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1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve a quadratic equation by graphing? 2) How do we solve a quadratic equation by factoring? 3) How do we solve a quadratic equation by the square root method? Warm up: 1) Solve the given equation 3 different ways. 2 −4=0 Graphically Factoring Square Root Method DOPS ( − 2)( + 2) = 0 − 2 = 0 + 2 = 0 = ±2 2 =4 2 = √4 = ±2 2) Express in simplest radical form: √−25 √−1 ∗ √25 5 √−32 √−16 ∗ √2 4√2 4 5 2 ∗ √5 √5 ∗ √5 2√5 5 Advice to solving Quadratic Equations: Express quadratic in standard form 2 + + = 0 Determine appropriate methods to solve quadratic equations.

Lesson 7.1: Solving Quadratic Equations by Graphing ......1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve

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Page 1: Lesson 7.1: Solving Quadratic Equations by Graphing ......1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve

1

Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and

Square Root Method

Learning Goals:

1) How do we solve a quadratic equation by graphing?

2) How do we solve a quadratic equation by factoring?

3) How do we solve a quadratic equation by the square root method?

Warm up:

1) Solve the given equation 3 different ways. 𝑥2 − 4 = 0

Graphically Factoring Square Root Method

DOPS (𝑥 − 2)(𝑥 + 2) = 0 𝑥 − 2 = 0 𝑥 + 2 = 0

𝑥 = ±2

𝑥2 = 4

√𝑥2 = √4 𝑥 = ±2

2) Express in simplest radical form:

√−25

√−1 ∗ √25 5𝑖

√−32

√−16 ∗ √2

4𝑖√2

√4

5

2 ∗ √5

√5 ∗ √5

2√5

5

Advice to solving Quadratic Equations:

Express quadratic in standard form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

Determine appropriate methods to solve quadratic equations.

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1. Solve the following equation by factoring: 𝑥2 − 48 = 2𝑥

Set = 0

𝑥2 − 2𝑥 − 48 = 0

(𝑥 − 8)(𝑥 + 6) = 0

𝑥 − 8 = 0 𝑥 + 6 = 0

𝑥 = 8 & − 6

2. Solve for 𝑥 and express in simplest form: 3𝑥2 + 9 = 0

3𝑥2 = −9

𝑥2 = −3

𝑥 = ±𝑖√3

3. Use graphing to find the roots of 𝑥2 − 2𝑥 = 8

1. Solve for 𝑎: 𝑎2 = 5𝑎

𝑎2 − 5𝑎 = 0 set = 0

𝑎(𝑎 − 5) = 0 GCF = 𝑎

𝑎 = 0 𝑎 − 5 = 0

𝑎 = 0 & 5

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2. Find the zeros of 𝑓(𝑥) = 5𝑥2 − 8𝑥 − 4

0 = 5𝑥2 − 8𝑥 − 4 AC method

0 = 5𝑥2 − 10𝑥 + 2𝑥 − 4 5(−4) = −20

0 = 5𝑥(𝑥 − 2) + 2(𝑥 − 2) −10, 2

0 = (5𝑥 + 2)(𝑥 − 2)

𝑥 = −2

5 & 2

3. Solve for 𝑥: (𝑥 + 3)2 = 16

√(𝑥 + 3)2 = √16

𝑥 + 3 = ±4

𝑥 = −3 ± 4

−3 + 4 − 3 − 4

𝑥 = 1 & − 7

4. Algebraically solve for 𝑥 in simplest form: 3𝑥2 + 4 = 0

3𝑥2 = −4

𝑥2 = −4

3

𝑥 = ±√−4

√3= ±

2𝑖

√3 ∗√3

√3 or ±

2𝑖√3

3

5. Find the roots of the equation: 2𝑥2 + 𝑥 − 3 = (𝑥 − 1)(𝑥 + 2)

2𝑥2 + 𝑥 − 3 = 𝑥2 + 2𝑥 − 𝑥 − 2

2𝑥2 + 𝑥 − 3 = 𝑥2 + 𝑥 − 2

𝑥2 − 1 = 0

𝑥 = ±1

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6. Find the zeros of 𝑥 − 4 =5

𝑥 Cross Multiply!

𝑥(𝑥 − 4) = 5

𝑥2 − 4𝑥 = 5

𝑥2 − 4𝑥 − 5 = 0

(𝑥 − 5)(𝑥 + 1) = 0

𝑥 = 5 & − 1

7. Solve for 𝑥: 1

4(𝑥 − 6)2 = 8

(𝑥 − 6)2 = 32

𝑥 − 6 = ±√32

𝑥 = 6 ± 4√2

8. Solve for 𝑥: (𝑥 + 1)2 − 87 = 𝑥2

(𝑥 + 1)(𝑥 + 1) − 87 = 𝑥2

𝑥2 + 2𝑥 + 1 − 87 = 𝑥2

2𝑥 − 86 = 0

2𝑥 = 86

𝑥 = 43

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Check for understanding…

1. On a test, Samantha says that the roots of the equation 𝑥2 + 2𝑥 − 3 = 0 are

−1 and 3. Before she hands in her test, she wants to check over her work.

Show the check that Samantha could do on her test and state whether or not she

is correct. If she is not correct, give the correct solution.

(−1)2 + 2(−1) − 3 = 0 (3)2 + 2(3) − 3 = 0

1 − 2 − 3 = 0 9 + 6 − 3 = 0

−4 ≠ 0 12 ≠ 0

Therefore she is not correct!

(𝑥 + 3)(𝑥 − 1) = 0

𝑥 = −3 & 1

2. Phil’s teacher gave the class the quadratic function 𝑓(𝑥) = (𝑥 − 2)2 − 4.

a) State two different methods Phil could use to solve the equation 𝑓(𝑥) = 0.

b) Using one of the methods stated in part a, solve 𝑓(𝑥) = 0 for 𝑥.

Factoring, Quadratic Formula

𝑓(𝑥) = (𝑥 − 2)(𝑥 − 2) − 4

𝑓(𝑥) = 𝑥2 − 4𝑥 + 4 − 4

𝑓(𝑥) = 𝑥(𝑥 − 4)

𝑥 = 0 & 4

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Homework Lesson 7.1: Solving Quadratic Equations by Graphing,

Factoring, and Square Root Method

1. Solve for 𝑥: 𝑥 − 4 =−3

𝑥

2. Find the roots of 3𝑥2 − 5𝑥 = 36 − 2𝑥

3. Solve for 𝑥 in simplest form: 3𝑥2 + 25 = 0

4. Solve for 𝑥: 5(𝑥 − 1)2 = 50

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5. Use graphing to solve the given

equation for 𝑥: 𝑥2 + 5𝑥 + 6 = 0

6. Find the zeros of 𝑓(𝑥) = 4𝑥2 − 16 − 20

7. Solve for 𝑥 in simplest form: (2𝑥 + 3)(𝑥 − 4) = 2𝑥2 + 13𝑥 + 15

8. Solve for 𝑥: 𝑥2 + 𝑥 − 1 = (−4𝑥 + 3)2

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Lesson 7.2: Solving Quadratics by Completing the Square

Learning Goal: How do we solve a quadratic equation by completing the

square?

Practice: Find the value of 𝑐 that makes the expression a perfect square

trinomial. Then write the expression as the square of a binomial.

𝑥2 + 14𝑥 + 𝑐

𝑐 = (14

2)2

= 72 = 49

𝑥2 + 14𝑥 + 49 (𝑥 + 7)(𝑥 + 7) (𝑥 + 7)2

𝑥2 − 9𝑥 + 𝑐

𝑐 = (−9

2)2

=81

4

𝑥2 − 9𝑥 +81

4

(𝑥 −9

2)2

Steps to Completing the Square

1. The "𝑎" coefficient must equal 1 (divide all terms by "𝑎").

2. Isolate the 𝑥-terms (move "𝑐" to the other side).

3. Take (𝑏

2)2 and add that number to each side.

4. Factor the perfect square trinomial and express it as (𝑥 + 𝑎)2.

5. Solve for 𝑥 by using the square root method.

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Completing the Square when 𝑎 = 1

Example 1: Solve 𝑥2 − 10𝑥 + 13 = 0 by completing the square.

Solution Steps

𝑥2 − 10𝑥 + 13 = 0 Write original equation.

𝑥2 − 10𝑥 = −13 Write left side in the form 𝑥2 + 𝑏𝑥, I say “isolate the 𝑥”

𝑥2 − 10𝑥 + 𝑐 = −13 + 𝑐

𝑐 = (−10

2)

2

= (−5)2 = 25

Complete the square, I say “divide by two and square it”

𝑥2 − 10𝑥 + 25 = −13 + 25

(𝑥 − 5)2 = 12

Write left side as a binomial squared.

√(𝑥 − 5)2 = √12 Take the square roots of each side.

𝑥 − 5 = ±√12

𝑥 = 5 ± √12

Solve for 𝑥.

𝑥 = 5 ± 2√3 Simplify

Practice: Solve 𝑥(𝑥 + 3) = −2 by completing the square, and express the results

in simplest form.

𝑥2 + 3𝑥 = −2

𝑥2 + 3𝑥 + 𝑐 = −2 + 𝑐

𝑐 = (3

2)2=9

4

𝑥2 + 3𝑥 +9

4= −2 +

9

4

(𝑥 +3

2)2=1

4

𝑥 +3

2= ±

1

2

𝑥 = −3

2±1

2

𝑥 = −1 & − 2

Page 10: Lesson 7.1: Solving Quadratic Equations by Graphing ......1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve

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Completing the Square when 𝑎 ≠ 1

Example 1: Solve 3𝑥2 − 12𝑥 + 27 = 0 by completing the square.

Solution Steps

3𝑥2 − 12𝑥 + 27 = 0 Write original equation.

𝑥2 − 4𝑥 + 9 = 0 Get 𝑎 = 1before you can complete the square! Divide each side by the coefficient

of 𝑥2. 𝑥2 − 4𝑥 = −9 Write left side in the form 𝑥2 + 𝑏𝑥.

𝑥2 − 4𝑥 + 𝑐 = −9 + 𝑐

𝑐 = (−4

2)2

= (−2)2 = 4

Complete the square.

𝑥2 − 4𝑥 + 4 = −9 + 4

(𝑥 − 2)2 = −5

Write left side as a binomial squared.

√(𝑥 − 2)2 = √−5 Take the square roots of each side.

𝑥 − 2 = ±√−5

𝑥 = 2 ± √−5

Solve for 𝑥.

𝑥 = 2 ± 𝑖√5 Simplify

Practice: Solve 4𝑥2 − 20𝑥 = −9 by completing the square, and express the

results in simplest form.

𝑥2 − 5𝑥 = −9

4 Isolate the 𝑥 and make sure 𝑎 = 1

𝑥2 − 5𝑥 + 𝑐 = −9

4+ 𝑐

𝑐 = (−5

2)2=25

4

𝑥2 − 5𝑥 +25

4= −

9

4+25

4

(𝑥 −5

2)2= 4

𝑥 −5

2= ±2

𝑥 =5

2± 2

𝑥 =9

2 &

1

2

Page 11: Lesson 7.1: Solving Quadratic Equations by Graphing ......1 Lesson 7.1: Solving Quadratic Equations by Graphing, Factoring, and Square Root Method Learning Goals: 1) How do we solve

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Practice: Solve the following equations by completing the square, and express

the results in simplest form.

a) 𝑥2 − 9𝑥 − 1 = 0 b) 5𝑥(𝑥 + 6) = −50

𝑥2 − 9𝑥 = 1 5𝑥2 + 30𝑥 = −50

𝑥2 − 9𝑥 + 𝑐 = 1 + 𝑐 𝑥2 + 6𝑥 = −10

𝑐 = (−9

4)2=81

16 𝑥2 + 6𝑥 + 𝑐 = −10 + 𝑐

𝑥2 − 9𝑥 +81

16= 1 +

81

16 𝑐 = (

6

2)2= (3)2 = 9

(𝑥 −9

4)2=97

16 𝑥2 + 6𝑥 + 9 = −10 + 9

𝑥 −9

4= ±

√97

4 (𝑥 + 3)2 = −1

𝑥 =9

4±√97

4 𝑥 + 3 = ±𝑖

𝑥 = −3 ± 𝑖

Analysis Question: Which equation has the same solutions as 𝑥2 + 6𝑥 − 7 = 0?

1. (𝑥 + 3)2 = 2 2. (𝑥 − 3)2 = 2 3. (𝑥 − 3)2 = 16 4. (𝑥 + 3)2 = 16

𝑥2 + 6𝑥 = 7

𝑥2 + 6𝑥 + 𝑐 = 7 + 𝑐

𝑐 = (6

2)2= (3)2 = 9

𝑥2 + 6𝑥 + 9 = 7 + 9

(𝑥 + 3)2 = 16

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Homework 7.2: Solving Quadratics by Completing the Square

1. Find the value of 𝑐 that makes the expression a perfect square trinomial.

Then write the expression as the square of a binomial.

𝑥2 + 10𝑥 + 𝑐

𝑦2 − 7𝑦 + 𝑐

2. Solve the following equations by completing the square and express the

results in simplest form.

a. 𝑥2 + 6𝑥 + 3 = 0 b. 𝑡(𝑡 − 8) = 5

c. 7𝑦2 + 28𝑦 + 56 = 0 d. 4𝑥2 − 30𝑥 = −9 − 10𝑥

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3. Determine whether you would use factoring or the square root method to

solve each equation. Then solve the equation.

a. 𝑥2 − 4𝑥 − 21 = 0 b. (𝑥 + 4)2 = 16

c. 5𝑥2 − 10𝑥 + 1 = 2𝑥2 − 5𝑥 + 13 d. (𝑥 + 3)2 = 17𝑥 + 21

4. Error Analysis: Describe and correct the error in solving the equation below

4𝑥2 + 24𝑥 − 11 = 0

4(𝑥2 + 6𝑥) = 11

4(𝑥2 + 6𝑥 + 9) = 11 + 9

4(𝑥 + 3)2 = 20

(𝑥 + 3)2 = 5

𝑥 + 3 = ±√5

𝑥 = −3 ± √5

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5. Determine the roots of the following equations graphically:

a. 𝑥2 − 5𝑥 − 14 = 0 b. 𝑥2 = 9

6. The zeros of the function 𝑓(𝑥) = (𝑥 + 2)2 − 25 are

1) -2 and 5 2) -3 and 7 3) -5 and 2 4) -7 and 3

7. A student was given the equation 𝑥2 + 6𝑥 − 13 = 0 to solve by completing the

square. The first step that was written is shown below.

𝑥2 + 6𝑥 = 13

The next step in the student’s process was 𝑥2 + 6𝑥 + 𝑐 = 13 + 𝑐. State the value

of 𝑐 that creates a perfect square trinomial. Explain how the value of 𝑐 is

determined.

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Lesson 7.3: Solving Quadratics by Using the Quadratic Formula

Learning Goals:

1) What is the quadratic formula?

2) How do we solve a quadratic equation by using the quadratic formula?

Do Now: Solve for 𝑥 by completing the square and express your answer in

simplest radical form: 2𝑥2 + 2 = 6𝑥

2𝑥2 − 6𝑥 + 2 = 0

𝑥2 − 3𝑥 + 1 = 0 Get 𝑎 = 1

𝑥2 − 3𝑥 = −1

𝑥2 − 3𝑥 + 𝑐 = −1 + 𝑐

𝑐 = (−3

2)2=9

4

𝑥2 − 3𝑥 +9

4= −1 +

9

4

(𝑥 −3

2)2=5

4

𝑥 −3

2= ±

√5

2

𝑥 =3

2±√5

2

The Quadratic Formula

Previously, you solved quadratic equations by completing the square.

In this lesson, you will learn about solving quadratic equations by using a

formula that is derived by completing the square for the general equation

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0.

The formula is called the Quadratic Formula. 𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎 On

Reference Sheet

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Example: Solve for 𝑥 by using the quadratic formula and express your answer in

simplest radical form: 2𝑥2 + 2 = 6𝑥

2𝑥2 − 6𝑥 + 2 = 0 Set = 0

𝑎 = 2, 𝑏 = −6, 𝑐 = 2

𝑥 =−𝑏±√𝑏2−4𝑎𝑐

2𝑎=−(−6)±√(−6)2−4(2)(2)

2(2)

𝑥 =6±√36−16

4=6±√20

4

𝑥 =6±2√5

4=3±√5

2

Steps to Using the Quadratic Formula

1. Set quadratic equation equal to zero (𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0)

2. Identify the 𝑎, 𝑏, and 𝑐 coefficients

3. Substitute 𝑎, 𝑏, and 𝑐 into the quadratic formula

4. Simplify the formula carefully

5. Look to simplify and reduce (always start with radical first)

Summary of Quadratic Formula

What is the quadratic formula? 𝒙 =−𝒃±√𝒃𝟐−𝟒𝒂𝒄

𝟐𝒂

When should you use the quadratic formula? When you cannot solve by factoring

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Model Problem: Solve for 𝑞: 2𝑞2 − 8 = 3𝑞

2𝑞2 − 3𝑞 − 8 = 0

𝑎 = 2, 𝑏 = −3, and 𝑐 = −8

𝑞 =−(−3)±√(−3)2−4(2)(−8)

2(2)=3±√9+64

4

𝑞 =3±√73

4

Directions: Use the quadratic formula to solve each equation.

1. Solve for 𝑚: 1

3𝑚2 + 2𝑚 + 8 = 5

1

3𝑚2 + 2𝑚 + 3 = 0

3 (1

3𝑚2 + 2𝑚 + 3 = 0) Easier to eliminate the fraction!

𝑚2 + 6𝑚 + 9 = 0

𝑎 = 1, 𝑏 = 6, and 𝑐 = 9

𝑚 =−(6)±√(6)2−4(1)(9)

2(1)=−6±√36−36

2

𝑚 = −3

2. Solve for 𝑥: 5𝑥 − 7𝑥2 = 3𝑥 + 4

−7𝑥2 + 2𝑥 − 4 = 0

−1(−7𝑥2 + 2𝑥 − 4 = 0) Easier to have a positive coefficient for 𝑎!

7𝑥2 − 2𝑥 + 4 = 0

𝑎 = 7, 𝑏 = −2, and 𝑐 = 4

𝑥 =−(−2)±√(−2)2−4(7)(4)

2(7)=2±√−108

14

𝑥 =2±6𝑖√3

14=1±3𝑖√3

7

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Application of the Quadratic Formula

Barb pulled the plug in her bathtub and it started to drain. The amount of water

in the bathtub as it drains is represented by the equation 𝐿 = −5𝑡2 − 8𝑡 + 120,

where 𝐿 represents the number of liters of water in the bathtub and 𝑡 represents

the amount of time, in minutes, since the plug was pulled.

How many liters of water were in the bathtub when Barb pulled the plug? Show

your reasoning. 𝐿 =? , 𝑡 = 0 because no time has passed!

𝐿 = −5(0)2 − 8(0) + 120

𝐿 = 120

Determine, to the nearest tenth of a minute, the amount of time it takes for all the

water in the bathtub to drain. 𝑡 =? , 𝐿 = 0 no more water!

0 = −5𝑡2 − 8𝑡 + 120

−1(0 = −5𝑡2 − 8𝑡 + 120)

0 = 5𝑡2 + 8𝑡 − 120

𝑎 = 5, 𝑏 = 8, and 𝑐 = −120

𝑡 =−(8)±√(8)2−4(5)(−120)

2(5)=−8±√64+2400

10=−8±√2464

10

𝑡 = 4.2 & − 5.8 (but omit because time can′t be negative)

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Homework 7.3: Solving Quadratics by Using the Quadratic Formula

1. Solve the following equations by using the quadratic formula and express the

results in simplest form.

a. 𝑥2 + 2𝑥 − 8 = 0 b. 2𝑦2 + 3𝑦 − 5 = 4

2. Matt’s rectangular patio measures 9 feet by 12 feet. He wants to increase the

patio’s dimensions so its area will be twice the area it is now. He plans to

increase both the length and the width by the same amount, 𝑥. Find 𝑥, to the

nearest hundredth of a foot.

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3. Determine the best method to solve each equation and use it to find all values

of 𝑥 in simplest form.

a. 𝑥2 − 2𝑥 = 12

b. (𝑥 − 2)2 = 8

c. 2𝑥2 − 54 = 12𝑥

d. 5𝑥2 + 38 = 3

4. Find all real solutions to the equation (𝑥2 − 6𝑥 + 3)(2𝑥2 − 4𝑥 − 7) = 0

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Lesson 7.4: Solving Systems of Equations Graphically

Learning Goals:

1) How can we change an equation of a circle from standard form to center-

radius form?

2) How can we solve a quadratic/linear system of equations graphically?

Warm-Up:

1. Suppose you were given an equation for a circle and an equation for a line.

What possibilities are there for the two figures to intersect? Sketch a graph for

each possibility.

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2. Suppose you were given an equation for a parabola and an equation for a

line. What possibilities are there for the two figures to intersect? Sketch a graph

for each possibility.

3. State the center and radius of the given circles:

a) (𝑥 − 1)2 + (𝑦 − 2)2 = 25 Center: (1, 2) & 𝑟 = √25 = 5

b) (𝑥 + 5)2 + (𝑦 − 2)2 = 29 Center: (−5, 2) & 𝑟 = √29

c) 𝑥2 + (𝑦 − 2)2 = 4 Center: (0, 2) & 𝑟 = √4 = 2

4. Rewrite 𝑥2 + 𝑦2 − 4𝑥 + 2𝑦 = −1 by completing the square in both 𝑥 and 𝑦.

Describe the circle represented by this equation.

𝑥2 − 4𝑥⏟ + 𝑦2 + 2𝑦⏟ = −1 (−4

2)2= (−2)2 = 4 (

2

2)2= 1

𝑥2 − 4𝑥 + 4 + 𝑦2 + 2𝑦 + 1 = −1 + 4 + 1

(𝑥 − 2)2 + (𝑦 + 1)2 = 4 Center: (2,−1) and Radius= 2

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5. The equation of a circle is 𝑥2 + 𝑦2 + 6𝑦 = 7. What are the coordinates of the

center and the length of the radius of the circle?

𝑥2 + 𝑦2 + 6𝑦 = 7 (6

2)2= (3)2 = 9

𝑥2 + 𝑦2 + 6𝑦 + 9 = 7 + 9

𝑥2 + (𝑦 + 3)2 = 16 Center: (0,−3) and Radius= 4

6. If the equation of a circle is 2𝑥2 + 2𝑦2 − 32𝑥 + 12𝑦 + 52 = 0, find the length of

the radius and the coordinates of the center of the circle.

2𝑥2 + 2𝑦2 − 32𝑥 + 12𝑦 + 52 = 0

2

𝑥2 + 𝑦2 − 16𝑥 + 6𝑦 + 26 = 0

𝑥2 − 16𝑥 + 𝑦2 + 6𝑦 = −26 (−16

2)2= (−8)2 = 64 (

6

2)2= (3)2 = 9

𝑥2 − 16𝑥 + 64⏟ + 𝑦2 + 6𝑦 + 9⏟ = −26 + 64 + 9

(𝑥 − 8)2 + (𝑦 + 3)2 = 47 Center: (8,−3) and Radius= √47

7. If the equation of a circle is 4𝑥2 + 4𝑦2 − 32𝑥 + 8𝑦 + 16 = 0, find the length of

the radius and the coordinates of the center of the circle.

4𝑥2 + 4𝑦2 − 32𝑥 + 8𝑦 + 16 = 0

4

𝑥2 + 𝑦2 − 8𝑥 + 2𝑦 + 4 = 0

𝑥2 − 8𝑥 + 𝑦2 + 2𝑦 = −4 (−8

2)2= (−4)2 = 16 (

2

2)2= (1)2 = 1

𝑥2 − 8𝑥 + 16⏟ + 𝑦2 + 2𝑦 + 1⏟ = −4 + 16 + 1

(𝑥 − 4)2 + (𝑦 + 1)2 = 13 Center: (−4, 1) and Radius= √13

Remember to put the 4 back in: 4(𝑥 − 4)2 + 4(𝑦 + 1)2 = 52

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Graphing Systems of Equations:

1. Graph the line given by 3𝑥 + 4𝑦 = 25 and the circle given by 𝑥2 + 𝑦2 = 25.

Find all solutions to the system of equations.

3𝑥 + 4𝑦 = 25

4𝑦 = −3𝑥 + 25

𝑦 = −3

4𝑥 +

25

4

𝑦 = −3

4𝑥 + 6.25

𝑥2 + 𝑦2 = 25

Center: (0, 0) 𝑟 = 5

Solution: (4, 3)

2. Graph the line given by 𝑥 +

𝑦 = −2 and the quadratic curve

given by

𝑦 = 𝑥2 − 4. Find all solutions to

the system of equations.

Solutions: (−2, 0) & (1,−3)

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3. Find all solutions to the

following system of equations.

5𝑦 − 5𝑥 = 30

𝑥2 + 𝑦2 + 4𝑥 − 2𝑦 − 4 = 0

𝑦 = 𝑥 + 6

𝑥2 + 4𝑥 + 𝑦2 − 2𝑦 = 4

(𝑥 + 2)2 + (𝑦 − 1)2 = 9

Center: (−2, 1) 𝑟 = 3

Solutions: (−5, 1) & (−2, 4)

4. Find all values of 𝑘 so that the following system has two solutions.

𝑥2 + 𝑦2 = 25

𝑦 = 𝑘

Center: (0, 0) 𝑟 = 5

−5 < 𝑘 < 5

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Real-World Application:

An asteroid is moving in a parabolic

arc that is modeled by the function

𝑝(𝑥) = 𝑥2 − 4𝑥 + 9 where 𝑥

represents time. A laser is on the

path of 𝑓(𝑥) = 2𝑥 + 4. When will

the laser first hit the asteroid?

(1) (0, 9) & (1, 6)

(2) (1, 6) & (5, 14)

(3) (1, 6)

(4) (2, 5)

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Homework 7.4: Solving Systems of Equations Graphically

1. Which of the following systems of equations has exactly one point of

intersection?

(1) 𝑦 = 𝑥2 − 5𝑥 + 7 and 𝑦 − 1 = 2𝑥

(2) 𝑦 − 4𝑥2 = 𝑥 − 3 and 𝑦 = 3

(3) 𝑦 − 𝑥2 = 2𝑥 + 4 and 𝑦 = 3

(4) 𝑦 + 9𝑥2 = −8 and 𝑦 = −1

2. Given the equation of a circle is 3𝑥2 + 3𝑦2 − 12𝑥 + 30𝑦 − 10 = 0, state the

length of the radius and coordinates of the center.

3. Find all values of 𝑘 so that the

following system has exactly one solution.

Illustrate with a graph.

𝑦 = 5 − (𝑥 − 3)2

𝑦 = 𝑘

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4. Find all solutions to the following

system of equations.

𝑦 + 2𝑥 = 3

𝑦 = 𝑥2 − 6𝑥 + 3

5. Solve the following system of

equations graphically.

{2𝑥 + 𝑦 = 15

(𝑥 − 2)2 + (𝑦 − 1)2 = 25

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Lesson 7.5: Solving Systems of Equations Algebraically

Learning Goals:

1) How can we solve a quadratic/linear system of equations algebraically?

2) How is the solution to a quadratic/linear system related to its graphical

solution?

Warm-Up: Solve the system 𝑥2 + 𝑦2 = 9 and 𝑥 − 3𝑦 = 3 graphically.

Center: (0, 0) & 𝑟 = 3

−3𝑦 = −𝑥 + 3

𝑦 =1

3𝑥 − 1

Solutions: (3, 0)& ?

What is difficult about solving this

graphically? Hard to draw a circle

without a compass

Steps to Solving a System of Equations Algebraically

1) Get one equation to be written as 𝑦 = or 𝑥 =

2) Substitute this equation into the other equation for 𝑥 or 𝑦

3) Simplify this equation and set it equal to zero.

4) Use one of the methods for solving a quadratic to solve for the variable

Factoring, square root method, completing the square, quadratic formula

5) Plug your answer back into one of the equations to solve for the other variable.

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1. Find all solutions of the system of equations algebraically:

𝑥2 + 𝑦2 = 9

𝑥 − 3𝑦 = 3

Substitute: 𝑥 = 3𝑦 + 3

(3𝑦 + 3)2 + 𝑦2 = 9 Now find 𝑥-values!

(3𝑦 + 3)(3𝑦 + 3) + 𝑦2 = 9 𝑥 − 3(0) = 3 𝑥 − 3 (−9

5) = 3

9𝑦2 + 18𝑦 + 9 + 𝑦2 = 9 𝑥 − 0 = 3 𝑥 +27

5= 3

10𝑦2 + 18𝑦 = 0 𝑥 = 3 𝑥 = −2.4 or −12

5

2𝑦(5𝑦 + 9) = 0

𝑦 = 0 and y = −9

5 Solutions: (3, 0) and (−

12

5, −

9

5)

How is the solution to a system or equations related to its graphical solution?

2. Find all solutions of the system of equations algebraically:

𝑦2 − 2𝑥2 = 6

𝑦 = −2𝑥

Substitute: 𝑦 = −2𝑥

(−2𝑥)2 − 2𝑥2 = 6 Now find 𝑦-values!

4𝑥2 − 2𝑥2 = 6 𝑦 = −2(√3) 𝑦 = −2(−√3)

2𝑥2 = 6 𝑦 = −2√3 𝑦 = 6√3

𝑥2 = 3

𝑥 = ±√3 Solutions: (√3,−2√3) and (−√3, 6√3)

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3. Given 𝑓(𝑥) = −2𝑥 + 3 and 𝑔(𝑥) = 𝑥2 − 6𝑥 + 3, find the 𝑥-value(s) that satisfy

𝑓(𝑥) = 𝑔(𝑥).

𝑓(𝑥) = 𝑔(𝑥) means they are equal when the graphs intersect!

−2𝑥 + 3 = 𝑥2 − 6𝑥 + 3

0 = 𝑥2 − 4𝑥 Now find 𝑦-values or 𝑓(𝑥)!

0 = 𝑥(𝑥 − 4) 𝑦 = −2(0) + 3 𝑦 = −2(4) + 3

𝑥 = 0 and 𝑥 = 4 𝑦 = 3 𝑦 = −5

Solutions: (0, 3) and (4, −5)

4. Algebraically, determine the points of intersection of (𝑥 − 1)2 + (𝑦 − 2)2 = 4

and 𝑦 − 2 = 2𝑥.

Substitute: 𝑦 = 2𝑥 + 2

(𝑥 − 1)2 + (2𝑥 + 2 − 2)2 = 4

(𝑥 − 1)(𝑥 − 1) + (2𝑥)2 = 4 Quadratic Formula?

𝑥2 − 2𝑥 + 1 + 4𝑥2 = 4 Completing the Square?

5𝑥2−2𝑥 − 3 = 0 AC Method: (5)(−3) = −15

5𝑥2−5𝑥 + 3𝑥 − 3 = 0 −5, 3

5𝑥(𝑥 − 1) + 3(𝑥 − 1) = 0 Grouping

(5𝑥 + 3)(𝑥 − 1) = 0 Now find 𝑦-values!

𝑥 = −3

5 and 1 𝑦 − 2 = 2 (−

3

5) 𝑦 − 2 = 2(1)

𝑦 = 1 𝑦 = 4

Solutions: (−3

5, 1) and (1, 4)

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5. Solve the following system of equatons algebraically:

𝑥 + 2𝑦 = 0

𝑥2 − 2𝑥 + 𝑦2 − 2𝑦 − 3 = 0

Substitute: 𝑥 = −2𝑦

(−2𝑦)2 − 2(−2𝑦) + 𝑦2 − 2𝑦 − 3 = 0

4𝑦2 + 4𝑦 + 𝑦2 − 2𝑦 − 3 = 0 Completing the Square?

5𝑦2 + 2𝑦 − 3 = 0 AC Method?

𝑦 =−(2)±√(2)2−4(5)(−3)

2(5) Quadratic Formula: 𝑎 = 5, 𝑏 = 2, 𝑐 = −3

𝑦 =−2±√4+60

10=−2±√64

10 Now find 𝑥-values!

𝑦 =−2±8

10 𝑥 + 2 (

3

5) = 0 𝑥 + 2(−1) = 0

𝑦 =3

5 and − 1 𝑥 = −

6

5 𝑥 = 2

Solutions: (−6

5,3

5) and (2, −1)

Push Yourself! Applications of Systems of Equations DECIMALS use Calculator

6. Sabrina is playing ball with her dog. She throws the ball in a parabolic path

that can be modeled by the function 𝑦 = −1

2(𝑥 − 3)2 + 7. Her brother, Bobby, is

playing in a tree next to her. Bobby shines his laser pointer from the tree in a line

that can be modeled by the function 𝑦 = −1

2𝑥 + 8.5. At what point(s) will the ball

and the laser beam intersect?

−1

2(𝑥 − 3)2 + 7 = −

1

2𝑥 + 8.5

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Homework Lesson 7.5: Solving Systems of Equations Algebraically

1. Find all solutions of the system of equation algebraically:

(𝑥 − 2)2 + (𝑦 + 3)2 = 4

𝑥 − 𝑦 = 3

2. Given 𝑓(𝑥) = −2𝑥 + 3 and 𝑔(𝑥) = 𝑥2 − 6𝑥 + 3, find the 𝑥-value(s) that satisfy

𝑓(𝑥) = 𝑔(𝑥).

3. Algebraically, determine the points of intersection of −𝑦2 + 6𝑦 + 𝑥 − 9 = 0

and 6𝑦 = 𝑥 + 27

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4. A boy standing on the top of a building in Albany throws a water balloon up

vertically. The height, ℎ (in feet) of the water balloon is given by the equation

ℎ(𝑡) = −16𝑡2 + 64𝑡 + 192, where 𝑡 is the time (in seconds) after he threw the

water balloon. What is the value of 𝑡 when the balloon hits the ground? Explain

and show how you arrived at the answer.

5. What is the total number of points of intersection of the graphs of the

equations 2𝑥2 − 𝑦2 = 8 and 𝑦 = 𝑥 + 2?

(1) 1 (2) 2 (3) 3 (4) 0

6. Amy solved the equation 2𝑥2 + 5𝑥 − 42 = 0. She stated that the solutions to

the equation were 7

2 and −6. Do you agree with Amy’s solutions? Explain why or

why not.