# If-x-5-2-6-then-x-1-x-

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
$${x}=\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}} \\$$$$\Rightarrow\frac{{x}−\mathrm{1}}{\:\sqrt{{x}}}\:=\:\frac{\mathrm{4}+\mathrm{2}\sqrt{\mathrm{6}}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}×\frac{\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\:\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}} \\$$$$=\frac{\mathrm{2}\left(\mathrm{2}+\sqrt{\mathrm{6}}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}×\frac{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}}{\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}} \\$$$$=\:\frac{\mathrm{2}\left(\mathrm{10}+\sqrt{\mathrm{6}}−\mathrm{12}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{25}−\mathrm{4}×\mathrm{6}} \\$$$$=\:\frac{\mathrm{2}\left(\sqrt{\mathrm{6}}−\mathrm{2}\right)\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}}}{\mathrm{1}} \\$$$$=\:\mathrm{2}\sqrt{\left(\mathrm{6}−\mathrm{4}\sqrt{\mathrm{6}}+\mathrm{4}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)} \\$$$$=\:\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{5}−\mathrm{2}\sqrt{\mathrm{6}}\right)\left(\mathrm{5}+\mathrm{2}\sqrt{\mathrm{6}}\right)} \\$$$$=\:\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{25}−\mathrm{24}\right)}\:=\mathrm{2}\sqrt{\mathrm{2}×\mathrm{1}}\:=\:\mathrm{2}\sqrt{\mathrm{2}} \\$$