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Question Number 161563 by HongKing last updated on 19/Dec/21

$$\mathrm{Solve}\mathrm{the}\mathrm{equation}:$$$$\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+\sqrt{\mathrm{x}+...}}}=\mathrm{x}\cdot \sqrt{\mathrm{x}\cdot \sqrt{\mathrm{x}\cdot \sqrt{\mathrm{x}\cdot ...}}}$$$$\mathrm{where},\mathrm{x}>0$$

Answered by MJS_new last updated on 20/Dec/21

$$\mathrm{lhs}={(\mathrm{lhs}-x)}^2\wedge \mathrm{lhs}⩾x\Rightarrow \mathrm{lhs}=\frac{2x+1+\sqrt{4x+1}}{2}$$$$\mathrm{rhs}=\frac{{\mathrm{rhs}}^2}{x^2}\wedge x>0\Rightarrow \mathrm{rhs}=x^2$$$$\mathrm{now}\mathrm{we}\mathrm{have}$$$$\frac{2x+1+\sqrt{4x+1}}{2}=x^2$$$$\sqrt{4x+1}=2x^2-2x-1$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}$$$$x^3(x-2)=0\Rightarrow x=2$$

Commented by HongKing last updated on 20/Dec/21

$$\mathrm{very}\mathrm{nice}\mathrm{dear}\mathrm{Sir}\mathrm{thank}\mathrm{you}$$

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