Question-179064

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

Commented by HeferH last updated on 24/Oct/22

$${is}\:{this}\:{correct}? \\ $$

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

I think it depends upon ′notation′: if we decide “(√( ))” denotes both roots then it may be correct.(The Hiper calculators decided this) But on the other hand our notebook math don′t accept this convention. According to our common math (that I have called here ′notebook math′): (√9) =3, −(√9) =−3 & ±(√9) = { (( 3)),((−3)) :}

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{depends}\:\mathrm{upon}\:'\mathrm{notation}': \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{decide}\:“\sqrt{\:\:\:\:}''\:\mathrm{denotes}\:\mathrm{both}\:\mathrm{roots} \\ $$$$\mathrm{then}\:\mathrm{it}\:\mathrm{may}\:\mathrm{be}\:\mathrm{correct}.\left(\mathrm{The}\:\mathrm{Hiper}\right. \\ $$$$\left.\mathrm{calculators}\:\mathrm{decided}\:\mathrm{this}\right) \\ $$$$\mathrm{But}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other}\:\mathrm{hand}\:\mathrm{our}\:\mathrm{notebook} \\ $$$$\mathrm{math}\:\mathrm{don}'\mathrm{t}\:\mathrm{accept}\:\mathrm{this}\:\mathrm{convention}. \\ $$$$\mathrm{According}\:\mathrm{to}\:\mathrm{our}\:\mathrm{common}\:\mathrm{math}\:\left(\mathrm{that}\right. \\ $$$$\left.\mathrm{I}\:\mathrm{have}\:\mathrm{called}\:\mathrm{here}\:'\mathrm{notebook}\:\mathrm{math}'\right): \\ $$$$\sqrt{\mathrm{9}}\:=\mathrm{3},\:−\sqrt{\mathrm{9}}\:=−\mathrm{3}\:\&\:\pm\sqrt{\mathrm{9}}\:=\begin{cases}{\:\:\:\:\mathrm{3}}\\{−\mathrm{3}}\end{cases} \\ $$

Commented by MJS_new last updated on 24/Oct/22

not correct. it doesn′t fit other rules. ∣a+bi∣≥0 for any numbers a, b ∈R ∣a+bi∣=(√(a^2 +b^2 )) ⇒ (√(a^2 +b^2 ))≥0 we do not solve an equation here, we simply follow this rule: z=re^(iθ) with r≥0∧−π≤θ<π ⇒ (√z)=(√r)e^(i(θ/2)) (z)^(1/3) =(r)^(1/3) e^(i(θ/3)) ... it′s a calculation but if we have to solve x^2 =4 we ask for all possible x which give 4 when squared since a few years this seems to get confused even by math teachers, not sure why... if (√4)=±2 how can you determine (√4)+(√4)=r you must get r_1 =−4 r_2 =0 r_3 =+4

$$\mathrm{not}\:\mathrm{correct}.\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{fit}\:\mathrm{other}\:\mathrm{rules}. \\ $$$$\mid{a}+{b}\mathrm{i}\mid\geqslant\mathrm{0}\:\mathrm{for}\:\mathrm{any}\:\mathrm{numbers}\:{a},\:{b}\:\in\mathbb{R} \\ $$$$\mid{a}+{b}\mathrm{i}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:\Rightarrow\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\geqslant\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{do}\:\mathrm{not}\:\mathrm{solve}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{here},\:\mathrm{we}\:\mathrm{simply} \\ $$$$\mathrm{follow}\:\mathrm{this}\:\mathrm{rule}: \\ $$$${z}={r}\mathrm{e}^{\mathrm{i}\theta} \:\mathrm{with}\:{r}\geqslant\mathrm{0}\wedge−\pi\leqslant\theta<\pi \\ $$$$\Rightarrow \\ $$$$\sqrt{{z}}=\sqrt{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{2}}} \\ $$$$\sqrt[{\mathrm{3}}]{{z}}=\sqrt[{\mathrm{3}}]{{r}}\mathrm{e}^{\mathrm{i}\frac{\theta}{\mathrm{3}}} \\ $$$$… \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{calculation} \\ $$$$\mathrm{but}\:\mathrm{if}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{solve} \\ $$$${x}^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{we}\:\mathrm{ask}\:\mathrm{for}\:\mathrm{all}\:\mathrm{possible}\:{x}\:\mathrm{which}\:\mathrm{give}\:\mathrm{4}\:\mathrm{when} \\ $$$$\mathrm{squared} \\ $$$$\mathrm{since}\:\mathrm{a}\:\mathrm{few}\:\mathrm{years}\:\mathrm{this}\:\mathrm{seems}\:\mathrm{to}\:\mathrm{get}\:\mathrm{confused} \\ $$$$\mathrm{even}\:\mathrm{by}\:\mathrm{math}\:\mathrm{teachers},\:\mathrm{not}\:\mathrm{sure}\:\mathrm{why}… \\ $$$$\mathrm{if}\:\sqrt{\mathrm{4}}=\pm\mathrm{2}\:\mathrm{how}\:\mathrm{can}\:\mathrm{you}\:\mathrm{determine} \\ $$$$\sqrt{\mathrm{4}}+\sqrt{\mathrm{4}}={r} \\ $$$$\mathrm{you}\:\mathrm{must}\:\mathrm{get}\:{r}_{\mathrm{1}} =−\mathrm{4}\:{r}_{\mathrm{2}} =\mathrm{0}\:{r}_{\mathrm{3}} =+\mathrm{4} \\ $$

Answered by thenxtkvng last updated on 24/Oct/22

since 3×3=9 ∴the square root of (√9) is 3

$${since}\:\mathrm{3}×\mathrm{3}=\mathrm{9} \\ $$$$\therefore{the}\:{square}\:{root}\:{of}\:\sqrt{\mathrm{9}} \\ $$$${is}\:\mathrm{3} \\ $$

Commented by MJS_new last updated on 24/Oct/22

but also (−3)×(−3)=9 your explanation is not good

$$\mathrm{but}\:\mathrm{also}\:\left(−\mathrm{3}\right)×\left(−\mathrm{3}\right)=\mathrm{9} \\ $$$$\mathrm{your}\:\mathrm{explanation}\:\mathrm{is}\:\mathrm{not}\:\mathrm{good} \\ $$

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