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Question Number 179064 by HeferH last updated on 24/Oct/22

Commented by HeferH last updated on 24/Oct/22

$$isthiscorrect?$$

Commented by Rasheed.Sindhi last updated on 24/Oct/22

$$\mathrm{I}\mathrm{think}\mathrm{it}\mathrm{depends}\mathrm{upon}{}^{\prime }{\mathrm{notation}}^{\prime }:$$$$\mathrm{if}\mathrm{we}\mathrm{decide}‘‘{\sqrt{}}^{″}\mathrm{denotes}\mathrm{both}\mathrm{roots}$$$$\mathrm{then}\mathrm{it}\mathrm{may}\mathrm{be}\mathrm{correct}.(\mathrm{The}\mathrm{Hiper}$$$$\mathrm{calculators}\mathrm{decided}\mathrm{this})$$$$\mathrm{But}\mathrm{on}\mathrm{the}\mathrm{other}\mathrm{hand}\mathrm{our}\mathrm{notebook}$$$$\mathrm{math}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{accept}\mathrm{this}\mathrm{convention}.$$$$\mathrm{According}\mathrm{to}\mathrm{our}\mathrm{common}\mathrm{math}(\mathrm{that}$$$$\mathrm{I}\mathrm{have}\mathrm{called}\mathrm{here}{}^{\prime }\mathrm{notebook}{\mathrm{math}}^{\prime }):$$$$\sqrt{9}=3,-\sqrt{9}=-3\mathrm{\&}\pm \sqrt{9}=\{\begin{array}{l}3\\ -3\end{array}$$

Commented by MJS_new last updated on 24/Oct/22

$$\mathrm{not}\mathrm{correct}.\mathrm{it}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{fit}\mathrm{other}\mathrm{rules}.$$$$\mid a+b\mathrm{i}\mid ⩾0\mathrm{for}\mathrm{any}\mathrm{numbers}a,b\in \mathbb{R}$$$$\mid a+b\mathrm{i}\mid =\sqrt{a^2+b^2}\Rightarrow \sqrt{a^2+b^2}⩾0$$$$\mathrm{we}\mathrm{do}\mathrm{not}\mathrm{solve}\mathrm{an}\mathrm{equation}\mathrm{here},\mathrm{we}\mathrm{simply}$$$$\mathrm{follow}\mathrm{this}\mathrm{rule}:$$$$z=r{\mathrm{e}}^{\mathrm{i}\theta }\mathrm{with}r⩾0\wedge -\pi ⩽\theta <\pi$$$$\Rightarrow$$$$\sqrt{z}=\sqrt{r}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{2}}$$$$\sqrt[3]{z}=\sqrt[3]{r}{\mathrm{e}}^{\mathrm{i}\frac{\theta }{3}}$$$$...$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{calculation}$$$$\mathrm{but}\mathrm{if}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{solve}$$$$x^2=4$$$$\mathrm{we}\mathrm{ask}\mathrm{for}\mathrm{all}\mathrm{possible}x\mathrm{which}\mathrm{give}4\mathrm{when}$$$$\mathrm{squared}$$$$\mathrm{since}\mathrm{a}\mathrm{few}\mathrm{years}\mathrm{this}\mathrm{seems}\mathrm{to}\mathrm{get}\mathrm{confused}$$$$\mathrm{even}\mathrm{by}\mathrm{math}\mathrm{teachers},\mathrm{not}\mathrm{sure}\mathrm{why}...$$$$\mathrm{if}\sqrt{4}=\pm 2\mathrm{how}\mathrm{can}\mathrm{you}\mathrm{determine}$$$$\sqrt{4}+\sqrt{4}=r$$$$\mathrm{you}\mathrm{must}\mathrm{get}{r}_1=-4r_2=0r_3=+4$$

Answered by thenxtkvng last updated on 24/Oct/22

$$since3\times 3=9$$$$\therefore thesquarerootof\sqrt{9}$$$$is3$$

Commented by MJS_new last updated on 24/Oct/22

$$\mathrm{but}\mathrm{also}(-3)\times (-3)=9$$$$\mathrm{your}\mathrm{explanation}\mathrm{is}\mathrm{not}\mathrm{good}$$

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