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Question Number 59764 by ANTARES VY last updated on 14/May/19

$$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}.$$$$\mathbf{simplify}.$$

Answered by Kunal12588 last updated on 14/May/19

$$\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$$$$=\frac{\sqrt{2}(2+\sqrt{3})}{3+\sqrt{3}}+\frac{\sqrt{2}(2-\sqrt{3})}{3-\sqrt{3}}$$$$=\sqrt{2}\left(\frac{6+\sqrt{3}-3+6-\sqrt{3}-3}{9-3}\right)$$$$=\sqrt{2}$$

Commented by Kunal12588 last updated on 14/May/19

$$findingrootsofsurds$$$$oftype\sqrt{a\pm \sqrt{b}}$$$$\sqrt{a\pm \sqrt{b}}=\sqrt{m}\pm \sqrt{n}$$$$\Rightarrow a\pm b=(m+n)\pm 2\sqrt{mn}$$$$\Rightarrow m+n=a,4mn=b$$$$\Rightarrow m-n=\sqrt{{(m+n)}^2-4mn}=\sqrt{a^2-b}$$$$\Rightarrow m=\frac{a+\sqrt{a^2-b}}{2},n=\frac{a-\sqrt{a^2-b}}{2}$$$$\therefore \sqrt{a\pm \sqrt{b}}=\frac{\sqrt{a+\sqrt{a^2-b}}\pm \sqrt{a-\sqrt{a^2-b}}}{\sqrt{2}}$$$$example$$$$\sqrt{2+\sqrt{3}}=\frac{\sqrt{2+\sqrt{4-3}}+\sqrt{2-\sqrt{4-3}}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}$$$$\sqrt{2-\sqrt{3}}=\frac{\sqrt{2+\sqrt{4-3}}-\sqrt{2-\sqrt{4-3}}}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}$$$$\sqrt{14-3\sqrt{3}}=\frac{\sqrt{14+\sqrt{196-27}}-\sqrt{14-\sqrt{196-27}}}{\sqrt{2}}=\frac{3\sqrt{3}-1}{\sqrt{2}}$$

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