Question Number 133981 by bemath last updated on 26/Feb/21
$$\mathrm{If}\:{x}\:=\:\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:\mathrm{then}\:\frac{{x}−\mathrm{1}}{\:\sqrt{{x}}}\:=? \\ $$
Answered by Dwaipayan Shikari last updated on 26/Feb/21
$${x}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\:=\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}−\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by malwan last updated on 26/Feb/21
$$\frac{\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{2}}\:=\:\sqrt{\mathrm{6}}\:=\:\sqrt{\mathrm{2}}×\sqrt{\mathrm{3}} \\ $$$${such}\:{that}\:\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:=\:\mathrm{5} \\ $$$${thank}\:{you}\:{sir}\:{shikari} \\ $$$${and}\:{this}\:{method}\:{can}\:{be}\:{used} \\ $$$${to}\:{find}\:{the}\:{square}\:{root}\:{of} \\ $$$${complex}\:{numbers}\: \\ $$$${very}\:{simple} \\ $$
Commented by Dwaipayan Shikari last updated on 26/Feb/21
$${General}\:{way} \\ $$$$\sqrt{{a}+\sqrt{{b}}}=\sqrt{{x}}+\sqrt{{y}} \\ $$$${a}+\sqrt{{b}}={x}+{y}+\mathrm{2}\sqrt{{xy}} \\ $$$${a}={x}+{y}\:\:\:\:\:\:\:\:\:\:{b}=\mathrm{4}{xy} \\ $$$${x}−{y}=\sqrt{{a}^{\mathrm{2}} −{b}} \\ $$$${x}=\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}\:\:\:{y}=\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}\:\:\: \\ $$$$\sqrt{{a}+\sqrt{{b}}}=\sqrt{\frac{{a}+\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}}+\sqrt{\frac{{a}−\sqrt{{a}^{\mathrm{2}} −{b}}}{\mathrm{2}}} \\ $$$$\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}=\sqrt{\mathrm{5}+\sqrt{\mathrm{24}}}=\sqrt{\frac{\mathrm{5}+\sqrt{\mathrm{25}−\mathrm{24}}}{\mathrm{2}}}+\sqrt{\frac{\mathrm{5}−\sqrt{\mathrm{25}−\mathrm{24}}}{\mathrm{2}}} \\ $$$$=\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}} \\ $$
Answered by malwan last updated on 26/Feb/21
$${x}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\frac{{x}−\mathrm{1}}{\:\sqrt{{x}}}\:=\:\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}×\frac{\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{6}}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}×\frac{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}} \\ $$$$=\:\frac{\mathrm{2}\left(\mathrm{10}+\sqrt{\mathrm{6}}−\mathrm{12}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{25}−\mathrm{4}×\mathrm{6}} \\ $$$$=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{6}}−\mathrm{2}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{1}} \\ $$$$=\:\mathrm{2}\sqrt{\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{6}}+\mathrm{4}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)} \\ $$$$=\:\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{25}−\mathrm{24}\right)}\:=\mathrm{2}\sqrt{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\ $$