Question Number 186662 by pascal889 last updated on 08/Feb/23
Answered by Frix last updated on 08/Feb/23
$${u}=\sqrt{{x}+\mathrm{1}}\geqslant\mathrm{0}\:\Leftrightarrow\:{x}={u}^{\mathrm{2}} −\mathrm{1} \\ $$$$\frac{{u}^{\mathrm{2}} }{\:\sqrt{{u}+\mathrm{2}}}=\mathrm{2}{u}^{\mathrm{2}} −\mathrm{1} \\ $$$${v}=\sqrt{{u}+\mathrm{2}}\geqslant\sqrt{\mathrm{2}}\:\Leftrightarrow\:{u}={v}^{\mathrm{2}} −\mathrm{1} \\ $$$$\frac{\left({v}^{\mathrm{2}} −\mathrm{2}\right)^{\mathrm{2}} }{{v}}=\mathrm{2}{v}^{\mathrm{4}} −\mathrm{8}{v}^{\mathrm{2}} +\mathrm{7} \\ $$$${v}^{\mathrm{5}} −\frac{{v}^{\mathrm{4}} }{\mathrm{2}}−\mathrm{4}{v}^{\mathrm{3}} +\mathrm{2}{v}^{\mathrm{2}} +\frac{\mathrm{7}{v}}{\mathrm{2}}−\mathrm{2}=\mathrm{0} \\ $$$$\left({v}−\mathrm{1}\right)\left({v}^{\mathrm{2}} +{v}−\mathrm{1}\right)\left({v}^{\mathrm{2}} −\frac{{v}}{\mathrm{2}}−\mathrm{2}\right)=\mathrm{0} \\ $$$${v}\geqslant\sqrt{\mathrm{2}}\:\Rightarrow\:{v}=\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{4}}\:\Rightarrow\:{u}=\frac{\mathrm{1}+\sqrt{\mathrm{33}}}{\mathrm{8}}\:\Rightarrow \\ $$$${x}=\frac{−\mathrm{15}+\sqrt{\mathrm{33}}}{\mathrm{32}} \\ $$