Lesson 4: Properties of Exponents and Radicals
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M3 Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA II Lesson 4: Properties of Exponents and Radicals Student Outcomes ο§ Students rewrite expressions involving radicals and rational exponents using the properties of exponents. Lesson Notes In Lesson 1, students reviewed the properties of exponents for integer exponents before establishing the meaning of the πth root of a positive real number and how it can be expressed as a rational exponent in Lesson 3. In Lesson 4, students extend properties of exponents that applied to expressions with integer exponents to expressions with rational 1 1 exponents. In each case, the notation π π specifically indicates the principal root (e.g., 22 is √2, as opposed to −√2). This lesson extends students’ thinking using the properties of radicals and the definitions from Lesson 3 so that they can see why it makes sense that the properties of exponents hold for any rational exponents (N-RN.A.1). Examples and exercises work to establish fluency with the properties of exponents when the exponents are rational numbers and emphasize rewriting expressions and evaluating expressions using the properties of exponents and radicals (N-RN.A.2). Classwork Opening (2 minutes) Students revisit the properties of square roots and cube roots studied in Module 1 to remind them that those were extended to any πth root in Lesson 3. So, they are now ready to verify that the properties of exponents hold for rational exponents. Scaffolding: ο§ Throughout the lesson, remind students of past properties of integer exponents and radicals either through an anchor chart posted on the wall or by recording relevant properties as they come up. Included is a short list of previous properties used in this module. ο§ For all real numbers π, π > 0, and all integers π, π for which the expressions are defined: π π ⋅ π π = π π+π Draw students’ attention to a chart posted prominently on the wall or to their notebooks where the properties of exponents and radicals are displayed, including those developed in Lesson 3. Remind students of the description of exponential expressions of the form which they make use of throughout the lesson: π ππ, Let π be any positive real number, and π, π be any integers with π > 0; then π π π π (π π )π = π ππ (ππ)π = ππ ⋅ π π 1 ππ Additionally, if π > 0, π −π = π π = √π π = ( √π) . 1 π √π = π π π π π √π ⋅ √π = √ππ π π π π π √π π = ( √π) = π π π √π π = ( √π) = π π . Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 62 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. M3 Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM ALGEBRA II Opening Exercise (5 minutes) These exercises briefly review content from Module 1 and the last lesson. Opening Exercise Write each exponential expression as a radical expression, and then use the definition and properties of radicals to write the resulting expression as an integer. π π ππ ⋅ ππ a. √π ⋅ √π = √ππ = π π π π ππ ⋅ ππ ⋅ ππ b. π π π π π π √π β √π β √π = √π β √π = √ππ = π π π πππ ⋅ ππ c. √ππ ⋅ √π = √ππ β π = √ππ = π π π π (πππ ) d. √ π√ππ = √π = π To transition from the Opening Exercise to Example 1, ask students to write parts (a) and (b) of the Opening Exercise in exponent form. Then, ask them to discuss with a partner whether or not it would be true in general that π π π π + π π π β ππ = π π ο§ 3 ο§ 3 3 How could you write √7 β √7 = 7 with rational exponents? How about √3 β √3 β √3 = 3? οΊ MP.3 & MP.8 for positive real numbers π where π, π, π, and π are integers with π ≠ 0 and π ≠ 0. 1 1 1 1 1 72 β 72 = 71 and 33 β 33 β 33 = 31 Based on these examples, does the exponent property π π β π π = π π+π appear to be valid when π and π are rational numbers? Explain how you know. οΊ Since the exponents on the left side of each statement add up to the exponents on the right side, it 1 1 2 appears to be true. However, the right side exponent was always 1. If we work with 83 β 83 = 83 and 3 3 3 3 write it in radical form, then √8 β √8 = √82 = √64. So, it may be true in general. Note that examples alone do not prove that a mathematical statement is always true. In the rest of this lesson, students make sense of these observations in general to extend the properties of exponents to rational numbers by applying the definition of the πth root of π and the properties of radicals introduced in Lesson 3. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 63 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Examples 1–3 (10 minutes) In the previous lesson, it was assumed that the exponent property π π π π = π π+π for positive real numbers π and 1 integers π and π would also hold for rational exponents when the exponents were of the form , where π was a positive π integer. This example helps students see that the property below makes sense for any rational exponent. π π π π + π, π π ⋅ ππ = π π where π, π, π, and π are integers with π > 0 and π > 0. Consider modeling Example 1 below and having students work with a partner on Example 2. Make sure students include a justification for each step in the problem. The following discussion questions can be used to guide students through Example 3. ο§ How can we write these expressions using radicals? 1 οΊ MP.7 & MP.8 ο§ π π How do we rewrite this expression in exponent form? π π In Lesson 3, we related radicals and rational exponents by π π = √π π . What makes Example 3 different from Examples 1 and 2? οΊ ο§ π The property of radicals that states √π β √π = √ππ for positive real numbers π and π and positive integer π, and the property of exponents that states π π β π π = π π+π for positive real numbers π and integers π and π οΊ ο§ π Which properties help us to write the expression as a single radical? οΊ ο§ π π In Lesson 3, we learned that π π = √π and π π = √π π for positive real numbers π and positive integers π and π. The exponents have different denominators, so when we write the expression in radical form, the roots π π π are not the same, and we cannot apply the property that √π β √π = √ππ . Can you think of a way to rewrite the problem so it looks more like Examples 1 and 2? οΊ We can write the exponents as equivalent fractions with the same denominator. Examples 1–3 π Write each expression in the form π π for positive real numbers π and integers π and π with π > π by applying the properties of radicals and the definition of πth root. 1. π π ππ β ππ By the definition of πth root, π π π π ππ β ππ = √π β √π. π = √π β π By the properties of radicals and properties of exponents π = √ ππ π = ππ The rational number π ππ ⋅ π ππ = π ππ π By the definition of π π π π π is equal to . Thus, π Scaffolding: ο§ Throughout the lesson, create parallel problems to demonstrate that these problems work with numerical values as well. ο§ For example, in part (a), substitute 4 for π. ο§ In part (b), substitute a perfect cube such as 8 or 27 for π. . Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 64 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II π π ππ β ππ 2. π π π π π ππ β ππ = √π β √ππ By the definition of ππ and π π = √π β π π = π ( ) π By the properties of radicals and properties of exponents π √π π π π = ππ By the definition of π π Thus, π π π π π ππ β ππ = ππ . ππ β ππ 3. Write the exponents as equivalent fractions with the same denominator. π π π ππ ππ β ππ = πππ β πππ Rewrite in radical form. ππ ππ = √ππ β √πππ Rewrite as a single radical expression. ππ = √ππ β πππ ππ = √πππ Rewrite in exponent form using the definition. ππ = πππ Thus, π π ππ ππ β ππ = πππ . ο§ Now, add the exponents in each example. What is 1 οΊ ο§ 4 + 1 4 1 1 = , 2 3 + 4 3 = 5 3 , and 1 5 + 3 4 = 1 4 1 1 4 3 + ? 4 1 3 5 + ? 3 + ? 4 19 20 What do you notice about these sums and the value of the exponent when we rewrote each expression? οΊ The sum of the exponents was equal to the exponent of the answer. Based on these examples, particularly the last one, it seems reasonable to extend the properties of exponents to hold when the exponents are any rational number. Thus, the following property can be stated: ο§ 1 1 For any integers π, π, π and π, with π > 0 and π > 0, and any real numbers π so that π π and π π are defined, π π π π + π π π β ππ = π π . Have students copy this property into their notes along with the ones listed below. Also, consider writing these properties on a piece of chart paper and displaying them in the classroom. These properties are listed in the Lesson Summary. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 65 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II In a similar fashion, the other properties of exponents can be extended to hold for any rational exponents as well. ο§ 1 1 1 For any integers π, π, π, and π, with π > 0 and π > 0, and any real numbers π and π so that ππ , π π , and π π are defined, π π π π = √π π 1 π (π π ) = π 1 (π π )π = π π π π (ππ) π = π π ⋅ π π π π π (π π ) ππ = π ππ 1 π π− π = π. ππ At this point, consider having the class look at the Opening Exercise again and asking them which property could be used to simplify each problem. For advanced learners, a derivation of the property explored in Example 1 is provided below. π π Rewrite π π and π π as equivalent exponential expressions in which the exponents have the same denominator, and π apply the definition of π π as the πth root. π By the definition of π π and then using properties of algebra, students can rewrite the exponent to be π ππ ⋅ π ππ = π π π + . π ππ ππ π ππ π ππ ππ ππ = √π ππ ⋅ √π ππ ππ = √π ππ ⋅ π ππ ππ = √π ππ+ππ =π ππ+ππ ππ ππ ππ + ππ = π ππ π π + π = ππ Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 66 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exercises 1–4 (6 minutes) Scaffolding: Have students work with a partner or in small groups to rewrite expressions with rational exponents using the properties presented above. As students work, emphasize that it is not necessary to write these expressions using radicals since it has just been established that the properties of exponents hold for rational numbers. In the last two exercises, students have to use their knowledge of radicals to rewrite the answers without exponents. Exercises 1–4 Write each expression in the form your answer without exponents. 1. π π ππ. If a numeric expression is a rational number, then write π ππ β ππ π π π ππ + π = ππ 2. ο§ When students get to these exercises, it may be necessary to remind them that it is often easier to π π π rewrite π π as ( √π ) when evaluating radical expressions. ο§ Provide a scientific calculator for students who struggle with arithmetic, but encourage them to use the radical and exponent properties to show the steps. π π π (π−π ) π π π π− π β π = π−ππ 3. π π πππ β πππ π π ππ πππ + π = ππ π ππ π = (√ππ) = πππ = ππππ π 4. π π π ( ππ ) π π ππ π ππ ( π) = π π π π π (√π) ππ ππ = ππ ππβπππ = ππ = Example 4 (5 minutes) ο§ We can rewrite radical expressions using properties of exponents. There are other methods for rewriting radical expressions, but this example models using the properties of exponents. Often, textbooks and exams give directions to simplify an expression, which is vague unless we specify what it means. We want students to develop fluency in applying the properties, so the directions here say to rewrite in a specific fashion. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 67 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Example 4 Rewrite the radical expression √ππππ ππ ππ so that no perfect square factors remain inside the radical. π √ππ ⋅ ππ ⋅ ππ ⋅ ππ = (ππ ⋅ π ⋅ ππ ⋅ ππ ⋅ ππ )π π π π π π = ππ ⋅ ππ ⋅ ππ ⋅ ππ ⋅ ππ π π = π ⋅ ππ ⋅ ππ+π ⋅ ππ ⋅ π π = πππ ππ π ⋅ (ππ)π = πππ ππ π√ππ ο§ Although this process may seem drawn out, once it has been practiced, most of the steps can be internalized, and expressions are quickly rewritten using this technique. Exercise 5 (5 minutes) For the values π₯ = 50, π¦ = 12, and π§ = 3, the expressions in Exercises 5(a) and (b) are difficult to evaluate. Students need to rewrite these expressions in a simpler form by minimizing fractions in the exponents before attempting to evaluate them. Do not allow calculators to be used on these exercises. Exercise 5 5. Use the definition of rational exponents and properties of exponents to rewrite each expression with rational exponents containing as few fractions as possible. Then, evaluate each resulting expression for π = ππ, π = ππ, and π = π. a. √πππ ππ π π π √πππ ππ = ππ ππ ππ π = πππ ⋅ (ππ)π π Evaluating, we get π(ππ)(ππ)(π ⋅ ππ)π = πππ ⋅ ππ ⋅ ππ = ππβπππ. b. π √ππππ ππ π π π π π √ππππ ππ = πππ ⋅ ππ ⋅ ππ ⋅ ππ π = πππ ⋅ (ππππ )π π π Evaluating, we get π(ππ)π (π ⋅ ππ ⋅ ππ )π = π(πππ)(πππ)π = π ⋅ πππ ⋅ π = ππππ. Exercise 6 (5 minutes) This exercise reminds students that rational numbers can be represented in decimal form and gives them a chance to work on their numeracy skills. Students should work on this exercise with a partner or in their groups to encourage dialogue and debate. Have a few students demonstrate their results to the entire class. There is more than one possible approach, so when debriefing, try to share different approaches that show varied reasoning. Conclude with one or two strong arguments. Students can confirm their reasoning using a calculator. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 68 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exercise 6 6. Order these numbers from smallest to largest. Explain your reasoning. πππ.π ππ.π πππ.π The number πππ.π is between πππ and π, πππ. We can rewrite πππ.π = (ππ )π.π , which is πππ, so πππ.π = πππ = ππππ. MP.3 The number πππ.π is between ππ and π, πππ. We can rewrite πππ.π = (ππ )π.π , which is ππ , so πππ.π = ππ = ππ. The number ππ.π is larger than ππ , so ππ.π is larger than πππ. Thus, πππ.π is clearly the smallest number, but we need to determine if ππ.π is greater than or less than π, πππ. To do this, we know that ππ.π = ππ+π.π = ππ β ππ.π . This means that ππ.π > ππ β ππ.π , and ππ β ππ.π = πππ β π, which is greater than π, πππ. Thus, the numbers in order from smallest to largest are πππ.π , πππ.π , and ππ.π . Closing (2 minutes) Have students summarize the definition and properties of rational exponents and any important ideas from the lesson by creating a list of what they have learned so far about the properties of exponents and radicals. Circulate around the classroom to informally assess understanding. Reinforce the properties of exponents listed below. Lesson Summary The properties of exponents developed in Grade 8 for integer exponents extend to rational exponents. π π π That is, for any integers π, π, π, and π, with π > π and π > π, and any real numbers π and π so that ππ , ππ , and ππ are defined, we have the following properties of exponents: π π π π + π 1. π π β ππ = π π 2. π π = √ππ 3. (ππ ) = π 4. (ππ )π = π 5. (ππ) π = π π ⋅ π π 6. (π π ) = π ππ 7. π− π = π π π π π π π π π π π π ππ π π . ππ Exit Ticket (5 minutes) Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 69 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Name Date Lesson 4: Properties of Exponents and Radicals Exit Ticket 11 2 1. Find the exact value of 910 β 95 without using a calculator. 2. Justify that √8 β √8 = √16 using the properties of exponents in at least two different ways. 3 3 Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 70 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II Exit Ticket Sample Solutions 1. ππ π Find the exact value of πππ β ππ without using a calculator. ππ π ππ π πππ β ππ = πππ+π ππ = πππ π = ππ π π = (√π) = ππ 2. π π Justify that √π β √π = √ππ using the properties of exponents in at least two different ways. π π π π π πππ = (π β π)π ππ β ππ = ππ π π π = (ππ )π = ππ β ππ = ππ =πβπ = π π ππ π = ππ β ππ π π π = √π β √π = (ππ )π = √ππ Problem Set Sample Solutions 1. Evaluate each expression for π = ππ and π = ππ. a. π √π√π b. π (π β π β π)π = ππππ √ππ β √ππ = π β π = ππ c. π π ( √π + π√π) π π (π √π√π) d. π π π−π + ππ (π + π β π)π = πππ π π π π + (√ππ) = (√ππ) e. π π −π (π−π β ππ ) f. −π π π ( β πππ) = π πππ Lesson 4: π π π π π + πππ = πππ π π π −π (π−π − ππ ) −π π π πππ −π π ( − β πππ) = (− ) =− π π π πππ Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 71 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 2. Rewrite each expression so that each term is in the form πππ , where π is a real number, π is a positive real number, and π is a rational number. π π π−π β ππ a. π π ππ−π π−π π −π π ππππ c. π πππ β ππ−π b. (πππ ) d. πππ π −π π π π π ππ−π π π π π± π (πππ − ) e. π πππ − √ π ππ−π ππ ππ ππ−π π π πππ −ππ −ππ π √π β √−πππ β √ππππ g. π f. h. √π π −πππ π πππ π π − ππ − πππ √π−ππ−π πππ i. π −π π −π π π− π π π 3. π Show that (√π + √π) is not equal to ππ + ππ when π = π and π = ππ. π When π = π and π = ππ, the two expressions are (√π + √ππ) and π + ππ. The first expression simplifies to ππ, and the second simplifies to ππ. The two expressions are not equal. 4. π −π π Show that (π± π + π² π ) is not equal to π π ππ + π π when π = π and π = ππ. ππ −π When π = π and π = ππ, the two expressions are (√π + √ππ) and second one is 5. π π + π π = π π √π + π π . The first expression is , and the π √ππ . The two expressions are not equal. ππ From these numbers, select (a) one that is negative, (b) one that is irrational, (c) one that is not a real number, and (d) one that is a perfect square: π π π π π π π π π π − ππ β ππ , πππ β ππππ , πππ − πππ , and (π π − ππ ) . π π π π The first number, ππ β ππ , is irrational; the second number, πππ β ππππ , is a perfect square; the third number, π π π π π π πππ − πππ , is negative; and the last number, (π−π − ππ ) , is not a real number. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 72 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 6. π π π Show that for any rational number π, the expression ππ β ππ+π β ( ) is equal to π. ππ β πππ+π β π−ππ = ππ = π 7. Let π be any rational number. Express each answer as a power of ππ. a. Multiply πππ by ππ. πππ β ππ = πππ+π b. Multiply √ππ by πππ . π π πππ β πππ = πππ+π c. Square πππ . (πππ )π = ππππ d. Divide πππ β πππ by ππππ . πππ β πππ = πππ+π−ππ = πππ−π ππππ e. Show that πππ = ππ β πππ − πππ+π . ππ β πππ − πππ+π = ππ β πππ − ππ β πππ = πππ (ππ − ππ) = πππ β π = πππ 8. Rewrite each of the following radical expressions as an equivalent exponential expression in which each variable occurs no more than once. a. √πππ π π π π √πππ π = ππ ππ (ππ)π π = ππ ⋅ (ππ)π π π = ππ π ππ b. π √ππππ πππ ππ π π √ππππ πππ ππ = (ππ ⋅ π ⋅ ππ ⋅ πππ ⋅ ππ )π π π π ππ π = πππ ⋅ ππ ⋅ ππ ⋅ π π ⋅ ππ π π π = π β ππ β ππ β ππ β ππ 9. Use properties of exponents to find two integers that are upper and lower estimates of the value of ππ.π . ππ.π < ππ.π < ππ ππ.π = ππ = π and ππ = ππ, so π < ππ.π < ππ. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 73 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. Lesson 4 NYS COMMON CORE MATHEMATICS CURRICULUM M3 ALGEBRA II 10. Use properties of exponents to find two integers that are upper and lower estimates of the value of ππ.π . π ππ < ππ.π < ππ+π π π π ππ = ππ and ππ = π, so ππ+π = ππ ⋅ ππ = πππ. Thus, ππ < ππ.π < πππ. 11. Kepler’s third law of planetary motion relates the average distance, π, of a planet from the sun to the time, π, it takes the planet to complete one full orbit around the sun according to the equation ππ = ππ . When the time, π, is measured in Earth years, the distance, π, is measured in astronomical units (AUs). (One AU is equal to the average distance from Earth to the sun.) a. Find an equation for π in terms of π and an equation for π in terms of π. ππ = ππ π π = ππ π π = ππ b. Venus takes about π. πππ Earth year to orbit the sun. What is its average distance from the sun? π Because π = (π. πππ)π ≈ π. πππ, the average distance from Venus to the sun is π. πππ ππ. c. Mercury is an average distance of π. πππ AU from the sun. About how long is its orbit in Earth years? π Because π = (π. πππ)π ≈ π. πππ, the length of Mercury's orbit is approximately π. πππ Earth years. Lesson 4: Properties of Exponents and Radicals This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org This file derived from ALG II-M3-TE-1.3.0-08.2015 74 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
