Slide 1 / 200 Quadratic Functions Slide 2 / 200 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 3 / 200 Slide 4 / 200 Axis of symmetry: The vertical line that divides a parabola into two symmetrical halves Key Terms Completing the square: Adding a term to x 2 + bx to form a trinomial that is a perfect square Return to Table of Contents Discriminant: b 2-4ac in a quadratic in standard form Slide 5 / 200 Slide 6 / 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 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 Slide 8 / 200 Identifying Quadratic Functions 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 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. Explain Characteristics of Quadratic Functions Return to Table of Contents 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 Slide 12 / 200 Practice writing quadratic equations in standard form: (Reduce if possible.) Write 2x 2 = x + 4 in standard form: Write 3x = -x 2 + 7 in standard form: 2x 2 - x - 4 = 0 x 2 + 3x - 7 = 0
Slide 13 / 200 Slide 14 / 200 Write 6x 2-6x = 12 in standard form: Write 3x - 2 = 5x in standard form: x 2 - x - 2 = 0 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. 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. 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 The range of this quadratic is -6 to 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 Slide 26 / 200 3 The vertical line that divides a parabola into two symmetrical halves is called... Quadratic Equations A B C D E discriminant perfect square axis of symmetry vertex slice Finding Zeros (x- intercepts) Slide 27 / 200 Slide 28 / 200 Vocabulary Solve Quadratic Equations by Graphing Return to Table of Contents 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. Slide 30 / 200 How many zeros do the parabolas have? What are the values of the zeros? A zero is the point at which the parabola intersects the x-axis. A quadratic may have one, two or no zeros. No zeroes 2 zeroes; x = -1 and x=3 1 zero; x=1
Vertex Slide 31 / 200 Slide 32 / 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 1 - Write the Related Function 2x 2-18 = 0 2x 2-18 = y y = 2x 2 + 0x - 18 Step 2 - Graph the related function. Step 3 - Find the zeros (or x intercepts) of the related function. Slide 33 / 200 Slide 34 / 200 Step 2 - Graph the Function Step 2 - Graph the Function Two Points 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) Y Intercept The ver substit x-coord Symme equatio The po passe This o x-valu 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 Slide 36 / 200 Step 3 - Find the zeros y = 2x 2 + 0x 18 Solve the equation by graphing the related function. Step 3 - Find the zeros y = 2x 2 + 0x 18 Substitute 3 and -3 for x in the quadratic equation. The zeros appear to be 3 and -3. # # x = 0 # (3,0) # (2,-10) # (0,-18) 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 Slide 38 / 200 4 Solve the equation by graphing the related function. 5 What is the axis of symmetry? y = -2x 2 + 12x - 18-12x + 18 = -2x 2 Step 1: What of these is the related function? A -3 B 3 Formula: -b 2a A B y = -2x 2-12x + 18 y = 2x 2-12x - 18 C 4 D -5 C y = -2x 2 + 12x - 18 Slide 39 / 200 Slide 40 / 200 6 y = -2x 2 + 12x - 18 7 y = -2x 2 + 12x - 18 What is the vertex? What is the y- intercept? A (3,0) A (0,0) B (-3,0) B (0, 18) C (4,0) C (0, -18) D (-5,0) D (0, 12) Slide 41 / 200 Slide 42 / 200 8 y = -2x 2 + 12x - 18 If two other points are (5,-8) and (4,-2), what does the graph look like? 9 A If two other points are (5,-8) and (4,-2), what does the graph of y = -2x 2 + 12x - 18 look like? B A B C D
Slide 43 / 200 Slide 44 / 200 10 y = -2x 2 + 12x - 18 What is(are) the zero(s)? A -18 B 4 C 3 D -8 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 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. 2 and 9 11 3 and 6 9 x 2 + 9x + 18 = (x + 3)(x + 6) Slide 47 / 200 Slide 48 / 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 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 (x + 3)(x + 2) = x 2 + 2x + 3x + 6 = x 2 + 5x + 6 F O I L
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 = 2-6 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 Slide 49 / 200 Use "FUSE"! Factor the trinomial using the FOIL method. Use the Zero property Slide 51 / 200 Example 3: Solve x 2-16x + 64 = 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. 11 Solve (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 52 / 200 (x - 8)(x - 8) = 0 x - 8 = 0 +8 +8 x = 8 8 2-16(8) + 64 = 0 64-128 + 64 = 0 Factor the trinomial using the FOIL method. Use the Zero property Substitue found value into original equation A -7 B -5 C -3 D -2 E 2 F 3 G 5 H 6 I 7 J 15-64 + 64 = 0 Equal - problem solved! 0 = 0 Slide 53 / 200 Slide 54 / 200 12 Solve 13 Solve A -7 F 3 A -12 F 3 B -5 G 5 B -4 G 4 C -3 H 6 C -3 H 6 D -2 I 7 D -2 I 8 E 2 J 15 E 2 J 12
Slide 55 / 200 Slide 56 / 200 14 Solve 15 Solve A -7 F 3 A - 3 / 4 F 3 / 4 B -5 G 5 B - 1 / 2 G 1 / 2 C -3 H 6 C - 4 / 3 H 4 / 3 D -2 I 7 D -2 I -3 E 12 J 35 E 2 J 3 Slide 57 / 200 Slide 58 / 200 16 The product of two consecutive even integers is 48. Find the smaller of the two integers. 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 59 / 200 Slide 60 / 200 You can solve a quadratic equation by the square root method if you can write it in the form: x² = c Solve Quadratic Equations Using Square Roots If x and c are algebraic expressions, then: x = c or x = - c written as: x = ± c Return to Table of Contents
Slide 61 / 200 Slide 62 / 200 Solve for z: z² = 49 z = ± 49 z = ±7 A quadratic equation of the form x 2 = c can be solved using the Square Root Property. Example: Solve 4x 2 = 20 The solution set is 7 and -7 4x 2 = 20 4 4 x 2 = 5 x = ± 5 Divide both sides by 4 to isolate x² The solution set is 5 and - 5 Slide 63 / 200 Slide 64 / 200 Solve 5x² = 20 using the square root method: Solve (2x - 1)² = 20 using the square root method. 5x 2 = 20 5 5 or x 2 = 4 x = 4 or x = - 4 x = ±2 Slide 65 / 200 Slide 66 / 200 18 When you take the square root of a real number, your answer will always be positive. 19 If x 2 = 16, then x = A 4 True False B 2 C -2 D 26 E -4
Slide 67 / 200 Slide 68 / 200 20 If y 2 = 4, then y = A 4 B 2 C -2 D 26 E -4 Slide 69 / 200 Slide 70 / 200 23 If (3g - 9) 2 + 7= 43, then g = A B C D E Slide 71 / 200 Slide 72 / 200 Solving Quadratic Equations by Completing the Square Return to Table of Contents 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.
Slide 73 / 200 Slide 74 / 200 24 Find ( b / 2) 2 if b = 14 25 Find ( b / 2) 2 if b = -12 Slide 75 / 200 Slide 76 / 200 26 Complete the square to form a perfect square trinomial x 2 + 18x +? 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: x 2 + 14x = 15 Slide 78 / 200 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 x = 1 or x = -15 x 2 + 14x = 15 Step 5 - Take the square root of both sides 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 Slide 82 / 200 29 Solve the following by completing the square : x 2-8x = 20 A -10 B -2 C -1 D 10 E 2 A more difficult example: Slide 83 / 200 Write as x 2 +bx=c Slide 84 / 200 31 Solve the following by completing the square : or Find (b 2) 2 Add 25/9 to both sides Factor and simplify Take the square root of both sides Write and solve two equations 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 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. Return to Table of Contents 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." Example 1 2x 2 + 3x - 5 = 0 2x 2 + 3x + (-5) = 0 x = -b ± b 2-4ac 2a x = -3 ± 3 2-4(2)(-5) 2(2) Slide 90 / 200 continued on next slide Identify values of a, b and c Write the Quadratic Formula Substitute the values of a, b and c
Slide 91 / 200 Slide 92 / 200 x = -3 ± 9 - (-40) 4 x = -3 ± 49 4 x = -3 + 7 4 or x = 1 or x = -5 2 = -3 ± 7 4 x = -3-7 4 Simplify Write as two equations Solve each equation 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 Slide 94 / 200 x 2 + (-2x) + (-3) = 0 1x 2 + (-2x) + (-3) = 0 Identify values of a, b and c x = 2 ± 4 - (-12) 2 Simplify x = -b ± b 2-4ac 2a Write the Quadratic Formula x = 2 ± 16 2 = 2 ± 4 2 x = -(-2) ± (-2) 2-4(1)(-3) 2(1) Substitute the values of a, b and c x = 2 + 4 2 or x = 2-4 2 Write as two equations Continued on next slide x = 3 or x = -1 Solve each equation Slide 95 / 200 Slide 96 / 200 33 Solve the following equation using the quadratic formula: 34 Solve the following equation using the quadratic formula: A -5 F 1 A -5 F 1 B -4 G 2 B -4 G 2 C -3 H 3 C -3 H 3 D -2 E -1 I 4 J 5 D -2 E -1 I 4 J 5
Slide 97 / 200 Slide 98 / 200 35 Solve the following equation using the quadratic formula: Example 3 x 2-2x - 4 = 0 A -5 B -4 C -3 D -2 E -1 F 1 G 2 H 3 I 4 J 5 1x 2 + (-2x) + (-4) = 0 x = -b ± b 2-4ac 2a x = -(-2) ± (-2) 2-4(1)(-4) 2(1) Continued on next slide Identify values of a, b and c Write the Quadratic Formula Substitute the values of a, b and c x = 2 ± 4 - (-16) 2 x = 2 ± 20 2 Slide 99 / 200 Simplify 36 Find the larger solution to Slide 100 / 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 Slide 101 / 200 Slide 102 / 200 37 Find the smaller solution to The Discriminant Return to Table of Contents
Slide 103 / 200 Slide 104 / 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 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 Slide 106 / 200 Example 4 = ± 2 (2) (2) = 4 and (-2)(-2) = 4 So BOTH 2 and -2 are solutions Slide 107 / 200 Slide 108 / 200
Slide 109 / 200 ax 2 + bx + c = 0 Slide 110 / 200 38 What is value of the discriminant of 2x 2-2x + 3 = 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 111 / 200 Slide 112 / 200 39 Find the number of solutions using the discriminant for 2x 2-2x + 3 = 0 40 What is value of the discriminant of x 2-8x + 4 = 0? A 0 B 1 C 2 Slide 113 / 200 Slide 114 / 200 Find the number of solutions using the discriminant for x 2-8x + 4 = 0 A 0 B 1 C 2 Solving Non-Quadratics Return to Table of Contents
Slide 115 / 200 Summary of Factoring Factor the Polynomial Slide 116 / 200 2 Terms Factor out GCF 3 Terms Difference Sum/Difference of Squares of Cubes Perfect Square Trinomial Factor the Trinomial 4 Terms Group and Factor out GCF. Look for a Common Binomial Sum and Difference of Cubes 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 117 / 200 Slide 118 / 200 Examples: Slide 119 / 200 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 Slide 122 / 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) 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 2: 6xy + 8x - 21y - 28 Slide 124 / 200 Example 1: ab - 4b + 6a - 24 (6xy + 8x) + (-21y - 28) Group 2x(3y + 4) + (-7)(3y + 4) Factor GCF (3y +4) (2x - 7) Factor common binomial (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 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 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) Remember to check each problem by using FOIL.
Slide 127 / 200 46 Factor 10m 2 n - 25mn + 6m - 15 Slide 128 / 200 47 Factor 20ab - 35b - 63 +36a A B C D (2m-5)(5mn-3) (2m-5)(5mn+3) (2m+5)(5mn-3) (2m+5)(5mn+3) A (4a - 7)(5b - 9) B (4a - 7)(5b + 9) C (4a + 7)(5b - 9) D (4a + 7)(5b + 9) Slide 129 / 200 Slide 130 / 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) Mixed Factoring Slide 131 / 200 Summary of Factoring Factor the Polynomial Slide 132 / 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.
49 Factor completely: A B C Slide 133 / 200 50 Factor completely A B C Slide 134 / 200 D D prime polynomial Slide 135 / 200 Slide 136 / 200 51 Factor A B C D prime polynomial Slide 137 / 200 Slide 138 / 200 53 Factor A B C D Prime Polynomial 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 Slide 142 / 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} Slide 143 / 200 Slide 144 / 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
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 Slide 148 / 200 56 Choose all of the solutions to: A B C D E F Slide 149 / 200 Slide 150 / 200 58 A ball is thrown with its height at any time given by When does the ball hit the ground? A -1 seconds B C D 0 seconds 9 seconds 10 seconds 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 Slide 153 / 200 Slide 154 / 200 Example: Slide 155 / 200 Check: x = 7 Check: x = -2 Slide 156 / 200 59 Solve the equation. Check to see it works.
Slide 157 / 200 Slide 158 / 200 60 Solve the equation. Check to see it works. Slide 159 / 200 Slide 160 / 200 Solving Radical Equations Return to Table of Contents Slide 161 / 200 Slide 162 / 200 To solve a radical equation: isolate the radical on one side of the equation Example use the index to determine the power to eliminate the radical solve the equation check to see if solution is extraneous
Slide 163 / 200 Slide 164 / 200 64 Find the solution to Slide 165 / 200 Slide 166 / 200 65 Find the solution to 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 Slide 172 / 200 70 If the distance between (3,5) and (x,9) is 7, find x. Quadratic & Rational Inequalities Return to Table of Contents Slide 173 / 200 Slide 174 / 200
Slide 175 / 200 Slide 176 / 200 Graph Step 1: Graph Points on the Bounds Graph Step 2: Solid or Dotted? X Y -5-7 -4-10 -3-11 -2-10 -1-7 0-2 Slide 177 / 200 Slide 178 / 200 Graph Step 3: Shade Graph Step 1: Graph Points on the Bounds X Y -3-2 -1 0 1 Slide 179 / 200 Slide 180 / 200 Graph Step 2: Solid or Dotted? Graph Step 3: Shade
Slide 181 / 200 Slide 182 / 200 Graph Slide 183 / 200 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 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
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: Slide 190 / 200 76 Solve the following inequality: A 1 < x< 4 A 1 < x< 4 B 1 < x < 4 B 1 < x < 4 C x < 1 or x > 4 D x < 1 or x > 4 1 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 Solve Slide 194 / 200 1) Rewrite inequality <0 2) Factor 3) Solve 2) Factor 3) Set each factor equal to zero and solve 4) Create a number line with the solution as the points -4-2 Test points: -10, -3, 0 x 4) Create number line 5) Test points 5) Test points in each region to see if they satisfy the inequality 6) Write the solution F T -4-2 F x 6) Write the solution Slide 195 / 200 Slide 196 / 200 Solve 1) Rewrite inequality <0 2) Factor 3) Solve Solve -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 Slide 197 / 200 Slide 198 / 200 79 Solve A x < -3 or x > 2 B -3 < x < 2 C All Reals D No Solution 80 Solve A x < -2 or x > 5 B -2 < x < 5 C All Reals D No Solution
Slide 199 / 200 Slide 200 / 200 81 Solve A x = -2 B x = 2 C All Reals D No Solution 82 Solve A -3 < x < -2 B x < -3 or x > -2 C All Reals D No Solution