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

find real solution of equation x^5 +x^4 +1 = 0

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

Answered by bemath last updated on 04/Jun/20

(x^2 +x+1)(x^3 −x+1)=0 first term nothing have real solution the second term x^3 −x+1 = 0 by Cardano x = ((((−q)/2)+(√((q^2 /4)+(p^3 /(27))))))^(1/(3 )) + ((((−q)/2)−(√((q^2 /4)+(p^3 /(27))))))^(1/(3 )) x = ((((−1)/2)+(√((1/4)−(1/(27))))))^(1/(3 )) + ((((−1)/2)−(√((1/4)−(1/(27))))))^(1/(3 ))

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

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