Hi Everyone!

On this page you will find some material about Lesson 11. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 11: Square Root Property and Completing the Square & Quadratic Formula**

Table of Contents

### Resources

*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.**

- Use the square root property.
- Solve quadratic equations by completing the square.
- Solve quadratic equations by using the quadratic formula.

**Topic**. This lesson covers

Section 7.1: Square Root Property and Completing the Square, and

Section 7.2: Quadratic Formula.

**WeBWorK**. There are two WeBWorK assignments on today’s material:

SquareRootProperty

QuadraticFormula

**Lesson Notes.**

**Video Lesson.**

Video Lesson 11 – part 1 (based on Lesson 11 Notes – part 1)

Video Lesson 11 – part 2 (based on Lesson 11 Notes – part 2)

### Warmup Questions I

*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.*

Simplify $\sqrt{72}$.

\begin{align*}\sqrt{72} &= \sqrt{36\cdot 2}\\ &= \sqrt{36}\sqrt{2}\\ &= 6 \sqrt 2\end{align*}

Solve $x^2=49$.

$$x^2=49$$

$$7^2=49 \quad\quad (-7)^2=49$$

So $x=-7, 7$.

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

Expand $(x+a)^2$.

\begin{align*} (x+a)^2&=(x+a)(x+a)\\&=x^2+2ax+a^2\end{align*}

Expand $(x-a)^2$.

\begin{align*}(x-a)^2&=(x-a)(x-a)\\ &=x^2-2ax+a^2\end{align*}

### 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???

### Quick Intro I

*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 the Square Root Property and Completing the Square**

**Key Words**. Square Root Property, solving a quadratic equation, completing the square

In the Warmup Question 2, we saw that the solutions to $x^2=49$ are $x=-7$ and $7$. We can think of these solutions as being $x=\pm\sqrt{49}=\pm 7$.

$\bigstar$ **Square Root Property**: If $x^2=a$ then $x=\pm\sqrt a$.

$\bigstar$ **Solving a quadratic equation**: The Square Root Property allows us to solve a quadratic equation as long as there is a square on one side and a number on the side.

$$\underbrace{x^2}_{\text{square}}=\underbrace{a}_{\text{number}}$$

The square does not have to be $x^2$. It could be $(x-5)^2$, for example. This will be the case when the equation involves a term with $x$ like in

$$x^2-10x=-21.$$

In this example, the term $-10x$ prevents us from isolating $x^2$ on one side so that a number is left on the other side. By adding a specific number to $x^2-10x$, we can make it a square. This process is called **completing the square**. Let’s find out which number we need to add. In the Warmup Question 4 we saw that

$$x^2-2ax+a^2 = \underbrace{(x-a)^2}_{\text{square}}.$$

Let’s compare the coefficients of $x$ in $x^2-2ax+a^2$ and $x^2-10x$:

$$-2a\quad\text{ and }\quad -10.$$

If we set them to be the same, we get

$$-2a=-10,$$

So $a=5$. The missing term is $a^2=5^2=25$. Let’s return to the original equation:

$$x^2-10x=-21$$

We add 25 to both sides:

$$x^2-10x+25 = -21+25$$

The left side now looks like $x^2-2ax+a^2$ which is $(x-a)^2$. So

$$\underbrace{(x-5)^2}_{\text{square}}=4$$

By the Square Root Property,

\begin{align*} x-5&=\pm \sqrt 4\\ x-5&=\pm 2\\x &=5\pm 2\\x&= 5-2, 5+2\\ x&=3,7\end{align*}

The solution set is $\{3,7\}$.

### Video Lesson I

*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 1**

**A description of the video**

In this video you will see the following equations:

- $x^2=4$
- $x^2=8$
- $(x-2)^2=4$
- $(x-2)^2=27$
- $(x-2)^2=-18$
- $x^2+4x+4=20$
- $x^2+6x-54=0$
- $2x^2+10x-5=0$
- $ax^2+bx+c=0$, $a\neq 0$

### 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.*

Solve $x^2=9$ by using the Square Root Property.

\begin{align*} x^2&=9 \\x&=\pm\sqrt 9\\x&=\pm 3\end{align*}

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

Solve $x^2=300$ by using the Square Root Property.

\begin{align*} x^2&=300\\x &=\pm \sqrt{300}\\x&=\pm\sqrt{100\cdot 3}\\x&=\pm\sqrt{100}\sqrt{3}\\x&=\pm10\sqrt{3}\end{align*}

The solution set is $\{-10\sqrt 3, 10\sqrt 3\}$.

Solve $(x+3)^2=9$ by using the Square Root Property.

\begin{align*} (x+3)^2&=9\\x+3 &=\pm \sqrt{9}\\x+3 &=\pm 3\\x&=-3\pm 3\\ x&=-6,0\end{align*}

The solution set is $\{-6,0\}$.

Solve $(x-7)^2=18$ by using the Square Root Property.

\begin{align*} (x-7)^2&=18\\x-7 &=\pm \sqrt{18} \\ x-7&=\pm 3\sqrt 2 \\ x&=7\pm 3\sqrt 2\end{align*}

The solution set is $\{7-3\sqrt 2,7+3\sqrt 2\}$.

Solve $(x+3)^2=-20$ by using the Square Root Property.

\begin{align*}(x+3)^2&=-20\\x+3 &=\pm \sqrt{-20}\\x+3&=\pm 2\sqrt 5i\\x&=-3\pm 2\sqrt 5i\end{align*}

The solution set is $\{-3- 2\sqrt 5i,-3+ 2\sqrt 5i\}$.

Solve $x^2+6x+9=5$ by completing the square and using the Square Root Property.

\begin{align*} x^2+6x+9&=5\\x^2+6x& =-4\\x^2+6x+9&=-4+9\\(x+3)^2&=5\\x+3&=\pm\sqrt 5\\x&=-3\pm\sqrt 5\end{align*}

The solution set is $\{-3-\sqrt 5,-3+\sqrt 5\}$.

Solve $x^2+10x+5=0$ by completing the square and using the Square Root Property.

\begin{align*}x^2+10x+5&=0 \\x^2+10x& =-5\\x^2+10x+25&=-5+25\\(x+5)^2&=20\\x+5&=\pm\sqrt {20}\\x&=-5\pm 2\sqrt 5\end{align*}

The solution set is $\{-5-2\sqrt 5,-5+2\sqrt 5\}$.

Solve $3x^2+6x+4=0$ by completing the square and using the Square Root Property.

\begin{align*}3x^2+6x+4&=0\\3(x^2+2x) &=-4\\3(x^2+2x+1)&=-4+3\\3(x+1)^2&=-1\\(x+1)^2&=-\dfrac{1}{3}\\x+1&=\pm\sqrt {-\dfrac{1}{3}}\\x&=-1\pm \dfrac{\sqrt 3}{3}i\end{align*}

The solution set is $\left\{-1- \dfrac{\sqrt 3}{3}i, -1+ \dfrac{\sqrt 3}{3}i\right\}$.

### WeBWorK I

*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** I

It is time to do the homework on WeBWork:

SquareRootProperty

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

### Exit Questions I

*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!*

$\bigstar$ (a) Solve by using the Square Root Property.

$$(5x-2)^2-3=4$$

$\bigstar$ (b) Solve by completing the square and applying the Square Root Property.

$$x^2+x-5=0$$

(a)

\begin{align*}(5x-2)^2-3&=4 \\(5x-2)^2&=7 \\5x-2& =\pm \sqrt 7 \\5x&= 2\pm\sqrt 7 \\x&=\dfrac{2\pm\sqrt 7}{5}\end{align*}

The solution set is $\left\{\dfrac{2+\sqrt 7}{5}, \dfrac{2-\sqrt 7}{5}\right\}$.

(b)

\begin{align*}x^2+x-5&=0\\x^2+x&=5 \\ x^2+x+\dfrac{1}{4} &=5 +\dfrac{1}{4} \\ \left(x+\dfrac{1}{2}\right)^2 &= \dfrac{21}{4}\\x+\dfrac{1}{2} &= \pm \dfrac{\sqrt{21}}{2}\\ x &= -\dfrac{1}{2} \pm \dfrac{\sqrt{21}}{2} \\ x &= \dfrac{-1\pm\sqrt{21}}{2} \end{align*}

The solution set is $\left\{\dfrac{-1-\sqrt{21}}{2}, \dfrac{-1+\sqrt{21}}{2}\right\}$.