Objectives Solve quadratic equations using the Quadratic Formula.

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Presentation transcript:

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

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 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.

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. 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.

Properties of Solving Quadratic Equations

Properties of Solving Quadratic Equations

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

Lesson Quiz: Part II 5. A pebble is tossed from the top of a cliff. The pebble’s height is given by y(t) = –16t2 + 200, where t is the time in seconds. Its horizontal distance in feet from the base of the cliff is given by d(t) = 5t. How far will the pebble be from the base of the cliff when it hits the ground? about 18 ft