Why is $\sqrt{x^2}$ always $x$? [duplicate]

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?)

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

• It's actually $|x|$. – Stefan Perko Feb 19 '16 at 10:49
• That makes a bit more sense.. but why? – Aayush Agrawal Feb 19 '16 at 10:51
• See en.wikipedia.org/wiki/Square_root#Properties_and_uses Square root function is defined to be non-negative – SS_C4 Feb 19 '16 at 10:51
• If $x\geq 0$, then $\sqrt{x^2} = x = |x|$. If $x<0$, then $\sqrt{x^2} = \sqrt{(-x)^2} = -x = |x|$. – Stefan Perko Feb 19 '16 at 10:52
• 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 $a\ge 0$). – user236182 Feb 19 '16 at 10:55

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 $\sqrt{xy}=\sqrt{|x|}\sqrt{|y|}$.

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.

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

Actually, $\sqrt{x^2} = |x|$.

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

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.

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