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Question Number 156361 by mr W last updated on 10/Oct/21

$$\mathrm{Solve}\mathrm{in}\mathbb{R}$$$$\frac{1}{\sqrt{{\mathrm{x}}^2-1}}=1-\frac{2}{\mathrm{x}}$$

Commented by cortano last updated on 11/Oct/21

Commented by mr W last updated on 11/Oct/21

$$x=-\frac{1}{\cos \left(\frac{1}{2}{\sin }^{-1}\frac{\sqrt{5}-1}{2}\right)}$$$$x=\frac{1}{\sin \left(\frac{1}{2}{\sin }^{-1}\frac{\sqrt{5}-1}{2}\right)}$$

Answered by MJS_new last updated on 10/Oct/21

$$\sqrt{x^2-1}\ne 0\wedge x^2-1>0\wedge x\ne 0\wedge 1-\frac{2}{x}>0$$$$\iff$$$$x<-1\vee -1<x<0\vee x>2$$$$\mathrm{squaring}\mathrm{\&}\mathrm{transforming}\Rightarrow$$$$x^4-4x^3+2x^2+4x-4=0$$$$x=t+1$$$$t^4-4t^2-1=0$$$$(t^2-2-\sqrt{5})(t^2-2+\sqrt{5})=0$$$$(x^2-2x-1-\sqrt{5})(x^2-2x-1+\sqrt{5})=0$$$$x=1\pm \sqrt{2+\sqrt{5}}$$$$\mathrm{both}\mathrm{solve}\mathrm{the}\mathrm{given}\mathrm{equation}$$

Commented by mr W last updated on 10/Oct/21

$$thankssir!$$

Commented by MJS_new last updated on 10/Oct/21

$$\mathrm{it}\mathrm{was}\mathrm{just}\mathrm{a}\mathrm{typo}\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{line}\mathrm{which}\mathrm{I}$$$$\mathrm{inserted}\mathrm{later}$$

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