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

Question

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.

Why don't you like the fourth-degree equation? Squaring both sides to eliminate square roots in equations is a general and useful technique. — user7530, Feb 23 '13 at 10:11
@wj32 : this is exactly what I got from the 4th degree polynomial, but guessing don't count. — , Feb 23 '13 at 10:11
What's the problem with guessing here, though? Anyway, it's easy to check that $x \in [0, \sqrt{3}]$; I'll try to derive further constraints from that. — Johannes Kloos, Feb 23 '13 at 10:13
The function $\sqrt{3-\sqrt{3+x}}-x$ is strictly decreasing on $[0,\sqrt{3}]$, so there's at most one zero (this is easy to check by differentiation). Since all solutions must lie in this interval, there can be at most one solution. I think that solving this equation without going to a fourth-order polynomial is impossible, but I'll try to check this now. — Johannes Kloos, Feb 23 '13 at 10:23
I meant answers :) There were plenty of them, and now they're gone. — Kaster, Feb 23 '13 at 10:34

score 35 · Accepted Answer · answered Feb 23 '13 at 11:37

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".

score 3 · Answer 2 · answered Feb 23 '13 at 11:50

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

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).

+1) This is the most general methods (of the ones available at this moment) since this method might, in principle, work for non-rational powers too. — AD - Stop Putin -, Feb 23 '13 at 13:29

score 2 · Answer 3 · answered Feb 23 '13 at 11:43

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 $\,x=1\,$

+1) This is what I would do except for the first step, I would argue " since $\sqrt{something}=x$ we must have $x\geq 0$". — AD - Stop Putin -, Feb 23 '13 at 13:17
+1 Good point, @AD. I think that way is clearer than pointing out this at the end. — DonAntonio, Feb 23 '13 at 13:24
By the way, this reminds me of a time when I as an auxiliary teacher taught about Taylor expansions. At some occasions an old teacher watch me in order to guide me through the pedagogics, after that lesson he told me that sometimes I do not need to do things the sharpest way, e.g. if we needed three terms of the Taylor expansion in order to say if an integral converges etc., but on the other hand it might be clearer to write out some more than what is actually needed and then, at a retrospective look at the calculation, do some remarks on it. — AD - Stop Putin -, Feb 23 '13 at 13:51

score 1 · Answer 4 · answered Feb 23 '13 at 16:20

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.

Finding the derivative should help to understand a behavior of the function, I think — Max Malysh, Feb 23 '13 at 18:37
Yes. This was covered in other answers already, in particular that of xaviermo2. I just wanted to throw this in as a tool that the OP might want to know about. — Kaz, Feb 23 '13 at 18:47

Stoleg · Answer 5 · 2013-05-24T13:49:22.757

It's easy to guess that the answer is $x=1$.

However, one should prove that this is the only root. Here is how.

Function $ \sqrt {3- \sqrt {3+x} } $ is strictly decreasing.
Function $x$ is strictly increasing.
It is proven (ask if you need a proof), that a strictly increasing and a strictly decreasing functions can cross at no more than 1 point.

Since we have a point of $x=1$, then there are no other roots for this equasion.

score 0 · Answer 6 · answered Aug 21 '23 at 13:09

For the sake of variety, I'll add one trick I really like. As it was mentioned in other answers, the domain of $x$ is $[0;6]$.

Here's the trick: Let $3=a$ (not the othey way around!), i.e. we'll substitute each instance of $3$ with $a$: $$\sqrt{a-\sqrt{a+x}}=x$$ Square both sides once and rearrange to get: $$a-x^2=\sqrt{a+x}$$ Here another constraint is imposed on $x$: The LHS needs to be positive, so $a-x^2\geq0\Leftrightarrow x\in\left[-\sqrt{3};\sqrt{3}\right]$. This updates the domain, of the solutions $x$ to $x\in D:[0;\sqrt{3}]$.

Square both sides again: $$a^2-2ax^2+x^4=a+x$$ This is quartic w.r.t. $x$, but is quadratic in $a$. We'll solve it w.r.t. $a$: $$f(a;x)=a^2-a(2x^2+1)+x^4-x=0$$ $$\Delta=(2x^2+1)^2-4(x^4-x)=4x^2+4x+1=(2x+1)^2$$ The discriminant is a perfect square! We'll be able to factor $f(a;x)$ nicely: $$a_{1,2}=\frac{2x^2+1\pm(2x+1)}{2}$$ $$a_1=x^2+x+1$$ $$a_2=x^2-x$$ $$\Rightarrow f(a;x)=\left(a-x^2-x-1\right)\left(a-x^2+x\right)$$ But remember, $a$ is nothing more than a $3$, so we can substiture it back. After simplification, we obtain: $$f(x)=\left(x^2+x-2\right)\left(x^2-x-3\right)=0$$ Finally, we equate each multiple to $0$ to find the four roots: $$x_1=1\to \text{valid root}$$ $$x_2=-2\notin D:\left[0;\sqrt{3}\right]\to\text{invalid root}$$ $$x_{3,4}=\frac{1\pm\sqrt{13}}{2}\notin D:\left[0;\sqrt{3}\right]\to\text{invalid roots}$$ Conclusion: The given equation has the single root $x=1$.

Remark: This method is practically useful only when the equation w.r.t. $a$ is solveable with no surds, so it cannot solve every irrational equation. Also, picking exactly $3=a$ and substituting it everywhere was obvious here, but it may not at all be. For example, have a look at this equation: $$\sqrt{9-\sqrt{x+3}}=x$$ Letting $3=a$ again seems reasonable, but how do we substitute $9$? Is it $a^2$? Or $a+6$? Or maybe $3a$? In theory it can be any function $f$, for which $f(3)=9$. So here we have a guess and check scenario: if the equation for $a$ is nice, we got the right substitution. If not, we should try with the next one on the list. The problem is that the equation may not be solvable by this method at all (especially if it has no roots). So this method, despite looking awesome, is rather limited: it's a funky tool to have on your disposal, but it may be useless in some cases.

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

6 Answers6