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Question Number 148268 by mathdanisur last updated on 26/Jul/21

Answered by Olaf_Thorendsen last updated on 26/Jul/21

z^4 −3z^2 +1 = (√((4/(4−z^2 ))−1)) (1) Let w = z^2 (1) : w^2 −3w+1 = (√((4/(4−w))−1)) ⇒ (∗) w^4 −6w^3 +11w^2 −6w+1 = (w/(4−w)) w^5 −10w^4 +35w^3 −50w^2 +26w−4 = 0 (w^5 −2w^4 )−8(w^4 −2w^3 )+19(w^3 −2w^2 ) −12(w^2 −2w)+2(w−2) = 0 w^4 (w−2)−8w^3 (w−2)+19w^2 (w−2) −12w(w−2)+2(w−2) = 0 (w−2)(w^4 −8w^3 +19w^2 −12w+2) = 0 (w−2)(w^2 +pw+2)(w^2 +qw+1) = 0 By identification p = q = −4 (w−2)(w^2 −4w+2)(w^2 −4w+1) = 0 (w−2)(w−(2−(√2)))(w−(2+(√2))(w−(2−(√3)))(w−(2+(√3))) = 0 w = 2, w = 2±(√2), w = 2±(√3) ⇒z = ±(√2), z = ±(√(2±(√2))), z = ±(√(2±(√3))) (∗) Because of ⇒ we have to verify each solution. ±(√2) and ±(√(2−(√2))) are not good. S = {±(√(2+(√2))), ±(√(2±(√3)))}

$${z}^{\mathrm{4}} −\mathrm{3}{z}^{\mathrm{2}} +\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{z}^{\mathrm{2}} }−\mathrm{1}}\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Let}\:{w}\:=\:{z}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\::\:{w}^{\mathrm{2}} −\mathrm{3}{w}+\mathrm{1}\:=\:\sqrt{\frac{\mathrm{4}}{\mathrm{4}−{w}}−\mathrm{1}} \\ $$$$\Rightarrow\:\left(\ast\right)\:{w}^{\mathrm{4}} −\mathrm{6}{w}^{\mathrm{3}} +\mathrm{11}{w}^{\mathrm{2}} −\mathrm{6}{w}+\mathrm{1}\:=\:\frac{{w}}{\mathrm{4}−{w}} \\ $$$${w}^{\mathrm{5}} −\mathrm{10}{w}^{\mathrm{4}} +\mathrm{35}{w}^{\mathrm{3}} −\mathrm{50}{w}^{\mathrm{2}} +\mathrm{26}{w}−\mathrm{4}\:=\:\mathrm{0} \\ $$$$ \\ $$$$\left({w}^{\mathrm{5}} −\mathrm{2}{w}^{\mathrm{4}} \right)−\mathrm{8}\left({w}^{\mathrm{4}} −\mathrm{2}{w}^{\mathrm{3}} \right)+\mathrm{19}\left({w}^{\mathrm{3}} −\mathrm{2}{w}^{\mathrm{2}} \right) \\ $$$$−\mathrm{12}\left({w}^{\mathrm{2}} −\mathrm{2}{w}\right)+\mathrm{2}\left({w}−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$${w}^{\mathrm{4}} \left({w}−\mathrm{2}\right)−\mathrm{8}{w}^{\mathrm{3}} \left({w}−\mathrm{2}\right)+\mathrm{19}{w}^{\mathrm{2}} \left({w}−\mathrm{2}\right) \\ $$$$−\mathrm{12}{w}\left({w}−\mathrm{2}\right)+\mathrm{2}\left({w}−\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$$$\left({w}−\mathrm{2}\right)\left({w}^{\mathrm{4}} −\mathrm{8}{w}^{\mathrm{3}} +\mathrm{19}{w}^{\mathrm{2}} −\mathrm{12}{w}+\mathrm{2}\right)\:=\:\mathrm{0} \\ $$$$\left({w}−\mathrm{2}\right)\left({w}^{\mathrm{2}} +{pw}+\mathrm{2}\right)\left({w}^{\mathrm{2}} +{qw}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{By}\:\mathrm{identification}\:{p}\:=\:{q}\:=\:−\mathrm{4} \\ $$$$\left({w}−\mathrm{2}\right)\left({w}^{\mathrm{2}} −\mathrm{4}{w}+\mathrm{2}\right)\left({w}^{\mathrm{2}} −\mathrm{4}{w}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\left({w}−\mathrm{2}\right)\left({w}−\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\right)\left({w}−\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)\left({w}−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right)\left({w}−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right)\:=\:\mathrm{0}\right. \\ $$$$ \\ $$$${w}\:=\:\mathrm{2},\:{w}\:=\:\mathrm{2}\pm\sqrt{\mathrm{2}},\:{w}\:=\:\mathrm{2}\pm\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\Rightarrow{z}\:=\:\pm\sqrt{\mathrm{2}},\:{z}\:=\:\pm\sqrt{\mathrm{2}\pm\sqrt{\mathrm{2}}},\:{z}\:=\:\pm\sqrt{\mathrm{2}\pm\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$\left(\ast\right)\:\mathrm{Because}\:\mathrm{of}\:\Rightarrow\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{verify} \\ $$$$\mathrm{each}\:\mathrm{solution}.\:\pm\sqrt{\mathrm{2}}\:\mathrm{and}\:\pm\sqrt{\mathrm{2}−\sqrt{\mathrm{2}}}\: \\ $$$$\mathrm{are}\:\mathrm{not}\:\mathrm{good}. \\ $$$$ \\ $$$$\mathcal{S}\:=\:\left\{\pm\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}},\:\pm\sqrt{\mathrm{2}\pm\sqrt{\mathrm{3}}}\right\} \\ $$$$ \\ $$

Commented by mathdanisur last updated on 26/Jul/21

Thank you Ser cool but ±2 or ±(√2) .?

$${Thank}\:{you}\:{Ser}\:{cool} \\ $$$${but}\:\pm\mathrm{2}\:{or}\:\pm\sqrt{\mathrm{2}}\:.? \\ $$

Commented by Olaf_Thorendsen last updated on 26/Jul/21

I corrected sir.

$$\mathrm{I}\:\mathrm{corrected}\:\mathrm{sir}. \\ $$

Commented by mathdanisur last updated on 26/Jul/21

Thanks Ser

$${Thanks}\:{Ser} \\ $$

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