นำเสนอจำนวนจริงเพิ่มเติม

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นำเสนอจำนวนจริงเพิ่มเติม

  1. 1. REAL NUMBERS by Nittaya Noinan Kanchanapisekwittayalai Phetchabun School
  2. 2. Real Numbers • Real numbers consist of all the rational and irrational numbers. • The real number system has many subsets: – – – – Natural Numbers Whole Numbers Integers Ect.
  3. 3. Natural Numbers • Natural numbers are the set of counting numbers. {1, 2, 3,…}
  4. 4. Whole Numbers • Whole numbers are the set of numbers that include 0 plus the set of natural numbers. {0, 1, 2, 3, 4, 5,…}
  5. 5. Integers • Integers are the set of whole numbers and their opposites. {…,-3, -2, -1, 0, 1, 2, 3,…}
  6. 6. Rational Numbers • Rational numbers are any numbers that can be expressed in the form of a , where a and b b are integers, and b ≠ 0. • They can always be expressed by using terminating decimals or repeating decimals.
  7. 7. The Rational Numbers The word ratio means fraction. Therefore rational numbers are any numbers which can be written as fractions. 2 3 3 4 5 1 1 5
  8. 8. Integers are Rational Numbers 2 3 3 4 5 1 1 5 Like the 5 in our example, any integer can be made into a fraction by putting it over 1. Since it can be a fraction, it is a rational number.
  9. 9. Changing fractions to decimals It’s easy to change a fraction to a decimal, so rational numbers can also be written as decimals. Rational numbers convert to two different types of decimals: Terminating decimals – which end Repeating decimals – which repeat
  10. 10. Terminating decimals To convert a fraction to a decimal, divide the top by the bottom. To convert ½ to a decimal you would do: .5 2 1.0 There is no remainder. The answer just ends – or terminates.
  11. 11. Terminating Decimals • Terminating decimals are decimals that contain a finite number of digits. • Examples:  36.8  0.125  4.5
  12. 12. Repeating decimals To convert a fraction to a decimal, divide the top by the bottom. To convert 1/3 to a decimal you would do: − .333 3 1.000 .3 =0.3333… There is a remainder. The answer just keeps repeating.
  13. 13. Repeating Decimals • Repeating decimals are decimals that contain a infinite number of digits. • Examples:  0.333…  0.19191919…  7.689689… FYI…The line above the decimals indicate that number repeats.
  14. 14. Repeating decimals − .3 .09 The bar tells us that it is a repeating decimal. The bar extends over the entire pattern that repeats.
  15. 15. Rational numbers as decimals Rational numbers can be converted from fractions to either • Terminating decimals or • Repeating decimals
  16. 16. Rational numbers The subsets of real numbers that we’ve discussed are “nested” like Russian dolls.
  17. 17. Examples of Rational Numbers •16 •1/2 •3.56 •-8 •1.3333… •- 3/4
  18. 18. To show how these number are classified, use the Venn diagram. Place the number where it belongs on the Venn diagram. Rational Numbers 4 9 Irrational Numbers Integers π Whole Numbers Natural Numbers -3 0 117 − 1 2 6.36 4 9 -12.64039…
  19. 19. Irrational Numbers • Irrational numbers are any numbers that cannot be expressed a as . b • They are expressed as non-terminating, nonrepeating decimals; decimals that go on forever without repeating a pattern. • Examples of irrational numbers: – 0.34334333433334… – 45.86745893… – π (pi) – 2
  20. 20. Other Vocabulary Associated with the Real Number System • …(ellipsis)—continues without end • { } (set)—a collection of objects or numbers. Sets are notated by using braces { }. • Finite—having bounds; limited • Infinite—having no boundaries or limits • Venn diagram—a diagram consisting of circles or squares to show relationships of a set of data.
  21. 21. Example • Classify all the following numbers as natural, whole, integer, rational, or irrational. List all that apply. a. 117 b. 0 c. -12.64039… d. -½ e. 6.36 f. π g. -3
  22. 22. Solution • Now that all the numbers are placed where they belong in the Venn diagram, you can classify each number: – 117 is a natural number, a whole number, an integer, and a rational number. – − 1 is a rational number. 2 – 0 is a whole number, an integer, and a rational number. – -12.64039… is an irrational number. – -3 is an integer and a rational number. – 6.36 is a rational number. – π is an irrational number. – 4 is a rational number. 9
  23. 23. To show how these number are classified, use the Venn diagram. Place the number where it belongs on the Venn diagram. Rational Numbers 4 9 Irrational Numbers Integers π Whole Numbers Natural Numbers -3 0 117 − 1 2 6.36 4 9 -12.64039…
  24. 24. FYI…For Your Information • When taking the square root of any number that is not a perfect square, the resulting decimal will be nonterminating and non-repeating. Therefore, those numbers are always irrational.
  25. 25. Irrational Numbers • An irrational number is a number that cannot be written as a ratio of two integers. • Irrational numbers written as decimals are non-terminating and non-repeating.
  26. 26. Examples of Irrational Numbers • Square roots of non-perfect “squares” 17 • Pi
  27. 27. Irrational Numbers In English, the word “irrational” means not rational illogical, crazy, wacky. In math, irrational numbers are not rational. They usually look wacky! 5 3 17 π …and their decimals never end or repeat!
  28. 28. Irrational Numbers There is one trick you need to watch out for! Numbers like 25 and 81 They look wacky but because the number in the house is a perfect square, they are really the integers 5 and 9 in disguise! Sort of like the wolf at Grandma’s house!
  29. 29. Rounding or truncating Some decimals are much longer than we need. There are two ways we can make them shorter. Truncating – just lop the extra digits off. Rounding – use the digit to the right of the one we want to end with to determine whether to round up or not. If that digit is 5 or higher, round up.
  30. 30. Truncating π = 3.1415926... Truncating – just lop the extra digits off. If we want to use π with just 4 decimal places. We’d just chop off the rest! 3.1415/926… 3.1415 Truncate ~ tree trunk ~ chop!
  31. 31. Rounding π = 3.1415926... If we want to round π to 4 decimal places. We’d look at the digit in the 5th place 9 is “5 or bigger” so the digit in the 4th spot goes up 3.14159 3.1416
  32. 32. CLASS WORK 1. Given the set, 13 15   1.001,0.333..., π , −11,11, , 16,3.14,  15 3  list the elements of the set that are: a) b) c) d) Natural numbers Integers Rational numbers Irrational numbers
  33. 33. Properties of Real Numbers For any real number a, b and c Addition Multiplication Close a+b∈R ab ∈ R Commutative a + b = b + a ab = ba Associative (a + b) + c = a + (b + c) (ab)c = a(bc) Identity A+0=a=0+a A1 = a = 1a Inverse A + (-a) = 0 = (-a) + a If a in note zero then a-1 .a = 1 =a. a-1
  34. 34. Properties of Real Numbers • Distributive property – For all real numbers a, b, and c a(b+c) = ab + ac and (b+c)a = ba + ca
  35. 35. Solving Equations; 5 Properties of Equality Reflexive For any real number a, a=a Symmetric Property For all real numbers a and b, if a=b, then b=a Transitive Property For all reals, a, b, and c, if a=b and b=c, then a=c
  36. 36. Solving Equations; 5 Properties of Equality Addition and Subtraction For any reals a, b, and c, if a=b then a+c=b+c and ac=b-c Multiplication and Division For any reals a, b, and c, if a=b then a*c=b*c, and, if c is not zero, a/c=b/c
  37. 37. Applications of Equations Problem Solving Plan 1. 2. 3. 4. Explore the Problem Plan the solution Solve the problem Examine the solution
  38. 38. Absolute Value Equations Absolute value: Distance from zero For any real number a: If a ≥ 0 , then a = a If a < 0 , then a = − a
  39. 39. Properties of Real Numbers Commutative Property: a+b=b+a ab = ba order doesn’t matter Associative Property: (a+b)+c = a+(b+c) (ab)c = a(bc) order doesn’t change
  40. 40. Distributive Property: a(b+c) = ab + ac you can add then multiply or multiply then add.
  41. 41. Theory about Real Numbers Theorem 1. Eliminated Rule for Addition when a , b and c are real numbers. (i) if a + c = b + c then a = b (ii) if a + b = a + c then b = c Theorem 2. Eliminated Rule for Multiplication when a , b and c are real numbers. (i) if ac = bc and c ≠ 0 then a = b (ii) if ab = ac and a ≠ 0 then b = c
  42. 42. Theory about Real Numbers Theorem 3. Theorem 4. Theorem 5. Theorem 6. When a is real numbers. a.0 = 0 When a is real numbers. (-1)a = -a When a and b are real numbers. if ab = 0 then a = 0 or b = 0 When a and b are real numbers. 1. a(-b) 2. (-a)b 3. (-a)(-b) = = = -ab -ab ab
  43. 43. Subtraction and Divisor of Real Numbers Definition. When a and b are real numbers. a – b = a + (-b) Definition. When a and b are real numbers. = a(b-1)
  44. 44. Theory about Real Numbers Theorem 7. When a , b and c are real numbers. 1. a(b – c) 2. (a – b)c 3. (-a)(b – c) Theorem 8. = = = ab – ac ac – bc -ab + ac When a ≠ 0 then a-1 ≠ 0
  45. 45. Theory about Real Numbers Theorem 9. When a , b and c are real numbers. 1. when b , c ≠ 0 2. when b , c ≠ 0 3. when b , d ≠ 0 4. when b , d ≠ 0
  46. 46. Theory about Real Numbers Theorem 9. When a , b and c are real numbers. 5. when b , c ≠ 0 6. when b , c ≠ 0 7. when b , d ≠ 0
  47. 47. CLASS WORK State the property of real numbers being used. 2. 2 ( 3 + 5 ) = ( 3 + 5 ) 2 3. 2 ( A + B ) = 2 A + 2 B 4. ( p + 2q ) + 3r = p + ( 2q + 3r )
  48. 48. TRUE OR FALSE 1. The set of WHOLE numbers is closed with respect to multiplication.
  49. 49. TRUE OR FALSE 2. The set of NATURAL numbers is closed with respect to multiplication.
  50. 50. TRUE OR FALSE 3. The product of any two REAL numbers is a REAL number.
  51. 51. TRUE OR FALSE 4. The quotient of any two REAL numbers is a REAL number.
  52. 52. TRUE OR FALSE 5. Except for 0, the set of RATIONAL numbers is closed under division.
  53. 53. TRUE OR FALSE 6. Except for 0, the set of RATIONAL numbers contains the multiplicative inverse for each of its members.
  54. 54. TRUE OR FALSE 7. The set of RATIONAL numbers is associative under multiplication.
  55. 55. TRUE OR FALSE 8. The set of RATIONAL numbers contains the additive inverse for each of its members.
  56. 56. TRUE OR FALSE 9. The set of INTEGERS is commutative under subtraction.
  57. 57. TRUE OR FALSE 10. The set of INTEGERS is closed with respect to division.
  58. 58. Solving polynomial Equations one variable. We can write polynomial Equations of x variable that is anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be coefficiants of the polynomail are real numbers by a ≠ 0 Then we can called anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 is polynomail of degree n. The symbol is p(x) , q(x) , r(x) and if p(a) that mean we instead x in p(x) by a . Example. P(x) = x3 – 4x2 + 3x + 2 p(1) = 13 – 4(1)2 + 3(1) + 2 = 2
  59. 59. Solving polynomial Equations one variable. Example. Find the answer of 3x3 + 2x2 - 12x – 8 = 0 Solve. by use Addition and Multiplication of real numbers. Then we can multiplied by the following factor. 3x3 + 2x2 - 12x – 8 = (3x3 + 2x2) – (12x + 8) = x2 (3x + 2) – 4(3x + 2) = (3x + 2)(x2 - 4) = (3x + 2)(x - 2)(x + 2) By Theory 5 then x = Answer {-2 , , 2} , or x = 2 or x = -2
  60. 60. Solving polynomial Equations by remainder Theorem Method. Remainder theorem. When p(x) is anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be real numbers by a ≠ 0 . if p(x) is divied by x – c when c is real number then the remainder is equal p(c)
  61. 61. Solving polynomial Equations by remainder Theorem Method. Proof. Give p(x) is divied by x – c then we get quotient q(x) And remainder be q(x) Thus p(x) = (x – c)q(x) + r(x) ……(1) Which r(x) is zero or polynomail of degree is less than x – c that mean degree 0. hence r(x) is constant. Give r(x) = d when d is constant. Thus p(x) = (x – c)q(x) + d ……(2) when instead x in (2) by c We get p(c) = (c – c)q(x) + d = d Hence remainder equal is p(c)
  62. 62. Solving polynomial Equations by remainder Theorem Method. Example 1. Find the remainder when 9x3 + 4x - 1 is divided by x - 2 Solve. 9x2 + 18x + 40 9x3 - 18x2 18x2 + 4x – 1 18x2 - 36x 40x – 1 40x – 80 79 Thus remainder is 79
  63. 63. Solving polynomial Equations by remainder Theorem Method. Example 1. Find the remainder when 9x3 + 4x - 1 is divided by x - 2 Solve. Give p(x) = thus p(2) = = = 9x3 + 4x - 1 9(2)3 + 4(2) - 1 72 + 8 – 1 79 Thus the remainder is 79.
  64. 64. Solving polynomial Equations by remainder Theorem Method. Example 2. Find the remainder when 2x4 - 7x3 + x2 + 7x – 3 is divided by x + 1 Solve. Give p(x) = 2x4 - 7x3 + x2 + 7x – 3 Since x +1 = x – (-1) thus c = -1 thus p(-1) = = = 2(-1)4 – 7(-1)3 + (-1)2 + 7(-1) – 3 2 + 7 + 1 – 7 – 3 0 Thus the remainder is 0.
  65. 65. Factor theorem. When p(x) is anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be real numbers by a ≠ 0 . p(x) there is x – c that is factor iff p(c) = 0
  66. 66. Factor theorem. Example. 1) Write x4 – x3 -2x2 – 4x – 24 be factor. Solve. Give p(x) = x4 – x3 -2x2 – 4x – 24 since integers that can divide -24 are ±1,±2,±3,±4,±6,±8,±12, ±24 Then consider p(1) ,p(-1) ,p(2) that is not equal zero. But p(-2) = (-2)4 – (-2)3 -2(-2)2 – 4(-2) – 24 = 16 + 8 – 8 + 8 – 24 = 0 Thus x + 2 is the factor of x4 – x3 -2x2 – 4x – 24
  67. 67. Factor theorem. Example. 1) Write x4 – x3 -2x2 – 4x – 24 be factor. Solve. Thus x + 2 is the factor of x4 – x3 -2x2 – 4x – 24 Take x + 2 divide x4 – x3 -2x2 – 4x – 24 then get x3 -3x2 + 4x – 12 Thuse x4 – x3 -2x2 – 4x – 24 = (x + 2)(x3 -3x2 + 4x – 12) = (x + 2){(x3 -3x2 )+ 4(x – 3)} = (x + 2){x2 (x -3) + 4(x – 3)} = (x + 2)(x – 3)(x2 + 4)
  68. 68. Factor theorem. Example. 2) Write x3 -5x2 + 2x + 8 be factor. Solve. Give p(x) = x3 -5x2 + 2x + 8 since integers that can divide 28 are ±1,±2,±4,±8 Then consider p(1) ,p(-1) ,p(-2) that is not equal zero. But p(2) = (2)3 -5(2)2 + 2(2) + 8 = 8 - 20 + 4 + 8 = 0 Thus x - 2 is the factor of x3 -5x2 + 2x + 28
  69. 69. Factor theorem. Example. 2) Write x3 -5x2 + 2x + 8 be factor. Solve. Thus x - 2 is the factor of x3 -5x2 + 2x + 8 Take x - 2 divide x3 -5x2 + 2x + 8 then get x2 - 3x – 4 Thuse x3 -5x2 + 2x + 8 = (x - 2)(x2 - 3x – 4) = (x - 2)(x – 4)(x + 1)
  70. 70. Factor theorem. Example. 3) Write x3 + 2x2 - 5x - 6 be factor. Solve. Give p(x) = x3 + 2x2 - 5x - 6 since integers that can divide 6 are ±1,±2,±3,±6 Then consider p(1) that is not equal zero. But p(-1) = (-1)3 + 2(-1)2 – 5(-1) - 6 = -1 + 2 + 5 - 6 = 0 Thus x + 1 is the factor of x3 + 2x2 - 5x - 6
  71. 71. Factor theorem. Example. 3) Write x3 + 2x2 - 5x - 6 be factor. Solve. Thus x + 1 is the factor of x3 + 2x2 - 5x - 6 Take x + 1 divide x3 + 2x2 - 5x - 6 then get x2 + x – 6 Thuse x3 + 2x2 - 5x - 6 = (x + 1)(x2 + x – 6) = (x + 1)(x – 2)(x + 3)
  72. 72. Solving polynomial Equations by remainder Theorem Method. Rational factorization theorem. When p(x) is anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 when n be positive integers and an ,an-1,an-2 ,…,a1,a0 be real numbers by a ≠ 0 . if x - is factor of polynomial p(x) by m and k are integer which m ≠ 0 and greatest common factor of m and k is equal 1 Then m can divied an and k can divided a0 .
  73. 73. Example. 1) Write 12x3 + 16x2 - 5x – 3 be factor. Solve. Give p(x) = 12x3 + 16x2 - 5x – 3 since integers that can divide -3 are ±1,±2,±3 And integers that can divide 12 are ±1,±2,±3, ±4,±6,±12 Then the rational number that p( ) = 0 in among ±1,±2,±3, Then consider p( ) that is equal zero.
  74. 74. Example. 1) Write 12x3 + 16x2 - 5x – 3 be factor. Solve. Thus x Take x Thuse is the factor of 12x3 + 16x2 - 5x – 3 divide 12x3 + 16x2 - 5x – 3 then get 12x2 + 22x + 6 12x3 + 16x2 - 5x – 3 = (x - = (x - )(12x2 + 22x + 6) )(2) (6x2 + 11x + 3) = (2x – 1)(6x2 + 11x + 3) = (2x – 1)(3x + 1)(2x + 3)
  75. 75. Example. 2) Solving Equation 6x3 - 11x2 + 6x = 1 Solve. Since 6x3 - 11x2 + 6x = 1 Then we get 6x3 - 11x2 + 6x -1 = 0 Give p(x) = 6x3 - 11x2 + 6x -1 p(1) = 6 – 11 + 6 – 1 = 0 Thus p(x) = (x -1)(6x2 - 5x + 1) = (x – 1)(2x – 1)(3x – 1) Since 6x3 - 11x2 + 6x -1 = (x – 1)(2x – 1)(3x – 1) = Then x – 1 = 0 or 2x -1 = 0 or 3x – Thus x = 1 or x = or Answer {1 , , } 0 0 1 = 0 x =
  76. 76. Example. 1) Write 12x3 + 16x2 - 5x – 3 be factor. Solve. Thus x Take x Thuse is the factor of 12x3 + 16x2 - 5x – 3 divide 12x3 + 16x2 - 5x – 3 then get 12x2 + 22x + 6 12x3 + 16x2 - 5x – 3 = (x - )(12x2 + 22x + 6) = (x - )(2) (6x2 + 11x + 3) = (2x – 1)(6x2 + 11x + 3) = (2x – 1)(3x + 1)(2x + 3)
  77. 77. Properties of inequalities. IN real number system we use symbol < , > , ≤ ,≥ , ≠ be less than , more than , less than or equal to , more than or equal to ,not equal sort by order if a and be be real number the symbol a < b that mean a less than b and a > b that mean a more than b Trichotomy property. if a and b be real number then a = b , a < b and a > b That was actually only one .
  78. 78. Definition. a a a a a a ≤ ≥ < ≤ < ≤ b b b b b b < ≤ ≤ < c c c c that that that that that that mean mean mean mean mean mean a a a a a a less than or equal to b more than or equal to b < b and b < c ≤ b and b ≤ c < b and b ≤ c ≤ b and b < c
  79. 79. Properties of inequalities. Give a , b , c be real numbers 1. Transitive Property. if a > b and b > c then a > c Such as 5 > 3 and 3 > 1 then 5 > 1 2. Properties of Addition and Subtraction. So adding (or subtracting) the same value to both a and b will not change the inequality if a > b then a + c > b + c Such as 4 > 2 then 4 + 1 > 2 + 1
  80. 80. Properties of inequalities. Give a , b , c be real numbers 3.Positive and Negative number when compare with zero a is positive number iff a > 0 a is negative number iff a < 0 4. Property of Multiplication but not with zero. Case 1 if a > b and c > 0 then ac > bc Case 2 if a > b and c < o then ac < bc
  81. 81. Properties of inequalities. Give a , b , c be real numbers 5. Properties excision for Addition. if a + c > b + c then a > b Such as 5 + 2 > 3 + 2 then 5 > 3 6. Properties excision for Multiplication. Case 1 if ac > bc and c > 0 then a > b Such as 6 × 3 > 4 × 3 and 3 > 0 then 6 > 4 Case 2 if ac > bc and c < 0 then a < b Such as 3 × (-3) > 4 × (-3) and -3 < 0 then 3 < 4
  82. 82. Intervals Notation Graph Set-builder Notation (a, b) [a, b] [a, b) (a, b] (a, ∞) [a, ∞) (-∞, b) (-∞, b] (-∞, ∞) b b
  83. 83. Solving inequalities. Exmaple. Solving inequalities following problems. 1. 3x + 5 < x – 7 Solve. Since 3x + 5 3x – x 2x x Answer < x–7 < -7 – 5 < -12 < -6 {x/x < -6} or (-∝ , -6)
  84. 84. Solving inequalities. Exmaple. Solving inequalities following problems. 2. 4y + 7 > 2(y + 1) Solve. Since 4y + 7 > 2(y + 1) 4y + 7 > 2y + 2 4y – 2y > 2 - 7 2y > -5 y> Answer {y/y > } or ( , ∝)
  85. 85. Solving inequalities. Exmaple. Solving inequalities following problems. 3. x2 – x – 6 ≤ 0 Solve. Since x2 – x – 6 ≤ 0 (x – 3)(x + 2) ≤ 0 Then x – 3 = 0 or x + 2 = 0 We get x = 3 or x = -2 Thus + + -2 3 Answer { x / -2 ≤ x ≤3 } or [-2 , 3]
  86. 86. Solving inequalities. Exmaple. Solving inequalities following problems. 4. 2x2 + 7x + 3 ≥ 0 Solve. Since 2x2 + 7x + (2x + 1)(x Then 2x + 1 = 0 or x + 3 We get x = or x Thus + -3 Answer (-∝ , -3) ∪ ( 3 ≥ 0 + 3) ≤ 0 = 0 = -3 + , ∝)
  87. 87. Solving inequalities. Exmaple. 5. Solve. Since Solving inequalities following problems.
  88. 88. Solving inequalities. Exmaple. Solving inequalities following problems. 5. Solve. Take (x – 4)2 multiplicate all sides Then
  89. 89. then we take (-1) multiplicate all sides (x2 – 3x – 10)(x – 4) ≥ 0 We get (x – 5)(x + 2)(x – 4) ≥ 0 Then x – 5 = 0 or x + 2 = 0 or x – 4 = 0 x = 5 or -2 x = -2 or + 4 x=4 + 5
  90. 90. And if the inequalities to a degree greater than two. We can use the Remainder Theorem Method .
  91. 91. Absolute value inequalities. Definition. give a be real number |a| = a 0 -a if a > 0 if a = 0 if a < 0
  92. 92. Absolute value inequalities. Theorem. when x and y be real number 1.| x | = | -x | 2.| xy | = | x || y | 3. = ,y≠0 4.| x - y | = |y-x| 2 2 5.| x | = x 6.| x + y | ≤ |x|+|y|
  93. 93. Solving equations and inequalities in Absolute value. Theory 11. when a be positive number set of the answers of |a| = a is {a , -a}
  94. 94. Example. Find the answer of the following equations. 1. |2x - 3| = 9 Solve. since |2x - 3| = 9 then 2x -3 = 9 or 2x – 3 = -9 x = 6 or x = -3 Answer {-3 , 6}
  95. 95. Example. Find the answer of the following equations. 2. |3x - 1| = |x + 5| Solve. since |3x - 1| then 3x -1 = 2x x Answer = |x + 5| x + 5 or 3x -1 = -(x + 5) = 6 or 4x = -4 = 3 or x = -1 {-1 , 3}
  96. 96. Example. Find the answer of the following equations. 3. |2x + 1| = 3x - 5 Solve. since |2x + 1| = 3x - 5 then 3x – 5 ≥ 0 and [2x +1 = 3x - 5 or 2x +1 = -(3x - 5) x ≥ Answer { 6 } and x = 6 0r x =
  97. 97. Solving equations and inequalities in Absolute value. Theory 12. when a be positive number 1. 2. 3. 4. mean mean mean mean |a| |a| |a| |a| < ≤ > ≥ a a a a that that that that -a < x -a ≤ x x <-a x ≤-a < a ≤ a or x > a or x ≥ a
  98. 98. Example. Find the answer of the following equations. 1. |4x - 3| < 1 Solve. since then Answer |4x - 3| < 1 -1 < 4x – 3 < 1 2 < 4x < 4 < x < 1 ( , 1)
  99. 99. Example. Find the answer of the following equations. 2. |x - 3| ≥ 2 Solve. since |x - 3| ≥ 2 then x -3 ≤ -2 or x - 3 ≥ 2 x ≤ 1 or x ≥ 5 Answer (-∝ , 1] ∪ [ 5 , ∝)
  100. 100. Example. Find the answer of the following equations. 3. |x + 1| ≤ 2x - 3 Solve. since |x + 1| ≤ 2x – 3 Then -(2x – 3) ≤ x + 1 ≤ 2x – 3 -(2x – 3) ≤ x + 1 and x + 1 ≤ 2x – 3 -3x ≤ -2 and -x ≤ -4 x ≥ and x ≥ 4 Answer [4 , ∝)
  101. 101. Example. Find the answer of the following equations. 4. |2x + 4| > x + 1 Solve. since |2x + 4| > x + 1 Then 2x + 4 <-(x + 1) or 2x + 4 > x + 1 3x < -5 or x > -3 x < or x > -3 Answer (-∝ , - ) ∪ (-3 , ∝)
  102. 102. Example. Find the answer of the following equations. 5. |x| < |2x - 1| Solve. since Then |x| < |2x - 1| 2 x 2 < (2x – 1) x2 < 4x2 – 4x + 1 0 < 3x2 – 4x + 1 0 < (3x -1)(x – 1)
  103. 103. Example. Find the answer of the following equations. 5. |x| < |2x - 1| Solve. 0 < (3x -1)(x – 1) Then 3x – 1 = 0 or x – 1 = 0 x = or x = 1 + - + 1 Answer (-∝ , ) ∪ (1 , ∝)

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