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Question Number 100193 by bobhans last updated on 25/Jun/20

$$\sqrt{7+2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7+...}}}}}?$$

Commented by bobhans last updated on 25/Jun/20

$$x=\sqrt{7+2y};\mathrm{y}=\sqrt{7-2x}$$$$\iff x^2=7+2y\mathrm{\_}\left(1\right)$$$$\iff {\mathrm{y}}^2=7-2x\mathrm{\_}\left(2\right)$$$$\Rightarrow {\left(\frac{x^2-7}{2}\right)}^2=7-2x$$$$x^4-14x^2+49=28-8x$$$$x^4-14x^2+8x+21=0$$$$\Rightarrow x=3\wedge \mathrm{y}=1;\mathrm{then}x=\sqrt{7+2\sqrt{7-2\sqrt{7+2\sqrt{7-...}}}}$$$$x=\sqrt{7+2.1}=3◼$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$Yesitis3$$

Commented by Dwaipayan Shikari last updated on 25/Jun/20

$$Ialsodidthis$$

Answered by Dwaipayan Shikari last updated on 25/Jun/20

$$\mathrm{Suppose}\mathrm{it}\mathrm{is}\mathrm{p}$$$$\sqrt{7+2\sqrt{7-2\sqrt{7+2\sqrt{7-2..}}}}\mathrm{\infty }=\mathrm{p}$$$$7+2\sqrt{7-2\sqrt{7+2\sqrt{7-2\sqrt{7+2..}}}}={\mathrm{p}}^2$$$$7-2\sqrt{7+2\sqrt{7-2\sqrt{7+2\sqrt{7-2...}}}}={\left(\frac{{\mathrm{p}}^2-7}{2}\right)}^2$$$$\sqrt{7+2\sqrt{7-2\sqrt{7+2\sqrt{7-2...}}}}=\frac{28-{({\mathrm{p}}^2-7)}^2}{8}=\mathrm{p}$$$$\mathrm{p}=\frac{28-{({\mathrm{p}}^2-7)}^2}{8}=\frac{28-{\mathrm{p}}^4+{14\mathrm{p}}^2-49}{8}$$$$\mathrm{so},{\mathrm{p}}^4-{14\mathrm{p}}^2+8\mathrm{p}+21=0$$$$\mathrm{p}=3◼$$$$$$

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