Question Number 145344 by mathdanisur last updated on 04/Jul/21
$$\boldsymbol{{z}}^{\mathrm{4}} \:=\:\mathrm{49}\:-\:\mathrm{20}\sqrt{\mathrm{6}}\:\:\Rightarrow\:\boldsymbol{{z}}=? \\ $$
Answered by liberty last updated on 04/Jul/21
$$\:{z}^{\mathrm{2}} \:=\:\sqrt{\mathrm{49}−\mathrm{2}\sqrt{\mathrm{600}}}\:=\sqrt{\left(\mathrm{25}+\mathrm{24}\right)−\mathrm{2}\sqrt{\mathrm{25}×\mathrm{24}}} \\ $$$${z}^{\mathrm{2}} \:=\:\sqrt{\mathrm{25}}−\sqrt{\mathrm{24}}\:=\:\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\: \\ $$$${z}\:=\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}\:=\sqrt{\left(\mathrm{3}+\mathrm{2}\right)−\mathrm{2}\sqrt{\mathrm{3}×\mathrm{2}}} \\ $$$$\:{z}\:\:=\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\: \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${thank}\:{you}\:{Ser} \\ $$
Answered by Olaf_Thorendsen last updated on 04/Jul/21
$${z}^{\mathrm{4}} \:=\:\mathrm{49}−\mathrm{20}\sqrt{\mathrm{6}} \\ $$$${z}\:=\:\pm\sqrt[{\mathrm{4}}]{\mathrm{49}−\mathrm{20}\sqrt{\mathrm{6}}}\:\mathrm{or}\:\pm\sqrt[{\mathrm{4}}]{\mathrm{49}−\mathrm{20}\sqrt{\mathrm{6}}}.{i} \\ $$$$ \\ $$$$\mid{z}\mid\:=\:\sqrt[{\mathrm{4}}]{\mathrm{49}−\mathrm{20}\sqrt{\mathrm{6}}} \\ $$$$\mid{z}\mid\:=\:\sqrt[{\mathrm{4}}]{\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)^{\mathrm{2}} } \\ $$$$\mid{z}\mid\:=\:\sqrt{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}} \\ $$$$\mid{z}\mid\:=\:\sqrt{\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$\mid{z}\mid\:=\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$$$ \\ $$$$\mathrm{Finally},\:{z}\:=\:\pm\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right),\:\pm\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right){i} \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${Thank}\:{you}\:{Ser},\:{answer}:\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}} \\ $$
Commented by mr W last updated on 04/Jul/21
$${if}\:\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\:{is}\:{a}\:{root},\:{then} \\ $$$$−\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right)\:{is}\:{also}\:{a}\:{root},\:{and} \\ $$$$\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right){i}\:{is}\:{also}\:{a}\:{root},\:{and} \\ $$$$−\left(\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}\right){i}\:{is}\:{also}\:{a}\:{root}. \\ $$
Commented by mathdanisur last updated on 04/Jul/21
$${Thanks}\:{Sir} \\ $$