# Algebra 1 Name Per _____ Solving Quadratic Equations Lesson 9.4

```Algebra 1
Name _____________________________________ Per _____
Lesson 9.4: You can solve by graphing. The solutions of a quadratic equation and the related x-intercepts are often
called the roots of the equation or zeros of the function. They are where the parabola crosses the x-axis (x-intercepts).
Leave in simplest radical form, if necessary.
You can solve equations of the form x2 = a by finding the square roots of each side.
1. 2x2 - 98 = 0
2. t2 - 25 = 0
3. 3n2 + 12 = 12
4. 2g2 + 32 = 0.
5. x2 - 9 = 0
9.
1 2
x
3
-3=0
13. 27 - y2 = 0
6. x2 + 5 = 0
7. 2x2 - 8 = 0
10. x2 - 72 = 0
11.
14. c2 + 25 = 25
15. x2 - 9 = -16
17. Find the side of a square with an area of 256 m2.
1 2
x
4
-1=0
8. 4x2 = 0
12. k2 = 49
16. 6x2 - 24 = 0
18. Find the side of a square with an area of 90 ft2.
Lesson 9.5 Solve by Factoring
The Zero-Product Property:
For every real number a and b, if ab = 0, then either a = 0 or b = 0.
If (x + 3)(x + 2) = 0, then either x + 3 = 0 or x + 2 = 0.
Make sure the quadratic equations are in standard form! If possible, factor GCF first!
19. x2 + 6x + 8 = 0
20. x2 - 8x - 48 = 0
21. 2x2 - 5x = 88
22. x2 - 12x = -36
23. x2 - 6x - 27 = 0
24. 2x2 - x - 36 = 0
25. t2 + t = 0
26. n2 + n - 12 = 0
27. k3 + 7k2 = -12k
28. b2 + 3b - 4 = 0
29. w2 - 8w = 0
30. m2 - 5m - 14 = 0
31. 2q2 + 22q = -60
32. x2 + 8x = -15
33. 3q2 + 16q = -5
34. t2 - 3t = 28
35. 4y2 = 25
36. k2 - 3k - 10 = 0
37. 4 = -5n + 6n2
38. 2x2 - 7x = 15
39. p3 - 2p2 - 48p = 0
40. 3t2 + 8t = t2 - 3t - 12
41. 4x2 + 20 = 10x + 3x2 - 4
42. 2c2 - 7c = -5
Lesson 9.6 You can also solve by Completing the Square. To find the number to add to each side of the equation
𝒃 𝟐
x2 + bx = c to complete the square, you must use the formula (𝟐) .
What number would you add to each equation to make a perfect square trinomial? (No coefficent can be on x2)
43. k2 + 14k = 0
44. r2 + 8r = 48
45. x2 + 10x + 17 = 0
46. p2 - 12p + 11 = 0
47. 2k2 + 5k = k + 10
Lesson 9.7
Solve equations ax2 + bx + c = 0 by using the quadratic formula x =
−𝒃 &plusmn; √𝒃𝟐 −𝟒𝒂𝒄
.
𝟐𝒂
Make sure you simplify all radicals! Make sure all equations are in standard form.
48. x + 6 - 5x = 0
49. x2 - 2x - 8 = 0
50. 2x2 + 5x + 3 = 0
51. 3x2 - 10x = 2
2
52. x2 + 2x = 13
53. 5x2 + 13x - 1 = 0
54. 3x2 - 4x = 2
55. 8x2 - 2x = 7
56. x2 + x = 5
57. x2 + 2x - 7 = 0
58. 2x2 - 7x = -6
59. x2 - 4x = -10
You can use the discriminant to determine the # of solutions of the equation ax2 + bx + c = 0
(x-intercepts, roots of the equation, zeros of the function). The discriminant is b2 - 4ac.
If b2 - 4ac &gt; 0, there are 2 solution; if b2 - 4ac = 0, there is 1 solution; if b2 - 4ac &lt; 0 there are no solutions.
Determine the # of solutions. DO NOT SOLVE! Make sure the equations are in standard form!
60. x2 - 6x + 3 = 0
61. x2 - 6x + 9 = 0
62. 3x2 - 5x - 1 = 0
63. x2 - 6x = -12
64. 3x2 - 4x = 7
Lesson 9.8
Mixed Review: Solve the following quadratic equations using any method you choose. If there is no solution, write
&quot;no solution.&quot;
65. 4x2 - 81 = 0
66. 3x2 - 5x + 9 = 0
67. x2 + 13x = -36
68. x2 = 100
69. 2t2 = 72
70. 5b2 - 10 = 0
71. 3m2 - 6m = 2
72. 2r2 - 24r + 72 = 0
73. x2 + 2x - 1 = 0
74. x2 - 8x = 33
75. x3 - 5x2 + 4x = 0
76. x2 + 64 = 0
77. 2c2 + 11c + 15 = 0
78. x3 - x2 = 0
79. x2 + 1 = 0
80. Which equation does
a. -9x + 5x = -7
9 &plusmn; √(−9)2 − 4(5)(−7)
2(5)
b. 5x2 - 9x = -7
give the solutions to?
c. 5x2 + 7x = 9
d. 5x2 - 9x = 7
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