Question Number 201864 by ajfour last updated on 14/Dec/23
$$\left({ct}^{\mathrm{2}} −\frac{\mathrm{1}}{{ct}^{\mathrm{2}} }+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}^{\mathrm{4}} +\frac{\mathrm{1}}{{t}^{\mathrm{4}} }\right) \\ $$$${Find}\:\:{t}={f}\left({c}\right). \\ $$
Commented by Frix last updated on 14/Dec/23
$$\mathrm{This}\:\mathrm{is}\:\mathrm{equivalent}\:\mathrm{to} \\ $$$$\left({t}^{\mathrm{2}} \right)^{\mathrm{3}} −\frac{\left({t}^{\mathrm{2}} \right)}{{c}^{\mathrm{2}} }+\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−{c}^{\mathrm{4}} \right)}{\mathrm{4}{c}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\left(\mathrm{method}\:\mathrm{depending}\right. \\ $$$$\left.\mathrm{on}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{27}{c}^{\mathrm{8}} −\mathrm{54}{c}^{\mathrm{4}} −\mathrm{5}\right) \\ $$
Commented by ajfour last updated on 14/Dec/23
$${Thanks},\:{c}^{\mathrm{4}} =\mathrm{1}−\mathrm{2}\sqrt{\mathrm{2}}{p}\:\:\:\:\&\:\:{p}^{\mathrm{2}} <\frac{\mathrm{4}}{\mathrm{27}} \\ $$$$\:{Sir}\:\:{how}\:{would}\:{you}\:{do}\:{this}: \\ $$$$\:\:{x}^{\mathrm{4}} −\mathrm{9}{x}^{\mathrm{2}} +\mathrm{6}{x}−\mathrm{1}=\mathrm{0} \\ $$$${I}\:{figued}\:{out}\:{a}\:{nice}\:{way}\:{for}\:{such} \\ $$$$\:\:{c}^{\mathrm{2}} =\mathrm{4}{bd}\:\:\:{holding}\:{biquadratics}. \\ $$