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Question Number 98596 by bemath last updated on 15/Jun/20

Commented by bobhans last updated on 15/Jun/20

$$\Rightarrow (x^3+x^2-2)+(x^2+3x-4)i=0$$$$\left(1\right)realparts\Rightarrow x^3+x^2-2=0$$$$(x-1)(x^2+2x+2)=0\Rightarrow \{\begin{array}{l}x=1\\ x=\frac{-2\pm 2i}{2}=-1\pm i\end{array}$$$$\left(2\right)imagineryparts\Rightarrow x^2+3x-4=0$$$$(x+4)(x-1)=0\Rightarrow \{\begin{array}{l}x=-4\\ x=1\end{array}$$$$thesolutionwegetx\in \left\{\mathrm{real}\mathrm{parts}\right\}\cap \left\{\mathrm{imaginery}\mathrm{parts}\right\}$$$$\Rightarrow x=1◼$$

Commented by MJS last updated on 15/Jun/20

$$\mathrm{your}\mathrm{method}\mathrm{works}\mathrm{only}\mathrm{if}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{real}\mathrm{root}$$$$\mathrm{try}\mathrm{this}\mathrm{one}:$$$$x^3-(4+4\mathrm{i})x^2+11\mathrm{i}x+(5-5\mathrm{i})=0$$

Answered by MJS last updated on 15/Jun/20

$$x^3+x^2-2=0\wedge x^2+3x-4=0$$$$\mathrm{easy}\mathrm{to}\mathrm{see}\mathrm{that}\mathrm{one}\mathrm{solution}\mathrm{is}x=1$$$$\Rightarrow$$$$(x-1)(x^2+(2+\mathrm{i})x+(2+4\mathrm{i}))=0$$$$x_1=1$$$$x_{2,3}=-\frac{2+\mathrm{i}}{2}\pm \sqrt{\frac{{(2+\mathrm{i})}^2}{4}-(2+4\mathrm{i})}=$$$$=-1-\frac{1}{2}\mathrm{i}\pm \sqrt{-\frac{5}{4}-3\mathrm{i}}=-1-\frac{1}{2}\mathrm{i}\pm (1-\frac{3}{2}\mathrm{i})$$$$\Rightarrow$$$$x_2=-2\mathrm{i}$$$$x_3=-2+\mathrm{i}$$

Commented by bobhans last updated on 15/Jun/20

$$\mathrm{oo}\mathrm{yes}\mathrm{sir}.\mathrm{if}\mathrm{x}\in \mathbb{C}.\mathrm{you}\mathrm{are}\mathrm{right}\mathrm{sir}$$

Commented by bemath last updated on 15/Jun/20

$$\mathrm{where}\mathrm{does}\mathrm{the}\mathrm{eq}\mathrm{come}\mathrm{from}$$$${\mathrm{x}}^2+(2+\mathrm{i})\mathrm{x}+2+4\mathrm{i}=0\mathrm{sir}?$$

Commented by MJS last updated on 15/Jun/20

$$\mathrm{you}\mathrm{can}\mathrm{use}\mathrm{long}\mathrm{division}$$

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