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Question Number 156338 by MathSh last updated on 10/Oct/21
Solve in R (1/( (√(x^2 - 1)))) = (2/x) - 1
$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:-\:\mathrm{1} \\ $$$$ \\ $$
Commented by mr W last updated on 10/Oct/21
x>1 f(x)=(1/( (√(x^2 −1)))) g(x)=(2/x)−1 both functions are decreasing. f(1)→+∞ g(1)=1 <f(1) f(2)=((√3)/3) f(2)=0 <f(2) for x>2: f(x)>0 g(x)<0 that means for x>1: f(x)>g(x) ⇒no solution for f(x)=g(x)!
$${x}>\mathrm{1} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}} \\ $$$${g}\left({x}\right)=\frac{\mathrm{2}}{{x}}−\mathrm{1} \\ $$$${both}\:{functions}\:{are}\:{decreasing}. \\ $$$${f}\left(\mathrm{1}\right)\rightarrow+\infty \\ $$$${g}\left(\mathrm{1}\right)=\mathrm{1}\:<{f}\left(\mathrm{1}\right) \\ $$$${f}\left(\mathrm{2}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{0}\:<{f}\left(\mathrm{2}\right) \\ $$$${for}\:{x}>\mathrm{2}: \\ $$$${f}\left({x}\right)>\mathrm{0} \\ $$$${g}\left({x}\right)<\mathrm{0} \\ $$$${that}\:{means}\:{for}\:{x}>\mathrm{1}:\:{f}\left({x}\right)>{g}\left({x}\right) \\ $$$$\Rightarrow{no}\:{solution}\:{for}\:{f}\left({x}\right)={g}\left({x}\right)! \\ $$
Commented by MathSh last updated on 10/Oct/21
Perfect my dear Ser, thank you
$$\mathrm{Perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Ser},\:\mathrm{thank}\:\mathrm{you} \\ $$
Answered by peter frank last updated on 10/Oct/21
(1/( (√(x^2 −1))))=((2−x)/x) x=(2−x)(√(x^2 −1)) ) x^2 =(2−x)^2 (x^2 −1) x^2 =[4−4x+x^2 ](x^2 −1) x^2 =4x^2 −4−4x^3 +4x+x^4 −x^2 x^2 =4x^2 −4−4x^3 +4x+x^4 −x^2 2x^2 −4x^3 +x^4 +4x−4=0 x^4 −4x^3 +2x^2 +4x−4=0 .....
$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}}=\frac{\mathrm{2}−\mathrm{x}}{\mathrm{x}} \\ $$$$\left.\mathrm{x}=\left(\mathrm{2}−\mathrm{x}\right)\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\left(\mathrm{2}−\mathrm{x}\right)^{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\left[\mathrm{4}−\mathrm{4x}+\mathrm{x}^{\mathrm{2}} \right]\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4x}^{\mathrm{2}} −\mathrm{4}−\mathrm{4x}^{\mathrm{3}} +\mathrm{4x}+\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{4x}^{\mathrm{2}} −\mathrm{4}−\mathrm{4x}^{\mathrm{3}} +\mathrm{4x}+\mathrm{x}^{\mathrm{4}} −\mathrm{x}^{\mathrm{2}} \\ $$$$\mathrm{2x}^{\mathrm{2}} −\mathrm{4x}^{\mathrm{3}} +\mathrm{x}^{\mathrm{4}} +\mathrm{4x}−\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{4x}^{\mathrm{3}} +\mathrm{2x}^{\mathrm{2}} +\mathrm{4x}−\mathrm{4}=\mathrm{0} \\ $$$$….. \\ $$$$ \\ $$

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