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Before we discuss simplifying radicals

The largest source of student error when dealing with radicals is actually the easiest to avoid. To see what it is, consider the following question: does $\sqrt {x^2 + 9} = x + 9$?

There’s a number of ways we could answer this - maybe we remember from a previous class whether this works or not? Maybe we remember how square roots undoes square powers, so we try to cancel them? Maybe we Google it and see what we get for an answer.

But, often, the easiest option is to just plug in a few numbers to get an idea of whether we think it’s likely true or not. Note that this isn’t the same as proving whether or not it is true - this is sort of like “drawing a picture” when trying to model a problem... it helps build intuition and context for you to determine what the actual answer is.

For example, let’s say we plug in $x=0$. Then we would get $\sqrt {(0)^2 + 9} = \sqrt {9} = 3 = 0 + 3$. In other words, the equality checks out for $x=0$. So, does this mean the equality is true? Sadly... not so much.

If we instead try plugging in, say, $x=4$ then we get $\sqrt {(4)^2 + 9} = \sqrt {16 + 9} = \sqrt {25} = 5 \neq 7 = (4) + 3$. In other words, the expression on the lefthand side of the equals sign (the radical) evaluates to $5$ whereas the expression on the righthand side of the equals sign evaluates to $7$, so they are not equal and this equality doesn’t hold for $x=4$. ^[1] (Interestingly if we plug in any number other than zero, we will get the same result - which shows that zero is quite a special number here! In general it’s best to check at least 3 or 4 numbers, trying to include negatives, positives, and something ’weird’ like a fraction or $\pi$ or something similar, before you start to build intuition.)

But, if the equality is true, then it would have to be true for every $x$ value, and it clearly doesn’t work when $x = 4$ as we just showed. So clearly this equality isn’t true. This leads us to the important observation, which generates the most errors when dealing with radicals from students throughout all of precalculus and the calculus sequence: You cannot split a root over multiple terms! In other words, if your radicand is more than one term, then you cannot evaluate the radical in any way.

This is step 0 of simplifying any radical: Before you can simplify a radical you must first factor the radicand! Think of this like we did polynomials - whenever you wanted to find zeros of a polynomial, you had to factor first. Similarly, if you want to do anything with a radical ^[2] (There is one exception, called ’lifting’ the radical, which we discuss toward the end of the radical topic) you must first factor the inside.

Algebraic Radicals and how to simplify them.

Whether you have managed to factor a radicand that had multiple terms, or were lucky enough to have a factored radicand to begin with, once the radicand is a product of terms, you can now attempt to simplify the radical. In this case, each term should be viewed as a ‘factor’ of the radicand in the same way that primes were ‘factors’ of numeric radicals earlier. Specifically, we want to write each term as a piece that has a power equal to the root value, and then the ‘remainder’. The following example may be helpful to understand what is meant by this.

The good news is that the simplification process is the same as the numeric examples we’ve seen (albeit with a looser definition of ‘prime factor’). It should not be a total surprise that the factored form treats each (multiplicative) factor as a ‘prime factor’ in terms of simplifying the radical, similar to how we simplify numeric radicals. Indeed, we’ve already seen that these kinds algebraic factors (like “$(x-2)$”) are analogous to ‘prime-numbers’ when it comes to decomposing polynomials to their ‘most basic parts’, so it shouldn’t be surprising that they end up being treated similarly in other situations (like radicals) as well!

Unfortunately, the bad news is that the result isn’t quite as straightforward as in the case of the numeric radical case. To motivate the next part, we will first proceed with an example.

If we look closely at the previous example it may become apparent that the reason that $x = -2$ is also a solution, is that the $x^2$ in the original equation kills the negative sign. Thus, since $2$ is a solution, and the even power nullifies the effect of the negative sign, then $-2$ is also a solution (that is to say; since $2^2 = 4$ and $(-2)^2 = 4$, there is no difference between the two once we square them). But understanding why the solution is valid is not quite the same thing as figuring out why we didn’t find it the first time. After all if our methodology were correct, it should have given us all the valid solutions the first time around right? ^[3] (Turns out, this isn’t actually true in mathematics in general - lots of valid “solution” methods are very clear that they don’t actually give you all possible solutions. But that is also important. Part of a good solution method isn’t necessarily giving all solutions (although that’s ideal!) but rather being very explicit about the nature of the solutions it does provide - like when the method only gives some of the solutions and not all of them.)

So what’s the deal with even roots?

To understand why we failed to detect the solution of $-2$ to $\sqrt {x^2} = 2$, it is helpful to realize that this question is equivalent to solving the equality $x^2 = 4$. At it’s heart, then, the answer to why we couldn’t detect the $-2$ as a solution to $x^2 = 4$ comes down to the fact that a square root is intended to be a tool to isolate $x$ in the equality $x^2 = c$, but we also want the mechanism for this (the “square root” itself) to be a function.

And this immediately presents a problem. In our example of $x^2 = 4$ we determined that both $2$ and $-2$ are solutions. So, if we use a square root to ‘undo’ the power on $x$ we get $x = \sqrt {4}$. But we know that $x = 2$ and $x = -2$ are both solutions. So, on the one hand, the right hand side could be either $2$ or $-2$... and in fact it would need to be both if we want to make sure to get all the valid solutions. On the other hand we want the process of using square roots to be a function, and a function can only output a single answer. So we have a pretty big problem, either we lose half our answers, or we square rooting isn’t a function!

So, what do we do? There are a number of legitimate ways to address this issue, but it turns out mathematicians decided long ago that being a function is just way too useful to give up easily. As a result, they decided to simply pick a default answer. Since positives are generally easier to deal with than negatives, we decide by convention to have the square root return the positive valued answer. ^[4] (This decision of a ‘default’ actually has a special name in math, called the “principle branch” of the square root. This isn’t something you would likely hear again unless you take senior level math courses or math graduate courses though, so no need to remember or worry about it for now.)

Now, you might object at this point by saying that this doesn’t actually get around our original problem... and you’d be right! On the one hand if we relax the requirement on the square root operation to give both positive and negative values, it is no longer a function (which is all kinds of bad). On the other hand, if we only use the positive value result, we lose half of the ‘valid answers’ - which is also bad.

For the moment however, we sacrifice half our answers on the altar of function worship - meaning that we decide to make the square root (or really, any even root) to only output the positive answer. We’ll discuss how to account for the negative answers in the next segment, as it turns out there is still a way to recover the half of answers that we potentially lost by picking this “default answer” for even roots.

In the meantime however, this choice also impacts the process of simplifying algebraic radicals.

Ok... but really, what’s the deal with the even roots?

In order to simplify algebraic roots, we must treat even root values differently than odd root values. For even root values, when we try to simplify a term by evaluating the root, we can represent the fact that the output is only the positive value by using another function that only outputs positive values - the absolute value. For example, to evaluate something like $\sqrt {x^2}$ we wouldn’t want to write $x$ as it is not clear that we are ‘choosing’ the positive output. Instead we write $\sqrt {x^2} = |x|$.

In fact, since this entire issue stems from the fact that a negative raised to an even power becomes positive, this comes up whenever we simplify any even root-value radicals. A good rule of thumb is that everything that is pulled out of an even radical gets an absolute value. Unfortunately, it doesn’t end there though. Consider, for example, the following: $\sqrt [4]{16x^9}$. To solve this, we can do the same method we did with algebraic radicals above, with the added step that - since the root value is 4 (and thus even), anything we pull out gets absolute values.

Now, the above isn’t wrong, it is true that $\sqrt [4]{16x^9} = |2x^2|\sqrt [4]{x}$. Unfortunately, for reasons that will be much clearer in calculus, absolute values are actually incredibly annoying to deal with - indeed, we will get a little glimpse of the annoyance when we discuss absolutely values explicitly in the future. But as a consequence, although the equality is correct, it is not considered simplified or finished.

To see why, first recall (or maybe learn, if you hadn’t seen this before) that “the product of absolute values is the absolute value of the product” - in other words, you can split up products of absolute values as so: $|2x^2| = |2||x^2|$. But with very little thought, we should be able to figure out what $|2|$ is - and no, it’s not a trick question... it’s just $2$! ^[5] (Amusingly, this works if you read that as $2$ with exclamation, or if you read it as $2$ factorial. It’s a math-language pun! Not something you see often, but glorious when it appears!) Importantly, this means we can simply remove the absolute value around the $2$, because we already know what it is, so we could rewrite our equality as $\sqrt [4]{16x^9} = 2|x^2|\sqrt [4]{x}$. But, it turns out, this isn’t the best we can do!

Remember that this whole mess occurs because raising numbers to even powers obliterates negative signs. So the $x^2$ inside the absolute values is having the same result - the power is even, and thus even if $x$ itself is negative, $x^2$ is going to be positive, and so the absolute value isn’t doing anything! This means we can actually rewrite the original problem as: $\sqrt [4]{16x^9} = |2x^2|\sqrt [4]{x} = 2x^2\sqrt {x}$. Notice however, that the absolute values were originally necessary - because the root value was even. It was only the fact that we could justify, for each term inside those absolute values, that we didn’t actually need the absolute values (because the inside was already positive) that let us then remove the absolute values afterward.

With this, we have the final idea for simplifying even root-values. When simplifying even roots, you put absolute values around everything that comes out of the radical. Once you have finished pulling everything out that you can, you then go through everything inside the absolute values and remove the absolute values from anything that you can - typically because the term must be positive for some reason (most commonly because it is being raised to an even power).

As a final demonstration, consider the following: