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Question Number 85649 by Hassen_Timol last updated on 23/Mar/20

$$\mathrm{Proove}\mathrm{that}:$$$$$$$$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

Commented by mr W last updated on 23/Mar/20

$$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{8-4\sqrt{3}}}{4}=\frac{\sqrt{{\left(\sqrt{6}\right)}^2-2\times \sqrt{6}\times \sqrt{2}+{\left(\sqrt{2}\right)}^2}}{4}$$$$=\frac{\sqrt{{(\sqrt{6}-\sqrt{2})}^2}}{4}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

Answered by MJS last updated on 23/Mar/20

$$2\sqrt{2-\sqrt{3}}=\sqrt{6}-\sqrt{2}$$$$\mathrm{square}\mathrm{both}\mathrm{sides}$$$$4(2-\sqrt{3})={(\sqrt{6}-\sqrt{2})}^2$$$$8-4\sqrt{3}=6-2\sqrt{12}+2$$$$8-4\sqrt{3}=8-2\sqrt{2^{2}3}$$$$8-4\sqrt{3}=8-4\sqrt{3}$$$$$$$$\mathrm{or}:$$$$\sqrt{2-\sqrt{3}}=a+b\sqrt{3}$$$$\mathrm{squaring}$$$$2-\sqrt{3}=a^2+3b^2+2ab\sqrt{3}$$$$\Rightarrow 2=a^2+3b^2\wedge -1=2ab$$$$\Rightarrow 2=a^2+3b^2\wedge b=-\frac{1}{2a}$$$$\Rightarrow 2=a^2+\frac{3}{4a^2}\Rightarrow a^4-2a^2+\frac{3}{4}=0\Rightarrow$$$$\Rightarrow a_{1,2}=\pm \frac{\sqrt{2}}{2}\vee a_{3,4}=\pm \frac{\sqrt{6}}{2}\Rightarrow b_j=-a_j$$$$\Rightarrow \sqrt{2-\sqrt{3}}=a_j-a_j\sqrt{3}=\pm (\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2})$$$$\mathrm{but}\mathrm{we}\mathrm{know}\sqrt{2-\sqrt{3}}>0\Rightarrow$$$$\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}$$$$\mathrm{and}$$$$\frac{\sqrt{2-\sqrt{3}}}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$$

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