MAT 171 Precalculus Algebra Dr. Claude Moore Cape Fear Community College CHAPTER 3: Quadratic Functions and Equations; Inequalities 3.1 The Complex Numbers 3.2 Quadratic Equations, Functions, Zeros, and Models 3.3 Analyzing Graphs of Quadratic Functions 3.4 Solving Rational Equations and Radical Equations 3.5 Solving Equations and Inequalities with Absolute Value This program graphs quadratic functions and shows two forms of equation. http://cfcc.edu/mathlab/geogebra/quadratic2ss.html This program graphs a quadratic function and shows the roots (solutions). http://cfcc.edu/mathlab/geogebra/quadratic_roots.html 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, by completing the square, and by using the quadratic formula. Solve equations that are reducible to quadratic. Solve applied problems using quadratic equations. Graphing Quadratic Function with TI Calculator Finding Zeros of Quadratic Function by Graphing with TI Calculator 1
Quadratic Equations A quadratic equation is an equation that can be written in the form ax 2 + 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. Quadratic Functions A quadratic function f is a function that can be written in the form f (x) = ax 2 + bx + c, a 0, where a, b, and c are real numbers. The zeros of a quadratic function f (x) = ax 2 + bx + c are the solutions of the associated quadratic equation ax 2 + bx + c = 0. Quadratic functions can have real number or imaginarynumber zeros and quadratic equations can have real number or imaginary number solutions. Equation Solving Principles The Principle of Zero Products: If ab = 0 is true, then a = 0 or b = 0, Example Solve 2x 2 x = 3. Solution and if a = 0 or b = 0, then ab = 0. 2
Example Checking the Solutions Check: x = 1 Equation Solving Principles The Principle of Square Roots: If x 2 = k, then TRUE TRUE Completing the Square To solve a quadratic equation by completing the square: Isolate the terms with variables on one side of the equation and arrange them in descending order. Divide by the coefficient of the squared term if that coefficient is not 1. Complete the square by taking half the coefficient of the firstdegree term and adding its square on both sides of the equation. Express one side of the equation as the square of a binomial. Use the principle of square roots. Solve for the variable. Example Solve 2x 2 1 = 3x. Solution 3
Quadratic Formula The solutions of ax 2 + bx + c = 0, a 0, are given by This formula can be used to solve any quadratic equation. Discriminant When you apply the quadratic formula to any quadratic equation, you find the value of b 2 4ac, which can be positive, negative, or zero. This expression is called the discriminant. For ax 2 + bx + c = 0, where a, b, and c are real numbers: b 2 4ac = 0 One real number solution; b 2 4ac > 0 Two different real number solutions; b 2 4ac < 0 Two different imaginary number solutions, complex conjugates. Equations Reducible to Quadratic Some equations can be treated as quadratic, provided that we make a suitable substitution. Example: x 4 5x 2 + 4 = 0 Knowing that x 4 = (x 2 ) 2, we can substitute u for x 2 and the resulting equation is then u 2 5u + 4 = 0. This equation can then be solved for u by factoring or using the quadratic formula. Then the substitution can be reversed by replacing u with x 2, and solving for x. Equations like this are said to be reducible to quadratic, or quadratic in form. 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? 4
Example (continued) 1. Familiarize. The formula s = 16t 2 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 = 16t 2. 3. Carry out. Use the principle of square roots. 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. 257/2. Solve. (5x 2)( 2x + 3) = 0 257/6. Solve. 10x 2 16x + 6 = 0 257/12. Solve. 4x 2 + 12 = 0 5
257/19. Solve. 3x 3 + x 2 12x 4 = 0 (Hint: Factor by grouping.) 257/22. Use the graph to find (a) The x intercepts (b) The zeros of the function The x intercepts should be written as ordered pairs (x, 0). The zeros (or roots, solutions) are x values when y = 0. 258/28. Solve by completing the square to obtain exact solutions. x 2 + 8x = 15 257/24. Use the graph to find (a) The x intercepts (b) The zeros of the function The x intercepts should be written as ordered pairs (x, 0). The zeros (or roots, solutions) are x values when y = 0. 6
258/34. Solve by completing the square to obtain exact solutions. 2x 2 5x 3 = 0 258/42. Use the quadratic formula to find exact solutions. x 2 + 1 = x 258/54. Use the quadratic formula to find exact solutions. 3x 2 + 3x = 4 258/56. Find discriminant and describe solutions. 4x 2 12x + 9 = 0 7
258/72. Find the zeros of the function algebraically. f(x) = 3x 2 + 8x + 2 = 0 258/64. Solve graphically. Round solutions to 3 decimal places. 10x 2 23x + 12 = 0 258/82. Find the zeros of the function algebraically. f(x) = 4x 2 4x 5 = 0 8