Slide 1 / 200. Quadratic Functions

Transcription

1 Slide 1 / 200 Quadratic Functions

2 Key Terms Slide 2 / 200 Table of Contents Identify Quadratic Functions Explain Characteristics of Quadratic Functions Solve Quadratic Equations by Graphing Solve Quadratic Equations by Factoring Solve Quadratic Equations Using Square Roots Solve Quadratic Equations by Completing the Square Solve Quadratic Equations by Using the Quadratic Formula The Discriminant Solving Non-Quadratics Solving Rational Equations Solving Radical Equations Quadratic & Rational Inequalities

3 Slide 3 / 200 Key Terms Return to Table of Contents

4 Slide 4 / 200 Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves Completing the square: Adding a term to x 2 + bx to form a trinomial that is a perfect square Discriminant: b 2-4ac in a quadratic in standard form

5 Slide 5 / 200 Maximum: The y-value of the vertex if a < 0 and the parabola opens downward Minimum: The y-value of the vertex if a > 0 and the parabola opens upward Parabola: The curve result of graphing a quadratic equation Min Max

6 Slide 6 / 200 Quadratic Equation: An equation that can be written in the standard form ax 2 + bx + c = 0. Where a, b and c are real numbers and a does not = 0. Quadratic Function: Any function that can be written in the form y = ax 2 + bx + c. Where a, b and c are real numbers and a does not equal 0. Vertex: The highest or lowest point on a parabola. Zero of a Function: An x value that makes the function equal zero.

7 Slide 7 / 200 Identifying Quadratic Functions Return to Table of Contents

8 Slide 8 / 200 Any function that can be written in the form y = ax 2 + bx + c Where a, b, and c are real numbers and a 0 Examples Question: Is 2x 2 = x + 4 a quadratic equation? Answer: Yes Question: Is 3x - 4 = x + 1 a quadratic equation? Answer: No

9 Slide 9 / 200 Explain Characteristics of Quadratic Functions Return to Table of Contents

10 Slide 10 / 200 A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a is not equal to 0. The form ax 2 + bx + c = 0 is called the standard form of the quadratic equation. The standard form is not unique. For example, x 2 - x + 1 = 0 can be written as the equivalent equation -x 2 + x - 1 = 0. Also, 4x 2-2x + 2 = 0 can be written as the equivalent equation 2x 2 - x + 1 = 0.

11 Slide 11 / 200 Practice writing quadratic equations in standard form: (Reduce if possible.) Write 2x 2 = x + 4 in standard form: 2x 2 - x - 4 = 0

12 Slide 12 / 200 Write 3x = -x in standard form: x 2 + 3x - 7 = 0

13 Slide 13 / 200 Write 6x 2-6x = 12 in standard form: x 2 - x - 2 = 0

14 Slide 14 / 200 Write 3x - 2 = 5x in standard form: Not a quadratic equation

15 Slide 15 / 200

16 Slide 16 / The graph of a quadratic is a parabola, a u-shaped figure. 3. The parabola from a polynomial function will open upward or downward.

17 Slide 17 / A parabola that opens upward contains a vertex that is a minimum point. A parabola that opens downward contains a vertex that is a maximum point.

18 Slide 18 / The domain of a quadratic function is all real numbers.

19 Slide 19 / To determine the range of a quadratic function, ask yourself two questions: Is the vertex a minimum or maximum? What is the y-value of the vertex? If the vertex is a minimum, then the range is all real numbers greater than or equal to the y-value. The range of this quadratic is -6 to

20 Slide 20 / 200 If the vertex is a maximum, then the range is all real numbers less than or equal to the y-value. The range of this quadratic is to 10

21 Slide 21 / An axis of symmetry (also known as a line of symmetry) will divide the parabola into mirror images. The line of symmetry is always a vertical line of the form

22 Slide 22 / The x-intercepts are the points at which a parabola intersects the x-axis. These points are also known as zeroes, roots or solutions and solution sets. Each quadratic function will have two, one or no real x-intercepts.

23 Slide 23 / True or False: The vertex is the highest or lowest value on the parabola. True False

24 Slide 24 / If a parabola opens upward then... A a>0 B a<0 C a=0

25 Slide 25 / The vertical line that divides a parabola into two symmetrical halves is called... A B C D E discriminant perfect square axis of symmetry vertex slice

26 Slide 26 / 200 Quadratic Equations Finding Zeros (x- intercepts)

27 Slide 27 / 200 Solve Quadratic Equations by Graphing Return to Table of Contents

28 Slide 28 / 200 Vocabulary Every quadratic function has a related quadratic equation. A quadratic equation is used to find the zeroes of a quadratic function. When a function intersects the x-axis its y-value is zero. When writing a quadratic function as its related quadratic equation, you replace y with 0. So y = 0. y = ax 2 + bx + c 0 = ax 2 + bx + c ax 2 + bx + c = 0

29 Slide 29 / 200 One way to solve a quadratic equation in standard form is find the zeros of the related function by graphing. A zero is the point at which the parabola intersects the x-axis. A quadratic may have one, two or no zeros.

30 Slide 30 / 200 How many zeros do the parabolas have? What are the values of the zeros? No zeroes 2 zeroes; x = -1 and x=3 1 zero; x=1

31 Slide 31 / 200 One way to solve a quadratic equation in standard form is to find the zeros or x-intercepts of the related function. Solve a quadratic equation by graphing: Step 1 - Write the related function. Step 2 - Graph the related function. Step 3 - Find the zeros (or x intercepts) of the related function.

32 Slide 32 / 200 Step 1 - Write the Related Function 2x 2-18 = 0 2x 2-18 = y y = 2x 2 + 0x - 18

33 Slide 33 / 200 Step 2 - Graph the Function y = 2x 2 + 0x 18 Use the same six step process for graphing The axis of symmetry is x = 0. The vertex is (0,-18). Find the y intercept -- It is -18. Find two other points (2,-10) and (3,0) Vertex The ver substit x-coord Symme equatio Two Points Y Intercept The po passe This o x-valu

34 Slide 34 / 200 Step 2 - Graph the Function y = 2x 2 + 0x 18 Graph the points and reflect them across the axis of symmetry. # x = 0 # (3,0) # # (2,-10) # (0,-18)

35 Slide 35 / 200 Step 3 - Find the zeros y = 2x 2 + 0x 18 Solve the equation by graphing the related function. The zeros appear # x = 0 # (3,0) to be 3 and -3. # # (2,-10) # (0,-18)

36 Slide 36 / 200 Step 3 - Find the zeros y = 2x 2 + 0x 18 Substitute 3 and -3 for x in the quadratic equation. Check 2x 2 18 = 0 2(3) 2 18 = 0 2(9) - 18 = = 0 0 = 0 ü 2(-3) 2 18 = 0 2(9) - 18 = = 0 0 = 0 ü The zeros are 3 and -3.

37 Slide 37 / Solve the equation by graphing the related function. -12x + 18 = -2x 2 Step 1: What of these is the related function? A y = -2x 2-12x + 18 B y = 2x 2-12x - 18 C y = -2x x - 18

38 Slide 38 / What is the axis of symmetry? y = -2x x - 18 A -3 B 3 Formula: -b 2a C 4 D -5

39 Slide 39 / y = -2x x - 18 What is the vertex? A (3,0) B (-3,0) C (4,0) D (-5,0)

40 Slide 40 / y = -2x x - 18 What is the y- intercept? A (0,0) B (0, 18) C (0, -18) D (0, 12)

41 Slide 41 / y = -2x x - 18 If two other points are (5,-8) and (4,-2), what does the graph look like? A B

42 Slide 42 / If two other points are (5,-8) and (4,-2), what does the graph of y = -2x x - 18 look like? A B C D

43 Slide 43 / y = -2x x - 18 What is(are) the zero(s)? A -18 B 4 C 3 D -8

44 Slide 44 / 200 Solve Quadratic Equations by Factoring Return to Table of Contents

45 Slide 45 / 200 Solving Quadratic Equations by Factoring Review of factoring - To factor a quadratic trinomial of the form x 2 + bx + c, find two factors of c whose sum is b. Example - To factor x 2 + 9x + 18, look for factors whose sum is 9. Factors of 18 1 and Sum 2 and and 6 9 x 2 + 9x + 18 = (x + 3)(x + 6)

46 Slide 46 / 200 When c is positive, its factors have the same sign. The sign of b tells you whether the factors are positive or negative. When b is positive, the factors are positive. When b is negative, the factors are negative.

47 Slide 47 / 200 Remember the FOIL method for multiplying binomials 1. Multiply the First terms (x + 3)(x + 2) x x = x 2 2. Multiply the Outer terms (x + 3)(x + 2) x 2 = 2x 3. Multiply the Inner terms (x + 3)(x + 2) 3 x = 3x 4. Multiply the Last terms (x + 3)(x + 2) 3 2 = 6 (x + 3)(x + 2) = x 2 + 2x + 3x + 6 = x 2 + 5x + 6 F O I L

48 Slide 48 / 200 Zero Product Property For all real numbers a and b, if the product of two quantities equals zero, at least one of the quantities equals zero. Numbers Algebra 3(0) = 0 If ab = 0, 4(0) = 0 Then a = 0 or b = 0

49 Slide 49 / 200 Example 1: Solve x 2 + 4x - 12 = 0 (x + 6) (x - 2) = 0 x + 6 = 0 or x - 2 = x = -6 x = 2 Use "FUSE"! Factor the trinomial using the FOIL method. Use the Zero property (-6) - 12 = (-24) - 12 = = 0 0 = 0 or (2) - 12 = = 0 0 = 0 Substitue found value into original equation Equal - problem solved! The solutions are -6 and 2.

50 Slide 50 / 200 Example 2: Solve x = 12x -12x -12x The equation has to be written in standard form x 2-12x + 36 = 0 (ax 2 + bx + c). So subtract 12x from both sides. (x - 6)(x - 6) = 0 x - 6 = x = = 12(6) = = 72 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved!

51 Slide 51 / 200 Example 3: Solve x 2-16x + 64 = 0 (x - 8)(x - 8) = 0 x - 8 = x = (8) + 64 = = = 0 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved! 0 = 0

52 Slide 52 / Solve A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15

53 Slide 53 / Solve A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15

54 Slide 54 / Solve A -12 B -4 C -3 D -2 E 2 F 3 G 4 H 6 I 8 J 12

55 Slide 55 / Solve A -7 B -5 C -3 D -2 E 12 F 3 G 5 H 6 I 7 J 35

56 Slide 56 / Solve A - 3 / 4 F 3 / 4 B - 1 / 2 G 1 / 2 C - 4 / 3 H 4 / 3 D -2 E 2 I -3 J 3

57 Slide 57 / The product of two consecutive even integers is 48. Find the smaller of the two integers. Hint: x(x+2) = 48

58 Slide 58 / The width of a rectangular swimming pool is 10 feet less than its length. The surface area of the pool is 600 square feet. What is the pool's width? Hint: (L)(L - 10) = 600.

59 Slide 59 / 200 Solve Quadratic Equations Using Square Roots Return to Table of Contents

60 Slide 60 / 200 You can solve a quadratic equation by the square root method if you can write it in the form: x² = c If x and c are algebraic expressions, then: written as: x = c or x = - c x = ± c

61 Slide 61 / 200 Solve for z: z² = 49 z = ± 49 z = ±7 The solution set is 7 and -7

62 Slide 62 / 200 A quadratic equation of the form x 2 = c can be solved using the Square Root Property. Example: Solve 4x 2 = 20 4x 2 = x 2 = 5 Divide both sides by 4 to isolate x² x = ± 5 The solution set is 5 and - 5

63 Slide 63 / 200 Solve 5x² = 20 using the square root method: 5x 2 = x 2 = 4 x = 4 or x = - 4 x = ±2

64 Slide 64 / 200 Solve (2x - 1)² = 20 using the square root method. or

65 Slide 65 / When you take the square root of a real number, your answer will always be positive. True False

66 Slide 66 / If x 2 = 16, then x = A 4 B 2 C -2 D 26 E -4

67 Slide 67 / If y 2 = 4, then y = A 4 B 2 C -2 D 26 E -4

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69 Slide 69 / 200

70 Slide 70 / If (3g - 9) 2 + 7= 43, then g = A B C D E

71 Slide 71 / 200 Solving Quadratic Equations by Completing the Square Return to Table of Contents

72 Slide 72 / 200 Form a perfect square trinomial with lead coefficient of 1 x 2 + bx +c where c = ( b / 2 ) 2 Find the value that completes the square.

73 24 Find ( b / 2 ) 2 if b = 14 Slide 73 / 200

74 25 Find ( b / 2 ) 2 if b = -12 Slide 74 / 200

75 Slide 75 / Complete the square to form a perfect square trinomial x x +?

76 Slide 76 / Complete the square to form a perfect square trinomial x 2-6x +?

77 Slide 77 / 200 Solving quadratic equations by completing the square: Step 1 - Write the equation in the form x 2 + bx = c Step 2 - Find (b 2) 2 Step 3 - Complete the square by adding (b 2) 2 to both sides of the equation. Step 4 - Factor the perfect square trinomial. Step 5 - Take the square root of both sides Step 6 - Write two equations, using both the positive and negative square root and solve each equation.

78 Let's look at an example to solve: Slide 78 / 200 x x = 15 x x = 15 Step 1 - Already done! (14 2) 2 = 49 Step 2 - Find (b 2) 2 x x + 49 = Step 3 - Add 49 to both sides (x + 7) 2 = 64 Step 4 - Factor and simplify x + 7 = ±8 Step 5 - Take the square root of both sides x + 7 = 8 or x + 7 = -8 x = 1 or x = -15 Step 6 - Write and solve two equations

79 Slide 79 / 200 Another example to solve: x 2-2x - 2 = 0 x 2-2x - 2 = x 2-2x = 2 Step 1 - Write as x 2 +bx=c (-2 2) 2 = (-1) 2 = 1 Step 2 - Find (b 2) 2 x 2-2x + 1 = Step 3 - Add 1 to both sides (x - 1) 2 = 3 Step 4 - Factor and simplify x - 1 = ± 3 x - 1 = 3 or x - 1 = - 3 x = or x = 1-3 Step 5 - Take the square root of both sides Step 6 - Write and solve two equations

80 Slide 80 / Solve the following by completing the square : x 2 + 6x = -5 A -5 B -2 C -1 D 5 E 2

81 Slide 81 / Solve the following by completing the square : x 2-8x = 20 A -10 B -2 C -1 D 10 E 2

82 Slide 82 / 200

83 A more difficult example: Slide 83 / 200 Write as x 2 +bx=c Find (b 2) 2 Add 25/9 to both sides Factor and simplify Take the square root of both sides or Write and solve two equations

84 Slide 84 / Solve the following by completing the square : A B C D E

85 Slide 85 / 200 Another example to solve: x 2-2x + 2 = 0 x 2-2x + 2 = x 2-2x = -2 Step 1 - Write as x 2 +bx=c (-2 2) 2 = (-1) 2 = 1 Step 2 - Find (b 2) 2 x 2-2x + 1 = Step 3 - Add 1 to both sides (x - 1) 2 = -1 Step 4 - Factor and simplify x - 1 = ± -1 = +i x - 1 = i or x - 1 = i x = 1 + i or x = 1 - i Step 5 - Take the square root of both sides Step 6 - Write and solve two equations

86 Slide 86 / Solve the following by completing the square : A B C D E

87 Slide 87 / 200 Solve Quadratic Equations by Using the Quadratic Formula Return to Table of Contents

88 Slide 88 / 200 At this point you have learned how to solve quadratic equations by: graphing factoring using square roots and completing the square Many quadratic equations may be solved using these methods; however, some cannot be solved using any of these methods. Today we will be given a tool to solve ANY quadratic equation. It ALWAYS works.

89 Slide 89 / 200 The Quadratic Formula The solutions of ax 2 + bx + c = 0, where a 0, are: x = -b ± b 2-4ac 2a "x equals the opposite of b, plus or minus the square root of b squared minus 4ac, all divided by 2a."

90 Slide 90 / 200 Example 1 2x 2 + 3x - 5 = 0 2x 2 + 3x + (-5) = 0 Identify values of a, b and c x = -b ± b 2-4ac 2a Write the Quadratic Formula x = -3 ± 3 2-4(2)(-5) 2(2) Substitute the values of a, b and c continued on next slide

91 Slide 91 / 200 x = -3 ± 9 - (-40) 4 Simplify x = -3 ± 49 4 = -3 ± 7 4 x = or x = Write as two equations x = 1 or x = -5 2 Solve each equation

92 Slide 92 / 200 Example 2 2x = x 2-3 Remember - In order to use the Quadratic Formula, the equation must be in standard form (ax 2 + bx +c = 0). First, rewrite the equation in standard form. 2x = x x -2x 0 = x 2 + (-2x) + (-3) x 2 + (-2x) + (-3) = 0 Use only addition for standard form Flip the equation Now you are ready to use the Quadratic Formula Solution on next slide

93 Slide 93 / 200 x 2 + (-2x) + (-3) = 0 1x 2 + (-2x) + (-3) = 0 Identify values of a, b and c x = -b ± b 2-4ac 2a Write the Quadratic Formula x = -(-2) ± (-2) 2-4(1)(-3) 2(1) Substitute the values of a, b and c Continued on next slide

94 Slide 94 / 200 x = 2 ± 4 - (-12) 2 Simplify x = 2 ± 16 2 = 2 ± 4 2 x = or x = Write as two equations x = 3 or x = -1 Solve each equation

95 Slide 95 / Solve the following equation using the quadratic formula: A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5

96 Slide 96 / Solve the following equation using the quadratic formula: A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5

97 Slide 97 / Solve the following equation using the quadratic formula: A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5

98 Slide 98 / 200 Example 3 x 2-2x - 4 = 0 1x 2 + (-2x) + (-4) = 0 Identify values of a, b and c x = -b ± b 2-4ac 2a Write the Quadratic Formula x = -(-2) ± (-2) 2-4(1)(-4) 2(1) Substitute the values of a, b and c Continued on next slide

99 Slide 99 / 200 x = 2 ± 4 - (-16) 2 Simplify x = 2 ± 20 2 x = or x = Write as two x = or x = equations x = or x = 1-5 x 3.24 or x Use a calculator to estimate x

100 36 Find the larger solution to Slide 100 / 200

101 Slide 101 / Find the smaller solution to

102 Slide 102 / 200 The Discriminant Return to Table of Contents

103 Slide 103 / 200 Discriminant - the part of the equation under the radical sign in a quadratic equation. x = -b ± b 2-4ac 2a b 2-4ac is the discriminant

104 Slide 104 / 200 Remember: The square root of a positive number has two solutions. The square root of zero is 0. The square root of a negative number has no real solution.

105 Slide 105 / 200 Example 4 = ± 2 (2) (2) = 4 and (-2)(-2) = 4 So BOTH 2 and -2 are solutions

106 Slide 106 / 200

107 Slide 107 / 200

108 Slide 108 / 200

109 Slide 109 / 200 ax 2 + bx + c = 0 The discriminant, b 2-4ac, or the part of the equation under the radical sign, may be used to determine the number of real solutions there are to a quadratic equation. If b 2-4ac > 0, the equation has two real solutions If b 2-4ac = 0, the equation has one real solution If b 2-4ac < 0, the equation has no real solutions

110 Slide 110 / What is value of the discriminant of 2x 2-2x + 3 = 0?

111 Slide 111 / Find the number of solutions using the discriminant for 2x 2-2x + 3 = 0 A 0 B 1 C 2

112 Slide 112 / What is value of the discriminant of x 2-8x + 4 = 0?

113 Slide 113 / 200 Find the number of solutions using the discriminant for x 2-8x + 4 = 0 A 0 B 1 C 2

114 Slide 114 / 200 Solving Non-Quadratics Return to Table of Contents

115 Slide 115 / 200 Summary of Factoring Factor the Polynomial 2 Terms Factor out GCF 3 Terms 4 Terms Difference of Squares Sum/Difference of Cubes Perfect Square Trinomial Factor the Trinomial Group and Factor out GCF. Look for a Common Binomial a = 1 a = 1 Check each factor to see if it can be factored again. If a polynomial cannot be factored, then it is called prime.

116 Slide 116 / 200 Sum and Difference of Cubes

117 Slide 117 / 200

118 Examples: Slide 118 / 200

119 Slide 119 / 200

120 Slide 120 / Factor A (3 + 2a)(9 + 4a + a 2 ) B (3-2a)(9 + 4a + a 2 ) C (3 + 2a)(9-4a + a 2 ) D (3-2a)(9-4a + a 2 )

121 Slide 121 / Factor completely A (a 2-1)(a 4 + 4a 2 + 1) B (a 2-1)(a 4-4a 2 + 1) C (a - 1)(a + 1)(a 4 + 4a 2 + 1) D (a + 1)(a - 1)(a 4-4a 2 + 1)

122 Slide 122 / 200 Factoring 4 Term Polynomials

123 Slide 123 / 200 Polynomials with four terms like ab - 4b + 6a - 24, can be factored by grouping terms of the polynomials. Example 1: ab - 4b + 6a - 24 (ab - 4b) + (6a - 24) b(a - 4) + 6(a - 4) Group terms into binomials that can be factored using the distributive property Factor the GCF (a - 4) (b + 6) Notice that a - 4 is a common binomial factor and factor!

124 Slide 124 / 200 Example 2: 6xy + 8x - 21y - 28 (6xy + 8x) + (-21y - 28) Group 2x(3y + 4) + (-7)(3y + 4) Factor GCF (3y +4) (2x - 7) Factor common binomial

125 Slide 125 / 200 You must be able to recognize additive inverses!!! (3 - a and a - 3 are additive inverses because their sum is equal to Remember 3 - a = -1(a - 3). Example 3: 15x - 3xy + 4y - 20 (15x - 3xy) + (4y - 20) Group 3x(5 - y) + 4(y - 5) Factor GCF 3x(-1)(-5 + y) + 4(y - 5) Notice additive inverses -3x(y - 5) + 4(y - 5) Simplify (y - 5) (-3x + 4) Factor common binomial Remember to check each problem by using FOIL.

126 Slide 126 / Factor 15ab - 3a + 10b - 2 A (5b - 1)(3a + 2) B (5b + 1)(3a + 2) C (5b - 1)(3a - 2) D (5b + 1)(3a - 1)

127 Slide 127 / Factor 10m 2 n - 25mn + 6m - 15 A B C D (2m-5)(5mn-3) (2m-5)(5mn+3) (2m+5)(5mn-3) (2m+5)(5mn+3)

128 Slide 128 / Factor 20ab - 35b a A (4a - 7)(5b - 9) B (4a - 7)(5b + 9) C (4a + 7)(5b - 9) D (4a + 7)(5b + 9)

129 Slide 129 / Factor a 2 - ab + 7b - 7a A (a - b)(a - 7) B (a - b)(a + 7) C (a + b)(a - 7) D (a + b)(a + 7)

130 Slide 130 / 200 Mixed Factoring

131 Slide 131 / 200 Summary of Factoring Factor the Polynomial 2 Terms Factor out GCF 3 Terms 4 Terms Difference of Squares Sum/Difference of Cubes Perfect Square Trinomial Factor the Trinomial Group and Factor out GCF. Look for a Common Binomial a = 1 a = 1 Check each factor to see if it can be factored again. If a polynomial cannot be factored, then it is called prime.

132 Slide 132 / 200

133 Slide 133 / Factor completely: A B C D

134 Slide 134 / Factor completely A B C D prime polynomial

135 Slide 135 / Factor A B C D prime polynomial

136 Slide 136 / 200

137 Slide 137 / Factor A B C D Prime Polynomial

138 Slide 138 / 200 Solving Equations by Factoring

139 Slide 139 / 200 Given the following equation, what conclusion(s) can be drawn? ab = 0 Since the product is 0, one of the factors, a or b, must be 0. This is known as the Zero Product Property.

140 Slide 140 / 200 Given the following equation, what conclusion(s) can be dr (x - 4)(x + 3) = 0 Since the product is 0, one of the factors must be 0. Therefore, either x - 4 = 0 or x + 3 = 0. x - 4 = 0 or x + 3 = x = 4 or x = -3 Therefore, our solution set is {-3, 4}. To verify the results, substitute solution back into the original equation. To check x = -3: (x - 4)(x + 3) = 0 To check x = 4: (x - 4)(x + 3) = 0 (-3-4)(-3 + 3) = 0 (-7)(0) = 0 0 = 0 (4-4)(4 + 3) = 0 (0)(7) = 0 0 = 0

141 Slide 141 / 200 What if you were given the following equation? How would you solve it? We can use the Zero Product Property to solve it. How can we turn this polynomial into a multiplication problem? Fac Factoring yields: x(x - 6)(x + 4) = 0 By the Zero Product Property: x = 0 x - 6 = 0 or x + 4 = 0 After solving each equation, we arrive at our solution: {0,-4, 6}

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143 Slide 143 / 200 Zero Product rule works only when the product of factors equals zero. If the equation equals some value other than zero subtract to make one side of the equation zero. Example

144 Slide 144 / 200

145 Slide 145 / Choose all of the solutions to: A B C D E F

146 Slide 146 / Choose all of the real solutions to: A B C D E F

147 Slide 147 / Choose all of the solutions to: A B C D E F

148 Slide 148 / 200

149 Slide 149 / A ball is thrown with its height at any time given by When does the ball hit the ground? -1 seconds A B C D 0 seconds 9 seconds 10 seconds

150 Slide 150 / 200 Solving Rational Equations Return to Table of Contents

151 Slide 151 / 200 Steps to Solving a Rational Equation 1) Find LCD 2) Multiply each term by LCD 3) Reduce 4) Solve 5) Verify answer works (Answer may make denominator = 0) Example:

152 Check: Slide 152 / 200

153 Slide 153 / 200

154 Example: Slide 154 / 200

155 Slide 155 / 200 Check: x = 7 Check: x = -2

156 Slide 156 / Solve the equation. Check to see it works.

157 Slide 157 / Solve the equation. Check to see it works.

158 Slide 158 / 200

159 Slide 159 / 200

160 Slide 160 / 200 Solving Radical Equations Return to Table of Contents

161 Slide 161 / 200 To solve a radical equation: isolate the radical on one side of the equation use the index to determine the power to eliminate the radical solve the equation check to see if solution is extraneous

162 Example Slide 162 / 200

163 Slide 163 / 200

164 64 Find the solution to Slide 164 / 200

165 65 Find the solution to Slide 165 / 200

166 Slide 166 / 200

167 67 Find the solution to Slide 167 / 200

168 Slide 168 / 200 If an equation has multiple roots, move them to opposite sides of the equal sign and then solve.

169 68 Solve the following: Slide 169 / 200

170 69 Solve the following: Slide 170 / 200

171 Slide 171 / If the distance between (3,5) and (x,9) is 7, find x.

172 Slide 172 / 200 Quadratic & Rational Inequalities Return to Table of Contents

173 Slide 173 / 200

174 Slide 174 / 200

175 Slide 175 / 200 Graph Step 1: Graph Points on the Bounds X Y

176 Slide 176 / 200 Graph Step 2: Solid or Dotted?

177 Slide 177 / 200 Graph Step 3: Shade

178 Slide 178 / 200 Graph Step 1: Graph Points on the Bounds X Y

179 Slide 179 / 200 Graph Step 2: Solid or Dotted?

180 Slide 180 / 200 Graph Step 3: Shade

181 Graph Slide 181 / 200

182 Slide 182 / 200

183 Slide 183 / 200

184 Slide 184 / Which equation is graphed? A f(x) > -4x 2 + 2x + 5 B f(x) > -4x 2 + 2x + 5 C f(x) < -4x 2 + 2x + 5 D f(x) < -4x 2 + 2x + 5

185 Slide 185 / Which equation is graphed? A f(x) > -4x 2 + 2x + 5 B f(x) > -4x 2 + 2x + 5 C f(x) < -4x 2 + 2x + 5 D f(x) < -4x 2 + 2x + 5

186 Slide 186 / 200 Solving Method 1: Graphically Graph the related function The solution is where the shaded region intersects the x-axis

187 Slide 187 / 200 Note: It is possible to have a solution of a point or the empty set.

188 Slide 188 / 200 Find the solution set given the graph of the related function

189 Slide 189 / Solve the following inequality: A 1 < x< 4 B 1 < x < 4 C x < 1 or x > 4 D x < 1 or x > 4 1 4

190 Slide 190 / Solve the following inequality: A 1 < x< 4 B 1 < x < 4 C x < 1 or x > 4 D x < 1 or x > 4 1 4

191 Slide 191 / Solve the following inequality: A -4 < x< 2 B -4 < x < 2 C x < -4 or x > 2 D x < -4 or x > 2

192 Slide 192 / Solve the following inequality: A 2 < x< 5 B 2 < x < 5 C x < 2 or x > 5 D x < 2 or x > 5

193 Slide 193 / 200 Steps to Solving Quadratic Inequalities Algebraically 1) Get inequality so that it is compared to zero ie. ax 2 + bx + c > 0 2) Factor 3) Set each factor equal to zero and solve 4) Create a number line with the solution as the points 5) Test points in each region to see if they satisfy the inequality 6) Write the solution

194 Slide 194 / 200 Solve -4-2 Test points: -10, -3, 0 x 1) Rewrite inequality <0 2) Factor 3) Solve 4) Create number line 5) Test points F T -4-2 F x 6) Write the solution

195 Slide 195 / 200 Solve Test points: -2, 0, 2 x 1) Rewrite inequality <0 2) Factor 3) Solve 4) Create number line 5) Test points T F T x 6) Write the solution

196 Solve Slide 196 / 200

197 Slide 197 / Solve A x < -3 or x > 2 B -3 < x < 2 C D All Reals No Solution

198 Slide 198 / Solve A x < -2 or x > 5 B -2 < x < 5 C D All Reals No Solution

199 Slide 199 / Solve A x = -2 B x = 2 C D All Reals No Solution

200 Slide 200 / Solve A -3 < x < -2 B x < -3 or x > -2 C D All Reals No Solution