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Question Number 20201 by khamizan833@gmail.con last updated on 24/Aug/17

$${}^4\sqrt{49-20\sqrt{6}}=$$

Answered by Einstein Newton last updated on 24/Aug/17

$$\sqrt{49-20\sqrt{6}}=\sqrt{5^2+{\left(2\sqrt{6}\right)}^2-2\left(5\right)\left(2\sqrt{6}\right)}$$$$=5-2\sqrt{6}$$$$\mathrm{Now}\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}$$$$\therefore \sqrt[4]{49-20\sqrt{6}}=\sqrt{3}-\sqrt{2}$$

Answered by $@ty@m last updated on 24/Aug/17

$$\therefore \sqrt{40-20\sqrt{6}}=5-\sqrt{24}$$$$\Rightarrow x^2+y=49\mathrm{\&}2x\sqrt{y}=20\sqrt{6}$$$$\Rightarrow x^2+y=49\mathrm{\&}{x}^{2}y=600$$$$\Rightarrow \frac{600}{y}+y=49$$$$\Rightarrow y^2-49y+600=0$$$$\Rightarrow y=\frac{49\pm \sqrt{2401-2400}}{2}$$$$\Rightarrow y=\frac{49\pm 1}{2}$$$$\Rightarrow y=25,24$$$$\Rightarrow x=\sqrt{24},\sqrt{25}$$$$\therefore \sqrt{40-20\sqrt{6}}=\pm (5-\sqrt{24})$$$$again,$$$$let\sqrt{5-\sqrt{24}}=u-\sqrt{v}$$$$proceedinginthesamemanner$$$$wegetu=\sqrt{3},\sqrt{2},v=2,3$$$$\therefore {}^4\sqrt{49-20\sqrt{6}}=\pm (\sqrt{3}-\sqrt{2)},\pm (\sqrt{3}-\sqrt{2)}$$$$(nthrootof49-20\sqrt{6}willhavenroots.)$$

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