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Question Number 141076 by ajfour last updated on 15/May/21
t+x=(c/2) t^2 +x^4 =(c^2 /4) find x or t . Given 0<c<(2/(3(√3)))
$${t}+{x}=\frac{{c}}{\mathrm{2}} \\ $$$$\:{t}^{\mathrm{2}} +{x}^{\mathrm{4}} =\frac{{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${find}\:{x}\:{or}\:{t}\:.\:{Given}\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\: \\ $$
Answered by Rasheed.Sindhi last updated on 15/May/21
t−(c/2)=−x t^2 −(c^2 /4)=−x^4 ⇒(t−(c/2))(t+(c/2))=−x^4 ⇒(−x)(t+(c/2))+x^4 =0 ⇒x(−t−(c/2)+x^3 )=0 ⇒x=0 ∣ −t−(c/2)+x^3 =0 x=0⇒t+0=(c/2)⇒t=(c/2) (x,t)=(0,(c/2)) −t−(c/2)+x^3 =0...........(i) t−(c/2)+x=0.............(ii) (i)+(ii) x^3 +x−c=0(★) (i)⇒ x^3 =t+(c/2)........(iii) (ii)⇒x^3 =(−t+(c/2))^3 ....(iv) t+(c/2)=(−t+(c/2))^3 t+(c/2)=−t^3 +(c^3 /8)+3(−t)((c/2))(−t+(c/2)) t+(c/2)=−t^3 +(c^3 /8)+(3/2)ct^2 −(3/2)ct t^3 −(3/2)ct^2 +(3/2)ct−(c^3 /8)+(c/2)=0(★) (★) Don′t know how to solve cubic equation. (x,t)=(0,(c/2)) ......
$${t}−\frac{{c}}{\mathrm{2}}=−{x} \\ $$$${t}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{4}}=−{x}^{\mathrm{4}} \Rightarrow\left({t}−\frac{{c}}{\mathrm{2}}\right)\left({t}+\frac{{c}}{\mathrm{2}}\right)=−{x}^{\mathrm{4}} \\ $$$$\Rightarrow\left(−{x}\right)\left({t}+\frac{{c}}{\mathrm{2}}\right)+{x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\Rightarrow{x}\left(−{t}−\frac{{c}}{\mathrm{2}}+{x}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{0}\:\mid\:\:−{t}−\frac{{c}}{\mathrm{2}}+{x}^{\mathrm{3}} =\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{0}\Rightarrow{t}+\mathrm{0}=\frac{{c}}{\mathrm{2}}\Rightarrow{t}=\frac{{c}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\left({x},{t}\right)=\left(\mathrm{0},\frac{{c}}{\mathrm{2}}\right) \\ $$$$−{t}−\frac{{c}}{\mathrm{2}}+{x}^{\mathrm{3}} =\mathrm{0}………..\left({i}\right) \\ $$$$\:\:\:{t}−\frac{{c}}{\mathrm{2}}+{x}=\mathrm{0}………….\left({ii}\right) \\ $$$$\left({i}\right)+\left({ii}\right) \\ $$$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0}\left(\bigstar\right) \\ $$$$ \\ $$$$\left({i}\right)\Rightarrow\:{x}^{\mathrm{3}} ={t}+\frac{{c}}{\mathrm{2}}……..\left({iii}\right) \\ $$$$\left({ii}\right)\Rightarrow{x}^{\mathrm{3}} =\left(−{t}+\frac{{c}}{\mathrm{2}}\right)^{\mathrm{3}} ….\left({iv}\right) \\ $$$${t}+\frac{{c}}{\mathrm{2}}=\left(−{t}+\frac{{c}}{\mathrm{2}}\right)^{\mathrm{3}} \\ $$$${t}+\frac{{c}}{\mathrm{2}}=−{t}^{\mathrm{3}} +\frac{{c}^{\mathrm{3}} }{\mathrm{8}}+\mathrm{3}\left(−{t}\right)\left(\frac{{c}}{\mathrm{2}}\right)\left(−{t}+\frac{{c}}{\mathrm{2}}\right) \\ $$$${t}+\frac{{c}}{\mathrm{2}}=−{t}^{\mathrm{3}} +\frac{{c}^{\mathrm{3}} }{\mathrm{8}}+\frac{\mathrm{3}}{\mathrm{2}}{ct}^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}{ct} \\ $$$${t}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{ct}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}{ct}−\frac{{c}^{\mathrm{3}} }{\mathrm{8}}+\frac{{c}}{\mathrm{2}}=\mathrm{0}\left(\bigstar\right) \\ $$$$\left(\bigstar\right)\:\mathcal{D}{on}'{t}\:{know}\:{how}\:{to}\:{solve}\:{cubic} \\ $$$${equation}. \\ $$$$\:\:\:\:\:\left({x},{t}\right)=\left(\mathrm{0},\frac{{c}}{\mathrm{2}}\right) \\ $$$$…… \\ $$$$\:\:\: \\ $$
Commented by mr W last updated on 15/May/21
x^3 +x−c=0 (★) has always one real root: x=(((√((1/(27))+(c^2 /4)))+(c/2)))^(1/3) −(((√((1/(27))+(c^2 /4)))−(c/4)))^(1/3)
$${x}^{\mathrm{3}} +{x}−{c}=\mathrm{0}\:\left(\bigstar\right) \\ $$$${has}\:{always}\:{one}\:{real}\:{root}: \\ $$$${x}=\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}+\frac{{c}}{\mathrm{2}}}−\sqrt[{\mathrm{3}}]{\sqrt{\frac{\mathrm{1}}{\mathrm{27}}+\frac{{c}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{c}}{\mathrm{4}}} \\ $$
Commented by Rasheed.Sindhi last updated on 15/May/21
Thαnks mr W sir!
$$\mathcal{T}{h}\alpha{nks}\:{mr}\:\mathcal{W}\:\:\boldsymbol{{sir}}! \\ $$

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