Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 156338 by MathSh last updated on 10/Oct/21

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

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

$$x>1$$$$f\left(x\right)=\frac{1}{\sqrt{x^2-1}}$$$$g\left(x\right)=\frac{2}{x}-1$$$$bothfunctionsaredecreasing.$$$$f\left(1\right)\to +\mathrm{\infty }$$$$g\left(1\right)=1<f\left(1\right)$$$$f\left(2\right)=\frac{\sqrt{3}}{3}$$$$f\left(2\right)=0<f\left(2\right)$$$$forx>2:$$$$f\left(x\right)>0$$$$g\left(x\right)<0$$$$thatmeansforx>1:f\left(x\right)>g\left(x\right)$$$$\Rightarrow nosolutionforf\left(x\right)=g\left(x\right)!$$

Commented by MathSh last updated on 10/Oct/21

$$\mathrm{Perfect}\mathrm{my}\mathrm{dear}\mathrm{Ser},\mathrm{thank}\mathrm{you}$$

Answered by peter frank last updated on 10/Oct/21

$$\frac{1}{\sqrt{{\mathrm{x}}^2-1}}=\frac{2-\mathrm{x}}{\mathrm{x}}$$$$\mathrm{x}=(2-\mathrm{x})\sqrt{{\mathrm{x}}^2-1})$$$${\mathrm{x}}^2={(2-\mathrm{x})}^2({\mathrm{x}}^2-1)$$$${\mathrm{x}}^2=[4-4\mathrm{x}+{\mathrm{x}}^2]({\mathrm{x}}^2-1)$$$${\mathrm{x}}^2={4\mathrm{x}}^2-4-{4\mathrm{x}}^3+4\mathrm{x}+{\mathrm{x}}^4-{\mathrm{x}}^2$$$${\mathrm{x}}^2={4\mathrm{x}}^2-4-{4\mathrm{x}}^3+4\mathrm{x}+{\mathrm{x}}^4-{\mathrm{x}}^2$$$${2\mathrm{x}}^2-{4\mathrm{x}}^3+{\mathrm{x}}^4+4\mathrm{x}-4=0$$$${\mathrm{x}}^4-{4\mathrm{x}}^3+{2\mathrm{x}}^2+4\mathrm{x}-4=0$$$$.....$$$$$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com