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 number $n > 0$ has the form $n = 2^k \cdot 5^l \cdot r$, where $k, l$ are non-negative integers and $r$ is a positive integer not divisible by $2$ and $5$. Let $m = \min(k,l)$. Then $n$ has exactly $m$ tagging zeros because each tagging zero corresponds to a factor $10 = 2\cdot 5$. Let $n$ have an integer square root $s$. Write $s = 2^{k'} \cdot 5^{l'} \cdot r'$ as above. Then $n = s^2 = 2^{2k'} \cdot 5^{2l'} \cdot (r')^2$. Since $r'$ is not divisible by $2$ and $5$, also $(r')^2$ is not divisible by $2$ and $5$. We conclude $k = 2k'$ and $l = 2l'$, hence $m = \min(2k',2l') = 2 \min(k',l')$ is even.
Best Answer
Your new teacher is wrong. $\sqrt{\cdot}$ is the principal square root operator. That means it returns only the principal root -- the positive one. $\sqrt{64}=8$. It does NOT equal $-8$.
On the other hand, the equation $64=x^2$ DOES have $2$ solutions: $x=8$ or $x=-8$. Thus both $8$ and $-8$ are square roots of $64$.
Let's see what happens when we take the principal square root of both sides of this equation: $$\begin{align}64 &= x^2 \\ \implies \sqrt{64} &= \sqrt{x^2} \\ \implies 8 &= |x| \\ \implies x&=8 \text{ or } x=-8\end{align}$$
Thus the fact that the principal square root operation throws out the negative root isn't much of a problem as the math still works out correctly.