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Question Number 96715 by bobhans last updated on 04/Jun/20

$$\mathrm{find}\mathrm{real}\mathrm{solution}\mathrm{of}\mathrm{equation}$$$$x^5+x^4+1=0$$

Answered by bemath last updated on 04/Jun/20

$$(x^2+x+1)(x^3-x+1)=0$$$$\mathrm{first}\mathrm{term}\mathrm{nothing}\mathrm{have}\mathrm{real}\mathrm{solution}$$$$\mathrm{the}\mathrm{second}\mathrm{term}$$$$x^3-x+1=0$$$$byC\mathrm{ardano}$$$$x=\sqrt[3]{\frac{-q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}+\sqrt[3]{\frac{-q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}$$$$x=\sqrt[3]{\frac{-1}{2}+\sqrt{\frac{1}{4}-\frac{1}{27}}}+\sqrt[3]{\frac{-1}{2}-\sqrt{\frac{1}{4}-\frac{1}{27}}}$$

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