Question Number 187233 by ajfour last updated on 15/Feb/23
$${Find}\:{x}\:{if}\::\:\:\:{x}^{\mathrm{4}} +{a}={x}\:\:\:\:\forall\:\left(\mathrm{0}<{a}<\frac{\mathrm{2}}{\mathrm{9}}\right)\:\:\: \\ $$
Commented by MJS_new last updated on 15/Feb/23
$$\mathrm{maybe}\:\mathrm{off}\:\mathrm{topic}\:\mathrm{but}\:\mathrm{it}\:\mathrm{could}\:\mathrm{be}\:\mathrm{of}\:\mathrm{interest} \\ $$$$\left({x}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}−\frac{\sqrt{\mathrm{2}−{p}^{\mathrm{6}} }}{\mathrm{2}{p}}\right)\left({x}−\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{\sqrt{\mathrm{2}−{p}^{\mathrm{6}} }}{\mathrm{2}{p}}\right)\left({x}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}−\frac{\sqrt{\mathrm{2}+{p}^{\mathrm{6}} }}{\mathrm{2}{p}}\mathrm{i}\right)\left({x}+\frac{{p}^{\mathrm{2}} }{\mathrm{2}}+\frac{\sqrt{\mathrm{2}+{p}^{\mathrm{6}} }}{\mathrm{2}{p}}\mathrm{i}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −{p}^{\mathrm{2}} {x}−\frac{\mathrm{1}−{p}^{\mathrm{6}} }{\mathrm{2}{p}^{\mathrm{2}} }\right)\left({x}^{\mathrm{2}} +{p}^{\mathrm{2}} {x}+\frac{\mathrm{1}+{p}^{\mathrm{6}} }{\mathrm{2}{p}^{\mathrm{2}} }\right)=\mathrm{0} \\ $$$${x}^{\mathrm{4}} −{x}−\frac{\mathrm{1}−{p}^{\mathrm{12}} }{\mathrm{4}{p}^{\mathrm{4}} }=\mathrm{0} \\ $$
Answered by ajfour last updated on 15/Feb/23
$$\left({x}^{\mathrm{2}} +{px}+{h}\right)\left({x}^{\mathrm{2}} −{px}+{k}\right)=\mathrm{0} \\ $$$${p}\left({k}−{h}\right)=−\mathrm{1} \\ $$$${h}+{k}={p}^{\mathrm{2}} \\ $$$${hk}={a} \\ $$$${p}^{\mathrm{4}} −\frac{\mathrm{1}}{{p}^{\mathrm{2}} }=\mathrm{4}{a} \\ $$$${p}^{\mathrm{6}} −\mathrm{4}{ap}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${D}=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{64}{a}^{\mathrm{3}} }{\mathrm{27}} \\ $$$${D}>\mathrm{0} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{3}} <\frac{\mathrm{27}}{\mathrm{256}}\:\: \\ $$$${or}\:\:\:\mathrm{0}<{a}<\frac{\mathrm{2}}{\mathrm{9}}<\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$${hence}\:{p}^{\mathrm{2}} \:{is}\:{unique}\:\:{and}\:>\mathrm{0} \\ $$$$\mathrm{2}{k}={p}^{\mathrm{2}} −\frac{\mathrm{1}}{{p}} \\ $$$$\mathrm{2}{h}={p}^{\mathrm{2}} +\frac{\mathrm{1}}{{p}} \\ $$$${x}=−\frac{{p}}{\mathrm{2}}\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{h}} \\ $$$$\:\:=−\frac{{p}}{\mathrm{2}}\pm\sqrt{−\frac{\mathrm{1}}{\mathrm{2}{p}}−\frac{{p}^{\mathrm{2}} }{\mathrm{4}}} \\ $$$${hence}\:\:{p}>\mathrm{0}\:\:{gives}\:\:{x}\in\mathbb{C} \\ $$$${while}\:\:{p}<\mathrm{0}\:\:{might}\:{yield}\:{x}\in\mathbb{R} \\ $$$${if}\:\:\:{p}^{\mathrm{3}} \:>−\mathrm{2} \\ $$$${hence}\:{x}\:{could}\:{turn}\:{real}\:{only}\:{if} \\ $$$$\:\:\:\:−\mathrm{2}^{\mathrm{1}/\mathrm{3}} \leqslant{p}<\mathrm{0} \\ $$