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Question Number 8073 by 314159 last updated on 29/Sep/16

Commented by 123456 last updated on 30/Sep/16

cosh(x)=((e^x +e^(−x) )/2) x^4 +x^(−4) =47 x=e^t e^(4t) +e^(−4t) =47 2∙((e^(4t) +e^(−4t) )/2)=47 2cosh(4t)=47 t=((argcosh(((47)/2)))/4) x^3 +x^(−3) =2∙((e^(3t) +e^(−3t) )/2)=2cosh(3t) =2cosh(((3argcosh(((47)/2)))/4))

$$\mathrm{cosh}\left({x}\right)=\frac{{e}^{{x}} +{e}^{−{x}} }{\mathrm{2}} \\ $$$${x}^{\mathrm{4}} +{x}^{−\mathrm{4}} =\mathrm{47} \\ $$$${x}={e}^{{t}} \\ $$$${e}^{\mathrm{4}{t}} +{e}^{−\mathrm{4}{t}} =\mathrm{47} \\ $$$$\mathrm{2}\centerdot\frac{{e}^{\mathrm{4}{t}} +{e}^{−\mathrm{4}{t}} }{\mathrm{2}}=\mathrm{47} \\ $$$$\mathrm{2cosh}\left(\mathrm{4}{t}\right)=\mathrm{47} \\ $$$${t}=\frac{\mathrm{argcosh}\left(\frac{\mathrm{47}}{\mathrm{2}}\right)}{\mathrm{4}} \\ $$$${x}^{\mathrm{3}} +{x}^{−\mathrm{3}} =\mathrm{2}\centerdot\frac{{e}^{\mathrm{3}{t}} +{e}^{−\mathrm{3}{t}} }{\mathrm{2}}=\mathrm{2cosh}\left(\mathrm{3}{t}\right) \\ $$$$=\mathrm{2cosh}\left(\frac{\mathrm{3argcosh}\left(\frac{\mathrm{47}}{\mathrm{2}}\right)}{\mathrm{4}}\right) \\ $$

Commented by Rasheed Soomro last updated on 30/Sep/16

Appreciations for this novel approach! Can we obtain the answer in more simplified form?

$${Appreciations}\:{for}\:{this}\:{novel}\:{approach}! \\ $$$${Can}\:{we}\:{obtain}\:{the}\:{answer}\:{in}\:{more}\:{simplified}\:{form}? \\ $$

Commented by prakash jain last updated on 30/Sep/16

This gives only one solution. x^4 +(1/x^4 ) gives 8 solutions which are be α,β,γ,θ,(1/α),(1/β),(1/γ),(1/θ) effectively we get 4 solutions for x^3 +(1/x^3 ) as given in Rasheed′s Answer? How do we get same 4 answer using above approach? Actual calculation of the value gives 18.

$$\mathrm{This}\:\mathrm{gives}\:\mathrm{only}\:\mathrm{one}\:\mathrm{solution}. \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\:\mathrm{gives}\:\mathrm{8}\:\mathrm{solutions}\:\mathrm{which}\:\mathrm{are}\:\mathrm{be} \\ $$$$\alpha,\beta,\gamma,\theta,\frac{\mathrm{1}}{\alpha},\frac{\mathrm{1}}{\beta},\frac{\mathrm{1}}{\gamma},\frac{\mathrm{1}}{\theta} \\ $$$$\mathrm{effectively}\:\mathrm{we}\:\mathrm{get}\:\mathrm{4}\:\mathrm{solutions}\:\mathrm{for}\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} } \\ $$$$\mathrm{as}\:\mathrm{given}\:\mathrm{in}\:\mathrm{Rasheed}'\mathrm{s}\:\mathrm{Answer}?\:\mathrm{How}\:\mathrm{do} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{same}\:\mathrm{4}\:\mathrm{answer}\:\mathrm{using}\:\mathrm{above}\:\mathrm{approach}? \\ $$$$\mathrm{Actual}\:\mathrm{calculation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{value}\:\mathrm{gives}\:\mathrm{18}. \\ $$

Commented by 123456 last updated on 30/Sep/16

y=cosh(t)=((e^t +e^(−t) )/2) x=e^t 2y=x+(1/x) 2xy=x^2 +1 x^2 −2xy+1=0 Δ=4y^2 −4=4(y^2 −1) x=y±(√(y^2 −1)) (x≥0) e^t =y+(√(y^2 −1)) t=ln(y+(√(y^2 −1))) argcosh(t)=ln(t+(√(t^2 −1)))

$${y}=\mathrm{cosh}\left({t}\right)=\frac{{e}^{{t}} +{e}^{−{t}} }{\mathrm{2}} \\ $$$${x}={e}^{{t}} \\ $$$$\mathrm{2}{y}={x}+\frac{\mathrm{1}}{{x}} \\ $$$$\mathrm{2}{xy}={x}^{\mathrm{2}} +\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{4}{y}^{\mathrm{2}} −\mathrm{4}=\mathrm{4}\left({y}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${x}={y}\pm\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\:\:\:\:\left({x}\geqslant\mathrm{0}\right) \\ $$$${e}^{{t}} ={y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}} \\ $$$${t}=\mathrm{ln}\left({y}+\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$$$\mathrm{argcosh}\left({t}\right)=\mathrm{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right) \\ $$

Commented by 123456 last updated on 30/Sep/16

we lost some roots y=x^2 ⇒x=±(√y) also cos(x+2πk)=cos(x) cosh(xi)=cos(x) so cosh are periodic too, but the period is 2πi

$$\mathrm{we}\:\mathrm{lost}\:\mathrm{some}\:\mathrm{roots} \\ $$$${y}={x}^{\mathrm{2}} \Rightarrow{x}=\pm\sqrt{{y}} \\ $$$$\mathrm{also} \\ $$$$\mathrm{cos}\left({x}+\mathrm{2}\pi{k}\right)=\mathrm{cos}\left({x}\right) \\ $$$$\mathrm{cosh}\left({xi}\right)=\mathrm{cos}\left({x}\right) \\ $$$$\mathrm{so}\:\mathrm{cosh}\:\mathrm{are}\:\mathrm{periodic}\:\mathrm{too},\:\mathrm{but}\:\mathrm{the}\:\mathrm{period} \\ $$$$\mathrm{is}\:\mathrm{2}\pi{i} \\ $$

Commented by prakash jain last updated on 30/Sep/16

Thanks. I understand now. 2cosh(4t)=47 Will give 4t=2inπ±cosh^(−1) (47/2) So we get 8 solution for t for n=0,1,2,3

$$\mathrm{Thanks}.\:\mathrm{I}\:\mathrm{understand}\:\mathrm{now}. \\ $$$$\mathrm{2cosh}\left(\mathrm{4}{t}\right)=\mathrm{47} \\ $$$$\mathrm{Will}\:{give}\:\mathrm{4}{t}=\mathrm{2}{in}\pi\pm\mathrm{cosh}^{−\mathrm{1}} \left(\mathrm{47}/\mathrm{2}\right) \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{get}\:\mathrm{8}\:\mathrm{solution}\:\mathrm{for}\:{t}\:\mathrm{for}\:{n}=\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$ \\ $$

Answered by Rasheed Soomro last updated on 29/Sep/16

x^4 +(1/x^4 )=47 , x^3 +(1/x^3 )=? x^4 +(1/x^4 )=47 x^4 +2+(1/x^4 )=47+2 (x^2 )^2 +2(x^2 )((1/x^2 ))+((1/x^2 ))=49 (x^2 +(1/x^2 ))^2 =(±7)^2 x^2 +(1/x^2 )=7 ∣ x^2 +(1/x^2 )=−7 x^2 +2+(1/x^2 )=7+2 ∣ x^2 +2+(1/x^2 )=−7+2 (x)^2 +2(x)((1/x))+((1/x))^2 =9 ∣ (x)^2 +2(x)((1/x))+((1/x))^2 =−5 (x+(1/x))^2 =(±3)^2 ∣ (x+(1/x))^2 =(±(√5) i)^2 x+(1/x)= ±3 ∣ x+(1/x)= ±(√5) i (x+(1/x))^3 = (±3)^3 ∣ (x+(1/x))^3 = (±(√5) i)^3 x^3 +(1/x^3 )+3(x+(1/x))=(±3)^2 (±3) ∣ x^3 +(1/x^3 )+3(x+(1/x))=(±(√5) i)^2 (±(√5) i) x^3 +(1/x^3 )+3(±3)=(9)(±3) ∣ x^3 +(1/x^3 )+3(±(√5) i)=(−5)(±(√5) i) x^3 +(1/x^3 )=(9)(±3)−3(±3) ∣ x^3 +(1/x^3 )=(−5)(±(√5) i)−3(±(√5) i) x^3 +(1/x^3 )=(9−3)(±3) ∣ x^3 +(1/x^3 )=(−5−3)(±(√5) i) x^3 +(1/x^3 )=6(±3) ∣ x^3 +(1/x^3 )=−8(±(√5) i) x^3 +(1/x^3 )=±18 ∣ x^3 +(1/x^3 )=∓8(√5) i x^3 +(1/x^3 )=18 , −18 , −8(√5) i , 8(√5) i

$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{47}\:,\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{47} \\ $$$${x}^{\mathrm{4}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{47}+\mathrm{2} \\ $$$$\left({x}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{2}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)+\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)=\mathrm{49} \\ $$$$\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left(\pm\mathrm{7}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{7}\:\mid\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=−\mathrm{7} \\ $$$${x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=\mathrm{7}+\mathrm{2}\:\mid\:{x}^{\mathrm{2}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }=−\mathrm{7}+\mathrm{2} \\ $$$$\left({x}\right)^{\mathrm{2}} +\mathrm{2}\left({x}\right)\left(\frac{\mathrm{1}}{{x}}\right)+\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\mathrm{9}\:\mid\:\left({x}\right)^{\mathrm{2}} +\mathrm{2}\left({x}\right)\left(\frac{\mathrm{1}}{{x}}\right)+\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =−\mathrm{5} \\ $$$$\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\pm\mathrm{3}\right)^{\mathrm{2}} \:\mid\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} =\left(\pm\sqrt{\mathrm{5}}\:{i}\right)^{\mathrm{2}} \\ $$$$\:{x}+\frac{\mathrm{1}}{{x}}=\:\pm\mathrm{3}\:\:\mid\:\:{x}+\frac{\mathrm{1}}{{x}}=\:\pm\sqrt{\mathrm{5}}\:{i} \\ $$$$ \\ $$$$\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\:\left(\pm\mathrm{3}\right)^{\mathrm{3}} \:\:\mid\:\:\left({x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} =\:\left(\pm\sqrt{\mathrm{5}}\:{i}\right)^{\mathrm{3}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\left(\pm\mathrm{3}\right)^{\mathrm{2}} \left(\pm\mathrm{3}\right)\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left({x}+\frac{\mathrm{1}}{{x}}\right)=\left(\pm\sqrt{\mathrm{5}}\:{i}\right)^{\mathrm{2}} \left(\pm\sqrt{\mathrm{5}}\:{i}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left(\pm\mathrm{3}\right)=\left(\mathrm{9}\right)\left(\pm\mathrm{3}\right)\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }+\mathrm{3}\left(\pm\sqrt{\mathrm{5}}\:{i}\right)=\left(−\mathrm{5}\right)\left(\pm\sqrt{\mathrm{5}}\:{i}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\mathrm{9}\right)\left(\pm\mathrm{3}\right)−\mathrm{3}\left(\pm\mathrm{3}\right)\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(−\mathrm{5}\right)\left(\pm\sqrt{\mathrm{5}}\:{i}\right)−\mathrm{3}\left(\pm\sqrt{\mathrm{5}}\:{i}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(\mathrm{9}−\mathrm{3}\right)\left(\pm\mathrm{3}\right)\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\left(−\mathrm{5}−\mathrm{3}\right)\left(\pm\sqrt{\mathrm{5}}\:{i}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{6}\left(\pm\mathrm{3}\right)\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=−\mathrm{8}\left(\pm\sqrt{\mathrm{5}}\:{i}\right) \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\pm\mathrm{18}\:\mid\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mp\mathrm{8}\sqrt{\mathrm{5}}\:{i} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18}\:,\:−\mathrm{18}\:,\:−\mathrm{8}\sqrt{\mathrm{5}}\:{i}\:,\:\mathrm{8}\sqrt{\mathrm{5}}\:{i} \\ $$

Answered by prof.kerdus last updated on 29/Sep/16

its 18 •⌣•

$${its}\:\mathrm{18} \\ $$$$\bullet\smile\bullet \\ $$

Answered by Rasheed Soomro last updated on 30/Sep/16

Another approach_(−) x^4 +(1/x^4 )=47 , x^3 +(1/x^3 )=? (x^4 +(1/x^4 ))^3 =(47)^3 (a+(1/a))^3 =a^3 +(1/a^3 )+3(a+(1/a)) (x^4 )^3 +((1/x^4 ))^3 +3(x^4 +(1/x^4 ))=103823 x^(12) +(1/x^(12) )+3(47)=103823 x^(12) +(1/x^(12) )=103823−141=103682 (x^6 )^2 +((1/x^6 ))^2 =103682 (x^6 )^2 +2+((1/x^6 ))^2 =103682+2 (x^6 )^2 +2(x^6 )((1/x^6 ))+((1/x^6 ))^2 =103684 (x^6 +(1/x^6 ))^2 =(±322)^2 x^6 +(1/x^6 )=±322 x^6 +2+(1/x^6 )=±322+2 (x^3 )^2 +2(x^3 )((1/x^3 ))+((1/x^3 ))^2 =324 , −320 (x^3 +(1/x^3 ))^2 =(±18)^(2 ) ∣ (x^3 +(1/x^3 ))^2 =(±8(√5) i)^2 x^3 +(1/x^3 )=±18 ∣ x^3 +(1/x^3 )=±8(√5) i x^3 +(1/x^3 )=18, −18, 8(√5) i, −8(√5) i

$$\underset{−} {\boldsymbol{\mathrm{Another}}\:\boldsymbol{\mathrm{approach}}} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{47}\:,\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$$\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{3}} =\left(\mathrm{47}\right)^{\mathrm{3}} \\ $$$$\left({a}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +\frac{\mathrm{1}}{{a}^{\mathrm{3}} }+\mathrm{3}\left({a}+\frac{\mathrm{1}}{{a}}\right) \\ $$$$\left({x}^{\mathrm{4}} \right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)^{\mathrm{3}} +\mathrm{3}\left({x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }\right)=\mathrm{103823} \\ $$$${x}^{\mathrm{12}} +\frac{\mathrm{1}}{{x}^{\mathrm{12}} }+\mathrm{3}\left(\mathrm{47}\right)=\mathrm{103823} \\ $$$${x}^{\mathrm{12}} +\frac{\mathrm{1}}{{x}^{\mathrm{12}} }=\mathrm{103823}−\mathrm{141}=\mathrm{103682} \\ $$$$ \\ $$$$\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)^{\mathrm{2}} =\mathrm{103682} \\ $$$$\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} +\mathrm{2}+\left(\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)^{\mathrm{2}} =\mathrm{103682}+\mathrm{2} \\ $$$$\left({x}^{\mathrm{6}} \right)^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{6}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)+\left(\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)^{\mathrm{2}} =\mathrm{103684} \\ $$$$\left({x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\right)^{\mathrm{2}} =\left(\pm\mathrm{322}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }=\pm\mathrm{322} \\ $$$${x}^{\mathrm{6}} +\mathrm{2}+\frac{\mathrm{1}}{{x}^{\mathrm{6}} }=\pm\mathrm{322}+\mathrm{2}\: \\ $$$$\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}\left({x}^{\mathrm{3}} \right)\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)+\left(\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} =\mathrm{324}\:,\:−\mathrm{320} \\ $$$$\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} =\left(\pm\mathrm{18}\right)^{\mathrm{2}\:\:} \mid\:\:\left({x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }\right)^{\mathrm{2}} =\left(\pm\mathrm{8}\sqrt{\mathrm{5}}\:{i}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\pm\mathrm{18}\:\:\mid\:\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\pm\mathrm{8}\sqrt{\mathrm{5}}\:{i} \\ $$$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{18},\:−\mathrm{18},\:\mathrm{8}\sqrt{\mathrm{5}}\:{i},\:−\mathrm{8}\sqrt{\mathrm{5}}\:{i} \\ $$

Answered by Rasheed Soomro last updated on 30/Sep/16

Slightly defferent version of my answer dated 29/9/16 x^4 +(1/x^4 )=47, x^3 +(1/x^3 )=? Let x+(1/x)=y ( x+(1/x))^2 =y^2 x^2 +(1/x^2 )=y^2 −2 (x^2 +(1/x^2 ))^2 =(y^2 −2)^2 x^4 +(1/x^4 )=(y^2 −2)^2 −2=47 (y^2 −2)^2 =47+2=49 (y^2 −2)^2 =(±7)^2 y^2 −2=±7 y^2 =±7+2 y^2 =7+2 , −7+2 y^2 =(±3)^2 ∣ y^2 =(±(√(−5)) )^2 y=3,−3 ∣ y=(√5) i ,−(√5) i x+(1/x)=3 , −3 , (√5) i , −(√5) i For remaining process see my answer dated 29/09/16

$${Slightly}\:\:\:{defferent}\:{version} \\ $$$${of}\:{my}\:{answer}\:\:{dated}\:\mathrm{29}/\mathrm{9}/\mathrm{16} \\ $$$${x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\mathrm{47},\:{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=? \\ $$$${Let}\:\:{x}+\frac{\mathrm{1}}{{x}}={y} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\:{x}+\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} ={y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }={y}^{\mathrm{2}} −\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{\mathrm{2}} =\left({y}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\mathrm{4}} +\frac{\mathrm{1}}{{x}^{\mathrm{4}} }=\left({y}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} −\mathrm{2}=\mathrm{47} \\ $$$$\:\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} =\mathrm{47}+\mathrm{2}=\mathrm{49} \\ $$$$\:\:\:\:\:\:\:\:\left({y}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} =\left(\pm\mathrm{7}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} −\mathrm{2}=\pm\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\pm\mathrm{7}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\mathrm{7}+\mathrm{2}\:,\:−\mathrm{7}+\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:{y}^{\mathrm{2}} =\left(\pm\mathrm{3}\right)^{\mathrm{2}} \:\:\:\mid\:\:\:\:{y}^{\mathrm{2}} =\left(\pm\sqrt{−\mathrm{5}}\:\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:{y}=\mathrm{3},−\mathrm{3}\:\:\:\:\:\:\:\:\:\mid\:\:\:\:{y}=\sqrt{\mathrm{5}}\:{i}\:,−\sqrt{\mathrm{5}}\:{i} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:,\:−\mathrm{3}\:,\:\sqrt{\mathrm{5}}\:{i}\:,\:−\sqrt{\mathrm{5}}\:{i} \\ $$$${For}\:{remaining}\:{process}\:{see}\:{my}\:{answer}\:{dated}\:\mathrm{29}/\mathrm{09}/\mathrm{16} \\ $$

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