Question Number 174153 by daus last updated on 26/Jul/22
Commented by mr W last updated on 26/Jul/22
$${the}\:{question}\:{meant}\:{maybe}\:\sqrt{{c}}+\sqrt{{d}}. \\ $$
Answered by mr W last updated on 26/Jul/22
$$\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{2}}+\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{2}−\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}=\sqrt{\mathrm{23}−\mathrm{6}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)}=\sqrt{\mathrm{11}+\mathrm{6}\sqrt{\mathrm{2}}}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{6}\sqrt{\mathrm{2}}+\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\sqrt{\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }=\mathrm{3}+\sqrt{\mathrm{2}} \\ $$$$\mathrm{3}+\sqrt{\mathrm{2}}=\sqrt{{c}}−\sqrt{{d}} \\ $$$${since}\:{it}\:{is}\:{not}\:{requested}\:{that}\:{c}\:{and}\:{d} \\ $$$${should}\:{be}\:{integer},\:{there}\:{are}\:{infinite} \\ $$$${solutions}. \\ $$
Answered by BaliramKumar last updated on 26/Jul/22
$${formula}\::−\:\:\:\:\:\sqrt{{a}\pm\sqrt{{b}}\:}\:=\:\sqrt{\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}}\:\pm\:\sqrt{\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}} \\ $$$$\sqrt{\mathrm{6}−\sqrt{\mathrm{32}}}\:=\:\sqrt{\frac{\mathrm{6}+\sqrt{\mathrm{36}−\mathrm{32}}}{\mathrm{2}}}\:−\:\sqrt{\frac{\mathrm{6}−\sqrt{\mathrm{36}−\mathrm{32}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}\:=\:\mathrm{2}\:−\:\sqrt{\mathrm{2}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:\sqrt{\mathrm{23}−\mathrm{6}\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:\sqrt{\mathrm{23}−\mathrm{12}+\mathrm{6}\sqrt{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{23}−\mathrm{6}\sqrt{\mathrm{6}−\mathrm{4}\sqrt{\mathrm{2}}}}\:=\:\sqrt{\mathrm{11}+\sqrt{\mathrm{72}}} \\ $$$$\sqrt{\frac{\mathrm{11}+\sqrt{\mathrm{121}−\mathrm{72}}}{\mathrm{2}}}\:+\:\sqrt{\frac{\mathrm{11}−\sqrt{\mathrm{121}−\mathrm{72}}}{\mathrm{2}}} \\ $$$$\sqrt{\frac{\mathrm{11}+\sqrt{\mathrm{49}}}{\mathrm{2}}}\:+\:\sqrt{\frac{\mathrm{11}−\sqrt{\mathrm{49}}}{\mathrm{2}}} \\ $$$$\sqrt{\frac{\mathrm{11}+\mathrm{7}}{\mathrm{2}}}\:+\:\sqrt{\frac{\mathrm{11}−\mathrm{7}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{9}}\:+\:\sqrt{\mathrm{2}}\:=\:\sqrt{{c}}−\sqrt{{d}} \\ $$$$ \\ $$$$ \\ $$