 # The Quadratic Formula 5-6 Warm Up Lesson Presentation Lesson Quiz

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

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

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.

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

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.

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 Check Solve 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
Check Solve 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
Check Solve by completing the square. x2 – 8x + 10 = 0 x2 – 8x = –10 x2 – 8x + 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.

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

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.

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.

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.

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) = 24 b2 – 4ac > 0 The equation has two distinct real solutions.