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Question Number 103321 by bemath last updated on 14/Jul/20

$$5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+...}}}}}}$$

Answered by Dwaipayan Shikari last updated on 14/Jul/20

$$\sqrt{5-\sqrt{5+\sqrt{5...}}}=p$$$$5-\sqrt{5+\sqrt{5-\sqrt{5}}}...=p^2$$$$5+p={(5-p^2)}^2$$$$p^4-10p^2+20-p=0$$$$(p^2+p-5)(p^2-p-4)=0$$$$p=\frac{-1\pm \sqrt{21}}{2}$$$$orp=\frac{1\pm \sqrt{17}}{2}$$$$5-\frac{1\pm \sqrt{21}}{2}or5+\frac{1\pm \sqrt{17}}{2}$$$$\mathrm{More}\mathrm{precisely}$$$$\mathrm{p}=\frac{10-1+\sqrt{17}}{2}=\frac{9+\sqrt{17}}{2}\left(\mathrm{A}\mathrm{possible}\mathrm{one}\right)$$

Answered by OlafThorendsen last updated on 14/Jul/20

$$x=5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+\sqrt{5-\sqrt{5+...}}}}}}$$$$x=5+\sqrt{5-\sqrt{x}}$$$$\Rightarrow {(x-5)}^2=5-\sqrt{x}$$$$x^2-10x+\sqrt{x}+20=0$$$$x={\mathrm{X}}^2$$$${\mathrm{X}}^4-{10\mathrm{X}}^2+\mathrm{X}+20=0$$$$\{\begin{array}{l}\mathrm{X}=\frac{1}{2}-\frac{\sqrt{17}}{2}\left(1\right)\\ \mathrm{X}=\frac{1}{2}+\frac{\sqrt{17}}{2}\left(2\right)\\ \mathrm{X}=-\frac{1}{2}-\frac{\sqrt{21}}{2}\left(3\right)\\ \mathrm{X}=-\frac{1}{2}+\frac{\sqrt{21}}{2}\left(4\right)\end{array}$$$$\left(1\right)\mathrm{and}\left(3\right)\mathrm{impossible}(\mathrm{X}>0)$$$$x>5\Rightarrow \mathrm{X}>\sqrt{5}\approx 2,23$$$$\Rightarrow \left(4\right)\mathrm{impossible}$$$$\mathrm{Then}\mathrm{X}=\frac{1}{2}+\frac{\sqrt{17}}{2}$$$$\mathrm{And}x=\frac{1+2\sqrt{17}+17}{4}=\frac{9+\sqrt{17}}{2}$$

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