\n50 = 2 \u00d7 52<\/sup>
\nSince 50 = 2 \u00d7 52<\/sup>, then \\(\\sqrt{50}\\) = \\(\\sqrt{2 \\times 5^{2}}\\). We can rewrite \\(\\sqrt{50}\\) as a product of its factors:
\n\\(\\sqrt{50}\\) = \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5^{2}}\\).<\/p>\n
\u2192 Obviously, 52<\/sup> is a perfect square. Therefore, \\(\\sqrt{5^{2}}\\) = 5, so \\(\\sqrt{50}\\) = 5 \u00d7 \\(\\sqrt{2}\\) = 5\\(\\sqrt{2}\\). Since \\(\\sqrt{2}\\) is not a perfect square, we leave it as it is. We have simplified this expression as much as possible because there are no other perfect square factors remaining in the square root.<\/p>\n \u2192 The number \\(\\sqrt{50}\\) is said to be in its simplified form when all perfect square factors have been simplified. Therefore, 5\\(\\sqrt{2}\\) is the simplified form of \\(\\sqrt{50}\\). Example 2. \u2192 What are the factors of 28? \u2192 Obviously, 22<\/sup> is a perfect square. Therefore, \\(\\sqrt{2^{2}}\\) = 2, and \\(\\sqrt{28}\\) = 2 \u00d7 \\(\\sqrt{7}\\) = 2\\(\\sqrt{7}\\). Since \\(\\sqrt{7}\\) is not a perfect square, we leave it as it is.<\/p>\n \u2192 The number \\(\\sqrt{28}\\) is said to be in its simplified form when all perfect square factors have been simplified. Therefore, 2\\(\\sqrt{7}\\) is the simplified form of \\(\\sqrt{28}\\).<\/p>\n Example 3. \u2192 Is the number 128 a perfect square? Explain. \u2192 What are the factors of 128? Example 4. \u2192 Is the number 288 a perfect square? Explain. \u2192 Then, \\(\\sqrt{288}\\) is equivalent to \u2192 What does this simplify to? Opening Exercise ii. What does 4 \u00d7 4 equal? iii. Does \\(\\sqrt{16}\\) = \\(\\sqrt{4 \\times 4}\\)? b. ii. What does 6 \u00d7 6 equal? iii. Does \\(\\sqrt{36}\\) = \\(\\sqrt{6 \\times 6}\\)? c. ii. What does 11 \u00d7 11 equal? iii. Does \\(\\sqrt{121}\\) = \\(\\sqrt{11 \\times 11}\\)? d. ii. What does 9 \u00d7 9 equal? iii. Does \\(\\sqrt{81}\\) = \\(\\sqrt{9 \\times 9}\\)? e. Rewrite \\(\\sqrt{20}\\) using at least one perfect square factor. f. Rewrite \\(\\sqrt{28}\\) using at least one perfect square factor. Exercises 1\u20134 Exercise 1. Exercise 2. Exercise 3. Exercise 4. Exercises 5\u20138<\/p>\n Exercise 5. Exercise 6. Exercise 7. Exercise 8. Simplify each of the square roots in Problems 1\u20135 as much as possible.<\/p>\n Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. Question 9. Question 10. Question 11. Simplify the square roots as much as possible. Question 2. Question 3. Question 4. Engage NY Eureka Math 8th Grade Module 7 Lesson 4 Answer Key Eureka Math Grade 8 Module 7 Lesson 4 Example Answer Key Example 1. Simplify the square root as much as possible. = Answer: \u2192 Is the number 50 a perfect square? Explain. The number 50 is not a perfect square because there is … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[6],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82230"}],"collection":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/comments?post=82230"}],"version-history":[{"count":1,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82230\/revisions"}],"predecessor-version":[{"id":137304,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/posts\/82230\/revisions\/137304"}],"wp:attachment":[{"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/media?parent=82230"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/categories?post=82230"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/ccssanswers.com\/wp-json\/wp\/v2\/tags?post=82230"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\nNow that we know \\(\\sqrt{50}\\) can be expressed as a product of its factors, we also know that we can multiply expressions containing square roots. For example, if we are given \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5^{2}}\\), we can rewrite the expression as \u221a(2 \u00d7 52<\/sup> ) = \\(\\sqrt{50}\\).<\/p>\n
\nSimplify the square root as much as possible.
\n\\(\\sqrt{28}\\) =
\nAnswer:
\n\u2192 Is the number 28 a perfect square? Explain.
\nThe number 28 is not a perfect square because there is no integer squared that equals 28.<\/p>\n
\n28 = 22<\/sup> \u00d7 7
\nSince 28 = 22<\/sup> \u00d7 7, then \\(\\sqrt{28}\\) = \\(\\sqrt{2^{2} \\times 7}\\). We can rewrite \\(\\sqrt{28}\\) as a product of its factors:
\n\\(\\sqrt{28}\\) = \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{7}\\).<\/p>\n
\nSimplify the square root as much as possible.
\n\\(\\sqrt{128}\\) =
\nAnswer:
\n\u2192 In this example, students may or may not recognize 128 as 64 \u00d7 2. The work below assumes that they do not. Consider showing students the solution below, as well as this alternative solution:
\n\\(\\sqrt{128}\\) = \\(\\sqrt{64 \\times 2}\\) = \\(\\sqrt{64}\\) \u00d7 \\(\\sqrt{2}\\) = 8 \u00d7 \\(\\sqrt{2}\\) = 8\\(\\sqrt{2}\\).<\/p>\n
\nThe number 128 is not a perfect square because there is no integer squared that equals 128.<\/p>\n
\n128 = 27<\/sup>
\n\u2192 Since 128 = 27<\/sup>, then \\(\\sqrt{128}\\) = \u221a(27<\/sup> ). We know that we can simplify perfect squares, so we can rewrite 27<\/sup> as 22<\/sup> \u00d7 22<\/sup> \u00d7 22<\/sup> \u00d7 2 because of what we know about the laws of exponents. Then,
\n\\(\\sqrt{128}\\) = \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\).
\nEach 22<\/sup> is a perfect square. Therefore, \\(\\sqrt{128}\\) = 2 \u00d7 2 \u00d7 2 \u00d7 \\(\\sqrt{2}\\) = 8\\(\\sqrt{2}\\).<\/p>\n
\nSimplify the square root as much as possible.
\n\\(\\sqrt{288}\\) =
\nAnswer:
\nIn this example, students may or may not recognize 288 as 144 \u00d7 2. The work below assumes that they do not. Consider showing students the solution below, as well as this alternative solution:
\n\\(\\sqrt{288}\\) = \\(\\sqrt{144 \\times 2}\\) = \\(\\sqrt{144}\\) \u00d7 \\(\\sqrt{2}\\) = 12 \u00d7 \\(\\sqrt{2}\\) = 12\\(\\sqrt{2}\\).<\/p>\n
\nThe number 288 is not a perfect square because there is no integer squared that equals 288.
\n\u2192 What are the factors of 288?
\n288 = 25<\/sup> \u00d7 32<\/sup>
\nSince 288 = 25<\/sup> \u00d7 32<\/sup>, then \\(\\sqrt{288}\\) = \u221a(25<\/sup> \u00d7 32<\/sup> ). What do we do next?
\nUse the laws of exponents to rewrite 25<\/sup> as 22<\/sup> \u00d7 22<\/sup> \u00d7 2.<\/p>\n
\n\\(\\sqrt{288}\\) = \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3^{2}}\\).<\/p>\n
\n\\(\\sqrt{288}\\) = \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3^{2}}\\)) = \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{3^{2}}\\) ) \u00d7 \\(\\sqrt{2}\\) = 2 \u00d7 2 \u00d7 3 \u00d7 \\(\\sqrt{2}\\) = 12\\(\\sqrt{2}\\)<\/p>\nEureka Math Grade 8 Module 7 Lesson 4 Exercise Answer Key<\/h3>\n
\na.
\ni. What does \\(\\sqrt{16}\\) equal?
\nAnswer:
\n4<\/p>\n
\nAnswer:
\n16<\/p>\n
\nAnswer:
\nYes<\/p>\n
\ni. What does \\(\\sqrt{36}\\) equal?
\nAnswer:
\n6<\/p>\n
\nAnswer:
\n36<\/p>\n
\nAnswer:
\nYes<\/p>\n
\ni. What does \\(\\sqrt{121}\\) equal?
\nAnswer:
\n11<\/p>\n
\nAnswer:
\n121<\/p>\n
\nYes<\/p>\n
\ni. What does \\(\\sqrt{81}\\) equal?
\nAnswer:
\n9<\/p>\n
\nAnswer:
\n81<\/p>\n
\nAnswer:
\nYes<\/p>\n
\nAnswer:
\n\\(\\sqrt{20}\\) = \\(\\sqrt{4 \\times 5}\\)<\/p>\n
\nAnswer:
\n\\(\\sqrt{28}\\) = \\(\\sqrt{4 \\times 7}\\)<\/p>\n
\nSimplify the square roots as much as possible.<\/p>\n
\n\\(\\sqrt{18}\\)
\nAnswer:
\n\\(\\sqrt{18}\\) = \\(\\sqrt{2 \\times 3^{2}}\\)
\n= \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3^{2}}\\)
\n= 3\\(\\sqrt{2}\\)<\/p>\n
\n\\(\\sqrt{44}\\)
\nAnswer:
\n\\(\\sqrt{44}\\) = \\(\\sqrt{2^{2} \\times 11}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{11}\\)
\n= 2\\(\\sqrt{11}\\)<\/p>\n
\n\\(\\sqrt{169}\\)
\nAnswer:
\n\\(\\sqrt{169}\\) = \\(\\sqrt{13^{2}}\\)
\n= 13<\/p>\n
\n\\(\\sqrt{75}\\)
\nAnswer:
\n\\(\\sqrt{75}\\) = \\(\\sqrt{3 \\times 5^{2}}\\)
\n= \\(\\sqrt{3}\\) \u00d7 \\(\\sqrt{5^{2}}\\)
\n= 5\\(\\sqrt{3}\\)<\/p>\n
\nSimplify \\(\\sqrt{108}\\).
\nAnswer:
\n\\(\\sqrt{108}\\) = \\(\\sqrt{2^{2} \\times 3^{3}}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{3}\\)
\n= 2 \u00d7 3\\(\\sqrt{3}\\)
\n= 6\\(\\sqrt{3}\\)<\/p>\n
\nSimplify \\(\\sqrt{250}\\).
\nAnswer:
\n\\(\\sqrt{250}\\) = \\(\\sqrt{2 \\times 5^{3}}\\)
\n= \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5^{2}}\\) \u00d7 \\(\\sqrt{5}\\)
\n= 5\\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5}\\)
\n= 5\\(\\sqrt{10}\\)<\/p>\n
\nSimplify \\(\\sqrt{200}\\).
\nAnswer:
\n\\(\\sqrt{200}\\) = \\(\\sqrt{2^{3} \\times 5^{2}}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5^{2}}\\)
\n= 2 \u00d7 5\\(\\sqrt{2}\\)
\n= 10\\(\\sqrt{2}\\)<\/p>\n
\nSimplify \\(\\sqrt{504}\\).
\nAnswer:
\n\\(\\sqrt{504}\\) = \\(\\sqrt{2^{3} \\times 3^{2} \\times 7}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{7}\\)
\n= 2 \u00d7 3 \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{7}\\)
\n= 6\\(\\sqrt{14}\\)<\/p>\nEureka Math Grade 8 Module 7 Lesson 4 Problem Set Answer Key<\/h3>\n
\n\\(\\sqrt{98}\\)
\n\\(\\sqrt{98}\\) = \\(\\sqrt{2 \\times 7^{2}}\\)
\n= \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{7^{2}}\\)
\n= 7\\(\\sqrt{2}\\)<\/p>\n
\n\\(\\sqrt{54}\\)
\n\\(\\sqrt{54}\\) = \\(\\sqrt{2 \\times 3^{3}}\\)
\n= \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3}\\) \u00d7 \\(\\sqrt{3^{2}}\\)
\n= 3\\(\\sqrt{6}\\)<\/p>\n
\n\\(\\sqrt{144}\\)
\n\\(\\sqrt{144}\\) = \\(\\sqrt{12^{2}}\\)
\n= 12<\/p>\n
\n\\(\\sqrt{512}\\)
\n\\(\\sqrt{512}\\) = \\(\\sqrt{2^{9}}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\)
\n= 2 \u00d7 2 \u00d7 2 \u00d7 2\\(\\sqrt{2}\\)
\n= 16\\(\\sqrt{2}\\)<\/p>\n
\n\\(\\sqrt{756}\\)
\n\\(\\sqrt{756}\\) = \\(\\sqrt{2^{2} \\times 3^{3} \\times 7}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{3}\\) \u00d7 \\(\\sqrt{7}\\)
\n= 2 \u00d7 3 \u00d7 \\(\\sqrt{21}\\)
\n= 6\\(\\sqrt{21}\\)<\/p>\n
\nWhat is the length of the unknown side of the right triangle? Simplify your answer, if possible.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-4-Problem-Set-Answer-Key-1.png\")
\nAnswer:
\nLet c units represent the length of the hypotenuse.
\n(\\(\\sqrt{27}\\))2 <\/sup>+ (\\(\\sqrt{48}\\))2<\/sup> = c2<\/sup>
\n27 + 48 = c2<\/sup>
\n75 = c2<\/sup>
\n\\(\\sqrt{75}\\) = \\(\\sqrt{c^{2}}\\)
\n\\(\\sqrt{5^{2}}\\) \u00d7 \\(\\sqrt{3}\\) = c
\n5\\(\\sqrt{3}\\) = c
\nThe length of the hypotenuse is 5\\(\\sqrt{3}\\) units.<\/p>\n
\nWhat is the length of the unknown side of the right triangle? Simplify your answer, if possible.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-4-Problem-Set-Answer-Key-2.png\")
\nAnswer:
\nLet c cm represent the length of the hypotenuse.
\n32 <\/sup>+ 82<\/sup> = c2<\/sup>
\n9 + 64 = c2<\/sup>
\n73 = c2<\/sup>
\n\\(\\sqrt{73}\\) = \\(\\sqrt{c^{2}}\\)
\n\\(\\sqrt{73}\\) = c
\nThe length of the unknown side is \\(\\sqrt{73}\\) cm.<\/p>\n
\nWhat is the length of the unknown side of the right triangle? Simplify your answer, if possible.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-4-Problem-Set-Answer-Key-3.png\")
\nAnswer:
\nLet c mm represent the length of the hypotenuse.
\n32 <\/sup>+ 32<\/sup> = c2<\/sup>
\n9 + 9 = c2<\/sup>
\n18 = c2<\/sup>
\n\\(\\sqrt{18}\\) = \\(\\sqrt{c^{2}}\\)
\n\\(\\sqrt{18}\\) = c
\n\\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{2}\\) = c
\n3\\(\\sqrt{2}\\) = c
\nThe length of the unknown side is 3\\(\\sqrt{2}\\) mm.<\/p>\n
\nWhat is the length of the unknown side of the right triangle? Simplify your answer, if possible.
\n![\"Eureka](\"https:\/\/ccssanswers.com\/wp-content\/uploads\/2021\/04\/Eureka-Math-8th-Grade-Module-7-Lesson-4-Problem-Set-Answer-Key-4.png\")
\nAnswer:
\nLet x in. represent the unknown length.
\nx2 <\/sup>+ 82<\/sup> = 122<\/sup>
\nx2 <\/sup>+ 64 = 144
\nx2 <\/sup>+ 64 – 64 = 144 – 64
\nx2<\/sup> = 80
\n\\(\\sqrt{x^{2}}\\) = \\(\\sqrt{80}\\)
\nx = \\(\\sqrt{80}\\)
\nx = \\(\\sqrt{2^{4} \\cdot 5}\\)
\nx = \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{2^{2}}\\) \u22c5 \\(\\sqrt{5}\\)
\nx = 2 \u22c5 2\\(\\sqrt{5}\\)
\nx = 4\\(\\sqrt{5}\\)
\nThe length of the unknown side is 4\\(\\sqrt{5}\\) in.<\/p>\n
\nJosue simplified \\(\\sqrt{450}\\) as 15\\(\\sqrt{2}\\) Is he correct? Explain why or why not.
\nAnswer:
\n\\(\\sqrt{450}\\) = \\(\\sqrt{2 \\times 3^{2} \\times 5^{2}}\\)
\n= \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{5^{2}}\\)
\n= 3 \u00d7 5 \u00d7 \\(\\sqrt{1}\\)
\n= 15\\(\\sqrt{1}\\)
\nYes, Josue is correct because the number 450 = 2 \u00d7 32<\/sup> \u00d7 52<\/sup>. The factors that are perfect squares simplify to 15 leaving just the factor of 2 that cannot be simplified. Therefore, \\(\\sqrt{450}\\) = 15\\(\\sqrt{2}\\).<\/p>\n
\nTiah was absent from school the day that you learned how to simplify a square root. Using \\(\\sqrt{360}\\), write Tiah an explanation for simplifying square roots.
\nAnswer:
\nTo simplify \\(\\sqrt{360}\\), first write the factors of 360. The number 360 = 23<\/sup> \u00d7 32<\/sup> \u00d7 5. Now, we can use the factors to write \\(\\sqrt{360}\\) = \\(\\sqrt{2^{3} \\times 3^{2} \\times 5}\\)), which can then be expressed as \\(\\sqrt{360}\\) = \\(\\sqrt{2^{3}}\\) \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{5}\\). Because we want to simplify square roots, we can rewrite the factor \\(\\sqrt{2^{3}}\\) as \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{2}\\) because of the laws of exponents. Now, we have
\n\\(\\sqrt{360}\\) = \\(\\sqrt{2^{2}}\\) \u00d7 \u221a2 \u00d7 \\(\\sqrt{3^{2}}\\) \u00d7 \\(\\sqrt{5}\\).
\nEach perfect square can be simplified as follows:
\n\\(\\sqrt{360}\\) = 2 \u00d7 \\(\\sqrt{2}\\) \u00d7 3 \u00d7 \\(\\sqrt{5}\\)
\n= 2 \u00d7 3 \u00d7 \\(\\sqrt{2}\\) \u00d7 \\(\\sqrt{5}\\)
\n= 6\\(\\sqrt{10}\\).
\nThe simplified version of \\(\\sqrt{360}\\) = 6\\(\\sqrt{10}\\).<\/p>\nEureka Math Grade 8 Module 7 Lesson 4 Exit Ticket Answer Key<\/h3>\n
\nQuestion 1.
\n\\(\\sqrt{24}\\)
\nAnswer:
\n\\(\\sqrt{24}\\) = \\(\\sqrt{2^{2} \\times 6}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{6}\\)
\n= 2\\(\\sqrt{6}\\)<\/p>\n
\n\\(\\sqrt{338}\\)
\nAnswer:
\n\\(\\sqrt{338}\\) = \\(\\sqrt{13^{2} \\times 2}\\)
\n= \\(\\sqrt{13^{2}}\\) \u00d7 \\(\\sqrt{24}\\)
\n= 13\\(\\sqrt{2}\\)<\/p>\n
\n\\(\\sqrt{196}\\)
\nAnswer:
\n\\(\\sqrt{196}\\) = \\(\\sqrt{14^{2}}\\)
\n= 14<\/p>\n
\n\\(\\sqrt{2420}\\)
\nAnswer:
\n\\(\\sqrt{2420}\\) = \\(\\sqrt{2^{2} \\times 11^{2} \\times 5}\\)
\n= \\(\\sqrt{2^{2}}\\) \u00d7 \\(\\sqrt{11^{2}}\\) \u00d7 \\(\\sqrt{5}\\)
\n= 2 \u00d7 11 \u00d7 \\(\\sqrt{5}\\)
\n= 22\\(\\sqrt{5}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"