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x-6-x-3-2-solve-for-x-m-x-3-m-2-x-3-2-m-2-m-2-m-2-m-2-0-1-1-2-and-1-2-1-2-1-2-2-1-2-1-2-2-1-4-2-2-2-2-1-4-9-4-9-4-3-

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Question Number 191220 by anr0h3 last updated on 21/Apr/23
x^6 −x^3 =2 solve for x m=x^3 m^2 =(x^3 )^2 m^2 −m=2 m^2 −m−2=0 α+β=1→α=(1/2)+μ and β=(1/2)−μ α∙β=((1/2)+μ)∙((1/2)−μ)=−2 α∙β=((1/2)−μ)∙((1/2)+μ)=−2 (1/4)−μ^2 =−2 μ^2 =2+(1/4)=(9/4) μ=(√(9/4))=(3/2) α=(1/2)+(3/2)=((−1+3)/2) β=(1/2)−(3/2)=((−1−3)/2) (m+((1+3)/2))(m+((1−3)/2))=0 m_1 =((1+3)/2) m_2 =−((1−3)/2) m_(1,2) =((1±3)/2) m=x^3 →x=(((1±3)/2))^(1/3)
$${x}^{\mathrm{6}} −{x}^{\mathrm{3}} =\mathrm{2}\:{solve}\:{for}\:{x} \\ $$$$ \\ $$$${m}={x}^{\mathrm{3}} \\ $$$${m}^{\mathrm{2}} =\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −{m}=\mathrm{2} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{2}=\mathrm{0} \\ $$$$\alpha+\beta=\mathrm{1}\rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\mu\:\:{and}\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}−\mu \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)=−\mathrm{2} \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)=−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\mu^{\mathrm{2}} =−\mathrm{2} \\ $$$$\mu^{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mu=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{−\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{−\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$$\left({m}+\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}\right)\left({m}+\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{2}} =−\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${m}={x}^{\mathrm{3}} \rightarrow{x}=\left(\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$
Commented by Tinku Tara last updated on 21/Apr/23
(−(1/2)+μ)(−(1/2)−μ)=−2 ⇒(1/4)−μ^2 =−2 You made a mistake in sign at this step
$$\left(−\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)=−\mathrm{2} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}−\mu^{\mathrm{2}} =−\mathrm{2} \\ $$$$\mathrm{You}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{in}\:\mathrm{sign}\:\mathrm{at} \\ $$$$\mathrm{this}\:\mathrm{step} \\ $$$$ \\ $$
Commented by anr0h3 last updated on 21/Apr/23
x^6 −x^3 =2 solve for x m=x^3 m^2 =(x^3 )^2 m^2 −m=2 m^2 −m−2=0 α+β=1→α=(1/2)+μ and β=(1/2)−μ α∙β=((1/2)+μ)∙((1/2)−μ)=−2 α∙β=((1/2)−μ)∙((1/2)+μ)=−2 (1/4)−μ^2 =−2 μ^2 =2+(1/4)=(9/4) μ=(√(9/4))=(3/2) α=(1/2)+(3/2)=((1+3)/2) β=(1/2)−(3/2)=((1−3)/2) (m+((1+3)/2))(m+((1−3)/2))=0 m_1 =((1+3)/2) m_2 =((1−3)/2) m_(1,2) =((1±3)/2) m=x^3 →x=(((1±3)/2))^(1/3)
$${x}^{\mathrm{6}} −{x}^{\mathrm{3}} =\mathrm{2}\:{solve}\:{for}\:{x} \\ $$$$ \\ $$$${m}={x}^{\mathrm{3}} \\ $$$${m}^{\mathrm{2}} =\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −{m}=\mathrm{2} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{2}=\mathrm{0} \\ $$$$\alpha+\beta=\mathrm{1}\rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\mu\:\:{and}\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}−\mu \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)=−\mathrm{2} \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)=−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\mu^{\mathrm{2}} =−\mathrm{2} \\ $$$$\mu^{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mu=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$$\left({m}+\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}\right)\left({m}+\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${m}={x}^{\mathrm{3}} \rightarrow{x}=\left(\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$
Commented by Tinku Tara last updated on 21/Apr/23
α+β=1 (not −1)
$$\alpha+\beta=\mathrm{1}\:\left({not}\:−\mathrm{1}\right) \\ $$
Commented by anr0h3 last updated on 21/Apr/23
x^6 −x^3 =2 solve for x m=x^3 m^2 =(x^3 )^2 m^2 −m=2 m^2 −m−2=0 α+β=1→α=(1/2)+μ and β=(1/2)−μ α∙β=((1/2)+μ)∙((1/2)−μ)=−2 α∙β=((1/2)−μ)∙((1/2)+μ)=−2 (1/4)−μ^2 =−2 μ^2 =2+(1/4)=(9/4) μ=(√(9/4))=(3/2) α=(1/2)+(3/2)=((1+3)/2) β=(1/2)−(3/2)=((1−3)/2) (m+((1+3)/2))(m+((1−3)/2))=0 m_1 =((1+3)/2) m_2 =((1−3)/2) m_(1,2) =((1±3)/2) m=x^3 →x=(((1±3)/2))^(1/3)
$${x}^{\mathrm{6}} −{x}^{\mathrm{3}} =\mathrm{2}\:{solve}\:{for}\:{x} \\ $$$$ \\ $$$${m}={x}^{\mathrm{3}} \\ $$$${m}^{\mathrm{2}} =\left({x}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$$${m}^{\mathrm{2}} −{m}=\mathrm{2} \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{2}=\mathrm{0} \\ $$$$\alpha+\beta=\mathrm{1}\rightarrow\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\mu\:\:{and}\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}−\mu \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)=−\mathrm{2} \\ $$$$\alpha\centerdot\beta=\left(\frac{\mathrm{1}}{\mathrm{2}}−\mu\right)\centerdot\left(\frac{\mathrm{1}}{\mathrm{2}}+\mu\right)=−\mathrm{2} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}−\mu^{\mathrm{2}} =−\mathrm{2} \\ $$$$\mu^{\mathrm{2}} =\mathrm{2}+\frac{\mathrm{1}}{\mathrm{4}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mu=\sqrt{\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{2}}=\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$$\left({m}+\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}}\right)\left({m}+\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}}\right)=\mathrm{0} \\ $$$${m}_{\mathrm{1}} =\frac{\mathrm{1}+\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{2}} =\frac{\mathrm{1}−\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1},\mathrm{2}} =\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${m}={x}^{\mathrm{3}} \rightarrow{x}=\left(\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$ \\ $$
Commented by anr0h3 last updated on 21/Apr/23
thnks, I was clearing my doubts whit the PO shen technique, and now I'm learning about the complex solution in this problem, isn't sarcasm, again thnks
Answered by Frix last updated on 21/Apr/23
What have you done??? m^2 −m−2=0 Use formula: X^2 +pX+q=0 ⇒ X=−(p/2)±(√((p^2 /4)−q)) Here p=−1∧q=−2 m=(1/2)±(√((1/4)+2))=(1/2)±(3/2) m_1 =−1 m_2 =2 x_(1, 3, 5) ^3 =−1 x_(2, 4, 6) ^3 =2 x_1 =−1 x_3 =−ω x_5 =−ω^2 x_2 =(2)^(1/3) x_4 =(2)^(1/3) ω x_5 =(2)^(1/3) ω^2 [ω=−(1/2)+((√3)/2)i]
$$\mathrm{What}\:\mathrm{have}\:\mathrm{you}\:\mathrm{done}??? \\ $$$${m}^{\mathrm{2}} −{m}−\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{Use}\:\mathrm{formula}:\:{X}^{\mathrm{2}} +{pX}+{q}=\mathrm{0}\:\Rightarrow\:{X}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$$\mathrm{Here}\:{p}=−\mathrm{1}\wedge{q}=−\mathrm{2} \\ $$$${m}=\frac{\mathrm{1}}{\mathrm{2}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\pm\frac{\mathrm{3}}{\mathrm{2}} \\ $$$${m}_{\mathrm{1}} =−\mathrm{1} \\ $$$${m}_{\mathrm{2}} =\mathrm{2} \\ $$$${x}_{\mathrm{1},\:\mathrm{3},\:\mathrm{5}} ^{\mathrm{3}} =−\mathrm{1} \\ $$$${x}_{\mathrm{2},\:\mathrm{4},\:\mathrm{6}} ^{\mathrm{3}} =\mathrm{2} \\ $$$${x}_{\mathrm{1}} =−\mathrm{1}\:\:{x}_{\mathrm{3}} =−\omega\:\:{x}_{\mathrm{5}} =−\omega^{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\:{x}_{\mathrm{4}} =\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega\:\:{x}_{\mathrm{5}} =\sqrt[{\mathrm{3}}]{\mathrm{2}}\omega^{\mathrm{2}} \\ $$$$\left[\omega=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\right] \\ $$
Commented by anr0h3 last updated on 21/Apr/23
sorry, I have a mistake sign but the corrections have done, thnks

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