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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 bymr 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 byRasheed.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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