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Roots and Radical Expressions Basic Introduction and Rules The basic definitions of radical expressions: index power ππ =π β Radical sign π radicand π=π What are the real-number roots of each radical expression? 3 a) 343 Because the prime factors of 343 are 7*7*7, 7 is the third (cube) root of 343. On the graphing calculator, press MATH, 4, 343, Enter to get 7. b) 4 1 625 What is the prime factorization of 1? We are looking for groups of 4 (the index), so we get 1*1*1*1. What is the prime factorization of 625? When we take the factor tree all the way to all prime numbers, we get 5*5*5*5. b) 4 1β1β1β1 5β5β5β5 1 5 = Since we have one GROUP of four 1βs in the numerator, and one 1 1 GROUP of four 5βs in the denominator, the fourth root of = 625 5 In this case, we must remember that since we took an even root, there will be both a positive and a negative answer. 4 1 625 = 1 ± 5 Another way to express the fact that there could be two answers is to use the absolute value sign. 1 5 is the same thing as 1 ± . 5 What are the real-number roots of each radical expression? b) 4 1 625 We MUST use absolute value signs EVERY time that we take an EVEN root and get an ODD powered answer. 1 th In this case, we took the 4 root (even) and got to the first 5 power, (odd). We can leave out the absolute value signs if we are told to assume that all variables are positive. What are the real-number roots of each radical expression? 3 c) β0.064 On the graphing calculator, press MATH, 4, -0.064, Enter to get -0.4. d) β25 Because 52 = 25 and (β5)2 = 25, neither 5 nor β5 are real number square roots of β25. There are no REAL-number square roots of -25. On the graphing calculator, press 2nd, π₯ 2 , -25, Enter to get ERR:NONREAL ANS. Again, there is no real square root of -25. Find the real-number roots of each radical expression. 1) 4) 3 169 ππ 1 β 8 π β π 5) π΅πΆπ΅π¬ 8) β 7) 4 25 10) β0.0001 π΅πΆπ΅π¬ 4 2) 11) 3 729 π 3) π ππ 6) 4 0.1296 π. π 9) 5 1 243 4 121 π π 12) 4 0.0016 3 125 216 3 β0.343 βπ. π 3 8 125 π. π π π π π π You can not assume that ππ = π. For example, (β6)2 = 36 = 6, and β6. This leads to the following property for any real number π. π If π is odd, ππ = π π If π is even, ππ = π . Again, we use absolute value signs if we took an EVEN power and got an ODD power answer. What is the simplified form of each radical expression? a) 3 1000π₯ 3 π¦ 9 b) 4 256π8 β4 π 16 What is the simplified form of each radical expression? a) 3 1000π₯ 3 π¦ 9 π ππ β ππ β ππ β π β π β π β π β π β π β π β π β π β π β π β π π ππ β ππ β ππ β π β π β π β π β π β π β π β π β π β π β π β π We have 1 group of 3 10βs, 1 group of 3 xβs, and 3 groups of 3 yβs. These all come out from inside the radical. ππ β π β π β π β π πππππ What is the simplified form of each radical expression? 4 b) 256π8 β4 π 16 π π πβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπ π π πβπβπβπβπβπβπβπβπβπβπβπ πβπβπβπβπβπβπβπβπβπβπβπ πβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπβπ πππ π ππ To convert from radical form to rational exponents, place the power of the radicand in the numerator, over the index. π π π ππ = π . 4 2 For example, π₯ 4 = π₯ = π₯ 2 . Rewrite the following radicals into rational exponents. 4 3 4 2 1) π₯ π¦π§ 2) π₯ 4 π¦ 5 π§ 7 π π π ππ ππ ππ 4) π π π ππ ππ ππ π5 ππ 2 π π π ππ ππ ππ = π π ππ ππ π = π π πππ ππ 3) 7 π₯ 14 π¦π§ 2 ππ π π π π ππ ππ = π π π π ππ ππ To convert from rational exponent form to radical form, reverse the process. π π π = π ππ . 3 2 For example, π₯ = π₯ 3 . Rewrite the following rational exponents into radicals. 3 7 1) π₯ π¦ π 5 7 ππ ππ 3 4 1 7 4 4 2) π₯ π¦ π§ π ππ πππ 4 5 3 3 10 5 3) π π π ππ ππ ππ ππ Classwork: Complete the worksheet on simplifying radicals. Homework: Complete the worksheet on converting radicals to rationals and vice-versa.