$$\sqrt{3+\sqrt{3+\sqrt{3+x}}}=x$$
Question is: How to find x?
Could you help me? Thanks in advance
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Plug in $x=\frac{1}{2}(1+\sqrt{13})$ and see what happens. – Daniel W. Farlow Mar 31 '15 at 20:41
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1You should try to find the solution without Wolfram|Alpha – Daniel W. Farlow Mar 31 '15 at 20:44
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Yep, so that's what i need, could you help me – Enes Mar 31 '15 at 20:45
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Check this sly trick out. Just keep plugging in for $x$.
$$\begin{align*} x&=\sqrt{3+x}\\ &=\sqrt{3+\sqrt{3+x}}\\ &=\sqrt{3+\sqrt{3+\sqrt{3+x}}}\\ &=\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}} \end{align*}$$
This recursive behavior implies that
$$\sqrt{3+\sqrt{3+\sqrt{3+x}}}=\sqrt{3+x}$$
So just cut yourself some slack and solve the equation
$$x=\sqrt{3+x}$$
Arturo don Juan
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The lack of $\pm$ in the square roots allows us to neglect it. For example, $x^2=4$ has two solutions, whereas $x=\sqrt{2}$ has one. $(x^2-3)^2=x$ has four solutions, $x^2-3=\sqrt{x}\rightarrow x^2=3+\sqrt{x}$ has two solutions, and $x=\sqrt{3+\sqrt{x}}$ has one solution. – Arturo don Juan Mar 31 '15 at 22:06
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The extra solutions from taking the square root of a square come from the essential $\pm$. – Arturo don Juan Mar 31 '15 at 22:07