Question Number 192936 by Abdullahrussell last updated on 31/May/23
Answered by Frix last updated on 31/May/23
$$\mathrm{Obviously}\:{x}^{\mathrm{2}} =\mathrm{3}\vee{x}^{\mathrm{2}} =\mathrm{4}\:\Rightarrow \\ $$$${x}=\pm\sqrt{\mathrm{3}}\vee{x}=\pm\mathrm{2} \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{use}\:\sqrt[{\mathrm{3}}]{−{z}}=−\sqrt[{\mathrm{3}}]{{z}}\:\mathrm{also}\:{x}^{\mathrm{2}} =\mathrm{12}\:\Rightarrow\:{x}=\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$
Answered by MM42 last updated on 31/May/23
$$\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}={u}\Rightarrow{x}^{\mathrm{2}} ={u}^{\mathrm{2}} +\mathrm{3} \\ $$$$\Rightarrow{u}+\sqrt[{\mathrm{3}}]{\mathrm{1}−{u}^{\mathrm{2}} }=\mathrm{1}\Rightarrow\mathrm{1}−{u}^{\mathrm{2}} =\left(\mathrm{1}−{u}\right)^{\mathrm{3}} \\ $$$$\Rightarrow{u}\left({u}−\mathrm{1}\right)\left({u}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow{u}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} =\mathrm{3}\Rightarrow{x}=\pm\sqrt{\mathrm{3}} \\ $$$$\&\:{u}=\mathrm{1}\Rightarrow{x}^{\mathrm{2}} =\mathrm{4}\Rightarrow{x}=\pm\mathrm{2} \\ $$$$\&\:{u}=\mathrm{3}\Rightarrow{x}^{\mathrm{2}} =\mathrm{12}\Rightarrow{x}=\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$
Commented by MM42 last updated on 31/May/23
$${the}\:{diagram}\:{below}\:{shows}\:{the}\:{state}\left[\right. \\ $$$${of}\:{the}\:{threads} \\ $$$${f}\left({x}\right)=\sqrt{{x}^{\mathrm{2}} −\mathrm{3}}+\sqrt[{\mathrm{3}}]{\mathrm{4}−{x}^{\mathrm{2}} }−\mathrm{1} \\ $$
Commented by MM42 last updated on 31/May/23
