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Transcript
LESSON 3.3 Name Finding Complex Solutions of Quadratic Equations 3.3 A MP.2 Reasoning B Work with a partner or small group to determine whether solutions to quadratic equations are real or not real and justify reasoning. 2x 2 + 4x + 2 = 0 2x + 4x = -2 f(x) = 2x + 4x g(x) = -2 2x 2 + 4x + 3 = 0 2x + 4x = -3 f(x) = 2x + 4x g(x) = -3 2 2 2 2 The graph of ƒ(x) = 2x 2 + 4x is shown. Graph each g(x). Complete the table. -4 © Houghton Mifflin Harcourt Publishing Company g(x) = -1 2x + 4x = -1 x 0 -2 2 4 Equation Number of Real Solutions 2x 2 + 4x + 1 = 0 2 2x + 4x + 2 = 0 1 2x 2 + 4x + 3 = 0 0 2 Repeat Steps A and B when ƒ(x) = -2x 2 + 4x. ax 2 + bx + c = 0 ax 2 + bx = -c f(x) = ax 2 + bx g(x) = -c -2x 2 + 4x - 1 = 0 -2x 2 + 4x = 1 f(x) = -2x 2 + 4x g(x) = 1 -2x 2 + 4x - 2 = 0 -2x 2 + 4x = 2 f(x) = -2x 2 + 4x g(x) = 2 -2x + 4x - 3 = 0 -2x + 4x = 3 f(x) = -2x + 4x g(x) = 3 2 2 2 y Equation Number of Real Solutions -2x 2 + 4x - 1 = 0 2 -2x 2 + 4x - 2 = 0 1 -2x 2 + 4x - 3 = 0 0 2 -4 Module 3 -2 0 -2 x 4 be ges must EDIT--Chan DO NOT Key=NL-A;CA-A Correction Lesson 3 139 gh “File info” made throu Date Class lutions mplex So Finding Co ic Equations of Quadrat Name 3.3 ion: How Quest Essential A2_MNLESE385894_U2M03L3.indd 139 find the can you equations quadratic N-CN.C.7 Solve A-REI.B.4b .2, Also N-CN.C COMMON CORE s of l Solution g Rea ax2 + bx =0 2 +3 2x + 4x . Graph each is shown 2 2x + 4x -2 A Repeat Steps and Publishin x2 + 4x. (x) = -2 B when ƒ Harcour t 2 -1 -2x + 4x 2 -2x + 4x © Houghto n Mifflin 2 -2x + ax2 + bx 2=0 0 g(x) = -c + bx 2 -2x + 4x g(x) = 1 g(x) = 2 g(x) = 3 of Real Number ions Solut 2 -1 -2x + 4x 4 2 x + 4x f(x) = -2 x2 + 4x f(x) = -2 x2 + 4x f(x) = -2 Equation x 0 -2 1 2 =3 2 -2x + 4x 0 4x - 3 = 2 -2 2 f(x) = ax = -c =1 2 -2x + 4x =2 2 -2x + 4x =0 y -4 of Real Number ions Solut =0 2 +3=0 2x + 4x +c=0 ax2 + bx g(x) = -3 =0 2 +2 2x + 4x x 4 2 Turn to these pages to find this lesson in the hardcover student edition. g(x) = -2 + 4x lete the table. g(x). Comp 2 +1 2x + 4x 0 g(x) = -c g(x) = -1 + 4x 2 f(x) = 2x Equation y 2 -4 2 f(x) = 2x = -3 2x + 4x 2 =0 of ƒ(x) = ns ic Equatio Quadrat 2 + bx f(x) = ax 2 + 4x f(x) = 2x = -c = -1 2 2x + 4x = -2 2 2x + 4x =0 2 +2 2x + 4x The graph ax2 + bx +c=0 2 +1 2x + 4x HARDCOVER PAGES 99108 Resource Locker ion? atic equat of any quadr ns. ex solutio have compl ients that the table. Complete solutions complex with real coeffic Investigatin Explore y g Compan View the Engage section online. Discuss the photo and how to solve a quadratic equation to determine how high a baseball will go after it is hit. Then preview the Lesson Performance Task. g(x) = -c 2x + 4x + 1 = 0 y C f(x) = ax 2 + bx f(x) = 2x 2 + 4x 2 2 ENGAGE PREVIEW: LESSON PERFORMANCE TASK ax 2 + bx = -c 2 Mathematical Practices Possible answer: You can factor, if possible, to find real solutions; approximate from a graph; find a square root (which may be part of completing the square); complete the square; or apply the quadratic formula. For the general equation ax 2 + bx + c = 0, you must either complete the square or use the quadratic formula to find the complex solutions of the equation. Investigating Real Solutions of Quadratic Equations ax 2 + bx + c = 0 Solve quadratic equations with real coefficients that have complex solutions. Also N-CN.C.2, A-REI.B.4b Language Objective =0 2=0 2 -2x + 4x 3=0 2 1 0 Lesson 3 139 Module 3 Lesson 3.3 L3.indd 4_U2M03 SE38589 A2_MNLE 139 Resource Locker Complete the table. N-CN.C.7 Essential Question: How can you find the complex solutions of any quadratic equation? Finding Complex Solutions of Quadratic Equations Explore The student is expected to: COMMON CORE Date Essential Question: How can you find the complex solutions of any quadratic equation? Common Core Math Standards COMMON CORE Class 139 19/03/14 12:05 PM 19/03/14 12:00 PM Reflect 1. 2. 3. Look back at Steps A and B. Notice that the minimum value of f(x) in Steps A and B is -2. Complete the table by identifying how many real solutions the equation ƒ(x) = g(x) has for the given values of g(x). Look back at Step C. Notice that the maximum value of ƒ(x) in Step C is 2. Complete the table by identifying how many real solutions the equation ƒ (x) = g(x) has for the given values of g(x). EXPLORE Value of g(x) Number of Real Solutions of f(x) = g(x) g(x) = -2 1 g(x) > -2 2 g(x) < -2 0 Value of g(x) Number of Real Solutions of f(x) = g(x) g(x) = 2 1 g(x) > 2 0 g(x) < 2 2 Investigating Real Solutions of Quadratic Equations INTEGRATE TECHNOLOGY Students can use a graphing calculator to graph f(x) and each function g(x) to verify the number of real solutions to each equation. QUESTIONING STRATEGIES If an equation is written in vertex form, what information can you use to find out if it has real solutions? The sign of a determines the direction of the opening and the maximum or minimum value tells you whether there are real solutions. You can generalize Reflect 1: For ƒ(x) = ax 2 + 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 2 b b2 - _ b2 - _ b2 - _ b2 = _ b2 = _ 2b 2 = -_ b + b -_ b =a _ b2 . = a -_ ƒ4a 4a 4a 2a 2a 4a 2a 2a 2a 4a 2 b2 . So, ƒ(x) = g(x) has real solutions when g(x) ≥ -_ 4a b2 Substitute -c for g(x). -c ≥ -_ 4a b2 - c ≥ 0 b2 to both sides. _ Add __ 4a 4a Multiply both sides by 4a, which is positive. b 2 - 4ac ≥ 0 ( _) ( ) ( ) ( ) How do you determine where the graph of a quadratic function crosses the x-axis? You can find the x-intercepts of the graph of a quadratic function in standard form by factoring the function to get its intercept form. If the function is not factorable, the x-intercepts can be found by using the quadratic formula to find the zeros of the function. In other words, the equation ax + bx + c = 0 where a > 0 has real solutions when b - 4ac ≥ 0. 2 2 For f(x) = ax 2 + bx where a < 0, f(x) = g(x) where g(x) = -c has real solutions when g(x) is ( 2a ) b b less than or equal to the maximum value of f(x). The maximum value of f(x) is f -___ = -___ . 4a b So, f(x) = g(x) has real solutions when g(x) ≤ -___ . 4a b2 __ Substitute -c for g(x). -c ≤ - 4a 2 2 b2 b2 Add ___ to both sides. __ -c≤0 4a 4a Multiply both sides by 4a, which is negative. b 2 - 4ac ≥ 0 Whether a > 0 or a < 0, b 2 - 4ac ≥ 0 tells when ax 2 + bx + c = 0 has real solutions. Module 3 140 © Houghton Mifflin Harcourt Publishing Company Generalize the results of Reflect 2 in a similar way. What do you notice? Lesson 3 PROFESSIONAL DEVELOPMENT A2_MNLESE385894_U2M03L3 140 5/22/14 10:47 AM Math Background 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. The sign of the expression b 2 - 4ac determines whether the quadratic equation has two real solutions, one real solution, or two nonreal solutions. For cubic equations of the form ax 3 + bx 2 + cx + d = 0, the sign of the discriminant b 2c 2 - 4ac 3 - 4b 3d - 27a 2d 2 determines whether the equation has three real solutions, two real solutions, or one real solution. Finding Complex Solutions of Quadratic Equations 140 Finding Complex Solutions by Completing the Square Explain 1 EXPLAIN 1 Finding Complex Solutions by Completing the Square 2 () 2 ( ) 2 trinomial x 2 + bx + __b2 , which you can factor as x + __b2 . Don’t forget that when x 2 + bx appears on one side of an () () 2 2 equation, adding __b2 to it requires adding __b2 to the other side as well. Example 1 QUESTIONING STRATEGIES How do you convert quadratic functions to vertex form? Explain. You can convert quadratic functions from standard form to vertex 2 form f(x) = a(x - h) + k by completing the square on ax 2 + bx. You have to add and subtract the same constant to keep the function value the same. Solve the equation by completing the square. State whether the solutions are real or non-real. 3x 2 + 9x - 6 = 0 1. Write the equation in the form x 2 + bx = c. 4. Solve for x. (x + _23 ) = 2 + _49 (x + _23 ) = _174 2 3x 2 + 9x - 6 = 0 3x 2 + 9x = 6 2 ―― x 2 + 3x = 2 (2) 17 3 = ±_ x+_ 4 2 _ √ 17 3 = ±_ x+_ 2 2 _ √ 17 3 ±_ x = -_ 2 2 _ -3 ± √17 x=_ 2 _ -3 + √17 There are two_real solutions: _ 2 -3 - √17 and _. 2 2 2. Identify b and __b . INTEGRATE MATHEMATICAL PRACTICES Focus on Technology MP.5 Discuss with students how to use the b=3 () () 2 2 9 b = _ 3 =_ _ 4 2 2 () 2 3. Add __b2 to both sides of the equation. 9 =2+_ 9 x 2 + 3x + _ 4 4 x 2 - 2x + 7 = 0 1. Write the equation in the form x 2 + bx = c. © Houghton Mifflin Harcourt Publishing Company graphing calculator to find a maximum or minimum value of a quadratic function. Students can solve problems algebraically and then use their graphing calculators to check their solutions. () Recall that completing the square for the expression x 2 + bx requires adding __b2 to it, resulting in the perfect square 4. Solve for x. x - 2x = -7 2 () x 2 + 2x 1 = -7 + 1 (x - 1 ) = -6 2 2 2. Identify b and __b2 . () b = -2 ( ) 2 -2 b = _ = 1 _ 2 2 () ―― x- 1 =± 2 -6 ―― x=1± -6 2 3. Add __b2 to both sides. ― There are two real/non-real solutions: √ and 1 - i 6 . x 2 - 2x + 1 = -7 + 1 Module 3 141 ― 1 + i √6 Lesson 3 COLLABORATIVE LEARNING A2_MNLESE385894_U2M03L3.indd 141 Peer-to-Peer Activity Have students work in pairs. Provide each pair with several quadratic equations written in various forms. Have one student verbally instruct the partner in how to find the nonreal solutions to the equation. Then have partners switch roles, repeating the activity for a different quadratic equation. Have students discuss how their steps for solving the equation were similar or different. 141 Lesson 3.3 19/03/14 12:00 PM Reflect EXPLAIN 2 How many complex solutions do the equations in Parts A and B have? Explain. Each equation has two complex solutions, because the set of complex numbers includes 4. Identifying Whether Solutions are Real or Non-real all real numbers as well as all non-real numbers. Your Turn Solve the equation by completing the square. State whether the solutions are real or non-real. x 2 + 8x + 17 = 0 5. 6. x + 8x = - 17 QUESTIONING STRATEGIES x 2 + 10x - 7 = 0 x 2 + 10x = 7 2 x 2 + 8x +16 = - 17 + 16 Does the discriminant give the solution of a quadratic equation? Explain. No, it gives the number of solutions and type of solution, but it does not give the actual solution. x 2 + 10x + 25 = 7 + 25 (x + 4) 2 = -1 (x + 5) 2 = 32 x + 4 = ± √―― -1 _ x + 5 = ±√32 x = -4 ± i _ x = -5 ± 4√2 There are two non-real solutions: -4 + i and -4 - i. There are_two non-real solutions: _ -5 + 4√2 and -5 - 4√ 2 . AVOID COMMON ERRORS Identifying Whether Solutions Are Real or Non-real Explain 2 Remind students that they must write the quadratic equation in standard form before applying the quadratic formula. By completing the square for the general quadratic equation ax 2 + bx + c = 0, you can obtain the quadratic __ √ -b ± b - 4ac , which gives the solutions of the general quadratic equation. In the quadratic formula, the formula, x = ___________ 2a expression under the radical sign, b 2 - 4ac, is called the discriminant, and its value determines whether the solutions of the quadratic equation are real or non-real. 2 Number and Type of Solutions b - 4ac > 0 Two real solutions b 2 - 4ac = 0 One real solution b 2 - 4ac < 0 Two non-real solutions Example 2 Answer the question by writing an equation and determining whether the solutions of the equation are real or non-real. A ball is thrown in the air with an initial vertical velocity of 14 m/s from an initial height of 2 m. The ball’s height h (in meters) at time t (in seconds) can be modeled by the quadratic function h(t) = -4.9t 2 + 14t + 2. Does the ball reach a height of 12 m? Set h(t) equal to 12. Subtract 12 from both sides. Find the value of the discriminant. CONNECT VOCABULARY © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Fred Fokkelman/Shutterstock Value of Discriminant 2 -4.9t 2 + 14t + 2 = 12 -4.9t 2 + 14t + 10 = 0 Review vocabulary related to quadratic functions, such as discriminant and real numbers, by having students label the parts of a quadratic function written in various forms. 14 2 - 4(-4.9)(-10) = 196 - 196 = 0 Because the discriminant is zero, the equation has one real solution, so the ball does reach a height of 12 m. Module 3 142 Lesson 3 DIFFERENTIATE INSTRUCTION A2_MNLESE385894_U2M03L3 142 6/8/15 1:22 PM Cognitive Strategies Some students have trouble completing the square because there are so many steps. Show them how to break the process into three parts: (1) Get the equation into the form needed for completing the square. (2) Complete the square. (3) Finish the solution by taking square roots of both sides and simplifying the results. When students make errors, analyze their work carefully to see what part of the process is giving them trouble, and give them extra practice on that part of the process. Finding Complex Solutions of Quadratic Equations 142 INTEGRATE MATHEMATICAL PRACTICES Focus on Reasoning MP.2 The discriminant can be used to distinguish A person wants to create a vegetable garden and keep the rabbits out by enclosing it with 100 feet of fencing. The area of the garden is given by the function A(w) = w(50 - w) where w is the width (in feet) of the garden. Can the garden have an area of 700 ft 2? Set A(w) equal to 700. Multiply on the left side. between rational and irrational solutions. Give students several quadratic equations for which b 2 - 4ac is positive, some with rational solutions, and some with irrational solutions. Ask them to make a conjecture about how the value of the discriminant is related to whether the solutions are rational or irrational. Students should be able to explain why the solutions will be rational when the value of the discriminant is a perfect square. Subtract 700 from both sides. w(50 - w) = 700 50w - w 2 = 700 -w 2 + 50w - 700 = 0 Find the value of the discriminant. 50 2 - 4(-1)(-700) = 2500 -2800 = -300 Because the discriminant is [positive/zero/negative], the equation has [two real/one real/two non-real] solutions, so the garden [can/cannot] have an area of 700 ft 2. Your Turn Answer the question by writing an equation and determining if the solutions are real or non-real. 7. EXPLAIN 3 A hobbyist is making a toy sailboat. For the triangular sail, she wants the height h (in inches) to be twice the length of the base b (in inches). Can the area of the sail be 10 in 2? 1 b(2b) = b 2 Write the area A of the sail as a function of b. A=_ 2 Substitute 10 for A. 10 = b 2 0 = b 2 - 10 Subtract 10 from both sides. QUESTIONING STRATEGIES Why are there always two solutions to a quadratic equation that has nonreal solutions? How are they related? Since √b 2 - 4ac is not zero, its value will be both added to and subtracted from -b in the numerator, resulting in two solutions; they are complex conjugates. ――― What is the general solution of a quadratic b equation with only one solution? x = - ___ 2a Find the discriminant. © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©David Burton/Alamy Finding Complex Solutions Using the Quadratic Formula 0 2 - 4(1)(-10) = 0 + 40 = 40 Because the discriminant is positive, the equation has two real solutions, so the area of the sail can be 10 in 2. Explain 3 Finding Complex Solutions Using the Quadratic Formula When using the quadratic formula to solve a quadratic equation, be sure the equation is in the form ax 2 + bx + c = 0. Example 3 Solve the equation using the quadratic formula. Check a solution by substitution. -5x 2 - 2x - 8 = 0 Write the quadratic formula. ――― -b ± b 2 - 4ac x = __ 2a ――――――― Substitute values. 2 -(-2) ± (-2) - 4(-5)(-8) = ___ ( 2 -5) Simplify. 1 ± i39 2 ± -156 = __ = _ -5 -10 Module 3 ―― 143 ― Lesson 3 LANGUAGE SUPPORT A2_MNLESE385894_U2M03L3 143 Communicate Math Students play “How do you know?” Give students several cards containing quadratic equations; some have real number solutions, others nonreal or complex solutions. In small groups, students draw a card and state whether the solution is real or not real. They then answer the question “How do you know?” Players take turns and sort cards into piles according to the kind of solution. By the end of the game, all players in a group must agree on card placement. 143 Lesson 3.3 5/22/14 10:47 AM _ _ i√39 i√39 1 -_ 1 +_ So, the two solutions are -_ and -_ . 5 5 5 5 AVOID COMMON ERRORS Check by substituting one of the values. Substitute. Square. ― ― ) ( ) 2i ― 39 _ 39 i ― 1 +_ 1 -_ - 39 ) - 2(-_ -8≟0 -5(_ 5 ) 5 25 25 25 ( Distribute. √ √ ― ― 2i √39 _ 2i √39 2 +_ 1 -_ -_ + 39 + _ -8≟0 5 5 5 5 5 40 - 8 ≟ 0 _ 5 0=0 Simplify. B Students may have difficulty remembering the quadratic formula. Encourage students to copy the formula and have it on hand when they are working. Caution them to write the equation in standard form before identifying the values of a, b, and c to be used in the formula. 2 i √39 i √39 1 -_ 1 -_ - 2 -_ -8≟0 -5 -_ 5 5 5 5 INTEGRATE MATHEMATICAL PRACTICES Focus on Communication MP.3 You may wish to point out that quadratic 7x 2 + 2x + 3 = -1 7x 2 + 2x + 4 = 0 Write the equation with 0 on one side. Write the quadratic formula. x= ――― -b ± √b - 4ac __ 2 2a ――――――――― √ ( ) ( )( ) equations always have two roots. However, when the value of the discriminant is 0, the two roots happen to be the same. In this case, the quadratic is said to have a double root. 2 4 2 7 -4 - 2 ± = ____ 7 2 ( ) Substitute values. ――― - 2 ±- 108 = __ 14 Simplify. ―― ―― √ √ _ _ Check by substituting one of the values. Square. Distribute. ( 3i √― 3 _ _ ) 7 -1 + 7 Substitute. (_ 7 2 7 _― _) 6i √3 27 1 49 49 49 2 _ _ ( 3i √― 3 _ _ ) + 2 -1 + 7 ( 2 7 3i √― 3 _ _ ) + 2 -1 + 7 7 +4≟0 6i √― 6i √― 3 _ 3 27 _ 2 _1 - _ - +_+4≟0 7 7 7 Simplify. A2_MNLESE385894_U2M03L3.indd 144 +4≟0 2 7 7 - Module 3 © Houghton Mifflin Harcourt Publishing Company 3 3 - 2 ± 6 i - 1 ± 3 i = __ = __ 7 14 3i ― 3i ― 3 3 -1 -1 + 7 7 7 7 So, the two solutions are and . 144 28 _ +4≟0 7 0=0 Lesson 3 19/03/14 11:59 AM Finding Complex Solutions of Quadratic Equations 144 Your Turn ELABORATE Solve the equation using the quadratic formula. Check a solution by substitution. 8. 6x 2 - 5x - 4 = 0 9. x 2 + 8x + 12 = 2x ――― -b ± √b - 4ac __ 2a ―――――― -(-5) ± √(-5) - 4(6)(-4) ___ = 2(6) 5 ± √―― 121 _ INTEGRATE MATHEMATICAL PRACTICES Focus on Critical Thinking MP.3 Emphasize that choosing which method to x 2 + 6x + 12 = 0 2 x= ――― -b ± √b - 4ac __ 2a ―――――― -(-6) ± √(6) - 4(1)(12) ___ = 2(1) -6 ± √―― -12 __ = 2 12 = 5 ± 11 =_ use to solve a quadratic equation is as important as being able to use each method. Have students discuss when each method might be preferred. _ _ (_) (_) AVOID COMMON ERRORS ― 32 _ 20 _ -4≟ 0 Students may sometimes make a mistake in sign when calculating the discriminant, particularly when the quantity 4ac is less than 0. Remind them that subtracting a negative number is the same as adding the opposite, or positive, number. If a and c are opposite signs, the discriminant will always be positive. 3 3 0=0 Elaborate __― ― ― 2 -6 ± 2i √3 = 2 = -3 ± i √3 So, the solutions are = -3 + i √3 and = -3 - i √3 . Check 2 (-3 + i √3 ) + 8(-3 + i √3 ) + 12 ≟ 2(-3 + i √3 ) 6 - 6i √3 - 24 + 8i √3 + 12 ≟ -6 + 2i √3 -6 + 2i √3 = -6 + 2i √3 12 5 + 11 4 So, the solutions are = 12 3 5 - 11 1. and = -_ 2 12 Check 2 4 4 6 -5 -4≟ 0 3 3 _ 2 x= 2 ― ― ― ― ― ― ― ― 10. Discussion Suppose that the quadratic equation ax 2 + bx + c = 0 has p + qi where q ≠ 0 as one of its solutions. What must the other solution be? How do you know? The other solution must be p − qi. The radical √b 2 - 4ac in the quadratic formula ――― ――― produces imaginary numbers when b - 4ac < 0. Since √b 2 - 4ac is both added to and 2 subtracted from –b in the numerator of the quadratic formula, one solution will have the SUMMARIZE THE LESSON When does a quadratic equation have nonreal solutions, and how do you find them? When the value of the discriminant is negative, the quadratic equation will have two nonreal solutions. You find the solutions by using the quadratic formula to solve the equation, and then writing the solutions as a pair of complex conjugates of the form a±bi. © Houghton Mifflin Harcourt Publishing Company form p + qi, and the other will have the form p − qi. 11. Discussion You know that the graph of the quadratic function ƒ(x) = ax 2 + bx + c has the vertical line b x = -__ as its axis of symmetry. If the graph of ƒ(x) crosses the x-axis, where do the x-intercepts occur 2a relative to the axis of symmetry? Explain. ――― -b ± √b 2 - 4ac by the The x-intercepts are the solutions of f(x) = 0, which are x = ___________ 2a b quadratic formula. Writing the x-intercepts as x = -__ ± 2a the x-intercepts are the same distance, √――― b - 4ac ________ 2 2a 2 2a , away from the axis of symmetry, b with one x-intercept on each side of the line: x = -__ 2a b x = -__ + 2a √――― b - 4ac ________ shows that √――― b - 4ac ________ on the other side. 2 √――― b - 4ac ________ on one side and 2 2a 2a 12. Essential Question Check-In Why is using the quadratic formula to solve a quadratic equation easier than completing the square? The quadratic formula is the result of completing the square on the general quadratic equation ax + bx + c = 0. As long as any particular equation is in the form 2 ax + bx + c = 0, you can simply substitute the values of a, b, and c into the quadratic 2 formula and obtain the solutions of the equation. Module 3 A2_MNLESE385894_U2M03L3 145 145 Lesson 3.3 145 Lesson 3 10/17/14 4:53 PM EVALUATE Evaluate: Homework and Practice 1. The graph of ƒ(x) = x 2 + 6x is shown. Use the graph to determine how many real solutions the following equations have: x 2 + 6x + 6 = 0, x 2 + 6x + 9 = 0, and x 2 + 6x + 12 = 0. Explain. -12 -8 For each equation, subtract the constant from both sides to obtain these equations: x 2 + 6x = -6, x 2 + 6x = -9, and x 2 + 6x = -12. x 0 4 -4 ASSIGNMENT GUIDE -12 The graph of g(x) = -6 intersects the graph of f(x) twice, so the equation x 2 + 6x + 6 = 0 has two real solutions. The graph of g(x) = -9 intersects the graph of f(x) once, so the equation x 2 + 6x + 9 = 0 has one real solution. The graph of g(x) = -12 doesn’t intersect the graph of f(x) , so the equation x 2 + 6x + 12 = 0 has no real solutions. 2. -4 • Online Homework • Hints and Help • Extra Practice y 4 The graph of ƒ(x) = -_12 x 2 + 3x is shown. Use the graph to determine how many real solutions the following equations have: -_21x 2 + 3x - 3 = 0, -_12 x 2 + 3x - _29 = 0, and -_12x 2 + 3x - 6 = 0. Explain. 6 5 4 3 2 1 For each equation, subtract the constant from both sides to 9 1 2 1 2 x + 3x = 3, -_ x + 3x = _ , and obtain these equations: -_ 2 2 2 1 2 -_ x + 3x = 6. The graph of g(x) = 3 intersects the graph of 2 0 y x 1 2 3 4 5 6 7 1 2 f(x) twice, so the equation -_ x + 3x - 3 = 0 has two real 2 9 solutions. The graph of g(x) = _ intersects the graph of f(x) 2 9 1 2 once, so the equation -_ x + 3x - _ = 0 has one real solution. 2 2 1 2 equation -_ x + 3x - 6 = 0 has no real solutions. 2 Solve the equation by completing the square. State whether the solutions are real or non-real. 3. x 2 + 4x + 1 = 0 4. x 2 + 4x = -1 x 2 + 2x = -8 x 2 + 4x + 4 = -1 + 4 (x + 2)2 = 3 ― x + 2 = ±√3 x 2 + 2x + 8 = 0 x + 2x + 1 = -8 + 1 2 (x + 1)2 = -7 ―― x + 1 = ± i √― 7 x + 1 = ± √-7 ― x = -2 ± √3 ― ― two real solutions: -2 + √3 and -2 - √3 . Module 3 Exercise Explore Investigating Real Solutions of Quadratic Equations Exercises 1–2 Example 1 Finding Complex Solutions by Completing the Square Exercises 3–8 Example 2 Identifying Whether Solutions are Real or Non-real Exercises 9–16 Example 3 Finding Complex Solutions Using the Quadratic Formula Exercises 17–20 What information does the value of the discriminant give about a quadratic equation? The value of the discriminant indicates the number and types of roots. ― x = -1 ± i √7 ― ― two non-real solutions: -1 + i √7 and -1 - i √7 . Lesson 3 146 A2_MNLESE385894_U2M03L3 146 Practice CONNECT VOCABULARY © Houghton Mifflin Harcourt Publishing Company The graph of g(x) = 6 doesn’t intersect the graph of f(x), so the Concepts and Skills Depth of Knowledge (D.O.K.) COMMON CORE Mathematical Practices 1–8 1 Recall of Information MP.2 Reasoning 9–16 1 Recall of Information MP.3 Logic 17–20 2 Skills/Concepts MP.2 Reasoning 21 1 Recall of Information MP.3 Logic 22 3 Strategic Thinking MP.3 Logic 23–24 3 Strategic Thinking MP.4 Modeling 5/22/14 10:47 AM Finding Complex Solutions of Quadratic Equations 146 VISUAL CUES 5. If students have difficulty evaluating the discriminant, have them organize the variables in a table. Ask students to create a table for each of the variables (a, b, c, b 2, 4ac, b 2 - 4ac) and have them predict the number and type of solutions based on the variables they list in the table. x 2 - 5x = -20 6. 25 25 x 2 - 5x + __ = -20 + __ 4 4 x 2 - 1.2x = 1.6 2 55 (x - _52) = -__ 4 ―― 5 55 x-_ = ± -__ √ 2 4 ― i √55 5 ____ x-_ = ± 2 2 i √― 55 5 x=_ ± ____ 2 x 2 - 1.2x + 0.36 = 1.6 + 0.36 (x - 0.6)2 = 1.96 x - 0.6 = ±1.4 x = 0.6 ± 1.4 2 two real solutions: 2 and -0.8. i √― 55 i √― 55 5 _5 + ____ and _ - ____. 7. 2 2 2 7x + 13x = 5 2 x 2 8. 13 5 + __ x=_ 7 -x 2 - 6x - 11 = 0 x 2 + 6x + 11 = 0 7 x 2 + 6x + 9 = -11 + 9 13 169 5 169 x + ___ =_ + ___ x 2 + __ 7 7 196 196 (x + 3)2 = -2 ―― x + 3 = ± i √― 2 2 13 309 ) = ___ (x + __ x + 3 = ± √-2 ―― 13 309 = ± √___ x + __ 14 196 14 ―― x - 0.6 = ± √1.96 two non-real solutions: 2 5x 2 - 6x = 8 196 √―― 309 13 = ± _____ x + __ √―― 309 13 ± _____ x = -__ 14 14 14 two real solutions: ―― 14 ― ― x = -3 ± i √2 ― two non-real solutions: -3 + i√2 and -3 - i√2 . ―― -13 + √309 -13 - √309 _________ and _________ . 14 14 Without solving the equation, state the number of solutions and whether they are real or non-real. 9. -16x 2 + 4x + 13 = 0 10. 7x 2 - 11x + 10 = 0 Find the discriminant. Find the discriminant. © Houghton Mifflin Harcourt Publishing Company 4 2 - 4 (-16)(13) = 16 + 832 = 848 Because the discriminant is positive, the equation has two real solutions. 2x = 1 11. -x 2 - _ 5 2 x-1=0 -x 2 - _ 5 Find the discriminant. 4 96 - 4 = -__ (-_25) - 4(-1)(-1) = __ 25 25 2 Because the discriminant is negative, the equation has two non-real solutions. Module 3 A2_MNLESE385894_U2M03L3.indd 147 147 Lesson 3.3 (-11)2 - 4(7)(10) = 121 - 280 = -159 Because the discriminant is negative, the equation has two non-real solutions. 12. 4x 2 + 9 = 12x 4x 2 - 12x + 9 = 0 Find the discriminant. (-12)2 - 4(4)(9) = 144 - 144 = 0 Because the discriminant is zero, the equation has one real solution. 147 Lesson 3 19/03/14 11:59 AM Answer the question by writing an equation and determining whether the solutions of the equation are real or non-real. 13. A gardener has 140 feet of fencing to put around a rectangular vegetable garden. The function A(w) = 70w - w 2 gives the garden’s area A (in square feet) for any width w (in feet). Does the gardener have enough fencing for the area of the garden to be 1300 ft 2? Write an equation by setting A(w) equal to 1300. Then rewrite the equation with 0 on one side. 70w - w 2 = 1300 -w 2 + 70w − 1300 = 0 Find the discriminant. 70 2 − 4(−1)(−1300) = 4900 − 5200 = − 300 Because the discriminant is negative, the equation has two non-real solutions, so the gardener does not have enough fencing. 14. A golf ball is hit with an initial vertical velocity of 64 ft/s. The function h(t) = -16t 2 + 64t models the height h (in feet) of the golf ball at time t (in seconds). Does the golf ball reach a height of 60 ft? Write an equation by setting h(t) equal to 60. Then rewrite the equation with 0 on one side. -16t 2 + 64t = 60 -16t 2 + 64t - 60 = 0 4t 2 - 16t + 15 = 0 Find the discriminant. (-16) − 4(4)(15) = 256 − 240 = 16 2 Because the discriminant is positive, the equation has two real © Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Aflo Foto Agency/Alamy solutions, so the golf ball does reach a height of 60 ft. 15. As a decoration for a school dance, the student council creates a parabolic arch with balloons attached to it for students to walk through as they enter the dance. The shape of the arch is modeled by the equation y = x(5 - x), where x and y are measured in feet and where the origin is at one end of the arch. Can a student who is 6 feet 6 inches tall walk through the arch without ducking? Write an equation by setting y equal to 6.5. Then rewrite the equation with 0 on one side. x(5 - x) = 6.5 5x - x 2 = 6.5 -x 2 + 5x - 6.5 = 0 Find the discriminant. 5 2 -4(-1)(-6.5) = 25 - 26 = -1 Because the discriminant is negative, the equation has two non-real solutions, so a student who is 6 feet 6 inches tall cannot walk through the arch without ducking. Module 3 A2_MNLESE385894_U2M03L3 148 148 Lesson 3 10/17/14 4:56 PM Finding Complex Solutions of Quadratic Equations 148 16. A small theater company currently has 200 subscribers who each pay $120 for a season ticket. The revenue from season-ticket subscriptions is $24,000. Market research indicates that for each $10 increase in the cost of a season ticket, the theater company will lose 10 subscribers. A model for the projected revenue R (in dollars) from season-ticket subscriptions is R(p) = (120 + 10p)(200 - 10p), where p is the number of $10 price increases. According to this model, is it possible for the theater company to generate $25,600 in revenue by increasing the price of a season ticket? PEER-TO-PEER DISCUSSION Ask students to discuss with a partner how the graphs of the following three parabolas would look: a parabola with two real solutions, a parabola with one real solution, and a parabola with two nonreal solutions. Students should say that a parabola with two solutions will have two x-intercepts, and the parabola will open from the vertex toward the x-axis; that a parabola with one solution will have one x-intercept with the vertex on the x-axis; and that a parabola with two nonreal solutions will open from the vertex away from the axis and have no x-intercept. Write an equation by setting R(p) equal to 25,600. Then rewrite the equation with 0 on one side. (120 + 10p) (200 - 10p) = 25,600 -100p 2 + 800p + 24, 000 = 25,600 -100p 2 + 800p - 1600 = 0 p 2 - 8p + 16 = 0 Find the discriminant. (-8) - 4(1)(16) = 64 - 64 = 0 2 Because the discriminant is zero, the equation has one real solution, so it is possible to generate $25,600 in revenue by increasing the price of a season ticket. Solve the equation using the quadratic formula. Check a solution by substitution. 17. x 2 - 8x + 27 = 0 ――― -b ± √b - 4ac x = __ 2a ―――――― -(-8) ± √(-8) - 4(1)(27) = ___ 2 18. x 2 - 30x+ 50 = 0 ――― -b ± √b - 4ac __ 2a ――――――― -(-30) ± √(-30) - 4(1)(50) = ___ 2 2 = = 8 ± √―― -44 _ 2 8 ± 2i √― 11 _ 2(1) = = 2 © Houghton Mifflin Harcourt Publishing Company 11 = 4 ± i √― 4 + i √― 11 and 4 - i √― 11 . Check (4 + i √― 11 ) - 8(4 + i √― 11 ) + 27 ≟ 0 5 + 8i √― 11 - 8(4 + i √― 11 ) + 27 ≟ 0 5 + 8i √― 11 - 32 - 8i √― 11 + 27 ≟ 0 2 5 - 32 + 27 ≟ 0 0=0 A2_MNLESE385894_U2M03L3 149 149 Lesson 3.3 ―― 30 ± √700 __ 2 30 ± 10 √― 7 __ 2 2(1) ― = 15 ± 5 √7 So, the solutions are Module 3 2 x= So, the solutions are ― ― 15 + 5 √7 and 15 - 5 √7 . ― ― ― ― ― 7 + 50 ≟ 0 400 + 150 √7 - 450 - 150 √― Check 2 (15 + 5 √7 ) - 30(15 + 5 √7 ) + 50 ≟ 0 400 + 150 √7 - 30(15 + 5 √7 ) + 50 ≟ 0 400 - 150 + 50 ≟ 0 0=0 149 Lesson 3 5/22/14 10:47 AM 19. x + 3 = x 2 20. 2x 2 + 7 = 4x Rewrite the equation with 0 on one side. 2x 2 - 4x + 7 = 0 x 2 - x -3 = 0 Use the quadratic formula. x= ――― -b ± √b - 4ac __ 2a ―――――― -(-1)± √(-1) - 4(1)(-3) = ___ 2(1) 1 ± √― 13 _ 2 x= ――― -b ± √b - 4ac __ 2a ―――――― -(-4) ± √(-4) - 4(2)(7) ___ = 2(2) 4 ± √―― -40 __ = 4 4 ± 2i √― 2 ± i √― 10 10 __ _ = 2 So, the two solutions are 2 and 1 - √― 13 ______ 2 Students need to be careful to avoid making sign errors when completing the square. Point out that when the rule representing vertex form is simplified, the result should be the original rule written in standard form. Students can use this fact to perform a quick check of the reasonableness of their results, and in order to catch any sign errors they may have made. 2 2 2 1 + √― 13 ______ AVOID COMMON ERRORS = = 4 2 So, the two solutions are . ― ― ) ) (_ (_ ― ― __ (_) ― _ __ ( ―) Check 2 1 + √13 1 + √13 -3≟0 2 2 14 + 2 √13 1 + √13 -3≟0 4 2 14 + 2 √13 2 + 2 √13 -3≟0 4 4 12 -3≟0 4 0=0 _ ― ― 1 + ____ and 1 - ____ . 2 2 i √10 i √10 Check. i √― 10 i √― 10 _ - 4(1 + _) + 7 ≟ 0 2 ) 2 i √― 10 3 2(-_ + i √― 10 ) - 4(1 + _) + 7 ≟ 0 2 2 -3 + 2i √― 10 - 4 - 2i √― 10 + 7 ≟ 0 ( 2 2 1+ -3 - 4 + 7 ≟ 0 0=0 21. Place an X in the appropriate column of the table to classify each equation by the number and type of its solutions. Equation x 2 - 3x + 1 = 0 Two Real Solutions One Real Solution Two Non-Real Solutions X x - 2x + 1 = 0 2 x - x +1 = 0 X x2 + 1 = 0 X x +x+1=0 X 2 x + 2x + 1 = 0 X 2 x + 3x + 1 = 0 2 Module 3 A2_MNLESE385894_U2M03L3.indd 150 © Houghton Mifflin Harcourt Publishing Company X 2 X 150 Lesson 3 19/03/14 11:59 AM Finding Complex Solutions of Quadratic Equations 150 JOURNAL H.O.T. Focus on Higher Order Thinking Have students summarize how to use the discriminant to help solve any quadratic equation. Have them include examples of quadratic equations with one or two real solutions and with two nonreal solutions. 22. Explain the Error A student used the method of completing the square to solve the equation -x 2 + 2x - 3 = 0. Describe and correct the error. -x 2 + 2x - 3 = 0 -x 2 + 2x = 3 -x 2 + 2x + 1 = 3 + 1 (x + 1) 2 = 4 _ x + 1 = ±√4 x + 1 = ±2 x = -1 ± 2 So, the two solutions are -1 + 2 = 1 and -1 - 2 = -3. The student did not divide both sides by –1 first to make the coefficient of the x 2-term be 1. The correct solution is as follows. x 2 - 2x + 3 = 0 x 2 - 2x = -3 x - 2x + 1 = -3 + 1 2 (x -1)2 = -2 ―― x - 1 = ± i √― 2 x = 1 ± i √― 2 x - 1 = ± √-2 ― ― So, the two solutions are 1 + i √2 and 1 - i √2 . 23. Make a Conjecture Describe the values of c for which the equation x 2 + 8x + c = 0 has two real solutions, one real solution, and two non-real solutions. Find the value of the discriminant. © Houghton Mifflin Harcourt Publishing Company b 2 - 4ac = 8 2 - 4(1) c = 64 - 4c The equation has two real solutions when the discriminant is positive, so solving 64 - 4c > 0 for c gives c < 16. The equation has one real solution when the discriminant is zero, so solving 64 - 4c = 0 for c gives c = 16. The equation has two non-real solutions when the discriminant is negative, so solving 64 - 4c < 0 for c gives c > 16. 24. Analyze Relationships When you rewrite y = ax 2 + bx + c in vertex form by ( ) b b , c - __ . completing the square, you obtain these coordinates for the vertex: -__ 4a 2a Suppose the vertex of the graph of y = ax 2 + bx + c is located on the x-axis. Explain 2 how the coordinates of the vertex and the quadratic formula are in agreement in this situation. When the vertex is on the x-axis, the y-coordinate of the vertex must be 0, b = 0, which can be rewritten as b 2 - 4ac = 0. When you set y equal so c - __ 4a 2 to 0 in y = ax 2 + bx + c and solve for x, you get one real solution, namely, b x = -__ , which is the x-coordinate of the vertex. 2a Module 3 A2_MNLESE385894_U2M03L3.indd 151 151 Lesson 3.3 151 Lesson 3 19/03/14 11:59 AM Lesson Performance Task AVOID COMMON ERRORS Students may sometimes make a mistake in sign when calculating the discriminant, particularly when the quantity 4ac is less than 0. Remind them that subtracting a negative number is the same as adding the opposite, or positive, number. If a and c have opposite signs, the discriminant will always be positive. Matt and his friends are enjoying an afternoon at a baseball game. A batter hits a towering home run, and Matt shouts, “Wow, that must have been 110 feet high!” The ball was 4 feet off the ground when the batter hit it, and the ball came off the bat traveling vertically at 80 feet per second. a. Model the ball’s height h (in feet) at time t (in seconds) using the projectile motion model h(t) = -16t 2 + v 0t + h 0 where v 0 is the projectile’s initial vertical velocity (in feet per second) and h 0 is the projectile’s initial height (in feet). Use the model to write an equation based on Matt’s claim, and then determine whether Matt’s claim is correct. b. Did the ball reach a height of 100 feet? Explain. INTEGRATE TECHNOLOGY c. Let h max be the ball’s maximum height. By setting the projectile motion model equal to h max, show how you can find h max using the discriminant of the quadratic formula. Students can use a graphing utility to graph a parabola and find the maximum value. d. Find the time at which the ball reached its maximum height. a. The ball’s height h at time t is given by h(t) = -16t 2 + 80t + 4. Matt’s claim is that h(t) = 110 at some time t. Applying the discriminant of the quadratic formula to the equation -16t 2 + 80t + 4 = 110, or -16t 2 + 80t - 106 = 0, gives b 2 - 4ac = 80 2 - 4(-16)(-106) = 6400 - 6784 = -384. Since the discriminant is negative, there are no real values of t that solve the equation, so Matt’s claim is incorrect. QUESTIONING STRATEGIES How can the symmetry of a parabola help you to find the maximum or minimum if you know two different points on the graph with the same y-coordinate? The x-coordinate of the maximum or minimum will be halfway between the x-coordinates of the two points on the graph. b. For the ball to reach of height of 100 feet, h(t) must equal 100. Applying the discriminant of the quadratic formula to the equation -16t 2 + 80t + 4 = 100, or -16t 2 + 80t - 96 = 0, gives b 2 - 4ac = 80 2 - 4(-16)(-96) = 6400 - 6144 = 256. Since the discriminant is positive, there are two real values of t that solve the equation, so the ball did reach a height of 100 feet at two different times (once before reaching its maximum height and once after). b 2 - 4ac = 0 80 - 4(-16)(4 - h max) = 0 2 6400 + 64(4 - h max) = 0 64(4 - h max )= -6400 4 - h max = -100 -h max = -104 h max = 104 © Houghton Mifflin Harcourt Publishing Company c. Setting h(t) equal to h max gives -16t 2 + 80t + 4 = h max = 0, or -16t 2 + 80t + 4 - h max = 0. Since the maximum height occurs for a single real value of t, the discriminant of the quadratic equation must equal 0. So, the ball reached a maximum height of 104 feet. d. Solve the equation -16t 2 + 80t + 4 = 104, or -16t 2 + 80t - 100 = 0, using the quadratic formula. You already know that the discriminant is 0 when the ball reached its maximum ― -80 = 2.5. So, the ball reached its maximum height 2.5 seconds height, so t = ________ = ___ -32 2(-16) after it was hit. -80 ± √0 Module 3 152 EXTENSION ACTIVITY A2_MNLESE385894_U2M03L3 152 _ Lesson 3 On the moon, the force of gravity is 1 Earth’s gravity, so the equation for simple 6 projectile motion for a ball hit at a height of 4 feet above the ground is 16 y = - t 2 + vt + 4. Have students determine if a baseball hit upward traveling 6 at an initial vertical velocity of v = 80 ft/s on the moon reaches a height of 200 feet. They should find the real solutions. t = 2.7 and 27.3 s, so the ball does reach a height of 200 feet. 6/8/15 1:23 PM _ Scoring Rubric 2 points: Student correctly solves the problem and explains his/her reasoning. 1 point: Student shows good understanding of the problem but does not fully solve or explain his/her reasoning. 0 points: Student does not demonstrate understanding of the problem. Finding Complex Solutions of Quadratic Equations 152