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Question Number 187233 by ajfour last updated on 15/Feb/23

$$Findxif:x^4+a=x\mathrm{\forall }(0<a<\frac{2}{9})$$

Commented byMJS_new last updated on 15/Feb/23

$$\mathrm{maybe}\mathrm{off}\mathrm{topic}\mathrm{but}\mathrm{it}\mathrm{could}\mathrm{be}\mathrm{of}\mathrm{interest}$$ $$(x-\frac{p^2}{2}-\frac{\sqrt{2-p^6}}{2p})(x-\frac{p^2}{2}+\frac{\sqrt{2-p^6}}{2p})(x+\frac{p^2}{2}-\frac{\sqrt{2+p^6}}{2p}\mathrm{i})(x+\frac{p^2}{2}+\frac{\sqrt{2+p^6}}{2p}\mathrm{i})=0$$ $$(x^2-p^{2}x-\frac{1-p^6}{2p^2})(x^2+p^{2}x+\frac{1+p^6}{2p^2})=0$$ $$x^4-x-\frac{1-p^{12}}{4p^4}=0$$

Answered by ajfour last updated on 15/Feb/23

$$(x^2+px+h)(x^2-px+k)=0$$ $$p(k-h)=-1$$ $$h+k=p^2$$ $$hk=a$$ $$p^4-\frac{1}{p^2}=4a$$ $$p^6-4{ap}^2-1=0$$ $$D=\frac{1}{4}-\frac{64a^3}{27}$$ $$D>0$$ $$\Rightarrow a^3<\frac{27}{256}$$ $$or0<a<\frac{2}{9}<\frac{3}{4}{\left(\frac{1}{4}\right)}^{1/3}$$ $$hence{p}^{2}isuniqueand>0$$ $$2k=p^2-\frac{1}{p}$$ $$2h=p^2+\frac{1}{p}$$ $$x=-\frac{p}{2}\pm \sqrt{\frac{p^2}{4}-h}$$ $$=-\frac{p}{2}\pm \sqrt{-\frac{1}{2p}-\frac{p^2}{4}}$$ $$hencep>0givesx\in \mathbb{C}$$ $$whilep<0mightyieldx\in \mathbb{R}$$ $$if{p}^3>-2$$ $$hencexcouldturnrealonlyif$$ $$-2^{1/3}⩽p<0$$

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