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This question has the potential to sound extremely stupid, but I've seen (and also used) countless times the idea that $\sqrt{x^2} = x$. However $x^2 = x\cdot x = (-x)\cdot(-x)$.

I know that when taking the square root of something we take both the positive and negative root. Yet when solving an equation and we're faced with $\sqrt{x^2y}$ we make it $x\sqrt{y}$. Why didn't we consider $(-x)\sqrt{y}$? Similarly, $\sqrt{x^3}$ is often changed to $x\sqrt{x}$ and not $(-x)\sqrt{-x}$ which would still give the same result if cubed? (I do understand that the latter is imaginary, but that shouldn't stop us from using it, should it?)

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marked as duplicate by Najib Idrissi, Pierre-Guy Plamondon, Watson, quid, Dan Feb 19 '16 at 19:19

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    $\begingroup$ It's actually $|x|$. $\endgroup$ – Stefan Perko Feb 19 '16 at 10:49
  • $\begingroup$ That makes a bit more sense.. but why? $\endgroup$ – Aayush Agrawal Feb 19 '16 at 10:51
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    $\begingroup$ See en.wikipedia.org/wiki/Square_root#Properties_and_uses Square root function is defined to be non-negative $\endgroup$ – SS_C4 Feb 19 '16 at 10:51
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    $\begingroup$ If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$. $\endgroup$ – Stefan Perko Feb 19 '16 at 10:52
  • $\begingroup$ If we had $\sqrt{x^2}$ equal to both $x$ and $-x$, then $\sqrt{}$ wouldn't be a function. Therefore people have defined $\sqrt{}$ to be non-negative (so that the output of $\sqrt{a}$ is always unique for all $a\ge 0$). $\endgroup$ – user236182 Feb 19 '16 at 10:55
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Consider $(-2)^2=4=2^2$. If one could devise a definition of the square root function such that $\sqrt{x^2}=x$ for any $x$, what would be the value of $$ \sqrt{4}=\sqrt{(-2)^2}=\sqrt{2^2} $$ without getting a contradiction?

This shows that it is not possible to define a function with the desired property, so we abandon the idea and define, for $x\ge0$,

$\sqrt{x}$ is the unique nonnegative number $y$ such that $y^2=x$

In particular, $\sqrt{x^2}=|x|$, because $|x|\ge0$ by definition and $|x|^2=x^2$.


As an aside, note that $\sqrt{x^3}$ makes sense only if $x\ge0$, so in this case $\sqrt{x^3}=x\sqrt{x}$ is correct. You could also say $$ \sqrt{x^3}=\sqrt{x^2\cdot x}=\sqrt{x^2}\cdot \sqrt{x} =|x|\sqrt{x} $$ which would be correct too, but $|x|=x$ as $x\ge0$.

Note, instead, that if $x<0$ and $y<0$, it would be very incorrect to write $$ \sqrt{xy}=\sqrt{x}\sqrt{y} $$ but you can surely say $\sqrt{xy}=\sqrt{|x|}\sqrt{|y|}$.

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The symbol $\sqrt{\mathstrut\quad}$ is defined to denote the non-negative square root. In other words: $\sqrt{x}$ is the number $y$ such that $y\geq 0$ and $y^2=x$. The number $y$ is then called the radical of $x$. This is a matter of definition. We need to choose one of the two values to make $\sqrt{\mathstrut\quad}$ into a function, and we like positive numbers better.

If $y = \sqrt{x}$ then $y^2 = (-y)^2 = x$. So now, what if $y = \sqrt{x^2}$?

Since $y \geq 0$ by definition, we have $y=x$ if $x\geq0$ or $y = -x$ if $x<0$. We write this as $y = |x|$ and say $y$ is the absolute value of $x$.

Also note that if $x$ is positive, $\sqrt{-x}$ is not well defined even if you can still find a number $y$ such that $y^2=-x$. You have to be careful: the radical is only defined for non-negative numbers.

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    $\begingroup$ that most people (including me) seem to say "the square root of $x$" when they actually mean "the positive square root of $x$", or $\sqrt{x}$. $\endgroup$ – Andrea Feb 19 '16 at 11:02
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Actually, $\sqrt{x^2} = |x|$.

If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$.

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As mentioned in comments: $\sqrt {x^2}=|x|$

Reason is that range of square root function is non-negative.

Assume that using $\sqrt {x^2}=x$ you write $\sqrt {(-2)^2}=-2$ which is clearly wrong.

I hope you get the point.

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  • $\begingroup$ Why is it wrong? $\endgroup$ – tatan Feb 19 '16 at 11:02
  • $\begingroup$ @tatan Because $\sqrt{(-2)^2}=\sqrt{2^2}$. Would you say that $\sqrt{2^2}=-2$? $\endgroup$ – egreg Feb 19 '16 at 11:04
  • $\begingroup$ @egreg but in the first answer,it is written $\sqrt {-x^2}=\sqrt {x^2}=-x$.... $\endgroup$ – tatan Feb 19 '16 at 11:06
  • $\begingroup$ @tatan I see nothing of that kind in any of the current answers. $\endgroup$ – egreg Feb 19 '16 at 11:08
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    $\begingroup$ @tatan You're wrong: “if $x<0$, then $\sqrt{x^2}=\sqrt{(-x)^2}=-x=|x|$”, which is fully correct. Parentheses are important. $\endgroup$ – egreg Feb 19 '16 at 11:11

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