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The question is to evaluate $\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$ $$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$ $$x^2=2+\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots }}}}$$ $$x^2=2+x$$ $$x^2-x-2=0$$ $$(x-2)(x+1)=0$$ $$x=2,-1$$

because $x$ is positive $x=2$ is the answer. but where did the $x=-1$ come from ?

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I wouldn't say it "comes" from somewhere. It just satisfies the same algebraic condition that $\sqrt{2+\sqrt{\cdots}}$ does. Multiple numbers can satisfy the same algebraic relation. –  whacka Aug 16 at 8:21
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Well, it should not be a surprise that squaring both sides of an equation the set of solutions can become bigger. –  enzotib Aug 16 at 8:21
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$x=1\Rightarrow x^2=1\Rightarrow x=\pm1$ Where did the $x=-1$ come from? –  drhab Aug 16 at 8:26
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$$x^2=x+2\implies x=\pm\sqrt{x+2}$$ $$x=-\sqrt{2+x}=-\sqrt{2-\sqrt{-2-\cdots}}$$ $$x=+\sqrt{2+x}=+\sqrt{2+\sqrt{2+\cdots}}$$ –  lab bhattacharjee Aug 16 at 8:35
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I'll also add links to some posts about convergence of this sequence to 2: math.stackexchange.com/questions/115501/… and math.stackexchange.com/questions/849274/… –  Martin Sleziak Aug 16 at 12:56

5 Answers 5

up vote 45 down vote accepted

$x=\sqrt{2+\underbrace{\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}_{\text{$x$}}}$, so we get $x=\sqrt{2+x}$.

Now there is only one solution. If we square both sides, we add the case $-x=\sqrt{2+x}$

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This is exactly what I was looking for.Thanks :) –  Shabbeh Aug 16 at 8:47

Well, if you set the square root to the negative one (instead positive by convention), the limit of the sequence $x = -\sqrt{2-\sqrt{2-\sqrt{2-...}}}$ will be $-1$. So by squaring you obtain answers for both of the possible square roots (the positive and the negative one)

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By squaring both sides, you turn an equation with one solution into an equation with two solutions; you can end up creating an extraneous solution. In this case, you come to find that $-1$ satisfies $x^2 = 2 + x$, but does not satisfy $x = \sqrt{2 + \sqrt{2 + \cdots}}$, and so is extraneous.

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so is there anyway to calculate this without adding unwanted results ? –  Shabbeh Aug 16 at 8:25
    
@shabbeh nope. You have to use logically deduction to eliminate. –  Chinny84 Aug 16 at 8:27
    
@Chinny84 But are "logical deductions" always this simple ? –  Shabbeh Aug 16 at 8:29
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Again, nope. But it gets easier with more practice and seeing similar results so you build upon that. Then when you see something, say at research level, the problems are less trivial, and require more. Though, this comment is more generally about problem solving than this particular type of problem. –  Chinny84 Aug 16 at 8:42
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$\newcommand{\calc}{\begin{align}\quad&}\newcommand{\calcop}[2]{\\#1\quad &\quad\text{"#2"}\\\quad&}\newcommand{\endcalc}{\end{align}}$Well, here _is_ a way to calculate this without adding unwanted results: in this case one simply can use that (for $\;y\geq0\;$) we have $\;x=\sqrt y\;\equiv\;x\geq0\land x^2=y\;$. This gives (for $\;x\geq-2\;$) $$\calc x=\sqrt{2+x} \calcop{\equiv}{square both sides, using $\;2+x\geq0\;$} x\geq0\land x^2=2+x \calcop{\equiv}{solve quadratic equation, as in your question} x\geq0\land(x=2\lor x=-1) \calcop{\equiv}{logic and arithmetic: simplify} x=2 \endcalc$$ –  Marnix Klooster Aug 16 at 11:01

It's because when you have $x=\sqrt{2+x}$, and assuming you're dealing with reals, you know that $x>0$. However, if you have $$stuff=\sqrt{things}\implies stuff^2=|things|$$ That's where the negative value arise.

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Generally speaking, when you manipulate an equation by "doing the same thing to both sides", all you can say is that if your original equation was a true statement, then so is the new one. That doesn't always mean the new equation contains enough information to uniquely identify a solution of the original equation!

In this example, $x$ is a number such that $x = \sqrt{2+\sqrt{2+\dots}}$. (We happen to know this means that $x$ has to be 2.) Your conclusion is that $x$ has to be either $2$ or $-1$: that is a true statement! But it doesn't mean that $-1$ was a solution of the original equation.

As an analogy, suppose I have a marble in my pocket, and I tell you: "I have either a marble or a rhinoceros in my pocket." Then I have spoken truly!

Of course, in this case, you can easily verify (as an additional step) that 2 is a solution of the original problem and $-1$ is not. Since you know $x$ must be either $2$ or $-1$, all other possibilities are ruled out, and so you know $x=2$ must be the only solution.

Other manipulations can result in true statements that are even less informative. Suppose, for instance, that you multiplied both sides of the original equation by 0. You would get the new equation $0=0$. That is definitely a true statement, but it isn't very useful in solving the problem. So that was not a helpful manipulation to make and you should try something different.

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