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The Quadratic Formula 5-6 Warm Up Lesson Presentation Lesson Quiz
Holt Algebra 2
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Evaluate b2 – 4ac for the given values of the variables.
Warm Up Evaluate b2 – 4ac for the given values of the variables. 1. Express in terms of i. 2. Solve the equation. 3x = 0 4. a = 1, b = 3, c = –3 3. a = 2, b = 7, c = 5 21 9
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Objectives Solve quadratic equations using the Quadratic Formula.
Classify roots using the discriminant.
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Vocabulary discriminant
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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.
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To subtract fractions, you need a common denominator.
Remember!
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The symmetry of a quadratic function is evident in the last step,
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.
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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.
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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.
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Example 1 Continued Check Solve by completing the square.
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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.
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Check It Out! Example 1a Continued
Check Solve by completing the square. x2 + 3x – 7 = 0 x2 + 3x = 7
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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.
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Check It Out! Example 1b Continued
Check Solve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 16 = – (x + 4)2 = 6
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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.
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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.
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The discriminant is part of the Quadratic Formula that you can use to determine the number of real roots of a quadratic equation.
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Make sure the equation is in standard form before you evaluate the discriminant, b2 – 4ac.
Caution!
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Example 3A: Analyzing Quadratic Equations by Using the Discriminant
Find the type and number of solutions for the equation. x = 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.
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Example 3B: Analyzing Quadratic Equations by Using the Discriminant
Find the type and number of solutions for the equation. x = 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.
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Example 3C: Analyzing Quadratic Equations by Using the Discriminant
Find the type and number of solutions for the equation. x = 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.
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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.
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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.
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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) = 24 b2 – 4ac > 0 The equation has two distinct real solutions.
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Properties of Solving Quadratic Equations
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Properties of Solving Quadratic Equations
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No matter which method you use to solve a quadratic equation, you should get the same answer.
Helpful Hint
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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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