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