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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
	$z^4 −3z^2 +1 = (√((4/(4−z^2 ))−1)) (1) Let w = z^2 (1) : w^2 −3w+1 = (√((4/(4−w))−1)) ⇒ (∗) w^4 −6w^3 +11w^2 −6w+1 = (w/(4−w)) w^5 −10w^4 +35w^3 −50w^2 +26w−4 = 0 (w^5 −2w^4 )−8(w^4 −2w^3 )+19(w^3 −2w^2 ) −12(w^2 −2w)+2(w−2) = 0 w^4 (w−2)−8w^3 (w−2)+19w^2 (w−2) −12w(w−2)+2(w−2) = 0 (w−2)(w^4 −8w^3 +19w^2 −12w+2) = 0 (w−2)(w^2 +pw+2)(w^2 +qw+1) = 0 By identification p = q = −4 (w−2)(w^2 −4w+2)(w^2 −4w+1) = 0 (w−2)(w−(2−(√2)))(w−(2+(√2))(w−(2−(√3)))(w−(2+(√3))) = 0 w = 2, w = 2±(√2), w = 2±(√3) ⇒z = ±(√2), z = ±(√(2±(√2))), z = ±(√(2±(√3))) (∗) Because of ⇒ we have to verify each solution. ±(√2) and ±(√(2−(√2))) are not good. S = {±(√(2+(√2))), ±(√(2±(√3)))}$
	$z^{4} - 3 z^{2} + 1 = \sqrt{\frac{4}{4 - z^{2}} - 1} (1)$ $Let w = z^{2}$ $(1) : w^{2} - 3 w + 1 = \sqrt{\frac{4}{4 - w} - 1}$ $\Rightarrow () w^{4} - 6 w^{3} + 11 w^{2} - 6 w + 1 = \frac{w}{4 - w}$ $w^{5} - 10 w^{4} + 35 w^{3} - 50 w^{2} + 26 w - 4 = 0$ $(w^{5} - 2 w^{4}) - 8 (w^{4} - 2 w^{3}) + 19 (w^{3} - 2 w^{2})$ $- 12 (w^{2} - 2 w) + 2 (w - 2) = 0$ $w^{4} (w - 2) - 8 w^{3} (w - 2) + 19 w^{2} (w - 2)$ $- 12 w (w - 2) + 2 (w - 2) = 0$ $(w - 2) (w^{4} - 8 w^{3} + 19 w^{2} - 12 w + 2) = 0$ $(w - 2) (w^{2} + p w + 2) (w^{2} + q w + 1) = 0$ $By identification p = q = - 4$ $(w - 2) (w^{2} - 4 w + 2) (w^{2} - 4 w + 1) = 0$ $(w - 2) (w - (2 - \sqrt{2})) (w - (2 + \sqrt{2}) (w - (2 - \sqrt{3})) (w - (2 + \sqrt{3})) = 0$ $w = 2, w = 2 \pm \sqrt{2}, w = 2 \pm \sqrt{3}$ $\Rightarrow z = \pm \sqrt{2}, z = \pm \sqrt{2 \pm \sqrt{2}}, z = \pm \sqrt{2 \pm \sqrt{3}}$ $() Because of \Rightarrow we have to verify$ $each solution . \pm \sqrt{2} and \pm \sqrt{2 - \sqrt{2}}$ $are not good .$ $S = {\pm \sqrt{2 + \sqrt{2}}, \pm \sqrt{2 \pm \sqrt{3}}}$
	Commented by mathdanisur last updated on 26/Jul/21

	$T h a n k y o u S e r c o o l$ $b u t \pm 2 o r \pm \sqrt{2} . ?$
	Commented by Olaf_Thorendsen last updated on 26/Jul/21

	$I corrected sir .$
	Commented by mathdanisur last updated on 26/Jul/21

	$T h a n k s S e r$
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