Adding and Subtracting Radicals

Before we begin to simplify radical expressions, recall the following:

Adding and subtracting radicals is similar to adding and subtracting other algebraic expressions.

Recall that, to simplify \(4x+5x\), we identify and collect like terms (the \(x\)'s) and add their coefficients.

So, \(4\class{hl1}{x}+5\class{hl1}{x}=9\class{hl1}{x}\).

We can apply this same process to radical expressions:

• Identify the like terms.
• Add the coefficients.

We will see how this works in the following examples.

Example 1

Solution

This question is similiar to the earlier example, except instead of having like terms containing \(x\), we have like terms containing \(\sqrt{2}\).

We will 

• collect the like terms (the \(\sqrt{2}\)'s), and
• add their coefficients. 

So, \(4\class{hl1}{\sqrt{2}}+5\class{hl1}{\sqrt{2}}=9\class{hl1}{\sqrt{2}}\).

Example 2

Solution

We can still

• collect the like terms (the \(\sqrt{5}\)'s), and
• add and subtract their coefficients. 

So, \(7\class{hl1}{\sqrt{5}}+3\class{hl1}{\sqrt{5}}-6\class{hl1}{\sqrt{5}}=4\class{hl1}{\sqrt{5}}\).

Example 3

Solution

Notice that there are two sets of like terms in this example.

To simplify, we

• collect the like terms (the \(\sqrt{3}\)'s and \(\sqrt{6}\)'s), and

• add and subtract their coefficients.

\(\begin{align*} 5\sqrt{3}+6\sqrt{6}+\sqrt{3}-4\sqrt{6}&=5\class{hl1}{\sqrt{3}}+6\class{hl2}{\sqrt{6}}+1\class{hl1}{\sqrt{3}}-4\class{hl2}{\sqrt{6}}\\ &=5\class{hl1}{\sqrt{3}}+1\class{hl1}{\sqrt{3}}+6\class{hl2}{\sqrt{6}}-4\class{hl2}{\sqrt{6}}\\ &=6\class{hl1}{\sqrt{3}}+2\class{hl2}{\sqrt{6}} \end{align*}\)

In the examples that we have seen so far, the radicals in each question have been completely simplified (they have been written in the form \(a\sqrt{x}\), where \(x\) is a positive integer that has no perfect square factors). Let's look at an example where the radicals are not completely simplified.

Example 4

Solution

At first glance, it may appear that this expression has already been simplified as there are no like terms. However, notice that some of the radicals in this expression are not completely simplified!

• Begin by simplifying \(\sqrt{27}\) and \(\sqrt{48}\). \(\sqrt{3}\) is already simplified.

• Rewrite the expression with mixed radicals. We now have all like terms (\(\sqrt{3}\)).

\(\sqrt{3}-\sqrt{27}+\sqrt{48}=\sqrt{3}-3\sqrt{3}+4\sqrt{3}\)

• Collect like terms. Add and subtract their coefficients.

\(\begin{align*} \sqrt{3}-3\sqrt{3}+4\sqrt{3}=2\sqrt{3} \end{align*}\)

Example 5

Solution

As in the previous example, some of the radicals in this expression are not completely simplified!

Although we now have cube roots and not square roots, the process stays the same.

• Begin by simplifying \(\sqrt[3]{24}\). Note that \(\sqrt[3]{3}\) and \(\sqrt[3]{2}\) are already simplified since \(3\) and \(2\) do not have any perfect cube factors.

\(\begin{align*} \sqrt[3]{24}&=\sqrt[3]{8\times3}\\ &=\sqrt[3]{8}\times\sqrt[3]{3}\\ &=2\sqrt[3]{3} \end{align*}\)

• Rewrite the expression with mixed radicals.

\(\sqrt[3]{3}+\sqrt[3]{24}-\sqrt[3]{2}=\sqrt[3]{3}+2\sqrt[3]{3}-\sqrt[3]{2}\)

• Collect like terms (the \(\sqrt[3]{3}\)'s). Add their coefficients.

\(\begin{align*} \sqrt[3]{3}+\sqrt[3]{24}-\sqrt[3]{2}&=\sqrt[3]{3}+2\sqrt[3]{3}-\sqrt[3]{2}\\ &=3\sqrt[3]{3}-\sqrt[3]{2} \end{align*}\)