Question Number 7000 by FilupSmith last updated on 05/Aug/16
$${x}^{\mathrm{2}} ={x}+\sqrt{{x}} \\ $$$${x}=\mathrm{0}\:\vee\:{x}=?? \\ $$
Commented by Yozzii last updated on 05/Aug/16
$${Let}\:{u}=\sqrt{{x}}\Rightarrow{u}^{\mathrm{2}} ={x}\Rightarrow{u}^{\mathrm{4}} ={x}^{\mathrm{2}} \\ $$$${u}^{\mathrm{4}} ={u}^{\mathrm{2}} +{u} \\ $$$${u}^{\mathrm{4}} −{u}^{\mathrm{2}} −{u}=\mathrm{0} \\ $$$${u}\left({u}^{\mathrm{3}} −{u}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{0}\:{or}\:{u}^{\mathrm{3}} −{u}−\mathrm{1}=\mathrm{0} \\ $$$${u}\left({u}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{1} \\ $$$${u}\left({u}−\mathrm{1}\right)\left({u}+\mathrm{1}\right)=\mathrm{1}>\mathrm{0} \\ $$$$\Rightarrow{u}>\mathrm{1} \\ $$$$ \\ $$