Question-196973

Question Number 196973 by sonukgindia last updated on 05/Sep/23

Answered by Frix last updated on 05/Sep/23

We don′t need the approximate solution. x^5 −5x−3=0 (x^2 −x−1)(x^3 +x^2 +2x+3)=0 x^2 −x−1=0 x=((1±(√5))/2) x^3 +x^2 +2x+3=0 x=−(1/3)+((−((65)/(54))+((5(√(21)))/(18))))^(1/3) −((((65)/(54))+((5(√(21)))/8)))^(1/3) x=−(1/3)+ω((−((65)/(54))+((5(√(21)))/(18))))^(1/3) −ω^2 ((((65)/(54))+((5(√(21)))/8)))^(1/3) x=−(1/3)+ω^2 ((−((65)/(54))+((5(√(21)))/(18))))^(1/3) −ω((((65)/(54))+((5(√(21)))/8)))^(1/3) ω=−(1/2)+((√3)/2)i

$$\mathrm{We}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{the}\:\mathrm{approximate}\:\mathrm{solution}. \\ $$$${x}^{\mathrm{5}} −\mathrm{5}{x}−\mathrm{3}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{18}}}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{8}}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}+\omega\sqrt[{\mathrm{3}}]{−\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{18}}}−\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{8}}} \\ $$$${x}=−\frac{\mathrm{1}}{\mathrm{3}}+\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{−\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{18}}}−\omega\sqrt[{\mathrm{3}}]{\frac{\mathrm{65}}{\mathrm{54}}+\frac{\mathrm{5}\sqrt{\mathrm{21}}}{\mathrm{8}}} \\ $$$$\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply