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Learn how to solve quadratic equations using FOIL method, factoring, and find solutions with 5-minute check quizzes.

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 4–2) CCSS Then/Now New Vocabulary Key Concept: FOIL Method for Multiplying Binomials Example 1: Translate Sentences into Equations Concept Summary: Zero Product Property Example 2: Factor GCF Example 3: Perfect Squares and Differences of Squares Example 4: Factor Trinomials Example 5: Real-World Example: Solve Equations by Factoring Lesson Menu

  3. Use the related graph of y = x2 – 4 to determine its solutions. A. 4, –4 B. 3, –2 C. 2, 0 D. 2, –2 5-Minute Check 1

  4. Use the related graph of y = x2 – 4 to determine its solutions. A. 4, –4 B. 3, –2 C. 2, 0 D. 2, –2 5-Minute Check 1

  5. Use the related graph of y = –x2 – 2x + 3 to determine its solutions. A. –3, 1 B. –3, 3 C. –1, 3 D. 3, 1 5-Minute Check 2

  6. Use the related graph of y = –x2 – 2x + 3 to determine its solutions. A. –3, 1 B. –3, 3 C. –1, 3 D. 3, 1 5-Minute Check 2

  7. Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located. A. 0 B. 0, between 2 and 3 C. between 1 and 2 D. 2, –2 5-Minute Check 3

  8. Solve –2x2 + 5x = 0. If exact roots cannot be found, state the consecutive integers between which the roots are located. A. 0 B. 0, between 2 and 3 C. between 1 and 2 D. 2, –2 5-Minute Check 3

  9. Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14. A. 10, –4 B. 5, –1 C. –2, 7 D. –5, 2 5-Minute Check 4

  10. Use a quadratic equation to find two real numbers that have a sum of 5 and a product of –14. A. 10, –4 B. 5, –1 C. –2, 7 D. –5, 2 5-Minute Check 4

  11. Which term is not another name for a solution to a quadratic equation? A. zero B.x-intercept C. root D. vertex 5-Minute Check 5

  12. Which term is not another name for a solution to a quadratic equation? A. zero B.x-intercept C. root D. vertex 5-Minute Check 5

  13. Content Standards A.SSE.2 Use the structure of an expression to identify ways to rewrite it. F.IF.8.a Use the process of factoring and completing the square in a quadratic function to show zeros, extreme values, and symmetry of the graph, and interpret these in terms of a context. Mathematical Practices 2 Reason Abstractly and quantitatively. CCSS

  14. You found the greatest common factors of sets of numbers. • Write quadratic equations in intercept form. • Solve quadratic equations by factoring. Then/Now

  15. factored form • FOIL method Vocabulary

  16. Concept

  17. Replace p withand q with –5. Translate Sentences into Equations (x – p)(x – q) = 0 Write the pattern. Simplify. Use FOIL. Example 1

  18. Translate Sentences into Equations Multiply each side by 2 so b and c are integers. Answer: Example 1

  19. Answer: Translate Sentences into Equations Multiply each side by 2 so b and c are integers. Example 1

  20. A. ans B. ans C. ans D. ans Example 1

  21. A. ans B. ans C. ans D. ans Example 1

  22. Concept

  23. y = 0 Solve each equation. Factor GCF A.Solve 9y2 + 3y = 0. 9y2 + 3y = 0 Original equation 3y(3y) + 3y(1) = 0 Factor the GCF. 3y(3y + 1) = 0 Distributive Property 3y = 0 3y + 1 = 0 Zero Product Property Answer: Example 2

  24. y = 0 Solve each equation. Answer: Factor GCF A.Solve 9y2 + 3y = 0. 9y2 + 3y = 0 Original equation 3y(3y) + 3y(1) = 0 Factor the GCF. 3y(3y + 1) = 0 Distributive Property 3y = 0 3y + 1 = 0 Zero Product Property Example 2

  25. Factor GCF B.Solve 5a2 – 20a = 0. 5a2 – 20a = 0 Original equation 5a(a) – 5a(4) = 0 Factor the GCF. 5a(5a – 4) = 0 Distributive Property 5a = 0 a – 4 = 0 Zero Product Property a = 0 a = 4 Solve each equation. Answer: Example 2

  26. Factor GCF B.Solve 5a2 – 20a = 0. 5a2 – 20a = 0 Original equation 5a(a) – 5a(4) = 0 Factor the GCF. 5a(5a – 4) = 0 Distributive Property 5a = 0 a – 4 = 0 Zero Product Property a = 0 a = 4 Solve each equation. Answer:0, 4 Example 2

  27. Solve 12x – 4x2 = 0. A. 3, 12 B. 3, –4 C. –3, 0 D. 3, 0 Example 2

  28. Solve 12x – 4x2 = 0. A. 3, 12 B. 3, –4 C. –3, 0 D. 3, 0 Example 2

  29. Perfect Squares and Differences of Squares A.Solve x2 – 6x + 9 = 0. x2 = (x)2; 9 = (3)2 First and last terms are perfect squares. 6x = 2(x)(3) Middle term equals 2ab. x2 – 6x + 9 is a perfect square trinomial. x2 + 6x + 9 = 0 Original equation (x – 3)2 = 0 Factor using the pattern. x – 3 = 0 Take the square root of each side. x = 3 Add 3 to each side. Answer: Example 3

  30. Perfect Squares and Differences of Squares A.Solve x2 – 6x + 9 = 0. x2 = (x)2; 9 = (3)2 First and last terms are perfect squares. 6x = 2(x)(3) Middle term equals 2ab. x2 – 6x + 9 is a perfect square trinomial. x2 + 6x + 9 = 0 Original equation (x – 3)2 = 0 Factor using the pattern. x – 3 = 0 Take the square root of each side. x = 3 Add 3 to each side. Answer:3 Example 3

  31. Perfect Squares and Differences of Squares B.Solve y2 = 36. y2 = 32 Original equation y2 – 36 = 0 Subtract 36 from each side. y2 – (6)2 = 0 Write in the form a2 – b2. (y + 6)(y – 6) = 0 Factor the difference of squares. y + 6 = 0 y – 6 = 0 Zero Product Property y = –6 y = 6 Solve each equation. Answer: Example 3

  32. Perfect Squares and Differences of Squares B.Solve y2 = 36. y2 = 32 Original equation y2 – 36 = 0 Subtract 36 from each side. y2 – (6)2 = 0 Write in the form a2 – b2. (y + 6)(y – 6) = 0 Factor the difference of squares. y + 6 = 0 y – 6 = 0 Zero Product Property y = –6 y = 6 Solve each equation. Answer:–6, 6 Example 3

  33. Solve x2 – 16x + 64 = 0. A. 8, –8 B. 8, 0 C. 8 D. –8 Example 3

  34. Solve x2 – 16x + 64 = 0. A. 8, –8 B. 8, 0 C. 8 D. –8 Example 3

  35. Factor Trinomials A.Solve x2 – 2x – 15 = 0. ac = –15 a = 1, c = –15 Example 4

  36. Factor Trinomials x2 – 2x – 15 = 0 Original equation x2 + mx + px – 15 = 0 Write the pattern. x2 + 3x – 5x – 15 = 0 m = 3 and p = –5 (x2 + 3x) – (5x + 15) = 0 Group terms with common factors. x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping. (x – 5)(x + 3) = 0 Distributive Property x – 5 = 0 x + 3 = 0 Zero Product Property x = 5 x = –3 Solve each equation. Answer: Example 4

  37. Factor Trinomials x2 – 2x – 15 = 0 Original equation x2 + mx + px – 15 = 0 Write the pattern. x2 + 3x – 5x – 15 = 0 m = 3 and p = –5 (x2 + 3x) – (5x + 15) = 0 Group terms with common factors. x(x + 3) – 5(x + 3) = 0 Factor the GCF from each grouping. (x – 5)(x + 3) = 0 Distributive Property x – 5 = 0 x + 3 = 0 Zero Product Property x = 5 x = –3 Solve each equation. Answer:5, –3 Example 4

  38. Factor Trinomials B.Solve 5x2 + 34x + 24 = 0. ac = 120 a = 5, c = 24 Example 4

  39. x = –6 Solve each equation. Factor Trinomials 5x2 + 34x + 24 = 0 Original equation 5x2 + mx + px + 24 = 0 Write the pattern. 5x2 + 4x + 30x + 24 = 0 m = 4 and p = 30 (5x2 + 4x) + (30x + 24) = 0 Group terms with common factors. x(5x + 4) + 6(x + 4) = 0 Factor the GCF from each grouping. (x + 6)(5x + 4) = 0 Distributive Property x + 6 = 0 5x + 4 = 0 Zero Product Property Example 4

  40. Factor Trinomials Answer: Example 4

  41. Answer: Factor Trinomials Example 4

  42. A. B. C. D. Solve 6x2 – 5x – 4 = 0. Example 4

  43. A. B. C. D. Solve 6x2 – 5x – 4 = 0. Example 4

  44. B.Factor 3s2 – 11s – 4. A. (3s + 1)(s – 4) B. (s + 1)(3s – 4) C. (3s + 4)(s – 1) D. (s – 1)(3s + 4) Example 4

  45. B.Factor 3s2 – 11s – 4. A. (3s + 1)(s – 4) B. (s + 1)(3s – 4) C. (3s + 4)(s – 1) D. (s – 1)(3s + 4) Example 4

  46. Solve Equations by Factoring ARCHITECTURE The entrance to an office building is an arch in the shape of a parabola whose vertex is the height of the arch. The height of the arch is given by h = 9 – x2, where x is the horizontal distance from the center of the arch. Both h and x are measured in feet. How wide is the arch at ground level? To find the width of the arch at ground level, find the distance between the two zeros. Example 5

  47. Solve Equations by Factoring 9 – x2 = 0 Original expression x2 – 9 = 0 Multiply both sides by –1. (x + 3)(x – 3) = 0 Difference of squares x + 3 = 0 or x – 3 = 0 Zero Product Property x = –3 x = 3 Solve. Answer: Example 5

  48. Solve Equations by Factoring 9 – x2 = 0 Original expression x2 – 9 = 0 Multiply both sides by –1. (x + 3)(x – 3) = 0 Difference of squares x + 3 = 0 or x – 3 = 0 Zero Product Property x = –3 x = 3 Solve. Answer:The distance between 3 and – 3 is3 – (–3) or 6 feet. Example 5

  49. ? ? 9 – (3)2 = 0 or 9 – (–3)2 = 0 ? ? 9 – 9 = 0 9 – 9 = 0 Solve Equations by Factoring Check 9 – x2 = 0   0 = 0 0 = 0 Example 5

  50. TENNIS During a match, Andre hit a lob right off the court with the ball traveling in the shape of a parabola whose vertex was the height of the shot. The height of the shot is given by h = 49 – x2, where x is the horizontal distance from the center of the shot. Both h and x are measured in feet. How far was the lob hit? A. 7 feet B. 11 feet C. 14 feet D. 25 feet Example 5

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