Adding and Subtracting Radical Expressions

Hello! Today we are going to take a look at adding and subtracting radical expressions. A radical expression is any expression that includes a radical, or a root. For example, some radical expressions are:

\(4\sqrt{2}, \sqrt{x+5}, \text{ and } \sqrt[3]{39}\)

 

Radical expressions can include numbers, variables, and fractions. For this video, we are going to focus on adding and subtracting simple square root radical expressions, but the same process applies to all radical expressions.

Let’s start by adding \(6\sqrt{3}\) and \(2\sqrt{3}\).

\(6\sqrt{3}+2\sqrt{3}\)

 

To add or subtract radical expressions, their radicals must be the same. If they aren’t, then the two terms are not like terms and cannot be added or subtracted.

In this example, our radicals are the same, so we can add their coefficients. \(6+2=8\), and our radical stays the same, so our answer is:

\(6\sqrt{3}+2\sqrt{3}=8\sqrt{3}\)

 

Let’s try another one.

\(14\sqrt{6}-9\sqrt{6}\)

 

Our radicals are the same for this problem as well, so all we have to do is subtract our coefficients:

\(14\sqrt{6}-9\sqrt{6}=5\sqrt{6}\)

 

But, what if we’re asked to solve something like \(25\sqrt{3}+6\sqrt{27}\)?

At first glance, it doesn’t look like this is possible because our radicals are not the same, but it turns out it is possible. All we have to do is simplify our second term. So, let’s look at this one for a second. We’re going to take a look at \(6\sqrt{27}\), specifically the root right here. What if we write this as, \(6\sqrt{9\times 3}\), because \(9\times 3=27\). So we break our 27 apart into factors.

\(6\sqrt{27}=6\sqrt{9\times 3}\)

 

We know that 9 is a perfect square, so we can pull this out of the square root and multiply it by our original coefficient. So we can do \(3\times 6\), because \(\sqrt{9}=3\), so make sure you don’t multiply by 9, but you multiply by \(\sqrt{9}\).

\(3\times 6\sqrt{3}=18\sqrt{3}\)

 

So now we can rewrite this problem as:

\(25\sqrt{3}+18\sqrt{3}\)

 

Now our radicals are the same, and we can simply add our coefficients.

\(25\sqrt{3}+18\sqrt{3}=43\sqrt{3}\)

 

Now let’s take a look at a subtraction problem.

\(8\sqrt{32}-2\sqrt{8}\)

 

Our radicals are not the same, so we need to start by simplifying them. In this case, both our radicals can be simplified. So let’s start with \(8\sqrt{32}\). We can break 32 apart into the factors 16 and 2, so we can write this as:

\(8\sqrt{32}=8\sqrt{16\times 2}\)

 

We can pull our 16 out because that’s a perfect square, the square root of 16 is 4, so we’ll have:

\(8\sqrt{16\times 2}=4\times 8\sqrt{2}\)

 

And then \(4\times 8=32\), so we have \(32\sqrt{2}\).

\(8\sqrt{32}=32\sqrt{2}\)

 

Now let’s take a look at \(2\sqrt{8}\). \(\sqrt{8}\) can be rewritten as \(\sqrt{4\times 2}\), so we’ll have \(2\sqrt{4\times 2}\).

\(2\sqrt{8}=2\sqrt{4\times 2}\)

 

4 is a perfect square, so we can square root this and get 2, and then we’ll multiply 2 times \(2\sqrt{2}\).

\(2\sqrt{4\times 2}=2\times 2\sqrt{2}\)

 

\(2\times 2=4\), so this is equal to \(4\sqrt{2}\).

\(2\sqrt{8}=4\sqrt{2}\)

 

And now we can subtract! So, we’ll take our \(8\sqrt{32}\), which became \(32\sqrt{2}\), minus our \(2\sqrt{8}\), which became \(4\sqrt{2}\).

\(32\sqrt{2}-4\sqrt{2}\)

 

And we subtract our coefficients, \(32-4=28,\) and our root stays the same. So we have \(28\sqrt{2}\).

\(32\sqrt{2}-4\sqrt{2}=28\sqrt{2}\)

 

And that’s all there is to it! Remember, two radicals must be the same in order to add or subtract their coefficients. I hope this video was helpful. Thanks for watching and happy studying!

Practice Questions

Question #1:

 
The expression \(5\sqrt{6}-8\sqrt{6}\) is equivalent to which of the following?

\(-\sqrt{6}\)
\(-3\sqrt{6}\)
\(3\sqrt{6}\)
\(2\sqrt{6}\)
Answer:

To combine two or more radical terms, the radicals for each term must be the same. Since both terms contain the radical \(\sqrt{6}\), we can combine their coefficients to get \(5-8=-3\). The radical portion of the result stays the same, so our answer is \(-3\sqrt{6}\).

Question #2:

 
The expression \(4\sqrt{5}+2\sqrt{45}\) is equivalent to which of the following?

\(6\sqrt{50}\)
\(38\sqrt{45}\)
\(10\sqrt{5}\)

The expression does not simplify.

Answer:

To combine two or more radical terms, the radicals for each term must be the same. The two radical terms do not contain the same radical. So, simplify the second term.

\(2\sqrt{45}=2\times\sqrt{9\times5}=2\times\sqrt9\times\sqrt5=2\times3\times\sqrt5=6\sqrt5\)

Now, we have \(4\sqrt{5}+6\sqrt{5}\). Since both expressions now contain the radical \(\sqrt{5}\), we can combine their coefficients to get [/latex]4+6=10[/latex]. The radical portion of the result stays the same, so our answer is \(10\sqrt{5}\).

Question #3:

 
The expression \(5\sqrt{63}+2\sqrt{28}\) is equivalent to which of the following?

\(19\sqrt{7}\)
\(7\sqrt{91}\)
\(34\sqrt{63}\)

The expression does not simplify.

Answer:

To combine two or more radical terms, the radicals for each term must be the same. The two radical terms do not contain the same radical. So, simplify each term.

\(5\sqrt{63}=5\times\sqrt{9\times7}=5\times\sqrt{9}\times\sqrt{7}=5\times3\times\sqrt{7}=15\sqrt{7}\)
\(2\sqrt{28}=2\times\sqrt{4\times7}=2\times\sqrt{4}\times\sqrt{7}=2\times2\times\sqrt{7}=4\sqrt{7}\)

Now, we have \(15\sqrt{7}+4\sqrt{7}\). Since both terms now contain the radical \(\sqrt{7}\), we can combine their coefficients to get \(15+4=19\). The radical portion of the result stays the same, so our answer is \(\).

Question #4:

 
A garden is in the shape of a rectangle. If the length of the garden is \(30\sqrt{10}\text{ feet}\) and the width is \(11\sqrt{40}\text{ feet}\), how much fence will be needed, in feet, to enclose the garden?

\(52\sqrt{10}\)
\(104\sqrt{10}\)
\(82\sqrt{50}\)
\(41\sqrt{50}\)
Answer:

To find out how much fence is needed to enclose the rectangular garden, we need to find the perimeter of garden. Use the formula for finding the perimeter, P, of a rectangle (where l is the length, and w is the width of the rectangle).

\(P=2l+2w\text{ or } P=l+l+w+w\)

Since the length is \(30\sqrt{10}\) feet and the width is \(11\sqrt{40}\) feet, we have:

\(P=30\sqrt{10}+30\sqrt{10}+11\sqrt{40}+11\sqrt{40}\)

To combine two or more radical terms, the radicals for each term must be the same. The radical terms do not all contain the same radical. So, simplify the width.

\(11\sqrt{40}=11\times\sqrt{4\times10}=11\times\sqrt4\times\sqrt{10}=11\times2\times\sqrt{10}=22\sqrt{10}\)

Now, we have:

\(P=30\sqrt{10}+30\sqrt{10}+22\sqrt{10}+22\sqrt{10}\)

Since each term now contains the radical \(\sqrt{10}\), we can combine their coefficients to get:

\(30+30+22+22=104\)

The radical portion of the result stays the same, so the perimeter is \(104\sqrt{10}\text{ feet}\). This means \(104\sqrt{10}\text{ feet}\) of fence is needed to enclose the garden.

Question #5:

 
Your friend calculates the distances you run on two consecutive days. The first day you ran \(3\sqrt8\text{ miles}\), and the second day you ran \(\sqrt{50}\text{ miles}\). What is the total number of miles you ran on both days?

\(14\sqrt{2}\)
\(4\sqrt{58}\)
\(3\sqrt{58}\)
\(11\sqrt{2}\)
Answer:

To find the total distance you ran, you need to combine the distance ran for both days:

\(3\sqrt{8}+\sqrt{50}\)

To combine two or more radical terms, the radicals for each term must be the same. The two radical terms do not contain the same radical. So, simplify each term.

\(3\sqrt{8}=3\times\sqrt{4\times2}=3\times\sqrt{4}\times\sqrt{2}=3\times2\times\sqrt{4}=6\sqrt{2}\)
\(\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\times\sqrt{2}=5\sqrt{2}\)

Now, we have \(6\sqrt{2}+5\sqrt{2}\) Since both expressions now contain \(\sqrt{2}\), we can combine their coefficients to get \(6+5=11\). The radical portion of the result stays the same, so you ran a total of \(11\sqrt{2}\text{ miles}\).

 

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by Mometrix Test Preparation | This Page Last Updated: May 13, 2022