1. Chapter 5 – Quadratic Functions
Algebra 2

2. Table of Contents 5.5 - Complex Numbers and Roots
5.6 - The Quadratic Formula
5.7 – Solving Quadratic Inequalities
5.9 - Operations with Complex Numbers

3. 5.5 - Complex Numbers and Roots
Algebra 2

4. 5.5
Algebra 2 (bell work)
Pg. 350

5. 5.5
Just Read
You can see in the graph of f(x) = x2 + 1 below that f has no real zeros.
If you solve the corresponding equation 0 = x2 + 1, you find that x = ,which has no real solutions.

6. Simplifying Square Roots of Negative Numbers
5.5
Example 1
Simplifying Square Roots of Negative Numbers
Express the number in terms of i.

7. Express the number in terms of i
5.5

8. Math Joke Q: What’s a mathematician's favorite dessert? A: -1 ce cream
5.5
Math Joke
Q: What’s a mathematician's favorite dessert?
A: -1 ce cream

9. 5.5
Example 2
Solving a Quadratic Equation with Imaginary Solutions
Solve the equation
5x = 0

10. 5.5
Solve the equation
x = 0
9x = 0
x2 = –48
9x2 = –25

11. 5.5
A complex number is a number that can be written in the form a + bi,
where a and b are real numbers and i = The set of real numbers is a subset of the set of complex numbers C.
Every complex number has a real part a and an imaginary part b.

12. Equate the imaginary parts. Equate the real parts. 10 = –4y 4x = 2
5.5
Example 3
Equating Two Complex Numbers
Find the values of x and y that make the equation true .
Real parts
4x + 10i = 2 – (4y)i
Imaginary parts
Equate the imaginary parts.
Equate the real parts.
10 = –4y
4x = 2
Solve for y.
Solve for x.

13. Equate the imaginary parts. 2x = –8 –6 = 20y
5.5
Find the values of x and y that make each equation true
2x – 6i = –8 + (20y)i
Real parts
2x – 6i = –8 + (20y)i
Imaginary parts
Equate the real parts.
Equate the imaginary parts.
2x = –8
–6 = 20y
x = –4
Solve for y.
Solve for x.

14. f(x) = x2 + 10x + 26 g(x) = x2 + 4x + 12 5.5 Example 4
Finding Complex Zeros of Quadratic Functions
Day 2
Find the zeros of the function.
f(x) = x2 + 10x + 26
g(x) = x2 + 4x + 12
x2 + 10x + 26 = 0
x2 + 4x + 12 = 0
x2 + 10x = –26 +
x2 + 4x = –12 +
x2 + 10x + 25 = –
x2 + 4x + 4 = –12 + 4
(x + 5)2 = –1
(x + 2)2 = –8

15. f(x) = x2 + 4x + 13 5.5 Find the zeros of the function.
x = –2 ± 3i

16. Find each complex conjugate.
5.5
The solutions and are related. These solutions are a complex conjugate pair.
Their real parts are equal and their imaginary parts are opposites.
**The complex conjugate of any complex number a + bi is the complex number a – bi.**
Example 5
Finding Complex Conjugates
Find each complex conjugate.
A i
B. 6i
8 + 5i
0 + 6i
0 – 6i
8 – 5i
–6i

17. Find each complex conjugate
5.5
Find each complex conjugate
A. 9 – i B.
9 + (–i)
9 – (–i)
9 + i
C. –8i
0 + (–8)i
0 – (–8)i 8i

18. HW pg. 353
5.5-
Day 1: 2-11, 18-25, 92-98
Day 2: 12-17, 29, 31, 36
Ch: 43, 58-65, 72, 75
HW Guidelines or ½ off

19. 5.6 - The Quadratic Formula
Algebra 2

20. 5.6
Algebra 2 (bell work)
You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.
*Once you have your zeros, can always plug them back in and see if you get zero
Videos

21. Don’t Copy
5.6

22. f(x)= 2x2 – 16x + 27 f(x) = x2 + 3x – 7 5.6 Example 1
Quadratic Functions with Real Zeros
Find the zeros using the Quadratic Formula
f(x)= 2x2 – 16x + 27
f(x) = x2 + 3x – 7
2x2 – 16x + 27 = 0
x2 + 3x – 7 = 0

23. 5.6
Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula.
Optional
x2 – 8x + 10 = 0

24. 5.6
Math Joke
Q: What do you get when you cross an algebra class with prom?
A: The Quadratic formal

25. f(x) = 4x2 + 3x + 2 g(x) = 3x2 – x + 8 5.6 Example 2
Quadratic Functions with Complex Zeros
Find the zeros using the Quadratic Formula
f(x) = 4x2 + 3x + 2
g(x) = 3x2 – x + 8
4x2 + 3x + 2 = 0

26. 5.6
Day 2
The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

27. 5.6
Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!

28. 5.6
Example 3
Analyzing Quadratic Equations by Using the Discriminant
Find the type and number of solutions for the equation.
x = 12x
x = 12x
x2 – 12x + 36 = 0
x2 – 12x + 40 = 0
b2 – 4ac
b2 – 4ac
(–12)2 – 4(1)(36)
(–12)2 – 4(1)(40)
144 – 144 = 0
144 – 160 = –16
b2 – 4ac = 0
b2 –4ac < 0
The equation has one distinct real solution.
The equation has two distinct nonreal complex solutions.

29. 5.6
Find the type and number of solutions for the equation.
x = 12x
x2 – 12x + 30 = 0
b2 – 4ac
(–12)2 – 4(1)(30)
144 – 120 = 24
b2 – 4ac > 0
The equation has two distinct real solutions.

30. 5.6
Example 4
Application
A pilot of a helicopter plans to release a bucket of water on a forest fire.
The height y in feet of the water t seconds after its release is modeled
by y = –16t2 – 2t
The horizontal distance x in feet between the water and its point of release is modeled by x = 91t.
The pilot’s altitude decreases, which changes the function describing the water’s height to y = –16t2 –2t
To the nearest foot, at what horizontal distance from the target should the pilot begin releasing the water?

31. 5.6
Step 1 Use the equation to determine how long it will take the water to hit the ground. Set the height of the water equal to 0 feet, and then use the quadratic formula for t.
y = –16t2 – 2t + 400
0 = –16t2 – 2t + 400
t ≈ – or t ≈ 4.937
The time cannot be negative, so the water lands on a target about seconds after it is released.

32. 5.6
Step 2 The horizontal distance x in feet between the water and its point of release is modeled by x = 91t. Find the horizontal distance that the water will have traveled in this time.
x = 91t
x ≈ 91(4.937)
x ≈
x ≈ 449
The water will have traveled a horizontal distance of about 449 feet. Therefore, the pilot should start releasing the water when the horizontal distance between the helicopter and the fire is 449 feet.

33. 5.6 Optional An athlete on a track team throws a shot put.
The height y of the shot put in feet t seconds after it is thrown is modeled by y = –16t t
The horizontal distance x in between the athlete and the shot put is modeled by x = 29.3t.
To the nearest foot, how far does the shot put land from the athlete?

34. 5.6
Step 1 Use the first equation to determine how long it will take the shot put to hit the ground. Set the height of the shot put equal to 0 feet, and the use the quadratic formula to solve for t.
y = –16t t + 6.5
0 = –16t t + 6.5
The time cannot be negative, so the shot put hits the ground about 1.8 seconds after it is released.

35. Step 2 Find the horizontal distance that the shot put will have traveled in this time.
x = 29.3t
x ≈ 29.3(1.77)
x ≈ 51.86
x ≈ 52
The shot put will have traveled a horizontal distance of about 52 feet.

36. Properties of Solving Quadratic Equations
5.6
Properties of Solving Quadratic Equations
Pg. 360

37. HW pg. 361 5.6- Day 1: 3-13 (Odd), 45-53 Day 2: 14-17, 54, 59, 76-78
Ch: 36, 37, 60
HW Guidelines or ½ off

38. 5.7 - Solving Quadratic Inequalities
Algebra 2

39. 5.7
Algebra 2 (Bell work)
Copy the steps below
Pg. 366

40. 5.7
Example 1
Graphing Quadratic Inequalities in Two Variables
Graph y ≥ x2 – 7x + 10

41. Graph y ≥ 2x2 – 5x – 2
5.7

42. Graph y < –3x2 – 6x – 7
5.7

43. 5.7
Math Joke
Q: How can a fisherman determine how many fish he needs to catch to make a profit?
A: By using a cod-ratic inequality

44. 5.7
Example 2
Solving Quadratic Inequalities by Using Tables and Graphs
Solve the inequality by using tables or graphs.
x2 + 8x + 20 ≥ 5
Use a graphing calculator to graph each side of the inequality.
Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of x for which Y1 ≥ Y2.

45. 5.7
The parabola is at or above the line when x is less than or equal to –5 or greater than or equal to –3.
So, the solution set is x ≤ –5 or x ≥ –3 or (–∞, –5] U [–3, ∞). The table supports your answer.
The number line shows the solution set.
–6 –4 –

46. 5.7
Solve the inequality by using tables and graph.
x2 + 8x + 20 < 5
Use a graphing calculator to graph each side of the inequality.
Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of which Y1 < Y2.

47. 5.7
The parabola is below the line when x is greater than –5 and less than –3.
So, the solution set is –5 < x < –3 or (–5, –3). The table supports your answer.
The number line shows the solution set.
–6 –4 –

48. 2x2 – 5x + 1 ≥ 1 5.7 Solve the inequality by using tables and graph.
Optional
2x2 – 5x + 1 ≥ 1
Use a graphing calculator to graph each side of the inequality.
Set Y1 equal to 2x2 – 5x + 1 and Y2 equal to 1. Identify the values of which Y1 ≥ Y2.

49. 5.7
The parabola is at or above the line when x is less than or equal to 0 or greater than or greater than or equal to 2.5.
So, the solution set is (–∞, 0] U [2.5, ∞) or x ≤ 0 or x ≥ 2.5
The number line shows the solution set.
–6 –4 –

50. Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra.
5.7
Example 3
Solving Quadratic Inequalities by Using Algebra
Day 2
Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra.
Step 1 Write the related equation.
x2 – 10x + 18 = –3
Step 2 Solve the equation for x to find the critical values.
x2 –10x + 21 = 0
(x – 3)(x – 7) = 0
Step 3 Test an x-value in each interval.
x – 3 = 0 or x – 7 = 0
x = 3 or x = 7

51. x2 – 10x + 18 ≤ –3 5.7 Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 10x + 18 ≤ –3
(2)2 – 10(2) + 18 ≤ –3
Try x = 2. x
(4)2 – 10(4) + 18 ≤ –3 
Try x = 4.
(8)2 – 10(8) + 18 ≤ –3 x
Try x = 8.

52. x2 – 6x + 10 ≥ 2 5.7 Solve the inequality by using algebra.
Step 1 Write the related equation.
x2 – 6x + 10 = 2
x2 – 6x + 8 = 0
Step 2 Solve the equation for x to find the critical values.
(x – 2)(x – 4) = 0
x – 2 = 0 or x – 4 = 0
x = 2 or x = 4

53. x2 – 6x + 10 ≥ 2 5.7 Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
x2 – 6x + 10 ≥ 2
(1)2 – 6(1) + 10 ≥ 2 
Try x = 1.
(3)2 – 6(3) + 10 ≥ 2 x
Try x = 3.
(5)2 – 6(5) + 10 ≥ 2 
Try x = 5.

54. –2x2 + 3x + 7 < 2 5.7 Solve the inequality by using algebra.
Optional
–2x2 + 3x + 7 < 2
Step 1 Write the related equation.
–2x2 + 3x + 7 = 2
Step 2 Solve the equation for x to find the critical values.
–2x2 + 3x + 5 = 0
(–2x + 5)(x + 1) = 0
–2x + 5 = 0 or x + 1 = 0
x = 2.5 or x = –1

55. –2x2 + 3x + 7 < 2 5.7 Step 3 Test an x-value in each interval.
–3 –2 –
Critical values
Test points
–2x2 + 3x + 7 < 2
Try x = –2.
–2(–2)2 + 3(–2) + 7 < 2 
–2(1)2 + 3(1) + 7 < 2 x
Try x = 1.
–2(3)2 + 3(3) + 7 < 2 
Try x = 3.

56. 5.7
Example 4
Application
Skip 15_16
The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x x – 4200
Where x is the average selling price of a helmet.
What range of selling prices will generate a monthly profit of at least $6000?
List the important information:
The profit must be at least $6000.
The function for the business’s profit is P(x) = –8x x – 4200.
Write the inequality.
–8x x – 4200 ≥ 6000
Find the critical values by solving the related equation.
–8x x – 4200 = 6000
–8x x – 10,200 = 0
–8(x2 – 75x ) = 0

57. 5.7
x ≈ or x ≈ 48.96
–8(25) (25) – 4200 ≥ 6000
Try x = 25. x
5800 ≥ 6000
–8(45) (45) – 4200 ≥ 6000
Try x = 45.
6600 ≥ 6000 
–8(50) (50) – 4200 ≥ 6000
Try x = 50.
5800 ≥ 6000 x
Write the solution as an inequality. The solution is approximately ≤ x ≤

58. HW pg.370 5.7- Day 1: 2-4 (By Hand), 5-7 (Calc), 48-50, 72-74
Ch: 11, 14, 47, 51-53

59. 5.9 - Operations With Complex Numbers
Algebra 2

60. Just Read Algebra 2 (bell work)
5.9
Algebra 2 (bell work)
The complex plane is a set of coordinate axes in which where x = real numbers, y = imaginary numbers
Just Read
The real axis corresponds to the x-axis, and the imaginary axis corresponds to the y-axis. Think of a + bi as x + yi.
Helpful Hint

61. • • • • A. 2 – 3i B. –1 + 4i C. 4 + i D. –i 5.9 Example 1
Graphing Complex Numbers
Graph each complex number.
A. 2 – 3i •
–1+ 4i
B. –1 + 4i •
4 + i
C. 4 + i • –i
D. –i •
2 – 3i

62. 5.9

63. A. |3 + 5i| B. |–13| C. |–7i| 5.9 Example 2
Determining the Absolute Value of Complex Numbers
Find each absolute value.
A. |3 + 5i|
B. |–13|
C. |–7i|
|–13 + 0i|
|0 +(–7)i| 7 13

64. Math Joke Q: Why did the imaginary number turn red?
5.9
Math Joke
Q: Why did the imaginary number turn red?
A: It ran out of i-drops

65. (4 + 2i) + (–6 – 7i) (5 –2i) – (–2 –3i) (1 – 3i) + (–1 + 3i) 5.9
Example 3
Adding and Subtracting Complex Numbers
Add or subtract. Write the result in the form a + bi.
(4 + 2i) + (–6 – 7i)
(5 –2i) – (–2 –3i)
(1 – 3i) + (–1 + 3i)
(4 – 6) + (2i – 7i)
(5 – 2i) i
(1 – 1) + (–3i + 3i)
(5 + 2) + (–2i + 3i)
–2 – 5i
7 + i

66. –2i(2 – 4i) (3 + 6i)(4 – i) 5.9 Example 5 Multiplying Complex Numbers
Simplify by using the fact i2 = –1.
Multiply. Write the result in the form a + bi.
–2i(2 – 4i)
(3 + 6i)(4 – i)
–4i + 8i2
i – 3i – 6i2
–4i + 8(–1)
i – 6(–1)
–8 – 4i i

67. 5.9
Multiply. Write the result in the form a + bi.
(2 + 9i)(2 – 9i)
(–5i)(6i)
4 – 18i + 18i – 81i2
–30i2
4 – 81(–1)
–30(–1) 85 30

68. Simplify –6i14 Simplify i63 5.9 Example 6 Evaluating Powers of i
Day 2
Example 6
Evaluating Powers of i
Simplify –6i14
Simplify i63
i63 = i  i62
–6i14 = –6(i2)7
= –6(–1)7
= i  (i2)31
= i  (–1)31 = i  –1 = –i
= –6(–1) = 6

69. 5.9
Simplify i42
Optional
Simplify
i42 = ( i2)21
= (–1)21 = –1

70. Dividing Complex Numbers
5.9
Example 7
Dividing Complex Numbers
Simplify
-Recall that expressions in simplest form cannot have square roots in the denominator (Lesson 1-3).
-Because the imaginary unit represents a square root, you must rationalize any denominator that contains an imaginary unit.
-To do this, multiply the numerator and denominator by the complex conjugate of the denominator.

71. Simplify
Simplify
5.9

72. HW pg.386
5.9-
Day 1: 2-5 (One Graph), 9-17 (Odd), (Odd), (Odd)
Day 2: (Odd), (Odd), 121, 123
Ch: