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Question Number 172360 by Mikenice last updated on 25/Jun/22

Commented by mr W last updated on 25/Jun/22

$$\left(1\right)$$$$4=\sqrt{16}=\sqrt{13+\sqrt{9}}=\sqrt{13+\sqrt{5+4}}=\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}$$

Commented by mr W last updated on 25/Jun/22

$$\left(2\right)$$$$\text{You can't use 'macro parameter character \#' in math mode}$$

Answered by Rasheed.Sindhi last updated on 26/Jun/22

$$Letx=\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}$$$$x^2-13=\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}$$$${(x^2-13)}^2=5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}$$$${(x^2-13)}^2-5=x$$$$SeebelowthecommentsofmrW\mathit{s}\mathit{i}\mathit{r}\downarrow$$

Commented by mr W last updated on 25/Jun/22

$$pleaserechecksir:$$$${(x^2-13)}^2-5=x$$$$x^4-26x^2-x+164=0$$$$(x-4)(x^3+4x^2-10x-41)=0$$$$...$$

Commented by Rasheed.Sindhi last updated on 25/Jun/22

$${You}^{\prime }reright\mathit{s}\mathit{i}\mathit{r}!$$

Commented by mr W last updated on 25/Jun/22

$$x=\sqrt{13+...}>\sqrt{13}>3.5$$$$allrootsof{x}^3+4x^2-10x-41=0$$$$are<3.5,thereforex=4istheonly$$$$suitablesolution.$$

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