Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Lesson 29: Solving Radical Equations Student Outcomes ο§ Students develop facility in solving radical equations. Lesson Notes In the previous lesson, students were introduced to the notion of solving radical equations and checking for extraneous solutions (A-REI.A.2). Students continue this work by looking at radical equations that contain variables on both sides. The main point to stress to students is that radical equations become polynomial equations through exponentiation. So we really have not left the notion of polynomials that have been studied throughout this module. This lesson also provides opportunities to emphasize MP.7 (look for and make use of structure). Classwork Discussion (5 minutes) Before beginning the lesson, remind students of past experiences by providing the following scenario, which illustrates a case when an operation performed to both sides of an equation has changed the set of solutions. Carlos and Andrea were solving the equation π₯ 2 + 2π₯ = 0. Andrea says that there are two solutions, 0 and β2. Carlos says the only solution is β2 because he divided both sides by π₯ and got π₯ + 2 = 0. Who is correct and why? ο§ Do both 0 and β2 satisfy the original equation? οΊ ο§ What happened when Carlos divided both sides of the equation by π₯? οΊ ο§ Yes. If we replace π₯ with either 0 or β2, the answer is 0. He changed the solutions from 0 and β2 to simply β2. He lost one solution to the equation. What does this say about the solution of equations after we have performed algebraic operations on both sides? οΊ Performing algebraic steps may alter the set of solutions to the original equation. Now, Carlos and Andrea are solving the equation βπ₯ = β3. Andrea says the solution is 9 because she squared both sides and got π₯ = 9. Carlos says there is no solution. Who is correct? Why? ο§ Was Andrea correct to square both sides? οΊ MP.3 ο§ Yes. To eliminate a radical from an equation, we raise both sides to an exponent. Is she correct that the solution is 9? οΊ No. Carlos is correct. If we let π₯ = 9, then we get β9 = 3, and 3 β β3, so 9 is not a solution. Lesson 29: Solving Radical 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 Scaffolding ο§ Use several examples to illustrate that if π > 0, then an equation of the form βπ₯ = βπ will not have a solution (e.g., βπ₯ = β4, βπ₯ = β5). ο§ Extension: Write an equation that has an extraneous solution of π₯ = 50. 323 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II ο§ What is the danger in squaring both sides of an equation? οΊ ο§ Because of this danger, what is the final essential step of solving a radical equation? οΊ ο§ It sometimes produces an equation whose solution set is not equivalent to that of the original equation. If both sides of βπ₯ = β3 are squared, the equation π₯ = 9 is produced, but 9 is not a solution to the original equation. The original equation has no solution. Checking the solution or solutions to ensure that an extraneous solution was not produced by the step of squaring both sides. How could we have predicted that the equation would have no solution? οΊ The square root of a number is never equal to a negative value, so there is no π₯-value so that βπ₯ = β3. Example 1 (5 minutes) While this problem is difficult, students should attempt to solve it on their own first, by applying their understandings of MP.1 radicals. Students should be asked to verify the solution they come up with and describe their solution method. Discuss Example 1 as a class once they have worked on it individually. Example 1 Solve the equation π = π + βπ. π β π = βπ (π β π)π = βπ π ππ β πππ + ππ = π ππ β πππ + ππ = π (π β π)(π β π) = π The solutions are π and π. Check π = π: Check π = π: π + βπ = π + π = ππ π + βπ = π + π = π π β ππ So, π is an extraneous solution. The only valid solution is π. ο§ How does this equation differ from the ones from yesterdayβs lesson? οΊ ο§ Explain how you were able to determine the solution to the equation above. οΊ ο§ Not really; we still eliminated the radical by squaring both sides. What type of equation were we left with after squaring both sides? οΊ ο§ Isolate the radical and square both sides. Solve the resulting equation. Did that change the way in which the equation was solved? οΊ ο§ There are two π₯βs; one inside and one outside of the radical. A quadratic polynomial equation Why did 9 fail to work as a solution? οΊ The square root of 9 takes only the positive value of 3. Lesson 29: Solving Radical 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 324 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exercises 1β4 (13 minutes) Allow students time to work the problems independently and then pair up to compare solutions. Use this time to informally assess student understanding by examining their work. Display student responses, making sure that students checked for extraneous solutions. Exercises 1β4 Solve. ππ = π + πβπ 1. 2. The only solution is π. π = πβπ β π The two solutions are π and π. π Note that is an extraneous solution. π βπ + π = π β π 3. 4. The only solution is π. βππ + π + πβπ β π = π There are no solutions. Note that βπ is an extraneous solution. ο§ When solving Exercise 1, what solutions did you find? What happened when you checked these solutions? οΊ ο§ 9 9 No. Both solutions satisfied the original equation. Looking at Exercise 4, could we have predicted that there would be no solution? οΊ MP.7 1 Did Exercise 2 have any extraneous solutions? οΊ ο§ 1 The solutions found were and 1. Only 1 satisfies the original equation, so is an extraneous solution. Yes. The only way the two square roots could add to zero would be if both of them produced a zero, 7 3 meaning that 3π₯ + 7 = 0 and π₯ β 8 = 0. Since π₯ cannot be both β and 8, both radicals cannot be simultaneously zero. Thus, at least one of the square roots will be positive, and they cannot sum to zero. Example 2 (5 minutes) What do we do when there is no way to isolate the radical? What is going to be the easiest way to square both sides? Give students time to work on Example 2 MP.7 independently. Point out that even though we had to square both sides twice, we were still able to rewrite the equation as a polynomial. Example 2 Solve the equation βπ + βπ + π = π. βπ + π = π β βπ What if we had squared both sides of the equation as it was presented? Have early finishers work out the solution this way and share with the class. Check: π (βπ + π) = (π β βπ)π π + π = π β πβπ + π π = βπ Scaffolding: βπ + βπ + π = π + π = π So the solution is π. π=π Lesson 29: Solving Radical 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 325 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exercises 5β6 (7 minutes) Allow students time to work the problems independently and then pair up to compare solutions. Circulate to assess understanding. Consider targeted instruction with a small group of students while others work independently. Display student responses, making sure that students check for extraneous solutions. Exercises 5β6 Solve the following equations. βπ β π + βπ + π = π 5. 6. π + βπ = βπ + ππ πππ π Closing (5 minutes) Ask students to respond to these prompts in writing or with a partner. Use these responses to informally assess their understanding of the lesson. ο§ How did these equations differ from the equations seen in the previous lesson? οΊ ο§ How were they similar to the equations from the previous lesson? οΊ ο§ Most of them contained variables on both sides of the equation or a variable outside of the radical. They were solved using the same process of squaring both sides. Even though they were more complicated, the equations could still be rewritten as a polynomial equation and solved using the same process seen throughout this module. Give an example where ππ = π π but π β π. οΊ We know that (β3)2 = 32 but β3 β 3. Lesson Summary If π = π and π is an integer, then ππ = ππ . However, the converse is not necessarily true. The statement ππ = ππ does not imply that π = π. Therefore, it is necessary to check for extraneous solutions when both sides of an equation are raised to an exponent. Exit Ticket (5 minutes) Lesson 29: Solving Radical 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 326 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Name Date Lesson 29: Solving Radical Equations Exit Ticket 1. Solve β2π₯ + 15 = π₯ + 6. Verify the solution(s). 2. Explain why it is necessary to check the solutions to a radical equation. Lesson 29: Solving Radical 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 327 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II Exit Ticket Sample Solutions 1. Solve βππ + ππ = π + π. Verify the solution(s). ππ + ππ = ππ + πππ + ππ π = ππ + πππ + ππ π = (π + π)(π + π) The solutions are βπ and βπ. Check π = βπ: Check π = βπ: βπ(βπ) + ππ = βπ = π βπ(βπ) + ππ = βπ = π βπ + π = π βπ + π = βπ So, βπ is a valid solution. Since βπ β π, we see that βπ is an extraneous solution. Therefore, the only solution to the original equation is βπ. 2. Explain why it is necessary to check the solutions to a radical equation. Raising both sides of an equation to a power can produce an equation whose solution set is not equivalent to that of the original equation. In the problem above, π = βπ does not satisfy the equation. Problem Set Sample Solutions Solve. 1. βππ β π β βπ + π = π 2. ππ 3. No solution βπ β π β βπ + π = π 4. βπ + π = π β βπ 6. ππ ππ 7. βπ + π = π + βπ No solution βπ + π = βππ + π β π 8. βππ + π = π β π π π 9. βππ β π β βπ + π = π ππ No solution 5. βππ β π + βπ + π = π βπ + ππ + βπ = π 10. πβπ = π β βππ β π π π π Lesson 29: Solving Radical 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 328 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 29 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA II 11. ππ = βππ β π 12. βππ β π = π β ππ π π π π 13. π + π = πβπ β π 14. βππ β π + βππ β ππ = π π π 15. π = πβπ β π + π 16. π β π = βππ β ππ π, π π, π 17. Consider the right triangle π¨π©πͺ shown to the right, with π¨π© = π and π©πͺ = π. a. Write an expression for the length of the hypotenuse in terms of π. π¨πͺ = βππ + ππ b. Find the value of π for which π¨πͺ β π¨π© = π. The solutions to the mathematical equation βππ + ππ β π = π are βππ and ππ. Since lengths must be positive, βππ is an extraneous solution, and π = ππ. 18. Consider the triangle π¨π©πͺ shown to the right where π¨π« = π«πͺ, and Μ Μ Μ Μ Μ π©π« is the altitude of the triangle. a. If the length of Μ Μ Μ Μ Μ π©π« is π ππ¦, and the length of Μ Μ Μ Μ π¨πͺ is ππ ππ¦, write Μ Μ Μ Μ in terms of π. Μ Μ Μ Μ and π©πͺ an expression for the lengths of π¨π© π¨π© = π©πͺ = βππ + ππ ππ¦ b. Write an expression for the perimeter of β³ π¨π©πͺ in terms of π. (πβππ + ππ + ππ) ππ¦ c. Find the value of π for which the perimeter of β³ π¨π©πͺ is equal to ππ ππ¦. βππ ππ¦ Lesson 29: Solving Radical 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 329 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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