Solve-the-equation-x-1-x-1-x-1-x-1-1-

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

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

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

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

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