Hi Everyone!

On this page you will find some material about Lesson 20. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

Lesson 20: Division of Radicals and Rationalization


In this section you will find some important information about the specific resources related to this lesson:

  • the learning outcomes,
  • the section in the textbook,
  • the WeBWorK homework sets,
  • a link to the pdf of the lesson notes,
  • a link to a video lesson.

Learning Outcomes.

  • Divide radical expressions.
  • Find the conjugate of a radical expression.
  • Rationalize an expression with a radical term in the denominator.
  • Simplify radical expressions.

Topic. This lesson covers Section 6.6: Division of Radicals and Rationalization.

WeBWorK. There is one WeBWorK assignment on today’s material:


Lesson Notes.  

Video Lesson.

Video Lesson 20 (based on Lesson 20 Notes)

Warmup Questions

These are questions on fundamental concepts that you need to know before you can embark on this lesson. Don’t skip them! Take your time to do them, and check your answer by clicking on the “Show Answer” tab.

Warmup Question 1



Show Answer 1

$$(2\sqrt{5a})(-4\sqrt{5a})= -8\sqrt{(5a)^2}=-8(5a)=-40a$$

Warmup Question 2


$$(\sqrt 3-6)(\sqrt 3+6).$$

Show Answer 2

$$(\sqrt 3-6)(\sqrt 3+6)= (\sqrt 3)^2-6^2 = 3 – 36 = -33$$


If you are not comfortable with the Warmup Questions, don’t give up! Click on the indicated lesson for a quick catchup. A brief review will help you boost your confidence to start the new lesson, and that’s perfectly fine.

Need a review? Check Lesson 19.

Quick Intro

This is like a mini-lesson with an overview of the main objects of study. It will often contain a list of key words, definitions and properties – all that is new in this lesson. We will use this opportunity to make connections with other concepts. It can be also used as a review of the lesson.

A Quick Intro to Division of Radicals and Rationalization

Key Words. Radicals, division of radicals, simplified form, rationalization, conjugate.

When dividing radical terms, the following property can be very helpful.

Division Property


If not, it may be necessary to rationalize the denominator. On Lesson 18 we listed three conditions for a radical expression to be in simplified form. The third one is:

There should be no radicals in the denominator of a fraction.

$\bigstar$ We gave $\dfrac{1}{\sqrt 2}$ as an example that fails this condition. To simplify it, we multiply both the numerator and the denominator by $\sqrt 2$.

$$\underbrace{\dfrac{1}{\sqrt 2}}_{\text{radical in the denominator}}=\dfrac{1}{\sqrt 2}\cdot\dfrac{\sqrt 2}{\sqrt 2}=\dfrac{1\cdot\sqrt 2}{\sqrt 2\cdot\sqrt 2}=\underbrace{\dfrac{\sqrt 2}{2}}_{\text{no radical in the denominator}}$$

$\bigstar$ The process of removing a radical from the denominator is called rationalization.

$\bigstar$ The key idea was to multiply the original denominator by another copy of it, since squaring eliminates the radical.

$$(\sqrt a)(\sqrt a) = (\sqrt a)^2=a.$$

$\bigstar$ But what if we have $\dfrac{1}{\sqrt 3 -6}$? Squaring $\sqrt{3}-6$ will not help (try it!). In the Warmup Question #2 we saw that

$$(\sqrt 3 -6)(\sqrt 3 +6)=-33$$

results in a number free of radical. So

$$\underbrace{\dfrac{1}{\sqrt 3 -6}}_{\text{radical in the denominator}}=\dfrac{1}{\sqrt 3 -6}\cdot\dfrac{\sqrt 3 +6}{\sqrt 3 +6}$$

$$=\dfrac{1\cdot(\sqrt 3 +6)}{(\sqrt 3 -6)\cdot(\sqrt 3 +6)}=\underbrace{-\dfrac{\sqrt 3 +6}{33}}_{\text{no radical in the denominator}}$$

$\bigstar$ This happens because the above product is a difference of squares


and squaring a single radical eliminates the radical.

$\bigstar$ We say that $a-b$ and $a+b$ are conjugates. So if the denominator is $\sqrt 3 -6$, we rationalize it by multiplying the numerator and the denominator by its conjugate $\sqrt 3+6$.

Video Lesson

Many times the mini-lesson will not be enough for you to start working on the problems. You need to see someone explaining the material to you. In the video you will find a variety of examples, solved step-by-step – starting from a simple one to a more complex one. Feel free to play them as many times as you need. Pause, rewind, replay, stop… follow your pace!

Video Lesson

A video lesson on Division of Radicals and Rationalization [9:59]

A description of the video

In the video you will see the following radical expressions.

  • $\dfrac{3\sqrt x+1}{2\sqrt x}$
  • $\dfrac{3\sqrt 5 + 1}{2\sqrt 7 -3}$

Try Questions

Now that you have read the material and watched the video, it is your turn to put in practice what you have learned. We encourage you to try the Try Questions on your own. When you are done, click on the “Show answer” tab to see if you got the correct answer.

Try Question 1

Simplify $$\dfrac{\sqrt{15}}{5\sqrt{20}}.$$

Show Answer 1

$$\dfrac{\sqrt{15}}{5\sqrt{20}} = \dfrac{\sqrt{15}}{5\sqrt{20}}\cdot \dfrac{\sqrt{20}}{\sqrt{20}}$$

$$= \dfrac{\sqrt{15}\sqrt{20}}{5\cdot 20} = \dfrac{\sqrt{15} \cdot 2\sqrt{5}}{5\cdot 20}$$

$$= \dfrac{\sqrt {5\cdot 15}}{5\cdot 10} = \dfrac{5\sqrt {3}}{50} = \dfrac{\sqrt 3}{10}$$

Try Question 2

Simplify $$\dfrac{3}{4+2\sqrt 5}.$$

Show Answer 2

$$\dfrac{3}{4+2\sqrt 5}=\dfrac{3}{4+2\sqrt 5}\cdot \dfrac{4-2\sqrt 5}{4-2\sqrt 5}$$

$$= \dfrac{3(4-2\sqrt 5)}{(4+2\sqrt 5)(4-2\sqrt 5)}= \dfrac{12-6\sqrt 5}{16-20}$$

$$=\dfrac{12-6\sqrt 5}{-4} = \dfrac{-6+3\sqrt 5}{2} $$

Try Question 3

Rationalize and simplify

$$\dfrac{1+3\sqrt 2}{1-\sqrt 2}+\sqrt 2.$$

Show Answer 3

$$\dfrac{1+3\sqrt 2}{1-\sqrt 2}+\sqrt 2= \dfrac{1+3\sqrt 2}{1-\sqrt 2}\cdot\dfrac{1+\sqrt 2}{1+\sqrt 2}+\sqrt 2$$

$$=\dfrac{(1+3\sqrt 2)(1+\sqrt 2)}{(1-\sqrt 2)(1+\sqrt 2)} +\sqrt 2 = \dfrac{1+\sqrt 2+3\sqrt 2 +3\cdot 2}{1-2}+\sqrt 2 $$

$$=\dfrac{7+4\sqrt 2}{-1}+\sqrt 2 = -7-4\sqrt 2+\sqrt 2 = -7-3\sqrt 2$$


You should now be ready to start working on the WeBWorK problems. Doing the homework is an essential part of learning. It will help you practice the lesson and reinforce your knowledge.


It is time to do the homework on WeBWork:


When you are done, come back to this page for the Exit Questions.

Exit Questions

After doing the WeBWorK problems, come back to this page. The Exit Questions include vocabulary checking and conceptual questions. Knowing the vocabulary accurately is important for us to communicate. You will also find one last problem. All these questions will give you an idea as to whether or not you have mastered the material. Remember: the “Show Answer” tab is there for you to check your work!

Exit Questions

  • What is the goal in rationalizing the denominator? 
  • Why does the ‘conjugate’ play a role in accomplishing this?

$\bigstar$ Simplify

(a) $\dfrac{3-3\sqrt{3a}}{4\sqrt{8a}}$

(b) $\dfrac{\sqrt{5}+3}{4-\sqrt 5}$

Show Answer

(a) $$\dfrac{3-3\sqrt{3a}}{4\sqrt{8a}}=\dfrac{3-3\sqrt{3a}}{4\cdot 2 \sqrt{2a}} $$

$$= \dfrac{3-3\sqrt{3a}}{8\sqrt{2a}}\cdot\dfrac{\sqrt{2a}}{\sqrt{2a}} =\dfrac{(3-3\sqrt{3a}) \sqrt{2a}}{8\cdot 2a} $$

$$= \dfrac{3\sqrt{2a}-3\sqrt{6a^2}}{16a} =\dfrac{3\sqrt{2a}-3a\sqrt 6}{16a}$$

(b) $$\dfrac{\sqrt{5}+3}{4-\sqrt 5}= \dfrac{\sqrt{5}+3}{4-\sqrt 5}\cdot \dfrac{4+\sqrt 5}{4+\sqrt 5} $$

$$= \dfrac{(\sqrt{5}+3)(4+\sqrt 5)}{(4-\sqrt 5)(4-\sqrt 5)}=\dfrac{4\sqrt{5}+5+12+3\sqrt 5}{16-5}$$

$$ =\dfrac{7\sqrt 5+17}{11}$$

Need more help?

Don’t wait too long to do the following.

  • Watch the additional video resources.
Additional video resources on Division of Radicals and Rationalization
  • Talk to your instructor.
  • Form a study group.
  • Visit a tutor. For more information, check the tutoring page.