Slide 1 / 175 Slide 2 / 175 Algebra I Quadratics 2015-11-04 www.njctl.org Key Terms Table of Contents Click on the topic to go to that section Slide 3 / 175 Characteristics of Quadratic Equations Transforming Quadratic Equations Graphing Quadratic Equations Solve Quadratic Equations by Graphing Solve Quadratic Equations by Factoring Solve Quadratic Equations Using Square Roots Solve Quadratic Equations by Completing the Square The Discriminant Solve Quadratic Equations by Using the Quadratic Formula Solving Application Problems
Slide 4 / 175 Key Terms Return to Table of Contents Axis of Symmetry Slide 5 / 175 Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves Parabolas Slide 6 / 175 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 (+ a) Min Max (- a)
Quadratics Slide 7 / 175 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. Vertex: The highest or lowest point on a parabola. Zero of a Function: An x value that makes the function equal zero. Slide 8 / 175 Characteristics of Quadratic Equations Return to Table of Contents Quadratics Slide 9 / 175 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. Why is this equivalent?
Writing Quadratic Equations Slide 10 / 175 Practice writing quadratic equations in standard form: (Simplify if possible.) Write 2x 2 = x + 4 in standard form: 1 Write 3x = -x 2 + 7 in standard form: Slide 11 / 175 A. x 2 + 3x-7= 0 B. x 2-3x +7=0 C. -x 2-3x -7= 0 2 Write 6x 2-6x = 12 in standard form: A. 6x 2-6x -12 = 0 Slide 12 / 175 B. x 2 - x - 2 = 0 C. -x 2 + x + 2 = 0
3 Write 3x - 2 = 5x in standard form: A. 2x + 2 = 0 Slide 13 / 175 B. -2x - 2 = 0 C. not a quadratic equation Characteristics of Quadratic Functions Slide 14 / 175 The graph of a quadratic is a parabola, a u-shaped figure. The parabola will open upward or downward. upward downward Characteristics of Quadratic Functions Slide 15 / 175 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. vertex vertex
Characteristics of Quadratic Functions Slide 16 / 175 The domain of a quadratic function is all real numbers. D = Reals Characteristics of Quadratic Functions To determine the range of a quadratic function, ask yourself two questions: Slide 17 / 175 > 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, ) Characteristics of Quadratic Functions Slide 18 / 175 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 (,10]
Characteristics of Quadratic Functions Slide 19 / 175 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 x = b 2a y = 2x 2 8x + 2 x = ( 8) 2(2) = 2 x=2 Characteristics of Quadratic Functions Slide 20 / 175 To find the axis of symmetry simply plug the values of a and b into the equation: Remember the form ax 2 + bx + c. In this example a = 2, b = -8 and c =2 a b c x = b 2a x=2 y = 2x 2 8x + 2 x = ( 8) 2(2) = 2 Characteristics of Quadratic Functions 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 equation will have two, one or no real x-intercepts. Slide 21 / 175
4 The vertical line that divides a parabola into two symmetrical halves is called... Slide 22 / 175 A discriminant B perfect square C axis of symmetry D vertex E slice Slide 23 / 175 6 The equation y = x 2 + 3x 18 is graphed on the set of axes below. Slide 24 / 175 Based on this graph, what are the roots of the equation x 2 + 3x 18 = 0? A B C D 3 and 6 0 and 18 3 and 6 3 and 18 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011.
7 The equation y = x 2 2x + 8 is graphed on the set of axes below. Slide 25 / 175 Based on this graph, what are the roots of the equation x 2 2x + 8 = 0? A 8 and 0 B 2 and 4 C 9 and 1 D 4 and 2 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. 8 What is an equation of the axis of symmetry of the parabola represented by y = x 2 + 6x 4? Slide 26 / 175 A x = 3 B y = 3 C x = 6 D y = 6 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. Slide 27 / 175 9 The height, y, of a ball tossed into the air can be represented by the equation y = x 2 + 10x + 3, where x is the elapsed time. What is the equation of the axis of symmetry of this parabola? A y = 5 B y = 5 C x = 5 D x = 5 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011.
Slide 28 / 175 Slide 29 / 175 Transforming Quadratic Equations Return to Table of Contents Quadratic Parent Equation y = 2 x 2 3 The quadratic parent equation is y = x 2. The graph of all other quadratic equations are transformations of the graph of y= x 2. x x 2 y = x 2-3 9-2 4-1 1 0 0 1 1 2 4 3 9 Slide 30 / 175
Quadratic Parent Equation Slide 31 / 175 The quadratic parent equation is y = x 2. How is y = x 2 changed into y = 2x 2? y = 2x 2 y = x 2 x 2-3 18-2 8-1 2 0 0 1 2 2 8 3 18 Quadratic Parent Equation Slide 32 / 175 The quadratic parent equation is y = x 2. How is y = x 2 changed into y =.5x 2? y = x 2 1 y = x 2 2 x 0.5-3 4.5-2 2-1 0.5 0 0 1 0.5 2 2 3 4.5 What Does "A" Do? What does "a" do in y = ax 2 + bx + c? How does a > 0 affect the parabola? How does a < 0 affect the parabola? Slide 33 / 175 y = x 2 y = x 2
What Does "A" Do? What does "a" also do in y =ax 2 + bx +c? Slide 34 / 175 y = How does your conclusion about "a" change as "a" changes? 1 2 x2 y = 3x 2 y = x 2 1 y = 1x 2 y = 3x 2 y = x 2 2 What Does "A" Do? Slide 35 / 175 What does "a" do in y = ax 2 + bx + c? If a > 0, the graph opens up. If a < 0, the graph opens down. If the absolute value of a is > 1, then the graph of the equation is narrower than the graph of the parent equation. If the absolute value of a is < 1, then the graph of the equation is wider than the graph of the parent equation. 11 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent equation. Slide 36 / 175 y =.3x 2 A B C D up, wider up, narrower down, wider down, narrower
12 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function. Slide 37 / 175 y = 4x 2 A B C D up, wider up, narrower down, wider down, narrower 13 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent equation. Slide 38 / 175 y = 2x 2 + 100x + 45 A B C D up, wider up, narrower down, wider down, narrower 14 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function. Slide 39 / 175 A B C D up, wider up, narrower down, wider down, narrower y = 2 x 3 2
15 Without graphing determine which direction does the parabola open and if the graph is wider or narrower than the parent function. Slide 40 / 175 A B C D up, wider up, narrower down, wider down, narrower y = 7 x 5 2 What Does "C" Do? Slide 41 / 175 What does "c" do in y = ax 2 + bx + c? y = x 2 + 6 y = x 2 + 3 y = x 2 y = x 2 2 y = x 2 5 y = x 2 9 What Does "C" Do? Slide 42 / 175 What does "c" do in y = ax 2 + bx + c? "c" moves the graph up or down the same value as "c." "c" is the y- intercept.
16 Without graphing, what is the y- intercept of the the given parabola? Slide 43 / 175 y = x 2 + 17 17 Without graphing, what is the y- intercept of the the given parabola? Slide 44 / 175 y = x 2 6 18 Without graphing, what is the y- intercept of the the given parabola? Slide 45 / 175 y = 3x 2 + 13x 9
19 Without graphing, what is the y- intercept of the the given parabola? Slide 46 / 175 y = 2x 2 + 5x 20 Choose all that apply to the following quadratic: Slide 47 / 175 f(x) =.7x 2 4 A opens up A y-intercept of y = 4 B opens down B y-intercept of y = 2 C wider than parent function C y-intercept of y = 0 D narrower than parent D y-intercept of y = 2 function E y-intercept of y = 4 F y-intercept of y = 6 21 Choose all that apply to the following quadratic: Slide 48 / 175 A B C D opens up opens down wider than parent function narrower than parent function f(x) = 4 x 2 6x 3 E y-intercept of y = 4 F y-intercept of y = 2 G y-intercept of y = 0 H y-intercept of y = 2 I y-intercept of y = 4 J y-intercept of y = 6
Slide 49 / 175 Slide 50 / 175 Graphing Quadratic Equations Return to Table of Contents Graph by Following Six Steps: Slide 51 / 175 Step 1 - Find Axis of Symmetry Step 2 - Find Vertex Step 3 - Find Y intercept Step 4 - Find two more points Step 5 - Partially graph Step 6 - Reflect
Step 1 - Find Axis of Symmetry Axis of Symmetry Slide 52 / 175 What is the Axis of Symmetry? Axis of Symmetry Slide 53 / 175 Slide 54 / 175
Step 3 - Find y intercept What is the y-intercept? Slide 55 / 175 y- intercept Step 3 - Find y intercept Slide 56 / 175 Graph y = 3x 2 6x + 1 The y- intercept is always the c value, because x = 0. y = ax 2 + bx + c y = 3x 2 6x + 1 c = 1 The y-intercept is 1 and the graph passes through (0,1). Step 4 - Find Two More Points Slide 57 / 175 Graph y = 3x 2 6x + 1 Find two more points on the parabola. Choose different values of x and plug in to find points. Let's pick x = 1 and x = 2 y = 3x 2 6x + 1 y = 3( 1) 2 6( 1) + 1 y = 3 + 6 + 1 y = 10 ( 1,10)
Step 4 - Find Two More Points (continued) Slide 58 / 175 Graph y = 3x 2 6x + 1 y = 3x 2 6x + 1 y = 3( 2) 2 6( 2) + 1 y = 3(4) + 12 + 1 y = 25 ( 2, 25) Step 5 - Graph the Axis of Symmetry Slide 59 / 175 Graph the axis of symmetry, the vertex, the point containing the y-intercept and two other points. Step 6 - Reflect the Points Slide 60 / 175 Reflect the points across the axis of symmetry. Connect the points with a smooth curve. (4,25)
23 What is the axis of symmetry for y = x 2 + 2x - 3 (Step 1)? Slide 61 / 175 24 What is the vertex for y = x 2 + 2x - 3 (Step 2)? Slide 62 / 175 A (-1, -4) B (1, -4) C (-1, 4) 25 What is the y-intercept for y = x 2 + 2x - 3 (Step 3)? Slide 63 / 175 A -3 B 3
Graph y= x 2 + 2x 3 Slide 64 / 175 axis of symmetry = 1 vertex = 1, 4 y intercept = 3 2 other points (step 4) (1,0) (2,5) Partially graph (step 5) Reflect (step 6) Graph Slide 65 / 175 y = 2x 2 6x + 4 Graph Slide 66 / 175 y = x 2 4x + 5
Graph Slide 67 / 175 y = 3x 2 7 Slide 68 / 175 Solve Quadratic Equations by Graphing Return to Table of Contents Find the Zeros Slide 69 / 175 One way to solve a quadratic equation in standard form is find the zeros by graphing. A zero is the point at which the parabola intersects the x-axis. A quadratic may have one, two or no zeros.
Find the Zeros How many zeros do the parabolas have? What are the values of the zeros? Slide 70 / 175 clickno zeroes (doesn't cross the "x" axis) click 2 zeroes; x = -1 and x=3 click 1 zero; x=1 Review Slide 71 / 175 To solve a quadratic equation by graphing follow the 6 step process we already learned. Step 1 - Find Axis of Symmetry Step 2 - Find Vertex Step 3 - Find Y intercept Step 4 - Find two more points Step 5 - Partially graph Step 6 - Reflect 26 Solve the equation by graphing. Slide 72 / 175 12x + 18 = 2x 2 Which of these is in standard form? A B C y = 2x 2 12x + 18 y = 2x 2 12x + 18 y = 2x 2 + 12x 18
27 What is the axis of symmetry? Slide 73 / 175 y = 2x 2 + 12x 18 A 3 B 3 C 4 D 5 Slide 74 / 175 28 y = 2x 2 + 12x 18 What is the vertex? A (3,0) B ( 3,0) C (4,0) D ( 5,0) Slide 75 / 175 29 y = 2x 2 + 12x 18 What is the y- intercept? A (0, 0) B (0, 18) C (0, 18) D (0, 12)
30 A If two other points are (5, 8) and (4, 2),what does the graph of y = 2x 2 + 12x 18 look like? B Slide 76 / 175 C D 31 y = 2x 2 + 12x 18 Slide 77 / 175 What is(are) the zero(s)? A 18 B 4 C 3 D 8 click for graph of answer Slide 78 / 175 Solve Quadratic Equations by Factoring Return to Table of Contents
Solving Quadratic Equations by Factoring Slide 79 / 175 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 Sum 1 and 18 19 2 and 9 11 3 and 6 9 x 2 + 9x + 18 = (x + 3)(x + 6) Solving Quadratic Equations by Factoring Slide 80 / 175 When c is positive, it's 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. Solving Quadratic Equations by Factoring Slide 81 / 175 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
Zero Product Property Slide 82 / 175 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 Zero Product Property Example 1: Solve x 2 + 4x 12 = 0 Use "FUSE"! Slide 83 / 175 (x + 6) (x 2) = 0 x + 6 = 0 or x 2 = 0 6 6 + 2 +2 x = 6 x = 2 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 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved! The solutions are -6 and 2. Zero Product Property Slide 84 / 175 Example 2: Solve x 2 + 36 = 12x 12x 12x x 2 12x + 36 = 0 (x 6)(x 6) = 0 x 6 = 0 +6 +6 x = 6 6 2 + 36 = 12(6) 36 + 36 = 72 72 = 72 The equation has to be written in standard form (ax 2 + bx + c). So subtract 12x from both sides. Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation Equal - problem solved!
Zero Product Property Example 3: Solve x 2 16x + 48= 0 Slide 85 / 175 (x 4)(x 12) = 0 x 4 = 0 x 12 = 0 +4 +4 +12 +12 x = 4 x = 12 4 2 16(4) + 48 = 0 16 64 + 48 = 0 48+48 = 0 0 = 0 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation 12 2 16(12) + 48 = 0 144 192 + 48 = 0 48 + 48 = 0 0 = 0 48 Equal - problem solved! 32 Solve x 2 5x + 6 = 0 Slide 86 / 175 A 7 B 5 C 3 D 2 E 2 F 3 G 5 H 6 I 7 J 15 33 Solve m 2 + 10m + 25 = 0 Slide 87 / 175 A 7 B 5 C 3 D 2 E 2 F 3 G 5 H 6 I 7 J 15
34 Solve h 2 h = 12 Slide 88 / 175 A 12 B 4 C 3 D 2 E 2 F 3 G 4 H 6 I 8 J 12 35 Solve d 2 35d = 2d Slide 89 / 175 A 7 B 5 C 3 D 35 E 12 F 0 G 5 H 6 I 7 J 37 36 Solve 8y 2 + 2y = 3 Slide 90 / 175 A 3 / 4 B 1 / 2 C 4 / 3 D 2 E 2 F 3 / 4 G 1 / 2 H 4 / 3 I 3 J 3
37 Which equation has roots of 3 and 5? Slide 91 / 175 A x 2 + 2x 15 = 0 B x 2 2x 15 = 0 C x 2 + 2x + 15 = 0 D x 2 2x + 15 = 0 Slide 92 / 175 Solve Quadratic Equations Using Square Roots Return to Table of Contents Square Root Method Slide 93 / 175 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: x = c or x = c written as: x = ± c
Square Root Method Slide 94 / 175 Solve for z: z² = 49 z = ± 49 z = ±7 The solution set is 7 and 7 Square Root Method Slide 95 / 175 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 The solution set is 5 and 5 x = ± 5 Square Root Method Slide 96 / 175 Solve 5x² = 20 using the square root method: 5x 2 = 20 5 5 x 2 = 4 x = 4 or x = 4 x = ± 2
Square Root Method Slide 97 / 175 Solve (2x 1)² = 20 using the square root method. click 2x 1 = 20 2x 1 = (4)(5) 2x 1 = 2 5 2x = 1 + 2 5 1 + 2 5 x = 2 or click 2x 1 = 20 2x 1 = (4)(5) 2x 1 = 2 5 2x = 1 2 5 1 2 5 x = 2 solution: x = 1 ± 2 5 click 2 38 When you take the square root of a real number, your answer will always be positive. Slide 98 / 175 True False 39 If x 2 = 16, then x = Slide 99 / 175 A 4 B 2 C 2 D 26 E 4
40 If y 2 = 4, then y = Slide 100 / 175 A 4 B 2 C 2 D 26 E 4 41 If 8j 2 = 96, then j = Slide 101 / 175 A 3 2 B 2 3 C 2 3 D 3 2 E ±12 42 If 4h 2 10= 30, then h = Slide 102 / 175 A 10 B 2 5 C 2 5 D 10 E ±10
43 If (3g 9) 2 + 7= 43, then g = Slide 103 / 175 A 1 B 9 5 2 3 C 9 + 5 2 3 D 5 E ±3 Slide 104 / 175 Solving Quadratic Equations by Completing the Square Return to Table of Contents Find the Missing Value of "C" Slide 105 / 175 Before we can solve the quadratic equation, we first have to find the missing value of C. To do this, simply take the value of b, divide it in 2 and then square the result. ax 2 +bx+c Find the value that completes the square. 8/2 = 4 4 2 = 16 (b/2) 2 x 2 + 8x + x 2 + 20x + 100 x 2 16x + 64 x 2 2x + 1
44 Find ( b / 2) 2 if b = 14 Slide 106 / 175 45 Find ( b / 2) 2 if b = 12 Slide 107 / 175 46 Complete the square to form a perfect square trinomial Slide 108 / 175 x 2 + 18x +?
47 Complete the square to form a perfect square trinomial Slide 109 / 175 x 2 6x +? Solving Quadratic Equations by Completing the Square Slide 110 / 175 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. Solving Quadratic Equations by Completing the Square Let's look at an example to solve: x 2 + 14x = 15 Slide 111 / 175 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 x + 7 = 8 or x + 7 = 8 Step 5 - Take the square root of both sides Step 6 - Write and solve two equations x = 1 or x = 15
Solving Quadratic Equations by Completing the Square Another example to solve: x 2 2x 2 = 0 Slide 112 / 175 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 48 Solve the following by completing the square : Slide 113 / 175 x 2 + 6x = 5 A 5 B 2 C 1 D 5 E 2 49 Solve the following by completing the square: Slide 114 / 175 x 2 8x = 20 A 10 B 2 C 1 D 10 E 2
50 Solve the following by completing the square : Slide 115 / 175 36x = 3x 2 + 108 A 6 B 6 C 0 D 6 E 6 A more difficult example: 3x 2 10x = 3 3 3 3 10x x 2 = 1 3 2 2 2 10 3 2 = 10 x 1 = 5 = 25 ) 3 2 3 9 ( ( ) ( ) 10x 25 25 x 2 + = 1 + 3 9 9 ( 2 ) 5 16 x = 3 9 5 4 x = ± 3 3 Solve 3x 2 10x = 3 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 Slide 116 / 175 51 Solve the following by completing the square: 4x 2 7x 2 = 0 Slide 117 / 175 A 1 4 B C 5 4 1 4 2 D 5 4 2 E 2
Slide 118 / 175 The Discriminant Return to Table of Contents The Discriminant Slide 119 / 175 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 The Discriminant Slide 120 / 175 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
The Discriminant Slide 121 / 175 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. The Discriminant Slide 122 / 175 Example 4 = ± 2 (2) (2) = 4 and ( 2)( 2) = 4 So BOTH 2 and 2 are solutions The Discriminant Slide 123 / 175 What is the relationship between the discriminant of a quadratic and its graph? y = x 2 8x + 10 y = 3x 2 + 8x 4 Discriminant (8) 2 4(1)(10) = 64 40 = 24 ( 6) 2 4(3)( 4) = 36 + 48 = 84
The Discriminant Slide 124 / 175 What is the relationship between the discriminant of a quadratic and its graph? y = 2x 2 4x + 2 y = x 2 + 6x + 9 Discriminant ( 4) 2 4(2)(2) = 16 16 = 0 (6) 2 4(1)(9) = 36 36 = 0 The Discriminant Slide 125 / 175 What is the relationship between the discriminant of a quadratic and its graph? y = x 2 + 5x + 9 y = 3x 2 3x + 4 Discriminant (5) 2 4(1)(9) = 25 36 = 11 ( 3) 2 4(3)(4) = 9 48 = 39 52 What is value of the discriminant of 2x 2 3x + 5 = 0? Slide 126 / 175
53 Find the number of solutions using the discriminant for 2x 2 3x + 5 = 0 Slide 127 / 175 A 0 B 1 C 2 54 What is value of the discriminant of Slide 128 / 175 x 2 8x + 4 = 0? 55 Find the number of solutions using the discriminant for x 2 8x + 4 = 0 Slide 129 / 175 A 0 B 1 C 2
Slide 130 / 175 Solve Quadratic Equations by Using the Quadratic Formula Return to Table of Contents Solve Any Quadratic Equation Slide 131 / 175 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. The Quadratic Formula Slide 132 / 175 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."
The Quadratic Formula Slide 133 / 175 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 The Quadratic Formula Slide 134 / 175 x = 3 ± 9 ( 40) 4 Simplify x = 3 ± 49 4 = 3 ± 7 4 Write as two equations x = 3 + 7 x = 3 7 4 or 4 x = 1 or x = 5 2 Solve each equation The Quadratic Formula Example 2 2x = x 2 3 Slide 135 / 175 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 Use only addition for standard form 0 = x 2 + (-2x) + ( 3) x 2 + ( 2x) + ( 3) = 0 Flip the equation Now you are ready to use the Quadratic Formula Solution on next slide
The Quadratic Formula Slide 136 / 175 x 2 + ( 2x) + ( 3) = 0 1x 2 + ( 2x) + ( 3) = 0 Identify values of a, b and c x = b ± b 2 4ac 2a x = ( 2) ± ( 2) 2 4(1)( 3) 2(1) Write the Quadratic Formula Substitute the values of a, b and c Continued on next slide The Quadratic Formula Slide 137 / 175 x = 2 ± 4 ( 12) 2a 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 56 Solve the following equation using the quadratic formula: Slide 138 / 175 x 2 5x + 4 = 0 A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5
57 Solve the following equation using the quadratic formula: Slide 139 / 175 x 2 = x + 20 A 5 B 4 C 3 D 2 E 1 F 1 G 2 H 3 I 4 J 5 58 Solve the following equation using the quadratic formula: Slide 140 / 175 2x 2 + 12 = 11x A 5 B 4 3 C 2 D 2 E 1 F 1 G 2 3 H 2 I 4 J 5 Example 3 x 2 2x 4 = 0 The Quadratic Formula Slide 141 / 175 1x 2 + ( 2x) + ( 4) = 0 Identify values of a, b and c x = -b ± b 2-4ac 2a x = ( 2) ± ( 2) 2 4(1)( 4) 2(1) Write the Quadratic Formula Substitute the values of a, b and c Continued on next slide
The Quadratic Formula Slide 142 / 175 x = 2 ± 4 ( 16) 2 Simplify x = 2 ± 20 2 x = 2 ± 20 2 or x = 2-20 2 x = 2 ± 2 5 2 or x = 2-2 5 2 Write as two equations x = 1 + 5 or x = 1 5 x 3.24 or x 1.24 Use a calculator to estimate x 59 Find the larger solution to Slide 143 / 175 x 2 + 6x 1 = 0 60 Find the smaller solution to Slide 144 / 175 x 2 + 6x 1 = 0
Slide 145 / 175 Application Problems Return to Table of Contents Quadratic Equations and Applications Slide 146 / 175 A sampling of applied problems that lend themselves to being solved by quadratic equations: Number Reasoning Distances Geometry: Dimensions Free Falling Objects Height of a Projectile Number Reasoning Slide 147 / 175 The product of two consecutive negative integers is 1,122. What are the numbers? Remember that consecutive integers are one unit apart, so the numbers are n and n + 1. Multiplying to get the product: n(n + 1) = 1122 n 2 + n = 1122 n 2 + n 1122 = 0 (n + 34)(n - 33) = 0 n = 34 and n = 33. STANDARD Form FACTOR The solution is either 34 and 33 or 33 and 34, since the direction ask for negative integers 34 and 33 are the correct pair.
Application Problems Slide 148 / 175 PLEASE KEEP THIS IN MIND When solving applied problems that lead to quadratic equations, you might get a solution that does not satisfy the physical constraints of the problem. For example, if x represents a width and the two solutions of the quadratic equations are 9 and 1, the value 9 is rejected since a width must be a positive number. 61 The product of two consecutive even integers is 48. Find the smaller of the two integers. Slide 149 / 175 Hint: x(x+2) = 48 Click to reveal hint Application Problems Slide 150 / 175 TRY THIS: The product of two consecutive integers is 272. What are the numbers?
62 The product of two consecutive even integers is 528. What is the smaller number? Slide 151 / 175 More of a challenge... Slide 152 / 175 The product of two consecutive odd integers is 1 less than four times their sum. Find the two integers. Let n = 1st number n + 2 = 2nd number n(n + 2) = 4[n + (n + 2)] 1 n 2 + 2n = 4[2n + 2] 1 n 2 + 2n = 8n + 8 1 n 2 + 2n = 8n + 7 n 2 6n - 7 = 0 (n 7)(n + 1) = 0 n = 7 and n = 1 Which one do you use? Or do you use both? More of a challenge... Slide 153 / 175 If n = 7 then n + 2 = 9 7 x 9 = 4[7 + (7 + 2)] 1 63 = 4(16) 1 63 = 64 1 63 = 63 If n = 1 then n + 2 = 1 + 2 = 1 ( 1) x 1 = 4[ 1 + ( 1 + 2)] 1 1 = 4[ 1 + 1] 1 1 = 4(0) 1 1 = 1 We get two sets of answers.
63 The product of a number and a number 3 more than the original is 418. What is the smallest value the original number can be? Slide 154 / 175 64 Find three consecutive positive even integers such that the product of the second and third integers is twenty more than ten times the first integer. Enter the value of the smaller even integer. Slide 155 / 175 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. 65 When 36 is subtracted from the square of a number, the result is five times the number. What is the positive solution? Slide 156 / 175 A 9 B 6 C 3 D 4 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011.
66 Tamara has two sisters. One of the sisters is 7 years older than Tamara.The other sister is 3 years younger than Tamara. The product of Tamara s sisters ages is 24. How old is Tamara? Slide 157 / 175 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. Distance Problems Example Two cars left an intersection at the same time, one heading north and one heading west. Some time later, they were exactly 100 miles apart. The car that headed north had gone 20 miles farther than the car headed west. How far had each car traveled? Step 1 - Read the problem carefully. Step 2 - Illustrate or draw your information. 100 x+20 Step 3 - Assign a variable Let x = the distance traveled by the car heading x west Then (x + 20) = the distance traveled by the car heading north Step 4 - Write an equation Does your drawing remind you of the Pythagorean Theorem? a 2 + b 2 = c 2 Continued on next slide Slide 158 / 175 Distance Problems Slide 159 / 175 Step 5 - Solve a 2 + b 2 = c 2 x 2 + (x+20) 2 = 100 2 100 x x+20 x 2 + x 2 + 40x + 400 = 10,000 2x 2 + 40x 9600 = 0 2(x 2 +20x 4800) = 0 x 2 + 20x 4800 = 0 Square the binomial Standard form Factor Divide each side by 2 Think about your options for solving the rest of this equation. Completing the square? Quadratic Formula? Continued on next slide
Distance Problems Slide 160 / 175 Did you try the quadratic formula? x = 20 ± 400 4(1)( 4800) 2 x = 20 ± 19,600 2 x = 60 or x = -80 Since the distance cannot be negative, discard the negative solution. The distances are 60 miles and 60 + 20 = 80 miles. Step 6 - Check your answers. 67 Two cars left an intersection at the same time,one heading north and the other heading east. Some time later they were 200 miles apart. If the car heading east traveled 40 miles farther, how far did the car traveling north go? Slide 161 / 175 Geometry Applications Slide 162 / 175 Area Problem The length of a rectangle is 6 inches more than its width. The area of the rectangle is 91 square inches. Find the dimensions of the rectangle. Step 1 - Draw the picture of the rectangle. Let the width = x and the length = x + 6 Step 2 - Write the equation using the formula Area = length x width x + 6 x
Geometry Applications Slide 163 / 175 Step 3 - Solve the equation x( x + 6) = 91 x 2 + 6x = 91 x 2 + 6x 91 = 0 (x 7)(x + 13) = 0 x = 7 or x = 13 Since a length cannot be negative... The width is 7 and the length is 13. 68 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? Slide 164 / 175 Hint: (L)(L 10) = 600. Click to reveal hint 69 A square's length is increased by 4 units and its width is increased by 6 units. The result of this transformation is a rectangle with an area that 195 square units. Find the area of the original square. Slide 165 / 175
70 The rectangular picture frame below is the same width all the way around. The photo it surrounds measures 17" by 11". The area of the frame and photo combined is 315 sq. in. What is the length of the outer frame? Slide 166 / 175 length x x 71 The area of the rectangular playground enclosure at South School is 500 square meters. The length of the playground is 5 meters longer than the width. Find the dimensions of the playground, in meters. [Only an algebraic solution will be accepted.] Slide 167 / 175 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. 72 Jack is building a rectangular dog pen that he wishes to enclose. The width of the pen is 2 yards less than the length. If the area of the dog pen is 15 square yards, how many yards of fencing would he need to completely enclose the pen? Slide 168 / 175
Slide 169 / 175 Free Falling Objects Problems 73 A person walking across a bridge accidentally drops an orange in the river below from a height of 40 ft. The function h = 16t 2 + 40 gives the orange's approximate height h above the water, in feet, after t seconds. In how many t seconds will the orange hit the water? (Round to the nearest tenth.) Slide 170 / 175 Hint: when it hits the water it is at 0. 74 Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 16t 2 where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? Slide 171 / 175
75 The height of a golf ball hit into the air is modeled by the equation h = 16t 2 + 48t, where h represents the height, in feet, and t represents the number of seconds that have passed since the ball was hit. What is the height of the ball after 2 seconds? Slide 172 / 175 A 16 ft B 32 ft C 64 ft D 80 ft From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011. Slide 173 / 175 Height of Projectiles Problems 76 A skyrocket is shot into the air. It's altitude in feet, h, after t seconds is given by the function h = 16t 2 + 128t. Slide 174 / 175 What is the rocket's maximum altitude?
77 A rocket is launched from the ground and follows a parabolic path represented by the equation y = x 2 + 10x. At the same time, a flare is launched from a height of 10 feet and follows a straight path represented by the equation y = x + 10. Using the accompanying set of axes, graph the equations that represent the paths of the rocket and the flare, and find the coordinates of the point or points where the paths intersect. Slide 175 / 175 From the New York State Education Department. Office of Assessment Policy, Development and Administration. Internet. Available from www.nysedregents.org/integratedalgebra; accessed 17, June, 2011.