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

Solve in R  (1/( (√(x^2  - 1)))) =1− (2/x)

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

Commented by cortano last updated on 11/Oct/21

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

x=−(1/(cos ((1/2) sin^(−1) (((√5)−1)/2))))   x=(1/(sin ((1/2) sin^(−1) (((√5)−1)/2))))

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

Answered by MJS_new last updated on 10/Oct/21

(√(x^2 −1))≠0∧x^2 −1>0∧x≠0∧1−(2/x)>0  ⇔  x<−1∨−1<x<0∨x>2  squaring & transforming ⇒  x^4 −4x^3 +2x^2 +4x−4=0  x=t+1  t^4 −4t^2 −1=0  (t^2 −2−(√5))(t^2 −2+(√5))=0  (x^2 −2x−1−(√5))(x^2 −2x−1+(√5))=0  x=1±(√(2+(√5)))  both solve the given equation

$$\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\neq\mathrm{0}\wedge{x}^{\mathrm{2}} −\mathrm{1}>\mathrm{0}\wedge{x}\neq\mathrm{0}\wedge\mathrm{1}−\frac{\mathrm{2}}{{x}}>\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$${x}<−\mathrm{1}\vee−\mathrm{1}<{x}<\mathrm{0}\vee{x}>\mathrm{2} \\ $$$$\mathrm{squaring}\:\&\:\mathrm{transforming}\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{4}{x}−\mathrm{4}=\mathrm{0} \\ $$$${x}={t}+\mathrm{1} \\ $$$${t}^{\mathrm{4}} −\mathrm{4}{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{2}−\sqrt{\mathrm{5}}\right)\left({t}^{\mathrm{2}} −\mathrm{2}+\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}−\sqrt{\mathrm{5}}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}+\sqrt{\mathrm{5}}\right)=\mathrm{0} \\ $$$${x}=\mathrm{1}\pm\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}} \\ $$$$\mathrm{both}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$

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

thanks sir!

$${thanks}\:{sir}! \\ $$

Commented by MJS_new last updated on 10/Oct/21

it was just a typo in the first line which I  inserted later

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

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