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Question Number 174153 by daus last updated on 26/Jul/22

Commented by mr W last updated on 26/Jul/22

$$thequestionmeantmaybe\sqrt{c}+\sqrt{d}.$$

Answered by mr W last updated on 26/Jul/22

$$\sqrt{6-4\sqrt{2}}=\sqrt{2^2-4\sqrt{2}+{\left(\sqrt{2}\right)}^2}=\sqrt{{(2-\sqrt{2})}^2}=2-\sqrt{2}$$$$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{23-6(2-\sqrt{2})}=\sqrt{11+6\sqrt{2}}=\sqrt{3^2+6\sqrt{2}+{\left(\sqrt{2}\right)}^2}=\sqrt{{(3+\sqrt{2})}^2}=3+\sqrt{2}$$$$3+\sqrt{2}=\sqrt{c}-\sqrt{d}$$$$sinceitisnotrequestedthatcandd$$$$shouldbeinteger,thereareinfinite$$$$solutions.$$

Answered by BaliramKumar last updated on 26/Jul/22

$$formula:-\sqrt{a\pm \sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\pm \sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$$$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{23-6\sqrt{6-\sqrt{32}}}$$$$\sqrt{6-\sqrt{32}}=\sqrt{\frac{6+\sqrt{36-32}}{2}}-\sqrt{\frac{6-\sqrt{36-32}}{2}}$$$$\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$$$$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{23-6(2-\sqrt{2})}$$$$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{23-12+6\sqrt{2}}$$$$\sqrt{23-6\sqrt{6-4\sqrt{2}}}=\sqrt{11+\sqrt{72}}$$$$\sqrt{\frac{11+\sqrt{121-72}}{2}}+\sqrt{\frac{11-\sqrt{121-72}}{2}}$$$$\sqrt{\frac{11+\sqrt{49}}{2}}+\sqrt{\frac{11-\sqrt{49}}{2}}$$$$\sqrt{\frac{11+7}{2}}+\sqrt{\frac{11-7}{2}}$$$$\sqrt{9}+\sqrt{2}=\sqrt{c}-\sqrt{d}$$$$$$$$$$

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