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Question Number 184873 by topgrace100 last updated on 13/Jan/23

$$\mathrm{Find}\mathrm{the}\mathrm{real}\mathrm{number}\mathrm{satisfying}$$$$\mathrm{x}=\sqrt{1+\sqrt{1+\sqrt{1+\mathrm{x}}}}$$

Answered by Rasheed.Sindhi last updated on 13/Jan/23

$$\mathrm{x}=\sqrt{1+\sqrt{1+\sqrt{1+\mathrm{x}}}}$$$$\mathrm{x}=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+...}}}}}$$$${\mathrm{x}}^2=1+\mathrm{x}$$$${\mathrm{x}}^2-\mathrm{x}-1=0$$$$\mathrm{x}=\frac{1+\sqrt{1+4}}{2}=\frac{1+\sqrt{5}}{2}✓[\because \frac{1-\sqrt{5}}{2}<0]$$

Commented by Frix last updated on 13/Jan/23

$$x=\sqrt{...}⩾0\Rightarrow x\ne \frac{1-\sqrt{5}}{2}$$

Commented by Rasheed.Sindhi last updated on 13/Jan/23

$$Right\mathit{s}\mathit{i}\mathit{r}!I^{\prime }mgoingtogetout\frac{1-\sqrt{5}}{2}$$$$fromthesolutionset.\mathcal{T}han\mathcal{X}!$$

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