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Question Number 145113 by imjagoll last updated on 02/Jul/21

Answered by liberty last updated on 02/Jul/21

$$\Rightarrow 5(x^4+x^2+1-2x^3+2x^2-2x)=(9x^2+9)(x^2-2x+1)$$$$\Rightarrow 5x^4-10x^3+15x^2-10x+5=$$$$9x^4-18x^3+18x^2-18x+9$$$$\Rightarrow 4x^4-8x^3+3x^2-8x+4=0$$$$\Rightarrow (x-2)(4x^3+3x-2)=0$$$$\Rightarrow x=2\left(solution\right)$$$$\Rightarrow 4x^3+3x-2=0$$$$\Rightarrow (2x-1)(2x^2+x+2)=0$$$$\Rightarrow x=\frac{1}{2}\left(solution\right)$$$$\Rightarrow 2x^2+x+2=0$$$$\Rightarrow x=\frac{-1\pm \sqrt{1-16}}{4}=\frac{-1\pm i\sqrt{15}}{4}\left(notsolution\right)$$

Commented by MJS_new last updated on 02/Jul/21

$$\mathrm{something}\mathrm{went}\mathrm{wrong}.\mathrm{the}\mathrm{answer}\mathrm{is}$$$$x=\frac{1}{2}\vee x=2[\vee x=-\frac{1}{4}\pm \frac{\sqrt{15}}{4}\mathrm{i}]$$

Commented by liberty last updated on 02/Jul/21

$$yes.$$

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