Slide 1 / 200 Quadratic Functions
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
Slide 3 / 200 Key Terms Return to Table of Contents
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
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
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.
Slide 7 / 200 Identifying Quadratic Functions Return to Table of Contents
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
Slide 9 / 200 Explain Characteristics of Quadratic Functions Return to Table of Contents
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.
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
Slide 12 / 200 Write 3x = -x 2 + 7 in standard form: x 2 + 3x - 7 = 0
Slide 13 / 200 Write 6x 2-6x = 12 in standard form: x 2 - x - 2 = 0
Slide 14 / 200 Write 3x - 2 = 5x in standard form: Not a quadratic equation
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Slide 16 / 200 2. The graph of a quadratic is a parabola, a u-shaped figure. 3. The parabola from a polynomial function will open upward or downward.
Slide 17 / 200 4. 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.
Slide 18 / 200 5. The domain of a quadratic function is all real numbers.
Slide 19 / 200 6. 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
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
Slide 21 / 200 7. 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
Slide 22 / 200 8. 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.
Slide 23 / 200 1 True or False: The vertex is the highest or lowest value on the parabola. True False
Slide 24 / 200 2 If a parabola opens upward then... A a>0 B a<0 C a=0
Slide 25 / 200 3 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
Slide 26 / 200 Quadratic Equations Finding Zeros (x- intercepts)
Slide 27 / 200 Solve Quadratic Equations by Graphing Return to Table of Contents
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
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.
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
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.
Slide 32 / 200 Step 1 - Write the Related Function 2x 2-18 = 0 2x 2-18 = y y = 2x 2 + 0x - 18
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
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)
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)
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 18-18 = 0 0 = 0 ü 2(-3) 2 18 = 0 2(9) - 18 = 0 18-18 = 0 0 = 0 ü The zeros are 3 and -3.
Slide 37 / 200 4 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 2 + 12x - 18
Slide 38 / 200 5 What is the axis of symmetry? y = -2x 2 + 12x - 18 A -3 B 3 Formula: -b 2a C 4 D -5
Slide 39 / 200 6 y = -2x 2 + 12x - 18 What is the vertex? A (3,0) B (-3,0) C (4,0) D (-5,0)
Slide 40 / 200 7 y = -2x 2 + 12x - 18 What is the y- intercept? A (0,0) B (0, 18) C (0, -18) D (0, 12)
Slide 41 / 200 8 y = -2x 2 + 12x - 18 If two other points are (5,-8) and (4,-2), what does the graph look like? A B
Slide 42 / 200 9 If two other points are (5,-8) and (4,-2), what does the graph of y = -2x 2 + 12x - 18 look like? A B C D
Slide 43 / 200 10 y = -2x 2 + 12x - 18 What is(are) the zero(s)? A -18 B 4 C 3 D -8
Slide 44 / 200 Solve Quadratic Equations by Factoring Return to Table of Contents
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 18 19 Sum 2 and 9 11 3 and 6 9 x 2 + 9x + 18 = (x + 3)(x + 6)
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.
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
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
Slide 49 / 200 Example 1: Solve x 2 + 4x - 12 = 0 (x + 6) (x - 2) = 0 x + 6 = 0 or x - 2 = 0-6 - 6 + 2 +2 x = -6 x = 2 Use "FUSE"! Factor the trinomial using the FOIL method. Use the Zero property -6 2 + 4(-6) - 12 = 0-6 2 + (-24) - 12 = 0 36-24 - 12 = 0 0 = 0 or 2 2 + 4(2) - 12 = 0 4 + 8-12 = 0 0 = 0 Substitue found value into original equation Equal - problem solved! The solutions are -6 and 2.
Slide 50 / 200 Example 2: Solve x 2 + 36 = 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 = 0 +6 +6 x = 6 6 2 + 36 = 12(6) 36 + 36 = 72 72 = 72 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved!
Slide 51 / 200 Example 3: Solve x 2-16x + 64 = 0 (x - 8)(x - 8) = 0 x - 8 = 0 +8 +8 x = 8 8 2-16(8) + 64 = 0 64-128 + 64 = 0-64 + 64 = 0 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved! 0 = 0
Slide 52 / 200 11 Solve A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15
Slide 53 / 200 12 Solve A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15
Slide 54 / 200 13 Solve A -12 B -4 C -3 D -2 E 2 F 3 G 4 H 6 I 8 J 12
Slide 55 / 200 14 Solve A -7 B -5 C -3 D -2 E 12 F 3 G 5 H 6 I 7 J 35
Slide 56 / 200 15 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
Slide 57 / 200 16 The product of two consecutive even integers is 48. Find the smaller of the two integers. Hint: x(x+2) = 48
Slide 58 / 200 17 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.
Slide 59 / 200 Solve Quadratic Equations Using Square Roots Return to Table of Contents
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
Slide 61 / 200 Solve for z: z² = 49 z = ± 49 z = ±7 The solution set is 7 and -7
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 = 20 4 4 x 2 = 5 Divide both sides by 4 to isolate x² x = ± 5 The solution set is 5 and - 5
Slide 63 / 200 Solve 5x² = 20 using the square root method: 5x 2 = 20 5 5 x 2 = 4 x = 4 or x = - 4 x = ±2
Slide 64 / 200 Solve (2x - 1)² = 20 using the square root method. or
Slide 65 / 200 18 When you take the square root of a real number, your answer will always be positive. True False
Slide 66 / 200 19 If x 2 = 16, then x = A 4 B 2 C -2 D 26 E -4
Slide 67 / 200 20 If y 2 = 4, then y = A 4 B 2 C -2 D 26 E -4
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Slide 70 / 200 23 If (3g - 9) 2 + 7= 43, then g = A B C D E
Slide 71 / 200 Solving Quadratic Equations by Completing the Square Return to Table of Contents
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.
24 Find ( b / 2 ) 2 if b = 14 Slide 73 / 200
25 Find ( b / 2 ) 2 if b = -12 Slide 74 / 200
Slide 75 / 200 26 Complete the square to form a perfect square trinomial x 2 + 18x +?
Slide 76 / 200 27 Complete the square to form a perfect square trinomial x 2-6x +?
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.
Let's look at an example to solve: Slide 78 / 200 x 2 + 14x = 15 x 2 + 14x = 15 Step 1 - Already done! (14 2) 2 = 49 Step 2 - Find (b 2) 2 x 2 + 14x + 49 = 15 + 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
Slide 79 / 200 Another example to solve: x 2-2x - 2 = 0 x 2-2x - 2 = 0 +2 +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 = 2 + 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 = 1 + 3 or x = 1-3 Step 5 - Take the square root of both sides Step 6 - Write and solve two equations
Slide 80 / 200 28 Solve the following by completing the square : x 2 + 6x = -5 A -5 B -2 C -1 D 5 E 2
Slide 81 / 200 29 Solve the following by completing the square : x 2-8x = 20 A -10 B -2 C -1 D 10 E 2
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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
Slide 84 / 200 31 Solve the following by completing the square : A B C D E
Slide 85 / 200 Another example to solve: x 2-2x + 2 = 0 x 2-2x + 2 = 0-2 -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 = -2 + 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
Slide 86 / 200 32 Solve the following by completing the square : A B C D E
Slide 87 / 200 Solve Quadratic Equations by Using the Quadratic Formula Return to Table of Contents
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.
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."
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
Slide 91 / 200 x = -3 ± 9 - (-40) 4 Simplify x = -3 ± 49 4 = -3 ± 7 4 x = -3 + 7 4 or x = -3-7 4 Write as two equations x = 1 or x = -5 2 Solve each equation
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 2-3 -2x -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
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
Slide 94 / 200 x = 2 ± 4 - (-12) 2 Simplify x = 2 ± 16 2 = 2 ± 4 2 x = 2 + 4 2 or x = 2-4 2 Write as two equations x = 3 or x = -1 Solve each equation
Slide 95 / 200 33 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
Slide 96 / 200 34 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
Slide 97 / 200 35 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
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
Slide 99 / 200 x = 2 ± 4 - (-16) 2 Simplify x = 2 ± 20 2 x = 2 + 20 2 or x = 2-20 2 Write as two x = 2 +2 5 2 or x = 2-2 5 2 equations x = 1 + 5 or x = 1-5 x 3.24 or x -1.24 Use a calculator to estimate x
36 Find the larger solution to Slide 100 / 200
Slide 101 / 200 37 Find the smaller solution to
Slide 102 / 200 The Discriminant Return to Table of Contents
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
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.
Slide 105 / 200 Example 4 = ± 2 (2) (2) = 4 and (-2)(-2) = 4 So BOTH 2 and -2 are solutions
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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
Slide 110 / 200 38 What is value of the discriminant of 2x 2-2x + 3 = 0?
Slide 111 / 200 39 Find the number of solutions using the discriminant for 2x 2-2x + 3 = 0 A 0 B 1 C 2
Slide 112 / 200 40 What is value of the discriminant of x 2-8x + 4 = 0?
Slide 113 / 200 Find the number of solutions using the discriminant for x 2-8x + 4 = 0 A 0 B 1 C 2
Slide 114 / 200 Solving Non-Quadratics Return to Table of Contents
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.
Slide 116 / 200 Sum and Difference of Cubes
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Examples: Slide 118 / 200
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Slide 120 / 200 43 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 )
Slide 121 / 200 44 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)
Slide 122 / 200 Factoring 4 Term Polynomials
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!
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
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.
Slide 126 / 200 45 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)
Slide 127 / 200 46 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)
Slide 128 / 200 47 Factor 20ab - 35b - 63 +36a A (4a - 7)(5b - 9) B (4a - 7)(5b + 9) C (4a + 7)(5b - 9) D (4a + 7)(5b + 9)
Slide 129 / 200 48 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)
Slide 130 / 200 Mixed Factoring
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.
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Slide 133 / 200 49 Factor completely: A B C D
Slide 134 / 200 50 Factor completely A B C D prime polynomial
Slide 135 / 200 51 Factor A B C D prime polynomial
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Slide 137 / 200 53 Factor A B C D Prime Polynomial
Slide 138 / 200 Solving Equations by Factoring
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.
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 = 0 + 4 + 4-3 - 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
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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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
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Slide 145 / 200 54 Choose all of the solutions to: A B C D E F
Slide 146 / 200 55 Choose all of the real solutions to: A B C D E F
Slide 147 / 200 56 Choose all of the solutions to: A B C D E F
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Slide 149 / 200 58 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
Slide 150 / 200 Solving Rational Equations Return to Table of Contents
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:
Check: Slide 152 / 200
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Example: Slide 154 / 200
Slide 155 / 200 Check: x = 7 Check: x = -2
Slide 156 / 200 59 Solve the equation. Check to see it works.
Slide 157 / 200 60 Solve the equation. Check to see it works.
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Slide 160 / 200 Solving Radical Equations Return to Table of Contents
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
Example Slide 162 / 200
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64 Find the solution to Slide 164 / 200
65 Find the solution to Slide 165 / 200
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67 Find the solution to Slide 167 / 200
Slide 168 / 200 If an equation has multiple roots, move them to opposite sides of the equal sign and then solve.
68 Solve the following: Slide 169 / 200
69 Solve the following: Slide 170 / 200
Slide 171 / 200 70 If the distance between (3,5) and (x,9) is 7, find x.
Slide 172 / 200 Quadratic & Rational Inequalities Return to Table of Contents
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Slide 175 / 200 Graph Step 1: Graph Points on the Bounds X Y -5-7 -4-10 -3-11 -2-10 -1-7 0-2
Slide 176 / 200 Graph Step 2: Solid or Dotted?
Slide 177 / 200 Graph Step 3: Shade
Slide 178 / 200 Graph Step 1: Graph Points on the Bounds X Y -3-2 -1 0 1
Slide 179 / 200 Graph Step 2: Solid or Dotted?
Slide 180 / 200 Graph Step 3: Shade
Graph Slide 181 / 200
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Slide 184 / 200 73 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
Slide 185 / 200 74 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
Slide 186 / 200 Solving Method 1: Graphically Graph the related function The solution is where the shaded region intersects the x-axis. -3 2-3 2
Slide 187 / 200 Note: It is possible to have a solution of a point or the empty set.
Slide 188 / 200 Find the solution set given the graph of the related function. -2 3-3 -.5 1 2 -.5 1
Slide 189 / 200 75 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
Slide 190 / 200 76 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
Slide 191 / 200 77 Solve the following inequality: A -4 < x< 2 B -4 < x < 2 C x < -4 or x > 2 D x < -4 or x > 2
Slide 192 / 200 78 Solve the following inequality: A 2 < x< 5 B 2 < x < 5 C x < 2 or x > 5 D x < 2 or x > 5
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
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
Slide 195 / 200 Solve -1.5 1 Test points: -2, 0, 2 x 1) Rewrite inequality <0 2) Factor 3) Solve 4) Create number line 5) Test points T F -1.5 1 T x 6) Write the solution
Solve Slide 196 / 200
Slide 197 / 200 79 Solve A x < -3 or x > 2 B -3 < x < 2 C D All Reals No Solution
Slide 198 / 200 80 Solve A x < -2 or x > 5 B -2 < x < 5 C D All Reals No Solution
Slide 199 / 200 81 Solve A x = -2 B x = 2 C D All Reals No Solution
Slide 200 / 200 82 Solve A -3 < x < -2 B x < -3 or x > -2 C D All Reals No Solution