Menu Close

x-3-x-2-x-c-0-find-x-




Question Number 175095 by Gbenga last updated on 18/Aug/22
x^3 +x^2 +x+c=0  find x
$$\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{c}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$
Answered by MJS_new last updated on 18/Aug/22
x=z−(1/3)  z^3 +(2/3)z+(c−(7/(27)))=0  ⇒  z_1 =u+v  z_2 =ωu+ω^2 v  z_3 =ω^2 u+ωv  ω=−(1/2)+((√3)/2)i  u=((−(q/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/3) ∧v=−(((q/2)(√((q^2 /4)+(p^3 /(27))))))^(1/3)   p=(2/3)∧q=c−(7/(27))  D=(p^3 /(27))+(q^2 /4)=(c^2 /4)−((7c)/(54))+(1/(36))>0∀c∈R  do the rest yourself
$${x}={z}−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{3}}{z}+\left({c}−\frac{\mathrm{7}}{\mathrm{27}}\right)=\mathrm{0} \\ $$$$\Rightarrow \\ $$$${z}_{\mathrm{1}} ={u}+{v} \\ $$$${z}_{\mathrm{2}} =\omega{u}+\omega^{\mathrm{2}} {v} \\ $$$${z}_{\mathrm{3}} =\omega^{\mathrm{2}} {u}+\omega{v} \\ $$$$\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$${u}=\sqrt[{\mathrm{3}}]{−\frac{{q}}{\mathrm{2}}+\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}}\wedge{v}=−\sqrt[{\mathrm{3}}]{\frac{{q}}{\mathrm{2}}\sqrt{\frac{{q}^{\mathrm{2}} }{\mathrm{4}}+\frac{{p}^{\mathrm{3}} }{\mathrm{27}}}} \\ $$$${p}=\frac{\mathrm{2}}{\mathrm{3}}\wedge{q}={c}−\frac{\mathrm{7}}{\mathrm{27}} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}}=\frac{{c}^{\mathrm{2}} }{\mathrm{4}}−\frac{\mathrm{7}{c}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{36}}>\mathrm{0}\forall{c}\in\mathbb{R} \\ $$$$\mathrm{do}\:\mathrm{the}\:\mathrm{rest}\:\mathrm{yourself} \\ $$
Commented by Gbenga last updated on 19/Aug/22
$$ \\ $$
Commented by Tawa11 last updated on 19/Aug/22
Great sir
$$\mathrm{Great}\:\mathrm{sir} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *