DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-A;CA-A LESSON3 . 3 Name Class Date Finding Complex 3.3 Finding Complex Solutions Solutions of Quadratic of Quadratic Equations Essential Question: How can you find the complex solutions of any quadratic equation? Resource Equations Locker Common Core Math Standards Explore Investigating Real Solutions of Quadratic Equations The student is expected to: A Complete the table. COMMON CORE N-CN.C.7 2 2 2 a x + bx + c = 0 a x + bx = -c f( x) = a x + bx g( x) = -c Solve quadratic equations with real coefficients that have complex 2 2 2 2 x 4x 1 f x 2 x 4x g x 1 solutions. Also N-CN.C.2, A-REI.B.4b 2 x + 4x + 1 = 0 + = - ( ) = + ( ) = - 2 2 2 2 x 4x 2 f x 2 x 4x g x 2 Mathematical Practices 2 x + 4x + 2 = 0 + = - ( ) = + ( ) = - 2 2 COMMON 2 x 2 4x 3 0 2 x + 4x = -3 f( x) = 2 x + 4x g( x) = -3 CORE MP.2 Reasoning + + = Language Objective 2 B The graph of ƒ( x) = 2 x + 4x is shown. Graph each g( x) . Complete the table. Work with a partner or small group to determine whether solutions to y Number of Real quadratic equations are real or not real and justify reasoning. Equation Solutions 2 2 x 2 x + 4x + 1 = 0 2 0 4 2 2 4 2 ENGAGE - - 2 x + 4x + 2 = 0 1 2 2 x + 4x + 3 = 0 0 Essential Question: How can you 2 find the complex solutions of any C Repeat Steps A and B when ƒ( x) = -2 x + 4x. quadratic equation? 2 2 2 a x + bx + c = 0 a x + bx = -c f( x) = a x + bx g( x) = -c Possible answer: You can factor, if possible, to find 2 2 2 -2 x + 4x - 1 = 0 -2 x + 4x = 1 f (x) = -2 x + 4x g( x) = 1 real solutions; approximate from a graph; find a 2 2 2 -2 x + 4x - 2 = 0 -2 x + 4x = 2 f (x) = -2 x + 4x g( x) = 2 square root (which may be part of completing the 2 x 2 4x 3 0 2 x 2 4x 3 f x 2 x 2 4x g x 3 square); complete the square; or apply the - + - = - + = ( ) = - + ( ) = quadratic formula. For the general equation © Houghton Mifflin Houghton © Company Harcourt Publishing y Equation Number of Real 2 Solutions a x + bx + c = 0, you must either complete the 2 2 square or use the quadratic formula to find the x -2 x + 4x - 1 = 0 2 0 -4 -2 4 2 complex solutions of the equation. -2 x + 4x - 2 = 0 1 -2 2 -2 x + 4x - 3 = 0 0 Module 3 through “File info” 139 Lesson 3 DO NOT EDIT--Changes must be made PREVIEW: LESSON CorrectionKey=NL-A;CA-A Date Class PERFORMANCE TASK Name 3 . 3 Finding Complex Solutions Resource of Quadratic Equations Locker complex solutions of any quadratic equation? HARDCOVER PAGES 99108 ts that have complex solutions. Essential Question: How can you find the COMMON CORE N-CN.C.7 Solve quadratic equations with real coefficien Also N-CN.C.2, A-REI.B.4b View the Engage section online. Discuss the photo Investigating Real Solutions of Quadratic Equations c Explore g x) = - 2 bx ( f x) = a x + 1 ( g x) = - A2_MNLESE385894_U2M03L3.indd 139 Complete the table. 2 bx c 2 ( 19/03/14 12:00 PM a x + = - x 4x f x) = 2 + c 0 ( 2 2 bx = 2 1 g x) = - a x + + x 4x = - 2 ( 2 + 2 x + 4x f( x) = 2 3 4x 1 0 2 2 g x) = - 2 x + + = x 4x = - 2 ( 2 + 2 x + 4x Turn to these pages to f( x) = and how to solve a quadratic equation to determine 2 2 x 4x 2 0 2 3 + + = 2 x + 4x = - 2 2 x + 4x + 3 = 0 (x) . Complete the table. Number of Real x 2 4x is shown. Graph each g Solutions x) = 2 + find this lesson in the ( Equation The graph of ƒ 2 y how high a baseball will go after it is hit. Then 2 4x 1 0 2 x + + = 1 2 x 2 2 x + 4x + 2 = 0 0 0 2 4 hardcover student 2 2 x2 4x 3 0 -4 - + + = 2 4x. g x c preview the Lesson Performance Task. 2 x + 2 ( ) = - x ƒ( ) = - a x + bx f( x) = g x) = 1 edition. Repeat Steps A and B when 2 c ( a x bx = - 2 4x + 2 x + f( x) = - 2 c 0 2 1 g x 2 a x bx + = 4x = 2 ( ) = + 2 x + x 4x - f x -2 + 2 ( ) = 3 x 4x 1 0 2 2 g x) = 2 + - = x 4x = 2 ( - -2 + 2 x + 4x f x) = - 2 4x 2 0 ( 2 x + - = x2 4x = 3 - -2 + Number of Real x2 4x 3 0 -2 + - = Solutions Equation 2 y 2 2 x + 4x - 1 = 0 © Houghton Mifflin Harcourt Publishing Company Publishing Harcourt Mifflin Houghton © 1 2 - x 2 x 4x 2 0 -2 + - = 0 0 4 2 x2 4x 3 0 Lesson 3 -4 - -2 + - = -2 139 19/03/14 12:05 PM Module 3 3L3.indd 139 A2_MNLESE385894_U2M0 139 Lesson 3 . 3 DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through “File info” DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-B;CA-B CorrectionKey=NL-B;CA-B Reflect 1. Look back at Steps A and B. Notice Number of Real Solutions EXPLORE that the minimum value of f(x) in Steps Value of g( x) of A and B is -2. Complete the table by f (x) = g( x) identifying how many real solutions Investigating Real Solutions of g( x) = -2 1 the equation ƒ( x) = g( x) has for the Quadratic Equations given values of g(x). g( x) > -2 2 g( x) < -2 0 INTEGRATE TECHNOLOGY 2. Look back at Step C. Notice that the Number of Real Solutions Students can use a graphing calculator to graph f( x) maximum value of ƒ( x) in Step C is 2. Value of g( x) of Complete the table by identifying how f( x) = g( x) and each function g( x) to verify the number of real many real solutions the equation ƒ g( x) = 2 1 solutions to each equation. (x) = g( x) has for the given values of g (x) . g( x) > 2 0 2 g( x) < 2 QUESTIONING STRATEGIES 2 If an equation is written in vertex form, what 3. You can generalize Reflect 1: For ƒ( x) = a x + bx where a > 0, ƒ( x) = g( x) where g( x) = -c has real solutions when g( x) is greater than or equal to the minimum value of ƒ( x) . The minimum value of ƒ( x) is information can you use to find out if it has 2 b b b b 2 b 2 b 2 b 2 b 2 2 b 2 b 2 _ _ _ _ _ _ _ _ _ _ ƒ a b a 2 . real solutions? The sign of a determines the - 2a = - 2a + - 2a = 4 a - 2a = 4a - 2a = 4a - 4a = - 4a ( ) ( ) ( ) ( ) 2 So, ƒ x g x has real solutions when g x _ b . direction of the opening and the maximum or ( ) = ( ) ( ) ≥ - 4a 2 Substitute c for g x . c _ b minimum value tells you whether there are real - ( ) - ≥ - 4a b 2 b 2 solutions. Add __ to both sides. _ c 0 4a 4a - ≥ 2 Multiply both sides by 4a, which is positive. b - 4ac ≥ 0 How do you determine where the graph of a 2 2 You In other words, the equation a x + bx + c = 0 where a > 0 has real solutions when b - 4ac ≥ 0. quadratic function crosses the x-axis? © Houghton Mifflin Houghton © Company Harcourt Publishing can find the x-intercepts of the graph of a quadratic Generalize the results of Reflect 2 in a similar way. What do you notice? 2 function in standard form by factoring the function For f( x) = a x + bx where a < 0, f (x) = g( x) where g( x) = -c has real solutions when g (x) is b b 2 to get its intercept form. If the function is not less than or equal to the maximum value of f x . The maximum value of f x is f ___ ___ . ( ) ( ) - 2a = - 4a b 2 ( ) factorable, the x-intercepts can be found by using So, f x g x has real solutions when g x ___ . ( ) = ( ) ( ) ≤ - 4a b 2 the quadratic formula to find the zeros of the Substitute c for g x . c __ - ( ) - ≤ - 4a b 2 b 2 function. Add ___ to both sides. __ c 0 4a 4a - ≤ 2 Multiply both sides by 4a, which is negative. b - 4ac ≥ 0 2 2 Whether a > 0 or a < 0, b - 4ac ≥ 0 tells when a x + bx + c = 0 has real solutions. Module 3 140 Lesson 3 PROFESSIONAL DEVELOPMENT A2_MNLESE385894_U2M03L3Math 140Background 5/22/14 10:47 AM In Algebra 1, students used the quadratic formula to find real solutions to a quadratic equation. Students now revisit the formula to extend its use to complex solutions. 2 The sign of the expression b - 4ac determines whether the quadratic equation has two real solutions, one real solution, or two nonreal solutions.