1. Real Numbers and Algebraic Expressions

2. A set is a collection of objects whose contents can be clearly determined. The set {1, 3, 5, 7, 9} has five elements. The objects in a set are called the elements of the set. We use braces to indicate a set and commas to separate the elements of that set. The set of counting numbers can be represented by {1, 2, 3, … }. The set of even counting numbers are {2, 4, 6, …}. The set of even counting numbers is a subset of the set of counting numbers, since each element of the subset is also contained in the set. For example, The Basics About Sets

3. -15, -7, -4, 0, 4, 7{…, -2, -1, 0, 1, 2, 3, …} Add the negative natural numbers to the whole numbers Integers Z 0, 4, 7, 15{0, 1, 2, 3, … } Add 0 to the natural numbers Whole Numbers W 4, 7, 15{1, 2, 3, …} These are the counting numbers Natural Numbers N ExamplesDescriptionName Important Subsets of the Real Numbers

4. This is the set of numbers whose decimal representations are neither terminating nor repeating. Irrational numbers cannot be expressed as a quotient of integers. Irrational Numbers I These numbers can be expressed as an integer divided by a nonzero integer: Rational numbers can be expressed as terminating or repeating decimals. Rational Numbers Q ExamplesDescriptionName Important Subsets of the Real Numbers

5. Rational numbers Irrational numbers Integers Whole numbers Natural numbers The set of real numbers is formed by combining the rational numbers and the irrational numbers. The Real Numbers

6. Negative numbers Units to the left of the origin are negative. Positive numbers Units to the right of the origin are positive. The real number line is a graph used to represent the set of real numbers. An arbitrary point, called the origin, is labeled 0; -4 -3 -2 -1 0 1 2 3 4 the Origin The Real Number Line

7. Real numbers are graphed on the number line by placing a dot at the location for each number. –3, 0, and 4 are graphed below. -4 -3 -2 -1 0 1 2 3 4 Graphing on the Number Line

8. On the real number line, the real numbers increase from left to right. The lesser of two real numbers is the one farther to the left on a number line. The greater of two real numbers is the one farther to the right on a number line. Since 2 is to the left of 5 on the number line, 2 is less than 5. 2 < 5 Since 5 is to the right of 2 on the number line, 5 is greater than 2. 5 > 2 -2 -1 0 1 2 3 4 5 6 Ordering the Real Numbers

9. Because -5 = -5-5 > -5 Because 7 > 37 > 3b is greater than or equal to a.b > ab > a Because 7 =77 < 7 Because 3 < 73 < 7a is less than or equal to b.a < ba < b ExplanationExampleMeaningSymbols Inequality Symbols

10. Absolute value describes the distance from 0 on a real number line. If a represents a real number, the symbol |a| represents its absolute value, read “the absolute value of a.” For example, the real number line below shows that |-3| = 3 and |5| = 5. -3 -2 -1 0 1 2 3 4 5 The absolute value of –3 is 3 because –3 is 3 units from 0 on the number line. |–3| = 3 The absolute value of 5 is 5 because 5 is 5 units from 0 on the number line. |5| = 5 Absolute Value

11. The absolute value of x is given as follows: |x| = x if x > 0 -x if x < 0 { Definition of Absolute Value

12. For all real number a and b, 1. |a| > 0 5. =, b not equal to 0 2. |-a| = |a|3. a < |a| 4. |ab| = |a||b| 6. |a + b| < |a| + |b| (the triangle inequality) |a| |b| a b Properties of Absolute Value

13. Example Find the following: |-3| and |3|. Solution:

14. If a and b are any two points on a real number line, then the distance between a and b is given by |a – b| or |b – a| Distance Between Two Points on the Real Number Line

15. Find the distance between –5 and 3 on the real number line. SolutionBecause the distance between a and b is given by |a – b|, the distance between –5 and 3 is |-5 – 3| = |-8| = 8. -5 -4 -3 -2 -1 0 1 2 3 We obtain the same distance if we reverse the order of subtraction: |3 – (-5)| = |8| = 8. 8 Text Example

16. A combination of variables and numbers using the operations of addition, subtraction, multiplication, or division, as well as powers or roots, is called an algebraic expression. Here are some examples of algebraic expressions: x + 6, x – 6, 6x, x/6, 3x + 5. Algebraic Expressions

17. 1.Perform operations within the innermost parentheses and work outward. If the algebraic expression involves division, treat the numerator and the denominator as if they were each enclosed in parentheses. 2.Evaluate all exponential expressions. 3.Perform multiplication or division as they occur, working from left to right. 4.Perform addition or subtraction as they occur, working from left to right. The Order of Operations Agreement

18. The algebraic expression 2.35x + 179.5 describes the population of the United States, in millions, x years after 1980. Evaluate the expression when x = 20. Describe what the answer means in practical terms. SolutionWe begin by substituting 20 for x. Because x = 20, we will be finding the U.S. population 20 years after 1980, in the year 2000. 2.35x + 179.5 Replace x with 20. = 2.35(20) + 179.5 = 47 + 179.5 Perform the multiplication. = 226.5 Perform the addition. Thus, in 2000 the population of the United States was 226.5 million. Example

19. 3 + ( 8 + x) = (3 + 8) + x = 11 + x If 3 real numbers are added, it makes no difference which 2 are added first. (a + b) + c = a + (b + c) Associative Property of Addition x · 6 = 6xTwo real numbers can be multiplied in any order. ab = ba Commutative Property of Multiplication 13 + 7 = 7 + 13 13x + 7 = 7 + 13x Two real numbers can be added in any order. a + b = b + a Commutative Property of Addition ExamplesMeaningName Properties of the Real Numbers

20. 0 + 6x = 6xZero can be deleted from a sum. a + 0 = a 0 + a = a Identity Property of Addition 5 · (3x + 7) = 5 · 3x + 5 · 7 = 15x + 35 Multiplication distributes over addition. a · (b + c) = a · b + a · c Distributive Property of Multiplication over Addition -2(3x) = (-2·3)x = -6xIf 3 real numbers are multiplied, it makes no difference which 2 are multiplied first. (a · b) · c = a · (b · c) Associative Property of Multiplication ExamplesMeaningName Properties of the Real Numbers

21. 2 · 1/2 = 1The product of a nonzero real number and its multiplicative inverse gives 1, the multiplicative identity. a · 1/a = 1 and 1/a · a = 1 Inverse Property of Multiplication (-6x) + 6x = 0The sum of a real number and its additive inverse gives 0, the additive identity. a + (-a) = 0 and (-a) + a = 0 Inverse Property of Addition 1 · 2x = 2xOne can be deleted from a product. a · 1 = a and 1 · a = a Identity Property of Multiplication ExamplesMeaningName Properties of the Real Numbers

22. Let a and b represent real numbers. Subtraction: a – b = a + (-b) We call –b the additive inverse or opposite of b. Division: a ÷ b = a · 1/b, where b = 0 We call 1/b the multiplicative inverse or reciprocal of b. The quotient of a and b, a ÷ b, can be written in the form a/b, where a is the numerator and b the denominator of the fraction. Definitions of Subtraction and Division

23. Simplify: 6(2x – 4y) + 10(4x + 3y). Solution 6(2x – 4y) + 10(4x + 3y) = 6 · 2x – 6 · 4y + 10 · 4x + 10 · 3y Use the distributive property. = 12x – 24y + 40x + 30y Multiply. = (12x + 40x) + (30y – 24y) Group like terms. = 52x + 6y Combine like terms. Text Example

24. Properties of Negatives Let a and b represent real numbers, variables, or algebraic expressions. 1.(-1)a = -a 2.-(-a) = a 3.(-a)(b) = -ab 4.a(-b) = -ab 5.-(a + b) = -a - b 6.-(a - b) = -a + b = b - a

25. Real Numbers and Algebraic Expressions

26. Exponents and Scientific Notation

27. Definition of a Natural Number Exponent If b is a real number and n is a natural number, b n is read “the nth power of b” or “ b to the nth power.” Thus, the nth power of b is defined as the product of n factors of b. Furthermore, b 1 = b

28. The Negative Exponent Rule If b is any real number other than 0 and n is a natural number, then

29. The Zero Exponent Rule If b is any real number other than 0, b 0 = 1.

30. The Product Rule b m · b n = b m+n When multiplying exponential expressions with the same base, add the exponents. Use this sum as the exponent of the common base.

31. The Power Rule (Powers to Powers) (b m ) n = b mn When an exponential expression is raised to a power, multiply the exponents. Place the product of the exponents on the base and remove the parentheses.

32. The Quotient Rule When dividing exponential expressions with the same nonzero base, subtract the exponent in the denominator from the exponent in the numerator. Use this difference as the exponent of the common base.

33. Example Find the quotient of 4 3 /4 2 Solution:

34. Products to Powers (ab) n = a n b n When a product is raised to a power, raise each factor to the power.

35. Simplify: (-2y) 4. (-2y) 4 = (-2) 4 y 4 = 16y 4 Text Example Solution

36. Quotients to Powers When a quotient is raised to a power, raise the numerator to that power and divide by the denominator to that power.

37. Example Simplify by raising the quotient (2/3) 4 to the given power. Solution:

38. Properties of Exponents

39. The number 5.5 x 10 12 is written in a form called scientific notation. A number in scientific notation is expressed as a number greater than or equal to 1 and less than 10 multiplied by some power of 10. It is customary to use the multiplication symbol, x, rather than a dot in scientific notation. Scientific Notation

40. Text Example Write each number in decimal notation: a. 2.6 X 10 7 b. 1.016 X 10 -8 a. 2.6 x 10 7 can be expressed in decimal notation by moving the decimal point in 2.6 seven places to the right. We need to add six zeros. 2.6 x 10 7 = 26,000,000. Solution: b. 1.016 x 10 -8 can be expressed in decimal notation by moving the decimal point in 1.016 eight places to the left. We need to add seven zeros to the right of the decimal point. 1.016 x 10 -8 = 0.00000001016.

41. To convert from decimal notation to scientific notation, we reverse the procedure. Move the decimal point in the given number to obtain a number greater than or equal to 1 and less than 10. The number of places the decimal point moves gives the exponent on 10; the exponent is positive if the given number is greater than 10 and negative if the given number is between 0 and 1. Scientific Notation

42. Write each number in scientific notation. a. 4,600,000 b. 0.00023 Solution a. 4,600,000 = 4.6 x 10 ? 4.6 x 10 6 Decimal point moves 6 places b. 0.00023 = 2.3 x 10 ? 2.3 x 10 -4 Decimal point moves 4 places Text Example

43. Exponents and Scientific Notation

44. Radicals and Rational Exponents

45. Definition of the Principal Square Root If a is a nonnegative real number, the nonnegative number b such that b 2 = a, denoted by b =  a, is the principal square root of a.

46. Square Roots of Perfect Squares

47. The Product Rule for Square Roots If a and b represent nonnegative real number, then The square root of a product is the product of the square roots.

48. Text Example Simplify a.  500b.  6x  3x Solution:

49. The Quotient Rule for Square Roots If a and b represent nonnegative real numbers and b does not equal 0, then The square root of the quotient is the quotient of the square roots.

50. Text Example Simplify: Solution:

51. Example Perform the indicated operation: 4  3 +  3 - 2  3. Solution:

52. Example Perform the indicated operation:  24 + 2  6. Solution:

53. Definition of the Principal nth Root of a Real Number If n, the index, is even, then a is nonnegative (a > 0) and b is also nonnegative (b > 0). If n is odd, a and b can be any real numbers.

54. Finding the nth Roots of Perfect nth Powers

55. The Product and Quotient Rules for nth Roots For all real numbers, where the indicated roots represent real numbers,

56. Definition of Rational Exponents

57. Example Simplify 4 1/2 Solution:

58. Definition of Rational Exponents The exponent m/n consists of two parts: the denominator n is the root and the numerator m is the exponent. Furthermore,

59. Radicals and Rational Exponents

60. Polynomials

61. The Degree of ax n If a does not equal 0, the degree of ax n is n. The degree of a nonzero constant is 0. The constant 0 has no defined degree.

62. Definition of a Polynomial in x A polynomial in x is an algebraic expression of the form a n x n + a n-1 x n-1 + a n-2 x n-2 + … + a 1 n + a 0 where a n, a n-1, a n-2, …, a 1 and a 0 are real numbers. a n = 0, and n is a non-negative integer. The polynomial is of degree n, an is the leading coefficient, and a 0 is the constant term.

63. Perform the indicated operations and simplify: (-9x 3 + 7x 2 – 5x + 3) + (13x 3 + 2x 2 – 8x – 6) Solution (-9x 3 + 7x 2 – 5x + 3) + (13x 3 + 2x 2 – 8x – 6) = (-9x 3 + 13x 3 ) + (7x 2 + 2x 2 ) + (-5x – 8x) + (3 – 6) Group like terms. = 4x 3 + 9x 2 – (-13x) + (-3) Combine like terms. = 4x 3 + 9x 2 + 13x – 3 Text Example

64. The product of two monomials is obtained by using properties of exponents. For example, (-8x 6 )(5x 3 ) = -8·5x 6+3 = -40x 9 Multiply coefficients and add exponents. Furthermore, we can use the distributive property to multiply a monomial and a polynomial that is not a monomial. For example, 3x 4 (2x 3 – 7x + 3) = 3x 4 · 2x 3 – 3x 4 · 7x + 3x 4 · 3 = 6x 7 – 21x 5 + 9x 4. monomialtrinomial Multiplying Polynomials

65. Multiplying Polynomials when Neither is a Monomial Multiply each term of one polynomial by each term of the other polynomial. Then combine like terms.

66. Using the FOIL Method to Multiply Binomials (ax + b)(cx + d) = ax · cx + ax · d + b · cx + b · d Product of First terms Product of Outside terms Product of Inside terms Product of Last terms first last inner outer

67. Multiply: (3x + 4)(5x – 3). Text Example

68. Multiply: (3x + 4)(5x – 3). Solution (3x + 4)(5x – 3)= 3x·5x + 3x(-3) + 4(5x) + 4(-3) = 15x 2 – 9x + 20x – 12 = 15x 2 + 11x – 12 Combine like terms. first last inner outer FOIL Text Example

69. The Product of the Sum and Difference of Two Terms The product of the sum and the difference of the same two terms is the square of the first term minus the square of the second term.

70. The Square of a Binomial Sum The square of a binomial sum is first term squared plus 2 times the product of the terms plus last term squared.

71. The Square of a Binomial Difference The square of a binomial difference is first term squared minus 2 times the product of the terms plus last term squared.

72. Let A and B represent real numbers, variables, or algebraic expressions. Special Product Example Sum and Difference of Two Terms (A + B)(A – B) = A 2 – B 2 (2x + 3)(2x – 3) = (2x) 2 – 3 2 = 4x 2 – 9 Squaring a Binomial (A + B) 2 = A 2 + 2AB + B 2 (y + 5) 2 = y 2 + 2·y·5 + 5 2 = y 2 + 10y + 25 (A – B) 2 = A 2 – 2AB + B 2 (3x – 4) 2 = (3x) 2 – 2·3x·4 + 4 2 = 9x 2 – 24x + 16 Cubing a Binomial (A + B) 3 = A 3 + 3A 2 B + 3AB 2 + B 3 (x + 4) 3 = x 3 + 3·x 2 ·4 + 3·x·4 2 + 4 3 = x 3 + 12x 2 + 48x + 64 (A – B) 3 = A 3 – 3A 2 B – 3AB 2 + B 3 (x – 2) 3 = x 3 – 3·x 2 ·2 – 3·x·2 2 + 2 3 = x 3 – 6x 2 – 12x + 8 Special Products

73. Multiply: a. (x + 4y)(3x – 5y)b. (5x + 3y) 2 Solution We will perform the multiplication in part (a) using the FOIL method. We will multiply in part (b) using the formula for the square of a binomial, (A + B) 2. a. (x + 4y)(3x – 5y) Multiply these binomials using the FOIL method. = (x)(3x) + (x)(-5y) + (4y)(3x) + (4y)(-5y) = 3x 2 – 5xy + 12xy – 20y 2 = 3x 2 + 7xy – 20y 2 Combine like terms. (5 x + 3y) 2 = (5 x) 2 + 2(5 x)(3y) + (3y) 2 (A + B) 2 = A 2 + 2AB + B 2 = 25x 2 + 30xy + 9y 2 FOIL Text Example

74. Example Multiply: (3x + 4) 2. ( 3x + 4 ) 2 = (3x) 2 + (2)(3x) (4) + 4 2 = 9x 2 + 24x + 16 Solution:

75. Polynomials

76. Factoring Polynomials

77. Factoring is the process of writing a polynomial as the product of two or more polynomials. The factors of 6x 2 – x – 2 are 2x + 1 and 3x – 2. In this section, we will be factoring over the integers. Polynomials that cannot be factored using integer coefficients are called irreducible over the integers or prime. The goal in factoring a polynomial is to use one or more factoring techniques until each of the polynomial’s factors is prime or irreducible. In this situation, the polynomial is said to be factored completely. Factoring

78. In any factoring problem, the first step is to look for the greatest common factor. The greatest common factor is a n expression of the highest degree that divides each term of the polynomial. The distributive property in the reverse direction ab + ac = a(b + c) can be used to factor out the greatest common factor. Common Factors

79. Factor: a. 18x 3 + 27x 2 b. x 2 (x + 3) + 5(x + 3) Solution a. We begin by determining the greatest common factor. 9 is the greatest integer that divides 18 and 27. Furthermore, x 2 is the greatest expression that divides x 3 and x 2. Thus, the greatest common factor of the two terms in the polynomial is 9x 2. 18x 3 + 27x 2 = 9x 2 (2x) + 9x 2 (3) Express each term with the greatest common factor as a factor. = 9x 2 (2 x + 3) Factor out the greatest common factor. b. In this situation, the greatest common factor is the common binomial factor (x + 3). We factor out this common factor as follows. x 2 (x + 3) + 5(x + 3) = (x + 3)(x 2 + 5) Factor out the common binomial factor. Text Example

80. A Strategy for Factoring ax 2 + bx + c If no such combinations exist, the polynomial is prime. (Assume, for the moment, that there is no greatest common factor.) 1. Find two First terms whose product is ax 2 : ( x + )( x + ) = ax 2 + bx + c 2. Find two Last terms whose product is c: (x + )(x + ) = ax 2 + bx + c I 3. By trial and error, perform steps 1 and 2 until the sum of the Outside product and Inside product is bx: ( x + )( x + ) = ax 2 + bx + c O (sum of O + I)

81. Factor: a. x 2 + 6x + 8b. x 2 + 3x – 18 Solution a. The factors of the first term are x and x: (x )( x ) To find the second term of each factor, we must find two numbers whose product is 8 and whose sum is 6. -6-969Sum of Factors -4, -2-8, -14, 28, 1Factors of 8 From the table above, we see that 4 and 2 are the required integers. Thus, x 2 + 6x + 8 = (x + 4)( x + 2) or (x + 2)( x + 4). This is the desired sum. Text Example

82. Factor: a. x 2 + 6x + 8b. x 2 + 3x – 18 Solution b. We begin with x 2 + 3x – 18 = (x )( x ). To find the second term of each factor, we must find two numbers whose product is –18 and whose sum is 3. From the table above, we see that 6 and –3 are the required integers. Thus, x 2 + 3x – 18 = (x + 6)(x – 3) or (x – 3)(x + 6). -33-77-1717Sum of Factors -6, 36, -3-9, 29, -2-18, 118, -1Factors of -8 This is the desired sum. Text Example cont.

83. Factor: 8x 2 – 10x – 3. Step 2Find two Last terms whose product is –3. The possible factors are 1(-3) and –1(3). Step 3Try various combinations of these factors. The correct factorization of 8x 2 – 10x – 3 is the one in which the sum of the Outside and Inside products is equal to –10x. Here is a list of possible factors. Solution Step 1Find two First terms whose product is 8x 2. 8x 2 – 10x – 3 (8x )(x ) 8x 2 – 10x – 3 (4x )(2x ) Text Example

84. -4x + 6x = 2x(4x + 3)(2x – 1) 12x – 2x = 10x(4x – 1)(2x + 3) 4x – 6x = -2x(4x – 3)(2x + 1) -12x + 2x = -10x(4x + 1)(2x – 3) -8x + 3x = -5x(8x + 3)(x – 1) 24x – x = 23x(8x – 1)(x +3) 8x – 3x = 5x(8x – 3)(x + 1) -24x + x = -23x(8x + 1)(x – 3) Sum of Outside and Inside Products (Should Equal –10x) Possible Factors of 8x 2 – 10x – 3 Thus, 8x 2 – 10x – 3 = (4x + 1)(2x – 3) or (2x – 3)(4x + 1). This is the required middle term. Text Example cont.

85. The Difference of Two Squares If A and B are real numbers, variables, or algebraic expressions, then A 2 – B 2 = (A + B)(A – B). In words: The difference of the squares of two terms factors as the product of a sum and the difference of those terms.

86. Text Example Factor: 81x 2 - 49 Solution: 81x 2 – 49 = (9x) 2 – 7 2 = (9x + 7)(9x – 7).

87. Factoring Perfect Square Trinomials Let A and B be real numbers, variables, or algebraic expressions, 1. A 2 + 2AB + B 2 = (A + B) 2 2. A 2 – 2AB + B 2 = (A – B) 2

88. Text Example Factor: x 2 + 6x + 9. x 2 + 6x + 9 = x 2 + 2 · x · 3 + 3 2 = (x + 3) 2 Solution:

89. Text Example Factor: 25x 2 – 60x + 36. Solution: 25x 2 – 60x + 36 = (5x) 2 – 2 · 5x · 6 + 6 2 = (x + 3) 2.

90. Factoring the Sum and Difference of 2 Cubes 64x 3 – 125 = (4x) 3 – 5 3 = (4x – 5)(4x) 2 + (4x)(5) + 5 2 ) = (4x – 5)(16x 2 + 20x + 25) A 3 – B 3 = (A – B)(A 2 + 2AB + B 2 ) x 3 + 8 = x 3 + 2 3 = (x + 2)( x 2 – x·2 + 2 2 ) = (x + 2)( x 2 – 2x + 4) A 3 + B 3 = (A + B)(A 2 – 2AB + B 2 ) ExampleType

91. A Strategy for Factoring a Polynomial 1.If there is a common factor, factor out the GCF. 2.Determine the number of terms in the polynomial and try factoring as follows: a)If there are two terms, can the binomial be factored by one of the special forms including difference of two squares, sum of two cuubes, or difference of two cubes? b)If there are three terms, is the trinomial a perfect square trinomial? If the trinomial is not a perfects square trinomial, try factoring by trial and error. c)If there are four or more terms, try factoring by grouping. 3.Check to see if any factors with more than one term in the factored polynomial can be factored further. If so, factor completely.

92. Factor: x 3 – 5x 2 – 4x + 20 Solution x 3 – 5x 2 – 4x + 20 = (x 3 – 5x 2 ) + (-4x + 20) Group the terms with common factors. = x 2 (x – 5) – 4(x – 5) Factor from each group. = (x – 5)(x 2 – 4) Factor out the common binomial factor, (x – 5). = (x – 5)(x + 2)(x – 2) Factor completely by factoring x 2 – 4 as the difference of two squares. Example

93. Factoring Polynomials

94. Rational Expressions

95. Find all the numbers that must be excluded from the domain of each rational expression. This denominator would equal zero if x = 2. This denominator would equal zero if x = -1. This denominator would equal zero if x = 1. SolutionTo determine the numbers that must be excluded from each domain, examine the denominators. Text Example

96. Simplifying Rational Expressions 1.Factor the numerator and denominator completely. 2.Divide both the numerator and denominator by the common factors.

97. Example Simplify: Solution:

98. Multiplying Rational Expressions 1.Factoring all numerators and denominators completely. 2.Dividing both the numerator and denominator by common factors. 3.Multiply the remaining factors in the numerator and multiply the remaining factors in the denominator.

99. Example Multiply and simplify: Solution:

100. Example Divide and simplify: Solution:

101. Example Add: Solution:

102. Finding the Least Common Denominator 1.Factor each denominator completely. 2.List the factors of the first denominator. 3.Add to the list in step 2 any factors of the second denominator that do not appear in the list. 4.Form the product of each different factor from the list in step 3. This product is the least common denominator.

103. Adding and Subtracting Rational Expressions That Have Different Denominators with Shared Factors 1.Find the least common denominator. 2.Write all rational expressions in terms of the least common denominator. To do so, multiply both the numerator and the denominator of each rational expression by any factor(s) needed to convert the denominator into the least common denominator. 3.Add or subtract the numerators, placing the resulting expression over the least common denominator. 4.If necessary, simplify the resulting rational expression.

104. Example Subtract: Solution:

105. Rational Expressions

106. Linear Equations

107. Terms Involving Equations 3x - 1 = 2 An equation consists of two algebraic expressions joined by an equal sign. 3x – 1 = 2 3x = 3 x = 1 1 is a solution or root of the equation Left Side Right Side

108. Definition of a Linear Equation A linear equation in one variable x is an equation that can be written in the form ax + b = 0 where a and b are real numbers and a = 0.

109. An equation can be transformed into an equivalent equation by one or more of the following operations. Example 1. Simplify an expression by removing grouping symbols and combining like terms. 3(x - 6) = 6x - x 3x - 18 = 5x -18 = 2x -9 = x Divide both sides of the equation by 2. 3. Multiply (or divide) on both sides of the equation by the same nonzero quantity. Subtract 3x from both sides of the equation. 3x - 18 = 5x 3x - 18 - 3x = 5x - 3x -18 = 2x 2. Add (or subtract) the same real number or variable expression on both sides of the equation. -9 = x x = -9 4. Interchange the two sides of the equation. Generating Equivalent Equations

110. Solving a Linear Equation Simplify the algebraic expression on each side. Collect all the variable terms on one side and all the constant terms on the other side. Isolate the variable and solve. Check the proposed solution in the original equation.

111. Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2). Solution Step 1 Simplify the algebraic expression on each side. 2(x - 3) – 17 = 13 – 3(x + 2) This is the given equation. 2x – 6 – 17 = 13 – 3x – 6 Use the distributive property. 2x – 23 = - 3x + 7 Combine like terms. Text Example

112. Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2). Solution Step 2 Collect variable terms on one side and constant terms on the other side. We will collect variable terms on the left by adding 3x to both sides. We will collect the numbers on the right by adding 23 to both sides. 2x – 23 + 3x = - 3x + 7 + 3x Add 3x to both sides. 5x – 23 = 7 Simplify. 5x – 23 + 23 = 7 + 23 Add 23 to both sides. 5x = 30 Simplify. Text Example

113. Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2). Solution Step 3 Isolate the variable and solve. We isolate the variable by dividing both sides by 5. 5x = 30 5x/5 = 30/5 Divide both sides by 5 x = 6 Simplify. Text Example

114. Solve the equation: 2(x - 3) - 17 = 13 - 3(x + 2). Solution The solution set is {6}. Step 4 Check the proposed solution in the original equation. Substitute 6 for x in the original equation. 2(x - 3) - 17 = 13 - 3(x + 2) This is the original equation. -11 = -11 This true statement indicates that 6 is the solution. 2(6 - 3) - 17 = 13 - 3(6 + 2) Substitute 6 for x. ? 2(3) - 17 = 13 - 3(8) Simplify inside parentheses. ? 6 – 17 = 13 – 24 Multiply. ? Text Example

115. Types of Equations Identity:An equation that is true for all real numbers. Conditional:An equation that is true for at least one real number. Inconsistent:An equation that is not true for any real number.

116. Determine whether the equation 3(x - 1) = 3x + 5 is an identity, a conditional equation, or an inconsistent equation. Solution To find out, solve the equation. 3(x – 1) = 3x + 5 3x – 3 = 3x + 5 -3 = 5 This equation is inconsistent. Example

117. Linear Equations

118. Quadratic Equations

119. Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard form ax 2  bx  c  0 where a, b, and c are real numbers with a not equal to 0. A quadratic equation in x is also called a second-degree polynomial equation in x.

120. The Zero-Product Principle If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero. If AB  0, then A  0 or B  0.

121. Solving a Quadratic Equation by Factoring 1.If necessary, rewrite the equation in the form ax 2  bx  c  0, moving all terms to one side, thereby obtaining zero on the other side. 2.Factor. 3. Apply the zero  product principle, setting each factor equal to zero. 4. Solve the equations in step 3. 5.Check the solutions in the original equation.

122. Text Example Solve 2x 2  7x  4 by factoring and then using the zero  product principle. Step 1 Move all terms to one side and obtain zero on the other side. Subtract 4 from both sides and write the equation in standard form. 2x 2  7x  4  4  4 2x 2  7x  4  Step 2 Factor. 2x 2  7x  4  (2x  1)(x  4)  0

123. Solution cont. Solve 2x 2  7x  4 by factoring and then using the zero  product principle. Steps 3 and 4 Set each factor equal to zero and solve each resulting equation. 2 x  1  or x  4  2 x  1x  4 x = 1/2 Steps 5 check your solution

124. Example (2x + -3)(2x + 1) = 5 4x 2 - 4x - 3 = 5 4x 2 - 4x - 8 = 0 4(x 2 -x-2)=0 4(x - 2)*(x + 1) = 0 x - 2 = 0, and x + 1 = 0 So x = 2, or -1

125. The Square Root Method If u is an algebraic expression and d is a positive real number, then u 2 = d has exactly two solutions. If u 2 = d, then u =  d or u = -  d Equivalently, If u 2 = d then u =  d

126. Completing the Square If x 2 + bx is a binomial, then by adding (b/2) 2, which is the square of half the coefficient of x, a perfect square trinomial will result. That is, x 2 + bx + (b/2) 2 = (x + b/2) 2

127. Text Example What term should be added to the binomial x 2 + 8x so that it becomes a perfect square trinomial? Then write and factor the trinomial. The term that should be added is the square of half the coefficient of x. The coefficient of x is 8. Thus, (8/2) 2 = 4 2. A perfect square trinomial is the result. x 2 + 8x + 4 2 = x 2 + 8x + 16 = (x + 4) 2

128. Quadratic Equation

129. Quadratic Formula

132. No x-intercepts No real solution; two complex imaginary solutions b 2 – 4ac < 0 One x-intercept One real solution (a repeated solution) b 2 – 4ac = 0 Two x-intercepts Two unequal real solutionsb 2 – 4ac > 0 Graph of y = ax 2 + bx + c Kinds of solutions to ax 2 + bx + c = 0 Discriminant b 2 – 4ac The Discriminant and the Kinds of Solutions to ax 2 + bx +c = 0

133. The Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse. If the legs have lengths a and b, and the hypotenuse has length c, then a 2 + b 2 = c 2

134. Quadratic Equations

135. Linear Inequalities

136. Graphs of Inequalities; Interval Notation There are infinitely many solutions to the inequality x >  4, namely all real numbers that are greater than  4. Although we cannot list all the solutions, we can make a drawing on a number line that represents these solutions. Such a drawing is called the graph of the inequality.

137. Graphs of Inequalities; Interval Notation Graphs of solutions to linear inequalities are shown on a number line by shading all points representing numbers that are solutions. Parentheses indicate endpoints that are not solutions. Square brackets indicate endpoints that are solutions.

138. Graph the solutions of a. x < 3 b. x   1 c.  1< x  3. Solution: a. The solutions of x < 3 are all real numbers that are less than 3. They are graphed on a number line by shading all points to the left of 3. The parenthesis at 3 indicates that 3 is not a solution, but numbers such as 2.9999 and 2.6 are. The arrow shows that the graph extends indefinitely to the left. -5 -4 -3 -2 -1 0 1 2 3 Text Example

139. Text Example cont. Graph the solutions of a. x < 3 b. x   1 c.  1< x  3. Solution: b. The solutions of x   1 are all real numbers that are greater than or equal to  1. We shade all points to the right of  1 and the point for  1 itself The bracket at  1 shows that  1 is a solution for the given inequality. The arrow shows that the graph extends indefinitely to the right. -5 -4 -3 -2 -1 0 1 2 3

140. Text Example cont. Graph the solutions of a. x < 3 b. x   1 c.  1< x  3. Solution: c. The inequality  1< x  3 is read "  1 is less than x and x is less than or equal to 3," or "x is greater than  1 and less than or equal to 3." The solutions of  1< x  3 are all real numbers between  1 and 3, not including  1 but including 3. The parenthesis at  1 indicates that  1 is not a solution. By contrast, the bracket at 3 shows that 3 is a solution. Shading indicates the other solutions. -5 -4 -3 -2 -1 0 1 2 3

141. Properties of Inequalities  4x  20 Divide by –4 and reverse the sense of the inequality:  4x   4  20  4 Simplify: x  5 if we multiply or divide both sides of an inequality by the same negative quantity and reverse the direction of the inequality symbol, the result is an equivalent inequality. Negative Multiplication and Division Properties If a  b and c is negative, then ac  bc. If a  b and c is negative, then a  c  b  c. 2x  4 Divide by 2: 2x  2  4  2 Simplify: x  2 If we multiply or divide both sides of an inequality by the same positive quantity, the resulting inequality is equivalent to the original one. Positive Multiplication and Division Properties If a  b and c is positive, then ac  bc. If a  b and c is positive, then a  c  b  c. 2x  3  7 subtract 3: 2x  3  3  7  3 Simplify: 2x  4. If the same quantity is added to or subtracted from both sides of an inequality, the resulting inequality is equivalent to the original one. Addition and Subtraction properties If a  b, then a  c  b  c. If a  b, then a  c  b  c. ExampleThe Property In WordsProperty

142. Example Solve and graph the solution set on a number line: 4x  5  9x  10. SolutionWe will collect variable terms on the left and constant terms on the right. The solution set consists of all real numbers that are greater than or equal to , expressed in interval notation as ( ,  ]. The graph of the solution set is shown as follows: x  3 Simplify. 4x  5  9x  10 This is the given inequality. 4x  5 – 9x  9x  10  9x Subtract 9x from both sides.  5x  5   10 Simplify.  5x  5  5   10  5 Subtract 5 from both sides.  5x   15 Simplify. -5x/5 > -15/5 Divide both sides by  5 and reverse the sense of the inequality.

143. If X is an algebraic expression and c is a positive number, 1. The solutions of |X| < c are the numbers that satisfy  c < X < c. 2. The solutions of |X| > c are the numbers that satisfy X c. These rules are valid if is replaced by  Solving an Absolute Value Inequality

144. Text Example Solve and graph: |x  4| < 3. Solution |x  4| < 3 means  3< x  4< 3 |X| < c means  c < X < c We solve the compound inequality by adding 4 to all three parts.  3 < x  4 < 3  3  4 < x  4  4 < 3  4 1 < x < 7 The solution set is all real numbers greater than 1 and less than 7, denoted by {x| 1 < x < 7} or (1, 7). The graph of the solution set is shown as follows:

145. Linear Inequalities