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- 1. nextcontents
- 2. next 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. VISIO N backcontents
- 3. back next 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. MISSIO N contents
- 4. back next 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 contents
- 5. back next 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. 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 back nextcontents
- 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 back nextcontents
- 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 back nextcontents
- 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. back nextcontents
- 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; and •use calculator in solving quadratic equation. back nextcontents
- 11. 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 (aleli) Demo (shane) Slide share (aleli) Slide share (shane) contents back next
- 13. Chapter I This chapter deals with equations which are classified according to the highest power of its variable. An equation in the variable x whose highest power is 2 is called a quadratic equation. It will be observed here that variable 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. 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 a quadratic 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 form x2 = 8x can be written as x2 - 8x = 0 where a=1, b= ˉ8, and c=0. Example B. Express x2 = 64 in standard form nextbackcontents
- 15. x2 = 64 can be written as x2 – 64 = 0 where a=1. B=0, and c=ˉ64. Example C. Express the fractional equation x = 1/x-3 as a quadratic equation. x = 1/x-3 x (x-3) = 1 multiply both sides by x-3 x² - 3x = 1 using the distributive property x² - 3x - 1 = 0 a=1, b=ˉ3, c=ˉ1 Exercises: Which of the following equations are quadratic? 1.3x = x² - 5 2. 2x =1 3. x² = 25 4. 2x - 3 = x + 5 5. 5x – 2y = 0 nextbackcontents
- 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 _____________________________________________ nextbackcontents
- 17. _____________________________________________ 8x = x2 _____________________________________________ 7. 8. 9. 10. 11. 12. 13. 14. 15. _____________________________________________ 2 = 6 _____________________________________________ x2 = x2 + _____________________________________________ (x + 1)(x-3) = 6 _____________________________________________ _____________________________________________ _____________________________________________ _____________________________________________ contents back next
- 18. A. Define each of the following terms. 1. Quadratic equation 2. Standard form of a quadratic equation 3. Real numbers B. Which of the following equations are quadratic? 1. 4x = 2x2 – 6 2. 3x = 1 3. 5x2 = 30 4. 3x – 2 = 2x + 6 5. 2x – 5y = 0 6. 4x + 2x2 – 3x3 = 0 7. 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 = 6x 2. 3x2 = 32 3. 2x2 = 5x – 1 4. 12 = 4x – x2 5. (x + 3)2 = 8 6. 4x2 = 56 7. 1/x + x = 6 8. x(x – 4) – 1 = 0 9. 9x = x2 10.(1/x)2 = 10 contents back next
- 19. Chapter II This chapter centers on the complex numbers which is a number comprising a real number and an imaginary number. Under this, we have the number i, the complex plane where the points are plotted and the 4 arithmetic operations such as addition and subtraction, multiplication and division of complex numbers. To round up the chapter, 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. Less on 2Defining Complex NumbersOBJECTIVES: 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 finding the x-intercepts of a graph; and, the graph of y = x2 - 1 crosses the x-axis at (-1,0) and (1,0). nextbackcontents
- 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 complex numbers, equation 2 will have two solutions. nextbackcontents
- 22. Name: ___________________ Section: _______ Instructor: ________________ Date: ________ Rating: _______ Solve each equation and graph. 1. x² + 4 = 0 _____________________________________________ 2. 2x² + 18 = 0 _____________________________________________ 3. 2x² + 14 = 0 _____________________________________________ 4. 3x² + 27 = 0 _____________________________________________ 5. x² - 3 = 0 _____________________________________________ 6. x² + 21 = 0 _____________________________________________ nextbackcontents
- 23. 7. 3x² - 5 = 0 _____________________________________________ 8. 5x² + 30 = 0 _____________________________________________ 9. 2x² + 3 = 0 _____________________________________________ 10. x² + 50 = 0 _____________________________________________ 11. x² - 2 = 0 _____________________________________________ 12. 3x² - 50 = 0 _____________________________________________ 13. x² - 3 = 0 _____________________________________________ 14. x² + 4 = 0 _____________________________________________ 15. 2x² + 14 = 0 _____________________________________________ contents back next
- 24. Less on 3The Number i OBJECTIVES: At the end of this lesson, students are expected to: •recognize the property of the number i; •discuss the powers of i; and •solve the high powers of imaginary unit. Equation 1 Equation 2 x2 - 1 = 0. x2 + 1 = 0. x2 = 1. x2 = -1. Consider Equations 1 and 2 again. Equation 1 has solutions because the number 1 has two square roots, 1 and -1. Equation 2 has no solutions because -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. Example i3 = 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. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ nextbackcontents
- 27. 6. 7. 8. 9. 10. 11. 12. 13. ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ ______________________________________________________ nextbackcontents
- 28. Lesson 4 The 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. nextbackcontents
- 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 + 2i 2. 1 – 4i 3. 4 + 3i 4. 2 – 5i 5. 4 – 3i contents back next
- 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 back next
- 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 Subtraction To 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. nextbackcontents
- 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*i2 = 8 + 14i + 12i + 21*(-1) = (8 - 21) + (14 + 12)i = -13 + 26i. nextbackcontents
- 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. 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) = (a2 + b2 ) + 0i = a2 + b2 . Example Number Conjugate Product 2 + 3i 2 - 3i 4 + 9 = 13 3 - 5i 3 + 5i 9 + 25 = 34 4i -4i 16 nextbackcontents
- 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 nextcontents
- 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 nextcontents
- 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. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ back nextcontents
- 38. 8. 9. 10. 11. 12. 13. 14. 15. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ nextbackcontents
- 39. A. Define and/or describe each of the following terms. • Imaginary part • Real number • Complex number • Complex plane • Imaginary unit • Commutative property • Complex conjugate B. 1. Simplify: 1. i15 2. i25 3. i106 4. i207 5. i21 2. Perform the indicated operation and express each answer. a. + b. + c. + d. + e. + f. + nextbackcontents
- 40. 3. Represent each complex numbers by a point in the plane. a.3 – i b.-2 + 4i c.-3 + 3i d.4 + 5i e.-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) nextbackcontents
- 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. back nextcontents
- 42. Lesson 6 Solving 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 to obtain 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 = 0 x = 3 or x = 4 Solve: 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) ?=? 0 0 = 0 back nextcontents
- 43. Checking x = 4 in (x – 3)(x – 4) = 0: ([4] – 3)([4] – 4) ?=? 0 (4 – 3)(4 – 4) ?=? 0 (1)(0) ?=? 0 0 = 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) = 0 Solve each factor: x + 2 = 0 or x + 3 = 0 x = –2 or x = – 3 The solution to x2 + 5x + 6 = 0 is x = –3, –2 Checking x = –3 and x = –2 in x2 + 5x + 6 = 0: [–3]2 + 5[–3] + 6 ?=? 0 9 – 15 + 6 ?=? 0 9 + 6 – 15 ?=? 0 back nextcontents
- 44. 15 – 15 ?=? 0 0 = 0 [–2]2 + 5[–2] + 6 ?=? 0 4 – 10 + 6 ?=? 0 4 + 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, 3 •Solve (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. nextbackcontents
- 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, 1 •Solve 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, –5 •Solve 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: nextbackcontents
- 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, 2 Note: This solution may also be formatted as "x = ± 2“ Exercises: Solve: •(x – 3)(x – 5) = 0. •x2 + 6x + 7 = 0. •x2 – 4 = 2x. •x2 – 6x = 0. •x2 – 8 = 0. nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 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; and •carefully 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 doesn’t 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". nextbackcontents
- 50. The answer can also be written in rounded form as nextbackcontents
- 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 back next
- 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 = 0 2. 2x2 -3x + 4 = 0 3. x2 – 8x + 16 = 0 4. x2 + 18x + 72 = 0 5. 2x2 – 6x + 1 = 0 contents back next
- 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 back next
- 54. 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 _____________________________________________________ 8. 9b2 – 6b – 1 = 0 _____________________________________________________ nextbackcontents
- 55. Lesson 8 Quadratic Formula OBJECTIVES: 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/a or (x+b/2a)2 = (b2 -4ac)/4a2 Step 3. Take the square roots of both sides of the last equation. (x+b/2a) = ± ( /2a nextbackcontents
- 56. Step 4. Solve for x. x = or Let 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=⁻5/2 Substituting these values in the quadratic formula we obtain x = = = = = The roots are and . 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) = 8 2x2 – 5x + 10 = 8 2x2 – 5x + 2 = 0 nextbackcontents
- 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 as 2x2 – 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 = 0 2. x2 – 4x – 21 = 0 3. x2 + 5x – 36 = 0 4. x2 + x – 30 = 0 5. x2 + 3x = 40 contents back next
- 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 back next
- 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 _____________________________________________________ contents back next
- 60. Lesson 9 Solving "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 back next
- 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 are messy (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: 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
- 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 back next
- 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 _____________________________________________________ contents back next
- 66. A. Solve by factoring. 1. x2 – 3x – 10 = 0 2. x2 + 2x = 8 3. x2 – x – 4 = 2 4. 2x2 – 6x – 36 = x2 – 15 5. 4x2 + 4x = 15 6. 6x2 + 11x – 2 = 8 7. 49x2 + 28x – 10 = 0 8. 6x4 – 4x3 – 10x2 = 0 9. 18 + 15x – 18x2 = 0 10. x4 – 4x2 + 3 = 0 B. Solve by completing the square. • x2 - 4x – 3 = 0 • x2 + 3x – 6 = 0 • x2 – 7x + 5 = 0 • 2x2 + 5x + 1 = 0 • 2x2 + 8x – 5 = 0 C. Solve for x by the quadratic formula. 1. x2 .- 4x – 7 = 0 2. x2 – 3x + 4 = 0 3. 2x2 + 4x + 5 = 0 4. x2 + 7x – 3 = 0 5. x2 – 7x + 2 = 0 6. x2 + 5x – 7 = 0 7. x2 + 9x – 3 = 0 8. 4x2 – 6x + 2 = 0 9. 9x2 – 9x – 10 = 0 10.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 in quadratic form, equation containing radicals and equation reducible to quadratic equation. They have their own steps 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. contents back next
- 68. Lesson 10 Equation in Quadratic Form OBJECTIVES: 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 Form An equation is quadratic in form when it can be written in this standard form where 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. contents back next
- 69. Solving Equations that are Quadratic in Form 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. Step 1: Write in Standard Form, , if needed. 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. Step 2: Substitute a variable in for the expression that follows b in the second term. You can use any method you want to solve the quadratic equation: factoring, completing the square or quadratic formula. Step 3: Solve the quadratic equation created in step 2. 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. Step 4: Find the value of the variable from the original equation. contents back next
- 70. 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. Step 5: Check your solutions. 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 nextbackcontents
- 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 nextbackcontents
- 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 nextbackcontents
- 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. nextbackcontents
- 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 nextbackcontents
- 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 nextbackcontents
- 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 statement Since 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 = 0 2. x2 – 3x + 2 = 0 3. s6 + 8s3 – 6 = 0 4. n2 – 6n + 10 = 0 5. g8 + 2g4 – g = 0 nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 79. Lesson 11 Equation 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 write instead of to indicate the square root of b, thus is read as “cube root of b”; as “4th 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. nextbackcontents
- 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. nextbackcontents
- 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- 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. = 3 2. = x + 1 3. = 4. = 5 5. = 10 = 5 contents back 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 12 Equations 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 ; and •solve 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 1 7 x+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 get 12(x+3) + 12(x+2) = 7(x+2) (x+3) 24x + 60 = 7(x2 + 5x + 6) or 7x2 + 11x – 18 = 0 Factoring, we get, (7x + 18)(x – 1) = 0 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 x2 - 16x + 64 = 4 (x +16) x2 – 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
- 87. Example: Solve: 1. 2x-2 – 7x-1 + 3 = 0 2. x4 – 2x2 – 2 = 0 3. 2 – Solutions a.Let u = x-1 . Then u2 = (x-1 )2 = x-2 and our equation becomes 2u2 – 7u + 3 = 0, a quadratic equation in u. To solve the equation, we factor the left-hand side. (2u – 1)( u – 3) = 0 U = ½ 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 solution If x = 1/3, 2(1/3)-2 – 7(1/3)-1 + 3 = 2(3)2 – 7(3) + 3 So, 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 = 0 contents back 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 are and - . c. Let u = .This substitution yields a quadratic equation in u. u2 – 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. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ contents back next
- 90. _____________________________________________________ 7. 8. 9. 10. 11. 12. 13. 14. 15. _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ _____________________________________________________ contents back next
- 91. A. Solve for x. 1. = 4 2. - 5 = 0 3. = 4. = 5. = x + 2 B. Reduce to quadratic equation. 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 C. Solve for x. 1. + = 2. + + = 0 3. + = 4. + = 2 5. + = nextbackcontents
- 92. Chapter V The discriminant gives additional information on the nature of the roots beyond simply whether there are any repeated roots: it also gives information on whether the roots are real or complex, and rational or irrational. More formally, it gives information on whether the roots are in the field over which the polynomial is defined, 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 or higher 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. nextbackcontents
- 93. Lesson 13 The Discriminant and the roots of a Quadratic 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. Example 1. Find the x-intercept of y = 3x² - 6x + 4. Solution: As already mentioned, the values of x for which 3x² - 6x + 4 = 0 give the x-intercepts of the function. We apply the quadratic formula in solving the equation. 3x² - 6x + 4 = 0 x = = Since is not a real number, the equation 3x² - 6x + 4 = 0 has no real root. This means that the parabola y = 3x² - 6x + 4 does not intersect the x-axis. Let us write the equation in the form y = a(x – h)²+ k. y = 3(x² – 2x)² + 4 = 3(x – 1)² + 1 nextbackcontents
- 94. ∆ = b² - 4ac Roots of ax² + bx + c = 0 Positive Real and distinct r = s = Zero Real and equal r = s = Negative No real roots 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. nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 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; and •find 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 + + = 0 If r and s are the roots of the quadratic equation ax2 + bx + c = 0, then from the quadratic formula r = and s = nextbackcontents
- 98. Adding the roots, we obtain r + s = + = = Multiplying the roots, we obtain rs = = 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 / 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 x2 + bx/a + c/a = 0. Then x - r)(x – s) = 0 Expanding gives, x2 – rx – sx + rs = 0 or x2 – (r + s)x + rs = 0 nextbackcontents
- 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 = 0 Solutions: 2x2 + 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 3x2 – 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 nextbackcontents
- 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 _____________________________________________________ nextbackcontents
- 101. 7. 2x2 – 3x – 9 = 0 _____________________________________________________ 8. x2 + x – 2 - _____________________________________________________ 9. 3x2 + 2x – 8 = 0 _____________________________________________________ 10. 16x2 – 24x + ½ = 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 real roots. Give reasons. 1. x2 – 5x – 5 = 0 2. x2 – 3x – 2 = 3x – 11 3. 5x2 – 9x = 2x – 7 4. 2x2 – 7x + 8 = 0 5. 3 – 4x – 2x2 = 0 6. 4x2 – 9x + 5 = 3x – 7 B. By using the relations between roots and coefficients, determine if the given #s are roots of the corresponding 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. A. x2 – 8x – 9 = 0; r= 1 B. 2x2 – 3x – 9 = 0; r= 3 C. x2 + x – 2 - = 0 r = contents back 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=b² - 4ac 2a 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 3x² + 5x – 7 = 0, first compute the discriminant. d = b² - 4ac = 5² - 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 and contents back next
- 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. • 3.5x2 + 1.2x – 3.2 = 0 • 7.6 -2.2x – 1.7x2 = 0 • 2.5x2 + 5.6x – 13.5 = 0 • x – 77.3 + 2.3x2 = 0 • x2 - 1000.5 + 32.3 = 0 contents 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 _____________________________________________________ contents back next
- 108. Instruction: Find the roots o the following equations. 1. 5.3f2 + 2.1f – 2.4 = 0 2. 6.7 x - 2.2x – 8.1x2 = 0 3. 5.2r2 + 6.5r – 5.13 = 0 4. s2 – 50.001 + 33.2 = 0 5. g – 7.73 + 2.3g2 = 0 6. 3.3o2 – 1.9o – 7.10 = 0 7. 3.1e – 9.1e2 – 7.10 = 0 8. 6.3r2 + 8.5r = 9.5 9. 5.9i – 9.5i2 = 8.03 10.3.2o2 + 2.3o = 23.32 11.9.9p – 7.7p2 – 8.8 = 0 12.6.3h + 5.3h2 – 3.8 = 0 13.6.3x2 – 3.9x – 8.10 = 0 14.3.4x – 9.1x2 – 4.10 = 0 15.5.2v2 + 3.6v – 4.7 = 0 contents back next
- 109. 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 contents back next
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