# Why is $\sqrt{x^2}= |x|$ rather than $\pm x$? [duplicate]

Shouldn't the square root of a number have both a negative and positive root? According to Barron's, $\displaystyle \sqrt{x^2} = |x|$. I don't understand how.

• Search on this site to find the answer. Commented Jun 26, 2015 at 23:43
• Why should $\sqrt{x^2}=x$? There are two things that square to $x^2$, there is only one thing we call $\sqrt{x}$, you've got to pick one, it may as well be $|x|$. Commented Jun 26, 2015 at 23:43
• By convention, when we write $\sqrt{x}$ for some real, nonnegative $x$, we mean the positive square root. Commented Jun 26, 2015 at 23:43
• Here, we don't know if you mean $\sqrt{x}^2$ or $\sqrt{x^2}$. Use latex. This statement is wrong for the first but right for the other. Commented Jun 26, 2015 at 23:45
• The square root symbol denotes a function. A function has a single value (by convention the positive one is taken).
– user65203
Commented Jun 27, 2015 at 0:02

It is conventional that the notation $\sqrt x$ means the non-negative square root of $x$.

There are indeed two square roots of $x$, and for non-negative numbers $x$, only one of the two is conventionally denoted $\sqrt x$.

Don not confuse $x^2=a^2$ which is an equation that has two roots of opposite sign, $\pm\sqrt{a^2}$, and the expression $\sqrt{a^2}$, which is a positive number, equal to $|a|$.

• I think it's indeed a good point to distinguish between root function and equation Commented Jun 27, 2015 at 0:14

This is a very common question. Basically, people like to think, for example, that $\sqrt{9}$ is "the number such that when you square it, you get 9". So people think this must be $\pm 3$. But that's not what we are asking with square root. With $\sqrt{9}$, we are asking for "the positive number such that it squared equals $9$". That means $\sqrt{9} = 3$ (or $\sqrt{3^{2}} = |3|$).

The first (wrong) question applied to the square root should actually be used for solving the equation $x^{2} = 9$. To solve this, we need to "find the number such that it squared equals $9$", which means the solution is $x = \pm 3$.

• Wow, a downvote and no comment explaining why. Lazy &@$#! Commented Jun 27, 2015 at 0:00 because even if$x$is negative$x^2$is positive, that's why when you do$\sqrt{x^2}$you first evaluate$x^2$which is positive no matter if$x$is negative or positive. Then you apply the square root which is also positive and so that's why$\sqrt{x^2} = |x|\$ also notice. We evaluated the expression inside out.

$$\sqrt{\color{red}{inside}}$$