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Question Number 191304 by Mingma last updated on 22/Apr/23

Answered by cortano12 last updated on 22/Apr/23

x=t^2 −2 (t^2 −2)^3 −3(t^2 −2)−t=0 (t−2)(2t+1−(√5))(2t+1+(√5))((1/4)t^3 +bt+c)=0 { ((t=2⇒x=2)),((t=(((√5)−1)/2)⇒x=((−1−(√5))/2))),((t=((−1−(√5))/2)⇒x=((−1+(√5))/2))) :}

$$\:\mathrm{x}=\mathrm{t}^{\mathrm{2}} −\mathrm{2}\: \\ $$$$\:\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{t}^{\mathrm{2}} −\mathrm{2}\right)−\mathrm{t}=\mathrm{0} \\ $$$$\:\left(\mathrm{t}−\mathrm{2}\right)\left(\mathrm{2t}+\mathrm{1}−\sqrt{\mathrm{5}}\right)\left(\mathrm{2t}+\mathrm{1}+\sqrt{\mathrm{5}}\right)\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{t}^{\mathrm{3}} +\mathrm{bt}+\mathrm{c}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{t}=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{2}}\\{\mathrm{t}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{x}=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}}\\{\mathrm{t}=\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\Rightarrow\mathrm{x}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}}\end{cases} \\ $$

Commented by Frix last updated on 22/Apr/23

Your 3^(rd) solution is false and 1 solution is missing...

$$\mathrm{Your}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{solution}\:\mathrm{is}\:\mathrm{false}\:\mathrm{and}\:\mathrm{1}\:\mathrm{solution}\:\mathrm{is} \\ $$$$\mathrm{missing}... \\ $$

Answered by Frix last updated on 22/Apr/23

(√(x+2))≥0 ⇒ x^3 −3x≥0 ⇒ D: −(√3)≤x≤0∨x≥(√3) (x^3 −3x)^2 −(x+2)=0 x^6 −6x^4 +9x^2 −x−2=0 (x−2)(x^2 +x−1)(x^3 +x^2 −2x−1)=0 D∧x−2=0 ⇒ x=2 D∧x^2 +x−1 ⇒ x=−((1+(√5))/2) D∧x^3 +x^2 −2x−1=0 ⇒ x=−((1+2(√7)sin ((sin^(−1) ((√7)/(14)))/3))/3)≈−.445041868

$$\sqrt{{x}+\mathrm{2}}\geqslant\mathrm{0}\:\Rightarrow\:{x}^{\mathrm{3}} −\mathrm{3}{x}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${D}:\:−\sqrt{\mathrm{3}}\leqslant{x}\leqslant\mathrm{0}\vee{x}\geqslant\sqrt{\mathrm{3}} \\ $$$$ \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)^{\mathrm{2}} −\left({x}+\mathrm{2}\right)=\mathrm{0} \\ $$$${x}^{\mathrm{6}} −\mathrm{6}{x}^{\mathrm{4}} +\mathrm{9}{x}^{\mathrm{2}} −{x}−\mathrm{2}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$${D}\wedge{x}−\mathrm{2}=\mathrm{0}\:\Rightarrow\:{x}=\mathrm{2} \\ $$$${D}\wedge{x}^{\mathrm{2}} +{x}−\mathrm{1}\:\Rightarrow\:{x}=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${D}\wedge{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:{x}=−\frac{\mathrm{1}+\mathrm{2}\sqrt{\mathrm{7}}\mathrm{sin}\:\frac{\mathrm{sin}^{−\mathrm{1}} \:\frac{\sqrt{\mathrm{7}}}{\mathrm{14}}}{\mathrm{3}}}{\mathrm{3}}\approx−.\mathrm{445041868} \\ $$

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