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

Question

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.

Answer 1

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 $\sqrt x$.

Answer 2

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 $x = \pm 3$.

Answer 3

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 $|a|$.

Answer 4

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.

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