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Question Number 58754 by azizullah last updated on 29/Apr/19

Answered by Rasheed.Sindhi last updated on 29/Apr/19

$$-x^3+x^2-2x^2+2=0$$$$-x^2(x-1)-2(x^2-1)=0$$$$-x^2(x-1)-2(x-1)(x+1)=0$$$$(x-1)(-x^2-2x-2)=0$$$$x-1=0\mid x^2+2x+2=0$$$$x=1\mid x=\frac{-2\pm \sqrt{4-4\left(1\right)\left(2\right)}}{2}$$$$\mid x=\frac{-2\pm 2\sqrt{1-2}}{2}=-1\pm i$$$$x=1,-1\pm i$$

Commented by azizullah last updated on 30/Apr/19

$$\mathit{T}\mathit{h}\mathit{n}\mathit{k}\mathit{s}\mathit{d}\mathit{e}\mathit{a}\mathit{r}\mathit{s}\mathit{i}\mathit{r}!$$

Answered by tanmay last updated on 29/Apr/19

$$x^2(1-x)+2(1-x^2)=0$$$$x^2(1-x)+2(1+x)(1-x)=0$$$$(1-x)(x^2+2+2x)=0$$$$(1-x)\{{(x+1)}^2+1\}=0$$$$(1-x)=0$$$$x=1$$$$x^2+2x+2=0$$$$x=\frac{-2\pm \sqrt{2^2-4\times 1\times 2}}{2}$$$$x=\frac{-2\pm \sqrt{-4}}{2}$$$$x=\frac{-2\pm 2i}{2}$$$$x=(-1+i)and(-1-i)$$

Commented by azizullah last updated on 30/Apr/19

$$\mathbf{Alot}\mathbf{of}\mathbf{thanks}\mathbf{sir}!$$

Commented by tanmay last updated on 30/Apr/19

$$mostwlcomesir$$

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