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CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS Chapter Objectives

By the end of this chapter, students should be able to: Apply the Square Root Property to solve quadratic equations Solve quadratic equations by completing the square and using the Quadratic Formula Solve applications by applying the quadratic formula or completing the square

Contents CHAPTER 13: QUADRATIC EQUATIONS AND APPLICATIONS .................................................................. 355

SECTION 13.1: THE SQUARE ROOT PROPERTY .................................................................................... 356

A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY ............................. 356

B. ISOLATE THE SQUARED TERM .................................................................................................. 358

C. USE THE PERFECT SQUARE FORMULA ..................................................................................... 359

EXERCISE ........................................................................................................................................... 360

SECTION 13.2: COMPLETING THE SQUARE .......................................................................................... 361

A. COMPLETE THE SQUARE .......................................................................................................... 361

B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1 .................................. 362

C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A ≠ 1 .................................. 363

EXERCISE ........................................................................................................................................... 365

SECTION 13.3: QUADRATIC FORMULA ................................................................................................ 366

A. DETERMINANT OF A QUADRATIC EQUATION ......................................................................... 366

B. APPLY THE QUADRATIC FORMULA .......................................................................................... 368

C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO .............................................................. 370

EXERCISE ........................................................................................................................................... 371

SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS .......................................................... 372

A. PYTHAGOREAN THEOREM ....................................................................................................... 372

B. PROJECTILE MOTION ................................................................................................................ 373

C. COST AND REVENUE ................................................................................................................. 374

EXERCISE ........................................................................................................................................... 376

CHAPTER REVIEW ................................................................................................................................. 377

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We might recognize a quadratic equation from the factoring chapter as a trinomial equation. Although, it may seem that they are the same, but they aren’t the same. Trinomial equations are equations with any three terms. These terms can be any three terms where the degree of each can vary. On the other hand, quadratic equations are equations with specific degree on each term.

Definition

A quadratic equation is a polynomial equation of the form 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 + 𝒄𝒄 = 𝟎𝟎

Where 𝒂𝒂𝒂𝒂𝟐𝟐 is called the leading term, 𝒃𝒃𝒂𝒂 is call the linear term, and 𝒄𝒄 is called the constant coefficient (or constant term). Additionally, 𝒂𝒂 ≠ 𝟎𝟎.

SECTION 13.1: THE SQUARE ROOT PROPERTY A. SOLVE BASIC QUADRATIC EQUATIONS USING SQUAREROOT PROPERTY

Square root property

Let 𝒂𝒂 ≥ 𝟎𝟎 and 𝒂𝒂 ≥ 𝟎𝟎. Then

𝒂𝒂𝟐𝟐 = 𝒂𝒂 if and only if 𝒂𝒂 = ±√𝒂𝒂 In other words,

𝒂𝒂𝟐𝟐 = 𝒂𝒂 if and only if 𝒂𝒂 = √𝒂𝒂 or 𝒂𝒂 = −√𝒂𝒂

MEDIA LESSON Solve basic quadratic equations using square root property (Duration 2:53)

View the video lesson, take notes and complete the problems below

Example: a) 8𝑥𝑥2 = 648 b) 𝑥𝑥2 = 75

YOU TRY

Solve. a) 𝑥𝑥2 = 81

b) 𝑥𝑥2 = 44

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MEDIA LESSON Solve equations with even exponents (Duration 4:26)

View the video lesson, take notes and complete the problems below

Consider: 52= ________________ and (−5)2 = ________________________

When we clear an even exponent, we have ________________________________________________.

Example: Solve.

a) (5𝑥𝑥 − 1)2 = 49 b) �(3𝑥𝑥 + 2)44 = 81

YOU TRY

Solve. a) (𝑥𝑥 + 4)2 = 25

b) (6𝑥𝑥 − 9)2 = 45

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B. ISOLATE THE SQUARED TERM Let’s look at examples where the leading term, or squared term, is not isolated. Recall, the squared term must be isolated to apply the square root property.

MEDIA LESSON Solve equations using square root property – Isolating the squared term 1st (Duration 5:00)

View the video lesson, take notes and complete the problems below

Before we can clear an exponent, it must first be _____________________________.

Example:

a) 4 − 2(2𝑥𝑥 + 1)2 = −46 b) 5(3𝑥𝑥 − 2)2 + 6 = 46

YOU TRY

Solve.

a) 5(3x − 6)2 + 7 = 27

b) 5(r + 4)2 + 1 = 626

Note: When we have the other side of the equation of a squared term negative, the equation does not have a real solution. For example, the equation 𝑥𝑥2 = −1 does not have a real solution. There is a complex solution for this equation but we will not discuss it in this class.

Example: Solve

2𝑛𝑛2 + 5 = 4 2𝑛𝑛2 = 4 − 5 2𝑛𝑛2 = −1

𝑛𝑛2 = −12

This equation does not have a real solution.

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C. USE THE PERFECT SQUARE FORMULA In order for us to be able to apply the square root property to solve a quadratic equation, we cannot have the 𝑥𝑥 term in the middle because if we apply the square root property to the 𝑥𝑥 term, we will make the equation more complicated to solve. However sometimes, we have special cases that we can apply the perfect square formula to get rid of the 𝑥𝑥 term in the middle and then apply the square root property to solve the equations. Recall: Perfect square formula 𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃+ 𝒃𝒃𝟐𝟐 = (𝒂𝒂 + 𝒃𝒃)𝟐𝟐 or 𝒂𝒂𝟐𝟐 − 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐 = (𝒂𝒂 − 𝒃𝒃)𝟐𝟐

MEDIA LESSON Solve equations using square root property – Perfect Square formula (Duration 4:09)

View the video lesson, take notes and complete the problems below Example:

a) 𝑥𝑥2 + 8𝑥𝑥 + 16 = 4 b) 9𝑥𝑥2 − 12𝑥𝑥 + 4 = 25

YOU TRY

Solve.

a) 𝑥𝑥2 − 6𝑥𝑥 + 9 = 81

b) 9𝑥𝑥2 + 30𝑥𝑥 + 25 = 4

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EXERCISE Solve by applying the square root property.

1) (𝑥𝑥 − 3)2 = 16

2) (𝑥𝑥 − 2)2 = 49

3) (𝑥𝑥 − 7)2 = 4

4) (𝑠𝑠 − 5)2 = 16

5) (𝑝𝑝 + 5)2 = 81

6) (𝑠𝑠 + 3)2 = 4

7) (𝑡𝑡 + 9)2 = 37

8) (𝑎𝑎 + 5)2 = 57

9) (𝑛𝑛 − 9)2 = 63

10) (𝑟𝑟 + 1)2 = 125

11) (9𝑟𝑟 + 1)2 = 9

12) (7𝑚𝑚 − 8)2 = 36

13) (3𝑠𝑠 − 6)2 = 25

14) 5(𝑘𝑘 − 7)2 − 6 = 369

15) 5(𝑔𝑔 − 5)2 + 13 = 103

16) 2𝑛𝑛2 + 7 = 5

17) (2𝑠𝑠 + 1)2 = 0

18) (𝑧𝑧 − 4)2 = 25

19) 3𝑛𝑛2 + 2𝑛𝑛 = 2𝑛𝑛 + 24

20) 8𝑛𝑛2 − 29 = 25 + 2𝑛𝑛2

21) 2(𝑟𝑟 + 9)2 − 19 = 37

22) 3(𝑛𝑛 − 3)2 + 2 = 164

23) 7(2𝑥𝑥 + 6)2 − 5 = 170

24) 6(4𝑥𝑥 − 4)2 − 5 = 145

Apply the perfect square formula and solve the equations by using the square root property.

25) 𝑥𝑥2 + 12𝑥𝑥 + 36 = 49

26) 𝑥𝑥2 + 6𝑥𝑥 + 9 = 2

27) 16𝑥𝑥2 − 40𝑥𝑥 + 25 = 16

28) 𝑥𝑥2 + 4𝑥𝑥 + 4 = 1

29) 𝑥𝑥2 − 14𝑥𝑥 + 49 = 9

30) 25𝑥𝑥2 + 10𝑥𝑥 + 1 = 49

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SECTION 13.2: COMPLETING THE SQUARE

When solving quadratic equations previously (then known as trinomial equations), we factored to solve. However, recall, not all equations are factorable. Consider the equation 𝑥𝑥2 − 2𝑥𝑥 − 7 = 0. This equation is not factorable, but there are two solutions to this equation: 1 + √2 and 1 − √2. Looking at the form of these solutions, we obtained these types of solutions in the previous section while using the square root property. If we can obtain a perfect square, then we can apply the square root property and solve as usual. This method we use to obtain a perfect square is called completing the square.

Recall. Special product formulas for perfect square trinomials:

(𝒂𝒂 + 𝒃𝒃)𝟐𝟐 = 𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐 𝒐𝒐𝒐𝒐 (𝒂𝒂 − 𝒃𝒃)𝟐𝟐 = 𝒂𝒂𝟐𝟐 − 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐

We use these formulas to help us solve by completing the square.

A. COMPLETE THE SQUARE We first begin with completing the square and rewriting the trinomial in factored form using the perfect square trinomial formulas.

MEDIA LESSON Complete the square (Duration 5:00)

View the video lesson, take notes and complete the problems below

Complete the square. Find 𝒄𝒄.

𝒂𝒂𝟐𝟐 + 𝟐𝟐𝒂𝒂𝒃𝒃 + 𝒃𝒃𝟐𝟐 is easily factored to ________________________________

To make 𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂+ 𝒄𝒄 a perfect square, 𝒄𝒄 = ___________________

Example:

a) 𝑥𝑥2 + 10𝑥𝑥 + 𝑐𝑐

b) 𝑥𝑥2 − 7𝑥𝑥 + 𝑐𝑐

c) 𝑥𝑥2 − 37𝑥𝑥 + 𝑐𝑐 d) 𝑥𝑥2 + 6

5𝑥𝑥 + 𝑐𝑐

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Note

To complete the square of any trinomial, we always square half of the linear term’s coefficient, i.e.,

�𝒃𝒃𝟐𝟐�𝟐𝟐

𝒐𝒐𝒐𝒐 �𝟏𝟏𝟐𝟐𝒃𝒃�

𝟐𝟐

We usually use the second expression when the middle term’s coefficient is a fraction.

YOU TRY

Complete the square by finding 𝒄𝒄:

a) 𝑥𝑥2 + 8𝑥𝑥 + 𝑐𝑐

b) 𝑥𝑥2 − 7𝑥𝑥 + 𝑐𝑐

c) 𝑥𝑥2 + 53𝑥𝑥 + 𝑐𝑐

B. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, a = 1

Steps to solving quadratic equations by completing the square

Given a quadratic equation 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 + 𝒄𝒄 = 𝟎𝟎, we can use the following method to solve for 𝒂𝒂.

Step 1. Rewrite the quadratic equation so that the coefficient of the leading term is one, and the original constant coefficient is on the opposite side of the equal sign from the leading and linear terms. 𝒂𝒂𝒂𝒂𝟐𝟐 + 𝒃𝒃𝒂𝒂 __________ = 𝒄𝒄 + _____________

Step 2. If 𝒂𝒂 ≠ 𝟏𝟏, divide both sides of the equation by 𝒂𝒂

Step 3. Complete the square, i.e., �𝒃𝒃𝟐𝟐�𝟐𝟐𝑜𝑜𝑟𝑟 �𝟏𝟏

𝟐𝟐𝒃𝒃�

𝟐𝟐 and add the result to both sides of the quadratic

equation. Step 4. Rewrite the perfect square trinomial in factored form. Step 5. Solve using the square root property. Step 6. Verify the solution(s).

MEDIA LESSON Solve quadratic equation by completing the square (Duration 8:40)

View the video lesson, take notes and complete the problems below

Solve the quadratic equation using the square root principle.

(𝑥𝑥 − 5)2 = 28

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Example: a) 𝑥𝑥2 + 6𝑥𝑥 − 9 = 0 b) 𝑥𝑥2 − 5𝑥𝑥 + 10 = 0

c) 3𝑥𝑥2 + 2𝑥𝑥 − 9 = 0

YOU TRY

Solve.

a) 𝑥𝑥2 + 10𝑥𝑥 + 24 = 0

b) 𝑛𝑛2 − 8𝑛𝑛 + 4 = 0

C. SOLVE QUADRATIC EQUATIONS BY COMPLETING THE SQUARE, A ≠ 1

MEDIA LESSON Solve quadratic equation by completing the square – a ≠1 (Duration 4:59)

View the video lesson, take notes and complete the problems below To complete the square: 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

1. Separate ____________________ and ______________________

2. Divide by ________________ (everything!)

3. Find _________ and _____________ to ______________________

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Example

a) 3𝑥𝑥2 − 15𝑥𝑥 + 18 = 0 b) 8𝑥𝑥 + 32 = 4𝑥𝑥2

YOU TRY

Solve.

a) 3𝑥𝑥2 − 36𝑥𝑥 + 60 = 0

b) 2𝑘𝑘2 + 𝑘𝑘 − 2 = 0

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EXERCISE Complete the square.

1) 𝑥𝑥2 − 30𝑥𝑥 + _____

2) 𝑚𝑚2 − 36𝑚𝑚 + _____

3) 𝑥𝑥2 − 15𝑥𝑥 + _____

4) 𝑦𝑦2 − 𝑦𝑦 + _____

5) 𝑎𝑎2 − 24𝑎𝑎 + _____

6) 𝑥𝑥2 − 34𝑥𝑥 + _____

7) 𝑟𝑟2 − 19𝑟𝑟 + _____

8) 𝑝𝑝2 − 17𝑝𝑝 + _____

Solve each equation by completing the square. If the solution is not a real solution, then state not a real solution.

9) 6𝑟𝑟2 + 12𝑟𝑟 − 24 = −6

10) 6𝑛𝑛2 − 12𝑛𝑛 − 14 = 4

11) 𝑎𝑎2 − 56 = 10𝑎𝑎

12) 𝑥𝑥2 + 8𝑥𝑥 + 15 = 8

13) 8𝑛𝑛2 + 16𝑛𝑛 = 64

14) 𝑛𝑛2 + 4𝑛𝑛 = 12

15) 𝑥𝑥2 − 16𝑥𝑥 + 55 = 0

16) 𝑛𝑛2 = −21 + 10𝑛𝑛

17) 4𝑏𝑏2 − 15𝑏𝑏 + 56 = 3𝑏𝑏2

18) 𝑏𝑏2 + 7𝑏𝑏 − 33 = 0

19) 𝑥𝑥2 + 10𝑥𝑥 − 57 = 4

20) 𝑛𝑛2 − 8𝑛𝑛 − 12 = 0

21) 𝑛𝑛2 − 16𝑛𝑛 + 67 = 4

22) 𝑥𝑥2 = −10𝑥𝑥 − 29

23) 5𝑘𝑘2 − 10𝑘𝑘 + 48 = 0

24) 7𝑛𝑛2 − 𝑛𝑛 + 7 = 7𝑛𝑛 + 6𝑛𝑛2

25) 2𝑥𝑥2 + 4𝑥𝑥 + 38 = −6 26) 8𝑏𝑏2 + 16𝑏𝑏 − 37 = 5

27) 5𝑥𝑥2 + 5𝑥𝑥 = −31− 5𝑥𝑥 28) 𝑣𝑣2 + 5𝑣𝑣 + 28 = 0

29) 𝑘𝑘2 − 7𝑘𝑘 + 50 = 3

30) 5𝑥𝑥2 + 8𝑥𝑥 − 40 = 8

31) 8𝑟𝑟2 + 10𝑟𝑟 = −55

32) −2𝑥𝑥2 + 3𝑥𝑥 − 5 = −4𝑥𝑥2

33) 8𝑎𝑎2 + 16𝑎𝑎 − 1 = 0

34) 𝑝𝑝2 − 16𝑝𝑝 − 52 = 0

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SECTION 13.3: QUADRATIC FORMULA The quadratic formula is derived from the method of completing the square. If we took a general quadratic equation

𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0

and solved for 𝑥𝑥 by completing the square, we would obtain the quadratic formula. Let’s try this.

MEDIA LESSON Deriving the Quadratic Formula (Duration 4:04)

Quadratic formula

Given the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0. Then

𝑥𝑥 =−𝑏𝑏 ± √𝑏𝑏2 − 4𝑎𝑎𝑐𝑐

2𝑎𝑎

is called the quadratic formula. The quadratic formula is a formula for solving quadratic equations in terms of the coefficients.

A. DETERMINANT OF A QUADRATIC EQUATION To make the quadratic formula easier to manage, we should calculate the discriminant first. Then substitute it into the quadratic formula to find 𝑥𝑥.

Discriminant

Given the quadratic equation 𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0, then its discriminant, denoted as the upper-case

Greek letter delta, Δ, is defined as 𝚫𝚫 = 𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄

√𝚫𝚫 = �𝒃𝒃𝟐𝟐 − 𝟒𝟒𝒂𝒂𝒄𝒄

Quadratic formula:

𝒂𝒂 =−𝒃𝒃 ± √𝚫𝚫

𝟐𝟐𝒂𝒂

Case 1: If Δ is positive, the equation has 2 solutions.

Case 2: If Δ is zero, the equation has 1 solution.

Case 3: If Δ is negative, the equation has no real solution. (The equation has two complex solutions

but we will not discuss complex numbers in this class).

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MEDIA LESSON Determinant (Duration 4:58)

View the video lesson, take notes and complete the problems below

𝑎𝑎𝑥𝑥2 + 𝑏𝑏𝑥𝑥 + 𝑐𝑐 = 0 𝑥𝑥 = _______________________________ • If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 ________________________________________________________________________

• If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 ________________________________________________________________________

• If 𝑏𝑏2 − 4𝑎𝑎𝑐𝑐 ________________________________________________________________________

Example: Determine the number of solutions to the quadratic equation.

𝑥𝑥2 + 14𝑥𝑥 + 49 = 0

MEDIA LESSON Determinant examples (Duration 4:58)

View the video lesson, take notes and complete the problems below

Example: Describe the type of solutions to the quadratic equation.

a) 𝑥𝑥2 − 3𝑥𝑥 − 28 = 0 𝑎𝑎 = ______ 𝑏𝑏 = ______ 𝑐𝑐 =______

a) 𝑥𝑥2 − 4𝑥𝑥 + 12 = 0 𝑎𝑎 = ______ 𝑏𝑏 = ______ 𝑐𝑐 =______

b) −2𝑥𝑥2 + 𝑥𝑥 + 5 = 0 𝑎𝑎 = ______ 𝑏𝑏 = ______ 𝑐𝑐 =______

c) 2𝑥𝑥2 + 8𝑥𝑥 + 8 = 0 𝑎𝑎 = ______ 𝑏𝑏 = ______ 𝑐𝑐 =______

Chapter 13

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YOU TRY

Find the determinant and determine how many solutions each of the following equation has.

a) 𝑥𝑥2 + 3𝑥𝑥 − 4 = 0

b) 2𝑥𝑥2 + 4𝑥𝑥 + 1 = 0

c) 𝑥𝑥2 − 4𝑥𝑥 + 4 = 0

d) 8𝑎𝑎2 + 5𝑎𝑎 + 1 = 0

B. APPLY THE QUADRATIC FORMULA

MEDIA LESSON Solve equations uing quadratic formula – two real rational solutions (Duration 3:14 )

View the video lesson, take notes and complete the problems below

Example: Solve using the quadratic formula.

6𝑥𝑥2 − 𝑥𝑥 − 15 = 0

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369

MEDIA LESSON Solve equations uing quadratic formula – two real irrational solutions (Duration 4:16 )

View the video lesson, take notes and complete the problems below

Example: Solve using the quadratic formula.

2𝑥𝑥2 − 4𝑥𝑥 − 3 = 0

MEDIA LESSON Solve equations uing quadratic formula – one solution (Duration 2:50 )

View the video lesson, take notes and complete the problems below

Example: Solve using the quadratic formula.

4𝑥𝑥2 − 12𝑥𝑥 + 9 = 0 YOU TRY

Solve.

a) 𝑥𝑥2 + 3𝑥𝑥 + 2 = 0

b) 3𝑥𝑥2 + 2𝑥𝑥 − 7 = 0

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C. MAKE ONE SIDE OF AN EQUATION EQUAL TO ZERO

MEDIA LESSON Set one side of an equation equal to 0 before solving (Duration 4:46)

View the video lesson, take notes and complete the problems below

Before using the quadratic formula, the equation must equal ________. Example:

a) 2𝑥𝑥2 = 15 − 7𝑥𝑥 b) 3𝑥𝑥2 − 5𝑥𝑥 + 2 = 7

YOU TRY

Solve.

a) 25𝑥𝑥2 = 30𝑥𝑥 + 11

b) 3𝑥𝑥2 + 4𝑥𝑥 + 8 = 2𝑥𝑥2 + 6𝑥𝑥 − 5

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EXERCISE Find the determinant and solve each equation by applying the quadratic formula. If the solution is not a real solution, then state not a real solution.

1) 2𝑥𝑥2 − 8𝑥𝑥 − 2 = 0 2) 3𝑟𝑟2 − 2𝑟𝑟 − 1 = 0

3) 2𝑥𝑥2 + 5𝑥𝑥 = −3

4) 𝑣𝑣2 − 4𝑣𝑣 − 5 = −8

5) 2𝑎𝑎2 + 3𝑎𝑎 + 14 = 6

6) 3𝑘𝑘2 + 3𝑘𝑘 − 4 = 7

7) 7𝑥𝑥2 + 3𝑥𝑥 − 16 = −2

8) 2𝑝𝑝2 + 6𝑝𝑝 − 16 = 4

9) 3𝑛𝑛2 + 3𝑛𝑛 = −3

10) 2𝑥𝑥2 = −7𝑥𝑥 + 49

11) 5𝑥𝑥2 = 7𝑥𝑥 + 7

12) 8𝑛𝑛2 = −3𝑛𝑛 − 8

13) 4𝑝𝑝2 + 5𝑝𝑝 − 36 = 3𝑝𝑝2

14) −5𝑛𝑛2 − 3𝑛𝑛 − 52 = 2 − 7𝑛𝑛2

15) 7𝑟𝑟2 − 12 = −3𝑟𝑟

16) 2𝑥𝑥2 + 4𝑥𝑥 + 12 = 8

17) 5𝑝𝑝2 + 2𝑝𝑝 + 6 = 0

18) 2𝑥𝑥2 − 2𝑥𝑥 − 15 = 0

19) 6𝑛𝑛2 − 3𝑛𝑛 + 3 = −4

20) 𝑚𝑚2 + 4𝑚𝑚 − 48 = −3

21) 4𝑛𝑛2 + 5𝑛𝑛 = 7

22) 3𝑟𝑟2 + 4 = −6𝑟𝑟

23) 3𝑏𝑏2 − 3 = 8𝑏𝑏

24) 6𝑣𝑣2 = 4 + 6𝑣𝑣

25) 6𝑎𝑎2 = −5𝑎𝑎 + 13

26) 2𝑘𝑘2 + 6𝑘𝑘 − 16 = 2𝑘𝑘

27) 6𝑏𝑏2 = 𝑏𝑏2 + 7 − 𝑏𝑏

28) 49𝑚𝑚2 − 28𝑚𝑚 + 4 = 0

29) 12𝑥𝑥2 + 𝑥𝑥 + 7 = 5𝑥𝑥2 + 5𝑥𝑥

30) 𝑥𝑥2 − 6𝑥𝑥 + 9 = 0

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SECTION 13.4: APPLICATIONS WITH QUADRATIC EQUATIONS There are many applications involving quadratic equations that it is almost challenging to pick the few equations to discuss for this section. Yet, we only choose a few applications that are common in most algebra classes. We start with the very famous Pythagorean Theorem.

A. PYTHAGOREAN THEOREM A long time ago, a Greek mathematician named Pythagoras discovered an interesting property about right triangles: the sum of the squares of the lengths of each of the triangle’s legs is the same as the square of the length of the triangle’s hypotenuse. This property- which has many applications in science, art, engineering, and architecture – is now called the Pythagorean Theorem.

Pythagorean Theorem

If 𝒂𝒂 and 𝒃𝒃 are the lengths of the legs of a right triangle and 𝒄𝒄 is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse:

(𝒇𝒇𝒇𝒇𝒐𝒐𝒇𝒇𝒇𝒇 𝒍𝒍𝒍𝒍𝒍𝒍)𝟐𝟐 + (𝒇𝒇𝒍𝒍𝒄𝒄𝒐𝒐𝒔𝒔𝒔𝒔 𝒍𝒍𝒍𝒍𝒍𝒍)𝟐𝟐 = (𝒉𝒉𝒉𝒉𝒉𝒉𝒐𝒐𝒇𝒇𝒍𝒍𝒔𝒔𝒉𝒉𝒇𝒇𝒍𝒍)𝟐𝟐

𝒂𝒂𝟐𝟐 + 𝒃𝒃𝟐𝟐 = 𝒄𝒄𝟐𝟐

MEDIA LESSON Pythagorean theorem applications (Duration 5:00)

View the video lesson, take notes and complete the problems below

Pythagorean Theorem- Find the hypotenuse.

Name the sides of the right triangle:

Pythagorean Theorem:

_____________________________

𝒄𝒄 is always the _______________________

Example: a) Find the missing side.

b) The base of a ladder is four feet from a building. The top of the ladder is eight feet up the building. How long is the ladder?

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YOU TRY

a) Find the length of the hypotenuse of a right triangle given lengths 6 cm and 13 cm for the legs. Round to two decimal places.

b) Find the length of the missing leg of a right triangle given the length of the hypotenuse is 15 km and the length of the other leg is 11 km. Round to two decimal places.

.

B. PROJECTILE MOTION

MEDIA LESSON Quadratic equation application – Projectile motion (Duration 9:26)

View the video lesson, take notes and complete the problems below

Example:

You launch a toy rocket from a height of 5 feet. The height (h, in feet) of the rocket 𝑡𝑡 seconds after taking off is given by the formula ℎ = −3𝑡𝑡2 + 14𝑡𝑡 + 5.

a) How long will it take for the rocket to hit the ground? b) Find the time when the rocket is 5 feet from hitting the ground.

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YOU TRY

a) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation ℎ = −4.9𝑡𝑡2 + 52𝑡𝑡 + 367

At what time does the rocket hit the ground? (Round answer to 2 decimal places.) Note: With applications, such as these, it doesn’t make sense to leave an exact answer, but better to

approximate. It is common language to say the rocket would hit the ground in decimal form.

C. COST AND REVENUE

Cost and Revenue

The cost refers to the amount of money the manufacturer should pay to produce a commodity. The revenue refers to the amount of money the consumer (or customer) pays for a commodity. Note, profit and revenue are often confused for one another. Be careful and note that revenue is the money “in the register” and profit is money after all costs are paid.

MEDIA LESSON Quadratic Applications – Cost (Duration 7:03 ) (Turn up the volume)

View the video lesson, take notes and complete the problems below

Example:

The cost C of producing 𝒂𝒂 “Total Cool Coolers” is modeled by the equation 𝑪𝑪 = 𝟎𝟎.𝟎𝟎𝟎𝟎𝟎𝟎𝒂𝒂𝟐𝟐 − 𝟎𝟎.𝟑𝟑𝒂𝒂 + 𝟏𝟏𝟏𝟏. How many coolers are produced when the cost is $19? (Round to the nearest whole number.)

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MEDIA LESSON Quadratic Applications –Revenue (Duration 7:03) (Turn up the volume)

View the video lesson, take notes and complete the problems below

Example: The revenue, R, of producing and selling 𝑥𝑥 “Awesome Hearing Aids” is modeled by the equation 𝑹𝑹 = −𝟒𝟒𝒂𝒂𝟐𝟐 + 𝟏𝟏𝟏𝟏𝟏𝟏𝒂𝒂. How many hearing aids need to be produced and sold in order to earn a revenue of $850? You may have more than one answer. (Round to the nearest whole number.) YOU TRY

a) The cost, C, of producing 𝑥𝑥 “Totally Cool Coolers” is modeled by the equation

𝐶𝐶 = 0.005𝑥𝑥2 − 0.5𝑥𝑥 + 21 How many coolers are produced when the cost is $22? (Round to the nearest whole number.)

b) The Revenue, R, of producing and selling x Awesome Hearing Aids is modeled by the equation

𝑅𝑅 = −3𝑥𝑥2 + 87𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $611? You may have more than one answer. (Round to the nearest whole number.)

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EXERCISE 1) Find the length of the missing leg given the

hypotenuse is 17 cm and the other leg is 12 cm. (Round to two decimal places.)

2) Find the length of the missing leg given the hypotenuse is 15 cm and the other leg is 12 cm. (Round to two decimal places.)

3) Find the length of the missing leg given the

hypotenuse is 8 cm and the other leg is 6 cm. (Round to two decimal places.)

4) Find the length of the hypotenuse if the length of the legs are 14 inches and 10 inches. Round to two decimal places.

5) Find the length of the hypotenuse if the length

of the legs are 10 inches and 12 inches. Round to two decimal places.

6) Find the length of the hypotenuse if the length of the legs are 14 inches and 12 inches. Round to two decimal places.

7) The cost, 𝐶𝐶 , of producing 𝑥𝑥 Totally Cool Coolers is modeled by the equation 𝐶𝐶 = 0.005𝑥𝑥2 − 0.3𝑥𝑥 + 18

How many coolers are produced when the cost is $19? (Round to the nearest whole number.)

8) The cost, 𝐶𝐶 , of producing 𝑥𝑥 Totally Cool Coolers is modeled by the equation

𝐶𝐶 = 0.005𝑥𝑥2 − 0.25𝑥𝑥 + 10 How many coolers are produced when the cost is $19? (Round to the nearest whole number.)

9) The revenue, 𝑅𝑅 , of producing and selling 𝑥𝑥 Awesome Hearing Aids is modeled by the equation

𝑅𝑅 = −5𝑥𝑥2 + 105𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $531? You may have more than one answer. (Round to the nearest whole number.)

10) The revenue, 𝑅𝑅 , of producing and selling 𝑥𝑥 Awesome Hearing Aids is modeled by the equation

𝑅𝑅 = −4𝑥𝑥2 + 120𝑥𝑥 How many hearing aids need to be produced and sold to earn a revenue of $880? You may have more than one answer. (Round to the nearest whole number.)

11) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation

ℎ = −4.9𝑡𝑡2 + 325𝑡𝑡 + 227 After how many seconds does the rocket hit the ground? (Round to two decimal places.)

12) A rocket is launched at 𝑡𝑡 = 0 seconds. Its height, in meters above sea-level, is given by the equation ℎ = −4.9𝑡𝑡2 + 148𝑡𝑡 + 374

After how many seconds does the rocket hit the ground? (Round to two decimal places.)

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CHAPTER REVIEW KEY TERMS AND CONCEPTS

Look for the following terms and concepts as you work through the workbook. In the space below, explain the meaning of each of these concepts and terms in your own words. Provide examples that are not identical to those in the text or in the media lesson.

Quadratic equation

Square root property

Discriminant

Quadratic formula

Pythagorean Theorem

Cost and Revenue