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Solving Quadratic Equations using Quadratic Formula and Discriminant

Learn how to solve quadratic equations using the quadratic formula and the discriminant. Explore examples with rational roots, irrational roots, and complex roots.

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Solving Quadratic Equations using Quadratic Formula and Discriminant

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 4–5) CCSS Then/Now New Vocabulary Key Concept: Quadratic Formula Example 1: Two Rational Roots Example 2: One Rational Root Example 3: Irrational Roots Example 4: Complex Roots Key Concept: Discriminant Example 5: Describe Roots Concept Summary: Solving Quadratic Equations Lesson Menu

  3. Solve x2 – 2x + 1 = 9 by using the Square Root Property. A. 4, 2 B. 4, –2 C. 2, –2 D. 2, 1 5-Minute Check 1

  4. Solve x2 – 2x + 1 = 9 by using the Square Root Property. A. 4, 2 B. 4, –2 C. 2, –2 D. 2, 1 5-Minute Check 1

  5. A.–3, 3 B. C.–1, 1 D. Solve 4c2 + 12c + 9 = 7 by using the Square Root Property. 5-Minute Check 2

  6. A.–3, 3 B. C.–1, 1 D. Solve 4c2 + 12c + 9 = 7 by using the Square Root Property. 5-Minute Check 2

  7. A. B. C. D.1; (x + 1)2 Find the value of c that makes the trinomial x2 + x + c a perfect square. Then write the trinomial as a perfect square. 5-Minute Check 3

  8. A. B. C. D.1; (x + 1)2 Find the value of c that makes the trinomial x2 + x + c a perfect square. Then write the trinomial as a perfect square. 5-Minute Check 3

  9. A.–12, 12 B.–2, 2 C. D. Solve x2 + 2x + 24 = 0 by completing the square. 5-Minute Check 4

  10. A.–12, 12 B.–2, 2 C. D. Solve x2 + 2x + 24 = 0 by completing the square. 5-Minute Check 4

  11. A.2, –1 B. C. D.4, 2 Solve 5g2 – 8 = 6g by completing the square. 5-Minute Check 5

  12. A.2, –1 B. C. D.4, 2 Solve 5g2 – 8 = 6g by completing the square. 5-Minute Check 5

  13. Find the value(s) of k in x2 + kx + 100 = 0 that would make the left side of the equation a perfect square trinomial. A. –10, 10 B. 5, 20 C. –20, 20 D. 4, 25 5-Minute Check 6

  14. Find the value(s) of k in x2 + kx + 100 = 0 that would make the left side of the equation a perfect square trinomial. A. –10, 10 B. 5, 20 C. –20, 20 D. 4, 25 5-Minute Check 6

  15. Content Standards N.CN.7 Solve quadratic equations with real coefficients that have complex solutions. A.SSE.1.b Interpret complicated expressions by viewing one or more of their parts as a single entity. Mathematical Practices 8 Look for and express regularity in repeated reasoning. CCSS

  16. You solved equation by completing the square. • Solve quadratic equations by using the Quadratic Formula. • Use the discriminant to determine the number and type of roots of a quadratic equation. Then/Now

  17. Quadratic Formula • discriminant Vocabulary

  18. Concept

  19. ax2 + bx + c = 0 x2 – 8x = 33 1x2 – 8x – 33 = 0 Two Rational Roots Solve x2 – 8x = 33 by using the Quadratic Formula. First, write the equation in the form ax2 + bx + c = 0 and identify a, b, and c. Then, substitute these values into the Quadratic Formula. Quadratic Formula Example 1

  20. Two Rational Roots Replace a with 1, b with –8, and cwith –33. Simplify. Simplify. Example 1

  21. or Write as two equations. Two Rational Roots x = 11 x = –3 Simplify. Answer: Example 1

  22. or Write as two equations. Two Rational Roots x = 11 x = –3 Simplify. Answer: The solutions are 11 and –3. Example 1

  23. Solve x2 + 13x = 30 by using the Quadratic Formula. A. 15, –2 B. 2, –15 C. 5, –6 D. –5, 6 Example 1

  24. Solve x2 + 13x = 30 by using the Quadratic Formula. A. 15, –2 B. 2, –15 C. 5, –6 D. –5, 6 Example 1

  25. One Rational Root Solve x2 – 34x + 289 = 0 by using the Quadratic Formula. Identify a, b, and c. Then, substitute these values into the Quadratic Formula. Quadratic Formula Replace a with 1, b with –34, and c with 289. Simplify. Example 2

  26. One Rational Root Answer: Example 2

  27. [–5, 25] scl: 1 by [–5, 15] scl: 1 One Rational Root Answer: The solution is 17. Check A graph of the related function shows that there is one solution at x = 17. Example 2

  28. Solve x2 – 22x + 121 = 0 by using the Quadratic Formula. A. 11 B. –11, 11 C. –11 D. 22 Example 2

  29. Solve x2 – 22x + 121 = 0 by using the Quadratic Formula. A. 11 B. –11, 11 C. –11 D. 22 Example 2

  30. or or Irrational Roots Solve x2 – 6x + 2 = 0 by using the Quadratic Formula. Quadratic Formula Replace a with 1, b with –6, and c with 2. Simplify. Example 3

  31. Irrational Roots Answer: Example 3

  32. [–10, 10] scl: 1 by [–10, 10] scl: 1 Irrational Roots Answer: Check Check these results by graphing the related quadratic function, y = x2 – 6x + 2. Using the ZERO function of a graphing calculator, the approximate zeros of the related function are 0.4 and 5.6. Example 3

  33. A. B. C. D. Solve x2 – 5x + 3 = 0 by using the Quadratic Formula. Example 3

  34. A. B. C. D. Solve x2 – 5x + 3 = 0 by using the Quadratic Formula. Example 3

  35. Complex Roots Solve x2 + 13 = 6x by using the Quadratic Formula. Quadratic Formula Replace a with 1, b with –6, and c with 13. Simplify. Simplify. Example 4

  36. Complex Roots Answer: Example 4

  37. [–5, 15] scl: 1 by [–5, 15] scl: 1 Complex Roots Answer: The solutions are the complex numbers 3 + 2iand 3 – 2i. A graph of the related function shows that the solutions are complex, but it cannot help you find them. Example 4

  38. ? (3 + 2i)2 + 13 = 6(3 + 2i)x = (3 + 2i) ? ? 9 + 12i + 4i2 + 13 = 18 + 12i Square of a sum; Distributive Property 22 + 12i – 4 = 18 + 12i Simplify. Complex Roots CheckTo check complex solutions, you must substitute them into the original equation. The check for 3 + 2i is shown below. x2 + 13 = 6x Original equation 18 + 12i = 18 + 12i  Example 4

  39. Solve x2 + 5 = 4x by using the Quadratic Formula. A. 2 ± i B. –2 ± i C. 2 + 2i D. –2 ± 2i Example 4

  40. Solve x2 + 5 = 4x by using the Quadratic Formula. A. 2 ± i B. –2 ± i C. 2 + 2i D. –2 ± 2i Example 4

  41. Concept

  42. Describe Roots A.Findthevalueofthediscriminantforx2+3x + 5 = 0. Then describe the number and type of roots for the equation. a = 1, b = 3, c = 5 b2 – 4ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify. = –11 Subtract. Answer: Example 5

  43. Describe Roots A.Findthevalueofthediscriminantforx2+3x + 5 = 0. Then describe the number and type of roots for the equation. a = 1, b = 3, c = 5 b2 – 4ac = (3)2 – 4(1)(5) Substitution = 9 – 20 Simplify. = –11 Subtract. Answer: The discriminant is negative, so there are two complex roots. Example 5

  44. Describe Roots B.Findthevalue of the discriminant forx2–11x+ 10 = 0. Then describe the number and type of roots for the equation. a = 1, b = –11, c = 10 b2 – 4ac = (–11)2 – 4(1)(10) Substitution = 121 – 40 Simplify. = 81 Subtract. Answer: Example 5

  45. Describe Roots B.Findthevalue of the discriminant forx2–11x+ 10 = 0. Then describe the number and type of roots for the equation. a = 1, b = –11, c = 10 b2 – 4ac = (–11)2 – 4(1)(10) Substitution = 121 – 40 Simplify. = 81 Subtract. Answer: The discriminant is 81, so there are two rational roots. Example 5

  46. A.Find the value of the discriminant for x2 + 8x + 16 = 0. Describe the number and type of roots for the equation. A. 0; 1 rational root B. 16; 2 rational roots C. 32; 2 irrational roots D. –64; 2 complex roots Example 5

  47. A.Find the value of the discriminant for x2 + 8x + 16 = 0. Describe the number and type of roots for the equation. A. 0; 1 rational root B. 16; 2 rational roots C. 32; 2 irrational roots D. –64; 2 complex roots Example 5

  48. B.Find the value of the discriminant for x2 + 2x + 7 = 0. Describe the number and type of roots for the equation. A. 0; 1 rational root B. 36; 2 rational roots C. 32; 2 irrational roots D. –24; 2 complex roots Example 5

  49. B.Find the value of the discriminant for x2 + 2x + 7 = 0. Describe the number and type of roots for the equation. A. 0; 1 rational root B. 36; 2 rational roots C. 32; 2 irrational roots D. –24; 2 complex roots Example 5

  50. Concept

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