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CHAPTER 6 – Algebra 2 – Selected Answers “Evens” p.1 Section 6.1 (p.261) Oral Ex/ 1. 6 2. -2 3. not real 4. -1/5 5. -2 6. 5 7. 2 8. 11 9. -13 10. 7 11. -6 12. 0 13. 7 14. 1 15. no solution 16. 6 17. 3 18. 2 19. True 20. True 21. True 22. False 23. False 24. True 25. True 26. False 27. False 28. True 29. True 30. True 31. 4 (1)4 1 x but 5 (1)5 1 x Written Exercises 2a. 8 b. not real c. 4 d. -4 4a. 12 b. not real c. -12 d. not real 6a. .2 b. -.2 c. .02 d. 8a. 5 b. c. -5 d. -5 10a. ¼ b. 9/4 c. ½ d. 3/2 12a. 1/10 b. 100 c. 103 d. 1010 14a. b. |a| c. |a| d. |a3| 16. 0 18. 7 20. none 22. 3 24. none 26. 1/5 28. a 2 30. all reals 32. b 1 34a. x 4 b. 2 x 2 c. all reals d. all reals 36. 0 x 6 38. x 1 Section 6.2 (p.267) Oral Ex/ 1. no 2. yes 3. no 4. no 5. 5 3 6. 2 6 7. 10 6 8. 7 2 9. 6 13. 2 3 14. 5 6 15. 3 16. 2 3 22. no, (x + y)2 = x2 + 2xy + y2 Written Exercises 5 2 3 5 2. 5 5 4. 11 3 6. 18 8. 10. 12. 2 7 5 3 3 28 6 3 22. 24. 2 26. 54 28. 2 35 30. 32a. 3 2 11. 2 12. 2 17. 2 3 5 18. 3 4 10. 5 5 23 2 19. -4 20. 2 3 3 3 3 14. 4 2 16. 7 15 18. 3 20. 33 5 5 3 4 6 3 6 12 b. c. d. 34. 29.50 4 2 2 2 y x 2 3 3cd 2 3 36. 19.40 38. 5.48 40. 2 x 3x 42. 2c 2c 44. 46. 48. 3 a2 b2 2 3 3d x 3 3 2b 4b 2 3 2 50. x 2 3 52. 54. 5/12 56. 3 6 58. 9 2 60. 62. a x 4 n na 2 3 6c 2 64. 66. 3x 68. n c2 b a b n 70. n b m n n b mn n b nm n n the only (real) root. If n is even, both n n n a a a n is the nth root of . If n is odd, it is b b b a and n n n b are nonnegative and so is n m b bm . Now this means of bm. If n is odd, this is the only (real) root. If n is even, then n n b n m a (b 0 ) b is an nth root b 0 and so is n b m . CHAPTER 6 – Selected Answers “Evens” p.2 Section 6.3 (p.271) Oral Ex/ 1. not possible 2. not possible 3. 3 5 2 4. 3 2 2 3 5. 3 2 6. 3 2 2 7. not possible 8. 3 9. 11 3 30 10. 3 3 2 11. 2 2 2 12. 2 3 13. 5 14. 3 7 15. 4 2 1 16. no, x x 2 x , but 2 x 2 x Written Exercises 2. 5 4. 13 3 6. not possible 8. 5 6 5 10. 3 2 6 12. 5 3 3 2 3 7 14. 2 6 7 10 33 2 16. 18. 20. 6 6 2 22. 21 2 7 3 24. 15 30 15 26. 6 2 10 2 7 2 53 2 3 28. 2 2 30. 32. 10 4 25 34. 2 36. 5 y 2 5 y 38. 3 a b2 2 8 2 5 6x 40. 42. 2t 3 5 5t 2 44. 102 6 Section 6.4 (p.275) Oral Ex/ 1a. x2 – y2 b. 3 c. 1 2a. x2 + 2xy + y2 b. 7 2 10 c. 7 4 3 x y x y 5 2 3a. x2 – 2xy + y2 b. 7 2 10 c. 7 4 3 4a. 2 3 b. 2 5a. b. 3 x y x y Written Exercises 6 3 2. 23 4. 9 4 5 6. 21 2 3 8. 10. 109 6 110 12. 9 13 2 14. 4 3 2 7 33 16. 10 18. 48 7 6 20. 3 2 2 3 22. 24 12 3 24. 43 26. 30 20 2 28. 3 17 3 21 3 24 30. 12 15 45 3 24 5 90 32. 34. 4 2 36. 2 1 2 2 41 1 21 2 2 2 2 12 4 2 2 1 2 2 2 1 4 2 2 1 0 5 1 38. The reciprocal of 5 1 is 2 also the conjugate of 1 y2 x y 5 1 . 40. 10 42. 4b b 44. 46. x y 2 1 y2 2 = 5 1 2 5 1 2 5 1 5 1 4 5 1 which is 2 CHAPTER 6 – Selected Answers “Evens” p.3 Section 6.5 (p.279) Oral Ex/ 1. 25 2. 8 3. 10 49 5. no solution 6. 4 7. 11 8. no solution 9. -8 3 2 10a. linear b. radical c. linear 11a. b. 9/4 c. 3 2 2 Written Exercises 2. 16. 4. no solution 6. -4, 4 8. 16 10. 3 12. 8 14. 3 16. 9 18. -3, -2 20a. 16 b. 4 3 22a. 4 b. 2 3 24a. 1/9, 4 b. 3 7 26. 16 28. 2, 6 30. no solution 32. 0 gT 2 48 34a. 0.9 sec b. 36a. 9 3 3 b. Area 12 x 2 3 24 x 2 16 3 2 4 3 x 16 3 4 4 3 38. 12 Section 6.6 (p.286) Oral Ex/ I like to use Q for the set of rationals (Q stands for quotient or ratio of integers). Use “Ir” for irrationals. 1. Q 2. Q 3. Q 4. Q 5. Q 6. Ir 7. Q 8. Ir 9. Ir 10. Ir 11. Q 12. Ir 13. Real, Irrational 14. Not Real 15. Real, Rational 16. Real, Rational Written Exercises (Again, using Q for rational #’s and ‘Ir’ for irrationals) 2a. Ir b. Q 4a. Ir b. Q 6a. Q b. Q c. Ir 8. 0.45 10. 3.25 12. 751/250 14. 11/80 16. 5/9 18. 1/12 20. 14/11 22. 1100/999 24a. 0.37255 b. .372515115111… 26a. 3.1 b. 32 28a. 5 109 b. .000000001010010001… 30a. 3 13 b. .01 24 32a. The arithmetic mean is rational. B. No, if the product xy is not a perfect square, then xy is irrational. 34. Repeat the process by finding the arithmetic mean (a rational number) of ‘x’ and the previous arithmetic mean (rational number). 36a. There can be no more than 16 digits. b. .058823529117647 38. Let x ba and y dc where b, d 0 . ad bc which is the ratio of 2 integers (Closure Property for Integers…). bd 40. Let x ba (rational number) and z is an irrational number. Indirect Proof: Assume x + z is rational (the opposite of what we’re trying to show). Then x + z is a ratio of 2 integers, say: x z dc where c, d Integers, d 0 . Then x + y = c a bc ad which is a ratio of integers, which means z is d b bd rational, but that is a contradiction of part of our hypothesis that z is irrational. So x + z cannot be rational, it must be irrational. Q.E.D. 42a. Multiplying by 10p shifts the digits but doesn’t effect the pattern after the decimal pt. b. By matching the patterns after the decimal point, subtraction results in a rational number. c. All terminating decimals (see part b) above are rational numbers. d. Division of rationals results in rationals (no zero in the denominator). Then z = (x + z) – x = CHAPTER 6 – Selected Answers “Evens” p.4 Section 6.7 (p.290) Oral Ex/ 1. 8i 2. 10i 3. 11i 4. -3i 5. 2 3 i 6. 5 3 i 7. -3 8. -9 9. -16 10. -12 11. 8 12. -5 13. -3 14. 11i 15. -2i 16. 4i 17. 11i 18. 2i 19. 7 i 20. i 21. 5 i 22. -i Written Exercises 2. 11i 4. -24i 6. 5 3 i 8. 15 3 i 10. -15 12. 2 3 i 14. 3 2 16. -64 18. -25 20. -45 22. -54 24. 83 i 26. -2i 28. –½ i 30. 6 i 3 32. 20i 34. 2i 36. 2 2 i 6 i b. 72 42a. 14 2 b. 98 44. -6c2i 46. –t 48. 0 38a. 4 3 i b. -9 40a. 50. 10a a i 52. 3y2 54a. 1 b. I c. -1 d. –i Section 6.8 (p.295) Oral Ex/ 1. 8 + 9i 2. 16 – i 3. -10 + 4i 4. 3 + I 5. 2 + 10i 6. -6 + 21i 7. 37 8. 13 9. 5 10. 10 11. 2i 12. 15 – 8i 13. imaginary 14. pure imaginary 15. rational 16. irrational 17. rational 18. rational 19. imaginary 20. irrational Written Exercises 2. -2 – 6i 4. 10 – 5i 6. -8 – 3i 8. 18 + 15i 10. -6 – 2i 12. 17 14. -1 + 21i 16. 34 – 2i 18. -18 + I 20. 7 22. -13 – 84i 24. 4 6 5 i 26. 169 28. 10 30. 6 + 3i 32. 15 75 i 34. 12 13 5 13 i 36. 2 7 3 75 i 38. 1 17 174 i 40. 5 7 2 i 7 42. 8 5 56 i 44. 1 – 2i 46. (1 – 3i)2 – 2(1 – 3i) + 10 = -8 – 6i – 2 + 6i + 10 = 0 48a. (1 3 i )3 8 b. yes 50. 1 2 3i 1 2 3 i 1 52. a2 + b2 = 1 54. (a + bi)(c + di) = (ac – bd) + (ad + bc)i = (ca – db) + (da + cb)i (Commutative Property of Multiplication for Reals) = (c + di)(a + bi) showing complex numbers have a commutative property for multiplication also. 56. Show (a + bi)[(c + di) + (e + fi)] = (a + bi)(c + di) + (a + bi)(e + fi). Starting with the left side, we have (a + bi)[(c + e) + (d + f)i] by the definition of addition for complex #’s. Now by the definition of multiplication for complex #’s, we get: [ac + ae – (bd + bf)] + [ad + af + bc + be]i = [(ac – bd) + (ae – bf) ] + [(ad + bd) + (af + be)]i = (ac – bd) + (ad + bd)i + (ae – bf) + (af + be)i = (a + bi)(c + di) + (a + bi)(e + fi) which we were trying to show. Chapter 6 Review Chapter 6 Test 2. c 4. a 6. d 8. c 10. a 12. d 14. c 16. d 1. -8 2. 49 3. -4 4. 3 5. 11. 4 6 12. 3 5 2 94 6. -2 7. 3 2 8. 5a 3 8b 6b 2 9. 7 5 20 2 10. 8 15 x 15 13. 7 14. 7 15. 41/55 16. 1.9995, 1.999099990… 17. -30 18. 3 3 19. 2 + 3i 20. 9 – 8i 21. 60 + 14i 22. 16 15 i 13 13 Preparing for College Entrance Exams 2. B 4. D 6. A 8. B