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Question Number 113901 by frc2crc last updated on 16/Sep/20

$$whendoIuse\mid x\mid$$$$for\sqrt{x^2}?$$

Commented by MJS_new last updated on 16/Sep/20

$$\mathrm{almost}\mathrm{always}$$$$p^2⩾0\mathrm{\forall }p\in \mathbb{R}\iff p^2\in {\mathbb{R}}^{+}\mathrm{\forall }p\in \mathbb{R}$$$$\sqrt{q}⩾0\mathrm{\forall }q\in {\mathbb{R}}^{+}$$$$\Rightarrow \sqrt{p^2}⩾0\mathrm{\forall }p\in \mathbb{R}$$$$\Rightarrow \sqrt{x^2}=\mid x\mid \mathrm{\forall }x\in \mathbb{R}$$$$$$$$\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{use}\mathrm{it}\mathrm{in}\mathrm{calculus}(\mathrm{integral}\mathrm{solving}$$$$\mathrm{by}\mathrm{substitution})$$

Commented by frc2crc last updated on 16/Sep/20

$$\sqrt{4}=2or\mid 2\mid ?$$

Commented by bobhans last updated on 16/Sep/20

$$\sqrt{4}=2and\mid 2\mid =2$$

Commented by MJS_new last updated on 16/Sep/20

$$\sqrt{2^2}=\mid 2\mid =2$$$$\sqrt{{(-2)}^2}=\mid -2\mid =2$$

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