Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

Last year in Pre-Algebra we learned about square roots. I was taught then that $\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get. But this year, with a new school and teacher in a different state, our teacher is telling us that: $\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is: $+\sqrt{64}=8$

And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as $y^2=n$ I see where he is coming from.

Upon researching this Wikipedia says:

For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$

And Wolfram MathWorld says:

Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$

But on the other side, Wolfram Alpha, when given "The square root of 9" gives only 3.

So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$?

Solution 1:

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.

Solution 2:

Both of your professors are right. It is just an issue of notation.

The first professor defines $\sqrt{x}$ as the non-negative number that when multiplied by itself is equal to $x$, if any.
Your second professor defines $\sqrt{x}$ as the numbers that when multiplied by themselves are equal to $x$, if any.

This means that they are using the same symbol $\sqrt{x}$ to convey different concepts.
It would be better if everyone used the same words and symbols for the same concepts. But in maths as in other issues in life you will find different people using the same word or symbol for different concepts.

Since he is the professor you will have to respect his authority regarding the choice of notation. There is no significative gain between one or another notation but it is very important to chose a notation so that everyone is on the same page. And the one chosing the notation in an academic environment will be the professor. It is unfortunate that different professors of the same institution chose different notations but you will have to live with it.

Solution 3:

$\sqrt{\cdot}:[0,\infty)\to [0,\infty)$ is a function that to each $x\ge0$ assigns a $y\ge 0$ such that $y^2=x$. A very different thing is the set of solutions of the equation $x^2=9$, for example. In fact the only reason we have a canonical square root function in $\mathbb{R}$ is because $\mathbb{R}$ is often considered to have a total order $<$ that let's you pick a solution of the equation $x^2=9$. If you were doing only algebra (i.e. no order relation), $\sqrt{\cdot}$ might not be definable.