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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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**The Quadratic Formula**5-6 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 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**Objectives**Solve quadratic equations using the Quadratic Formula. Classify roots using the discriminant.**Vocabulary**discriminant**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.**Remember!**To subtract fractions, you need a common denominator.**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.**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.**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.**Example 1 Continued**CheckSolve by completing the square. **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.**Check It Out! Example 1a Continued**CheckSolve by completing the square. x2 + 3x – 7 = 0 x2 + 3x = 7 **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.**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 **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.**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.**The discriminant is part of the Quadratic Formula that you**can use to determine the number of real roots of a quadratic equation.**Caution!**Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.**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.**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.**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.**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.**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.**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.**Helpful Hint**No matter which method you use to solve a quadratic equation, you should get the same answer.**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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