Regarding that $\sqrt{x^2}=|x|$, why $\sqrt{4} = 2$?

Question

I have a question regarding this video

https://www.khanacademy.org/math/algebra/exponent-equations/exponent-properties-algebra/v/adding-and-simplifying-radicals

He said that $\sqrt {x^2}=|x|$ has to be the absolute value of $x$ because $x$ can be positive or negative, which makes sense but why is $\sqrt 4= 2$? Shoudln't it be $|2|$ too? Because $-2 \cdot -2 = 4$ and $2 \cdot 2 = 4$.

Answer 1

$\left | 2 \right |=2$. Indeed $\sqrt{x^2}=\left | x \right |$ since $\sqrt{x^2}=x$ only for non-negative values of $x$. Consider for example this case:

$\sqrt{9}=\sqrt{(-3)^2}\neq -3$. But $\left | -3 \right |=3$ which is the correct answer.

Answer 2

I'd make this a comment if I had the reputation, but as it stands, just think about what $|2|$ is.

Answer 3

Remember the definition of $|x|$: $$ |x| = \left\{ \begin{array}{ll} x & \quad x \geq 0 \\ -x & \quad x < 0 \end{array} \right. $$ So what is $|-2|$ according to this definition? It is $|-2|=-(-2)=2$