Here is a short answer to the question in the title of OP:
Well, if we don't do so, what could a better alternative be?
What is the notation $\sqrt{}$?
The confusion seems to be from understanding of the notation $\sqrt{}$. When writing, for instance $\sqrt{16}$, one pronounces it as "square root of $16$". However, what one really means is "the principal square root of $16$".
Let's go back to the definitions. A square root of a real number $a$ is a number $y$ such that $y^2 = a$; in other words, a number $y$ whose square is $a$. For example, $4$ and $−4$ are square roots of $16$ because $4^2=(-4)^2=16$. Note carefully that the notation $\sqrt{}$ is not involved in this definition at all.
Now, for every given positive real number, say $16$ again, there are two "square roots" (note carefully again that we don't write $\sqrt{x}$ for "square roots of $x$" yet) of it. What if one wants specifically to refer to the positive one? Instead of explicitly saying "I'm refering to the positive square root of $16$", one uses the notation $\sqrt{}$ to define $\sqrt{16}$ as the positive square root of $16$. Here comes the notation $\sqrt{}$. Of course you are losing "information" when you write $\sqrt{16}$ to mean "the positive square root of $16$". Because it is by definition so. What does one do for the "lost information"? One naturally has $-\sqrt{16}$ as the negative square root of $16$.
One can put two definitions together to see what is really going on:
A "square root" of a real number $a$ is a number $y$ such that $y^2=a$;
Given a positive real number $x$, the notation $\sqrt{x}$ is defined as a positive real number $y$ such that $y^2=x$. And in this case, we write $y=\sqrt{x}$.
Why is $\sqrt{}$ defined in the way above?
If one does not define $\sqrt{a}$ as the positive square root of $a$ and instead as the "square roots of $a$", then one would have $\sqrt{16}=\pm 4$. Now how would you write the answer to the following question?
What is the positive real number $x$ such that $x^2=\pi$?
[Added: ]Compare the following two possible definitions for the notation $\sqrt{}$:
- I. For any positive real number $a$, define $\sqrt{a}$ as the square roots of $a$;
- II. For any positive real number $a$, define $\sqrt{a}$ as the positive square root of $a$;
Now, if one uses definition I, then $\sqrt{16}=\pm4$. With this definition, you have perfectly what you might want:
$$
x^2=16\Rightarrow x=\pm 4;\quad\text{and }x=\sqrt{16}=\pm4.
$$
If one uses definition II instead, on the other hand, one would have $\sqrt{16}=4$.
You might be happier with definition I and ask why on earth one prefers II. Here is "why". Suppose you are asked to solve the following problem.
Find the solution to the equation $x^2-\pi=0$ such that $x>0$.
If one uses definition II, then one immediately has $x=\sqrt{\pi}$.
Now if one uses definition I, $x=\sqrt{\pi}$ would be the WRONG answer.
One more lesson from Terry Tao:
It’s worth bearing in mind that notation is ultimately an artificial human invention, rather than an innate feature of the mathematics one is working on; sometimes, two writers happen to use the same symbol to denote two rather different concepts, but this does not necessarily mean that these concepts have any deeper connection to them.
The definition (or one possible definition) of the absolute value of the complex number $a+bi$ (where $a$ and $b$ are real) is $\sqrt{a^2+b^2}$. So you ask, why is it not $\sqrt{a^2+(bi)^2}$ instead? The answer is that this is simply how we choose to define it. You could define a different quantity which is $\sqrt{a^2+(bi)^2}$, but you would have to give a different name to it because everyone else has already agreed that "absolute value" means to take $\sqrt{a^2+b^2}$ instead.
Now, a more interesting question is why everyone else decided on that definition. One reason is that you can represent complex numbers as points in the plane by letting $a+bi$ correspond to the point $(a,b)$, and then $|a+bi|$ is the distance from this point to the origin. Note that when you do this, $(a,b)$ is just a point in the ordinary Euclidean plane: it is a point that is $a$ units to the right of the origin and $b$ units above the origin. It doesn't make sense to say that the vertical distance is $bi$, since in geometry when we measure distances they are always positive real numbers. The vertical distance is $b$ because you have moved $b$ units vertically in the plane. (Actually, this is only accurate if $b$ is positive: if $b$ is negative, you have moved $-b$ units down, and so the distance is $-b$ rather than $b$. But you end up squaring this quantity when you use the Pythagorean theorem, so it doesn't matter if it's negative.)
Now ultimately this explanation is not very satisfying, because it doesn't explain why we chose to represent complex numbers in the plane this way. For instance, why don't we choose to represent them such that the complex number $i$ corresponds to a vertical distance different from $1$, or a distance in some direction other than vertical? One answer is that choosing $i$ to mean "go up one unit" happens to make distance have nice algebraic properties we would like absolute values to have. For instance, for real numbers, it is true that $|xy|=|x||y|$. Defining absolute values of complex numbers by $|a+bi|=\sqrt{a^2+b^2}$, it turns out that this is true for complex numbers as well. As a simple example, if we want $|xy|=|x||y|$ to be true for complex numbers, then we should have $|i|^2=|i^2|=|{-1}|=1$. So we should define $|i|=1$ or $-1$, and it is sensible to define it to be $1$ instead of $-1$ since absolute values are supposed to be positive.
Best Answer
Yes, $\vert x\vert$ means the outcome is always non-negative. This requires clarification because $x$ itself may be negative. Here are a few examples:
$$\sqrt{\color{blue}{+4}^2} = \sqrt{16} = \color{blue}{+4}$$
$$\sqrt{\color{blue}{+9}^2} = \sqrt{81} = \color{blue}{+9}$$
$$\sqrt{(\color{blue}{-4}^2)} = \sqrt{16} = \color{blue}{+4} = -(\color{blue}{-4})$$
$$\sqrt{(\color{blue}{-9}^2)} = \sqrt{81} = \color{blue}{+9} = -(\color{blue}{-9})$$
So, when $x > 0$, then $\sqrt{x^2}$ will return $x > 0$, so it equals $x$. However, when $x < 0$, then $\sqrt{x^2}$ will return $x > 0$ as well, so it equals $-x$. Hence,
$$\sqrt{x^2} = \vert x\vert = \begin{cases} x; \quad x \geq 0 \\ -x; \quad x < 0 \end{cases}$$
This is all just a way of saying $\sqrt{x^2}$ will always return a non-negative value.