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Question Number 179063 by sciencestudent last updated on 24/Oct/22

$$whyisnotapolynomial\sqrt{25x^8}?$$

Commented by sciencestudent last updated on 24/Oct/22

$$?$$

Commented by sciencestudent last updated on 26/Oct/22

$$pleasehelpme!$$

Answered by Tokugami last updated on 26/Oct/22

$$\sqrt{25x^8}\mathrm{can}\mathrm{be}\mathrm{reduced}\mathrm{to}5x^4,$$$$\mathrm{but}\mathrm{it}\mathrm{can}\mathrm{also}\mathrm{be}-5x^4,$$$$\mathrm{because}{\left(5x^4\right)}^2=25x^8\underset{―}{\mathrm{and}}{(-5x^4)}^2=25x^8$$$$\sqrt{25x^8}\mathrm{is}\mathrm{multivalued},\mathrm{so}\mathrm{it}\mathrm{is}$$$$\mathrm{not}\mathrm{a}\mathrm{polynomial}.$$

Commented by mr W last updated on 26/Oct/22

$$\sqrt{25x^8}=5x^4$$$$\sqrt{25x^8}\ne -5x^4$$

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