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Question Number 8600 by Chantria last updated on 17/Oct/16

$$\mathit{T}\mathit{e}\mathit{s}\mathit{t}$$$$1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\cdot \sqrt{-1}$$$$=i\cdot i=i^2=-1$$$$so1=-1$$$$Findtheerror.$$

Commented by prakash jain last updated on 17/Oct/16

$${\left(ab\right)}^x=a^x\cdot b^x$$$$\mathrm{only}\mathrm{for}a,b,x\in \mathbb{R}\mathrm{and}a,b,x\mathrm{are}+\mathrm{ve}.$$$${(-1)}^{1/2}{(-1)}^{1/2}\ne {(-1\times -1)}^{1/2}\mathrm{since}-1<0$$

Answered by malwaan last updated on 18/Oct/16

$$\sqrt{1}=\pm 1$$$$so1=\sqrt{1}isonepartfromthesolution$$$$and\sqrt{1}=-1isthe2ndpart$$$$1=\sqrt{1}=-1iswrong$$$$\sqrt{1}=\pm 1isright$$

Commented by Rasheed Soomro last updated on 19/Oct/16

$$‘‘\sqrt{}{}^{″}\mathrm{is}\mathrm{used}\mathrm{only}\mathrm{for}+\mathrm{ve}\mathrm{root}.$$$$\mathrm{Hence}\sqrt{1}=1\mathrm{not}\sqrt{1}=\pm 1$$$$\mathrm{We}\mathrm{can}\mathrm{write}\pm \sqrt{1}=\pm 1\mathrm{but}$$$$\mathrm{we}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{write}\sqrt{1}=\pm 1$$

Commented by FilupSmith last updated on 19/Oct/16

$$\mathrm{but}\sqrt{x^2}=\pm x$$$${(+x)}^2=x^2$$$${(-x)}^2={(-1)}^2{\left(x\right)}^2=x^2$$

Commented by nume1114 last updated on 19/Oct/16

$$\sqrt{x^2}=\mid x\mid$$$$example$$$$ifx=3,\sqrt{x^2}=\sqrt{9}=3,\mid x\mid =3$$$$x=-1,\sqrt{x^2}=\sqrt{1}=1,\mid x\mid =1$$$$x=0,\sqrt{x^2}=\sqrt{0}=0,\mid x\mid =0$$

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