Sometimes a simple question like what is the square root of 64 has an answer that can confuse a few. In this case, we will dispel a couple of myths.

The main objective in this tutorial is to learn a few things about square roots and radicals, so that you are able to answer questions about it without hesitation.

First thing is first. Let us spell out the definition of the square root:

The square root of a given number is the positive number (or zero) so that when squared results in that given number .

That is it. So, given a number \(x\), its square root is a number \(b\) so that \(b \ge 0\) and

By looking at the above expression, we can see that if \(b\) is going to be the square root of \(x\), then \(x = b^2\), and since a square number cannot be negative, \(x\) can only be non-negative (if we want to be able to find its square root).

Conclusion : We can only compute square roots of non-negative values \(x\). Or said differently, the domain of the function \(\sqrt x\) is \([0,+\infty)\).

So then, answering our initial question: What is the Square Root of 64?

Based on what we defined, we need to find a non-negative value \(b\) so that \(b^2 = 64\). Any number meeting those properties come to mind?

Well, yes, what if we tried with \(b = 8\)? Ok, so \(b = 8\) is non-negative, and \(b^2 = 8^2 = 64\).

So then, we have found the square root of 64, which is 8, because 8 is non-negative, and \(8^2 = 64\). We write this as:

The Myth About the Square Root Function

Now we go to the topic that motivated this tutorial... The above definition given of the square root allows us to discard the common statement that "the square root of 64 is plus or minus 8" , which is wrong. Indeed

Now, we can understand why such myth carries on. Indeed, both 8 and -8 have the property that \(8^2 = 64\) and \((-8)^2 = 64\). So then, why is -8 NOT the square root of 64?

Because by definition, we said that the square root needs to be that non-negative number that has the property that when squared they equal the given number. And -8 fails the condition of being non-negative.

The Graph of the Square Root Function

Look at the graph of the square root function below:

As you can see, that function only takes non-negative values, and it actually passes the vertical line test, so it is a function.

So in the end, the definition of the square root as the non-negative \(b\) so that \(b^2 = x\) makes the square root a function.

If indeed we had that \(\sqrt{64} = \pm 8\), then \(\sqrt x\) would not be a function, would be a relation instead, because the vertical line at \(x = 64\) would cross the graph twice (at 8 and -8).

What About Other Radical Functions?

There are other types of radical functions. For example, the cubic root \(\sqrt[3] x\). In this case, there is no need to make a rule for what radical to choose from, because the cubic root of a given number \(x\) is the number \(b\) so that \(b^3 = x\).

Cubic Root

For the cubic root case, there is no need to make distinctions because for a given \(x\) there will be just ONE number \(b\) such that \(b^3 = x\).

For example

simply because \(4^3 = 64\). Or

simply because \((-4)^3 = -64\). This is, there is no ambiguity like in the case of the square root.

Quartic Root

For the quartic root case, it is similar to the square root. We will have that \(\sqrt[4] x = b\) if \(b \ge 0\) and \(b^4 = x\).

For example

because \(2^4 = 16\) and \(2 \ge 0\). But

because although \((-2)^4 = -16\), we have that \(-2 < 0\) so then the non-negativity condition is not met.

How about for the n-th root \(\sqrt[n] x\) in general???.

I am sure you guessed it.

For \(n\) even, the situation is like the square root: \(\sqrt[n] x = b\) if \(b \ge 0\) and \(b^n = x\).

For \(n\) odd, the situation is like the square root: \(\sqrt[n] x = b\) if \(b^n = x\).