Hi Everyone!

On this page you will find some material about Lesson 8. Read through the material below, watch the videos, and follow up with your instructor if you have questions.

**Lesson 8: Quadratic Formula and Applications**

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.**

- Identify the coefficients of a quadratic polynomial.
- Solve quadratic equations by using the quadratic formula.
- Apply knowledge of algebra to solve verbal problems.

**Topic**. This lesson covers

Section 7.2: Quadratic Formula,

Section 4.8.2: Applications of Quadratic Equations, and

Section 7.2.3: Using the Quadratic Formula in Applications.

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

QuadraticFormula

**Lesson Notes.**

**Video Lesson.**

Video Lesson 8 – part 1 (based on Lesson 8 Notes – part 1)

Video Lesson 8 – part 2 (based on Lesson 8 Notes – part 2)

Video Lesson 8 – part 3 (based on Lesson 8 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 $$\sqrt{72}.$$

#### Show Answer 1

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

Solve by completing the square.

$$x^2+bx+c=0$$

\begin{align*}x^2+bx+c &=0\\x^2+bx&=-c\\x^2+bx+\dfrac{b^2}{4}& = -c+\dfrac{b^2}{4}\\ \left(x+\dfrac{b}{2}\right)^2 &= \dfrac{b^2-4c}{4}\\x+\dfrac{b}{2}& =\pm\sqrt{\dfrac{b^2-4c}{4}}\\x &=- \dfrac{b}{2}\pm\dfrac{\sqrt{b^2-4c}}{2}\\x&= \dfrac{-b\pm\sqrt{b^2-4c}}{2}\end{align*}

The solution set is $\left\{\dfrac{-b-\sqrt{b^2-4c}}{2},\dfrac{-b+\sqrt{b^2-4c}}{2}\right\}$.

### 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 7.

### 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 Quadratic Formula

**Key Words.** Coefficient, standard form, quadratic formula.

Consider

$$ax^2+bx+c=0.$$

In the Video Lesson 1, we saw that if we complete the square and use the Square Root Property, we obtain that

$$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}.$$

This is the **Quadratic Formula**. It depends on $a$, $b$ and $c$, which are the **coefficients** of the quadratic polynomial $ax^2+bx+c$. The first step in using the quadratic formula is to make sure that the polynomial is set to be zero. This is called the **standard form**.

$$-5x^2=-3x\quad\text{is not in standard form.}$$

$$x^2+3x-5=0\quad\text{is in standard form.}$$

In this case, $a=1$, $b=3$ and $c=-5$.

### 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 the video you will see how to solve the following equations by using the quadratic formula.

- $2x^2-5x-3=0$
- $4x^2-8x+1=0$
- $x(4x-3)=11+x$
- $x^2+2x+5=0$

### 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 $2x^2+4=-4x$ by using the quadratic formula.

The equation written in standard form is $2x^2+4x+4=0$. Let’s divide both sides by $2$. We get

$x^2 + 2x +2 =0$. So $a=1$, $b=2$ and $c=2$. The quadratic formula gives

\begin{align*}x &=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\& = \dfrac{-2\pm \sqrt{2^2-4\cdot 1 \cdot 2}}{2\cdot 1}\\&=\dfrac{-2\pm \sqrt{-4}}{2}\\& = \dfrac{-2\pm \sqrt{4}i}{2}\\ &= \dfrac{-2\pm 2i}{2}\\&= \dfrac{-2}{2}\pm \dfrac{2i}{2}\\& = -1\pm i\end{align*}

The solution set is $\{-1+ i,-1-i\}$.

### 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:

QuadraticFormula

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!*

- How is the quadratic formula related to the process of completing the square?
- Why would we ever need or want to factor if we have the quadratic formula?
- Suppose we use the quadratic formula to solve a quadratic equation, what does what is under the radical symbol tell us about the solution(s) ?

$\bigstar$ Solve $3x^2-4=2x$ by using the quadratic formula.

The equation written in standard form is $3x^2-2x+4=0$. So $a=3$, $b=-2$ and $c=-4$. The quadratic formula gives

\begin{align*} x & =\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\x &= \dfrac{2\pm \sqrt{(-2)^2-4\cdot 3 \cdot (-4)}}{2\cdot 3} \\x&= \dfrac{2\pm \sqrt{4+48}}{6}\\x&= \dfrac{2\pm \sqrt{52}}{6}\\x &= \dfrac{2\pm 2\sqrt{13}}{6} \\x&= \dfrac{2(1\pm \sqrt{13}}{2\cdot 3} \\x&= \dfrac{1\pm\sqrt{13}}{3}\end{align*}

The solution set is $\left\{\dfrac{1-\sqrt{13}}{3}, \dfrac{1+\sqrt{13}}{3}\right\}$.