Question Number 103321 by bemath last updated on 14/Jul/20
$$\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+…}}}}}} \\ $$
Answered by Dwaipayan Shikari last updated on 14/Jul/20
$$\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}…}}}={p} \\ $$$$\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}}}}…={p}^{\mathrm{2}} \\ $$$$\mathrm{5}+{p}=\left(\mathrm{5}−{p}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$${p}^{\mathrm{4}} −\mathrm{10}{p}^{\mathrm{2}} +\mathrm{20}−{p}=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{p}−\mathrm{5}\right)\left({p}^{\mathrm{2}} −{p}−\mathrm{4}\right)=\mathrm{0} \\ $$$${p}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}} \\ $$$${or}\:{p}=\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\mathrm{5}−\frac{\mathrm{1}\pm\sqrt{\mathrm{21}}}{\mathrm{2}}\:\:\:{or}\:\:\mathrm{5}+\frac{\mathrm{1}\pm\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\mathrm{More}\:\mathrm{precisely} \\ $$$$\mathrm{p}=\frac{\mathrm{10}−\mathrm{1}+\sqrt{\mathrm{17}}}{\mathrm{2}}=\frac{\mathrm{9}+\sqrt{\mathrm{17}}}{\mathrm{2}}\left(\mathrm{A}\:\mathrm{possible}\:\mathrm{one}\right) \\ $$
Answered by OlafThorendsen last updated on 14/Jul/20
$${x}\:=\:\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{\mathrm{5}+…}}}}}} \\ $$$${x}\:=\:\mathrm{5}+\sqrt{\mathrm{5}−\sqrt{{x}}} \\ $$$$\Rightarrow\:\left({x}−\mathrm{5}\right)^{\mathrm{2}} \:=\:\mathrm{5}−\sqrt{{x}} \\ $$$${x}^{\mathrm{2}} −\mathrm{10}{x}+\sqrt{{x}}+\mathrm{20}\:=\:\mathrm{0} \\ $$$${x}\:=\:\mathrm{X}^{\mathrm{2}} \\ $$$$\mathrm{X}^{\mathrm{4}} −\mathrm{10X}^{\mathrm{2}} +\mathrm{X}+\mathrm{20}\:=\:\mathrm{0} \\ $$$$\begin{cases}{\mathrm{X}\:=\:\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\:\left(\mathrm{1}\right)}\\{\mathrm{X}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}}\:\left(\mathrm{2}\right)}\\{\mathrm{X}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{21}}}{\mathrm{2}}\:\left(\mathrm{3}\right)}\\{\mathrm{X}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{21}}}{\mathrm{2}}\:\left(\mathrm{4}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{3}\right)\:\mathrm{impossible}\:\left(\mathrm{X}>\mathrm{0}\right) \\ $$$${x}>\mathrm{5}\:\Rightarrow\:\mathrm{X}>\sqrt{\mathrm{5}}\:\approx\:\mathrm{2},\mathrm{23} \\ $$$$\Rightarrow\:\left(\mathrm{4}\right)\:\mathrm{impossible} \\ $$$$\mathrm{Then}\:\mathrm{X}\:=\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$$$\mathrm{And}\:{x}\:=\:\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{17}}+\mathrm{17}}{\mathrm{4}}\:=\:\frac{\mathrm{9}+\sqrt{\mathrm{17}}}{\mathrm{2}} \\ $$