1

• This Slideshow was developed to accompany the textbook

▪ Big Ideas Algebra 2

▪ By Larson, R., Boswell

▪ 2022 K12 (National Geographic/Cengage)

• Some examples and diagrams are taken from the textbook.

Slides created by Richard Wright, Andrews Academy rwright@andrews.edu

Slides created by Richard Wright, Andrews Academy rwright@andrews.edu

2

3

• Simplify Square Roots

• 12 = 4 · 3

• = 4 · 3

• = 2 3

• 87 #1-3, 88 #1-14

• Factor

• Difference of Square

• 𝑎2 − 𝑏2 = 𝑎 − 𝑏 𝑎 + 𝑏

• Perfect Square

• 𝑎2 ± 2𝑎𝑏 + 𝑏2 = 𝑎 ± 𝑏 2

4

Objectives: By the end of the lesson,• I can solve quadratic equations by graphing.• I can solve quadratic equations algebraically.• I can use quadratic equations to solve real-life problems.

5

• Work with a partner.• a. Match each quadratic equation

with the graph of its related function. Then use the graph to fi nd the real solutions (if any) of each equation. Explain your reasoning.▪ i. x2 − 2x = 0 ▪ ii. x2 − 2x + 1 = 0▪ iii. x2 − 2x + 2 = 0 ▪ iv. −x2 + 2x = 0▪ v. −x2 + 2x − 1 = 0 ▪ vi. −x2 + 2x − 2 = 0

• b. How can you use a graph to determine the number of real solutions of a quadratic equation?

a. i. E; 0 and 2; The x-intercepts are 0 and 2.ii. D; 1; The x-intercept is 1.iii. C; no real solution; There are no x-intercepts.iv. A; 0 and 2; The x-intercepts are 0 and 2.v. F; 1; The x-intercept is 1.vi. B; no real solution; There are no x-intercepts.b. The number of x-intercepts is the number of real solutions.

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• Quadratic Equation

▪ 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

• Root of an equation

▪ Solution

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• Solving Quadratic Equations

• Graphing

▪ Solutions are x-intercepts of graph

• Square Roots

▪ When only x2 and no x

▪ Solve for the square and take a square root

• Factoring

▪ Set = 0

▪ Factor

▪ Each factor = 0 and solve

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• Solve by graphing

• 𝑥2 − 2𝑥 − 3 = 0

GraphSolutions are x-interceptsx = 3, x = −1

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• Solve by using square roots.

• 2x2 + 14 = 70

• 4x2 + 20 = 16

2𝑥2 + 14 = 702𝑥2 = 56𝑥2 = 28

𝑥 = ± 28 = ± 4 · 7 = ±2 7

4𝑥2 + 20 = 164𝑥2 = −4𝑥2 = −1

𝑥 = ± −1No real solution

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•3

4𝑥 + 1 2 = 10

3

4𝑥 + 1 2 = 10

𝑥 + 1 2 =40

3

𝑥 + 1 = ±40

3

𝑥 + 1 = ±40

3

𝑥 + 1 = ±2 10

3= ±

2 10

3·

3

3= ±

2 30

3

𝑥 = −1 ±2 30

3

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• Solve x2 + 2x = 48 by factoring.

• Find the zeros of 𝑓 𝑥 =4𝑥2 − 13𝑥 + 3.

𝑥2 + 2𝑥 = 48𝑥2 + 2𝑥 − 48 = 0𝑥 + 8 𝑥 − 6 = 0

𝑥 + 8 = 0 𝑜𝑟 𝑥 − 6 = 0𝑥 = −8 𝑜𝑟 𝑥 = 6

𝑓 𝑥 = 4𝑥2 − 13𝑥 + 30 = 4𝑥2 − 13𝑥 + 30 = (4𝑥 − 1)(𝑥 − 3)

4𝑥 − 1 = 0 𝑜𝑟 𝑥 − 3 = 0

𝑥 =1

4𝑜𝑟 𝑥 = 3

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• The height of an object in feet dropped from height s0 can be modeled by ℎ 𝑡 = −16𝑡2 + 𝑠0.

• For a science competition, students must design a container that prevents an egg from breaking when dropped from a height of 60 feet.

• a. Write a function to model this situation. How long does it take the container to hit the ground?

• b. Find and interpret h(0.5) − h(1.3).

• 95 #1, 5, 9, 13, 17, 21, 25, 29, 31, 35, 39, 43, 47, 51, 63, 79, 81, 83, 85, 91

ℎ 𝑡 = −16𝑡2 + 600 = −16𝑡2 + 60−60 = −16𝑡2

60

16= 𝑡2

𝑡 ≈ 1.9 𝑠

about 23.0; The container fell about 23 feet between 0.5 and 1.3 seconds.

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Objectives: By the end of the lesson,

• I can define the imaginary unit i and use it to rewrite the square root of a negative number.

• I can add, subtract, and multiply complex numbers.

• I can find complex solutions of quadratic equations and complex zeros of quadratic functions.

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• Work with a partner.• a. A student solves the equations

below as shown. Justify each solution step.

• b. In your study of mathematics, you have probably worked with only real numbers, which can be represented graphically on the real number line. Describe the solutions of the equation x2 = c when c > 0, when c = 0, and when c < 0.

•

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• Imaginary Number (imaginary unit)▪ i

▪ i = −1

▪ i2 = −1

• Examples

▪ −9

▪ −12

• Complex Number▪ Includes real numbers and

imaginary numbers

▪ Imaginary numbers are any number with i

▪ (a + bi) where a and b are real

• Find the values of x and y that satisfy the equation 5x − 20i = 35 + yi.

3i

i43 → 2i3

5x = 35 → x = 7-20i = yi→ y = -20

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• Adding and Subtracting Complex Numbers

▪ Add the same way you add (x + 4) + (2x – 3) = 3x+1

▪ Combine like terms

• Simplify

▪ (−1 + 2i) + (3 + 3i)

▪ (2 – 3i) – (3 – 7i)

▪ 2i – (3 + i) + (2 – 3i)

2 + 5i-1 + 4i-1 − 2i

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• Multiplying complex numbers▪ FOIL

▪ Remember i2 = −1

• Multiply▪ –i(3 + i)

▪ (2 + 3i)(-6 – 2i)

▪ (1 + 2i)(1 – 2i)

-3i – i2 = 1 – 3i-12 – 4i – 18i – 6i2 = -6 – 22i1 – 2i + 2i – 4i2 = 5

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▪ Notice on the last example that the answer was just real

• Complex conjugate → same numbers just opposite sign on the imaginary part

▪ When you multiply complex conjugates, the product is real

• Solve 𝑥2 + 121 = 0

• 3𝑥2 + 25 = −416

105 #1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 47, 49, 53, 57, 61, 65, 69, 75, 79, 95, 97, 99, 101, 103

𝑥2 + 121 = 0𝑥2 = −121

𝑥 = ± −121𝑥 = ±11𝑖

3𝑥2 + 25 = −4163𝑥2 = −441𝑥2 = −147

𝑥 = ± −147 = ±7 3𝑖

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Objectives: By the end of the lesson,• I can solve quadratic equations using square roots.• I can solve quadratic equations by completing the square.• I can apply completing the square to write quadratic functions in vertex form.

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• The Perfect Square

• 𝑥 + 3 2

• 𝑥 + 𝑘 2 = 𝑥2 + 2𝑘𝑥 + 𝑘2

• = 𝑎𝑥2 + 𝑏𝑥 + 𝑐

• In a perfect square,

𝑐 =1

2𝑏

2

=(x + 3)(x + 3) = x2 + 2(3x) + 32 = x2 + 6x + 9

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• Complete the square and then factor.

▪ 𝑥2 + 8𝑥

you have to add (8/2)2 = 42 = 16 to get a perfect square or (x + 4)2

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• Rewrite the quadratic so x terms on one side and constant on other.

• If the leading coefficient is not 1, divide everything by it.

• Complete the square: add 𝑏

2

2to both

sides.

• Rewrite the left hand side as a square (factor)

• Square root both sides

• Solve

𝑥2 + 6𝑥 = 16

• It already is

• 𝑥2 + 6𝑥 +6

2

2= 16 +

6

2

2

• 𝑥2 + 6𝑥 + 32 = 16 + 9• 𝑥 + 3 2 = 25• 𝑥 + 3 = ±5• 𝑥 = −3 ± 5• 𝑥 = 2,−8

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• Solve 𝑥2 − 18𝑥 + 5 = 0

𝑥2 − 18𝑥 + 5 = 0𝑥2 − 18𝑥 = −5

𝑥2 − 18𝑥 + 92 = −5 + 92

𝑥 − 9 2 = 76

𝑥 − 9 = ± 76

𝑥 = 9 ± 2 19

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Solve 2𝑥2 − 11𝑥 + 12 = 0

2𝑥2 − 11𝑥 = −12

𝑥2 −11

2𝑥 = −6

𝑥2 −11

2𝑥 + −

1122

2

= −6 + −

1122

2

→ −

1122

2

= −11

4

2

=121

16

𝑥 −11

4

2

=25

16

𝑥 −11

4= ±

5

4

𝑥 =11

4±5

4=11 ± 5

4= 4,

3

2

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• Writing quadratic functions in Vertex Form

• 𝑦 = 𝑎 𝑥 − ℎ 2 + 𝑘▪ (h, k) is the vertex

1. Start with standard form

2. Group the terms with the x

3. Factor out any number in front of the x2

4. Add (b/2)2 to both sides (inside the group on the right)

5. Rewrite as a perfect square

6. Subtract to get the y by itself

𝑦 = 2𝑥2 + 12𝑥 + 16

114 #1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 61, 73, 77, 79, 81, 87

2. 𝑦 = (2𝑥2 + 12𝑥) + 163. 𝑦 = 2(𝑥2 + 6𝑥) + 164. 𝑦 + 2 9 = 2 𝑥2 + 6𝑥 + 9 + 165. 𝑦 + 18 = 2 𝑥 + 3 2 + 166. 𝑦 = 2 𝑥 + 3 2 − 2

Vertex is at (−3, −2)−2 is the minimum for this functionFind the max or min by completing the square to find the vertex

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Objectives: By the end of the lesson,

• I can solve quadratic equations using the Quadratic Formula.

• I can find and interpret the discriminant of an equation.

• I can write quadratic equations with different numbers of solutions using the discriminant.

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• Work with a Partner: Solve 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

• 𝑎𝑥2 + 𝑏𝑥 = −𝑐

• 𝑥2 +𝑏

𝑎𝑥 = −

𝑐

𝑎

• 𝑥2 +𝑏

𝑎𝑥 +

𝑏

2𝑎

2=

𝑏

2𝑎

2−

𝑐

𝑎

• 𝑥 +𝑏

2𝑎

2=

𝑏2

4𝑎2−

4𝑎𝑐

4𝑎2

• 𝑥 +𝑏

2𝑎

2=

𝑏2−4𝑎𝑐

4𝑎2

• 𝑥 +𝑏

2𝑎= ±

𝑏2−4𝑎𝑐

4𝑎2

• 𝑥 = −𝑏

2𝑎±

𝑏2−4𝑎𝑐

2𝑎

• 𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎

𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0

𝑥2 +𝑏

𝑎𝑥 +

𝑐

𝑎= 0 Divide by a to get a = 1

𝑥2 +𝑏

𝑎𝑥 = −

𝑐

𝑎Add –(c/a) to get x’s by self

𝑥2 +𝑏

𝑎𝑥 +

𝑏

2𝑎

2= −

𝑐

𝑎+

𝑏

2𝑎

2Add the square of half of middle to get

perfect square

𝑥 +𝑏

2𝑎

2

= −𝑐

𝑎+

𝑏2

4𝑎2Factor left

𝑥 +𝑏

2𝑎= ±

−4𝑎𝑐+𝑏2

4𝑎2Square root

𝑥 = −𝑏

2𝑎±

𝑏2−4𝑎𝑐

4𝑎2Subtract

𝑥 =−𝑏± 𝑏2−4𝑎𝑐

2𝑎Simplify

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𝑥 =−𝑏 ± 𝑏2 − 4𝑎𝑐

2𝑎• This is called the quadratic formula and always works for

quadratic equations. • The part under the square root, discriminant, tells you

what kind of solutions you are going to have. ▪ b2 – 4ac > 0 → two distinct real solutions▪ b2 – 4ac = 0 → exactly one real solution (a double solution)▪ b2 – 4ac < 0 → two distinct imaginary solutions

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• What types of solutions to 5𝑥2 + 3𝑥 − 4 = 0?

• Solve 5𝑥2 + 3𝑥 = 4

𝑏2 − 4𝑎𝑐 = 32 − 4(5)(−4) = 9 + 80 = 89→ two distinct real roots

Put in standard form → 5𝑥2 + 3𝑥 − 4 = 0

Quadratic formula → 𝑥 =−3± 32−4 5 −4

2 5

Simplify →−3± 89

10

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• Solve 4𝑥2 − 6𝑥 + 3 = 0

4𝑥2 − 6𝑥 + 3 = 0

𝑥 =6 ± −6 2 − 4 4 3

2 4

𝑥 =6 ± 36 − 48

8

𝑥 =6 ± −12

8

𝑥 =6 ± 2 3𝑖

8

𝑥 =3 ± 3𝑖

4(𝑟𝑒𝑑𝑢𝑐𝑒 𝑡𝑜𝑝 𝑎𝑛𝑑 𝑏𝑜𝑡𝑡𝑜𝑚 𝑏𝑦 2)

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• Find a possible pair of integer values for a and c so that the equation 𝑎𝑥2 − 12𝑥 + 𝑐 = 0 has the given number and type of solution(s). Then write the equation.

• a. one real solution

• b. two imaginary solutions

𝑎𝑥2 − 12𝑥 + 𝑐 = 0Discriminant with 1 real solution

𝑏2 − 4𝑎𝑐 = 0−12 2 − 4𝑎𝑐 = 0

144 = 4𝑎𝑐36 = 𝑎𝑐

Sample Answer: 𝑎 = 9, 𝑐 = 4 → 9𝑥2 − 12𝑥 + 4 = 0

Discriminant with two imaginary solutions𝑏2 − 4𝑎𝑐 < 0

−12 2 − 4𝑎𝑐 < 0144 < 4𝑎𝑐36 < 𝑎𝑐

Sample Answer: 𝑎 = 10, 𝑐 = 5 → 10𝑥2 − 12𝑥 + 5 = 0

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• Real life problems

• The height of an object that is hit or thrown up or down can be modeled by

ℎ 𝑡 = −16𝑡2 + 𝑣0𝑡 + 𝑠0• Where v0 is the initial velocity (up +, down −), and s0 is the initial

height

• 123 #1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49, 53, 61, 75, 79, 81, 83, 91

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Objectives: By the end of the lesson,• I can describe what a nonlinear system of equations is.• I can solve nonlinear systems using graphing, substitution, or elimination.• I can solve quadratic equations by graphing each side of the equation.

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• Work with a partner.• a. Graph the equation 𝑥2 + 𝑦2 = 4. Make

several observations about the graph.• b. How many intersection points can the

graphs of a line and a circle have? Use graphs to support your answers. What do the intersection points represent?

• c. Consider the system below. Can you use a graph to solve the system? Explain.

𝑥2 + 𝑦2 = 4

𝑦 = −1

2𝑥 + 1

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• Nonlinear System

▪ System of equations where at least one is not a line

• Solve by Graphing

▪ Graph both equations

▪ Points of intersection are solutions

36

• Solve the system by graphing.

• 𝑦 = 𝑥2 − 8𝑥 + 12

• 𝑦 = 4𝑥 − 24

(6, 0)

37

• Solve the system by substitution.

• 𝑥2 + 𝑦2 = 20

• 𝑦 = 𝑥 + 2

𝑥2 + 𝑥 + 2 2 = 20𝑥2 + 𝑥2 + 4𝑥 + 4 = 202𝑥2 + 4𝑥 − 16 = 0𝑥2 + 2𝑥 − 8 = 0𝑥 + 4 𝑥 − 2 = 0𝑥 = −4 𝑜𝑟 𝑥 = 2

Use x=−4𝑦 = −4 + 2 = −2

(−4, −2)Use x=2

𝑦 = 2 + 2 = 4(2, 4)Solutions are (−4, −2) and (2, 4)

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• Solve the system by elimination.

• 3𝑥2 + 2𝑥 − 2𝑦 = 10

• 𝑥2 − 6𝑥 + 2𝑦 = −12

Add to eliminate y (try to eliminate all of 1 variable)4𝑥2 − 4𝑥 = −24𝑥2 − 4𝑥 + 2 = 02𝑥2 − 2𝑥 + 1 = 0

𝑥 =2 ± −2 2 − 4 2 1

2 2

𝑥 =2 ± −4

4

𝑥 =2 ± 2𝑖

4=1

2±1

2𝑖

No real solution

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• Solve by graphing.

• 𝑥 + 4 𝑥 − 1 = −𝑥2 + 3𝑥 + 4

• 132 #1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 47, 49, 55, 59, 63, 69, 71, 73

Two ways: graph both sides and find x-coordinates of intersectionOr make =0 and graph and find x-intercepts (zeros)x = −2, 2

41

Objectives: By the end of the lesson,

• I can describe the graph of a quadratic inequality.• I can graph quadratic inequalities.• I can graph systems of quadratic inequalities.• I can solve quadratic inequalities algebraically and graphically.

42

• Work with a partner. The figure shows the graph of 𝑓(𝑥) = 𝑥2 + 2𝑥 −3.

• a. Explain how you can use the graph to solve the inequality 0 > 𝑥2 + 2𝑥 −3. Then graph the solutions of the inequality.

• b. Explain how the inequality 𝑦 >𝑥2 + 2𝑥 − 3 is different from the inequality in part (a).

• c. Explain how you can use the graph above to represent the solutions of 𝑦 > 𝑥2 + 2𝑥 − 3. Then graph the inequality.

43

• Graphing a Quadratic Inequality in Two Variables• Graph the function as if it was =• Dashed or Solid line

▪ <, > → Dashed▪ ≤, =, ≥ → Solid

• Shade▪ Test a point not on the line

➢ If it is a solution, shade that side➢ If it is NOT a solution, shade the other side

▪ OR▪ Solve for y

➢ y >, shade above➢ y <, shade below

44

• Graph 𝑦 ≥ 𝑥2 − 10𝑥 + 25

Vertex at x = 10/2 = 5Table of values to include x = {1, 2, 3, 4, 5, 6, 7, 8, 9}Pick (0, 0) as test point → 0 25 false shade other side of line.

45

• Graph the system of quadratic inequalities.

• 𝑦 > 𝑥2 − 4

• 𝑦 ≤ −𝑥2 + 3𝑥 + 4

46

Solve inequalities in one variable. 1. Make = 02. Factor or use the quadratic

formula to find the zeros3. Graph the zeros on a number

line (notice it cuts the line into three parts)

4. Pick a number in each of the three parts as test points

5. Test the points in the original inequality to see true or false

6. Write inequalities for the regions that were true

❑ p2 – 4p 5

0

1.p2 – 4p – 5 02.(p – 5)(p + 1) 0 Zeros are 5, -14.test points = -2, 0, 6–2 → 4 – 4(-2) 5 → 12 5 false0 → 0 – 4(0) 5 → 0 5 true6 → 36 – 4(6) 5 → 12 5 false6.middle region true so –1 p 5

47

• Or You could also solve the quadratic inequality in one variable by graphing the quadratic

▪ Make the equation = 0

▪ Graph

▪ When the graph is below

the x-axis; ≤ 0

▪ When the graph is above

the x-axis; ≥ 0

140 #1, 5, 9, 13, 15, 17, 21, 25, 27, 29, 33, 37, 41, 45, 49, 57, 61, 63, 65, 67

2x2 + 4x – 6 ≤ 0 when -3 ≤ x ≤ 12x2 + 4x – 6 ≥ 0 when x ≤ -3 or x ≥ 1

48