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Question Number 49359 by peter frank last updated on 06/Dec/18

Answered by MJS last updated on 06/Dec/18

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{wrong}$$$$y=\frac{1}{x}\Rightarrow \frac{{(x^2-x+1)}^2}{x{(x-1)}^2}=\frac{{(y^2-y+1)}^2}{y{(y-1)}^2};\frac{y^2}{y-1}=\frac{1}{x(1-x)}$$$$\mathrm{anyway}\mathrm{the}\mathrm{equation}\mathrm{is}$$$${(x^2-3x+1)}^2=0\Rightarrow x=\frac{3}{2}\pm \frac{\sqrt{5}}{2}$$

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