Question Number 142971 by 0731619 last updated on 08/Jun/21
Commented by wassel last updated on 11/Jun/21
$$\left(\sqrt{{x}}−\frac{\mathrm{1}}{\sqrt{{x}}}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}=−\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{\mathrm{3}}{\mathrm{2}}{i}^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\sqrt{{x}}}=\pm{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}\:\:}\: \\ $$
Answered by Rasheed.Sindhi last updated on 08/Jun/21
$${x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{1}}{\mathrm{2}}\:;\:\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=? \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\left(\sqrt{{x}}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\sqrt{{x}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\left(\sqrt{{x}}\right)^{\mathrm{2}} }−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}=\frac{\mathrm{1}−\mathrm{4}}{\mathrm{2}} \\ $$$$\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}=−\frac{\mathrm{6}}{\mathrm{4}}=\left(\pm\frac{{i}\sqrt{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=\pm\frac{{i}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$
Commented by 0731619 last updated on 08/Jun/21
$${tanks} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Jun/21
$$\left(\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}\right)^{\mathrm{2}} ={x}+\frac{\mathrm{1}}{{x}}−\mathrm{2}=\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{−\mathrm{6}}{\mathrm{4}}=\left(\pm\frac{{i}\sqrt{\mathrm{6}}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\sqrt{{x}}−\frac{\mathrm{1}}{\:\sqrt{{x}}}=\pm\frac{{i}\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$