Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 165776 by ZiYangLee last updated on 08/Feb/22

Solve the equation (√(((√x)+1^ )/( (√x)−1 )))−(√(((√x)−1^ )/( (√x)+1 ))) = 1

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\:\sqrt{\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}}−\sqrt{\frac{\sqrt{{x}}−\mathrm{1}^{} }{\:\sqrt{{x}}+\mathrm{1}\:}}\:=\:\mathrm{1} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Feb/22

(√(((√x)+1^ )/( (√x)−1 )))−(√(((√x)−1^ )/( (√x)+1 ))) = 1 (√(((√x)+1^ )/( (√x)−1 ))) =y (let) y−(1/y)=1 y^2 −y−1=0 y=((1±(√(1+4)))/2)=((1±(√5))/2) (√(((√x)+1^ )/( (√x)−1 )))=((1±(√5))/2) (((√x)+1^ )/( (√x)−1 ))=((1+5±2(√5) )/4)=((3±(√5))/2) (((√x)+1^ )/( (√x)−1 ))−1=((3±(√5))/2)−1 (2/( (√x)−1 ))=((1±(√5))/2) (((√x) −1)/2)=(2/(1±(√5))) (√x) −1=(4/(1±(√5)))×((1∓(√5))/(1∓(√5)))=((4∓4(√5))/(1−5)) =−1±(√5) (√x)=±(√5) ⇒x=5

$$\:\:\:\sqrt{\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}}−\sqrt{\frac{\sqrt{{x}}−\mathrm{1}^{} }{\:\sqrt{{x}}+\mathrm{1}\:}}\:=\:\mathrm{1} \\ $$$$\sqrt{\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}}\:={y}\:\left({let}\right) \\ $$$${y}−\frac{\mathrm{1}}{{y}}=\mathrm{1} \\ $$$${y}^{\mathrm{2}} −{y}−\mathrm{1}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}=\frac{\mathrm{1}+\mathrm{5}\pm\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{4}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{\sqrt{{x}}+\mathrm{1}^{} }{\:\sqrt{{x}}−\mathrm{1}\:}−\mathrm{1}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}−\mathrm{1} \\ $$$$\frac{\mathrm{2}}{\:\sqrt{{x}}−\mathrm{1}\:}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\frac{\sqrt{{x}}\:−\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{2}}{\mathrm{1}\pm\sqrt{\mathrm{5}}} \\ $$$$\sqrt{{x}}\:−\mathrm{1}=\frac{\mathrm{4}}{\mathrm{1}\pm\sqrt{\mathrm{5}}}×\frac{\mathrm{1}\mp\sqrt{\mathrm{5}}}{\mathrm{1}\mp\sqrt{\mathrm{5}}}=\frac{\mathrm{4}\mp\mathrm{4}\sqrt{\mathrm{5}}}{\mathrm{1}−\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{1}\pm\sqrt{\mathrm{5}} \\ $$$$\sqrt{{x}}=\pm\sqrt{\mathrm{5}}\:\Rightarrow{x}=\mathrm{5} \\ $$

Commented by Tawa11 last updated on 10/Feb/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Answered by Rasheed.Sindhi last updated on 08/Feb/22

(√(((√x)+1)/( (√x)−1 )))−(√(((√x)−1)/( (√x)+1 ))) = 1 (√((((√x)+1)^2 )/( ((√x)−1)((√x) +1) )))−(√((((√x)−1)^2 )/( ((√x)+1)((√x) −1) ))) = 1 (((√x) +1)/( (√(x−1))))−(((√x) −1)/( (√(x−1))))=1 (√x) +1−(√x) +1=(√(x−1)) (√(x−1))=2 x−1=4 x=5

$$\:\:\:\sqrt{\frac{\sqrt{{x}}+\mathrm{1}}{\:\sqrt{{x}}−\mathrm{1}\:}}−\sqrt{\frac{\sqrt{{x}}−\mathrm{1}}{\:\sqrt{{x}}+\mathrm{1}\:}}\:=\:\mathrm{1} \\ $$$$\:\:\:\sqrt{\frac{\left(\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} }{\:\left(\sqrt{{x}}−\mathrm{1}\right)\left(\sqrt{{x}}\:+\mathrm{1}\right)\:}}−\sqrt{\frac{\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }{\:\left(\sqrt{{x}}+\mathrm{1}\right)\left(\sqrt{{x}}\:−\mathrm{1}\right)\:}}\:=\:\mathrm{1} \\ $$$$\frac{\sqrt{{x}}\:+\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}−\frac{\sqrt{{x}}\:−\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}=\mathrm{1} \\ $$$$\cancel{\sqrt{{x}}\:}+\mathrm{1}−\cancel{\sqrt{{x}}\:}+\mathrm{1}=\sqrt{{x}−\mathrm{1}} \\ $$$$\sqrt{{x}−\mathrm{1}}=\mathrm{2} \\ $$$${x}−\mathrm{1}=\mathrm{4} \\ $$$${x}=\mathrm{5} \\ $$

Commented by ZiYangLee last updated on 08/Feb/22

Nice sir.

$$\mathrm{Nice}\:\mathrm{sir}. \\ $$

Commented by Tawa11 last updated on 10/Feb/22

Great sir

$$\mathrm{Great}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 10/Feb/22

Thanks miss!

$$\mathbb{T}\mathrm{han}\Bbbk\mathrm{s}\:\mathrm{miss}! \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com