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Question Number 165751 by cortano1 last updated on 07/Feb/22

 x^3 +6x−9=0 ⇒x=?

$$\:\mathrm{x}^{\mathrm{3}} +\mathrm{6x}−\mathrm{9}=\mathrm{0}\:\Rightarrow\mathrm{x}=? \\ $$

Answered by MJS_new last updated on 08/Feb/22

x_1 =(((9+(√(113)))/2))^(1/3) −(((−9+(√(113)))/2))^(1/3) ≈1.20695981419  x_2 =ω(((9+(√(113)))/2))^(1/3) −ω^2 (((−9+(√(113)))/2))^(1/3) ≈−.603479907094+2.66318681185i  x_3 =ω^2 (((9+(√(113)))/2))^(1/3) −ω(((−9+(√(113)))/2))^(1/3) ≈−.603479907094−2.66318681185i

$${x}_{\mathrm{1}} =\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\frac{−\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}\approx\mathrm{1}.\mathrm{20695981419} \\ $$$${x}_{\mathrm{2}} =\omega\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}−\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\frac{−\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}\approx−.\mathrm{603479907094}+\mathrm{2}.\mathrm{66318681185i} \\ $$$${x}_{\mathrm{3}} =\omega^{\mathrm{2}} \sqrt[{\mathrm{3}}]{\frac{\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}−\omega\sqrt[{\mathrm{3}}]{\frac{−\mathrm{9}+\sqrt{\mathrm{113}}}{\mathrm{2}}}\approx−.\mathrm{603479907094}−\mathrm{2}.\mathrm{66318681185i} \\ $$

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