Multiplying Root Expressions

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Lesson Priority: Normal

 
Objectives
  • Review (once again) the root product property
  • Learn best practices for multiplying root expressions, with and without needing the distributive property
  • Understand why you cannot easily do any simplifying until the end, and practice simplifying answers completely
Lesson Description

We can use the root product property once again to better understand how to multiply root expressions. However, simplifying them is just as important as correctly multiplying them. Additionally, it is often only possible to simplify after you multiply, which is backward from many other variable multiplication processes in algebra, where it is often much much easier to simplify items on their own before multiplying. In this lesson we will practice both multiplying root expressions and simplifying our answers completely.

 
Practice Problems

Practice problems and worksheet coming soon!

 
You Should Know
This lesson covers square roots only. If you are looking for lessons on advanced roots of higher degrees, such as cube roots, you will find them in Algebra Two ».

Multiplying Roots by Roots

First and foremost, it's important to remember that multiplication is commutative in algebra - which means the order doesn't matter.Second, keep in mind that when you have a coefficient in front of the root, that each piece is a factor. e.g.$$2\sqrt{6} = 2 \times \sqrt{6}$$With those in mind, let's learn how to multiply roots and root expressions.The underlying rule that guides the process is called the root product property, which we saw when we learned how to simplify square roots ». We'll use it forward and backward.
The Root Product PropertyThe square root of a product is equal to the product of the square root of each factor, and vice versa. e.g.$$\sqrt{xy} = \sqrt{x}\sqrt{y}$$and$$\sqrt{x}\sqrt{y} = \sqrt{xy}$$
In fact, the root product property is just that - multiplication! That is indeed how you multiply roots - you combine them under the radical.
 
Example 1Multiply $\sqrt{2} \cdot \sqrt{3}$.$\blacktriangleright$ Use the root product property:$$\sqrt{2} \cdot \sqrt{3} = \sqrt{2 \cdot 3}$$$$=\sqrt{6}$$
 
Example 2Multiply $\sqrt{7a} \cdot \sqrt{5b}$.$\blacktriangleright$ Once again, combine the expressions underneath one radical, as the root product property tells us to do.$$\sqrt{7a} \cdot \sqrt{5b} = \sqrt{35ab}$$
 
Of course, if it stayed that simple, we'd hardly need a whole lesson on it.

Simplifying Root Multiplication

Most of the work involved with multiplying roots has to do with simplifying. It's common that we will multiply root expressions that cannot simplify on their own, but can once they are multiplied together.
Pro Tip
Whenever we're asked to multiply involving root expressions, we are supposed to give simplified form answers. Many times the instructions will tell us to do this, but some teachers simply expect it.
First multiply the roots using the root product property. Then simplify the resulting root expression, if possible.
 
Example 3Simplify $\sqrt{10}\cdot \sqrt{2}$.$\blacktriangleright$ Using the root product property,$$\sqrt{10}\cdot \sqrt{2} = \sqrt{20}$$However, we don't get full credit for this answer unless we try to reduce the radical to simplest form.$$\sqrt{20} = 2\sqrt{5}$$
 
I Used To Know That!
Simplifying radicals is a matter of looking for the largest perfect square factors to be able to pull out from under the radical. If you need a refresher on this, go back to the lessons on simplifying number-based roots » and simplifying variable-based roots ».
The complexity ramps up even more when we start to look at variable expressions. Teachers usually favor problems with multiple variables, and they also tend to load quizzes and tests with more of these problems than simpler ones with one or none.
 
Example 4Simplify $\sqrt{5xy^2} \cdot \sqrt{20x^3y^3}$.$\blacktriangleright$ Don't try to simplify each root on its own first. Use the root product property and then work on simplifying the resulting radical.$$\sqrt{5xy^2} \cdot \sqrt{20x^3y^3}$$$$=\sqrt{100x^4 y^5}$$To finish, we work on reducing the result to simplest form, like we recently learned » how to do. The coefficient happens to be a perfect square, so that will come out as $10$. The $x^4$ term is also a perfect square that will come out as $x^2$, but the $y^5$ term needs to have its biggest perfect square, $y^4$, factored from it.$$\sqrt{100x^4 y^5} = \sqrt{100x^4 y^4 \cdot y}$$$$=10x^2 y^2 \sqrt{y} \;\;\; \blacksquare$$
 

Distributing Roots

This isn't as common, but sometimes we'll need to clean up expressions that involve roots and the distributive property. As long as you're paying attention, everything you need to do should look familiar.
 
Example 5Multiply.$$\sqrt{3} \left( \sqrt{2} + \sqrt{6} \right)$$We have to distribute the $\sqrt{3}$ term (think about it as $a(b+c)$ if the roots are throwing you off), but we also have to simplify if possible.$$\sqrt{3} \left( \sqrt{2} + \sqrt{6} \right) = \sqrt{3 \cdot 2} + \sqrt{3 \cdot 6}$$$$= \sqrt{6} + \sqrt{18}$$$$=\sqrt{6} + 3\sqrt{2}$$
 
Example 6Multiply.$$\left( 4 + 3\sqrt{2} \right) \left(\sqrt{14} + \sqrt{7} \right)$$$\blacktriangleright$ Let's start with FOIL to fully distribute the multiplication.$$\longrightarrow 4\sqrt{14} + 4\sqrt{7} + 3\sqrt{28} + 3\sqrt{14}$$Right away we can combine some terms:$$\longrightarrow 7\sqrt{14} + 4\sqrt{7} + 3\sqrt{28}$$Next, we can simplify the last radical expression down, since it has a perfect square factor:$$\longrightarrow 7\sqrt{14} + 4\sqrt{7} + 3\sqrt{4 \cdot 7}$$$$\longrightarrow 7\sqrt{14} + 4\sqrt{7} + 6\sqrt{7}$$Lastly, we now have another like term to combine.$$\longrightarrow 7\sqrt{14} + 10\sqrt{7}$$
 

Put It To The Test

 
Example 7Simplify.$$\sqrt{5} \cdot \sqrt{30}$$
Show solution
$\blacktriangleright$ Multiply under the root.$$\sqrt{5} \cdot \sqrt{30} = \sqrt{150}$$We need to simplify this radical for the final answer:$$ = \sqrt{25 \cdot 6} = 5\sqrt{6}$$
 
Example 8Simplify.$$\sqrt{4xy} \cdot \sqrt{2xy^2}$$
Show solution
$\blacktriangleright$ Multiply under the root.$$\sqrt{4xy} \cdot \sqrt{2xy^2} = \sqrt{8x^2 y^3}$$Simplifying:$$= 2xy\sqrt{2y}$$
 
Example 9Simplify.$$\sqrt{3x^4y^3} \cdot \sqrt{21x^4 y}$$
Show solution
$\blacktriangleright$$$\longrightarrow \sqrt{63x^8 y^4}$$$$=3x^4 y^2 \sqrt{7}$$
 
Example 10Simplify.$$\sqrt{a} \left(\sqrt{a^3} + \sqrt{a^2 b^8}\right)$$
Show solution
$\blacktriangleright$ Distribute first, then simplify and combine like terms until you can't anymore.$$\sqrt{a^4} + \sqrt{a^3 b^8}$$$$\longrightarrow a^2 + ab^4 \sqrt{a}$$
 
Example 11Simplify.$$(2+\sqrt{5})^2$$
Show solution
$\blacktriangleright$ Make sure you FOIL here, and don't fall into the trap of incorrectly squaring a binomial. If it helps, write it out in expanded form first:$$(2+\sqrt{5})^2 = (2+\sqrt{5})(2+\sqrt{5})$$$$= 4 + 4\sqrt{5} + 5$$$$= 9 + 4\sqrt{5}$$
 
Lesson Takeaways
  • Continue practicing the root product property
  • Know how to multiply root expressions with other root expressions
  • Make sure radical answers are fully simplified
  • Be able to distribute and simplify FOIL-like expressions

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Special Notes

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