Let $n \in \mathbb N$ be a natural number and $a \in \mathbb R$ be a real number. The $n$-th root of the number $a$ is defined as follows:
Case I: $n$ is an odd number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \in \mathbb R$ such that $b^n = a$.
Case II: $n$ is an even number. In this case the $n^{\text{th}}$ root of $a$ is defined to be that number $b \geq 0$ such that $b^n = a$.
Why is it that when $n$ is even, we only consider $b \geq 0$. For example, both $+2$ and $-2$ squared equal $4$, but when we say the square root of $4$ is $2$. Is there a reason for this?
To make complete sense of this, we need to look at things in in the complex plane. When we choose to take the positive square root, we are just choosing a particular branch cut of the complex logarithm. In general, given a number $x\neq 0$, $x$ will have $n$ distinct $n^{th}$ roots.
You noticed that $-2$ and $2$ are both square roots of $4$. What about the cube roots of $2^{\frac{3}{2}}$? Well we have $\sqrt{2}$, but is that the only one? No, we also have $-1+i$ and $-1-i$. Similarly for the fourth root of $16$. We have the roots $2$ and $-2$, but we also have $2i,-2i$.
In general, when we take the $n^{th}$ root of a real number, we get $n$ different possibilities in the complex plane. In other words, taking $n^{th}$ roots is a multivalued function, so we have to make a choice, and this corresponds to choosing a branch cut for the logarithm function.
For most situations we choose the branch which sends the positive real line to the positive real line, but there are sometimes where we need to choose a different branch. (For example, evaluating certain integrals by complex methods.)
Hope that helps,