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)))}

Commented bymathdanisur last updated on 26/Jul/21

Thank you Ser cool but ±2 or ±(√2) .?

Commented byOlaf_Thorendsen last updated on 26/Jul/21

I corrected sir.

Commented bymathdanisur last updated on 26/Jul/21

Thanks Ser