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if-x-2-x-2-2-2-2-x-16-x-16-




Question Number 138367 by KwesiDerek last updated on 12/Apr/21
if x^2 +x^(−2) =(√(2+(√(2+(√2)))))  x^(16) +x^(−16) =?
$$\boldsymbol{\mathrm{if}}\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}^{−\mathrm{2}} =\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{16}} +\boldsymbol{\mathrm{x}}^{−\mathrm{16}} =? \\ $$
Commented by Kamel last updated on 12/Apr/21
x is a complex number (4 solutions)
$${x}\:{is}\:{a}\:{complex}\:{number}\:\left(\mathrm{4}\:{solutions}\right) \\ $$
Answered by Kamel last updated on 12/Apr/21
Answered by cherokeesay last updated on 12/Apr/21
(x^2 +(1/x^2 ))^2  = 2+(√(2+(√2) )) ⇔  x^4 +(1/x^4 ) + 2 = 2 + (√(2+(√2))) ⇒   (x^4 +(1/x^4 ))^2  = 2+ (√2) ⇔ x^8 +(1/x^8 ) + 2 = 2+(√2)  (x^8 +(1/x^8 ))^2  = 2 ⇔ x^(16)  + (1/x^(16) ) + 2 = 2 ⇒  x^(16)  + (1/x^(16) ) = 0
$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} \:=\:\mathrm{2}+\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}\:}\:\Leftrightarrow \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:+\:\mathrm{2}\:=\:\mathrm{2}\:+\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}\:\Rightarrow\: \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{2}} \:=\:\mathrm{2}+\:\sqrt{\mathrm{2}}\:\Leftrightarrow\:{x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\:+\:\mathrm{2}\:=\:\mathrm{2}+\sqrt{\mathrm{2}} \\ $$$$\left({x}^{\mathrm{8}} +\frac{\mathrm{1}}{{x}^{\mathrm{8}} }\right)^{\mathrm{2}} \:=\:\mathrm{2}\:\Leftrightarrow\:{x}^{\mathrm{16}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{16}} }\:+\:\mathrm{2}\:=\:\mathrm{2}\:\Rightarrow \\ $$$${x}^{\mathrm{16}} \:+\:\frac{\mathrm{1}}{{x}^{\mathrm{16}} }\:=\:\mathrm{0} \\ $$

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