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Question Number 140271 by EnterUsername last updated on 06/May/21

$$\mathrm{The}\mathrm{number}\mathrm{of}\mathrm{common}\mathrm{roots}\mathrm{of}\mathrm{the}\mathrm{equations}$$$${\mathrm{x}}^5-{\mathrm{x}}^3+{\mathrm{x}}^2-1=0\mathrm{and}{\mathrm{x}}^4-1=0\mathrm{is}\mathrm{\_}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Answered by liberty last updated on 06/May/21

$$\left(1\right)({\mathrm{x}}^2-1)({\mathrm{x}}^2+1)=0$$$$\Rightarrow (\mathrm{x}+1)(\mathrm{x}-1)(\mathrm{x}+i)(\mathrm{x}-i)=0$$$$\Rightarrow \mathrm{x}=\pm 1,\mathrm{x}=\pm i$$$$\left(2\right)(\mathrm{x}-1)({\mathrm{x}}^4+{\mathrm{x}}^3+\mathrm{x}+1)=0$$$$\Rightarrow (\mathrm{x}-1)(\mathrm{x}+1)({\mathrm{x}}^3+1)=0$$$$\Rightarrow (\mathrm{x}-1)(\mathrm{x}+1)(\mathrm{x}+1)({\mathrm{x}}^2-\mathrm{x}+1)=0$$$$\Rightarrow (\mathrm{x}-1){(\mathrm{x}+1)}^2({(\mathrm{x}-\frac{1}{2})}^2+\frac{3}{4})=0$$$$\Rightarrow \mathrm{x}=\pm 1;\mathrm{x}=\pm \frac{\sqrt{3}}{2}i+\frac{1}{2}$$$$\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{common}\mathrm{roots}\mathrm{is}2$$

Commented by EnterUsername last updated on 06/May/21

$$\mathrm{Thank}\mathrm{you}!$$

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