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2nd Quarter, Math Level 2 (10th grade math)

00.00 Start Here (Math Level 2)

 

Course Description

The fundamental purpose of Mathematics II is to formalize and extend the mathematics that students learned in 9th grade.  Students will focus on quadratic expressions, equations, and functions but also be introduced to radical, piecewise, absolute value, inverse and exponential functions. They will extend the set of rational numbers to the set of complex numbers and link probability and data through conditional probability and counting methods. Finally, they will study similarity and right triangle trigonometry and circles with their quadratic algebraic representations. The course ties together the algebraic and geometric ideas studied. Students should experience mathematics as a coherent, useful, and logical subject that makes use of their ability to make sense of problem situations.

Class Overview

This integrated Secondary Mathematics II course is based on the New Utah State Standards Initiative.

Credit

This course is worth .25 credits, or nine weeks of Mathematics I. There are four Mathematics II quarter classes available. Taking all four will add up to one credit or one year of Mathematics I. In order to earn credit for each quarter, you must commit to following the EHS Honor Code: "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect." This course is for tenth grade students. After completing the work for the class, students must pass a proctored final exam to earn credit. There is not a paper-based textbook assigned for this course. If you find that having a textbook is useful, you can check out a textbook from most local libraries. You can also search for topics on the Internet to find many useful resources.

Prerequisites

You should have successfully completed 9th grade math.

Supplies needed

  • Graph paper (this can also be downloaded)
  • Scientific or Graphing Calculator (You can download scientific and graphing calculator simulators or find online versions, but as you will need a graphing calculator for the rest of your high school career, you may consider buying one now.)
  • Access to a printer to print the daily assignments is vital. Most assignments are NOT interactive and must be printed out to complete.
  • Ability to scan or photograph a completed assignment to submit electronically.

Organization of Secondary Math Level 2

Units: There are 11 units for the full credit of Secondary Math 2.

Quarter 1 has four units.
Quarter 2 has one unit.
Quarter 3 has three units.
​Quarter 4 has three units.

Schedule: When you enroll in a quarter class, you are given 10 weeks to finish all the requirements needed to earn the credit for a quarter. There are no "required" due dates for the assignments in this course. However, there is a pacing guide provided for you that will help you stay on track to being successful and finishing the course within the 10 week time frame. The pacing guide is located in the Syllabus in Module 1. Before you begin, go over the pacing guide to help you set up your own due dates for the assignments. Give your parents permission to nag you about it. You don't want to be one of those students who does a whole lot of work, but never finishes the course.

This Quarter Class

The units in this class have lessons, assignments, quizzes and a unit test.

Lessons: Each lesson provides instruction on a given topic. Many include instructional videos (hosted on YouTube) and one or more assignments for independent practice.

Assignments: Print, then complete the practice worksheet, showing how you arrived at your answer. In order to earn full points on each problem, you are required to show step by step the process needed to arrive at your answer. Then circle your answer so it is easily seen. I do give partial credit for showing your work--so ALWAYS show your work even if you think it is obvious. Assignments are submitted by uploading the digital assignment through the course website by following the instructions within each assignment. Under some circumstances, you may snail-mail a hard copy of the assignment to the instructor. If you choose this option, be sure to make a copy for yourself, as the instructor will NOT return your assignment. Also, please send the instructor an e-mail if you must mail an assignment.

Quizzes: Some quizzes are taken online. Others are more like the assignments. Print, then complete the quiz, showing all of your work as to how you arrived at your answer. In order to earn full points on each problem, you are required to show step by step the process needed to arrive at your answer. Then circle your answer so it is easily seen. I do give partial credit for showing your work--so ALWAYS show your work even if you think it is obvious. Quizzes are submitted by uploading the digital assignment through the course website by following the instructions within each quiz.

Proctored Final

Each quarter class has a proctored final exam and is worth 25% of the final grade.

Information about the Final Exam

  1. You must get approval from your teacher before you are allowed to take the final exam.
  2. You must complete every assignment and have an overall grade of C in the course to be approved to take the final exam.
  3. The final exam is a comprehensive exam that must be taken with an approved proctor.
  4. You are allowed to have a page of notes. You also will need a calculator and scratch paper.
  5. You must pass the final exam with a 60% in order to pass the class.
  6. The final exam is worth 25% of your final grade.
  7. The exam is timed. You will have 2.0 hours to complete the exam. You must finish it in one attempt.

Final Grade

Assignments and quizzes are worth 75% of the final grade. The proctored final test is worth 25% of the final grade.

Grading Scale

You earn a grade based on a modified total points percentage method. This means that the total number of points you earned is divided by the total number of points possible, times 100%. That will make up 75% of your final grade. The final exam is the remaining 25%. These scores are combined for a total percentage of the class. This percentage is translated into a grade based on this standard scale:

94-100% A
90-93% A-
87-89% B+
83-86% B
80-82% B-
77-79% C+
73-76% C
70-72% C-
67-69% D+
63-66% D
60-62% D-
0-59% no credit

00.00 *Student supplies for Secondary Math 2 (Math2)

There is no textbook assigned for this course. If you find that having a textbook is useful, you will want to select a textbook that aligns with the Common Core State Standards Initiative. You can purchase textbooks online, at some bookstores (selection is limited), or you can check out a textbook from most local libraries. Supplies needed:

  • Graph paper (this can also be downloaded)
  • Tracing paper or transparency film (if you use transparencies you will need transparency markers).
  • Straight edge (A ruler is an example of a straight edge. I have used a scrap of mat board. A piece of paper will not work! My favorite is a drafting ruler with the cork on the bottom; this gives a very clean line, and doesn't slide around.)
  • Compass (There is a very wide range of compass prices. You don't need a drafting quality compass, but you will be happier if you don't buy the cheapest compass you can find. You want something that doesn't change size as you draw with it.)
  • Graphing Calculator (You can download graphing calculator simulators, but as you will need a graphing calculator for the rest of your high school career, you may as well buy one now.)
  • Geometric software.

 

00.01 Curriculum Standards (Math Level 2)

Overview information on the Utah Mathematics Level II Core is here.

00.01.01 Student Software Needs

 

Students need access to a robust internet connection and a modern web browser.

This class may also require the Apple QuickTime plug-in to view media.

For students using a school-issued Chromebook, ask your technical support folks to download the QuickTime plug-in and enable the plug-in for your Chromebook.

$0.00

00.02 About Me (Math Level 2)

teacher-scored 10 points possible 10 minutes

About Me Assignment: This assignment gives me, as your teacher, a chance to get to know you better! To complete and submit this assignment copy the material between the asterisks into a blank word-processing document. Answer the questions using complete sentences, appropriate punctuation and sentence structure. Please write your answers in either BOLD or a {\color{Blue}DIFFERENT } {\color{Blue}COLOR }. Save the document. Finally, select all, copy, then paste the entire document into the box that opens when you click to submit this assignment.

**********************************************************************************

1. What is your full name, what name do you prefer to go by, your parent's/guardian's names, and contact information for both you and your parents? (email addresses and phone numbers.)

2. What high school do you attend and what grade are you in? What is the name of the last math class you completed?

3. Why have you chosen to take this math class with EHS?

4. What is your counselor's full name and contact information?

5. Have you read the EHS Honor Code and do you commit to following it? EHS Honor Code "As a student of the Electronic High School, I agree to turn in my assignments in a timely manner, do my own work, not share my work with others, and treat all students, teachers, and staff with respect."

6. Are you committed to finishing the class within the 10 week time frame, completing your final exam in week 9?

7. Now tell me about you! What are your likes/dislikes etc. Please be sure to include anything you think I need to know as your teacher.

*************************************************************************

I am excited to learn more about you!

Grading criteria:

1. All requested information is included.

2. Complete sentences, correct punctuation and correct grammar are used.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


00.03 Basic Algebra Skills (Math Level 2)

Just like learning to use the English or Spanish language, you learn the language of mathematics by acquiring its vocabulary and grammar and by practicing until you can "think" fluently in the language.

This pre-requisite lesson helps you verify you have a basic understanding of multiplying polynomials, factoring polynomials, and simplifying radical expressions, some of the basic building blocks of the "mathematics algebra language."

If you need to review any of these topics, please view the following instructional videos:

00.03 Basic Algebra Skills - Extra Videos (Math Level 2)

00.03 Basic Algebra Skills - Quiz (Math Level 2)

computer-scored 48 points possible 30 minutes
Pacing: complete this by the end of Week 1 of your enrollment date for this class.


05.00 Quadratics (Math Level 2)

The goal of all math classes is for students to become mathematically proficient. In particular, students need to be able to:

  • Make sense of problems and persevere in solving them.
  • Reason abstractly and quantitatively.
  • Construct viable arguments and critique the reasoning of others.
  • Model with mathematics.
  • Use appropriate tools strategically.
  • Attend to precision.
  • Look for and make use of structure.
  • Look for and express regularity in repeated reasoning.

As you work through this unit, keep these proficiencies in mind!

This unit provides you an indepth study of quadratics. You will learn to graph them and use a variety of methods to solve them. In addition, you will learn to solve quadratic inequalities. Finally, you will study the terms and formulas for parabolas.

At the conclusion of this unit you must be able to state without hesitation that:

  • I can graph quadratic functions.
  • I can solve quadratic equations using factoring techniques and express the solution(s) as a set.
  • I can solve quadratic equations by taking square roots.I can solve quadratic equations by completing the squares.
  • I can convert a quadratic equation to vertex form.
  • I can solve quadratic equations using the quadratic formula.
  • I can determine there are two roots, one root or zero roots.
  • I can solve systems of equations involving linear, quadratic, and other non-linear functions.
  • I can solve quadratics that have complex roots.
  • I can compare methods for solving quadratics.
  • I can use the Fundamental Theorem of Algebra to determine the number of roots and/or factors in a polynomial function.
  • I can solve quadratic inequalities.
  • I can derive the equation of a parabola given a focus and directrix.
  • I understand the properties of quadratics.
  • I can solve quadratics problems.

 

05.01 Graphs of Quadratic Equations (Math Level 2)

Graph quadratic functions.

In quarter one you learned that one of the most common types of functions we work with in math is the quadratic function. In this lesson, we will learn several different ways to graph quadratic functions. Before beginning, let’s review the various forms of writing them.

 

A quadratic function can be described by an equation of the form:

y = ax2 + bx + c, where a ≠ 0.

This is called the standard or general form of a quadratic equation. You also learned about the vertex form:

f(x) = a(x – h)2 + k, where (h, k) is the vertex.

Image: http://www.ck12.org/saythanks

There is another useful form of a quadratic equation. It is the intercept or factored form:

y = a(x – p)(x – q), where p and q are the x-intercepts.

You will find that each form has its usefulness as you move through the quarter.

Graph Quadratics by Finding Ordered Pairs

You are very familiar with graphing linear functions by finding ordered pairs. Let’s begin graphing quadratics by finding ordered pairs for this function f(x) = x2.

We begin by creating a table of ordered pairs:

Next, we plot the points:

Now, you need to know the shape of a graph of a quadratic is not made up of straight lines! Here’s the finished product:

This is one way to graph a quadratic when it is in Standard Form. Another strategy to graph a quadratic in Standard Form is to first find the vertex and then find at least 2 other points (as shown above) and plot them.

The vertex can be found by first finding the x value of the ordered pair using this formula: x = -\frac{b}{2a}   To find the y value of the ordered pair, substitute the x value you find into the equation and work through the math.

One easy point to find is the y-intercept. It is found by substituting a 0 in for the x's in the equation. When you do that, you will be left with just the constant value "c". For example: Given the quadratic function y = 2x2 - 3x + 7 find the y-intercept.

y = 2(0)2 - 3(0) + 7

   = 2(0) - 0 + 7

   = 0 - 0 + 7

   = 7

So, the y-intercept is (0,7)! Notice how the constant value is 7. In Standard Form, the y-intercept will always be the constant value, c.

The advantage of having the quadratic in Standard Form is that you can easily find the y-intercept.

Graph Quadratics Written in Vertex Form or General Form

f(x) = a(x – h)2 + k, where (h, k) is the vertex.

To graph a quadratic that is in Vertex Form, you will again need the vertex and then two other points found by substituting values in for x in the equation. The advantage of Vertex Form is that you can easily find the vertex of the quadratic, (h,k).

Graph Quadratics Written in Intercept Form or Factored Form

y = a(x – p)(x – q), where p and q are the x-intercepts.

The advantage to having the quadratic equation in Intercept Form is that you can easily find the x-intercepts, or the roots of the function. The neat thing about a quadratic is that its vertex will always be equidistant from each intercept Once you know the x-intercepts, you will be able to find the vertex by finding the x value directly between them. To find that x value between the x-intercepts, you will add the x intercepts together and divide by 2. (Sounds a lot like finding a midpoint!)

For example: Given the quadratic function in Intercept Form: y = (x - 2)(x + 4) First find the x-intercepts by setting each factor equal to zero and solving for x.

x - 2 = 0 Add 2 to both sides.

x = 2

x + 4 = 0  Subtract 4 from both sides.

x = -4 

The x intercepts are 2 and -4. Now we have to find the x value that is directly between these two intercepts. We add them and divide by 2.

\frac{2+(-4))}{2}=\frac{-2}{2}=-1 

That tells us that the vertex of this quadratic function will have an x value of -1. To find the y value of the vertex, substitute -1 in for the x's in the equation.

y = (-1 - 2)(-1 + 4) = (-3)(3) = -9. The vertex is at (-1,-9) You can now plot each x-intercept and the vertex to graph the parabola.

Watch the videos below to learn to graph quadratics in this form.

05.01 Graphs of Quadratic Equations - Videos (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can graph quadratic functions.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.01 Graphs of Quadratic Equations - Worksheet (Math Level 2)

teacher-scored 68 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


05.02 Solve Quadratic Equations by Factoring (Math Level 2)

Solve quadratic equations using factoring techniques and express the solution(s) as a set.

We can find the solutions, or roots, of quadratic equations by setting one side equal to zero, factoring the polynomial, and then applying the Zero Product Property.

The Zero Product Property states that if ab = 0, then either a = 0 or b = 0, or both a and b are 0. When the product of factors equals zero, one or more of the factors must also equal zero. Once the polynomial is factored, set each factor equal to zero and solve them separately. The answers will be the set of possible solutions to the original equation.

Click on the link below for the lesson material.

05.02 Solve Quadratic Equations by Factoring - External Link (Math Level 2)

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

I highly recommend that you click on the links below and watch the videos after completing the topic.

05.02 Solve Quadratic Equations by Factoring - Extra Videos (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can solve quadratic equations using factoring techniques and express the solution(s) as a set.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.02 Solve Quadratic Equations by Factoring - Worksheet (Math Level 2)

teacher-scored 80 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 1 of your enrollment date for this class.


05.02.01 Checkpoint Quiz - Quiz (Math Level 2)

computer-scored 30 points possible 40 minutes
Pacing: complete this by the end of Week 1 of your enrollment date for this class.


05.03 Solve Quadratics Equations by Taking Square Roots (Math Level 2)

Solve quadratic equations by taking square roots.

 

 

 

 

 

 

 

Now that you are familiar with square roots, we will use them to solve quadratic equations. Keep in mind, that square roots cannot be used to solve every type of quadratic. In order to solve a quadratic equation by using square roots, an x-term cannot be present. Solving a quadratic equation by using square roots is very similar to solving a linear equation. In the end, you must isolate the x2 or whatever is being squared.

Example A

Solve 2x2 – 3 = 15.

Solution: Start by isolating the x2.

2x2 – 3 = 15

2x2 = 18

x2 = 9

At this point, you can take the square root of both sides.

\sqrt{x^{2}} = \pm \sqrt{9}

x = \pm 3

Notice that x has two solutions; 3 or – 3. When taking the square root, always put the \pm (plus or minus sign) in front of the square root. This indicates that the positive or negative answer will be the solution.

Check:

Example B

Solve:

\frac{x^{2}}{16}+3=27.

Solution: Isolate x2 and then take the square root.

\frac{x^{2}}{16}+3=27

\frac{x^{2}}{16}=24

x^{2}=384

x=\pm \sqrt{384}=\pm 8\sqrt{6}

Example C

Solve: 3(x-5)2+7 = 43.

Solution: In this example, x is not the only thing that is squared. Isolate the (x – 5)2 and then take the square root.

3(x-5)^{2}+7 = 43

3(x-5)^{2} = 36

(x-5)^{2} = 12

(x-5) = \pm \sqrt{12} or \pm 2\sqrt{3}

Now that the square root is gone, add 5 to both sides.

x-5=\pm 2\sqrt{3}

x=5\pm 2\sqrt{3}

x=5+2\sqrt{3} or x=5-2\sqrt{3}

We can estimate these solutions as decimals; 8.46 or 1.54. Remember, that the most accurate answer includes the radical numbers.

Guided Practice

Solve the following quadratic equations.

1. \frac{2}{3}x^{2}-14 = 38

2. 11 + x^{2} = 4x^{2} + 5

3. (2x + 1)^{2} - 6 = 19

Answers

1. Isolate x2 and take the square root:

\frac{2}{3}x^{2}-14 = 38

\frac{2}{3}x^{2} = 52

x^{2} =78

x=\pm \sqrt{78}

2. Combine all like terms, then isolate x2:

11+x^{2}=4x^{2}+5

-3x^{2}=-6

x^{2}=2

x=\pm \sqrt{2}

3. Isolate what is being squared, take the square root, and then isolate x.

(2x+1)^{2}=25

2x+1 = \pm 5

2x=-1\pm 5

x=\frac{-1\pm 5}{2} or

x=\frac{-1+5}{2} or x=\frac{-1-5}{2}

x=2 or x=-3

I highly recommend that you click on the links below and watch the videos before continuing.

05.03 Solve Quadratics Equations by Taking Square Roots - Videos (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can solve quadratic equations by taking square roots.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.03 Solve Quadratics Equations by Taking Square Roots - Worksheet (Math Level 2)

teacher-scored 96 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.


05.04 Complete the Square (Math Level 2)

Solve quadratic equations by completing the squares and convert a quadratic equation to vertex form.

Sometimes a quadratic equation is impossible to factor. To solve this kind of quadratic equation, different strategies are needed. Completing the square is such a strategy. It turns a polynomial into a perfect square trinomial, which is easier to graph and to solve.

    \Rightarrow  

Images licensed under Creative Commons Attribution-ShareAlike 3.0 License

Completing the square is used to change a binomial of the form x^{2}+bx to a perfect square trinomial

x^{2}+bx+\left ( \frac{b}{2} \right )^{2},

which can be factored to

\left (x+\frac{b}{2} \right )^{2}.

When solving quadratic equations, completing the square is used to find the roots of the quadratic equation, but we must be careful to add:

\dpi{100} \left ( \frac{b}{2} \right )^{2}

to both sides of the equation to maintain equality.

Completing the square also helps us convert a quadratic equation into vertex form, a(x - h)2 + k. In the vertex form of a quadratic equation, the coordinates of the vertex are given by (h, k).

The following content is taken from “Solving Equations and Inequalities”, chapter 6 from the book Advanced Algebra (v. 1.0). It is licensed under a Creative Commons by-nc-sa 3.0 license.

In this section, we will devise a method for rewriting any quadratic equation of the form

ax2+bx+c=0

as an equation of the form (x−p)2=q

This is the process is called completing the square. As we have seen, quadratic equations in this form can be easily solved by extracting roots. We begin by examining perfect square trinomials:

(x+3)^{2}=x^{2}+6x + 9

 \left ( \frac{6}{2} \right )^{2}=(3)^{2}=9

The last term, 9, is the square of one-half of the coefficient of x. In general, this is true for any perfect square trinomial of the form x2+bx+c.

(x+\frac{b}{2})^{2}=x^{2}+2\cdot \frac{b}{2}x+(\frac{b}{2})^{2}

=x^{2}+bx+\left ( \frac{b}{2} \right )^{2}

In other words, any trinomial of the form x2+bx+c will be a perfect square trinomial if

c=\left ( \frac{b}{2} \right )^{2}

Note: It is important to point out that the leading coefficient must be equal to 1 for this to be true.

Example 1

Complete the square: x^{2}-6x+{\color{Blue} ?}=\left ( x+{\color{Blue} ?} \right )^{2}

Solution:

In this example, the coefficient b of the middle term is −6. Find the value that completes the square as follows:

\left ( \frac{b}{2} \right )^{2}=\left ( \frac{-6}{2} \right )^{2}=(-3)^{2}={\color{Blue} 9}

The value that completes the square is 9.

x^{2}-6x{\color{Blue} +9}=(x-3)(x-3)

=(x-3)^{2}

Answer: x^{2}-6x+9=(x-3)^{2}

EXAMPLE 2

Complete the square: x^{2}+x+{\color{Blue} ?}=(x+{\color{Blue} ?})^{2}

Solution:

Here b = 1. Find the value that will complete the square as follows:

\left ( \frac{b}{2} \right )^{2}=\left ( \frac{1}{2} \right )^{2}={\color{Blue} \frac{1}{4}}

The value \frac{1}{4} completes the square:

x^{2}+x{\color{Blue} +\frac{1}{4}}=\left ( x+\frac{1}{2} \right )\left ( x+\frac{1}{2} \right )

=\left ( x+\frac{1}{2} \right )^{2}

Answer: x^{2}+x+\frac{1}{4}=\left ( x+\frac{1}{2} \right )^{2}

 

We can use this technique to solve quadratic equations. The idea is to take any quadratic equation in standard form and complete the square so that we can solve it by extracting roots. The following are general steps for solving a quadratic equation with leading coefficient 1 in standard form by completing the square.

EXAMPLE 3

Solve by completing the square: x^{2}-8x-2=0

Solution:

It is important to notice that the leading coefficient is 1.

Step 1: Add or subtract the constant term to obtain an equation of the form x2 + bx = c. Here we add 2 to both sides of the equation.

x^{2}-8x-2=0

x^{2}-8x=2

Step 2: Use \left ( \frac{b}{2} \right )^{2}to determine the value that completes the square. In this case, b = −8:

\left ( \frac{b}{2} \right )^{2}=\left ( \frac{-8}{2} \right )^{2}=(-4)^{2}={\color{Blue} 16}

Step 3: Add \left ( \frac{b}{2} \right )^{2}to both sides of the equation and complete the square.

x^{2}-8x=2

x^{2}-8x{\color{Blue} +16}=2{\color{Blue} +16}

(x-4)(x-4)=18

(x-4)^{2}=18

Step 4: Solve by extracting roots.

(x-4)^{2}=18

x-4=\pm \sqrt{18}

x=4\pm \sqrt{9\cdot 2}

x=4\pm3\sqrt{2}

Answer: The solutions are 4-3\sqrt{2} and 4+3\sqrt{2}

So far, all of the examples have had a leading coefficient of 1. The formula \left ( \frac{b}{2} \right )^{2} determines the value that completes the square only if the leading coefficient is 1. If this is not the case, then simply divide both sides by the leading coefficient before beginning the steps outlined for completing the square.

EXAMPLE 4

Solve by completing the square: 2x^{2}+5x-1=0

Solution:

Notice that the leading coefficient is 2. Therefore, divide both sides by 2 before beginning the steps required to solve by completing the square.

\frac{2x^{2}+5x-1}{{\color{Blue} 2}}=\frac{0}{{\color{Blue} 2}}

\frac{2x^{2}}{2}+\frac{5x}{2}-\frac{1}{2}=0

x^{2}+\frac{5}{2}x-\frac{1}{2}=0

Add \frac{1}{2} to both sides of the equation.

x^{2}+\frac{5}{2}x-\frac{1}{2}=0

x^{2}+\frac{5}{2}x=\frac{1}{2}

Here b=\frac{5}{2}, and we can find the value that completes the square as follows:

\left ( \frac{b}{2} \right )^{2}=\left ( \frac{{\color{Green} 5/2}}{2} \right )^{2}=\left ( \frac{5}{2}\cdot \frac{1}{2} \right )^{2}=\left ( \frac{5}{4} \right )^{2}={\color{Blue} \frac{25}{16}}

To complete the square, add \frac{25}{16} to both sides of the equation.

x^{2}+\frac{5}{2}x=\frac{1}{2}

x^{2}+\frac{5}{2}x{\color{Blue} +\frac{25}{16}}=\frac{1}{2}{\color{Blue} +\frac{25}{16}}

\left ( x+\frac{5}{4} \right )\left ( x+\frac{5}{4} \right )=\frac{8}{16}+\frac{25}{16}

\left ( x+\frac{5}{4} \right )^{2}=\frac{33}{16}

Next, solve by extracting square roots.

\left ( x+\frac{5}{4} \right )^{2}=\frac{33}{16}

x+\frac{5}{4}=\pm \sqrt{\frac{33}{16}}

x+\frac{5}{4}=\pm \frac{\sqrt{33}}{4}

x=-\frac{5}{4}\pm \frac{\sqrt{33}}{4}

x=\frac{-5\pm \sqrt{33}}{4}

Answer: The solutions are x=\frac{-5+\sqrt{33}}{4} and x=\frac{-5-\sqrt{33}}{4}

 

Click on the link below for the lesson material.

 

05.04 Complete the Square - External Link (Math Level 2)

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

I highly recommend that you click on the links below and watch the videos after completing the topic.

05.04 Complete the Square - Extra Videos (Math Level 2)

Eichhorn, Sarah; Lehman, Rachel Pre-calculus 1A/1B (UCI OpenCourseWare: University of California, Irvine), http://ocw.uci.edu/courses/math_1a1b_precalculus.html. [Access date]. License: Creative Commons Attribution-ShareAlike 4.0 United States License.

If after completing this topic you can state without hesitation that...

  • I can solve quadratic equations by completing the squares.
  • I can convert a quadratic equation to vertex form.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.04 Complete the Square - Worksheet (Math Level 2)

teacher-scored 84 points possible 45 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 2 of your enrollment date for this class.


05.04.01 Checkpoint Quiz - Quiz (Math Level 2)

computer-scored 32 points possible 40 minutes
Pacing: complete this by the end of Week 2 of your enrollment date for this class.


05.05: Solving Quadratic Equations Using the Quadratic Formula (Math Level 2)

Solve quadratic equations using the quadratic formula and determine if there are two roots, one root or zero roots.

We can solve any quadratic equation by completing the square—turning a polynomial into a perfect square trinomial. If we complete the square on the generic equation ax2+bx+c = 0 and then solve for x, we find that

x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}

This rather awkward looking equation is known as the quadratic formula.

This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The quadratic formula can be used to solve any quadratic equation of the form ax2+bx+c = 0.

The quadratic formula is found by completing the square of the quadratic equation ax2+bx+c = 0.  The formula can be used to find the solution to a quadratic equation and to identify any possible roots, or x-intercepts, of the function.

The discriminant of the quadratic formula is the quantity under the radical, b2 - 4ac. It determines how many solutions there are to the quadratic equation. If the discriminant is positive, there are 2 roots. If it is zero, there is 1 root. If the discriminant is negative, there are 0 roots.

  

Images licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License

Click on the link below for the lesson material.

05.05: Solving Quadratic Equations Using the Quadratic Formula - External Link (Math Level 2)

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

I highly recommend that you click on the links below and watch the videos after completing the topic.

05.05: Solving Quadratic Equations Using the Quadratic Formula - Extra LInks (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can solve quadratic equations by completing the squares.
  • I can convert a quadratic equation to vertex form.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.05: Solving Quadratic Equations Using the Quadratic Formula - Worksheet (Math Level 2)

teacher-scored 82 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.


05.06 Solve Quadratics with Complex Solutions (Math Level 2)

Solve quadratic that have complex roots.

Lesson material is from Beginning Algebra http://2012books.lardbucket.org that has a Creative Commons by-nc-sa 3.0 license.

We complete our study of solutions to quadratic equations by including solutions involving imaginary roots

Let's review a little bit before moving on. We know that \sqrt{-9} is not a real number.

\sqrt{-9}={\color{Blue} ?}      or      \left ( {\color{Blue} ?} \right )^{2}=-9

There is no real number that when squared results in a negative number. We begin to resolve this issue by defining the imaginary unit, i, as the square root of −1.

i=\sqrt{-1}    and    i^{2}=-1

To express a square root of a negative number in terms of the imaginary unit i, we use the following property where a represents any non-negative real number:

\sqrt{-a}=\sqrt{-1\cdot a}=\sqrt{-1}\cdot \sqrt{a}=i\sqrt{a}

With this we can write

\sqrt{-9}=\sqrt{-1\cdot 9}=\sqrt{-1}\cdot \sqrt{9}=i\cdot 3=3i

If \sqrt{-9}=3i, then we would expect that 3i squared will equal −9:

(3i)^{2}=9i^{2}=9\cdot -1=-9

In this way any square root of a negative real number can be written in terms of the imaginary unit. Such a number is often called an imaginary number.

http://www.onemathematicalcat.org

Example 1

Rewrite in terms of the imaginary unit i.

a. \sqrt{-7}

b. \sqrt{-25}

c. \sqrt{-72}

Solution:

a. \sqrt{-7}=\sqrt{-1\cdot 7}=\sqrt{-1}\cdot \sqrt{7}=i\sqrt{7}

b. \sqrt{-25}=\sqrt{-1\cdot 25}=\sqrt{-1}\cdot \sqrt{25}=i\cdot 5=5i

c. \sqrt{-72}=\sqrt{-1\cdot36\cdot 2}=\sqrt{-1}\cdot \sqrt{36}\cdot \sqrt{2}=i\cdot6\cdot \sqrt{2} =6i\sqrt{2}

Notation Note: When an imaginary number involves a radical, we place i in front of the radical. Consider the following:

6i\sqrt{2}=6\sqrt{2}i

Since multiplication is commutative, these numbers are equivalent. However, in the form 6\sqrt{2}i the imaginary unit i is often misinterpreted to be part of the radicand. To avoid this confusion, it is a best practice to place i in front of the radical and use 6i\sqrt{2}.

Now let's apply this to solving quadratic equations that involve including solutions involving imaginary roots

Example 1: Solve using the quadratic formula: x2−2x + 5 = 0

Solution: Begin by identifying a, b, and c: a = 1, b = -2, and c = 5

Substitute these values into the quadratic formula and then simplify.

x=\frac{-{\color{Red} b}\pm\sqrt{{\color{Red} b}^{2}-4{\color{Blue} a}{\color{Green} c}} }{2{\color{Blue} a}}

x=\frac{-({\color{Red} -2})\pm\sqrt{({\color{Red} -2})^{2}-4({\color{Blue} 1})({\color{Green} 5})} }{2({\color{Blue} 1})}

x=\frac{2\pm \sqrt{4-20}}{2}

x=\frac{2\pm \sqrt{-16}}{2}

x=\frac{2\pm 4i}{2}

x=\frac{2}{3}\pm \frac{4i}{2}

x=1\pm 2i

Check these solutions by substituting them into the original equation.

Answer: The solutions are 1−2i and 1 + 2i.

The equation may not be given in standard form. The general steps for solving using the quadratic formula are outlined in the following example.

Example 2: Solve: (2x + 1)(x − 3) = x − 8.

Solution:

Step 1: Write the quadratic equation in standard form:

(2x+1)(x-3) = -8

2x^{2}-6x+x-3 = x-8

2x^{2}-5x-3 = x-8

2x^{2}-6x+5=0

Step 2: Identify a, b, and c for use in the quadratic formula: a = 2, b = - 6, c = 5

Step 3: Substitute the appropriate values into the quadratic formula and then simplify.

x=\frac{-{\color{Red} b}\pm\sqrt{{\color{Red} b}^{2}-4{\color{Blue} a}{\color{Green} c}} }{2{\color{Blue} a}}

x=\frac{-{\color{Red} (-6)}\pm\sqrt{{\color{Red} (-6)}^{2}-4{\color{Blue} (2)}{\color{Green} (5)}} }{2{\color{Blue} (2)}}

x=\frac{6\pm \sqrt{36-40}}{4}

x=\frac{6\pm \sqrt{-4}}{4}

x=\frac{6\pm 2i}{4}

x=\frac{6}{4}\pm \frac{2i}{4}x=\frac{6}{4}\pm \frac{1}{2}i

Answer: The solution is \frac{3}{2} \pm \frac{1}{2} i. The check is optional.

Example 3: Solve: x(x + 2) = - 19

x(x+2)=-19

x^{2}+2x=-19

x^{2}+2x+19 = 0

Here a = 1, b = 2, and c = 19. Substitute these values into the quadratic formula:

x=\frac{-{\color{Red} b}\pm\sqrt{{\color{Red} b}^{2}-4{\color{Blue} a}{\color{Green} c}} }{2{\color{Blue} a}}

x=\frac{-{\color{Red} (-2)}\pm\sqrt{{\color{Red} 2}^{2}-4{\color{Blue} (1)}{\color{Green} (19)}} }{2{\color{Blue} (1)}}

x=\frac{-2\pm \sqrt{4-76}}{2}

x=\frac{-2\pm \sqrt{-72}}{2}

x=\frac{-2\pm \sqrt{-1\cdot 36\cdot 2}}{2}

x=\frac{-2\pm 6i\sqrt{ 2}}{2}

x=\frac{-2}{2}\pm 6i\sqrt{2}

x=-1\pm 3i\sqrt{2}

Answer: The solutions are x=-1\pm 3i\sqrt{2}

Notation Note

Consider the following: -1+3i\sqrt{2}= -1+3\sqrt{2}i

Both numbers are equivalent and -1+3\sqrt{2}i is in standard form, where the real part is -1 and the imaginary part is 3\sqrt{2}. However, this number is often expressed as -1+3i\sqrt{2} to avoid the possibility of misinterpreting the imaginary unit as part of the radicand.

Try this! Solve: (2x+3)(x+5) = 5x+4.

Answer: \frac{-4\pm i\sqrt{6}}{2} = -2\pm \frac{\sqrt{6}}{2}i

Click on the link below for a video solution to this problem.

05.06 Solve Quadratics with Complex Solutions - Video Solution (Math Level 2)

I highly recommend that you click on the links below and watch the videos before continuing:

05.06 Solve Quadratics with Complex Solutions - Videos (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can solve quadratic that have complex roots.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.06 Solve Quadratics with Complex Solutions - Worksheet (Math Level 2)

teacher-scored 80 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 3 of your enrollment date for this class.


05.07 Comparing Methods for Solving Quadratics (Math Level 2)

Compare methods for solving quadratics.

This material is from .

In mathematics, you’ll need to solve quadratic equations that describe application problems or that are part of more complicated problems. You’ve learned four ways of solving a quadratic equation:

  • Factoring
  • Taking the square root
  • Quadratic formula
  • Completing the square

Usually you’ll have to decide for yourself which method to use. However, here are some guidelines as to which methods are better in different situations.

Factoring is always best if the quadratic expression is easily factorable. It is always worthwhile to check if you can factor because this is the fastest method. Many expressions are not factorable so this method is not used very often in practice.

Taking the square root is best used when there is no x-term in the equation.

Quadratic formula is the method that is used most often for solving a quadratic equation. When solving directly by taking square root and factoring does not work, this is the method that most people prefer to use.

Completing the square can be used to solve any quadratic equation. This is usually not any better than using the quadratic formula (in terms of difficult computations), but it is very useful if you need to rewrite a quadratic function in vertex form. It’s also used to rewrite the equations of circles, ellipses and hyperbolas in standard form (something you’ll do in math classes in the future).

If you are using factoring or the quadratic formula, make sure that the equation is in standard form.

Example A

Solve each quadratic equation.

a. x2 - 4x - 5 = 0

b. x2 = 8

c. - 4x2 + x = 2

d. 25x2 - 9 = 0

e. 3x2 = 8x

Solution

a. x2 – 4x – 5 = 0. This expression if easily factorable so we can factor and apply the zero-product property:

       x2 – 4x - 5 = 0
Factor:   (x – 5)(x + 1) = 0
Apply zero-product property:   x – 5 = 0 and x + 1 = 0
Solve:   x = 5 and x = –1

 

Answer:  x = 5 and x = – 1.

b. x2 = 8. Since the expression is missing the x term we can take the square root:

      x2 = 8
Take the square root of both sides:    x = \pm \sqrt{8}
Solve:   x = \sqrt{8} and x = -\sqrt{8}

 

Answer: x = \sqrt{8} and x = -\sqrt{8}

c. – 4x2 + x = 2. Rewrite the equation in standard form: – 4x2 + x - 2 = 0.

It is not apparent right away if the expression is factorable so we will use the quadratic formula:

        -4x^{2}+x - 2 = 0
Quadratic Formula:   x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a}
Plug in the values a = – 4, b = 1, c = 2:   x=\frac{-1\pm \sqrt{b^{2}-4(-2)(-2)}}{2(-4)}
Simplify:   x=\frac{-1\pm \sqrt{1-32}}{-8}
    x=\frac{-1\pm \sqrt{-31}}{-8}
    x=\frac{-1\pm i\sqrt{31}}{-8}
    x=\frac{1\pm i\sqrt{31}}{8}

 

Answer: \frac{1\pm i\sqrt{31}}{8}

d. 25x2 – 9 = 0. This problem can be solved easilty either with factoring or taking the square root. Let's take the square root in this case:

       25x^{2} - 9 = 0
Add 9 to both sides of the equation:   25x^{2} = 9
Divide both sides by 25:   x^{2} = \frac{9}{25}
Take the square root of both sides:   x = \pm \sqrt{\frac{9}{25}}
Simplify:   x = \pm {\frac{3}{5}}

 

Answer: x ={\frac{3}{5}}  and  x =-{\frac{3}{5}}

e. 3x2 = 8x.

      3x^{2}=8x
Re-write the equation in standard form:   3x^{2}-8x=0
Factor out common x term:   x(3x-8)=0
Apply zero-product property:   x=0 and 3x=8
Solve:   x=0 and x=\frac{8}{3}

 

Answer: x = 0 and x=\frac{8}{3}.

Solving Real-World Problems by Completing the Square

An object that is dropped falls under the influence of gravity. The equation for its height with respect to time is given by

y=\frac{1}{2}gt^{2}+v_{_0{y}}+y_{0}

y_{0} where represents the initial height of the object and g is the coefficient of gravity on earth, which equals -9.8\frac{m}{s^{2}}  or -32\frac{ft}{s^{2}} .

On the other hand, if an object is thrown straight up or straight down in the air, it has an initial vertical velocity. This term is usually represented by the notation v_{o_{y}}. Its value is positive if the object is thrown up in the air and is negative if the object is thrown down. The equation for the height of the object in this case is:

y=\frac{1}{2}gt^{2}+v_{o_{y}}+y_{0}

Plugging in the appropriate value for g turns this equation into:

y=-4.9t^{2}+v_{o_{y}}+y_{0} if you wish to have the height in meters, or:

y=-16t^{2}+v_{o_{y}}+y_{0} if you wish to have the height in feet.

Example B

An arrow is shot straight up from a height of 2 meters with a velocity of 50 m/s.

a. How high will the arrow be 4 seconds after being shot? After 8 seconds?
b. At what time will the arrow hit the ground again?
c. What is the maximum height that the arrow will reach and at what time will that happen?

Solution

Since we are given the velocity in m/s, use: y=-4.9t^{2}+v_{o_{y}}+y_{0}

We know that v_{o_{y}} = 50 \frac{m}{s} and y_{0} = 2 meters so: y=-4.9t^{2}+50t + 2

a. To find how high the arrow will be 4 seconds after being shot we plug in t =4:

y=-4.9(8)^{2}+50(8) + 2=-4.9(64)+400+2=-313.6+402= 88.4

b. The height of the arrow on the ground is y = 0, so 0=-4.9t^{2}+50t + 2

Solve for t by completing the square:

-4.9t^{2}+50t = -2

-4.9(t^{2}-10.2t) = -2

t^{2}-10.2t = 0.41

t^{2}-2(5.1)t +(5.1)^{2}=0.41 + (5.1)^{2}

(t-5.1)^{2}=26.43

t-5.1 = \pm 5.14

t-5.1 = 5.14 and t-5.1 = -5.14

t=10.2 sec and t=-0.04 sec

The arrow will hit the ground about 10.2 seconds after it is shot.

c. If we graph the height of the arrow with respect to time we would get an upside down parabola (a > 0). The maximum height and the time when this occurs is really the vertex of this parabola: (t, h).

Rewrite the equation in vertex form:   y=-4.9t^{2}+ 50t + 2
      y-2=4.9t^{2}+50t
    y-2=4.9(t^{2}-10.2t)
Complete the square:   y-2-4.9(5.1)^{2}=-4.9(t^{2}-10.25+(5.1)^{2})
    y-129.45=-4.9(t-5)^{2}

 

The vertex is (5.1, 129.45). In other words, when t = 5.1 seconds, the height is y = 129 meters.

Another type of application problem that can be solved using quadratic equations is one where two objects are moving away from each other in perpendicular directions. Here is an example of this type of problem.
 

Example C

Two cars leave an intersection. One car travels north; the other travels east. When the car traveling north had gone 30 miles, the distance between the cars was 10 miles more than twice the distance traveled by the car heading east. Find the distance between the cars at that time.

Solution

Let's make a sketch:

We can use the Pythagorean Theorem to find an equation for x:   [x^{2}+30^{2}=(2x+10)^{2}]
Expand the parentheses and simplify:     x^{2}+900=4x^{2}+40x+100
    800 = 3x^{2}+40x
Solve by completing the square:   \dpi{100} \frac{2400}{3}+ (\frac{20}{3})^{2}=x^{2}+ 2(\frac{20}{3})x+(\frac{20}{3})^2
    \frac{2800}{9}=(x+\frac{20}{3})^{2}
    x+\frac{20}{3}=17.6 and x+\frac{20}{3}=-17.6
    x=11 and x=-24.3

 

Since only positive distances make sense here, the distance between the two cars is:

2x+10=2(11)+10 = 32 miles.

Solve Applications Using Quadratic Functions by any Method

Here is an application problem that arises from number relationships and geometry applications.

Example D

The product of two positive consecutive integers is 156. Find the integers.

Solution

Define: Let x = the smaller integer. Then x + 1 = the next integer

Translate: The product of the two numbers is 156. We can write the equation:

x(x+1)=156

Solve:

       x^{2}+x-156=0

 

Apply the quadratic formula with: a = 1, b = 1, c = – 156  

 

 

x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a}

x=\frac{-1\pm\sqrt{1^{2}-4(1)(-156)} }{2(1)}

Simplify:   x=\frac{-1\pm\sqrt{1+624} }{2}
    x=\frac{-1\pm\sqrt{625} }{2}
    x=\frac{-1\pm 25}{2}

 

x=\frac{24}{12}=2 and     x=\frac{-26}{12}=-3

Since we are looking for positive integers, we want [x = 12] . So the numbers are 12 and 13.

Check: 12 x 13 = 156. The answer checks out.

Example E

Suzie wants to build a garden that has three separate rectangular sections. She wants to fence around the whole garden and between each section as shown. The plot is twice as long as it is wide and the total area is 200 sq ft. How much fencing does Suzie need?  

Solution

Define: Let x = the width of the plot. Then 2x =  the length of the plot

Translate: area of a rectangle is A = length x width, so:

x(2x) = 200

2x^{2} = 200 or x^{2} = 100

    x^{2} = 100
Solve by taking the square root:     x=\pm \sqrt{100}=\pm 10
    x=10 and x=-10

 

We take x = 10 since only positive dimensions make sense. The plot of land is 10 feet x 20 feet. To fence the garden the way Suzie wants, we need 2 lengths and 4 widths = 2(20) + 4(10) = 80 feet of fence.  

Check: 10 x 20 = 200 ft2 and 2(20) + 4(10) = 80 feet. The answer checks out.

If after completing this topic you can state without hesitation that...

  • I can compare methods for solving quadratics.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.07 Comparing Methods for Solving Quadratics - Quiz (Math Level 2)

computer-scored 60 points possible 40 minutes
Pacing: complete this by the end of Week 4 of your enrollment date for this class.


05.08 Quadratic Models--Using Quadratic Word Problems (Math Level 2)

Apply quadratic functions to real world situations in order to solve problems.

Quadratic functions are more than algebraic curiosities—they are widely used in science, business, and engineering. The U-shape of a parabola can describe the trajectories of water jets in a fountain and a bouncing ball, or be incorporated into structures like the parabolic reflectors that form the base of satellite dishes and car headlights. Quadratic functions help forecast business profit and loss, plot the course of moving objects, and assist in determining minimum and maximum values. Most of the objects we use every day, from cars to clocks, would not exist if someone, somewhere hadn't applied quadratic functions to their design.

We commonly use quadratic equations in situations where two things are multiplied together and they both depend of the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, we use a quadratic equation. Because the quantity of a product sold often depends on the price, we sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.

Using the Parabola


A very common and easy-to-understand application of a quadratic function is the trajectory followed by objects thrown upward at an angle. In these cases, the parabola represents the path of the ball (or rock, or arrow, or whatever is tossed).  If we plot distance on the x-axis and height on the y-axis, the distance of the throw will be the x value when y is zero. This value is one of the roots of a quadratic equation, or x-intercepts, of the parabola.  We know how to find the roots of a quadratic equation—by either factoring, completing the square, or by applying the quadratic formula.

Let’s look at a throw made by a shot-putter. Notice that x = 0 when the shot-putter has the shot (a heavy metal ball) in his hand—the shot hasn't gone anywhere yet. The shot-putter usually starts with the shot at his shoulder, so y (height) is not 0 when x = 0:

Example 1

Problem:

A shot-put throw can be modeled using the equation y=-0.0241x^{2}+x+5.5, where x is distance traveled (in feet) and y is the height (also in feet).  How long was the throw?

y=-0.0241x^{2}+x+5.5. The throw ends when the shot hits the ground. The height y at that point is 0, so set the equation equal to zero.

This equation is difficult to factor or to complete the square, so we'll solve by applying the quadratic formula,

y=-0.0241x^{2}+x+5.5

y=-0.0241x^{2}+x+5.5

Simplify.

y=-0.0241x^{2}+x+5.5

y=-0.0241x^{2}+x+5.5

Find both roots.

y=-0.0241x^{2}+x+5.5 or y=-0.0241x^{2}+x+5.5

Do the roots make sense? The parabola described by the quadratic function has two x-intercepts. But the shot only traveled along part of that curve.

x ≈ 46.4 or -4.9

One solution, –4.9, cannot be the distance traveled because it is a negative number.

The other solution, 46.4 feet, must give the distance of the throw.

Answer: Approximately 46.4 feet

EXAMPLE 2

Although a stadium field of synthetic turf appears to be flat, its surface is actually shaped like a parabola. This is so that rainwater runs off to the sides. If we take a cross section of the turf, the surface can be modeled by y=-0.000234(x-80)^{2}+1.5, where x is the distance from the left end of the field and y is the height of the field. What is the width of the field?

A) 80 ft

B) 1.5 ft

C) 234 ft

D) 160 ft

A) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.

B) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.

C) Incorrect. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.

D) Correct. The width of the field is the distance between the roots of the equation. Set the equation equal to 0:  –0.000234(x – 80)2 + 1.5 = 0. Subtracting 1.5 from both sides and then dividing both sides by -0.000234 gives (x – 80)2 = 6410 (approximately). Taking the square root of both sides gives x – 80 = ±80, so the roots are x = 0 and x = 160. The correct answer is 160 feet.

Finding the Maximum and Minimum


Another common use of the quadratic equation in real world applications is to find maximum (the most or highest) or minimum (the least or lowest) values of something.  Recall that the vertex is the turning point of a parabola. For a parabola opening downward, the vertex is the high point, which occurs at the maximum possible y value. For a parabola opening upward, the vertex is the lowest point of the parabola, and occurs at the minimum y value.

To use a quadratic equation to find a maximum or minimum, we usually want to put the quadratic equation into the vertex form of a quadratic equation, y=a(x-h)^{2}+k. This lets us quickly identify the coordinates of the vertex (h, k).

Let's see how this works by trying a motion problem. The equation h=-16t^{2}+v_{0}t+h_{0} is commonly used to model an object that is launched or thrown. The variable h represents the height in feet, and t represents the time in seconds. The other two values are usually given numbers: h0 is the initial height in feet and v0 is the initial velocity in feet/second.

When working with this equation, we assume the object is in “free fall”, which means it is moving under the sole influence of gravity.  There is no air resistance or any other interference of any kind (not so likely in the real world, but still, these equations are quite helpful).

Example

Problem

A ball is launched upward at 48 ft/s from a platform that is 100 ft. high. Find the maximum height the ball reaches and how long it will take to get there.   

Start with the equation that models an object being launched or thrown.

h=-16t^{2}+v_{0}t+h_{0}

Substitute the initial velocity v0 = 48 and height h0 = 100.

h=-16t^{2}+48t+100

Since we want to get the vertex form of the equation, a(x – h)2, factor -16 out of the first two terms. The value of a is -16 and we will use t for x. That way we can complete the square on t2 – 3t to get the equation in vertex form.

h=-16(t^{2}-3t)+100

Remember that when we complete the square, we add a value to the expression. Because of the coefficient on the t2 term, this can get a little confusing, so we are going to prepare to complete the square for t2 – 3t by adding c to t2 – 3t, inside the parentheses.

When we add a quantity to one side of the equation, we must also add it to the other side. Because the quantity added, c, is inside the parenthesis on the right, we are actually adding -16c. This means when we add the quantity to the left side, we must add -16c.

h+(-16c)=-16(t^{2}-3t+c)+100

To complete the square on x2 + bx, we add \left ( \frac{b}{2} \right )^{2}, so c=\left ( \frac{b}{2} \right )^{2}=\left ( \frac{-3}{2} \right )^{2}=\left ( \frac{3}{2} \right )^{2}

Substitute this value in for c on both sides of the equation.

h+-16\left ( \frac{3}{2} \right )^{2}=-16\left [ t^{2}-3t+\left ( \frac{3}{2} \right )^{2} \right ]+100

Simplify by  writing the square of the binomial on the right, and simplify-16\left ( \frac{3}{2} \right )^{2}=-16\left ( \frac{9}{4} \right )=-36 on the left.

h+-36=-16\left ( t-\frac{3}{2} \right )^{2}+100

Add 36 to both sides. We now have the vertex form.

h=-16\left ( t-\frac{3}{2} \right )^{2}+136

We can identify the vertex as, \left ( \frac{3}{2},136 \right )

The x-coordinate is t in this equation, which stands for time. The y-coordinate represents height.

ANSWER:

The maximum height is 136 feet and it will take 1.5 seconds to reach this height

(We could have found the vertex by using other methods, such as by graphing or by using the formula x=-\frac{b}{2a} to find the x-coordinate of the vertex, then substitute that x-value into the original formula to get the y-value of the vertex).

Give this one a try.

A farmer has 1000 feet of fencing and a very big field. She can enclose a rectangular area with dimensions x ft and 500 – x ft. What is the largest rectangular area she can create?

A) 62,500 ft2

B) 250,000 ft2

C) 1,000 ft2

D) 500 ft2

A) Correct. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.

B) Incorrect. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). When completing the square to find the vertex form of this equation, the binomial to be squared is x – 250, not x – 500. You probably forgot that the number is \frac{b}{2}, not b. The vertex form is y = -(x – 250)2 + 62,500 and the vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.

C) Incorrect. The perimeter of the rectangular area will be 1,000 feet, because that’s the amount of fencing you have. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.

D) Incorrect. The area y is y = x(500 – x), or y = -x2 + 500x (in ft2). You can read the maximum area using the vertex form of the equation, but this not the vertex form. The vertex form of a quadratic equation is y = a(x – h)2 + k. Completing the square to find the vertex form of this equation gives y = -(x – 250)2 + 62,500. The vertex of the parabola will be (250, 62,500). The maximum area possible is the y coordinate of the vertex, or 62,500 ft2.

Modeling a Situation


Quadratic equations are sometimes used to model situations and relationships in business, science and medicine. A common use in business is to maximize profit, that is, the difference between the total revenue (money taken in) and the production costs (money spent).

The relationship between the cost of an item and the quantity sold is often linear. In other words, for each $1 increase in price there is a corresponding decrease in quantity sold. (Think about it: if the price of something goes up, do you buy more or less? Hopefully less!) Once we determine a relationship between the selling price of an item and the quantity sold, we can think about how to generate the most profit.  At what selling price do we make the most money?

The amount of profit will be found by taking the total revenue (the quantity sold multiplied by the selling price) and subtracting the cost to produce the all the items: Profit = Total Revenue – Production Costs.  We can integrate the linear relationship of selling price to quantity and the Profit formula and create a quadratic equation, which we can then maximize.  Let’s look at an example:

Here is a sample data set:

Selling Price $(s)

Quantity Sold in 1 year (q)

10 1000
15 900
20 800
25 700

 

In order to calculate the profit, we also need to know how much it costs to produce each item. For this example, the cost to produce each item is $10.

Example

Problem

Using the data given above, determine the selling price s, which produces the maximum yearly profit.

Plot s on the horizontal axis and q on the vertical axis. Use any two of the points on the straight line of the plot to find the slope of the line to be -20.  Read the y-intercept as 1200.

Put these values into slope-intercept form (y = mx + b):

q = -20s + 1200

q = -20s + 1200

q = quantity sold

s = selling price of the item

Profit formula is P = Total Revenue – Production Costs

Total Revenue = price • quantity sold

Production Costs = cost per item • quantity sold

So P = sq – 10q

Substitute -20s + 1200 for q in profit formula.

P = s(-20s + 1200) – 10(-20s + 1200)

Multiply the expressions and combine like terms. We now have a quadratic function. 

P = -20s2 + 1200s + 200s – 12000

P = -20s2 + 1400s – 12000

By finding the vertex of the parabola, we will find the selling price that will generate the most profit. The x-axis represents selling price, so the value of the x-coordinate at the vertex represents the best price.

The y-value at the vertex will give the amount of profit made.

Find the x-coordinate of the vertex by applying the formula, x=-\frac{b}{2a}

In this case, the variable is s rather than x. The other values are a = -20, the coefficient in the s2 term, and 1400, the coefficient in the s term.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Answer

The selling price that generates the maximum profit is $35.

Here is a graph of the profit function, showing the vertex:

Next is a word problem you may not think is a quadratic equation.  The area problem below does not include a quadratic formula of any type and the problem seems to be something you have solved many times before by simply multiplying.  But in order to solve it, you will need to use a quadratic equation.

Example

Problem

Bob made a quilt that is 4 ft x 5 ft.  He has 10 sq. ft. of fabric to create a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)

Sketch the problem. Since we don’t know the width of the border, we give it the variable x.

Since each side of  the original 4 x 5 quilt has the border of width x added, the length of the quilt with the border will be 5 + 2x, and the width will be 4 + 2x.

(This is where you may start to think “Ah ha, this may be a quadratic equation after all. We have both dimensions written in the same variable, and will multiply them to get an area!")

We are only interested in the area of the border strips. Write an expression for the area of the border.

Area of border = Area of the blue rectangle minus the area of the red rectangle

Area of border = (4 + 2x)(5 + 2x) – (4)(5)

We have 10 sq ft of fabric for the border, so set the area of the border to be 10.

10 = (4 + 2x)(5 + 2x) – 20

Multiply (4 + 2x)(5 + 2x).

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Simplify.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Subtract 10 from both sides so that we have a quadratic equation set equal to zero and can apply the quadratic formula to find the roots of the equation.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Use the quadratic formula. In this case, a = 4, b = 18, and c = -10.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Simplify.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Find the solutions, making sure that the ± is evaluated for both values.

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35  or

s=-\frac{1400}{2(-20)}=\frac{-1400}{-40}=35

Ignore the solution x = -5, as the width could not be negative.

Answer

The width of the border should

be 0.5 ft.

Summary


Quadratic functions are used in many types of real world situations. They are useful in describing the trajectory of a ball, determining the height of a thrown object and in optimizing profit for businesses.  When solving a problem using a quadratic function it may be necessary to find the vertex or to describe a section of the parabola.

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

I highly recommend that you click on the links below and watch the videos after completing the topic.

05.08 Quadratic Models--Using Quadratic Word Problems - Worksheet (Math Level 2)

teacher-scored 56 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 4 of your enrollment date for this class.


05.09 Fundamental Theorem for Quadratic Polynomials (Math Level 2)

Use the Fundamental Theorem of Algebra to determine the number of roots and/or factors in a polynomial function.

The material for this lesson is from the Mathematics Vision Project (MVP), licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

In your earlier math classes you spent a great deal of time solving linear equations. For example:

Problem Solution
x+7=11 5x=7
3x-4=2 x=2
5x=7 x = \frac{7}{5}

 

How many solutions are there for each linear equation?

Yes, there is one.

What degree is the binomial x + 7? Remember the degree of a polynomial is the value of the highest exponent. Though we don’t write x1, the degree of this binomial is one.

We have spent quite some time solving quadratic equations this quarter. Let’s look at some of the problems and solutions we have worked on.

Problem Solution
x^{2}-4x-5=0 x=5 and x=-1
x^{2}=8 x=\sqrt{8} and x=-\sqrt{8}
-4x^{2}+x=2 x=\frac{1\pm i\sqrt{31}}{8}

 

How many solutions are there for each quadratic?

Yes, there are two.

How about this one? x^{2}+4x+4=0

Let's factor it: (x+2)(x+2)=0

Now solve: x+2=0 and x+2=0

So, the solution is x = –2. We don’t write duplicate solutions. However, there technically are still two solutions.

What is the degree of the polynomial: -4x^{2}+x-2?

Yes, it is two.

The following material is from EngageNY:Lesson 40: Obstacles Resolved—A Surprising Result licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

Example 1

Consider the polynomial P(x) = x3 + 3x2 + x = 5 whose graph is shown below:

  1. Looking at the graph, how do we know that there is only one real solution? (The graph has only one x-intercept)
  2. Is it possible for a cubic polynomial function to have no zeros? (No. Since the opposite ends of the graph of a cubic function go in opposite directions, the graph must cross the x-axis at some point. Since the graph must have an y-intercept, the function must have a zero).
  3. From the graph, what appears to be one solution to the equation x3 + 3x2 + x - 5 = 0? (The only real solution appears to be 1. Written as a factor this would be (x – 1))

Let’s divide this polynomial by x – 1:

Now, let’s see if this quotient has solutions. We’ll use the quadratic formula with a = 1, b = 4, and c = 5.

x=\frac{-b\pm\sqrt{b^{2}-4ac} }{2a}

x=\frac{-1\pm\sqrt{4^{2}-4(1)(5)} }{2(1)}

x=\frac{-4\pm \sqrt{16-20}}{2}

x=\frac{-4\pm \sqrt{-4}}{2} = \frac{-4\pm 2i}{2} = -2\pm i

So, x=-2+i and x=-2-i

Now, how many solutions are there for the degree 3 polynomial or a cubic?

Yes, there are three.

Are you beginning to see a pattern?

The Fundamental Theorem of Algebra

A polynomial function is a function of the form:

y=a_{0}x^{n}+ a_{1}x^{n-1}+ a_{2}x^{n-2}+ ... +a_{n-3}x^{3}+ a_{n-2}x^{2}+ a_{n-1}x+a_{n}

where all of the exponents are positive integers and all of the coefficients a_{0} . . . a_{n} are constants.

As the theory of finding roots of polynomial functions evolved, a 17th century mathematician, Girard

(1595-1632) made the following claim which has come to be known as the Fundamental Theorem of Algebra

An nth degree polynomial function has n roots.  (Mathematics Vision Project)

Let’s go back to some of our examples from earlier in this lesson but add a column:

Problem Solution Factored Form
x^{2}-4x-5=0 x=5 and x = -1 (x-5)(x+1)=0
x^{2}=8 x=\sqrt{8} and x=-\sqrt{8} (x-\sqrt{8})(x+\sqrt{8})=0
x^{3}+3x^{2}+x-5=0 x=1 and x=-2\pm i

(x-1)(x-(-2+i))(x-(-2-i)) or

(x-1)(x+2-i)(x+2-+)

 

All of the factors are called linear pairs. To generalize:

If P is a polynomial function of degree n ≥ 1, given by P(x) = a_{0}x^{n}+ a_{1}x^{n-1}+ a_{2}x^{n-2}+ ... +a_{n-3}x^{3}+ a_{n-2}x^{2}+ a_{n-1}x+a_{n} with real or complex coefficients ai, then P has exactly n zeros r1, r2, … rn (not all necessarily distinct), such that: P(x) = a(x – r1)(x – r2)…(x – rn).

So, when you know the factors, you can find the roots. Likewise, when you know the roots, you can find the factors.

If after completing this topic you can state without hesitation that...

  • I can use the Fundamental Theorem of Algebra to determine the number of roots and/or factors in a polynomial function.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

I highly recommend that you click on the links below and watch the videos before continuing.

05.09 Fundamental Theorem for Quadratic Polynomials - Worksheet (Math Level 2)

teacher-scored 66 points possible 35 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.


05.10 Solve Quadratic Inequalities (Math Level 2)

Solve quadratic inequalities.

The material for this lesson is from MathCentre: Solving Inequalities. Resources are licensed under Creative Commons License. (See link below). This authors are from Australia. They use the word “factorise.” We would say the “expression can be factored.”

Quadratic inequalities need handling with care.

Example

Suppose we wish to solve x2 − 3x + 2 > 0.

The quadratic expression on the left will factorise to give (x − 2)(x − 1) > 0. If this was a quadratic equation we would simply state x − 2 = 0 and x − 1 = 0 and hence x = 2 and x = 1.

Unfortunately with inequalities the situation is more complicated and we have a bit more work to do.

Whether (x − 2)(x − 1) is greater than zero or not depends upon the signs of the two factors (x − 2) and (x − 1). We investigate the possibilities using a grid as shown in Figure 10.

On the top line of the grid we have indicated the places where (x − 2)(x − 1) is equal to zero, that is when x is 1 or 2.

We write the two factors (x − 1) and (x − 2) in the first column on the left. We write their product at the bottom left.

The second column corresponds to where x is less than 1. When x < 1 both x − 1 and x – 2 will be negative and so we have inserted − signs to show this. The product (x − 1)(x − 2) will therefore be positive, and hence the + sign.

The third column corresponds to where x is greater than 1 but less than 2. In this interval x−1 is positive, but x−2 is negative, and hence the corresponding signs. The product will then be negative.

The fourth column shows what happens when x is greater than 2. Both factors are positive.

Hence their product is positive too.

We are looking for where (x − 2)(x − 1) > 0 and our grid shows us that this is true when x < 1 and when x > 2. The solution of the inequality is therefore x < 1 or x > 2. The solution is shown on the number line in Figure 11.

Example

Suppose we wish to solve the inequality −2x2 + 5x + 12 ≥ 0.

It will be easier to deal with this if the coefficient of x2 is positive rather than negative and so we multiply every term by −1 remembering to reverse the inequality.

The problem then is to solve 2x2 − 5x − 12 ≥ 0.

The quadratic expression can be factorised to give (2x + 3)(x − 4) ≥ 0.

Again we produce a grid. The first factor is zero when x = −3/2. The second factor is zero when x = 4. We write these two numbers on the top row of the grid as shown in Figure 12.

When x is less than −3/2 both factors are negative and hence their product is positive as indicated.

When x is greater than −3/2 but less than 4, 2x + 3 is positive, but x − 4 is negative. Hence the product of the two factors is negative.

When x is greater than 4, both factors are positive, and hence their product is positive.

We are looking for where 2x2 − 5x − 12 ≥ 0. From the grid we see that this occurs when

-\frac{3}{2}\leq x \leq 4

Note that since the quadratic expression is zero at the points x = −3/2 and x = 4 these must be included in the solution.

The range of values of x satisfying the inequality is shown on the number line in Figure 13.

Example

Suppose we wish to solve x2 − 3x + 2 > 0.

We consider the graph of y = x2 − 3x + 2 which has been drawn in Figure 14. Note that the quadratic expression factorises to give y = (x − 1)(x − 2) and so the graph crosses the x axis when x = 1 and when x = 2. We are looking for where x2 − 3x + 2 is greater than zero so we look at that part of the graph which is above the x axis. So the solution is

x < 1 or x > 2

We can mark this solution using the x axis as the number line.

Example

Suppose we wish to solve x2 − x − 6 ≤ 0.

The quadratic expression factorises to (x − 3)(x + 2) and the graph of y = (x − 3)(x + 2) is shown in Figure 15.

The graph crosses the x axis at x = −2 and at x = 3.

We are looking for where x2 − x − 6 lies on or below the x axis. By inspection the solution is −2 ≤ x ≤ 3

Again this solution is indicated on the graph.

I highly recommend that you click on the links below and watch the videos before continuing:

05.10 Solve Quadratic Inequalities - Videos (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I can solve quadratic inequalities.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.10 Solve Quadratic Inequalities - Worksheet (Math Level 2)

teacher-scored 72 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 5 of your enrollment date for this class.


05.10.01 Checkpoint Quiz - Quiz (Math Level 2)

computer-scored 22 points possible 30 minutes
Pacing: complete this by the end of Week 6 of your enrollment date for this class.


05.11 Derive the Equation of a Parabola Given a Focus and Directrix (Math Level 2)

Derive the equation of a parabola given a focus and directrix.

The material for this lesson is from EngageNY: Lesson 33: The Definition of a Parabola licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.

Suppose you are viewing the cross-section of a mirror. Where would the incoming light be reflected in each type of design? Sketch your ideas.

Hopefully, your drawings look something like this:

When Newton designed his reflector telescope he understood two important ideas. Figure 1 shows a diagram of this type of telescope.

http://en.wikipedia.org/wiki/File:Newton01.png#filelinks

  • The curved mirror needs to focus all the light to a single point that we will call the focus. An angled flat mirror is placed near this point and reflects the light to the eyepiece of the telescope.
  • The reflected light needs to arrive at the focus at the same time, otherwise the image is distorted.

As you learned in the previous lesson, the graphs of quadratic functions are in the shape of a parabola. We are going to learn more details about the elements of the parabola.

Definition: A parabola with directrix and focus point is the set of all points in the plane that are equidistant from the point and line.

Figure 2 above illustrates this definition of a parabola. In this diagram, FP1 = P1Q1, FP2 = P2Q2 = P3Q3 showing that for any point P on the parabola, the distance between P and F is equal to the distance between P and the line L.

All parabolas have the reflective property illustrated in Figure 3. Rays parallel to the axis will reflect off the parabola and through the focus point, F.

Thus, a mirror shaped like a rotated parabola would satisfy Newton’s requirements for his telescope design.

Figure 4 below shows several different line segments representing the reflected light with one endpoint on the curved mirror that is a parabola and the other endpoint at the focus. Anywhere the light hits this type of curved surface, it always reflects to the focus, F, at exactly the same time.

Figure 5 shows the same image with a directrix. Imagine for a minute that the mirror was not there. Then, the light would arrive at the directrix all at the same time. Since the distance from each point on the parabolic mirror to the directrix is the same as the distance from the point on the mirror to the focus, and the speed of light is constant, it takes the light the same amount of time to travel to the focus as it would have taken it to travel to the directrix. In the diagram, this means that AF = AFA, BF = BFB, and so on. Thus, the light rays arrive at the focus at the same time, and the image is not distorted.

Terminology

Example 1

Given a focus and a directrix, create an equation for a parabola.

Focus:  (0, 2)

Directrix: x-axis

Parabola: {(x, y) | (x, y) is equidistant to and to  and to the x-axis.}

Let A be any point (x, y) on the parabola P. Let F′ be a point on the directrix with the same x-coordinate as point A.

What is the length of AF′?

Distance Formula = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{^{}2}}

AF' = \sqrt{(x-x)^{2}+(y-0)^{2}} = \sqrt{y^{2}} = y

Use the distance formula to create an expression that represents the length of AF.

AF = \sqrt{(x-0)^{2}+(y-2)^{2}} = \sqrt{x^{2}+(y-2)^{2}}

We know that AF = AF' so:

y = \sqrt{x^{2}+(y-2)^{2}}

Now, square both sides:

y^{2} = {x^{2}+(y-2)^{2}} = x^{2}+y^{2} - 4y + 4

Solve for y:

4y = x^{2} + 4

y = \frac{1}{4}x^{2} + 1

Thus, the parabola P is the graph of the equation:

y = \frac{1}{4}x^{2} + 1

Now we probably do not want to have to use this process of using the distance formula every time we write an equation of the parabola. Let’s see if we can figure out a “general formula” that we can use to write the equation of a parabola given the vertex and the focus or directrix.

Consider the general graph of a parabola with the given focus (0, p), any point on the parabola, (x, y) and a point on the directrix, (x,-p).

Let’s use the same process shown in the example above. We know that the distance from the focus to any point on the parabola is the same distance from that point to the directrix as well. That means in this case that FM = MD. To find the equation of the parabola, we can use the distance formula to find the distance of FM and MD.

FM= \sqrt{(x-0)^{2}+(y-p)^{2}}

Since FM = MD, we can set these two equations equal to each other.

\sqrt{(x-0)^{2}+(y-p)^{}2} = \sqrt{(x-x)^{2}+(y+p)^{}2}

To solve this equation, we need to square both sides.

(x-0)^{2}+(y-p)^{}2 =(x-x)^{2}+(y+p)^{}2

Next, we simplify:

(x)^{2}+(y-p)^{}2 =(0)^{2}+(y+p)^{2}

Now, we will expand all the squares:

(x)^{2}+(y)^{2}-2py+p^{2} =(0)^{2}+(y)^{2}+2py + p^{2}

Now, we combine like terms:

(x)^{2}=4py

We can now solve for y:

y=\frac{1}{4p}x^{2} or in general, y=\frac{1}{4p}(x-h)^{2}+k where h and k are the coordinates of the vertex and p is the distance from the focus to the vertex or the distance from the vertex to the directrix.

The formula we can use to find the equation of a parabola when given the vertex and focus, or directrix is this:

\dpi{100} \fn_phv y=\frac{1}{4p}(x-h)^{2}+k

Let’s try an example.

Given the following information, write the vertex form of the equation of the parabola:

Vertex: (2, 3)

Focus: (2, 1)

In the general equation we found above, we need the vertex and the value of p. We have been given the vertex, so we need to find the value of p. p is the distance from the focus to the vertex.

To find that distance between the focus and vertex, you can plot each point and find the vertical distance between the two.

As you can see, the vertical distance between these two points is 2. That tells us that p = 2. Now we have all the information we need to write the equation of this parabola.

\dpi{100} \fn_phv y=\frac{1}{4p}(x-h)^{2}+k

\dpi{100} \fn_phv y=\frac{1}{4(2)}(x-2)^{2}+3

Now simplify:

y=\frac{1}{8}(x-2)^{2}+3

This is the equation of the parabola!!

Let’s try another example.

Given the following information, write the vertex form of the equation of the parabola:

Vertex: (-3,1) and Directrix: y = 6.

We have the vertex and now we need to find the value of p, which is the distance between the vertex and the directrix. To find the distance between the vertex and directrix, you can graph both and find the vertical distance between them.

Count the distance between the vertex point and the directrix line. It is a distance of 5. With the directrix above the vertex, that tells us the parabola is opening downward. When the parabola opens downward, the value of p will be negative. p = -5 Now we have enough information to write the equation of the parabola.

\dpi{100} \fn_phv y=\frac{1}{4p}(x-h)^{2}+k

y=\frac{1}{4(-5)}(x-(-3))^{2}+1

Now simplify:

y=\frac{1}{-20}(x+3)^{2}+1

This is the equation of the parabola!

We can also find the equation of a parabola given the vertex and another point on the parabola.

Given the vertex (-1,6) and another point (0,3) [This is also the y-intercept.]

This time we have to use a different method to find the value of "p" because we do not know the focus or the directrix. We will use both the vertex and the given point on the parabola to substitute into the equation of a parabola and then solve that for p.

\dpi{100} \fn_phv y=\frac{1}{4p}(x-h)^{2}+k

3=\frac{1}{4p}(0-(-1))^{2}+6

3=\frac{1}{4p}(0+1)^{2}+6 

Now subtract 6 from both sides.

-3=\frac{1}{4p}(1)^{2}

Square the 1.

-3=\frac{1}{4p}(1)

Multiply the 1/4p and the 1.

-3=\frac{1}{4p}    

Multiply both sides by 4p.

-12p=1   

Divide both sides by -12.

p=-\frac{1}{12}    

Now we have our value of p!! We can use it and the vertex to fill in the equation of the parabola.

y=\frac{1}{4(-\frac{1}{12})}(x-(-1))^{2}+6

Multiply the 4(-1/12).

y=\frac{1}{(-\frac{1}{3})}(x+1)^{2}+6       

To simplify the \frac{1}{-\frac{1}{3}}, we multiply the numerator by the reciprocal of -\frac{1}{3} which is -\frac{3}{1}=-3

y=-3(x+1)^{2}+6       

There we have the equation of the parabola!!

So far we have only been dealing with "vertical parabolas", parabolas that open up or down. We can also have "horizontal parabolas" which are parabolas that open right or left.

The equation for a horizontal parabola is very similar to the equation of a vertical parabola. The only difference is the x and y terms are exchanged, along with their corresponding vertex values.

\dpi{100} \fn_phv x=\frac{1}{4p}(y-k)^{2}+h

The process of finding the equation of a horizontal parabola is the same as demonstrated above for vertical parabolas. You just have to pay attention to where the values of the vertex now go.

Let's try an example!

Given Vertex: (-2,1), Focus: (-3,1)

We have been given the vertex, so we need to find the value of p. p is the distance from the focus to the vertex. We can find that distance by plotting each point on a graph.

Count the distance between the vertex point and the focus point. It is a distance of 1. With the focus to the left of the vertex, that tells us the parabola is opening to the left. When the parabola opens to the left, the value of p will be negative. p = -1 Now we have enough information to write the equation of the parabola.

x=\frac{1}{4p}(y-k)^{2}+h

x=\frac{1}{4(-1)}(y-1)^{2}-2

x=-\frac{1}{4}(y-1)^{2}-2

That's it!! You did it!!

Lesson Summary

Parabola: A parabola with directrix line L and focus point F is the set of all points in the plane that are equidistant from the point F and line L.

Axis of symmetry: The axis of symmetry of a parabola given by a focus point and a directrix is the perpendicular line to the directrix that passes through the focus.

Vertex of a parabola: The vertex of a parabola is the point where the axis of symmetry intersects the parabola.

In the Cartesian plane, the distance formula can help us to derive an analytic equation for the parabola.

If after completing this topic you can state without hesitation that...

  • I can derive the equation of a parabola given a focus and directrix.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

I highly recommend that you click on the links below and watch the videos before continuing:

05.11 Derive the Equation of a Parabola Given a Focus and Directrix - Videos (Math Level 2)

05.11 Derive the Equation of a Parabola Given a Focus and Directrix - Worksheet (Math Level 2)

teacher-scored 72 points possible 45 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 6 of your enrollment date for this class.


05.12 Solve Literal Equations With Squares (Math Level 2)

Solve literal equations involving squares.

Lesson material is from “Solving Linear Equations: Part II”, section 2.4 from the book Beginning Algebra (v. 1.0) at http://2012books.lardbucket.org. This book is licensed under a Creative Commons by-nc-sa 3.0 license.

Literal Equations

Algebra lets us solve whole classes of applications using literal equations, or formulas. Formulas often have more than one variable and describe, or model, a particular real-world problem. For example, the familiar formula D=rt describes the distance traveled in terms of the average rate and time; given any two of these quantities, we can determine the third. Using algebra, we can solve the equation for any one of the variables and derive two more formulas.

If we divide both sides by r, we obtain the formula t=\frac{D}{r}. Use this formula to find the time, given the distance and the rate.

If we divide both sides by t, we obtain the formula r=\frac{D}{t}. Use this formula to find the rate, given the distance traveled and the time it takes to travel that distance. Using the techniques learned up to this point, we now have three equivalent formulas relating distance, average rate, and time:


When given a literal equation, it is often necessary to solve for one of the variables in terms of the others. Use the properties of equality to isolate the indicated variable.

 

Example 1: Solve for a: P=2a+b

Solution: The goal is to isolate the variable a.

Answer: a=\frac{p-b}{2}

 

Example 2: Solve for y:z=\frac{x+y}{2}

Solution: The goal is to isolate the variable y.

Answer: y=2z-x

 

Try this! Solve for b: 2a-3b=c

 

Answer: b=\frac{2a-c}{3}

End of lesson material from “Solving Linear Equations: Part II”, section 2.4 from the book Beginning Algebra (v. 1.0) at http://2012books.lardbucket.org.

So far we have only dealt with "linear" literal equations where there are not any exponents greater than 1. Let's explore some literal equations that have exponents greater than 1.

A key thing to remember is when solving any type of literal equation, what you do to one side of the equation, you have to do it to the other side as well. You can add, subtract, multiply or divide something on both sides of the equation. Likewise, you can square both sides, or even square root both sides of the equation if necessary.

EXAMPLE 1: Solve for w: p=\sqrt{w}

Solution: We need to get w by itself. To do this, we will need a way to "undo" square rooting. We can do this by squaring both sides of the equation.

p^{2}=w

Answer: w=p^{2} We were able to get w by itself! To "undo" a square root, you simply square both sides of the equation.

 

EXAMPLE 2: Solve for x: w=\sqrt{x-y}

Solution:

w=\sqrt{x-y}      Square both sides of the equation.

w^{2}=x-y       Add y to both sides of the equation.

w^{2}+y=x    

Answer: x=w^{2}+y

 

EXAMPLE 3: Solve for n: k=\frac{m}{\sqrt{n}}

Solution:

k=\frac{m}{\sqrt{n}}      Undo the square root first by squaring both sides of the equation.

k^{2}=\frac{m^{2}}{n}     Multiply both sides by n.

nk^{2}=m^{2}    Divide both sides by k2.

n=\frac{m^{2}}{k^{2}}

EXAMPLE 4: Solve for x: y=5-x^{2}

Solution:

y=5-x^{2}    Subtract 5 from both sides of the equation.

y-5=-x^{2}  Divide everything by -1.

-y+5=x^{2} or 5-y=x^{2}   Square root both sides to solve for x.

\sqrt{5-y}=x

Answer: x=\sqrt{5-y}

SUMMARY

To solve literal equations, you apply the techniques you know for solving any type of equation. You just need to remember that what you do to one side of the equation, you must do to the other side as well in order to truly keep the equation balanced.

I highly recommend that you click on the links below and watch the videos before continuing:

05.12 Solve Literal Equations With Squares - videos (Math Level 2)

05.12 Solve Literal Equations With Squares - Worksheet (Math Level 2)

teacher-scored 72 points possible 45 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 7 of your enrollment date for this class.


05.13 Summary of Quadratic Equations (Math Level 2)

Understand the properties of quadratics and solve quadratics problems.

The material for this lesson is from onemathematicalcat.org licensed under a Creative Commons Attribution-NonCommerical 2.5 License.

Definition: Quadratic function (standard form; vertex form)

Let a, b and c be real numbers, with a≠0.

A quadratic function is a function that can be written in the form y = ax2+bx+c; this is called the standard form of the quadratic function.

Every quadratic function can also be written in the form y = a(x−h)2 + k; this is called the vertex form of the quadratic function.

Click on the Wolfram Alpha widgets below to graph and compute properties of quadratic functions. After pressing ‘Submit’ be sure to scroll down to see the graphs!

There are also sample exercises at the bottom of the Resource Website link below you can complete to practice the concepts. Do not key in your answer. Scroll down to the bottom until you see this section:

Work through as many problems as you choose to make certain you understand the concept.

05.13 Summary of Quadratic Equations - External Practice (Math Level 2)

05.13 Summary of Quadratic Equations - Resource Website (Math Level 2)

If after completing this topic you can state without hesitation that...

  • I understand the properties of quadratics.
  • I can solve quadratics problems.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.13.01 Summary of Quadratic Equations - Quiz (Math Level 2)

teacher-scored 31 points possible 40 minutes
Pacing: complete this by the end of Week 7 of your enrollment date for this class.


05.14 Solve Systems of Equations with a Linear and Quadratic (Math Level 2)

Solve systems of equations involving linear, quadratic, and other non-linear functions.

A system of equations is a set of two or more equations that share two or more unknowns. The solutions to a system of equations are all the values that make all of the equations true, or the points where the graphs of the equations intersect.

We can solve a system of linear equations through graphing, substitution and linear combination. Systems of nonlinear functions, such as quadratic or exponential equations, can be handled with the same techniques.

To illustrate how to solve these systems, we'll focus on linear and quadratic systems with just two equations.  But be aware that systems can be larger and more complex than these examples.

Just like with systems of linear equations, when we find solutions to systems of non-linear equations, we are finding the intersection of their graphs or the places where the equations have the same variable values.

Systems of Linear and Quadratic Equations


Let’s begin by talking about two linear equations. The solution to this kind of system is the point of intersection between the two lines, or the place where the two equations have the same x and y values. There may be one solution, no solution, or an infinite number of solutions to a system of two linear equations:

One Solution No Solutions Infinite Solutions
If the graphs of the equations intersect, then there is one solution that is true for both equations. If the graphs of the equations do not intersect (for example, if they are parallel), then there are no solutions that are true for both equations. If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

 

In quarter 1, we solved systems of linear equations using three different methods; graphing, substitution or elimination. Let's do a quick review.

This portion of the lesson comes from “Solving Linear Systems by Graphing”, section 4.1 from the book Beginning Algebra (v. 1.0). This book is licensed under a Creative Commons by-nc-sa 3.0 license.

Solve by Graphing

Geometrically, a linear system consists of two lines, where a solution is a point of intersection. To illustrate this, we will graph the following linear system with a solution of (3, 2):

First, rewrite the equations in slope-intercept form so that we may easily graph them.

If we graph both of the lines on the same set of axes, then we can see that the point of intersection is indeed (3, 2), the solution to the system.

 

The Substitution Method

The idea is to solve one equation for one of the variables and substitute the result into the other equation. After performing this substitution step, we will be left with a single equation with one variable, which can be solved using algebra. This is called the substitution method, and the steps are outlined in the following example.

 

Example 1: Solve by substitution: \left \{ 2x+y=7

                                                  \left \{ 3x-2y=-7

Solution:

Step 1: Solve for either variable in either equation. If you choose the first equation, you can isolate y in one step.

Step 2: Substitute the expression -2x+7 for the y variable in the other equation.

This leaves you with an equivalent equation with one variable, which can be solved using the techniques learned up to this point.

Step 3: Solve for the remaining variable. To solve for x, first distribute −2:

Step 4: Back substitute to find the value of the other coordinate. Substitute x = 1 into either of the original equations or their equivalents. Typically, we use the equivalent equation that we found when isolating a variable in step 1.

The solution to the system is (1, 5). Be sure to present the solution as an ordered pair.

Step 5: Check. Verify that these coordinates solve both equations of the original system:

The Elimination Method

In this section, the goal is to develop another completely algebraic method for solving a system of linear equations. We begin by defining what it means to add equations together. In the following example, notice that if we add the expressions on both sides of the equal sign, we obtain another true statement.

For the system

we add the two equations together:

The sum of y and −y is zero and that term is eliminated. This leaves us with a linear equation with one variable that can be easily solved:

At this point, we have the x coordinate of the simultaneous solution, so all that is left to do is back substitute to find the corresponding y-value.

Hence the solution to the system is (3, 2). This process describes the elimination (or addition) method for solving linear systems. Of course, the variable is not always so easily eliminated. Typically, we have to find an equivalent system by applying the multiplication property of equality to one or both of the equations as a means to line up one of the variables to eliminate. The goal is to arrange that either the x terms or the y terms are opposites, so that when the equations are added, the terms eliminate. The steps for the elimination method are outlined in the following example.

 

Example 1: Solve by elimination: \left \{ 2x+y=7

                                                 \left \{ 3x-2y=-7

Solution:

Step 1: Multiply one, or both, of the equations to set up the elimination of one of the variables. In this example, we will eliminate the variable y by multiplying both sides of the first equation by 2. Take care to distribute.

This leaves us with an equivalent system where the variable y is lined up to eliminate.

Step 2: Add the equations together to eliminate one of the variables.

Step 3: Solve for the remaining variable.

Step 4: Back substitute into either equation or its equivalent equation.

Step 5: Check. Remember that the solution must solve both of the original equations.

End of lesson.


To solve a system of a linear equation and a quadratic equation, we will do the same thing, find the point—or points—of intersection between the two graphs:

One Solution No Solutions Two Solutions
If the parabola and the line touch at a single point, then there is one solution that is true for both equations. If the graphs of the equations do not intersect, then there are no solutions that are true for both equations. If the line intersects the parabola in two places, then there are two solutions that are true for both equations.

 

It does not make sense to consider the case where the two equations represent the same set of points, since a straight line can never be a parabola, and vice versa.

Notice that this means the possible number of solutions for a system of two linear equations is 0 (never touch), 1 (cross in one place), or infinity (the lines are identical). The number of solutions for a system with one linear equation and one quadratic equation is either 0 (never touch), 1 (touch in one place), or 2 (cross in two places).

Let's solve a system of a linear and a quadratic equation by graphing:

Example

Problem:  Solve the system by graphing the equations y=x^{2}-5 and y=3x+7

Graph each equation and locate the points of intersection.

Answer

This system has two solutions. We cannot determine the exact location of the intersection points from the graph, but they are approximately (-2,0) and (5,22)

Notice that while we can tell approximately where the two graphs intersect, it is difficult to find an exact location.

Now let’s solve the same system using substitution. When solving using substitution, use the following steps

  1. Select one equation and isolate a variable. (Choose an equation and variable which are easy to isolate!)
  2. Substitute the resulting expression for a variable into the other equation, every time that variable appears.
  3. Solve the second equation for the second variable.
  4. Substitute the solution to step 3 into the expression in step 1, to find the other variable.

 

Example

Problem:  Solve the system using the substitution method y=x^{2}-5 and y=3x+7

In this case both equations have y isolated, so we can set them equal.

y=x^{2}-5

Subtract 3x from both sides and subtract 7 from both sides.

y=x^{2}-5

Now there is a quadratic equation set equal to 0 and we can use the quadratic formula, to find the solution.

Substitute the values of a, b, and c into the formula.

Simplify.

Simplify some more, remembering to evaluate both and .

or

Evaluate either function for each x to find the corresponding y value.

              

         

Answer:

(5.27, 22.82) and (-2.27, 0.18)

By using substitution we have arrived at a more precise answer than we did when graphing this system, however, approximations were made when we took the square root of 57. This still is not an exact solution!

As the algebra of substitution can become quite involved, it is always a good idea to check the result in the original equations.

Here is a check of the point (5.27, 22.82):

                     

          

Notice that the check doesn’t result in perfect equality, but it's close.

Systems of Two Quadratic Equations


Now let’s look at two quadratic equations.  Picture for a moment how the graphs of two quadratic equations can intersect (or not).

One Solution

Two quadratic equations that have a single point in common, such as a shared vertex, have one solution.

Two Solutions

Two quadratic equations that overlap but have different equations have two solutions.

No Solutions

Two quadratic functions that have no overlap in range (no shared y-values) have no solutions.

Infinite Solutions

If the graphs of the equations are the same, then there are an infinite number of solutions that are true for both equations.

 

we can solve a system of quadratic equations by substitution:

Example

Problem

Solve the system using the substitution method: y=x^{2}+4 and y=-2x^{2}-6

In this case both equations have y isolated, so we can set them equal.

Add 2x2 and 6 to both sides to bring all the variables to one side of the equation.

Apply the quadratic formula.  a = 3, b = 0, and c = 10

Simplify, noticing that the quantity underneath the square root symbol is a negative value - this is the [discriminant] - no solution, and the graphs will not intersect.

Answer: no solution

We can also use linear combination to solve systems of equations, following these steps:

  1. Rearrange equations so the terms line up.
  2. Multiply none, or one, or both equations by a constant so that the coefficients of one of the variables are opposites
  3. Add the equations to eliminate one of the variables
  4. Solve the resulting equation
  5. Substitute the solution in step 4 back into an original equation to find the other variable

Ready to give that a try?

Example

Problem   Solve the system by using linear combination. y=x^{2}+2 and y=-x^{2}+8

Line up the equations.

Since there are already two variables that are opposites (x2 and –x2), we can add the two equations.

Isolate y by dividing both sides of the equation by 2

y = 5

Substitute for y into one of the equations to find the values of x.

Answer and

Summary


Solving systems of non-linear equation can be done using the techniques of graphing, substitution and linear combination. Just like with systems of linear equations, when we find solutions to systems of non-linear equations, we are finding the intersection of their graphs or the places where the equations have the same variable values.

Click on the link below for the lesson material.

05.14 Solve Systems of Equations with a Linear and Quadratic - External Link (Math Level 2)

Each topic is divided into sections that include the following:

  • Warm Up - questions to answer to see if you are ready for the lesson.
  • Presentation - high quality video with excellent illustrations that teaches the topic.
  • Worked Examples - examples that are worked out step-by-step with narration.
  • Practice - quiz problems on the topic covered.
  • Review - practice test to check your knowledge before moving on.

You are not required to complete every section. However, REMEMBER the goal is to MASTER the material!!

If after completing this topic you can state without hesitation that...

  • I can solve quadratic equations by completing the squares.
  • I can convert a quadratic equation to vertex form.

…you are ready for the assignment! Otherwise, go back and review the material before moving on.

05.14 Solve Systems of Equations with a Linear and Quadratic - Worksheet (Math Level 2)

teacher-scored 78 points possible 40 minutes

Activity for this lesson

Complete the attached worksheet.

  1. Print the worksheet and complete the assignment in the space provided. You may use additional paper if needed. Work all the problems showing ALL your steps.
  2. Once you have completed the assignment, digitize (scan or take digital photo, up close and clear) and save it to the computer and convert it to an image file such as .pdf or .jpg.
  3. Finally, upload the image using the assignment submission window under the assignment link on your math home page for this assignment.

Pacing: complete this by the end of Week 8 of your enrollment date for this class.