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Question Number 26224 by ktomboy1992 last updated on 22/Dec/17

find the value of x−(1/x).when x^4 +(1/x^4 )=332

$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}.\mathrm{when}\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\mathrm{332} \\ $$

Commented by kaivan.ahmadi last updated on 22/Dec/17

x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2=322⇒  x^2 +(1/x^2 )=(√(324))=18⇒since  (x−(1/x))^2 =x^2 +(1/x^2 )−2=18−2=16⇒  x−(1/x)=∓4

$$\mathrm{x}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{4}} }=\left(\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{322}\Rightarrow \\ $$$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\sqrt{\mathrm{324}}=\mathrm{18}\Rightarrow\mathrm{since} \\ $$$$\left(\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} =\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{18}−\mathrm{2}=\mathrm{16}\Rightarrow \\ $$$$\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}=\mp\mathrm{4} \\ $$$$ \\ $$

Answered by $@ty@m last updated on 23/Dec/17

x^4 +(1/x^4 )=(x^2 +(1/x^2 ))^2 −2  (x^2 +(1/x^2 ))^2 =324  x^2 +(1/x^2 )=(√(324))  x^2 +(1/x^2 )−2=18−2  (x−(1/x))^2 =16  x−(1/x)=±4

$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} −\mathrm{2} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\mathrm{324} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\sqrt{\mathrm{324}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\mathrm{2}=\mathrm{18}−\mathrm{2} \\ $$$$\left({x}−\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{16} \\ $$$${x}−\frac{\mathrm{1}}{{x}}=\pm\mathrm{4} \\ $$

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