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Question Number 156921 by cortano last updated on 17/Oct/21
x^3 −4x^2 −3=0 x∈R
$${x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\:{x}\in\mathbb{R} \\ $$
Answered by mr W last updated on 17/Oct/21
path 1: ((1/x))^3 +(4/3)((1/x))−(1/3)=0 applying cardano: (1/x)=(((√(((1/6))^2 +((4/9))^3 ))+(1/6)))^(1/3) −(((√(((1/6))^2 +((4/9))^3 ))−(1/6)))^(1/3) ⇒x=(1/( ((((√(337))/(54))+(1/6)))^(1/3) −((((√(337))/(54))−(1/6)))^(1/3) ))≈4.1723 path 2: let x=t+(4/3) t^3 +3×(4/3)t^2 +3×((16)/9)t+(4^3 /3^3 )−4(t^2 +2×(4/3)t+(4^2 /3^2 ))−3=0 t^3 −((16)/3)t−((209)/(27))=0 applying cardano: t=(((√((((209)/(54)))^2 −(((16)/9))^3 ))+((209)/(54))))^(1/3) −(((√((((209)/(54)))^2 −(((16)/9))^3 ))−((209)/(54))))^(1/3) x=(4/3)+((((√(337))/6)+((209)/(54))))^(1/3) −((((√(337))/6)−((209)/(54))))^(1/3) ≈4.1723
$${path}\:\mathrm{1}: \\ $$$$\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{3}} +\frac{\mathrm{4}}{\mathrm{3}}\left(\frac{\mathrm{1}}{{x}}\right)−\frac{\mathrm{1}}{\mathrm{3}}=\mathrm{0} \\ $$$${applying}\:{cardano}: \\ $$$$\frac{\mathrm{1}}{{x}}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{1}}{\mathrm{6}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{9}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{337}}}{\mathrm{54}}+\frac{\mathrm{1}}{\mathrm{6}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{337}}}{\mathrm{54}}−\frac{\mathrm{1}}{\mathrm{6}}}}\approx\mathrm{4}.\mathrm{1723} \\ $$$$ \\ $$$${path}\:\mathrm{2}: \\ $$$${let}\:{x}={t}+\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${t}^{\mathrm{3}} +\mathrm{3}×\frac{\mathrm{4}}{\mathrm{3}}{t}^{\mathrm{2}} +\mathrm{3}×\frac{\mathrm{16}}{\mathrm{9}}{t}+\frac{\mathrm{4}^{\mathrm{3}} }{\mathrm{3}^{\mathrm{3}} }−\mathrm{4}\left({t}^{\mathrm{2}} +\mathrm{2}×\frac{\mathrm{4}}{\mathrm{3}}{t}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{3}^{\mathrm{2}} }\right)−\mathrm{3}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{16}}{\mathrm{3}}{t}−\frac{\mathrm{209}}{\mathrm{27}}=\mathrm{0} \\ $$$${applying}\:{cardano}: \\ $$$${t}=\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{209}}{\mathrm{54}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{16}}{\mathrm{9}}\right)^{\mathrm{3}} }+\frac{\mathrm{209}}{\mathrm{54}}}−\sqrt[{\mathrm{3}}]{\sqrt{\left(\frac{\mathrm{209}}{\mathrm{54}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{16}}{\mathrm{9}}\right)^{\mathrm{3}} }−\frac{\mathrm{209}}{\mathrm{54}}} \\ $$$${x}=\frac{\mathrm{4}}{\mathrm{3}}+\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{337}}}{\mathrm{6}}+\frac{\mathrm{209}}{\mathrm{54}}}−\sqrt[{\mathrm{3}}]{\frac{\sqrt{\mathrm{337}}}{\mathrm{6}}−\frac{\mathrm{209}}{\mathrm{54}}}\approx\mathrm{4}.\mathrm{1723} \\ $$
Commented by cortano last updated on 17/Oct/21
cardano by cortano
$${cardano}\:{by}\:{cortano} \\ $$

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