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 93170 by ar247 last updated on 11/May/20

$$x^2+\frac{1}{x^2}=27$$$$\sqrt{x}+\frac{1}{\sqrt{x}}=....$$

Commented by mr W last updated on 11/May/20

$${(x+\frac{1}{x})}^2=x^2+\frac{1}{x^2}+2=27+2=29$$$$x+\frac{1}{x}=\pm \sqrt{29}$$$$\sqrt{x}+\frac{1}{\sqrt{x}}=t>0$$$$t^2=x+\frac{1}{x}+2=\sqrt{29}+2$$$$\Rightarrow t=\sqrt{\sqrt{29}+2}$$

Commented by i jagooll last updated on 11/May/20

$$\mathrm{let}\sqrt{\mathrm{x}}+\frac{1}{\sqrt{\mathrm{x}}}=\mathrm{t}$$$$\mathrm{x}+\frac{1}{\mathrm{x}}={\mathrm{t}}^2-2$$$${\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}+2={({\mathrm{t}}^2-2)}^2$$$$29={({\mathrm{t}}^2-2)}^2$$$${\mathrm{t}}^2=2+\sqrt{29}$$$$\mathrm{t}=\sqrt{2+\sqrt{29}}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com