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The Quadratic Formula. 5-6. Warm Up. Lesson Presentation. Lesson Quiz. Holt Algebra 2. 1. Express in terms of i . Warm Up Evaluate b 2 – 4 ac for the given values of the variables. 2. Solve the equation. 3 x 2 + 96 = 0. 4. a = 1, b = 3, c = –3.

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5-6

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  1. The Quadratic Formula 5-6 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

  2. 1. Express in terms of i. Warm Up Evaluate b2 – 4ac for the given values of the variables. 2. Solve the equation. 3x2 + 96 = 0 4.a = 1, b = 3, c = –3 3.a = 2, b = 7, c = 5 21 9

  3. Objectives Solve quadratic equations using the Quadratic Formula. Classify roots using the discriminant.

  4. Vocabulary discriminant

  5. You have learned several methods for solving quadratic equations: graphing, making tables, factoring, using square roots, and completing the square. Another method is to use the Quadratic Formula, which allows you to solve a quadratic equation in standard form. By completing the square on the standard form of a quadratic equation, you can determine the Quadratic Formula.

  6. Remember! To subtract fractions, you need a common denominator.

  7. The symmetry of a quadratic function is evident in the last step, . These two zeros are the same distance, , away from the axis of symmetry, ,with one zero on either side of the vertex.

  8. You can use the Quadratic Formula to solve any quadratic equation that is written in standard form, including equations with real solutions or complex solutions.

  9. Example 1: Quadratic Functions with Real Zeros Find the zeros of f(x)= 2x2 – 16x + 27 using the Quadratic Formula. 2x2– 16x + 27 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 2 for a, –16 for b, and 27 for c. Simplify. Write in simplest form.

  10. Example 1 Continued CheckSolve by completing the square. 

  11. Check It Out! Example 1a Find the zeros of f(x) = x2 + 3x –7 using the Quadratic Formula. x2 + 3x–7 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, 3 for b, and –7 for c. Simplify. Write in simplest form.

  12. Check It Out! Example 1a Continued CheckSolve by completing the square. x2 + 3x – 7 = 0 x2 + 3x = 7 

  13. Check It Out! Example 1b Find the zeros of f(x)= x2 – 8x + 10 using the Quadratic Formula. x2 – 8x + 10 = 0 Set f(x) = 0. Write the Quadratic Formula. Substitute 1 for a, –8 for b, and 10 for c. Simplify. Write in simplest form.

  14. Check It Out! Example 1b Continued CheckSolve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 16 = –10 + 16 (x + 4)2 = 6 

  15. Example 2: Quadratic Functions with Complex Zeros Find the zeros of f(x) = 4x2 + 3x + 2 using the Quadratic Formula. f(x)= 4x2 + 3x + 2 Set f(x) = 0. Write the Quadratic Formula. Substitute 4 for a, 3 for b, and 2 for c. Simplify. Write in terms of i.

  16. Check It Out! Example 2 Find the zeros of g(x) = 3x2 – x + 8 using the Quadratic Formula. Set f(x) = 0 Write the Quadratic Formula. Substitute 3 for a, –1 for b, and 8 for c. Simplify. Write in terms of i.

  17. The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.

  18. Caution! Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.

  19. Example 3A: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 36 = 12x x2–12x + 36 = 0 b2 – 4ac (–12)2 – 4(1)(36) 144 – 144 = 0 b2 – 4ac = 0 The equation has one distinct real solution.

  20. Example 3B: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 40 = 12x x2–12x + 40 = 0 b2 – 4ac (–12)2 – 4(1)(40) 144 – 160 = –16 b2 –4ac < 0 The equation has two distinct nonreal complex solutions.

  21. Example 3C: Analyzing Quadratic Equations by Using the Discriminant Find the type and number of solutions for the equation. x2 + 30 = 12x x2–12x + 30 = 0 b2 – 4ac (–12)2 – 4(1)(30) 144 – 120 = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

  22. Check It Out! Example 3a Find the type and number of solutions for the equation. x2 – 4x = –4 x2–4x + 4 = 0 b2 – 4ac (–4)2 – 4(1)(4) 16 – 16 = 0 b2 – 4ac = 0 The equation has one distinct real solution.

  23. Check It Out! Example 3b Find the type and number of solutions for the equation. x2 – 4x = –8 x2–4x + 8 = 0 b2 – 4ac (–4)2 – 4(1)(8) 16 – 32 = –16 b2 – 4ac < 0 The equation has two distinct nonreal complex solutions.

  24. Check It Out! Example 3c Find the type and number of solutions for each equation. x2 – 4x = 2 x2–4x– 2 = 0 b2 – 4ac (–4)2 – 4(1)(–2) 16 + 8 = 24 b2 – 4ac > 0 The equation has two distinct real solutions.

  25. Properties of Solving Quadratic Equations

  26. Properties of Solving Quadratic Equations

  27. Helpful Hint No matter which method you use to solve a quadratic equation, you should get the same answer.

  28. Lesson Quiz: Part I Find the zeros of each function by using the Quadratic Formula. 1. f(x) = 3x2 – 6x – 5 2. g(x) = 2x2 – 6x + 5 Find the type and member of solutions for each equation. 3. x2 – 14x + 50 4. x2 – 14x + 48 2 distinct real 2 distinct nonreal complex

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