Modulus and square root

Question

How is this statement true? "For any real number $x$ we have $\sqrt{x^2} = |x|$"? Because putting $x=2$ $\sqrt{x^2}$ gives BOTH $2$ and $-2$ But $|x|$ only gives $2$

Answer 1

Remember that by definition and for $x \in \mathbb{R}$ $$|x| = \begin{cases} x, & \mbox{if } x \ge 0 \\ -x, & \mbox{if } x < 0. \end{cases}$$

Answer 2

Let a number $n \in \mathbb{R}$, Then
$$\sqrt{n} \geq 0.$$

Answer 3

Remember that the square root of $y\ge 0$ is the $x\ge 0$ such that $x^2=y$, not only the $x\in\mathbb{R}$ such that $x^2=y$.

For example:

$$\sqrt{3^2}=\sqrt{9}=3=|3|,$$

but

$$\sqrt{(-3)^2}=\sqrt{9}=3=|-3|.$$