Intermediate Algebra – Student Workbook – Second Edition 2013

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$Lesson 2a – Assessment - Scottsdale Community College - Math Blog$

Lesson 5b - Solving Quadratic Equations Mini-Lesson Problem 5 WORKED EXAMPLE–SOLVE QUADRATIC EQUNS BY TRIAL/ERROR FACTORING Use the Trial and Error Factoring Method to solve each of the quadratic equations below. Verify your result by graphing the quadratic part of the equation and looking at where it crosses the x- axis. a) Solve x 2 + x - 6 = 0 Step 1: Make sure the equation is in standard form (check!). Step 2: Can we use GCF method? If no common factor other than one then no (can’t use here). Step 3: Try to factor the left side by Trial and Error Factor x 2 + x – 6 = (x + 3)(x – 2) Check by Foiling to be sure your answer is correct. Check: (x + 3)(x – 2) = x 2 – 2x + 3x – 6 = x 2 + x – 6 (checks!) Step 4: Write the factored form of the quadratic then set each factor to 0 and solve for x. (x + 3)(x – 2) = 0 so x + 3 = 0 or x – 2 = 0 x = -3 or x = 2 are the solutions to x 2 + x – 6 = 0. Can also write as x = -3, 2 Note: Graph x 2 + x - 6 and verify it crosses the x-axis at -3 and at 2. b) Solve x 2 – 4x – 32 = 0 Step 1: In standard form (check!). Step 2: No GCF other than 1 (check!). Scottsdale Community College Page 216 Intermediate Algebra
Lesson 5b - Solving Quadratic Equations Mini-Lesson Step 3: Try to factor the left side by Trial and Error Factor: x 2 – 4x – 32 = (x – 8)(x + 4) Check by Foiling to be sure your answer is correct. Check: (x – 8)(x + 4) = x 2 + 4x – 8x – 32 = x 2 – 4x – 32 (checks!) Step 4: Write the factored form of the quadratic then set each factor to 0 and solve for x. (x – 8)(x + 4)= 0 so x – 8 = 0 or x + 4 = 0. Solutions to x 2 – 4x – 32 = 0 are x = 8, -4. Note: Verify by graphing. c) Solve 2x 2 + 7x =15 Step 1: Convert the equation to Standard Form Step 2: No GCF other than 1 (check!). Step 3: Try to factor the left side by Trial and Error Factor: 2x 2 + 7x – 15 = (2x – 3)(x + 5) Check by Foiling. ( )( ) Step 4: Write the factored form of the quadratic then set each factor to 0 and solve for x. (2x – 3)(x + 5)= 0 so 2x – 3 = 0 or x + 5 = 0. Solutions to 2x 2 + 7x =15 are x = 3/2, -5. Verify by graphing. Scottsdale Community College Page 217 Intermediate Algebra
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Scottsdale Community College Interm
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Table of Contents Lesson 1a - Intro
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Lesson 1a - Introduction to Functio
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Lesson 1b - Linear Equations In the
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Lesson 1b - Linear Equations Mini-L
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Lesson 4a - Introduction to Logarit
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Lesson 4b - More on Logarithms This
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Appendix A: You Try Answers Lesson
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Appendix A: You Try Answers You Try
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Intermediate Algebra – Student Workbook – Second Edition 2013

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