z^4 + 12z + 3 = 0 (z=?)


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Question Number 142582 by mathdanisur last updated on 02/Jun/21

$z^{4} + 12 z + 3 = 0 (z = ?)$
	Answered by ajfour last updated on 02/Jun/21
	$(z^2 +pz+q)(z^2 −pz+(3/q))=0 ⇒ q+(3/q)=p^2 p(q−(3/q))=−12 ⇒ (q+(3/q))(q−(3/q))^2 =144 (q+(3/q))^3 −6(q+(3/q))−144=0 ⇒ q+(3/q)={72+(√((72)^2 −8))}^(1/3) +{72−(√((72)^2 −8))}^(1/3) =k ≈5.62243 q^2 −kq+3=0 q=(k/2)±(√((k^2 /4)−3)) z=−(p/2)±(√((p^2 /4)−q)) z_1 , z_2 =−((√k)/2)±(√(−(k/4)+(√((k^2 /4)−3))))$
	$(z^{2} + p z + q) (z^{2} - p z + \frac{3}{q}) = 0$ $\Rightarrow$ $q + \frac{3}{q} = p^{2}$ $p (q - \frac{3}{q}) = - 12$ $\Rightarrow (q + \frac{3}{q}) {(q - \frac{3}{q})}^{2} = 144$ ${(q + \frac{3}{q})}^{3} - 6 (q + \frac{3}{q}) - 144 = 0$ $\Rightarrow q + \frac{3}{q} = {72 + \sqrt{{(72)}^{2} - 8}}^{1 / 3}$ $+ {72 - \sqrt{{(72)}^{2} - 8}}^{1 / 3}$ $= k \approx 5.62243$ $q^{2} - k q + 3 = 0$ $q = \frac{k}{2} \pm \sqrt{\frac{k^{2}}{4} - 3}$ $z = - \frac{p}{2} \pm \sqrt{\frac{p^{2}}{4} - q}$ $z_{1}, z_{2} = - \frac{\sqrt{k}}{2} \pm \sqrt{- \frac{k}{4} + \sqrt{\frac{k^{2}}{4} - 3}}$
	Answered by MJS_new last updated on 02/Jun/21

	$z^{4} + 12 z + 3 = (z^{2} - \sqrt{6} z + 3 + \sqrt{6}) (z^{2} + \sqrt{6} z + 3 - \sqrt{6})$ $now {it}^{'} s easy to solve$
	Commented by MJS_new last updated on 02/Jun/21

	${you}^{'} re welcome!$
	Commented by mr W last updated on 02/Jun/21

	$g r e a t!$ $i w i s h i a l s o c o u l d . . .$
	Commented by MJS_new last updated on 02/Jun/21
	$but this is easy x^4 +px^2 +qx+r=0 (x^2 −αx−β)(x^2 +αx−γ)=0 ⇒ { ((a=−α^2 −β−γ)),((b=−α(β−γ))),((c=βγ)) :} solve (1) and (2) for β and γ and insert into (3) ⇒ { ((β=((−aα−b−α^3 )/(2α)))),((γ=((−aα+b−α^3 )/(2α)))),((α^6 +2aα^4 +(a^2 −4c)α^2 −b^2 =0)) :} let α=(√(t−((2a)/3))) t^3 −(((a^2 +12c))/3)t−((2a^3 −72ac+27b^2 )/(27))=0 solve this for t if it has got no “nice” solution it makes no sense to exactly solve the given equation if it has 2 or 3 “nice” solutions take one to get α∈R$
	$but this is easy$ $x^{4} + {p x}^{2} + q x + r = 0$ $(x^{2} - α x - β) (x^{2} + α x - γ) = 0$ $\Rightarrow$ ${\begin{cases} a = - α^{2} - β - γ \\ b = - α (β - γ) \\ c = β γ \end{cases}$ $solve (1) and (2) for β and γ and insert into (3)$ $\Rightarrow$ ${\begin{cases} β = \frac{- a α - b - α^{3}}{2 α} \\ γ = \frac{- a α + b - α^{3}}{2 α} \\ α^{6} + 2 a α^{4} + (a^{2} - 4 c) α^{2} - b^{2} = 0 \end{cases}$ $let α = \sqrt{t - \frac{2 a}{3}}$ $t^{3} - \frac{(a^{2} + 12 c)}{3} t - \frac{2 a^{3} - 72 a c + 27 b^{2}}{27} = 0$ $solve this for t$ $if it has got no ‘ ‘ {nice}^{″} solution it makes no$ $sense to exactly solve the given equation$ $if it has 2 or 3 ‘ ‘ {nice}^{″} solutions take one to$ $get α \in R$
	Commented by mr W last updated on 02/Jun/21

	$t h a n k s s i r!$
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