3-y} \end{array} $$ Looks kind of cool. If we square everything and then subtract equations, we get $$ y^2 - x^2 = x + y. $$ Since both $x$ and $y$ are positive, this means that $y-x=1$. Then from the first equation we have $$ x+1 = \sqrt{3+x}. $$ Now we square everything and solve the quadratic, which gives $x=1$.

I know this is practically the same as the straightforward approach, but this looks a bit more "contained".

$\sqrt{3-\sqrt{3+x}} = x$

Let $S$ be the set of all real solutions.

You need $\sqrt{3+x}$ to be defined, that is $x\ge -3$

You need $\sqrt{3-\sqrt{3+x}}$ to be defined, that is $\sqrt{3+x}\le 3$

$3+x\le 9$

$x\le 6$

So $S \subset [-3,6]$

$\sqrt{3-\sqrt{3+(-3)}} = \sqrt{3}\not= -3$

$\sqrt{3-\sqrt{3+6}} = 0\not= 6$

So $S \subset ]-3,6[$

Let $f:]-3,6[\to \mathbb{R}$

$f(x)=\sqrt{3-\sqrt{3+x}}-x$

$f'(x)=-\cfrac{1}{4\sqrt{3+x}\sqrt{3-\sqrt{3+x}}}-1 < 0$

So $f$ is strictly decreasing on $]-3,6[$

$f(-3)=\sqrt{3}+3>0$

$f(6)=0-6<0$

So $\exists ! x \in ]-3,6[, f(x)=0$

In your case, you can easily find that $f(1)=0$

So you find $S=\{1\}$

Otherwise, you would know there's one solution but you wouldn't be able to find it... You would only be able to approximate it with Newton's_method or something similar.

Whereas if you square everything to remove square-roots, you can find all rational solutions with the Rational root theorem (and in this case, you would even be able to find other real solutions since the degree is only $4$ (Quartic) but in the general case, you wouldn't be able to).

First, it must be $\,3+x\ge 0\Longleftrightarrow x\ge -3\,$ , for the inner-most square root to be defined on the real field.

Then it also must be

$$3-\sqrt{3+x}\ge 0\Longrightarrow9\ge 3+x\Longrightarrow x\le 6$$

for the outer square root to be defined, thus our definition domain is $\,-3\le x \le 6\,$ . Now directly, by successive squaring:

$$x=\sqrt{3-\sqrt{3+x}}\Longrightarrow x^2=3-\sqrt{3+x}\Longrightarrow x^4-6x^2+9=3+x\Longrightarrow$$

$$x^4-6x^2-x+6=0\Longleftrightarrow x^2(x^2-6)-(x-6)=0$$

We get by inspection the zero $\,x=\,$ , so

$$x^4-6x^2-x+6=(x-1)(x^3+x^2-5x-6)$$

Checking for rational solutions to the above we have that $\,x=-2\,$ is a zero, too, so

$$x^4-6x^2-x+6=(x-1)(x+2)(x^2-x-3)$$

Finally, the solutions to that quadratic are

$$x_{1,2}=\frac{1\pm\,\sqrt{13}}{2}=\begin{cases}\frac{1+\sqrt{13}}{2}\cong 2.3\\{}\\\frac{1-\sqrt{13}}{2}\cong -1.3\end{cases}$$

So the roots of the quartic are $\,-2\,,\,-1.3\,,\,1\,,\,2.3\,$ , but since we squared the original irrational equation to get the quartic, we now must check each of the above solutions in the original equation; we get at once that the negative ones must be dropped as we've a square root in one of the sides.

A quick calculator check also rules out the root $\,2.3\,$ of the quartic, so the only real solution is $\,x=1\,$

I'd like to add that since this problem is of the form

$$f(x) = x$$

this means that the solutions to the equation are fixed points of the function $f$.

If a fixed point is attractive, it can be found by fixed point iteration.

This simply means that we begin with some reasonable guess for $x$, then evaluate $f(x)$, numerically. If the result is not equal to $x$, we then just repeat with that result by evaluating $f(f(x))$ and so on. Let us try it starting with the guess 2:

$$f(2) = \sqrt{3 - \sqrt{2 + 3}} \approx 0.874032$$ $$f(f(2)) \approx 1.031744$$

Aha! It seems to be attracting toward the solution that we already know. Let's keep going:

$$f(f(f(2))) \approx 0.966032$$

After that I get $1.000496$, $0.999938$, $1.000001$, $0.999999$ and $1.000000$. The oscillations about the fixed point converge on it to six figures within just a few iterations.