z-4-12z-3-0-z-

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

$$\boldsymbol{{z}}^{\mathrm{4}} \:+\:\mathrm{12}\boldsymbol{{z}}\:+\:\mathrm{3}\:=\:\mathrm{0}\:\left(\boldsymbol{{z}}=?\right) \\ $$

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))))

$$\left({z}^{\mathrm{2}} +{pz}+{q}\right)\left({z}^{\mathrm{2}} −{pz}+\frac{\mathrm{3}}{{q}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${q}+\frac{\mathrm{3}}{{q}}={p}^{\mathrm{2}} \\ $$$${p}\left({q}−\frac{\mathrm{3}}{{q}}\right)=−\mathrm{12} \\ $$$$\Rightarrow\:\left({q}+\frac{\mathrm{3}}{{q}}\right)\left({q}−\frac{\mathrm{3}}{{q}}\right)^{\mathrm{2}} =\mathrm{144} \\ $$$$\left({q}+\frac{\mathrm{3}}{{q}}\right)^{\mathrm{3}} −\mathrm{6}\left({q}+\frac{\mathrm{3}}{{q}}\right)−\mathrm{144}=\mathrm{0} \\ $$$$\Rightarrow\:\:{q}+\frac{\mathrm{3}}{{q}}=\left\{\mathrm{72}+\sqrt{\left(\mathrm{72}\right)^{\mathrm{2}} −\mathrm{8}}\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left\{\mathrm{72}−\sqrt{\left(\mathrm{72}\right)^{\mathrm{2}} −\mathrm{8}}\right\}^{\mathrm{1}/\mathrm{3}} \\ $$$$={k}\:\approx\mathrm{5}.\mathrm{62243} \\ $$$${q}^{\mathrm{2}} −{kq}+\mathrm{3}=\mathrm{0} \\ $$$${q}=\frac{{k}}{\mathrm{2}}\pm\sqrt{\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}} \\ $$$${z}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$${z}_{\mathrm{1}} ,\:{z}_{\mathrm{2}} =−\frac{\sqrt{{k}}}{\mathrm{2}}\pm\sqrt{−\frac{{k}}{\mathrm{4}}+\sqrt{\frac{{k}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{3}}} \\ $$$$ \\ $$

Answered by MJS_new last updated on 02/Jun/21

z^4 +12z+3=(z^2 −(√6)z+3+(√6))(z^2 +(√6)z+3−(√6)) now it′s easy to solve

$${z}^{\mathrm{4}} +\mathrm{12}{z}+\mathrm{3}=\left({z}^{\mathrm{2}} −\sqrt{\mathrm{6}}{z}+\mathrm{3}+\sqrt{\mathrm{6}}\right)\left({z}^{\mathrm{2}} +\sqrt{\mathrm{6}}{z}+\mathrm{3}−\sqrt{\mathrm{6}}\right) \\ $$$$\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$

Commented by MJS_new last updated on 02/Jun/21

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}! \\ $$

Commented by mr W last updated on 02/Jun/21

$${great}! \\ $$$${i}\:{wish}\:{i}\:{also}\:{could}… \\ $$

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

$$\mathrm{but}\:\mathrm{this}\:\mathrm{is}\:\mathrm{easy} \\ $$$${x}^{\mathrm{4}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\alpha{x}−\beta\right)\left({x}^{\mathrm{2}} +\alpha{x}−\gamma\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\begin{cases}{{a}=−\alpha^{\mathrm{2}} −\beta−\gamma}\\{{b}=−\alpha\left(\beta−\gamma\right)}\\{{c}=\beta\gamma}\end{cases} \\ $$$$\mathrm{solve}\:\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{for}\:\beta\:\mathrm{and}\:\gamma\:\mathrm{and}\:\mathrm{insert}\:\mathrm{into}\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow \\ $$$$\begin{cases}{\beta=\frac{−{a}\alpha−{b}−\alpha^{\mathrm{3}} }{\mathrm{2}\alpha}}\\{\gamma=\frac{−{a}\alpha+{b}−\alpha^{\mathrm{3}} }{\mathrm{2}\alpha}}\\{\alpha^{\mathrm{6}} +\mathrm{2}{a}\alpha^{\mathrm{4}} +\left({a}^{\mathrm{2}} −\mathrm{4}{c}\right)\alpha^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{0}}\end{cases} \\ $$$$\mathrm{let}\:\alpha=\sqrt{{t}−\frac{\mathrm{2}{a}}{\mathrm{3}}} \\ $$$${t}^{\mathrm{3}} −\frac{\left({a}^{\mathrm{2}} +\mathrm{12}{c}\right)}{\mathrm{3}}{t}−\frac{\mathrm{2}{a}^{\mathrm{3}} −\mathrm{72}{ac}+\mathrm{27}{b}^{\mathrm{2}} }{\mathrm{27}}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{this}\:\mathrm{for}\:{t} \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{has}\:\mathrm{got}\:\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no} \\ $$$$\mathrm{sense}\:\mathrm{to}\:\mathrm{exactly}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{has}\:\mathrm{2}\:\mathrm{or}\:\mathrm{3}\:“\mathrm{nice}''\:\mathrm{solutions}\:\mathrm{take}\:\mathrm{one}\:\mathrm{to} \\ $$$$\mathrm{get}\:\alpha\in\mathbb{R} \\ $$

Commented by mr W last updated on 02/Jun/21

$${thanks}\:{sir}! \\ $$

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