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Question Number 145344 by mathdanisur last updated on 04/Jul/21

$${\mathit{z}}^4=49-20\sqrt{6}\Rightarrow \mathit{z}=?$$

Answered by liberty last updated on 04/Jul/21

$$z^2=\sqrt{49-2\sqrt{600}}=\sqrt{(25+24)-2\sqrt{25\times 24}}$$$$z^2=\sqrt{25}-\sqrt{24}=5-2\sqrt{6}$$$$z=\sqrt{5-2\sqrt{6}}=\sqrt{(3+2)-2\sqrt{3\times 2}}$$$$z=\sqrt{3}-\sqrt{2}$$

Commented by mathdanisur last updated on 04/Jul/21

$$thankyouSer$$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

$$z^4=49-20\sqrt{6}$$$$z=\pm \sqrt[4]{49-20\sqrt{6}}\mathrm{or}\pm \sqrt[4]{49-20\sqrt{6}}.i$$$$$$$$\mid z\mid =\sqrt[4]{49-20\sqrt{6}}$$$$\mid z\mid =\sqrt[4]{{(5-2\sqrt{6})}^2}$$$$\mid z\mid =\sqrt{5-2\sqrt{6}}$$$$\mid z\mid =\sqrt{{(\sqrt{3}-\sqrt{2})}^2}$$$$\mid z\mid =\sqrt{3}-\sqrt{2}$$$$$$$$\mathrm{Finally},z=\pm (\sqrt{3}-\sqrt{2}),\pm (\sqrt{3}-\sqrt{2})i$$

Commented by mathdanisur last updated on 04/Jul/21

$$ThankyouSer,answer:\sqrt{3}-\sqrt{2}$$

Commented by mr W last updated on 04/Jul/21

$$if\sqrt{3}-\sqrt{2}isaroot,then$$$$-(\sqrt{3}-\sqrt{2})isalsoaroot,and$$$$(\sqrt{3}-\sqrt{2})iisalsoaroot,and$$$$-(\sqrt{3}-\sqrt{2})iisalsoaroot.$$

Commented by mathdanisur last updated on 04/Jul/21

$$ThanksSir$$

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