The document provides examples of using the quadratic formula to solve quadratic equations. It defines key terms like the quadratic formula and discriminant. The quadratic formula is given as x = −b ±√(b2 - 4ac)/2a. The discriminant, which is part of the formula inside the square root, determines the number and type of roots. Several examples are worked through step-by-step to demonstrate solving quadratic equations algebraically using the quadratic formula.

1. Section 3-6
The Quadratic Formula and the
Discriminant

2. Essential Questions
• How do you solve quadratic equations by
using the Quadratic Formula?
• How do you use the discriminant to determine
the number and type of roots of a quadratic
equation?

3. Vocabulary
1. Quadratic Formula:
2. Discriminant:

4. Vocabulary
1. Quadratic Formula: Finds the solution of a
quadratic when a ≠ 0
2. Discriminant:

5. Vocabulary
1. Quadratic Formula:
x =
−b ± b2
− 4ac
2a
Finds the solution of a
quadratic when a ≠ 0
2. Discriminant:

6. Vocabulary
1. Quadratic Formula:
x =
−b ± b2
− 4ac
2a
Finds the solution of a
quadratic when a ≠ 0
2. Discriminant: The part of the quadratic formula
inside the square root which determines the
number and type of roots a quadratic
equation has

7. Vocabulary
1. Quadratic Formula:
x =
−b ± b2
− 4ac
2a
Finds the solution of a
quadratic when a ≠ 0
2. Discriminant: The part of the quadratic formula
inside the square root which determines the
number and type of roots a quadratic
equation has
discriminant = b2
− 4ac

8. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33

9. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0

10. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33

11. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a

12. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)

13. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2

14. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2

15. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2
x =
8 ±14
2

16. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2
x =
8 ±14
2
x =
8 +14
2

17. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2
x =
8 ±14
2
x =
8 +14
2
x =
8 −14
2

18. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2
x =
8 ±14
2
x =
8 +14
2
x =
8 −14
2
x = 11

19. Example 1
Solve by using the quadratic formula.
x 2
− 8x = 33
x 2
− 8x − 33 = 0
a = 1, b = −8, c = −33
x =
−b ± b2
− 4ac
2a
x =
8 ± (−8)2
− 4(1)(−33)
2(1)
x =
8 ± 64 +132
2
x =
8 ± 196
2
x =
8 ±14
2
x =
8 +14
2
x =
8 −14
2
x = 11 x = −3

20. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0

21. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289

22. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a

23. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a
x =
34 ± (−34)2
− 4(1)(289)
2(1)

24. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a
x =
34 ± (−34)2
− 4(1)(289)
2(1)
x =
34 ± 1156 −1156
2

25. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a
x =
34 ± (−34)2
− 4(1)(289)
2(1)
x =
34 ± 1156 −1156
2
x =
34 ± 0
2

26. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a
x =
34 ± (−34)2
− 4(1)(289)
2(1)
x =
34 ± 1156 −1156
2
x =
34 ± 0
2
x =
34
2

27. Example 2
Solve by using the quadratic formula.
x 2
− 34x + 289 = 0
a = 1, b = −34, c = 289
x =
−b ± b2
− 4ac
2a
x =
34 ± (−34)2
− 4(1)(289)
2(1)
x =
34 ± 1156 −1156
2
x =
34 ± 0
2
x =
34
2
x = 17

28. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0

29. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2

30. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a

31. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)

32. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2

33. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2

34. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2
x =
6 ± 4 7
2

35. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2
x =
6 ± 4 7
2
x =
6 ± 2 7
2

36. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2
x =
6 ± 4 7
2
x =
6 ± 2 7
2
x = 3 ± 7

37. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2
x =
6 ± 4 7
2
x =
6 ± 2 7
2
x ≈ 5.65
x = 3 ± 7

38. Example 3
Solve by using the quadratic formula. Give both the
exact and approximate answers.
x 2
− 6x + 2 = 0
a = 1, b = −6, c = 2
x =
−b ± b2
− 4ac
2a
x =
6 ± (−6)2
− 4(1)(2)
2(1)
x =
6 ± 36 − 8
2
x =
6 ± 28
2
x =
6 ± 4 7
2
x =
6 ± 2 7
2
x ≈ 5.65 x ≈ .35
x = 3 ± 7

39. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x

40. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0

41. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26

42. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a

43. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a
x =
12 ± (−12)2
− 4(2)(26)
2(2)

44. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a
x =
12 ± (−12)2
− 4(2)(26)
2(2)
x =
12 ± 144 − 208
4

45. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a
x =
12 ± (−12)2
− 4(2)(26)
2(2)
x =
12 ± 144 − 208
4
x =
12 ± −64
4

46. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a
x =
12 ± (−12)2
− 4(2)(26)
2(2)
x =
12 ± 144 − 208
4
x =
12 ± −64
4
x =
12 ± 8i
4

47. Example 4
Solve by using the quadratic formula.
2x 2
+ 26 = 12x
2x 2
−12x + 26 = 0
a = 2, b = −12, c = 26
x =
−b ± b2
− 4ac
2a
x =
12 ± (−12)2
− 4(2)(26)
2(2)
x =
12 ± 144 − 208
4
x =
12 ± −64
4
x =
12 ± 8i
4
x = 3 ± 2i

48. Using the Discriminant

49. Using the Discriminant

50. Using the Discriminant
Value of the
Discriminant
Type and Number of
Roots

51. Using the Discriminant
Value of the
Discriminant
Type and Number of
Roots
Positive and a perfect square 2 real, rational roots

52. Using the Discriminant
Value of the
Discriminant
Type and Number of
Roots
Positive and a perfect square 2 real, rational roots
Positive and NOT a perfect square 2 real, irrational roots

53. Using the Discriminant
Value of the
Discriminant
Type and Number of
Roots
Positive and a perfect square 2 real, rational roots
Positive and NOT a perfect square 2 real, irrational roots
Equal to 0 1 real, rational double root

54. Using the Discriminant
Value of the
Discriminant
Type and Number of
Roots
Positive and a perfect square 2 real, rational roots
Positive and NOT a perfect square 2 real, irrational roots
Equal to 0 1 real, rational double root
Less than zero 2 complex roots

55. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0

56. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac

57. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
32
− 4(1)(5)

58. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
32
− 4(1)(5)

59. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
32
− 4(1)(5)

60. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
Two complex roots; use
the quadratic formula
32
− 4(1)(5)

61. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
Two complex roots; use
the quadratic formula
x =
−b ± b2
− 4ac
2a
32
− 4(1)(5)

62. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
Two complex roots; use
the quadratic formula
x =
−b ± b2
− 4ac
2a
x =
−3 ± −11
2
32
− 4(1)(5)

63. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
Two complex roots; use
the quadratic formula
x =
−b ± b2
− 4ac
2a
x =
−3 ± −11
2
x =
−3 ± i 11
2
32
− 4(1)(5)

64. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
a. x 2
+ 3x + 5 = 0
b2
− 4ac
9 − 20
−11
Two complex roots; use
the quadratic formula
x =
−b ± b2
− 4ac
2a
x =
−3 ± −11
2
x =
−3 ± i 11
2
32
− 4(1)(5)
x =
−3
2
±
11
2
i

65. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0

66. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac

67. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
(−11)2
− 4(1)(10)

68. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
(−11)2
− 4(1)(10)

69. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
(−11)2
− 4(1)(10)

70. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
(−11)2
− 4(1)(10)

71. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
(−11)2
− 4(1)(10)
1i10

72. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
(−11)2
− 4(1)(10)
1i10 = 10

73. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
(−11)2
− 4(1)(10)
1i10 = 10
−10

74. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1

75. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1

76. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0

77. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0

78. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0
(x −10)(x −1) = 0

79. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0
(x −10)(x −1) = 0
x −10 = 0

80. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0
(x −10)(x −1) = 0
x −10 = 0 x −1= 0

81. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0
(x −10)(x −1) = 0
x −10 = 0 x −1= 0
x = 10

82. Example 5
Find the value of the discriminant. Then
describe the number and type of roots.
Determine the best method to solve the
quadratic by and solve it.
b. x 2
−11x +10 = 0
b2
− 4ac
121− 40
81
Two real, rational roots;
use factoring
x 2
−10x − x +10 = 0
(−11)2
− 4(1)(10)
1i10 = 10
−10 −1
(x 2
−10x )+ (−x +10) = 0
x(x −10)−1(x −10) = 0
(x −10)(x −1) = 0
x −10 = 0 x −1= 0
x = 10 x = 1