Hi Everyone!

On this page you will find some material about Lesson 21. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 21: Solving Radical Equations**

**Learning Outcomes.**

- Solve a radical equation.
- Check the potential solutions.

**Topic**. This lesson covers Section 6.7: Solving Radical Equations.

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

RadicalEquations

**Lesson Notes.**

**Video Lesson. **

Video Lesson 21 (based on Lesson 21 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

Expand

$$(3-x)^2.$$

#### Show Answer 1

$$(3-x)^2=(3-x)(3-x)=9-3x-3x+x^2=9-6x+x^2$$

#### Warmup Question 2

(a) If $\sqrt{x}=7$, what is $x$?

Which arithmetic operation did you perform to isolate $x$?

(b) Use the same arithmetic operation to solve

$$\sqrt{x} = 2a$$

where $a$ is positive.

#### Show Answer 2

(a) Squaring $\sqrt{x}=7$ isolates $x$.

$$\sqrt{x} = 7$$

$$(\sqrt{x})^2 =7^2$$

$$x=49$$

The solution set is $\{49\}$.

(b) Squaring

$$\sqrt{x} = 2a$$

gives

$$(\sqrt{x})^2 =( 2a)^2$$

$$x=4a^2$$

The solution set is $\{4a^2\}$.

#### Warmup Question 3

Solve

$$x^2=-2x+35.$$

#### Show Answer 3

$$x^2=-2x+35$$

$$x^2+2x-35=0$$

$$(x+7)(x-5)=0$$

$$x+7 = 0 \quad\text{ or }\quad x-5=0$$

$$x=-7\quad\text{ or }\quad x=5$$

The solution set is $\{-7,5\}$.

### Review

*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 ?? and Lesson ???.

### 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 Radical Equations**

**Key Words.** Radical equation, solution, squaring.

The first step in solving a radical equation is to get rid of the radical term. We can do that by using the following property from Lesson???:

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

So if the radical equation is

$$\sqrt{x-1}=2,$$

we simply square both sides

$$(\sqrt{x-1})^2=2^2$$

$$x-1=4$$

$$x=5$$

The potential solution is $x=5$. We will check it.

$\bullet$ **Check**:

$$\sqrt{x-1}\stackrel{?}{=}2$$

$$\sqrt{5-1}\stackrel{?}{=}2$$

$$\sqrt{4}\stackrel{?}{=}2\quad\checkmark$$

The solution set is $\{5\}$.

$\bigstar$ The squaring method works best if the radical term is **isolated**. For example, for the equation

$$\sqrt x + 1 =3,$$

squaring $\sqrt x +1$ gives

$$(\sqrt{x}+1)^2 = x+2\sqrt x +1,$$

and we still have a radical term; so that is not helpful. Watch the video for more examples of this type.

### Try Questions I

*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

Solve

$$\sqrt x + 1 =3.$$

#### Show Answer 1

$$\sqrt x + 1 =3$$

$$\sqrt x =3-1$$

$$\sqrt x =2$$

$$(\sqrt x )^2 =2^2$$

$$x=4$$

The potential solution is $x=4$. We will check it.

$\bullet$ **Check**:

$$\sqrt 4 + 1 \stackrel{?}{=}3$$

$$2+1 \stackrel{?}{=}3$$

$$3 =3\quad\checkmark$$

The solution set is $\{4\}$.

### 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 description of the video**

In the video you will see how to solve

- $\sqrt x = 4$
- $\sqrt {x-1}=2$
- $\sqrt {x-1}+3=15$
- $x-\sqrt{3x-5}=1$
- $\sqrt x =-2$
- $2=x-\sqrt{6-5x}$
- $x-2=2\sqrt{x+1}-x$

### Try Questions II

*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 2

Solve $$\sqrt{y-4}+y=4.$$

#### Show Answer 2

$$\sqrt{y-4}+y=4 $$

$$\sqrt{y-4}=4-y $$

$$(\sqrt{y-4})^2=(4-y)^2 $$

$$y-4= (4-y)(4-y) $$

$$y-4=16 – 8y + y^2 $$

$$0 = y^2-9y+20 $$

$$y^2-9y+20 = 0 $$

$$(y-4)(y-5) = 0 $$

$$y=4 \quad\text{ or } \quad y=5$$

The potential solutions are $y=4$ and $y=5$. We will check them now.

$\bullet$ **Check:** $y=4$

$$\sqrt{y-4}+y\stackrel{?}{=}4 $$

$$\sqrt{4-4}+4\stackrel{?}{=}4 $$

$$0+4\stackrel{?}{=} 4 $$

$$4 = 4 \qquad \checkmark$$

$\bullet$ **Check**: $y=5$

$$\sqrt{y-4}+y\stackrel{?}{=}4 $$

$$\sqrt{5-4}+5\stackrel{?}{=}4 $$

$$1+5 \stackrel{?}{=} 4 $$

$$6 \neq 4 \qquad \text{ False }$$

So $y=5$ is not a solution. The solution set is $\{4\}$.

### WeBWorK

*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.*

**WeBWork**

It is time to do the homework on WeBWork:

RadicalEquations

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

- When we are trying to solve a radical equation, what are we trying to accomplish?
- Why do we isolate the term with the radical?
- Is it necessary to check our answers even if we know we did not make a mistake? Give an example to support your answer.

$\bigstar$ Solve the equation $2\sqrt{p-3} -3 = -p$.

#### Show Answer

$$2\sqrt{p-3} -3 = -p$$

$$2\sqrt{p-3} = -p+3$$

$$(2\sqrt{p-3})^2 = (-p+3)^2$$

$$2^2(\sqrt{p-3})^2 = (-p+3)^2$$

$$4(p-3) = p^2-6p+9$$

$$4p-12 = p^2-6p+9 $$

$$-p^2+10p-21=0 $$

$$p^2-10p+21=0 $$

$$(p-3)(p-7) =0$$

$$p-3=0 \quad \text{or} \quad p-7 =0 $$

$$p=3 \quad \text{or} \quad p=7$$

The potential solutions are $p=3$ and $p=7$. We will check them.

$\bullet$ **Check:** $p=3$

$$2\sqrt{p-3} -3 = -p$$

$$2\sqrt{3-3} -3 \stackrel{?}{=} -3$$

$$2\cdot 0 -3 \stackrel{?}{=} -3$$

$$-3 \stackrel{?}{=} -3 \qquad\checkmark$$

$\bullet$ **Check**: $p=7$

$$2\sqrt{p-3} -3 = -p $$

$$2\sqrt{7-3} -3 \stackrel{?}{=} -7$$

$$2\cdot 2 -3 \stackrel{?}{=} -7$$

$$1 \stackrel{?}{=} -7 \qquad \text{False}$$

The solution set is $\{3\}$.

### Need more help?

*Don’t wait too long to do the following.*

- Watch the additional video resources.

- Talk to your instructor.
- Form a study group.
- Visit a tutor. For more information, check the tutoring page.