3.2 Quadratic Equations, Functions, Zeros, and Models Find zeros of quadratic functions and solve quadratic equations by using the principle of zero products, by using the principle of square roots, and by using the quadratic formula. Solve applied problems using quadratic equations. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Quadratic Equations A quadratic equation is an equation that can be written in the form ax2 + bx + c = 0, a  0, where a, b, and c are real numbers. A quadratic equation written in this form is said to be in standard form. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Quadratic Functions A quadratic function f is a function that can be written in the form f (x) = ax2 + bx + c, a  0, where a, b, and c are real numbers. The zeros of a quadratic function f (x) = ax2 + bx + c are the solutions of the associated quadratic equation ax2 + bx + c = 0. Quadratic functions can have real-number or imaginary-number zeros and quadratic equations can have real-number or imaginary-number solutions. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Equation-Solving Principles The Principle of Zero Products: If ab = 0 is true, then a = 0 or b = 0, and if a = 0 or b = 0, then ab = 0. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Equation-Solving Principles The Principle of Square Roots: If x2 = k, then Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Solve 2x2  x = 3. Solution Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Example - Checking the Solutions Check: x = – 1 TRUE The solutions are –1 and TRUE Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Solve 2x2  10 = 0. Solution Check: TRUE The solutions are and Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Quadratic Formula The solutions of ax2 + bx + c = 0, a  0, are given by This formula can be used to solve any quadratic equation. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example Solve 3x2 + 2x = 7. Find exact solutions and approximate solutions rounded to the thousandths. Solution: 3x2 + 2x  7 = 0 a = 3, b = 2, c = 7 The exact solutions are: The approximate solutions are –1.897 and 1.230. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Discriminant When you apply the quadratic formula to any quadratic equation, you find the value of b2  4ac, which can be positive, negative, or zero. This expression is called the discriminant. For ax2 + bx + c = 0, where a, b, and c are real numbers: b2  4ac = 0 One real-number solution; b2  4ac > 0 Two different real-number solutions; b2  4ac < 0 Two different imaginary-number solutions, complex conjugates. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Applications Some applied problems can be translated to quadratic equations. Example Time of Free Fall. The Petronas Towers in Kuala Lumpur, Malaysia are 1482 ft tall. How long would it take an object dropped from the top to reach the ground? Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 1. Familiarize. The formula s = 16t2 is used to approximate the distance s, in feet, that an object falls freely from rest in t seconds. 2. Translate. Substitute 1482 for s in the formula: 1482 = 16t2. 3. Carry out. Use the principle of square roots. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley

Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley Example (continued) 4. Check. In 9.624 seconds, a dropped object would travel a distance of 16(9.624)2, or about 1482 ft. The answer checks. 5. State. It would take about 9.624 sec for an object dropped from the top of the Petronas Towers to reach the ground. Copyright © 2012 Pearson Education, Inc. Publishing as Addison Wesley