Question Number 179063 by sciencestudent last updated on 24/Oct/22
$${why}\:{is}\:{not}\:{a}\:{polynomial}\:\sqrt{\mathrm{25}{x}^{\mathrm{8}} }\:\:? \\ $$
Commented by sciencestudent last updated on 24/Oct/22
$$? \\ $$
Commented by sciencestudent last updated on 26/Oct/22
$${please}\:{help}\:{me}! \\ $$
Answered by Tokugami last updated on 26/Oct/22
$$\sqrt{\mathrm{25}{x}^{\mathrm{8}} }\:\mathrm{can}\:\mathrm{be}\:\mathrm{reduced}\:\mathrm{to}\:\mathrm{5}{x}^{\mathrm{4}} , \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{can}\:\mathrm{also}\:\mathrm{be}\:−\mathrm{5}{x}^{\mathrm{4}} , \\ $$$$\mathrm{because}\:\left(\mathrm{5}{x}^{\mathrm{4}} \right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{8}} \:\underline{\mathrm{and}}\:\left(−\mathrm{5}{x}^{\mathrm{4}} \right)^{\mathrm{2}} =\mathrm{25}{x}^{\mathrm{8}} \\ $$$$\sqrt{\mathrm{25}{x}^{\mathrm{8}} }\:\mathrm{is}\:\mathrm{multivalued},\:\mathrm{so}\:\mathrm{it}\:\mathrm{is} \\ $$$$\mathrm{not}\:\mathrm{a}\:\mathrm{polynomial}. \\ $$
Commented by mr W last updated on 26/Oct/22
$$\sqrt{\mathrm{25}{x}^{\mathrm{8}} }=\mathrm{5}{x}^{\mathrm{4}} \\ $$$$\sqrt{\mathrm{25}{x}^{\mathrm{8}} }\neq−\mathrm{5}{x}^{\mathrm{4}} \\ $$