Slide 1 / 200 Quadratic Functions Table of Contents Key Terms 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 2 / 200 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 Maximum: The y-value of the vertex if a < 0 and the parabola opens downward Slide 5 / 200 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 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. Slide 6 / 200 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
A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a is not equal to 0. Slide 10 / 200 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. Practice writing quadratic equations in standard form: (Reduce if possible.) Write 2x 2 = x + 4 in standard form: Slide 11 / 200 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 Slide 15 / 200
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. 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 17 / 200 Slide 18 / 200 5. The domain of a quadratic function is all real numbers.
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? Slide 19 / 200 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 If the vertex is a maximum, then the range is all real numbers less than or equal to the y-value. Slide 20 / 200 The range of this quadratic is to 10 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 21 / 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 22 / 200 1 True or False: The vertex is the highest or lowest value on the parabola. Slide 23 / 200 True False 2 If a parabola opens upward then... Slide 24 / 200 A a>0 B a<0 C a=0
3 The vertical line that divides a parabola into two symmetrical halves is called... Slide 25 / 200 A B C D E discriminant perfect square axis of symmetry vertex slice Quadratic Equations Slide 26 / 200 Finding Zeros (x- intercepts) Slide 27 / 200 Solve Quadratic Equations by Graphing Return to Table of Contents
Vocabulary Slide 28 / 200 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. How many zeros do the parabolas have? What are the values of the zeros? Slide 30 / 200 No zeroes 2 zeroes; x = -1 and x=3 1 zero; x=1
Vertex One way to solve a quadratic equation in standard form is to find the zeros or x-intercepts of the related function. Slide 31 / 200 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) 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 to be 3 and -3. # # x = 0 # (3,0) # (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 5 What is the axis of symmetry? Slide 38 / 200 y = -2x 2 + 12x - 18 A -3 B 3 Formula: -b 2a C 4 D -5 6 y = -2x 2 + 12x - 18 What is the vertex? Slide 39 / 200 A (3,0) B (-3,0) C (4,0) D (-5,0)
7 y = -2x 2 + 12x - 18 What is the y- intercept? Slide 40 / 200 A (0,0) B (0, 18) C (0, -18) D (0, 12) 8 y = -2x 2 + 12x - 18 If two other points are (5,-8) and (4,-2), what does the graph look like? Slide 41 / 200 A B 9 If two other points are (5,-8) and (4,-2), what does the graph of y = -2x 2 + 12x - 18 look like? Slide 42 / 200 A B C D
10 y = -2x 2 + 12x - 18 What is(are) the zero(s)? Slide 43 / 200 A -18 B 4 C 3 D -8 Slide 44 / 200 Solve Quadratic Equations by Factoring Return to Table of Contents Solving Quadratic Equations by Factoring Slide 45 / 200 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. Remember the FOIL method for multiplying binomials Slide 47 / 200 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
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 Slide 49 / 200-6 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. 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. Slide 50 / 200 (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! Example 3: Solve x 2-16x + 64 = 0 Slide 51 / 200 (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
11 Solve Slide 52 / 200 A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15 12 Solve Slide 53 / 200 A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15 13 Solve Slide 54 / 200 A -12 B -4 C -3 D -2 E 2 F 3 G 4 H 6 I 8 J 12
14 Solve Slide 55 / 200 A -7 B -5 C -3 D -2 E 12 F 3 G 5 H 6 I 7 J 35 15 Solve Slide 56 / 200 A - 3 / 4 F 3 / 4 B - 1 / 2 C - 4 / 3 D -2 E 2 G 1 / 2 H 4 / 3 I -3 J 3 16 The product of two consecutive even integers is 48. Find the smaller of the two integers. Slide 57 / 200 Hint: x(x+2) = 48
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 58 / 200 Slide 59 / 200 Solve Quadratic Equations Using Square Roots Return to Table of Contents You can solve a quadratic equation by the square root method if you can write it in the form: x² = c Slide 60 / 200 If x and c are algebraic expressions, then: x = c or x = - c written as: 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 Solve 5x² = 20 using the square root method: Slide 63 / 200 5x 2 = 20 5 5 x 2 = 4 x = 4 or x = - 4 x = ±2
Solve (2x - 1)² = 20 using the square root method. Slide 64 / 200 or 18 When you take the square root of a real number, your answer will always be positive. Slide 65 / 200 True False 19 If x 2 = 16, then x = Slide 66 / 200 A 4 B 2 C -2 D 26 E -4
20 If y 2 = 4, then y = Slide 67 / 200 A 4 B 2 C -2 D 26 E -4 Slide 68 / 200 Slide 69 / 200
23 If (3g - 9) 2 + 7= 43, then g = Slide 70 / 200 A B C D E Slide 71 / 200 Solving Quadratic Equations by Completing the Square Return to Table of Contents Form a perfect square trinomial with lead coefficient of 1 Slide 72 / 200 x 2 + bx +c where c = ( b / 2) 2 Find the value that completes the square.
Slide 73 / 200 24 Find ( b / 2) 2 if b = 14 Slide 74 / 200 25 Find ( b / 2) 2 if b = -12 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 +? Solving quadratic equations by completing the square: Slide 77 / 200 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: 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 2 + 14x = 15 Slide 78 / 200 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
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 Slide 79 / 200 (-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 28 Solve the following by completing the square : Slide 80 / 200 x 2 + 6x = -5 A -5 B -2 C -1 D 5 E 2 29 Solve the following by completing the square : Slide 81 / 200 x 2-8x = 20 A -10 B -2 C -1 D 10 E 2
Slide 82 / 200 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 31 Solve the following by completing the square : Slide 84 / 200 A B C D E
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 Slide 85 / 200 (-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 32 Solve the following by completing the square : Slide 86 / 200 A B C D E Slide 87 / 200 Solve Quadratic Equations by Using the Quadratic Formula Return to Table of Contents
At this point you have learned how to solve quadratic equations by: graphing factoring using square roots and completing the square Slide 88 / 200 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 89 / 200 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." Example 1 Slide 90 / 200 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) continued on next slide Substitute the values of a, b and c
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 Example 2 2x = x 2-3 Slide 92 / 200 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) Use only addition for standard form x 2 + (-2x) + (-3) = 0 Flip the equation Now you are ready to use the Quadratic Formula Solution on next slide x 2 + (-2x) + (-3) = 0 Slide 93 / 200 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 33 Solve the following equation using the quadratic formula: Slide 95 / 200 A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5 34 Solve the following equation using the quadratic formula: Slide 96 / 200 A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5
35 Solve the following equation using the quadratic formula: Slide 97 / 200 A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5 Example 3 Slide 98 / 200 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) Continued on next slide Substitute the values of a, b and c x = 2 ± 4 - (-16) 2 x = 2 ± 20 2 Simplify Slide 99 / 200 x = 2 + 20 2 x = 2 +2 5 2 or or x = 2-20 2 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
36 Find the larger solution to Slide 100 / 200 37 Find the smaller solution to Slide 101 / 200 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. Example Slide 105 / 200 4 = ± 2 (2) (2) = 4 and (-2)(-2) = 4 So BOTH 2 and -2 are solutions
Slide 106 / 200 Slide 107 / 200 Slide 108 / 200
ax 2 + bx + c = 0 Slide 109 / 200 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 38 What is value of the discriminant of 2x 2-2x + 3 = 0? Slide 110 / 200 Slide 111 / 200 39 Find the number of solutions using the discriminant for 2x 2-2x + 3 = 0 A 0 B 1 C 2
40 What is value of the discriminant of x 2-8x + 4 = 0? Slide 112 / 200 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
Summary of Factoring Factor the Polynomial Slide 115 / 200 2 Terms Factor out GCF 3 Terms 4 Terms Difference Sum/Difference of Squares 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 Slide 117 / 200
Examples: Slide 118 / 200 Slide 119 / 200 43 Factor Slide 120 / 200 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 )
44 Factor completely Slide 121 / 200 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 Polynomials with four terms like ab - 4b + 6a - 24, can Slide 123 / 200 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!
Example 2: 6xy + 8x - 21y - 28 Slide 124 / 200 (6xy + 8x) + (-21y - 28) Group 2x(3y + 4) + (-7)(3y + 4) Factor GCF (3y +4) (2x - 7) Factor common binomial 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). Slide 125 / 200 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. 45 Factor 15ab - 3a + 10b - 2 Slide 126 / 200 A (5b - 1)(3a + 2) B (5b + 1)(3a + 2) C (5b - 1)(3a - 2) D (5b + 1)(3a - 1)
46 Factor 10m 2 n - 25mn + 6m - 15 Slide 127 / 200 A B C D (2m-5)(5mn-3) (2m-5)(5mn+3) (2m+5)(5mn-3) (2m+5)(5mn+3) 47 Factor 20ab - 35b - 63 +36a Slide 128 / 200 A (4a - 7)(5b - 9) B (4a - 7)(5b + 9) C (4a + 7)(5b - 9) D (4a + 7)(5b + 9) 48 Factor a 2 - ab + 7b - 7a Slide 129 / 200 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 Summary of Factoring Factor the Polynomial Slide 131 / 200 2 Terms Factor out GCF 3 Terms 4 Terms Difference Sum/Difference of Squares 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 132 / 200
49 Factor completely: Slide 133 / 200 A B C D 50 Factor completely Slide 134 / 200 A B C D prime polynomial 51 Factor Slide 135 / 200 A B C D prime polynomial
Slide 136 / 200 53 Factor Slide 137 / 200 A B C D Prime Polynomial Slide 138 / 200 Solving Equations by Factoring
Given the following equation, what conclusion(s) can be drawn? ab = 0 Slide 139 / 200 Since the product is 0, one of the factors, a or b, must be 0. This is known as the Zero Product Property. 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 Slide 140 / 200 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 What if you were given the following equation? Slide 141 / 200 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}
Slide 142 / 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. Slide 143 / 200 Example Slide 144 / 200
54 Choose all of the solutions to: Slide 145 / 200 A B C D E F 55 Choose all of the real solutions to: Slide 146 / 200 A B C D E F 56 Choose all of the solutions to: Slide 147 / 200 A B C D E F
Slide 148 / 200 58 A ball is thrown with its height at any time given by Slide 149 / 200 When does the ball hit the ground? A -1 seconds B C D 0 seconds 9 seconds 10 seconds Slide 150 / 200 Solving Rational Equations Return to Table of Contents
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: Slide 151 / 200 Check: Slide 152 / 200 Slide 153 / 200
Example: Slide 154 / 200 Check: x = 7 Check: x = -2 Slide 155 / 200 59 Solve the equation. Check to see it works. Slide 156 / 200
60 Solve the equation. Check to see it works. Slide 157 / 200 Slide 158 / 200 Slide 159 / 200
Slide 160 / 200 Solving Radical Equations Return to Table of Contents To solve a radical equation: isolate the radical on one side of the equation Slide 161 / 200 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
Slide 163 / 200 64 Find the solution to Slide 164 / 200 65 Find the solution to Slide 165 / 200
Slide 166 / 200 67 Find the solution to Slide 167 / 200 If an equation has multiple roots, move them to opposite sides of the equal sign and then solve. Slide 168 / 200
68 Solve the following: Slide 169 / 200 69 Solve the following: Slide 170 / 200 70 If the distance between (3,5) and (x,9) is 7, find x. Slide 171 / 200
Slide 172 / 200 Quadratic & Rational Inequalities Return to Table of Contents Slide 173 / 200 Slide 174 / 200
Graph Step 1: Graph Points on the Bounds Slide 175 / 200 X Y -5-7 -4-10 -3-11 -2-10 -1-7 0-2 Graph Step 2: Solid or Dotted? Slide 176 / 200 Graph Step 3: Shade Slide 177 / 200
Graph Step 1: Graph Points on the Bounds Slide 178 / 200 X Y -3-2 -1 0 1 Graph Step 2: Solid or Dotted? Slide 179 / 200 Graph Step 3: Shade Slide 180 / 200
Graph Slide 181 / 200 Slide 182 / 200 Slide 183 / 200
73 Which equation is graphed? Slide 184 / 200 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 74 Which equation is graphed? Slide 185 / 200 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 Solving Slide 186 / 200 Method 1: Graphically Graph the related function The solution is where the shaded region intersects the x-axis. -3 2-3 2
Note: It is possible to have a solution of a point or the empty set. Slide 187 / 200 Find the solution set given the graph of the related function. Slide 188 / 200-2 3-3 -.5 1 2 -.5 1 75 Solve the following inequality: Slide 189 / 200 A 1 < x< 4 B 1 < x < 4 C x < 1 or x > 4 D x < 1 or x > 4 1 4
76 Solve the following inequality: Slide 190 / 200 A 1 < x< 4 B 1 < x < 4 C x < 1 or x > 4 D x < 1 or x > 4 1 4 77 Solve the following inequality: Slide 191 / 200 A -4 < x< 2 B -4 < x < 2 C x < -4 or x > 2 D x < -4 or x > 2 78 Solve the following inequality: Slide 192 / 200 A 2 < x< 5 B 2 < x < 5 C x < 2 or x > 5 D x < 2 or x > 5
Steps to Solving Quadratic Inequalities Algebraically Slide 193 / 200 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 Solve 1) Rewrite inequality <0 2) Factor 3) Solve Slide 194 / 200-4 -2 Test points: -10, -3, 0 x 4) Create number line 5) Test points F T -4-2 F x 6) Write the solution Solve 1) Rewrite inequality <0 2) Factor 3) Solve Slide 195 / 200-1.5 1 Test points: -2, 0, 2 x 4) Create number line 5) Test points T F -1.5 1 T x 6) Write the solution
Solve Slide 196 / 200 79 Solve Slide 197 / 200 A x < -3 or x > 2 B -3 < x < 2 C All Reals D No Solution 80 Solve Slide 198 / 200 A x < -2 or x > 5 B -2 < x < 5 C D All Reals No Solution
81 Solve Slide 199 / 200 A x = -2 B x = 2 C D All Reals No Solution 82 Solve Slide 200 / 200 A -3 < x < -2 B x < -3 or x > -2 C D All Reals No Solution