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How does anyone solve this equation on $x$ ($x$ is a real number)?

$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{\cdots}}}$$

Thank you in advance.

ajotatxe
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Gooble
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2 Answers2

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$$\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{\cdots}}}}=y;\quad y=x+\cfrac{x}{x+\cfrac{x}{x+\cfrac{x}{\cdots}}}$$

System $\sqrt{x+y}=y;\quad y=x+\dfrac{x}{y}$

both $x$ and $y$ are positive

$x+y=y^2\to x=y^2-y$

plug in the second equation

$y=y^2-y+\dfrac{y^2-y}{y}\to y=y^2-y+y-1$

$y^2-y-1=0\to y=\dfrac{1+\sqrt{5}}{2}$

and $x=\left(\dfrac{1+\sqrt{5}}{2}\right)^2-\dfrac{1+\sqrt{5}}{2}$

$x=1$

Hope it helps

Raffaele
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  • Excellent answer. Thank you very much. And y is... the Golden Ratio! The number 1 makes it possible for the equation to yield the Golden Number. – Gooble Aug 31 '17 at 14:15
  • If $\sqrt{x + y} = y$ and $x + \frac{x}{y} = y$, then surely you could make this shorter by saying $x + y = y^2$ and $xy + x = y^2$. You can then equate these two equations getting: $x + y = xy + x \Rightarrow xy = y \Rightarrow x = 1$? – Beta Decay Aug 31 '17 at 16:05
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Note that if $L(x)$ and $R(x)$ are the left-hand side and the right-hand side respectively, then $$L(x)=\sqrt{x+L(x)}=x+\frac{x}{R(x)}=R(x).$$ Hence $L(x)> 0$, $R(x)>0$, $$L^2(x)=L(x)+x\quad\mbox{and}\quad R^2(x)=xR(x)+x.$$ Finally, from $L(x)=R(x)$, we obtain $$L(x)+x=xL(x)+x\implies (1-x)L(x)=0\implies x=1.$$

P.S. Note that $$L(x)=\frac{1+\sqrt{1+4x}}{2}\quad\mbox{and}\quad R(x)=\frac{x+\sqrt{x^2+4x}}{2}.$$

Robert Z
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