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Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Lesson 38: Complex Numbers as Solutions to Equations Student Outcomes ο§ Students solve quadratic equations with real coefficients that have complex solutions (N-CN.C.7). They recognize when the quadratic formula gives complex solutions and write them as π + ππ for real numbers π and π. (A-REI.B.4b) Lesson Notes This lesson models how to solve quadratic equations over the set of complex numbers. Students relate the sign of the discriminant to the nature of the solution set for a quadratic equation. Continue to encourage students to make connections between graphs of a quadratic equation, π¦ = ππ₯ 2 + ππ₯ + π, and the number and type of solutions to the equation ππ₯ 2 + ππ₯ + π = 0. Classwork Opening (2 minutes) In Algebra I, students learned that when the quadratic formula resulted in an expression that contained a negative number in the radicand, the equation would have no real solution. Now, students understand the imaginary unit π as a number that satisfies π 2 = β1, which allows them to solve quadratic equations over the complex numbers. Thus, they can see that every quadratic equation has at least one solution. Opening Exercises (5 minutes) Have students work on this opening exercise alone or in pairs. In this exercise, students apply the quadratic formula to three different relatively simple quadratic equations: one with two real roots, one with one real repeated root, and one with two complex roots. Students are then asked to explain the results in terms of the discriminant. Afterward, go over the answers with the class. Scaffolding: Advanced students may be able to handle a more abstract framing ofβin essenceβthe same exercise. The exercise below offers the advanced student an opportunity to discover the discriminant and its significance on his or her own. βRecall that a quadratic equation can have exactly two distinct real solutions, exactly one distinct real solution, or exactly two distinct complex solutions. What is the quadratic formula that we can use to solve an equation in the form ππ₯ 2 + ππ₯ + π₯ = 0, where π, π, and π are real numbers and π β 0? Analyze this formula to decide when the equation will have two, one, or no real solutions.β Solution: βπ ± βπ 2 β 4ππ 2π The type of solutions to a quadratic equation hinges on the expression under the radical in the quadratic formula, namely, π 2 β 4ππ. When π 2 β 4ππ < 0, both solutions will have imaginary parts. When π 2 β 4ππ > 0, the quadratic equation has two distinct real solutions. When π 2 β 4ππ = 0, the quadratic formula π₯= simplifies to π₯ = β π . In this case, there is only one real solution, 2π which we call a zero of multiplicity two. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 446 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Review the quadratic formula π₯ = 2 βπ±βπ β4ππ before beginning this exercise, and define the discriminant as the 2π number under the radical; that is, the discriminant is the quantity π 2 β 4ππ. Opening Exercises The expression under the radical in the quadratic formula, ππ β πππ, is called the discriminant. 1. Use the quadratic formula to solve the following quadratic equations. Calculate the discriminant for each equation. a. ππ β π = π The equation ππ β π = π has two real solutions: π = π and π = βπ. The discriminant of ππ β π = π is ππ. b. ππ β ππ + π = π The equation ππ β ππ + π = π has one real solution: π = π. The discriminant of ππ β ππ + π = π is π. c. ππ + π = π The equation ππ + π = π has two complex solutions: π = ππ and π = βππ. The discriminant of ππ + π = π is βππ. 2. How does the value of the discriminant for each equation relate the number of solutions you found? If the discriminant is negative, the equation has complex solutions. If the discriminant is zero, the equation has one real solution. If the discriminant is positive, the equation has two real solutions. Discussion (8 minutes) ο§ ο§ Why do you think we call π 2 β 4ππ the discriminant? οΊ In English, a discriminant is a characteristic that allows something (e.g., an object, a person, a function) among a group of other somethings to be distinguished. οΊ In this case, the discriminant distinguishes a quadratic equation by its number and type of solutions: one real solution (repeated), two real solutions, or two complex solutions. English language learners may benefit from a Frayer diagram or other vocabulary exercise for the word discriminant. Letβs examine the situation when the discriminant is zero. Why does a quadratic equation with discriminant zero have only one real solution? οΊ ο§ Scaffolding: When the discriminant is zero, the quadratic formula gives the single solution β π±0 π =β . 2π 2π Why is the solution when π 2 β 4ππ = 0 a repeated zero? οΊ If π 2 β 4ππ = 0, then π = 2 π , and we can factor the quadratic expression ππ₯ 2 + ππ₯ + π as follows: 4π ππ₯ 2 + ππ₯ + π = ππ₯ 2 + ππ₯ + π2 π π2 π 2 = π (π₯ 2 + π₯ + 2 ) = π (π₯ + ) . 4π π 4π 2π π is a repeated zero. 2π π+0 πβ0 π Analytically, the solutions can be thought of as β and β , which are both β . So, there are 2π 2π 2π From what we know of factoring quadratic expressions from Lesson 11, β two solutions that are the same number. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 447 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II οΊ Geometrically, a quadratic equation represents a parabola. If the discriminant is zero, then the equation of the parabola is π¦ = π (π₯ + π 2 π ) , so the vertex of this parabola is (β , 0), meaning the 2π 2π vertex of the parabola lies on the π₯-axis. Thus, the parabola is tangent to the π₯-axis and intersects the π₯-axis only at the point (β π , 0). 2π ο§ For example, the graph of π¦ = π₯ 2 + 6π₯ + 9 intersects the π₯-axis only at (β3, 0), as follows. ο§ Describe the graph of a quadratic equation with positive discriminant. ο§ οΊ If the discriminant is positive, then the quadratic formula gives two different real solutions. οΊ Two real solutions mean the graph intersects the π₯-axis at two distinct real points. For example, the graph of π¦ = π₯ 2 + π₯ β 6 intersects the π₯-axis at (β3, 0) and (2, 0), as follows. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 448 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II ο§ ο§ Describe the graph of a quadratic equation with negative discriminant. οΊ Since the discriminant is negative, the quadratic formula will give two different complex solutions. οΊ Since there are no real solutions, the graph does not cross or touch the π₯-axis in the real plane. For example, the graph of π¦ = π₯ 2 + 4, shown below, does not intersect the π₯-axis. Example 1 (5 minutes) Example 1 Consider the equation ππ + ππ = βπ. What does the value of the discriminant tell us about number of solutions to this equation? Solve the equation. Does the number of solutions match the information provided by the discriminant? Explain. Consider the equation 3π₯ + π₯ 2 = β7. ο§ ο§ What does the value of the discriminant tell us about number of solutions to this equation? οΊ The equation in standard form is π₯ 2 + 3π₯ + 7 = 0, so we have π = 1, π = 3, π = 7. οΊ The discriminant is 32 β 4(1)(7) = β19. The negative discriminant indicates that no real solutions exist. There are two complex solutions. Solve the equation. Does the number of solutions match the information provided by the discriminant? Explain. οΊ Using the quadratic formula, π₯= β3+ββ19 β3+ββ19 or π₯ = . 2 2 3 2 β19 3 2 The solutions, in π + ππ form, are β + οΊ The two complex solutions are consistent with the rule for a negative discriminant. Lesson 38: 2 π and β β β19 οΊ Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 2 π. 449 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exercise (15 minutes) Have students work individually on this exercise; then, have them work with a partner or in a small group to check their solutions. This exercise could also be conducted by using personal white boards and having students show their answers to each question after a few minutes. If many students struggle, invite them to exchange papers with a partner to check MP.3 for errors. Having students identify errors in their work or the work of others will help them to build fluency when working with these complicated expressions. Debrief this exercise by showing the related graph of the equation in the coordinate plane, and verify that the number of solutions corresponds to the number of π₯-intercepts. Exercise Compute the value of the discriminant of the quadratic equation in each part. Use the value of the discriminant to predict the number and type of solutions. Find all real and complex solutions. a. ππ + ππ + π = π We have π = π, π = π, and π = π. Then ππ β πππ = ππ β π(π)(π) = π. Note that the discriminant is zero, so this equation has exactly one real solution. π= β(π) ± βπ = βπ π(π) Thus, the only solution is βπ. b. ππ + π = π We have π = π, π = π, and π = π. Then ππ β πππ = βππ. Note that the discriminant is negative, so this equation has two complex solutions. π= βπ ± ββππ π(π) Thus, the two complex solutions are ππ and β ππ. c. πππ β ππ β ππ = π We have π = π, π = βπ, and π = βππ. Then ππ β πππ = (βπ)π β π(π)(βππ) = ππ + πππ = πππ. Note that the discriminant is positive, so this equation has two distinct real solutions. Using the quadratic formula, π= So, the two real solutions are Lesson 38: π+βπππ π and β(βπ) ± πβπππ . π(π) πββπππ π . Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 450 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II d. πππ + ππ + π = π We have π = π, π = π, and π = π. Then ππ β πππ = ππ β π(π)(π) = ππ β ππ = βπ. The discriminant is negative, so there will be two complex solutions. Using the quadratic formula, π= π π So, the two complex solutions are β + e. βπ π βπ ± ββπ . π(π) π π π and β β βπ π π. π = πππ + π We can rewrite this equation in standard form with π = π, π = βπ, and π = π: πππ β π + π = π. Then ππ β πππ = (βπ)π β π(π)(π) = π β ππ = βππ. The discriminant is negative, so there will be two complex solutions. Using the quadratic formula, The two solutions are π π + π= β(βπ) ± ββππ π(π) π= π ± πβππ . π π βππ βππ π and β π. π π π Scaffolding: f. π ππ + ππ + ππ = π We can factor π from the left side of this equation to obtain π(πππ + π + π) = π, and we know that a product is zero when one of the factors are zero. Since π β π, we must have πππ + π + π = π. This is a quadratic equation with π = π, π = π, and π = π. Then ππ β πππ = ππ β π(π)(π) = βππ. The discriminant is negative, so there will be two complex solutions. Using the quadratic formula, π= βπ ± ββππ π(π) π= βπ ± ππβπ . π π π The complex solutions are β + Lesson 38: πβπ π πβπ π and β β π . π π π Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 Consider assigning advanced students to create quadratic equations that have specific solutions. For example, request a quadratic equation that has only the solution β5. Answer: One such equation is π₯ 2 β 10π₯ + 25 = 0. This follows from the expansion of the left side of (π₯ + 5)2 = 0. Also consider requesting a quadratic equation with solutions 3 + π and 3 β π. One answer is π₯ 2 β 6π₯ + 10 = 0. 451 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Closing (5 minutes) As this lesson is summarized, ask students to create a graphic organizer that allows them to compare and contrast the nature of the discriminant, the number and types of solutions to ππ₯ 2 + ππ₯ + π = 0, and the graphs of the equation π¦ = ππ₯ 2 + ππ₯ + π. Have them record a problem of each type from the previous exercise as an example in their graphic organizer. Lesson Summary ο§ A quadratic equation with real coefficients may have real or complex solutions. ο§ Given a quadratic equation πππ + ππ + π = π, the discriminant ππ β πππ indicates whether the equation has two distinct real solutions, one real solution, or two complex solutions. β If ππ β πππ > π, there are two real solutions to πππ + ππ + π = π. β If ππ β πππ = π, there is one real solution to πππ + ππ + π = π. β If ππ β πππ < π, there are two complex solutions to πππ + ππ + π = π. Exit Ticket (5 minutes) The Exit Ticket gives students the opportunity to demonstrate their mastery of this lessonβs content. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 452 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Name Date Lesson 38: Complex Numbers as Solutions to Equations Exit Ticket Use the discriminant to predict the nature of the solutions to the equation 4π₯ β 3π₯ 2 = 10. Then, solve the equation. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 453 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exit Ticket Sample Solutions Use the discriminant to predict the nature of the solutions to the equation ππ β πππ = ππ. Then, solve the equation. πππ β ππ + ππ = π We have π = π, π = βπ, and π = ππ. Then ππ β πππ = (βπ)π β π(π)(ππ) = ππ β πππ = βπππ. The value of the discriminant is negative, indicating that there are two complex solutions. Thus, the two solutions are π π + π= β(βπ) ± ββπππ π(π) π= π ± ππβππ π π βππ βππ π and β π. π π π Problem Set Sample Solutions The Problem Set offers students more practice solving quadratic equations with complex solutions. 1. Give an example of a quadratic equation in standard form that has β¦ a. Exactly two distinct real solutions. Since (π + π)(π β π) = ππ β π, the equation ππ β π = π has two distinct real solutions, π and βπ. b. Exactly one distinct real solution. Since (π + π)π = ππ + ππ + π, the equation ππ + ππ + π = π has only one real solution, π. c. Exactly two complex (non-real) solutions. Since ππ + π = π has no solutions in the real numbers, this equation must have two complex solutions. They are π and βπ. 2. Suppose we have a quadratic equation πππ + ππ + π = π so that π + π = π. Does the quadratic equation have one solution or two distinct solutions? Are they real or complex? Explain how you know. If π + π = π, then either π = π = π, π > π and π < π, or π < π and π > π. The definition of a quadratic polynomial requires that π β π, so either π > π and π < π or π < π and π > π. In either case, πππ < π. Because ππ is positive and πππ is negative, we know ππ β πππ > π. Therefore, a quadratic equation πππ + ππ + π = π always has two distinct real solutions when π + π = π. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 454 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 3. Solve the equation πππ β ππ + π = π. We have a quadratic equation with π = π, π = βπ, and π = π. π= So, the solutions are 4. π π πβππ + π and π π β πβππ π β(βπ) ± πββππ π(π) . Solve the equation πππ + ππ = βπ. In standard form, this is the quadratic equation πππ + ππ + π = π with π = π, π = π, and π = π. βπ ± πββπ βπ ± πβπ = π(π) π π= Thus, the solutions are π + 5. πβπ πβπ and π β . π π Solve the equation ππ β πππ = π + π + ππ . In standard form, this is the quadratic equation ππππ β ππ + π = π with π = ππ, π = βπ, and π = π. π=β Thus, the solutions are 6. π π + πβππ ππ and π π β β(βπ) ± πββππ π ± ππβππ = π(ππ) ππ πβππ ππ . Solve the equation πππ β π + π = π. This is a quadratic equation with π = π, π = βπ, and π = π. π=β Thus, the solutions are 7. π π + πβππ π and π π β β(βπ) ± ββππ π ± πβππ = π(π) π πβππ π . Solve the equation πππ + πππ β ππ + π = πππ (πππ β π). When expanded, this is a quadratic equation with π = π, π = βπ, and π = π. πππ + πππ β ππ + π = πππ β πππ πππ β ππ + π = π π= So, the solutions are Lesson 38: π π + πβππ ππ and π π β β(βπ) ± β{βππ} π(π) πβππ . ππ Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 455 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 8. Solve the equation ππππ + ππππ + πππ = π. We can factor ππ from the left side of this equation to obtain ππ(ππ + ππ + π) = π, and we know that a product is zero when one of the factors is zero. Since ππ β π, we must have ππ + ππ + π = π. This is a quadratic equation with π = π, π = π, and π = π. Then π= βπ ± πββπ , π and the solutions are βπ + ππ and βπ β ππ. 9. Write a quadratic equation in standard form such that βπ is its only solution. (π + π)π = π π π + πππ + ππ = π 10. Is it possible that the quadratic equation πππ + ππ + π = π has a positive real solution if π, π, and π are all positive real numbers? No. The solutions are MP.2 βπ+βππ βπππ ππ and βπββππ βπππ . If π is positive, the second one of these will be negative. ππ So, we need to think about whether or not the first one can be positive. If βπ + βππ β πππ > π, then βππ β πππ > π; so, ππ β πππ > ππ , and βπππ > π. This means that either π or π must be negative. So, if all three coefficients are positive, then there cannot be a positive solution to πππ + ππ + π = π. 11. Is it possible that the quadratic equation πππ + ππ + π = π has a positive real solution if π, π, and π are all negative real numbers? No. If π, π, and π are all negative, then βπ, βπ, and βπ are all positive. The solutions of πππ + ππ + π = π are the same as the solutions to βπππ β ππ β π = π, and by Problem 10, this equation has no positive real solution since it has all positive coefficients. Extension: 12. Show that if π > π. π, the solutions of πππ β ππ + π = π are not real numbers. We have π = π, π = βπ, and π = π; then ππ β πππ = (βπ)π β π β π β π = ππ β πππ. When the discriminant is negative, the solutions of the quadratic function are not real numbers. ππ β πππ = ππ β πππ π < π. π ππ β πππ < ππ β ππ(π. π) ππ β πππ < π π > π. π Thus, if π > π. π, then the discriminant is negative and the solutions of πππ β ππ + π = π are not real numbers. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 456 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 13. Let π be a real number, and consider the quadratic equation (π + π)ππ + πππ + π = π. a. Show that the discriminant of (π + π)ππ + πππ + π = π defines a quadratic function of π. The discriminant of a quadratic equation written in the form πππ + ππ + π = π is ππ β πππ. Here, π = π + π, π = ππ, and π = π. We get ππ β πππ = (ππ)π β π β (π + π) β π = ππππ β π(π + π) = ππππ β ππ β π. With π unknown, we can write π(π) = ππππ β ππ β π, which is a quadratic function of π. b. Find the zeros of the function in part (a), and make a sketch of its graph. If π(π) = π, then we have π = ππππ β ππ β π = πππ β π β π = πππ β ππ + π β π = ππ(π β π) + π(π β π) = (π β π)(ππ + π). Then, π β π = π or ππ + π = π. π π So, π = π or π = β . c. For what value of π are there two distinct real solutions to the original quadratic equation? The original quadratic equation has two distinct real solutions when the discriminant given by π(π) is π π positive. This occurs for all real numbers π such that π < β or π > π. d. For what value of π are there two complex solutions to the given quadratic equation? π π There are two complex solutions when π(π) < π. This occurs for all real numbers π such that β < π < π. e. For what value of π is there one solution to the given quadratic equation? π π There is one solution when π(π) = π. This occurs at π = β and π = π. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 457 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 14. We can develop two formulas that can help us find errors in calculated solutions of quadratic equations. a. Find a formula for the sum πΊ of the solutions of the quadratic equation πππ + ππ + π = π. The zeros of the quadratic equation are given by π = π βπ±βπ βπππ . Then ππ πΊ= βπ + βππ β πππ βπ β βππ β πππ + ππ ππ = βπ + βππ β πππ + βπ β βππ β πππ ππ βπ + βπ + βππ β πππ β βππ β πππ ππ βππ = ππ π =β . π = π π Thus, πΊ = β . b. Find a formula for the product πΉ of the solutions of the quadratic equation πππ + ππ + π = π. πΉ= βπ + βππ β πππ βπ β βππ β πππ β ππ ππ Note that the numerators differ only in that one is a sum, and one is a difference. The difference of squares formula applies where π = βπ and π = βππ β πππ. Then, π (βπ)π β (βππ β πππ) ππ β ππ ππ β ππ + πππ = πππ πππ = πππ π = . π πΉ= π π So, the product is πΉ = . c. June calculated the solutions π and βπ to the quadratic equation ππ β ππ + π = π. Do the formulas from parts (a) and (b) detect an error in her solutions? If not, determine if her solution is correct. The sum formula agrees with Juneβs calculations. From Juneβs zeros, π + βπ = π, and from the formula, πΊ= π = π. π However, the product formula does not agree with her calculations. From Juneβs zeros, π β βπ = βπ, and from the formula, πΉ= π = π. π Juneβs solutions are not correct: (π)π β π(π) + π = ππ β ππ + π = ππ; so, π is not a solution to this quadratic equation. Likewise, π β π + π = π, so π is also not a solution to this equation. Thus, the formulas caught her error. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 458 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 38 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II d. Paul calculated the solutions π β πβπ and π + πβπ to the quadratic equation ππ β ππ + π = π. Do the formulas from parts (a) and (b) detect an error in his solutions? If not, determine if his solutions are correct. In part (c), we calculated that πΉ = π and πΊ = π. From Paulβs zeros, πΊ = π + πβπ + π β πβπ = π, and for the product, πΉ = (π + πβπ) β (π β πβπ) π = ππ β (πβπ) = πβ π β π = ππ. This disagrees with the calculated version of πΉ. So, the formulas do find that he made an error. e. Joy calculated the solutions π β βπ and π + βπ to the quadratic equation ππ β ππ + π = π. Do the formulas from parts (a) and (b) detect an error in her solutions? If not, determine if her solutions are correct. Joyβs zeros will have the same sum as Paulβs, so πΊ = π, which agrees with the sum from the formula. For the product of her zeros we get πΉ = (π β βπ)(π + βπ) = πβπ = π, which agrees with the formulas. Checking her solutions in the original equation, we find π (π β βπ) β π(π β βπ) + π = (π β πβπ + π) β ππ + πβπ + π = π, π (π + βπ) β π(π + βπ) + π = (π + πβπ + π) β ππ β πβπ + π = π. Thus, Joy has correctly found the solutions of this quadratic equation. f. MP.2 If you find solutions to a quadratic equation that match the results from parts (a) and (b), does that mean your solutions are correct? Not necessarily. We only know that if the sum and product of the solutions do not match πΊ and πΉ, then we have not found a solution. Evidence suggests that if the sum and product of the solutions do match πΊ and πΉ, then we have found the correct solutions, but we do not know for sure until we check. g. Summarize the results of this exercise. π π For a quadratic equation of the form πππ + ππ + π = π, the sum of the solutions is given by πΊ = β , and the π π product of the solutions is given by πΉ = . So, multiplying and adding the calculated solutions will identify if we have made an error. Passing these checks, however, does not guarantee that the numbers we found are the correct solutions. Lesson 38: Complex Numbers as Solutions to Equations This work is derived from Eureka Math β’ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M1-TE-1.3.0-07.2015 459 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.