1.contents next

2. VISIONA 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. contents back next 3. MISSION The University shall primarily provide advanced education, professional, technological and vocational instruction in agriculture, fisheries, forestry, science, engineering, in dustrial technologies, teacher education, medicine, law, artsand sciences, information technologies and other related fields. It shall also undertake research and extension services and provide progressive leadership in its areas of specialization. contents back next 4. GOALS 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. contents back next 5. OBJECTIVES OF BSEDProduce 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. contents back next 6. This Teachers 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 groups 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 contents back next 7. This Teachers 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 groups 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. SANDOVALComputer Instructor / AdviserEducational Technology 2DELIA F. MERCADO Module Consultant / Instructor 3 Principal of Laboratory High SchoolLYDIA R. CHAVEZ Dean College of Education contentsback next 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 AUTHORScontentsbacknext 9. A quadratic equation is a second-order polynomial equation 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. contents backnext 10. At the end of this module, students are expected to:distinguish what is quadratic equation and complex numbers;recognize property of the number I and solutions in solving the four arithmetic operations;solve quadratic equation by factoring, completing the square, quadratic formula and solving by graphing;learn the technique on how to use any method for solving quadratic equation, taking the square root and transforming quadratic equations by appropriate substitution;determine discriminant, and relations between roots and coefficient; anduse calculator in solving quadratic equation. contentsback next 11. VMGOs of BSEd Foreword Acknowledgement Introduction General Objectives Table of ContentsChapter 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 Graphingcontents 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 CalculatorReferencesDemo Slide sharecontents back next 13. Chapter I This chapter deals with equations which are classified according to the highest power of its variable. Anequation in the variable x whose highest power is 2 is called a quadratic equation. It will be observed here thatvariable a, b and c are real numbers and a cannot be 0. TARGET SKILLS: At the end of this chapter, students are expected to: identify quadratic equation; discuss real numbers and standard form of the quadratic equation; express quadratic equation in standard form; and apply distributive property in solving quadratic equation. contents back next 14. Lesson 1 Identifying the quadratic equation OBJECTIVES:At the end of this lesson, students are expected to: define the quadratic equation; discuss real numbers in quadratic equation; and improve writing the standard form of the quadratic equation. Polynomials are classified according to the highest power of its variable. A first degree polynomial, like 2x + 5 is linear; a second degree polynomial, like x2 + 2 3 is quadratic; a third degree polynomial, like x3 + 4x2 3x + 12 is cubic. Similarly, equation and inequalities are classified according to the highest power of its variable. An equation in the variable x whose highest power is two is called a quadratic equation. Some examples are x2 64, 4n2 = 25, 3x2 4x + 1 = 0. An equation of the form ax2 + bx + c = 0, where a, b and c are constant and a not equal to 0, a id aquadratic equation. to 0, a id a quadratic equation.Any quadratic equation can be written in the form ax2 + bx + c = 0. This is also called the standard form of the quadratic equation. Here, a, b and c are real numbers and a cannot be 0.Example A. Express x2 = 8x in standard formx2 = 8x can be written as x2 - 8x = 0where a=1, b= 8, and c=0.Example B. Express x2 = 64 in standard formcontentsbacknext 15. x2 = 64 can be written as x2 64 = 0 where a=1. B=0, and c=64.Example C. Express the fractional equation x = 1/x-3 as a quadratic equation.x = 1/x-3x (x-3) = 1 multiply both sides by x-3x - 3x = 1using the distributive propertyx - 3x - 1 = 0a=1, b=3, c=1Exercises: Which of the following equations are quadratic?1.3x = x - 5 2. 2x =1 3. x = 25 4. 2x - 3 = x + 5 5. 5x 2y = 0 contents back next 16. Name: ___________________ Section: _______ Instructor: ________________Date: _______Rating: _______ Instruction: Write the following equations in the form ax2 + bx + c = 0, and give the value of a, b, and c.1. x2 = 6x _____________________________________________2. 2x2 = 32 _____________________________________________3. 3x2 = 5x 1 _____________________________________________4. 10 = 3x x2 _____________________________________________5. (x + 2)2 = 9_____________________________________________6. 4x2 = 64 _____________________________________________ contentsbacknext 17. 7. _____________________________________________8. _____________________________________________9.8x = x2 _____________________________________________2 10.=6 _____________________________________________11. _____________________________________________ 12. x2 = _____________________________________________13. _____________________________________________14. x2 + _____________________________________________15. (x + 1)(x-3) = 6 _____________________________________________contents back next 18. A. Define each of the following terms.1. Quadratic equation2. Standard form of a quadratic equation3. Real numbers B. Which of the following equations are quadratic?1. 4x = 2x2 62. 3x = 13. 5x2 = 304. 3x 2 = 2x + 65. 2x 5y = 06. 4x + 2x2 3x3 = 07. 12x2 x = 11 C. Write the following equations in the form ax2 + bx + c = 0, and give the values of a, b and c.1. 3x2 = 6x2. 3x2 = 323. 2x2 = 5x 14. 12 = 4x x25. (x + 3)2 = 86. 4x2 = 567. 1/x + x = 68. x(x 4) 1 = 09. 9x = x210. (1/x)2 = 10 contentsback next 19. Chapter II This chapter centers on the complex numbers which is a number comprising a real number and an imaginarynumber. Under this, we have the number i, the complex plane where the points are plotted and the 4 arithmeticoperations such as addition and subtraction, multiplication and division of complex numbers. To round up thechapter, simple equation involving complex numbers will be studied and solved.TARGET SKILLS:At the end of this chapter, students are expected to: identify complex numbers; differentiate the real pat and imaginary part of complex numbers; and explore solving of the 4 arithmetic operations on the complex numbers.contents back next 20. Lesson 2 Defining Complex Numbers OBJECTIVES:At the end of this lesson, students are expected to:identify complex numbers; differentiate the real number and standard imaginary unit; and extend the ordinary real number.A complex number, in mathematics, is a number comprising a real number and an imaginary number; it can be written in the form a + bi, where a and b are real numbers, and i is the standard imaginary unit, having the property that i2 = 1. The complex numbers contain the ordinary real numbers, but extend them by adding in extra numbers and correspondingly expanding the understanding of addition and multiplication.Equation 1: x2 - 1 = 0. Equation 1 has two solutions, x = -1 and x = 1. We know that solving an equation in x is equivalent to findingthe x-intercepts of a graph; and, the graph of y = x2 - 1 crosses the x-axis at (-1,0) and (1,0). contentsback next 21. Equation 2: x2 + 1 = 0 Equation 2 has no solutions, and we can see this by looking at the graph of y = x2 + 1.Since the graph has no x-intercepts, the equation has no solutions. When we define complexnumbers, equation 2 will have two solutions.contents back next 22. Name: ___________________Section: _______ Instructor: ________________ Date: ________Rating: _______Solve each equation and graph.1. x + 4 = 0_____________________________________________ 2. 2x + 18 = 0 _____________________________________________ 3. 2x + 14 = 0 _____________________________________________ 4. 3x + 27 = 0 _____________________________________________ 5. x - 3 = 0_____________________________________________ 6. x + 21 = 0_____________________________________________contents back next 23. 7. 3x - 5 = 0 _____________________________________________8. 5x + 30 = 0_____________________________________________9. 2x + 3 = 0 _____________________________________________10. x + 50 = 0_____________________________________________11. x - 2 = 0_____________________________________________12. 3x - 50 = 0_____________________________________________13. x - 3 = 0 _____________________________________________14. x + 4 = 0 _____________________________________________15. 2x + 14 = 0_____________________________________________ contents back next 24. Lesson 3 The Number iOBJECTIVES:At the end of this lesson, students are expected to: recognize the property of the number i;discuss the powers of i; andsolve the high powers of imaginary unit.Consider Equations 1 and 2 again.Equation 1 Equation 2 x2 - 1 = 0.x2 + 1 = 0. x2 = 1.x2 = -1.Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has no solutionsbecause -1 does not have a square root. In other words, there is no number such that if we multiply it by itself we get -1. If Equation 2 is to be given solutions, then we must create a square root of -1.The imaginary unit i is defined by The definition of i tells us that i2 = -1. We can use this fact to find other powers of i.contents back next 25. Examplei3 = i2 * i = -1*i = -i. i4 = i2 * i2 = (-1) * (-1) = 1.Exercise:Simplify i8 and i11. 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, 3i 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 + 3i also cannot be simplified to a real number.However, (-i)2 can be simplified. (-i)2 = (-1*i)2 = (-1)2 * i2 = 1 * (-1) = -1.Because i2 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.______________________________________________________ contents backnext 27. 6. ______________________________________________________ 7. ______________________________________________________ 8. ______________________________________________________ 9. ______________________________________________________ 10. ______________________________________________________ 11. ______________________________________________________ 12. ______________________________________________________ 13. ______________________________________________________contents back next 28. Lesson 4The Complex Plane OBJECTIVES: At the end of this lesson, students are expected to:distinguish the points on the plane; differentiate the real and imaginary part; and draw from memory the figure form by the plot points on the complex plane. A complex number is one of the form a + bi, where a and b are real numbers. a is called the real part of the complex number, and b is called the imaginary part. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. I.e., a+bi = c+di if and only if a = c, and b = d.Example.2 - 5i. 6 + 4i. 0 + 2i = 2i. 4 + 0i = 4.The last example above illustrates the fact that every real number is a complex number (with imaginary part 0). Another example: the real number -3.87 is equal to the complex number -3.87 + 0i. It is often useful to think of real numbers as points on a number line. For example, you can define the order relation c < d, where c and d are real numbers, by saying that it means c is to the left of d on the number line.contents backnext 29. We can visualize complex numbers by associating them with points in the plane. We do this by letting the number a + bi correspond to the point (a,b), we use x for a and y for b.Exercises: Represent each of the following complex number by a point in the plane. 1. 3 + 2i2. 1 4i3. 4 + 3i4. 2 5i5. 4 3i contents backnext 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. ______________________________________________________ contents backnext 31. 9. _____________________________________________________10. ____________________________________________________11. _____________________________________________________12. _____________________________________________________13. _____________________________________________________14. _____________________________________________________ 15. _____________________________________________________contents back next 32. Lesson 5 Complex Arithmetic OBJECTIVES: At the end of this lesson, students are expected to:define the four arithmetic operations on complex numbers; comply with the steps in solving the different operations; and solve the four arithmetic operations. When a number system is extended the arithmetic operations must be defined for the new numbers, and the important properties of the operations should still hold. For example, addition of whole numbers is commutative. This means that we can change the order in which two whole numbers are added and the sum is the same: 3 + 5 = 8 and 5 + 3 = 8. We need to define the four arithmetic operations on complex numbers.Addition and SubtractionTo add or subtract two complex numbers, you add or subtract the real parts and the imaginary parts. (a + bi) + (c + di) = (a + c) + (b + d)i. (a + bi) - (c + di) = (a - c) + (b - d)i. Example (3 - 5i) + (6 + 7i) = (3 + 6) + (-5 + 7)i = 9 + 2i.(3 - 5i) - (6 + 7i) = (3 - 6) + (-5 - 7)i = -3 - 12i. contents back next 33. NoteThese 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*i2 = 8 + 14i + 12i + 21*(-1) = (8 - 21) + (14 + 12)i = -13 + 26i.contentsback next 34. Notice that in the second line of the example, the i2 has been replaced by -1.Using the formula for multiplication, we would have gone directly to the third line.ExercisePerform 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) = (a2 + b2) + 0i = a2 + b2.ExampleNumberConjugateProduct2 + 3i2 - 3i 4 + 9 = 133 - 5i3 + 5i 9 + 25 = 344i-4i16contentsback next 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. contents back next 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.contents back next 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. _____________________________________________________ contents back next 38. 8. _____________________________________________________9. _____________________________________________________10. _____________________________________________________11. _____________________________________________________12. _____________________________________________________13. _____________________________________________________ 14. _____________________________________________________ 15. _____________________________________________________contents back next 39. A. Define and/or describe each of the following terms.1. Imaginary part2. Real number3. Complex number4. Complex plane5. Imaginary unit6. Commutative property7. Complex conjugate B. 1. Simplify:1. i152. i253. i1064. i2075. i212. Perform the indicated operation and express each answer.a. + b. + c. + d. + e. + f. +contents back next 40. 3. Represent each complex numbers by a point in the plane. a. 3ib. -2 + 4ic. -3 + 3id. 4 + 5ie. -3 + 5i 4. Give the real part and the imaginary part of each complex numbers in #3.5. Perform the indicated operations.a. (3 2i) + (-7 + 3i) b. (-4 + 7i) + (9 2i) c. (14 9i) + (7 6i) d. (5 + i) (3 + 2i) e. (7 2i) (4 6i) f. (8 + 3i) (-4 2i) g. (3 2i) (3 +2i) h. (5 + 3i) (4 i) i. (11 + 2i)2 (5 2i) j. (5 + 4i) / (3 2i) k. (4 + i) (3 5i) / (2 3i) l. (7 + 3i) / (3 3i / 4)contents back next 41. Chapter III In this chapter, the different ways of solving the quadratic equation are recalled. There are by using the factoring, completing the square, by quadratic formula and solving by graphing. Students are given guides to determine the most appropriate method to use. TARGET SKILLS:At the end of this chapter, students are expected to: distinguish appropriate method in solving quadratic equation; discuss and follow the steps in such different method; and resolve quadratic equation using any method you want. contents back next 42. Lesson 6Solving by factoring OBJECTIVES: At the end of this lesson, students are expected to:define what is factoring; discuss the Zero Factor Principle; and solve equation by using the factoring method. Factoring rearrange the equation; factor the left member; equate each factor to zero toobtain the two roots. Solve (x 3)(x 4) = 0. The Zero Factor Principle tells me that at least one of the factors must be equal to zero. Since at least one of the factors must be zero, I'll set them each equal to zero:x 3 = 0 or x 4 = 0x = 3 or x = 4Solve: x = 3, 4 Note that "x = 3, 4" means the same thing as "x = 3 or x = 4"; the only difference is the formatting. The "x = 3, 4" format is more-typically used.Checking x = 3 in (x 3)(x 4) = 0:([3] 3)([3] 4) ?=? 0 (3 3)(3 4) ?=? 0(0)(1) ?=? 00 = 0 contents back next 43. Checking x = 4 in (x 3)(x 4) = 0: ([4] 3)([4] 4) ?=? 0(4 3)(4 4) ?=? 0 (1)(0) ?=? 00 = 0 Solve x2 + 5x + 6 = 0.This equation is already in the form "(quadratic) equals (zero)" but, unlike the previous example, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.So the first thing I have to do is factor: x2 + 5x + 6 = (x + 2)(x + 3)Set this equal to zero: (x + 2)(x + 3) = 0Solve each factor:x + 2 = 0 or x + 3 = 0 x = 2 or x = 3 The solution to x2 + 5x + 6 = 0 is x = 3, 2Checking x = 3 and x = 2 in x2 + 5x + 6 = 0: [3]2 + 5[3] + 6 ?=? 0 9 15 + 6 ?=? 0 9 + 6 15 ?=? 0contentsbacknext 44. 15 15 ?=? 0 0 = 0[2] 2 + 5[2] + 6 ?=? 0 4 10 + 6 ?=? 04 + 6 10 ?=? 0 10 10 ?=? 0 0 = 0 So both solutions "check".Solve x2 3 = 2x.This equation is not in "(quadratic) equals (zero)" form, so I can't try to solve it yet. The first thing I need to do is get all the terms over on one side, with zero on the other side. Only then can I factor and solve:x2 3 = 2x x2 2x 3 = 0 (x 3)(x + 1) = 0 x 3 = 0 or x + 1 = 0 x = 3 or x = 1 Then the solution to x2 3 = 2x is x = 1, 3Solve (x + 2)(x + 3) = 12. The (10 + 2)(9 + 3) does not equal 12, you should never forget that you must have "(quadratic) equals (zero)" before you can solve.So, tempting though it may be, the factors above equal to the other side of the equation and "solve". Instead, multiply out and simplify the left-hand side, then subtract the 12 over to the left-hand side, and re-factor. contents backnext 45. (x + 2)(x + 3) = 12 x2 + 5x + 6 = 12 x2 + 5x 6 = 0(x + 6)(x 1) = 0 x + 6 = 0 or x 1 = 0 x = 6 or x = 1 Then the solution to (x + 2)(x + 3) = 12 is x = 6, 1Solve x(x + 5) = 0. To "solve" the equation for "x + 5 = 0", divide it by x. But it can't divide by zero; dividing off the x makes the implicit assumption that x is not zero. Used the variable factors having variables and numbers (like the other factor, x + 5), a factor can contain only a variable, so "x" is a perfectly valid factor. So set the factors equal to zero, and solve:x(x + 5) = 0 x = 0 or x + 5 = 0 x = 0 or x = 5 Then the solution to x(x + 5) = 0 is x = 0, 5Solve x2 5x = 0.Factor the x out of both terms, taking the x out front.x(x 5) = 0 x = 0 or x 5 = 0 x = 0 or x = 5 Then the solution to x2 5x = 0 is x = 0, 5 There is one other case of two-term quadratics that you can factor: contents backnext 46. Solve x2 4 = 0. This equation is in "(quadratic) equals (zero)" form, it's ready to solve. The quadratic itself is a difference of squares, then apply the difference-of-squares formula:x2 4 = 0 (x 2)(x + 2) = 0 x 2 = 0 or x + 2 = 0 x = 2 or x = 2 Then the solution is x = 2, 2Note: This solution may also be formatted as "x =2Exercises: Solve:1.(x 3)(x 5) = 0.2.x2 + 6x + 7 = 0.3.x2 4 = 2x.4.x2 6x = 0.5.x2 8 = 0.contents backnext 47. Name: ___________________Section: _______ Instructor: ________________ Date: ________ Rating: _______ Instruction: Solve the following Quadratic Equation by Factoring Method. 1. x2 36 = 0_____________________________________________________2. x2= 25_____________________________________________________3. x2 12x + 35 = 0 _____________________________________________________4. x2 3x 40 = 0 _____________________________________________________5. 2x2 5x = 3 _____________________________________________________6. 3x2 + 25x = 18 _____________________________________________________7. 15x2 2x 8 = 0 _____________________________________________________contents back next 48. 8. 3x2 x = 10 _____________________________________________________9. x2 + 6x 27 = 0 _____________________________________________________10. y2 2y 3 = y 3_____________________________________________________11. 4y2 + 4y = 3 _____________________________________________________12. 3a2 + 10a = -3_____________________________________________________13. a2 2a 15 = 0 _____________________________________________________14. r2 + 6r 27 = 0_____________________________________________________ 15. 2z2 2 1 = 0 _____________________________________________________ contents back next 49. Lesson 7 Solving by Completing the Square OBJECTIVES:At the end of this lesson, students are expected to: analyze the techniques in completing the square;comply with the techniques of completing the square; andcarefully change the exact signs for every equation. Some quadratics is fairly simple to solve because they are of the form "something-with-x squared equals some number", and then you take the square root of both sides. An example would be: (x 4)2 = 5 x 4 = sqrt(5) x = 4 sqrt(5) x = 4 sqrt(5) and x = 4 + sqrt(5)Unfortunately, most quadratics doesnt come neatly squared like this. For your average everyday quadratic, you first have to use the technique of "completing the square" to rearrange the quadratic into the neat "(squared part) equals (a number)" format demonstrated above. For example:Find the x-intercepts of y = 4x2 2x 5. First off, remember that finding the x-intercepts means setting y equal to zero and solving for the x-values, so this question is really asking you to "Solve 4x2 2x 5 = 0". contentsbacknext 50. The answer can also be written in rounded form ascontents back next 51. You will need rounded form for "real life" answers to word problems, and for graphing. But (warning!) in most other cases, you should assume that the answer should be in "exact" form, complete with all the square roots.When you complete the square, make sure that you are careful with the sign on the x-term when you multiply by one-half. If you lose that sign, you can get the wrong answer in the end, because you'll forget what goes inside the parentheses. Also, don't be sloppy and wait to do the plus/minus sign until the very end. On your tests, you won't have the answers in the back, and you will likely forget to put the plus/minus into the answer. Besides, there's no reason to go ticking off your instructor by doing something wrong when it's so simple to do it right. On the same note, make sure you draw in the square root sign, as necessary, when you square root both sides. Don't wait until the answer in the back of the book "reminds" you that you "meant" to put the square root symbol in there. If you get in the habit of being sloppy, you'll only hurt yourself! Solve x2 + 6x 7 = 0 by completing the square. Do the same procedure as above, in exactly the same order. (Study tip: Always working these problems in exactly the same way will help you remember the steps when you're taking your tests.)contents backnext 52. If you are not consistent with remembering to put your plus/minus in as soon as you square-root both sides, then this is an example of the type of exercise where you'll get yourself in trouble. You'll write your answer as "x = 3 + 4 = 1", and have no idea how they got "x = 7", because you won't have a square root symbol "reminding" you that you "meant" to put the plus/minus in. That is, if you're sloppy, these easier problems will embarrass you!Exercise: 1. 3x2 4x 6 = 02. 2x2 -3x + 4 = 03. x2 8x + 16 = 04. x2 + 18x + 72 = 05. 2x2 6x + 1 = 0 contentsbacknext 53. Name: ___________________ Section: _______ Instructor: ________________Date: _______Rating: ____ Instruction: Solve the following Quadratic Equation by Completing the Square.1. x2 + 3x = 4_____________________________________________________2. x2 2x = 24 _____________________________________________________3. x2 + 4 = 4x_____________________________________________________4. 2x2 6 = x_____________________________________________________5. 4a2 + 12a + 9 = 0 _____________________________________________________6. 3a2 5 = 14a _____________________________________________________7. 16b2 + 1 = 16b _____________________________________________________ contents backnext 54. 8. 9b2 6b 1 = 0 _____________________________________________________9. 9z2 + 30z + 20 = 0 _____________________________________________________10. 2a2 + a = 10a _____________________________________________________11. 2x2 + 17 = 10x _____________________________________________________12. 2a2 + 6a + 9 = 0 _____________________________________________________13. 5x2 2x + 1 = 0 _____________________________________________________14. 3x2 + 2x + 1= 0 _____________________________________________________15. 2y2 + 5y = 42 _____________________________________________________contents back next 55. Lesson 8Quadratic FormulaOBJECTIVES:At the end of this lesson, students are expected to:follow the step in solving quadratic formula; distinguish the roots of the quadratic equation; and perform substituting the values in the quadratic formula. The following steps will serve as guide in solving this method.Step 1. First subtract c from both sides of the equation and then, divide both sides by (a # 0 by hypothesis) to obtain the equivalent equation, x2 + =Step 2. Complete the left-hand side in to the perfect square. x2 + bx/a + (b/2a)2 = (b/2a)2 c/aor (x+b/2a)2 = (b2-4ac)/4a2Step 3. Take the square roots of both sides of the last equation. (x+b/2a) =( /2a contentsback next 56. Step 4. Solve for x.x == orLet a, b and c be real constant, where a 0. Then the roots of ax2 + bx + c = 0 are= x= The above formula is referred to as the quadratic formula.Example: Solve a. 3x2 x 5/2 = 0 Solutions: Here a=3, b=1, c=5/2Substituting these values in the quadratic formulawe obtain x == = The roots areand.a. 2x2 5 (x-2) = 8 To be able to apply the formula, we must first put the given equation in standard form. 2x2 5 (x-2) = 82x2 5x + 10 = 82x2 5x + 2 = 0 contents back next 57. Here a=2, b=5 c=2. By the quadratic formula. x== The roots are 2 and.Note that the expression 2x2 5x + 2 can be factored as2x2 5x + 2 = (2x 1) (x 2) The roots of the quadratic equation x = 1/2 and x = 2. This example shown that if we can see that the given equation in factorable, it will be quicker to solve it by factoring.Exercises: Solve each equation by quadratic formula.1. x2 14x + 49 = 02. x2 4x 21 = 03. x2 + 5x 36 = 04. x2 + x 30 = 05. x2 + 3x = 40contentsbacknext 58. Name: ___________________Section: _______ Instructor: ________________ Date: ________Rating: ______ Instruction: Solve the following equations by the Quadratic Formula.1. 2a2 10 = 9 _____________________________________________________2. 6b2 b = 12 _____________________________________________________3. 3x2 + x = 14 _____________________________________________________4. 10a2 + 3 = 11a _____________________________________________________5. 2x2 + 5x = 12 _____________________________________________________6. 4x2 + 5x = 21 _____________________________________________________contents backnext 59. 7. 2x2 7x + 3 = 0 _____________________________________________________8. 3a2 6a + 2 = 0 _____________________________________________________9. 3b2 2b 4 = 0 _____________________________________________________10 a2 3a 40 = 0 _____________________________________________________11. 3y2 11y + 10 = 0 _____________________________________________________12. 3w2 = 9 + 2w_____________________________________________________13. 15z2 + 22z = 48_____________________________________________________14. 9a2 + 14 = 24a _____________________________________________________15. 16m2 = 24m + 19_____________________________________________________ contentsback next 60. Lesson 9Solving "by Graphing OBJECTIVES:At the end of this lesson, students are expected to:define graphing; resolve the equation by graphing; and draw the points from the equations given. To be honest, solving "by graphing" is an achingly trendy but somewhat bogus topic. The basic idea behind solving by graphing is that, since the "solutions" to "ax2 + bx + c = 0" are the x-intercepts of "y = ax2 + bx + c", you can look at the x-intercepts of the graph to find the solutions to the equation. There are difficulties with "solving" this way, though....When you graph a straight line like "y = 2x + 3", you can find the x-intercept (to a certain degree of accuracy) by drawing a really neat axis system, plotting a couple points, grabbing your ruler and drawing a nice straight line, and reading the (approximate) answer from the graph with a fair degree of confidence.On the other hand, a quadratic graphs as a wiggly parabola. If you plot a few non-x-intercept points and then draw a curvy line through them, how do you know if you got the x-intercepts even close to being correct? You don't. The only way you can be sure of your x-intercepts is to set the quadratic equal to zero and solve. But the whole point of this topic is that they don't want you to do the (exact) algebraic solving; they want you to guess from the pretty pictures. So "solving by graphing" tends to be neither "solving" nor "graphing". That is, you don't actually graph anything, and you don't actually do any of the "solving". Instead, you are told to punch some buttons on your graphing calculator and look at the pretty picture, and then you're told which other buttons to hit so the software can compute the contents back next 61. intercepts (or you're told to guess from the pretty picture in the book, hoping that the printer lined up the different print runs for the different ink colors exactly right). I think the educators are trying to "help" you "discover" the connection between x-intercepts and solutions, but the concept tends to get lost in all the button-pushing. Okay, enough of my ranting... To "solve" by graphing, the book may give you a very neat graph, probably with at least a few points labeled; the book will ask you to state the points on the graph that represent solutions. Otherwise, it will give you a quadratic, and you will be using your graphing calculator to find the answer. Since different calculator models have different key-sequences, I cannot give instruction on how to "use technology" to find the answers, so I will only give a couple examples of how to solve from a picture that is given to you. Solve x2 8x + 15 = 0 by using the following graph. The graph is of the related quadratic, y = x2 8x + 15, with the x-intercepts being where y = 0. The point here is to look at the picture (hoping that the points really do cross at whole numbers, as it appears), and read the x- intercepts (and hence the solutions) from the picture. The solution is x = 3, Since x2 8x + 15 factors as (x 3)(x 5), we know that our answer is correct.contents backnext 62. Solve 0.3x2 0.5x 5/3 = 0 by using the following graph. For this picture, they labeled a bunch of points. Partly, this was to be helpful, because the x-intercepts aremessy (so I could not have guessed their values without the labels), but mostly this was in hopes of confusing me,in case I had forgotten that only the x-intercepts, not the vertices or y-intercepts, correspond to "solutions". The x-values of the two points where the graph crosses the x-axis are the solutions to the equation. The solution is x = 5/3, 10/3 Find the solutions to the following quadratic:contentsback 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.contentsbacknext 64. Name: ___________________ Section: _______ Instructor: ________________Date: _______Rating: ______ Instruction: Solve each equation by graphing.1. x2 6x + 9 = 0 _____________________________________________________2. x2 5x + 10 = 0_____________________________________________________3. 2x2 6x + 8 = 0_____________________________________________________4. x2 7x + 12 = 0_____________________________________________________5. 2x2 8x + 10 = 0 _____________________________________________________6. 3x2 + 6x 9 = 0 _____________________________________________________7. x2+ 8x 12 = 0 _____________________________________________________contents backnext 65. 8. 2 + 4x 3 = 0 _____________________________________________________9. x2 2x 2 = 0 _____________________________________________________10. 2x2 4x 2 = 0_____________________________________________________11. 4x2 8x 16 = 0_____________________________________________________12. x2 9x + 21 = 0_____________________________________________________13. x2 + 10x + 18 = 0_____________________________________________________14. 2x2 16x + 8 = 0_____________________________________________________15. 3x2 12x 9 = 0_____________________________________________________ contentsback next 66. A. Solve by factoring.1. x2 3x 10 = 02. x2 + 2x = 83. x2 x 4 = 24. 2x2 6x 36 = x2 155. 4x2 + 4x = 156. 6x2 + 11x 2 = 87. 49x2 + 28x 10 = 08. 6x4 4x3 10x2 = 09. 18 + 15x 18x2 = 010. x4 4x2 + 3 = 0B. Solve by completing the square.1. x2 - 4x 3 = 02. x2 + 3x 6 = 03. x2 7x + 5 = 04. 2x2 + 5x + 1 = 05. 2x2 + 8x 5 = 0C. Solve for x by the quadratic formula.1. x2.- 4x 7 = 02. x2 3x + 4 = 03. 2x2 + 4x + 5 = 04. x2 + 7x 3 = 05. x2 7x + 2 = 06. x2 + 5x 7 = 07. x2 + 9x 3 = 08. 4x2 6x + 2 = 09. 9x2 9x 10 = 010. x2 + 5x + 8 = 0 contents back next 67. Chapter IV Much of the study in quadratic equation consist of different solving equation, we have equation inquadratic form, equation containing radicals and equation reducible to quadratic equation. They have their ownsteps and procedures to be followed in order to solve the given equation.TARGET SKILLS: At the end of this chapter, students are expected to: discuss solving equation on quadratic; determine the index and its radicals; interpret the solution of the original equation; and select appropriate method in solving quadratic equation. contentsback next 68. Lesson 10Equation in Quadratic FormOBJECTIVES: At the end of this lesson, students are expected to:identify equation in quadratic form; select appropriate method in solving quadratic equation; and change the equation in standard form.Quadratic in FormAn equation is quadratic in form when it can be written in this standard formwhere the same expression is inside both ( )'s.In other words, if you have a times the square of the expression following b plus b times that same expression not squared plus c equal to 0, you have an equation that is quadratic in form. If we substitute what is in the ( ) with a variable like t, then the original equation will become a quadratic equation. contentsback next 69. Solving Equations that are Quadratic in Form Step 1: Write in Standard Form,, if needed. If it is not in standard form, move any term(s) to the appropriate side by using theaddition/subtraction property of equality. Also, make sure that the squared term is written first left to right, the expression not squared issecond and the constant is third and it is set equal to 0.Step 2: Substitute a variable in for the expression that follows b in the second term. In other words, substitute your variable for what is in the ( ) when it is in standard form, . Im going to use t for my substitution, but really you can use any variable as long as it is not thevariable that is used in the original equation.Step 3: Solve the quadratic equation created in step 2. You can use any method you want to solve the quadratic equation: factoring, completing the square orquadratic formula. Step 4: Find the value of the variable from the original equation.Keep in mind that you are finding a solution to the original equation and that the variable yousubstituted 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 backnext 70. . Step 5: Check your solutions. 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. Example 1: Solve the equation that is quadratic in form: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 contentsbacknext 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 contents back next 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 solutionLet'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 contentsback next 73. . Example 2: Solve the equation that is quadratic in form:Standard Form, *Inverse of add. 3 is sub. 3 *Equation in standard formNote 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 inform*Note that (x to the 1/3 power) squared = x to the 2/3power *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.contentsbacknext 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 solvecontents back next 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 powerLet'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 statementcontentsback next 76. Since we got a true statement, x = 27 is a solution. Let's double check to see if x = -1 is a solution to the original equation. *Plugging in -1 for x*True statementSince we got a true statement, x = -1 is a solution. There are two solutions to this equation: x = 27 and x = -1. Exercises:1.a4 + 2a2 5 = 02.x2 3x + 2 = 03.s6 + 8s3 6 = 04.n2 6n + 10 = 05.g8 + 2g4 g = 0 contents back next 77. Name: ___________________Section: _______ Instructor: ________________ Date: _______Rating: ______ Instruction: Solve the equation that is in quadratic form. 1. a8 + 2a4 8 = 0 _____________________________________________________2. l2 + 4l2 6 = 0_____________________________________________________3. e4 8e2 3 = 0 _____________________________________________________4. l6 10l 5 = 0_____________________________________________________5. i10 8i5 4 = 0 _____________________________________________________6. s6 5s3 25 = 0 _____________________________________________________ contents back next 78. 7. h2/4 + 8h1/4 12 = 0_____________________________________________________8. a6- 5a4 15 = 0_____________________________________________________9. n8 + 12n2 8 = 0_____________________________________________________10. e9 3n3 10 = 0_____________________________________________________11. x2/3 2x 1/3 = 8_____________________________________________________12. x3/6 3x1/2 = 9_____________________________________________________13. y2- 8y = 5_____________________________________________________14. y4 + 2y2 = 6_____________________________________________________15. x6 9x2 + 8 = 0 _____________________________________________________contentsback next 79. Lesson 11Equation Containing Radicals OBJECTIVES:At the end of this lesson, students are expected to:determine the index and its radicals; positively respond to the note to be remembered; and perform isolation of one radical if there are two radicals in the equation.In the radicals which is read the nth root of b, the positive integer n is called the index or order of the radical, and b is called its radicand. When n is 2, 2 is no longer written, just simply writeinstead ofto indicate the square root of b, thusis read as cube root of b; as 4 th of b.Note: In order to solve for x, you must isolate x. In order to isolate x, you must remove it from under the radial. If there are two radicals in the equation, isolate one of the radicals. Then raise both sides of the equation to a power equal to the index of the isolated radical. Isolate the the remaining radical. Raise both sides of the equation to a power equal to the index of the isolated radical. You should now have a polynomial equation. Solve it. Remember that you did not start out with a polynomial; therefore, there may be extraneous solutions. Therefore, you must check your answers.contents backnext 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 ifIsolate the termSquare 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. contentsbacknext 81. Left side: Right Side:1 Since the left side of the original equation does not equal the right side of the original equation after we substituted our solution for x, then there is no solution.You can also check the answer by graphing the equation: .The graph represents the right side of the original equation minus the left side of the original equation.. The x- =5 intercept(s) of this graph is (are) the solution(s). Since there are no x-intercepts, there are no solutions.Exercises:Solve each of the following equation.1.=32.=x+13.=4.=55.= 10 contentsback next 82. Name: ___________________ Section: _______ Instructor: ________________Date: _______ Rating: ______Instruction: Solve each of the following equation.1. _____________________________________________________ 2. _____________________________________________________ 3. _____________________________________________________ 4. _____________________________________________________ 5. _____________________________________________________ 6. _____________________________________________________ contents back next 83. 7.____________________________________________________ 8.____________________________________________________ 9.____________________________________________________ 10. ____________________________________________________ 11. ____________________________________________________12. ____________________________________________________13. ____________________________________________________14. ____________________________________________________15. ____________________________________________________contents back next 84. Lesson 12Equations Reducible to Quadratic Equations OBJECTIVES: At the end of this lesson, students are expected to: interpret the solution of the original equation;organize the equation if it is quadratic equation; andsolve the equation by factoring or quadratic formula.A variety of equations can be transformed into quadratic equations and solved by methods that we have discussed in the previous section. We will consider fractional equations, equations involving radicals and equation that can be transformed into quadratic equations by appropriate substitutions. Since the transformation process may introduce extraneous roots which are not solutions of the original equation, we must always check the solution in the original equation.Example: Solve 1 17x+2 + x+3 = 12 Solution: First note that neither -2 nor -3 can be a solution since at either of these points the equation is meaningless.Multiplying by the LCD, 12(x+2) (x+3), we get12(x+3) + 12(x+2) = 7(x+2) (x+3) 24x + 60 = 7(x2 + 5x + 6) or 7x2 + 11x 18 = 0Factoring, we get, (7x + 18)(x 1) = 0 contents backnext 85. x = 1 or -18 7If x = 1, _1_ _1_ _1_ _1_ _7_ 1+2 1+3 = 3 + 4 = 12Therefore 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_4312Therefore, x = _-18_ is a solution.7Example 2. =-2Solution: squaring both sides of the equation, we obtain,= +4 2x 16 = 4 contents back next 86. Dividing both sides by 2 gives, x 8 = -2Squaring both sides of the equation we get 2(x - 8)= x2 - 16x + 64 = 4 (x +16) x2 20x = 0 x(x 20) =0 x = 0 or x = 20 Check: if x = 0,= = 8=6284 Therefore x = 20 is not a solution of the original equation.Thus the only root ofMany equations are not quadratics equations. However, we can transform them by means of appropriatesubstitutions into quadratics equations and then solve these by techniques that we know.contentsback next 87. Example: Solve: 1. 2x-2 7x-1 + 3 = 0 2. x4 2x2 2 = 0 3. 2 Solutionsa. Let u = x-1. Then u2 = (x-1)2 = x-2 and our equation becomes2u2 7u + 3 = 0, a quadratic equation in u.To solve the equation, we factor the left-hand side.(2u 1)( u 3) = 0 U = or u = 3 Since u = x-1, x-1=or x-1 = 3, from which x = 2 or x =Check: if x = 2, 2(2-2) 7(2-1) + 3 =Thus x = 2 is solutionIf x = 1/3, 2(1/3)-2 7(1/3)-1 + 3 = 2(3)2 7(3) + 3So, x = 1/3 is a solution.b. Let u = x2. Then u2 = x4 and the given equation becomes a quadratic equation in u. u2 2u 2 = 0contentsback next 88. u== . u=1+ or u =Since u = x2 and u = < 0, we have to discard this solution. . u = x2 =implies x=It is simple to verify that both values of x satisfy the original equation. The roots of x4 x2 2 = 0 areand - c. Let u =.This substitution yields a quadratic equation in u.u2 u 2 = 0(u 2)(u + 1) = 0u = 2 0r u = 1 u==2 implies x = 2(4x + 1) or x=u==1 implies x = 4x 1orx= Again, it can easily be verified that both solutions check in the original equation. The roots are andcontentsbacknext 89. Name: ___________________Section: _______ Instructor: ________________ Date: _______ Rating: ______Instruction: Solve the following equation. 1. _____________________________________________________ 2. _____________________________________________________3. _____________________________________________________ 4. _____________________________________________________ 5._____________________________________________________ 6._____________________________________________________ contents back next 90. 7. _____________________________________________________8. _____________________________________________________9. _____________________________________________________ 10. _____________________________________________________ 11. _____________________________________________________12. _____________________________________________________ 13._____________________________________________________ 14._____________________________________________________15._____________________________________________________ contents back next 91. A. Solve for x.C. Solve for x. 1. + = 1.=42. -5=02. + + =0 3.=3. + = 4.=5. =x+24. + =2B. Reduce to quadratic equation.5. + = 1. x4 5x + 4 = 0 2. 4(x + 3) + 5 = 21 3. x2/3 5x1/3 6 = 0 4. (x2 + 4x)2 (x2 + 4x) = 20 5. 2x4 9x2 + 7 = 0 contents back next 92. Chapter VThe discriminant gives additional information on the nature of the roots beyond simply whether thereare any repeated roots: it also gives information on whether the roots are real or complex, and rational orirrational. More formally, it gives information on whether the roots are in the field over which the polynomial isdefined, or are in an extension field, and hence whether the polynomial factors over the field of coefficients.This is most transparent and easily stated for quadratic and cubic polynomials; for polynomials of degree 4 orhigher this is more difficult to state. TARGET SKILLS:At the end of this chapter, students are expected to: determine discriminant, roots and coefficient; discuss the relation the roots and coefficient; find the sum and product of the roots; and change quadratic equation to discriminant formula.contentsbacknext 93. Lesson 13The Discriminant and the roots of aQuadratic Equation OBJECTIVES: At the end of this lesson, students are expected to:determine discriminant and the roots; compare discriminant and the nature of the roots; and change quadratic equation to discriminant using the nature of the roots. Example1. Find the x-intercept of y = 3x - 6x + 4.Solution: As already mentioned, the values of x for which 3x - 6x + 4 = 0 give the x-intercepts of the function. We apply the quadratic formula in solving the equation.3x - 6x + 4 = 0 x= = Since is not a real number, the equation 3x - 6x + 4 = 0 has no real root. This means that theparabola y = 3x - 6x + 4 does not intersect the x-axis. Let us write the equation in the form y = a(x h)+ k. y = 3(x 2x) + 4= 3(x 1) + 1 contents backnext 94. = b - 4acRoots of ax + bx + c = 0 Positive Real and distinctr=s= Zero Real and equalr=s= Negative No real rootsExample 2. Use the disciminant to determine the nature of the roots of thefollowing quadratic equation.a. x - x + = 0a = 1, b = 1, c = b - 4ac = (1) - 4 (1)()=11=0 There is only one solution, that is, a double root. Note that x - x = = (x - ), so that double root is b/2a = . b. 5x - 4x + 1 = 0a = 5, b = 4, c = 1b - 4ac = (4) - 4 (5)(1)= 16 20= 4 < 0 There are no real roots since a negative number has no real square root.contents back next 95. Name: ___________________ Section: _______ Instructor: ________________Date: _______ Rating: ______ Instruction: Use the Discriminant to determine the nature of the root of the following Quadratic Equations.1.x2- 2x 3=0 _____________________________________________________2. 6x2 x 1 = 0 _____________________________________________________3. 2x2 50 = 0 _____________________________________________________4. x2 8x + 12 = 0_____________________________________________________5. x2 + 5x 14 = 0 _____________________________________________________6. -4x2 4x + 1 = 0 _____________________________________________________7. 7x2 + 2x 1 = 0 _____________________________________________________contents backnext 96. 8.x2 + 3x = 40_____________________________________________________9. 3x2= 5x 1_____________________________________________________10. 3x2+ 12 1=0_____________________________________________________11. (x-2)(x-3) = 4_____________________________________________________12.2x2 + 2x + 1 = 0_____________________________________________________13. 7x2 + 3 6x = 0_____________________________________________________14. 5x2 6x + 4 = 0_____________________________________________________15. 3x2 + 2x + 2 = 0_____________________________________________________ contents back next 97. Lesson 14 Relation between roots and coefficient OBJECTIVES: At the end of this lesson, students are expected to: classify roots and coefficient;discuss relations between the roots and coefficient of the quadratic equation; andfind the sum and product of the roots of a given quadratic equation There are some interesting relations between the sum and the product of the roots of a quadratic equation. To discover these, consider the quadratic equation ax2 + bx + c = 0, where a 0. Multiply both sides of this equation by 1/a so that the coefficient of x2 is 1. (ax2 + bx + c) = We obtain an equivalent quadratic equation in the form x2 + + =0If r and s are the roots of the quadratic equation ax2 + bx + c = 0, then from the quadratic formular= and s =contents backnext 98. Adding the roots, we obtainr+s= + ==Multiplying the roots, we obtainrs = = c/a Observe the coefficient in the quadratic equation x2 + 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 / a2. The product of the roots is equal to the constant term rs = c / aAn alternate way of arriving at these relations is as followsLet r and s be the roots of x2 + bx/a + c/a = 0. Thenx - r)(x s) = 0Expanding gives, x2 rx sx + rs = 0or x2 (r + s)x + rs = 0 contents back next 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. 2x2 + x 6 = 0Solutions: 2x2 + x 6 = (2x 3)(x + 2) = 0x = 3/2 or x = 2The 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 3x2 6x + 8 = 0 without having to first determine the roots.Solution: The sum of the roots is r + s = -c/a = -(-6)/3 = 2and their product is rs = c/a = 8/3contentsbacknext 100. Name: ___________________Section: _______ Instructor: ________________ Date: _______Rating: ______ Instruction: Without solving the roots, find the sum and product of the roots of the following.1. 6x2 5x + 2 = 0 _____________________________________________________2. x2 + x 182 = 0 _____________________________________________________3. x2 5x 14 = 0 _____________________________________________________4. 2x2 9x + 8 = 0 _____________________________________________________5. 3x2 - 5x 2 = 0 _____________________________________________________6. x2 8x 9 = 0 _____________________________________________________contentsbacknext 101. 7. 2x2 3x 9 = 0 _____________________________________________________8. x2 + x 2 - _____________________________________________________9. 3x2 + 2x 8 = 0 _____________________________________________________10. 16x2 24x + = 0 _____________________________________________________11. x2 6x + 25 = 0_____________________________________________________12. 3x2 + x 2 = 0_____________________________________________________13. 5x2 + 11x 8 = 0_____________________________________________________14. x2 8x + 16 = 0_____________________________________________________15. 4x2 16x + 10 = 0_____________________________________________________ contents back next 102. A. Use the discriminant to determine which of the following quadratic equations have two, one or no realroots. Give reasons. 1. x2 5x 5 = 02. x2 3x 2 = 3x 113. 5x2 9x = 2x 74. 2x2 7x + 8 = 05. 3 4x 2x2 = 06. 4x2 9x + 5 = 3x 7B. By using the relations between roots and coefficients, determine if the given #s are roots of thecorresponding given equation. 1. 6x2 5x + 3 = 0(1/6, -1)2. x2 + x _ 182 = 0(13, -4)3. x2 5x 14 =0 (2, -7)4. 2x 9x + 8 = 0 (3, 4/3)5. 3x2 5x 2 = 0(2, -1/3)C. Given one roots of the equation, find the other. 1. x2 8x 9 = 0;r= 12. 2x2 3x 9 = 0; r= 33. x2 + x 2 - =0 r= contentsback next 103. Chapter VI In this chapter, students will be taught how to find solutions to quadratic equations. This lesson assumes students are already familiar with solving simple quadratic equations by hand, and that they have become relatively comfortable using their graphing calculator for solving arithmetic problems and simple algebra problems. Students will also be shown strategies on how to use the keys on the graphing calculator to show a complete graph. TARGET SKILLS: At the end of this chapter, students are expected to: use calculator in solving quadratic equation; solve equation on a calculator; and. improve skills using calculator by solving quadratic equation. contents back next 104. Lesson 15 Equation on a Calculator OBJECTIVES:At the end of this lesson, students are expected to: acquire knowledge using calculator in solving quadratic equation; resolve equations on a calculator; and improve skills on solving quadratic equation using a calculator.The simplest way to solve a quadratic equation on a calculator is to use the quadratic formula. x=-b where d=b - 4ac2a As we have seen, if d < 0, there are no real solutions. But f d 0, then we can use calculator to get the solutions.To solve 3x + 5x 7 = 0, first compute the discriminant. d = b - 4ac = 5 - 4(3)(-7).On an arithmetic calculator, the keystroke sequence for d is,[AC][MC] 4 [x] 3 [x] 7 [=][M+] 5 [x][=][+][MR][=]The display will show the value of the discriminant to be 109, and so the quadratic equation has two distinct roots. To compute the roots, proceed as follows:[AC][MC] 109 [][M+][MR][-] 5 [+] 2 [x] 3 [=] the first root andcontentsbacknext 105. 0 [-][MR][-] 5 [+] 2 [ x] 3 [=] the second root. On an algebraic calculator, the keystroke sequence is easier. Recall that the actual roots are: x= -5 + 2(3) x= -5 - 2(3)First, we compute the square root of the discriminant and store it is memory.[AC] 5 [2] 4 [x] 3 [x] 7 [+/-][=][][Min]Then, compute the first root as:x1 = (5 [+/-][+][MR])[ ] (2 [x] 3) [=];and then compute the second root as:x2 = (5[+/-][-][MR])[+] (2 [x] 3)[=].Exercise: Find the roots of the following equations. 1. 3.5x2 + 1.2x 3.2 = 02. 7.6 -2.2x 1.7x2 = 03. 2.5x2 + 5.6x 13.5 = 04. x 77.3 + 2.3x2 = 05. x2 - 1000.5 + 32.3 = 0contents back next 106. Name: ___________________ Section: _______ Instructor: ________________Date: _______Rating: ______ Instruction: Find the roots of the following equations.1. 5.3x2 + 2.1v 2.3 = 0_____________________________________________________2. 6.7 v - 2.2x 7.1x2 = 0 _____________________________________________________3. 5.2x2 + 6.5x 5.13 = 0_____________________________________________________4. x2 50.001 + 33.2 = 0_____________________________________________________5. x 7.73 + 2.3x2 = 0_____________________________________________________6. 3.3x2 1.9x 7.10 = 0_____________________________________________________contents back next 107. 7. 3.1x 9.1x2 7.10 = 0 _____________________________________________________8. 6.3x2+ 8.5x = 9.5 _____________________________________________________9. 5.9x 9.5x2 = 8.03_____________________________________________________10. 3.2x2 + 2.3x = 23.32 _____________________________________________________11. 9.9x 7.7x2 8.8 = 0 _____________________________________________________12. 6.3x + 5.3x2 3.4 = 0_____________________________________________________13. 6.3x2 2.9x 8.10 = 0_____________________________________________________14. 3.4x 8.1x2 4.10 = 0_____________________________________________________15. 6.2x2 + 3.6v 3.7 = 0_____________________________________________________ contentsback next 108. 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 Erdlyi, 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_I0chttp://www.youtube.com/watch?v=ug7IDDBRO94contentsbacknext 109. contents back