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Question Number 182459 by HeferH last updated on 09/Dec/22

Commented by HeferH last updated on 09/Dec/22

$$howisthiscalculatorgettinthesecomplex$$$$values?$$

Commented by MJS_new last updated on 09/Dec/22

$${\mathrm{It}}^{\prime }\mathrm{s}\mathrm{wrong}.{\mathrm{It}}^{\prime }\mathrm{s}not\mathrm{calculating}\mathrm{the}\mathrm{root},{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{instead}\mathrm{solving}{x}^{16}=2^{15}.\mathrm{You}\mathrm{should}\mathrm{get}\mathrm{a}$$$$\mathrm{better}\mathrm{calculator}.$$

Commented by Acem last updated on 10/Dec/22

$$Whatshouldbetheanswer?$$

Commented by MJS_new last updated on 10/Dec/22

$$1.915...$$$$\sqrt[16]{2^{15}}=\sqrt[16]{32768}\in {\mathbb{R}}^{+}$$

Commented by Acem last updated on 10/Dec/22

$$YesSir,but{}^{16}\sqrt{32768}=1.915...$$$$so{what}^{\prime }stheissue?$$

Commented by Acem last updated on 10/Dec/22

Commented by Acem last updated on 10/Dec/22

$${what}^{\prime }stheproblemwiththisresult?$$

Commented by MJS_new last updated on 10/Dec/22

$$\mathrm{no}\mathrm{problem}\mathrm{with}this\mathrm{result},\mathrm{but}\mathrm{the}\mathrm{calculator}$$$$\mathrm{shows}\mathrm{a}\mathrm{complex}\mathrm{solution}\mathrm{too}\mathrm{which}\mathrm{is}\mathrm{wrong}$$

Commented by mr W last updated on 10/Dec/22

$$thisisnotaproblemoftheapp.$$$$itdisplaysallroots,becauseinthis$$$$casetheuserhaschosentoletit$$$$displayallrootsincomplexmode.$$$$ifthisinformationconfusesyou,$$$$youshouldnotlettheapptodisplay$$$$it.insettingsyouonlyneedtoselect$$$$‘‘OnlyPrincipal{Root}^{″}insteadof$$$$‘‘All{Roots}^{″}.$$

Commented by mr W last updated on 10/Dec/22

Commented by mr W last updated on 10/Dec/22

$$yougetwith‘‘OnlyPrincipal{Root}^{″}:$$

Commented by mr W last updated on 10/Dec/22

Commented by mr W last updated on 10/Dec/22

$$yougetwith‘‘All{Roots}^{″}:$$

Commented by mr W last updated on 10/Dec/22

Commented by mr W last updated on 10/Dec/22

$$bydefaulttheapp{doesn}^{\prime }tuse$$$$complexmodeand{doesn}^{\prime }tdisplay$$$$allroots.soi{don}^{\prime }tunderstand$$$$thosepeoplewhoexplicitlywantthe$$$$appusecomplexmodeanddisplay$$$$allroots,andthenareconfusedwhen$$$$theappindeeddoesthis.$$

Commented by MJS_new last updated on 10/Dec/22

$$\mathrm{I}\mathrm{hereby}\mathrm{give}\mathrm{up}.$$$$\mathrm{people}\mathrm{started}\mathrm{to}\mathrm{mix}\mathrm{up}x=\sqrt{4}\mathrm{and}{x}^2=4$$$$\mathrm{and}\mathrm{now}\mathrm{the}\mathrm{app}\mathrm{makers}\mathrm{also}\mathrm{do}\mathrm{this}.\mathrm{use}\mathrm{it}$$$$\mathrm{as}\mathrm{you}\mathrm{like},\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{against}\mathrm{the}\mathrm{definitions}.$$$$\mathrm{if}\sqrt[16]{2^{15}}\notin \mathbb{R}\mathrm{then}\mathrm{you}\mathrm{also}\mathrm{have}\mathrm{to}\mathrm{accept}$$$$\sqrt{4}=-2\mathrm{with}\mathrm{all}\mathrm{kinds}\mathrm{of}\mathrm{weird}\mathrm{logical}$$$$\mathrm{problems}.$$$$\mathrm{i}.\mathrm{e}.\mid z\mid ⩾0\mathrm{\forall }z\in \mathbb{C}\mathrm{and}\mid a+b\mathrm{i}\mid =\sqrt{a^2+b^2}$$$$\mathrm{but}\mathrm{what}\mathrm{if}z=\sqrt[4]{16}+\sqrt[4]{81}\mathrm{i}?\mathrm{how}\mathrm{many}\mathrm{values}$$$$\mathrm{is}z\mathrm{at}\mathrm{once}\mathrm{or}\mathrm{which}\mathrm{value}\mathrm{is}\mathrm{the}‘‘\mathrm{main}{\mathrm{value}}^{″}?$$$$\mathrm{and}\sqrt{{\left(\sqrt[4]{16}\right)}^2+{\left(\sqrt[4]{81}\right)}^2}\mathrm{can}\mathrm{not}\mathrm{only}\mathrm{be}<0\mathrm{but}$$$$\mathrm{also}\mathrm{be}\mathrm{a}\mathrm{complex}\mathrm{number}???$$$$\mathrm{or}\mathrm{simply}:$$$$\sqrt{4}+\sqrt{4}=0$$$$\mathrm{proof}:$$$$\sqrt{4}=-2\vee \sqrt{4}=2$$$$\Rightarrow \sqrt{4}+\sqrt{4}=-2+2=0$$$${\mathrm{I}}^{\prime }\mathrm{ll}\mathrm{keep}\mathrm{following}\mathrm{the}\mathrm{rules}\mathrm{I}\mathrm{learned}35\mathrm{years}$$$$\mathrm{ago},\mathrm{everybody}\mathrm{else}\mathrm{is}\mathrm{free}\mathrm{to}\mathrm{use}\mathrm{whatever}$$$$\mathrm{they}\mathrm{want}.$$

Commented by MJS_new last updated on 10/Dec/22

$$\mathrm{if}\mathrm{these}‘‘\mathrm{new}{\mathrm{rules}}^{″}\mathrm{are}\mathrm{used}$$$$\sqrt[4]{x+3}-\sqrt{x+4}=3$$$$\mathrm{has}2\mathrm{solutions}:$$$$x_1\approx -.816094359$$$$x_2\approx 23.8367388$$$$\mathrm{someone}\mathrm{prove}\mathrm{the}\mathrm{opposite}...$$

Commented by mr W last updated on 10/Dec/22

$$itotallyagreewithsir!basicallyit$$$$isclearand100\mathrm{\%}correctwhatyou$$$$said.$$$$butasabouttheapp,i{don}^{\prime }tthinkthat$$$$theappmakesanerror.itstates$$$$clearlythatitdisplaysallrootsof$$$$anumberincomplexplane.$$$$incomplexmodethisappmeans$$$$with\sqrt[3]{3}simplytherootsofthenumber$$$$3incomplexplane.$$

Commented by Acem last updated on 10/Dec/22

$$YesIagreewithyouSirMJS$$

Commented by Acem last updated on 10/Dec/22

$$Thewrongisnotwiththeapplicationbutwith$$$$howitisused.TheoptionCPLXmustchosenjust$$$$for{\mathbb{R}}^{-}$$

Commented by Acem last updated on 10/Dec/22

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