$|x|=\sqrt{x^2}$ as Wolfram|Alpha shows. But, as $(x^2)^\frac12=x$, I can't understand where am I wrong interpreting Square-root.
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1The square root function gives nonnegative numbers. – anon Dec 15 '13 at 08:59
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Anyway, the question remains for me: Does this mean $\sqrt x\neq x^\frac12$? – Silent Dec 15 '13 at 09:04
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See also: http://math.stackexchange.com/questions/518487/how-is-it-that-sqrtx2-is-not-x-but-x – Martin Sleziak Sep 10 '14 at 11:36
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You know that $x^2$ is not an invertible function. To find an inverse you must restrict it. The convention is to restrict the function to positive numbers, hence the $\sqrt y$ function always gives the non negative solution to the equation $x^2=y$.
This being said you understand that the rule: $$ (x^p)^q = x^{pq} $$ is not valid when $x<0$, but is only valid for $x>0$.
Emanuele Paolini
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$$\sqrt{x^2}=|x|$$ so if $x\ge 0$ then $\sqrt{x^2}=+x$ and if $x<0$ then $\sqrt{x^2}=-x$. In both cases the square root should be held positive. Many students have troubles in making difference between the question you asked and this fact that $(\pm x)^2=x^2$.
Mikasa
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1Because the statement $(x^2)^{\frac{1}{2}}=x^{2\times\frac{1}{2}}$ is true for only positive numbers $x$. – Tigran Hakobyan Dec 15 '13 at 09:19
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1@Sush: Think of this $x=(x^2)^{1/2}=((\pm x)^2)^{1/2}=\pm x$. Do u this this is correct? To avoid this event, we define $\sqrt{x^2}=|x|$. – Mikasa Dec 15 '13 at 09:23
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1Thank you so much, So should we use the rule $(x^2)^{\frac{1}{2}}=x^{2\times\frac{1}{2}}$ only with positive $x$ as shown by Tigran? – Silent Dec 15 '13 at 09:28
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By convention, the square-root sign $\sqrt{}$ denotes the non-negative square root. So $\sqrt{x^2} = |x|$.
TonyK
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