Wronsonia8g

2022-07-03

How to solve ${x}^{4}+6{x}^{3}-14{x}^{2}+160=0$

Charlize Manning

Expert

The equation has two real roots and two imaginary ones. The real ones are:
${x}_{1}=-\sqrt{R}-\sqrt{T+\sqrt{S}}-c$
${x}_{2}=-\sqrt{R}+\sqrt{T+\sqrt{S}}-c$
the parameter values are:
$R={A}^{1/3}+{a}^{1/3}+b$
$S=h+{g}^{1/3}+{G}^{1/3}-{f}^{1/3}-{F}^{1/3}$
$T=j-{k}^{1/3}-{K}^{1/3}$
further:
$F=\frac{3237158375}{5832}-\frac{166375\sqrt{19765}}{54}$
$f=\frac{3237158375}{5832}+\frac{166375\sqrt{19765}}{54}$
$G=\frac{609113809}{729}-\frac{155656\sqrt{19765}}{27}$
$g=\frac{609113809}{729}+\frac{155656\sqrt{19765}}{27}$
$h=\frac{101}{4}$
$K=\frac{19457}{216}-\frac{\sqrt{19765}}{2}$
$b=\frac{55}{12}$
$c=\frac{3}{2}$

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