Question Number 194389 by mathlove last updated on 05/Jul/23
$$\sqrt{\sqrt{\mathrm{49}+\mathrm{20}\sqrt{\mathrm{6}}}}=? \\ $$
Answered by som(math1967) last updated on 05/Jul/23
$$\sqrt{\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} +\mathrm{2}.\mathrm{5}.\mathrm{2}\sqrt{\mathrm{6}}}} \\ $$$$\sqrt{\sqrt{\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} }} \\ $$$$=\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}} \\ $$$$=\sqrt{\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$
Answered by AST last updated on 05/Jul/23
$${a}=\sqrt[{\mathrm{4}}]{\mathrm{49}+\mathrm{20}\sqrt{\mathrm{6}};}{b}=\sqrt[{\mathrm{4}}]{\mathrm{49}−\mathrm{20}\sqrt{\mathrm{6}}}\left({a}>{b}\right) \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} =\mathrm{98};{ab}=\mathrm{1}\Rightarrow{a}+{b}=\mathrm{2}\sqrt{\mathrm{3}};\Rightarrow{a}\:{and}\:{b}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}=\mathrm{0}\Rightarrow{x}=\frac{\mathrm{2}\sqrt{\mathrm{3}}\underset{−} {+}\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}}=\sqrt{\mathrm{3}}\underset{−} {+}\sqrt{\mathrm{2}} \\ $$$${Since}\:{a}>{b},\:{a}=\sqrt[{\mathrm{4}}]{\mathrm{49}+\mathrm{20}\sqrt{\mathrm{6}}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$