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x-2-x-x-x-0-x-




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

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