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Module in solving quadratic equation
1. Module in Solving Quadratic Equations In Fourth Year High School Aleli M. Ariola Shane Maureen D. Atendido next contents
2. A premier university in CALABARZON, offering academic programs and related services designed to respond to the requirements of the Philippines and the global economy, particularly in Asian Countries. VISION next back contents
3. The University shall primarily provide advanced education, professional, technological and vocational instruction in agriculture, fisheries, forestry, science, engineering, industrial technologies, teacher education, medicine, law, arts and sciences, information technologies and other related fields. It shall also undertake research and extension services and provide progressive leadership in its areas of specialization. MISSION back next contents
4. In pursuit of the college vision/mission the College of Education is committed to develop the full potentials of the individuals and equip them with knowledge, skills and attitudes in Teacher Education allied fields to effectively respond to the increasing demands, challenges and opportunities of changing time for global competitiveness . GOALS back next contents
5. OBJECTIVES OF BSED Produce graduate who can demonstrate and practice the professional and ethical requirement for the Bachelor of Secondary Education such as: 1.To serve as positive and powerful role models in the pursuit of the learning thereby maintaining high regards to professional growth. 2. Focus on the significance of providing wholesome and desirable learning environment. 3. Facilitate learning process in diverse type of learners. 4. Used varied learning approaches and activities, instructional materials and learning resources. 5. Used assessment data, plan and revise teaching – learning plans. 6. Direct and strengthen the links between school and community activities. 7. Conduct research and development in Teacher Education and other related activities. back next contents
6. This Teacher’s “ MODULE IN SOLVING QUADRATIC EQUATION” is part of the requirements in Educational Technology 2 under the revised curriculum based on CHED Memorandum Order (CMO)-30, Series of 2004. Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and innovative technologies to facilitate and foster meaningful and effective learning where students are expected to demonstrate a sound understanding of the nature, application and production of the various types of educational technologies. Students are provided with guidance and assistance of selected faculty The members of the College on the selection, production and utilization of appropriate technology tools in developing technology-based teacher support materials. Through the role and functions of computers especially the Internet, the student researchers and the advisers are able to design and develop various types of alternative delivery systems. These kinds of activities offer a remarkable learning experience for the education students as future mentors especially in the preparation and utilization of instructional materials. The output of the group’s effort may serve as a contribution to the existing body instructional materials that the institution may utilize in order to provide effective and quality education. The lessons and evaluations presented in this module may also function as a supplementary reference for secondary teachers and students. reference for secondary teachers and students. Aleli M. Ariola Module Developer Shane Maureen D. Atendido Module Developer FOREWORD back next contents
7. This Teacher’s “ MODULE IN SOLVING QUADRATIC EQUATION” is part of the requirements in Educational Technology 2 under the revised curriculum based on CHED Memorandum Order (CMO)-30, Series of 2004. Educational Technology 2 is a three (3)-unit course designed to introduce both traditional and innovative technologies to facilitate and foster meaningful and effective learning where students are expected to demonstrate a sound understanding of the nature, application and production of the various types of educational technologies. Students are provided with guidance and assistance of selected faculty The members of the College through the selection, production and utilization of appropriate technology tools in developing technology-based teacher support materials. Through the role and functions of computers especially the Internet, the student researchers and the advisers are able to design and develop various types of alternative delivery systems. These kinds of activities offer a remarkable learning experience for the education students as future mentors especially in the preparation and utilization of instructional materials. The output of the group’s effort may serve as a contribution to the existing body instructional materials that the institution may utilize in order to provide effective and quality education. The lessons and evaluations presented in this module may also function as a supplementary reference for secondary teachers and students. reference for secondary teachers and students. FOR-IAN V. SANDOVAL Computer Instructor / Adviser Educational Technology 2 DELIA F. MERCADO Module Consultant / Instructor 3 Principal of Laboratory High School LYDIA R. CHAVEZ Dean College of Education FOREWORD back next contents
8. The authors would like to acknowledge with deep appreciation and gratitude the invaluable help of the following persons: Mr. For-Ian V. Sandoval our module adviser, Computer Instructor / Adviser Educational Technology 2 for giving us opportunity to participate on this project, and for guiding us and pursue us to finish this module. Mrs. Delia F. Mercado, Instructor III and Director of Laboratory High School, for being our Teacher Consultant for the completion of this modular workbook. Mrs. Corazon San Agustin, our instructor in Educational Technology I, for giving us guidance and encouragement us in completing the requirement. Mrs. Lydia Chavez, Dean of Education for the support and guidance. We also wish to thank our family and friends as an inspiration and understand us they were robbed of many precious moments as we looked ourselves in our rooms when our minds went prolific and our hands itched to write. And finally, we thank Almighty God, the source of all knowledge, understanding and wisdom. From him we owe all that we have and all that we are! Once again, we thank all those who have encourage and helped us in preparing this module for publication and who have extended us much understanding, patience, and support. THE AUTHORS Acknowledgement back next contents
9. A quadratic equation is a second-order polynomial equatio n in a single variable x. The general form is where x represents a variable, and a , b , and c , represent coefficients and constants, with a ≠ 0. (If a = 0, the equation becomes a linear equation.) Among his many other talents, Major General Stanley in Gilbert and Sullivan's operetta the Pirates of Penzance impresses the pirates with his knowledge of quadratic equations in "The Major General's Song" as follows: "I am the very model of a modern Major-General, I've information vegetable, animal, and mineral, I know the kings of England, and I quote the fights historical, From Marathon to Waterloo, in order categorical; I'm very well acquainted too with matters mathematical, I understand equations, both the simple and quadratic, About binomial theorem I'm teeming with a lot o' news-- With many cheerful facts about the square of the hypotenuse." The constants a , b , and c , are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term. Quadratic comes from quadratus , which is the Latin word for "square." Quadratic equations can be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula. One common use of quadratic equations is computing trajectories in projectile motion. This module centers on the different ways of solving the quadratic equation by factoring, by finding square roots, by completing the square, and by using the quadratic formula. Students are given guides to determine the most appropriate method to use. Identifying the disciminant of the quadratic equation and finding the relationship between the coefficient and the root of the quadratic equation are also discussed. Introduction back next contents
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11. Table of Contents VMGO ’ s of BSEd Foreword Acknowledgement Introduction General Objective ’ s Table of Contents Chapter I. Identify the Quadratic Equation Lesson 1. Quadratic Equation Chapter II. Complex Number Lesson 2. Defining Complex Number Lesson 3. Number i Lesson 4. Complex Plane Lesson 5. Complex Arithmetic Chapter III. Solving Quadratic Equation Lesson 6. Factoring Lesson 7. Completing the Square Lesson 8. Quadratic Formula Lesson 9. Solving by Graphing contents back next
12. Chapter IV. Solving Equation on Quadratic Lesson 10. Equation in Quadratic Form Lesson 11. Equation Containing Radicals Lesson 12. Equation Reducible to Quadratic Equation Chapter V. The Discriminant, Roots and Coefficient Lesson 13. Discriminant and the Roots of a Quadratic Equation Lesson 14. Relation between Roots and Coefficient Chapter VI. Solving Quadratic Equation on a Calculator Lesson 15. Equation on a Calculator References Demo Slide share contents back next
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16. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: _______ Instruction : Write the following equations in the form ax 2 + bx + c = 0, and give the value of a, b, and c. 1. x 2 = 6x _____________________________________________ 2. 2x 2 = 32 _____________________________________________ 3. 3 x2 = 5x – 1 _____________________________________________ 4. 10 = 3x – x 2 _____________________________________________ 5. (x + 2) 2 = 9 _____________________________________________ 6. 4x 2 = 64 _____________________________________________ Activity 1.1 Identifying Quadratic Equation next back contents
17. _____________________________________________ 8x = x 2 _____________________________________________ 7. 8. 9. 10. 11. 12. 13. 14. 15. _____________________________________________ 2 = 6 _____________________________________________ x 2 = x 2 + _____________________________________________ (x + 1)(x-3) = 6 _____________________________________________ _____________________________________________ _____________________________________________ _____________________________________________ contents back next
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21. Equation 2: x 2 + 1 = 0 Equation 2 has no solutions, and we can see this by looking at the graph of y = x 2 + 1. Since the graph has no x-intercepts, the equation has no solutions. When we define complex numbers, equation 2 will have two solutions. next back contents
25. Example i 3 = i 2 * i = -1* i = - i. i 4 = i 2 * i 2 = (-1) * (-1) = 1. Exercise: Simplify i 8 and i 11 . We treat i like other numbers in that we can multiply it by numbers, we can add it to other numbers, etc. The difference is that many of these quantities cannot be simplified to a pure real number. For example, 3 i just means 3 times i , but we cannot rewrite this product in a simpler form, because it is not a real number. The quantity 5 + 3 i also cannot be simplified to a real number. However, (- i ) 2 can be simplified. (- i ) 2 = (-1* i ) 2 = (-1) 2 * i 2 = 1 * (-1) = -1. Because i 2 and (- i ) 2 are both equal to -1, they are both solutions for Equation 2 above. contents back next
26. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Express each number in terms of i and simplify. 1. 2. 3. 4. 5. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ Activity 2.2 Number i next back contents
27. 6. 7. 8. 9. 10. 11. 12. 13. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ next back contents
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30. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Represent each of the following Complex Numbers by a point in the plane. 1. 2. 3. 4. 0 5. 3 6. 7. 1/2 8. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ Activity 2.3 Complex Plane contents back next
31. 9. 10. 11. 12. 13. 14. 15. _____________________________________________________ _____________________________________________________ _____________________________________________________ ____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ contents back next
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33. Note These operations are the same as combining similar terms in expressions that have a variable. For example, if we were to simplify the expression (3 - 5x) + (6 + 7x) by combining similar terms, then the constants 3 and 6 would be combined, and the terms -5x and 7x would be combined to yield 9 + 2x. The Complex Arithmetic applet below demonstrates complex addition in the plane. You can also select the other arithmetic operations from the pull down list. The applet displays two complex numbers U and V, and shows their sum. You can drag either U or V to see the result of adding other complex numbers. As with other graphs in these pages, dragging a point other than U or V changes the viewing rectangle. Multiplication The formula for multiplying two complex numbers is (a + bi) * (c + di) = (ac - bd) + (ad + bc)i. You do not have to memorize this formula, because you can arrive at the same result by treating the complex numbers like expressions with a variable, multiply them as usual, then simplify. The only difference is that powers of i do simplify, while powers of x do not. Example (2 + 3i)(4 + 7i) = 2*4 + 2*7i + 4*3i + 3*7*i 2 = 8 + 14i + 12i + 21*(-1) = (8 - 21) + (14 + 12)i = -13 + 26i. next back contents
34. Notice that in the second line of the example, the i 2 has been replaced by -1. Using the formula for multiplication, we would have gone directly to the third line. Exercise Perform the following operations. (a) (-3 + 4i) + (2 - 5i) (b) 3i - (2 - 4i) (c) (2 - 7i)(3 + 4i) (d) (1 + i)(2 - 3i) Division The conjugate (or complex conjugate) of the complex number a + bi is a - bi. Conjugates are important because of the fact that a complex number times its conjugate is real; i.e., its imaginary part is zero. (a + bi)(a - bi) = (a 2 + b 2 ) + 0i = a 2 + b 2 . Example Number Conjugate Product 2 + 3i 2 - 3i 4 + 9 = 13 3 - 5i 3 + 5i 9 + 25 = 34 4i -4i 16 next back contents
35. Suppose we want to do the division problem (3 + 2i) ÷ (2 + 5i). First, we want to rewrite this as a fractional expression . Even though we have not defined division, it must satisfy the properties of ordinary division. So, a number divided by itself will be 1, where 1 is the multiplicative identity; i.e., 1 times any number is that number. So, when we multiply by, , we are multiplying by 1 and the number is not changed. Notice that the quotient on the right consists of the conjugate of the denominator over itself. This choice was made so that when we multiply the two denominators, the result is a real number. Here is the complete division problem, with the result written in standard form. back next contents
36. Exercise: Write (2 - i) ÷ (3 + 2i) in standard form. We began this section by claiming that we were defining complex numbers so that some equations would have solutions. So far we have shown only one equation that has no real solutions but two complex solutions. In the next section we will see that complex numbers provide solutions for many equations. In fact, all polynomial equations have solutions in the set of complex numbers. This is an important fact that is used in many mathematical applications. Unfortunately, most of these applications are beyond the scope of this course. See your text (p. 195) for a discussion of the use of complex numbers in fractal geometry. back next contents
37. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Perform the indicated operations and express the result in the form . 1. 2. 3. 4. 5. 6. 7. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ Activity 2.4 Complex Arithmetic back next contents
38. 8. 9. 10. 11. 12. 13. 14. 15. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ next back contents
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47. Name: ___________________ Section: _______ Instructor: ________________ Date: ________ Rating: _______ Instruction: Solve the following Quadratic Equation by Factoring Method. 1. x 2 – 36 = 0 _____________________________________________________ 2. x 2 = 25 _____________________________________________________ 3. x 2 – 12x + 35 = 0 _____________________________________________________ 4. x 2 – 3x – 40 = 0 _____________________________________________________ 5. 2x 2 – 5x = 3 _____________________________________________________ 6. 3x 2 + 25x = 18 _____________________________________________________ 7. 15x 2 – 2x – 8 = 0 _____________________________________________________ Activity 3.1 Solving Quadratic Equation contents back next
48. 8. 3x 2 – x = 10 _____________________________________________________ 9. x 2 + 6x – 27 = 0 _____________________________________________________ 10. y 2 – 2y – 3 = y – 3 _____________________________________________________ 11. 4y 2 + 4y = 3 _____________________________________________________ 12. 3a 2 + 10a = -3 _____________________________________________________ 13. a 2 – 2a – 15 = 0 _____________________________________________________ 14. r 2 + 6r – 27 = 0 _____________________________________________________ 15. 2z 2 – 2 – 1 = 0 _____________________________________________________ next back contents
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50. The answer can also be written in rounded form as next back contents
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53. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ____ Instruction: Solve the following Quadratic Equation by Completing the Square. 1. x 2 + 3x = 4 _____________________________________________________ 2. x 2 – 2x = 24 _____________________________________________________ 3. x 2 + 4 = 4x _____________________________________________________ 4. 2x 2 – 6 = x _____________________________________________________ 5. 4a 2 + 12a + 9 = 0 _____________________________________________________ 6. 3a 2 – 5 = 14a _____________________________________________________ 7. 16b 2 + 1 = 16b _____________________________________________________ Activity 3.2 Completing the Square contents back next
63. They haven't given me the quadratic equation, so I can't check my work algebraically. (And, technically, they haven't even given me a quadratic to solve; they have only given me the picture of a parabola from which I am supposed to approximate the x -intercepts, which really is a different question....) I ignore the vertex and the y -intercept, and pay attention only to the x -intercepts. The "solutions" are the x -values of the points where the pictured line crosses the x -axis: The solution is x = –5.39, 2.76 "Solving" quadratics by graphing is silly in "real life", and requires that the solutions be the simple factoring-type solutions such as " x = 3", rather than something like " x = –4 + sqrt(7)". In other words, they either have to "give" you the answers (by labeling the graph), or they have to ask you for solutions that you could have found easily by factoring. About the only thing you can gain from this topic is reinforcing your understanding of the connection between solutions and x -intercepts: the solutions to "(some polynomial) equals (zero)" correspond to the x -intercepts of " y equals (that same polynomial)". If you come away with an understanding of that concept, then you will know when best to use your graphing calculator or other graphing software to help you solve general polynomials; namely, when they aren't factorable. contents back next
69. Solving Equations that are Quadratic in Form Step 1: Write in Standard Form, , if needed. Step 2: Substitute a variable in for the expression that follows b in the second term. Step 3: Solve the quadratic equation created in step 2. Step 4: Find the value of the variable from the original equation. If it is not in standard form, move any term(s) to the appropriate side by using the addition/subtraction property of equality. Also, make sure that the squared term is written first left to right, the expression not squared is second and the constant is third and it is set equal to 0. In other words, substitute your variable for what is in the ( ) when it is in standard form, . I’m going to use t for my substitution, but really you can use any variable as long as it is not the variable that is used in the original equation. You can use any method you want to solve the quadratic equation: factoring, completing the square or quadratic formula. Keep in mind that you are finding a solution to the original equation and that the variable you substituted in for in step 2 is not your original variable. Use the substitution that was used to set up step 2 and then solve for the original variable. contents back next
70. Step 5: Check your solutions. Example 1: Solve the equation that is quadratic in form: . In some cases, you will be working with rational exponents and square roots in your problems. Those types of equations can cause extraneous solutions. Recall that an extraneous solution is one that is a solution to an equation after doing something like raising both sides of an equation by an even power, but is not a solution to the original problem. Even though not all of the quadratic in form equations can cause extraneous solutions, it is better to be safe than sorry and just check them all. Standard Form, *Rewriting original equation to show it is quadratic in form *Note that (y squared) squared = y to the fourth *When in stand. form, let t = the expression following b. Next, we need to substitute t in for y squared in the original equation. *Original equation *Substitute t in for y squared next back contents
71. Note how we ended up with a quadratic equation when we did our substitution. From here, we need to solve the quadratic equation that we have created. Solve the quadratic equation: factoring, completing the square or quadratic formula. *Factor the trinomial *Use Zero-Product Principle *Set 1st factor = 0 and solve *Set 2nd factor = 0 and solve next back contents
72. Let's find the value(s) of y when t = -4: *Plug in - 4 for t *Use square root method to solve for y *First solution *Second solution Let's find the value(s) of y when t = 1: *Plug in 1 for t *Use square root method to solve for y *First solution *Second solution next back contents
73. Example 2: Solve the equation that is quadratic in form: . Standard Form, *Inverse of add. 3 is sub. 3 *Equation in standard form Note how when you square x to the 1/3 power you get x to the 2/3 power, which is what you have in the first term. * Rewriting original equation to show it is quadratic in form *Note that (x to the 1/3 power) squared = x to the 2/3 power *When in stand. form, let t = the expression following b. Next, we need to substitute t in for x to the 1/3 power in the original equation. next back contents
74. *Original equation *Substitute t in for x to the 1/3 power You can use any method you want to solve the quadratic equation: factoring, completing the square or quadratic formula. *Factor the trinomial *Use Zero-Product Principle *Set 1st factor = 0 and solve *Set 2nd factor = 0 and solve next back contents
75. Let's find the value(s) of x when t = 3: *Plug in 3 for t *Solve the rational exponent equation *Inverse of taking it to the 1/3 power is raising it to the 3rd power Let's find the value(s) of x when t = -1: *Plug in -1 for t *Solve the rational exponent equation *Inverse of taking it to the 1/3 power is raising it to the 3rd power Let's double check to see if x = 27 is a solution to the original equation. *Plugging in 27 for x *True statement next back contents
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77. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Solve the equation that is in quadratic form. 1. a 8 + 2a 4 – 8 = 0 _____________________________________________________ 2. l 2 + 4l 2 – 6 = 0 _____________________________________________________ 3. e 4 – 8e 2 – 3 = 0 _____________________________________________________ 4. l 6 – 10l – 5 = 0 _____________________________________________________ 5. i 10 – 8i 5 – 4 = 0 _____________________________________________________ 6. s 6 – 5s 3 – 25 = 0 _____________________________________________________ Activity 4.1 Solving Equation on Quadratic next back contents
78. 7. h 2/4 + 8h 1/4 – 12 = 0 _____________________________________________________ 8. a 6 - 5a 4 – 15 = 0 _____________________________________________________ 9. n 8 + 12n 2 – 8 = 0 _____________________________________________________ 10. e 9 – 3n 3 – 10 = 0 _____________________________________________________ 11. x 2/3 – 2x 1/3 = 8 _____________________________________________________ 12. x 3/6 – 3x 1/2 = 9 _____________________________________________________ 13. y 2 - 8y = 5 _____________________________________________________ 14. y 4 + 2y 2 = 6 _____________________________________________________ 15. x 6 – 9x 2 + 8 = 0 _____________________________________________________ next back contents
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80. Example 1: First make a note of the fact that you cannot take the square root of a negative number. Therefore, the term is valid only if and the second term is valid if Isolate the term Square both sides of the equation. Isolate the term Square both sides of the equation. Check the solution by substituting 9 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer. next back contents
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82. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Solve each of the following equation. 1. 2. 3. 4. 5. 6. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ Activity 4.2 Equation Containing Radicals contents back next
85. x = 1 or -18 7 If x = 1, _ 1_ _ 1_ _ 1_ _ 1_ _ 7_ 1+2 1+3 = 3 + 4 = 12 Therefore x = 1 is a solution. If x = -18 , __ 1__ + __ 1__ 7 -18/7 + 2 -18/7 + 3 = __ 7 __ + __ 7 __ -18 + 14 -18 + 21 = _ -7 _ + _ 7 _ = _ 7 _ 4 3 12 Therefore, x = _-18 _ is a solution. 7 Example 2. = - 2 Solution: squaring both sides of the equation, we obtain, = + 4 2x – 16 = 4 contents back next
86. Dividing both sides by 2 gives, x – 8 = -2 Squaring both sides of the equation we get (x - 8) = 2 x 2 - 16x + 64 = 4 (x +16) x 2 – 20x = 0 x(x – 20) =0 x = 0 or x = 20 Check: if x = 0, = = 8 = 6 – 2 8 ≠ 4 Therefore x = 20 is not a solution of the original equation. Thus the only root of Many equations are not quadratics equations. However, we can transform them by means of appropriate substitutions into quadratics equations and then solve these by techniques that we know. contents back next
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88. u = = u = 1 + or u = Since u = x 2 and u = < 0, we have to discard this solution. u = x 2 = implies x = ± It is simple to verify that both values of x satisfy the original equation. The roots of x 4 – x 2 – 2 = 0 are and - . c. Let u = .This substitution yields a quadratic equation in u. u 2 – u – 2 = 0 (u – 2)(u + 1) = 0 u = 2 0r u = ˉ1 u = = 2 implies x = 2(4x + 1) or x = u = = 1 implies x = ˉ4x – 1 or x = Again, it can easily be verified that both solutions check in the original equation. The roots are and . contents back next
89. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Solve the following equation. 1. 2. 3. 4. 5. 6. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ Activity 4.3 Equation Riducible to Quadratic Equation contents back next
91. Chapter Test A. Solve for x. 1. = 4 2. - 5 = 0 3. = 4. = 5. = x + 2 B. Reduce to quadratic equation. 1. x 4 – 5x + 4 = 0 2. 4(x + 3) + 5 = 21 3. x 2/3 – 5x 1/3 – 6 = 0 4. (x 2 + 4x) 2 – (x 2 + 4x) = 20 5. 2x 4 – 9x 2 + 7 = 0 C. Solve for x. 1. + = 2. + + = 0 3. + = 4. + = 2 5. + = next back contents
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94. Example 2. Use the disciminant to determine the nature of the roots of the following quadratic equation. a. x² - x + ¼ = 0 a = 1, b = ˉ1, c = ¼ b² - 4ac = (ˉ1)² - 4 (1)(¼) = 1 – 1 = 0 There is only one solution, that is, a double root. Note that x² - x = ½ = (x - ½), so that double root is ˉb/2a = ½. b. 5x² - 4x + 1 = 0 a = 5, b = ˉ4, c = 1 b² - 4ac = (ˉ4)² - 4 (5)(1) = 16 – 20 = ˉ4 < 0 There are no real roots since a negative number has no real square root. ∆ = b² - 4ac Roots of ax² + bx + c = 0 Positive Real and distinct r = s = Zero Real and equal r = s = Negative No real roots next back contents
95. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Use the Discriminant to determine the nature of the root of the following Quadratic Equations. 1.x 2 - 2x – 3=0 _____________________________________________________ 2. 6x 2 – x – 1 = 0 _____________________________________________________ 3. 2x 2 – 50 = 0 _____________________________________________________ 4. x 2 – 8x + 12 = 0 _____________________________________________________ 5. x 2 + 5x – 14 = 0 _____________________________________________________ 6. -4x 2 – 4x + 1 = 0 _____________________________________________________ 7. 7x 2 + 2x – 1 = 0 _____________________________________________________ Activity 5.1 The Discriminant, Roots and Coefficient next back contents
98. Adding the roots, we obtain r + s = + = = Multiplying the roots, we obtain rs = = c / a Observe the coefficient in the quadratic equation x 2 + bx/a + c/a = 0. How do they compare with the sum and the product of the roots? Did you observe the following? 1. The sum of the roots is equal to the negative of the coefficient of x. r + s = -b / a 2. The product of the roots is equal to the constant term rs = c / a An alternate way of arriving at these relations is as follows Let r and s be the roots of x 2 + bx/a + c/a = 0. Then x - r)(x – s) = 0 Expanding gives, x 2 – rx – sx + rs = 0 or x 2 – (r + s)x + rs = 0 next back contents
99. Comparing the coefficients of the corresponding terms, we obtain r + s = -b / a and rs c / a The above relations between the roots and the coefficients provide a fast and convenient means of checking the solutions of a quadratic equation. Example: Solve and check. 2x 2 + x – 6 = 0 Solutions: 2x 2 + x – 6 = (2x – 3)(x + 2) = 0 x = 3/2 or x = 2 The roots are 3/2 and 2. To check, we add the roots, 3/2 = (-2) = -1/2 = -b/a. and multiply them 3/2 = (-2) = -3 = c/a Example: Find the sum and the product of the roots of 3x 2 – 6x + 8 = 0 without having to first determine the roots. Solution: The sum of the roots is r + s = -c/a = -(-6)/3 = 2 and their product is rs = c/a = 8/3 next back contents
100. Name: ___________________ Section: _______ Instructor: ________________ Date: _______ Rating: ______ Instruction: Without solving the roots, find the sum and product of the roots of the following. 1. 6x 2 – 5x + 2 = 0 _____________________________________________________ 2. x 2 + x – 182 = 0 _____________________________________________________ 3. x 2 – 5x – 14 = 0 _____________________________________________________ 4. 2x 2 – 9x + 8 = 0 _____________________________________________________ 5. 3x 2 - 5x – 2 = 0 _____________________________________________________ 6. x 2 – 8x – 9 = 0 _____________________________________________________ Activity 5.2 Relation between Roots and Coefficient next back contents
108. References Salamat, Lorina G., College Algebra, National Book Store 1988 pg. 151 – 159. Coronel, Iluminada C. F.M.M, Mathematics 3 An Integrated Approach, Bookmark Inc. 1991 pg. 77 – 79, 134 – 158. Coronel, Iluminada C. F.M.M, Mathematics 4 An Integrated Approach, Bookmark Inc. 1992 pg. 276 – 297. Borwein, P. and Erdélyi, T. "Quadratic Equations." §1.1.E.1a in Polynomials and Polynomial Inequalities. New York: Springer-Verlag, p. 4, 1995. URL – Images http://www.mathwarehouse.com/algebra/complex-number/absolute-value-complex-number.php http://commons.wikimedia.org/wiki/File:Quadratic_equation_coefficients.png http://mathworld.wolfram.com/PolynomialRoots.html http://www.livephysics.com/shop/tools-and-gadgets.html http://www.youtube.com/watch?v=vAhuSVu_I0c http://www.youtube.com/watch?v=ug7IDDBRO94 contents back next
