Question Number 85358 by jagoll last updated on 21/Mar/20
$$\sqrt[{\mathrm{3}\:\:}]{\mathrm{x}\:}\:+\:\sqrt{\mathrm{x}}\:=\:\mathrm{12} \\ $$
Answered by john santu last updated on 21/Mar/20
$${x}\:=\:{t}^{\mathrm{6}} \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} +{t}^{\mathrm{3}} −\mathrm{12}=\mathrm{0} \\ $$$$\left({t}−\mathrm{2}\right)\left({t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{6}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}\:=\:\mathrm{2}\:\Rightarrow\:{x}\:=\:\mathrm{2}^{\mathrm{6}} \:=\:\mathrm{64} \\ $$$$\Rightarrow{t}\:=\:\frac{−\mathrm{3}\:\pm\:{i}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$$\Rightarrow{x}\:=\:\left(\frac{−\mathrm{3}\:\pm\:{i}\sqrt{\mathrm{15}}}{\mathrm{2}}\right)^{\mathrm{6}} \\ $$
Commented by john santu last updated on 21/Mar/20
$${hahaha}…{upside}\:{down}\:{sir} \\ $$
Commented by MJS last updated on 21/Mar/20
$$\mathrm{sorry},\:\mathrm{but}\:\mathrm{you}\:\mathrm{made}\:\mathrm{a}\:\mathrm{weird}\:\mathrm{mistake} \\ $$$${x}={t}^{\mathrm{6}} \\ $$$$\Rightarrow\:{t}^{\mathrm{3}} +{t}^{\mathrm{2}} −\mathrm{12}=\mathrm{0}\:\Rightarrow\:{t}=\mathrm{2}\:\Rightarrow\:{x}=\mathrm{2}^{\mathrm{6}} =\mathrm{64} \\ $$
Commented by MJS last updated on 21/Mar/20
$$\mathrm{yes},\:\mathrm{I}\:\mathrm{also}\:\mathrm{tend}\:\mathrm{to}\:\mathrm{this}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{error}\:\mathrm{sometimes} \\ $$