# 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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 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

## 4 Answers

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