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Question-142971




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

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