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 13029 by chux last updated on 11/May/17

Commented by chux last updated on 12/May/17

$$\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{out}....$$$$$$$$\mathrm{find}\mathrm{x}\mathrm{and}\mathrm{y}$$

Answered by ajfour last updated on 12/May/17

$${(x-\frac{1}{x})}^2+\frac{1}{x^2}+\frac{x}{y}=0.....\left(1\right)$$$$or{[{(x-\frac{1}{x})}^2+\frac{1}{x^2}]}^2=\frac{x^2}{y^2}....\left(i\right)$$$$further3x^2+\frac{1}{y^2}=4......\left(2\right)$$$$\Rightarrow 3x^4+\frac{x^2}{y^2}=4x^2......\left(ii\right)$$$$so{[{(x-\frac{1}{x})}^2+\frac{1}{x^2}]}^2=4x^2-3x^4$$$${(x^2-2+\frac{2}{x^2})}^2=4x^2-3x^4$$$$x^4+4+\frac{4}{x^4}+2(-2x^2+2-\frac{4}{x^2})=4x^2-3x^4$$$$x^4+4+\frac{4}{x^4}-4x^2+4-\frac{8}{x^2}=4x^2-3x^4$$$$4x^4+8+\frac{4}{x^4}-8x^2-\frac{8}{x^2}=0$$$$4{(x^2+\frac{1}{x^2})}^2=8(x^2+\frac{1}{x^2})$$$$\Rightarrow x^2+\frac{1}{x^2}=2$$$$x^2-2+\frac{1}{x^2}=0$$$${(x-\frac{1}{x})}^2=0$$$$x=\frac{1}{x}\Rightarrow x^2=1$$$$x=\pm 1$$$$substitutingforxinequation..\left(1\right)$$$$whichbecomesy=-x^3,as(x-\frac{1}{x})=0$$$$soy=\mp 1$$$$solutionis(1,-1)and(-1,1).$$

Commented by chux last updated on 13/May/17

$$\mathrm{this}\mathrm{is}\mathrm{wonderful}$$

Commented by RasheedSindhi last updated on 12/May/17

$$\mathrm{\Sigma }\mathrm{xcelent}!$$

Commented by ajfour last updated on 12/May/17

$$thanks.$$

Commented by mrW1 last updated on 13/May/17

$$great!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com