Exercises 105-120: Complete the following. (a) Write the equation as ax² + bx + e = 0 with a > 0. (b) Calculate the discriminant b² – 4ac and determine the number of real solutions. (c) Solve the equation. 105. 3x² = 12 106. 8x - 2 = 14 107. x² – 2x = -1 108. 6x² = 4x 109. 4x = x? 110. 16x + 9 = 24x 111. x² + 1 = x 112. 2x² + x = 2 113. 2x² + 3x = 12 – 2x 114. 3x² + 3 = 5x 115. x(x – 4) = -4 116. + 3x = x – 4 117. x(x + 2) = -13 118. 4x = 6 + x? 119. 3x = 1- x 120. x(5x – 3) = 1
Exercises 105-120: Complete the following. (a) Write the equation as ax² + bx + e = 0 with a > 0. (b) Calculate the discriminant b² – 4ac and determine the number of real solutions. (c) Solve the equation. 105. 3x² = 12 106. 8x - 2 = 14 107. x² – 2x = -1 108. 6x² = 4x 109. 4x = x? 110. 16x + 9 = 24x 111. x² + 1 = x 112. 2x² + x = 2 113. 2x² + 3x = 12 – 2x 114. 3x² + 3 = 5x 115. x(x – 4) = -4 116. + 3x = x – 4 117. x(x + 2) = -13 118. 4x = 6 + x? 119. 3x = 1- x 120. x(5x – 3) = 1
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter9: Quadratic Functions And Equations
Section9.6: Solving Quadratic Equations By Using The Quadratic Formula
Problem 57HP
Related questions
Question
![Exercises 105-120: Complete the following.
(a) Write the equation as ax² + bx + e = 0 with a > 0.
(b) Calculate the discriminant b² – 4ac and determine the
number of real solutions.
(c) Solve the equation.
105. 3x² = 12
106. 8x - 2 = 14
107. x² – 2x = -1
108. 6x² = 4x
109. 4x = x?
110. 16x + 9 = 24x
111. x² + 1 = x
112. 2x² + x = 2
113. 2x² + 3x = 12 – 2x 114. 3x² + 3 = 5x
115. x(x – 4) = -4
116. + 3x = x – 4
117. x(x + 2) = -13
118. 4x = 6 + x?
119. 3x = 1- x
120. x(5x – 3) = 1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5c922dfb-68a6-4469-9d51-c56b46c273fd%2F71fae471-4a88-4f26-8f16-345192a852cc%2Fqukv1f_processed.png&w=3840&q=75)
Transcribed Image Text:Exercises 105-120: Complete the following.
(a) Write the equation as ax² + bx + e = 0 with a > 0.
(b) Calculate the discriminant b² – 4ac and determine the
number of real solutions.
(c) Solve the equation.
105. 3x² = 12
106. 8x - 2 = 14
107. x² – 2x = -1
108. 6x² = 4x
109. 4x = x?
110. 16x + 9 = 24x
111. x² + 1 = x
112. 2x² + x = 2
113. 2x² + 3x = 12 – 2x 114. 3x² + 3 = 5x
115. x(x – 4) = -4
116. + 3x = x – 4
117. x(x + 2) = -13
118. 4x = 6 + x?
119. 3x = 1- x
120. x(5x – 3) = 1
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