Lecture Notes: Intermediate Algebra

Transcription

1 Lecture Notes: Intermediate Algebra Joseph Lee Metropolitan Community College Contents 1 Linear Equations Linear Equations Linear Equations, Continued Functions Systems of Linear Equations Systems of Linear Equations, Continued Systems of Linear Equations: Applications Polynomials.1 Polynomials Polynomials, Continued Factoring Factoring, Continued Factoring III Factoring IV: An Application Rational Expressions 5.1 Rational Expressions Rational Expressions, Continued Rational Expressions III: Complex Rational Expressions Rational Expressions IV Radicals Radicals: An Introduction Radicals: Rational Exponents Radicals III Radicals IV Radicals V Radicals VI Quadratic Equations The Complex Number System Quadratic Equations Quadratic Equations: The Grand Finale Quadratic Equations III Prelude To College Algebra: Graphing Quadratic Equations

2 Preface I created this set of notes during the winter quarter 01 for my intermediate algebra students. I had taught this course for the two preceeding years, and honestly, intermediate algebra became my favorite course to teach at Metropolitan Community College. Often times I find my appreciation for this course is not shared by my students 1, but I think its regrettable if a student misses the beauty of the algebra presented. The question I receive most often, regardless of the course, is, When am I ever going to use this? I think the question misses the point entirely. While I don t determine which classes students need to get their degree, I do think it s a good policy that students are required to take my course for more reasons than just my continued employment, which I support as well. If a student asked an English instructor why he or she had to read Willa Cather s My Ántonia, the instructor wouldn t argue that understanding nineteenth century prairie life was essential to becoming a competent tax specialist or licensed nurse. The instructor would not argue that reading My Ántonia would benefit the student directly through a future application. Instead, the benefit of reading this beautiful piece of American literature is entirely intrinsic. The mere enjoyment and appreciation is enough to justify its place in a post-secondary education. Moreover, the result arrived to at the end of my course is as beautiful as any prose or poetry a student will encounter in his or her studies here at Metro or any other college. Background The spirit of this study was first undertaken by Greek mathematicians, most notably Diophantus. While these ideas would continue to be explored independently by Greek, Indian, and Chinese mathematicians, it was the Arab mathematician Al-Khwarizmi s 9th century treatise Al-Kitab almukhtasar fi hisab al-jabr wa l-muqabala that inspired the term algebra. Al-Khwarizmi s solution to a quadratic equation was not entirely complete, but it largely resembles the techniques we will apply in Section 5.. About the Course The course is divided into five units. The first unit covers linear equations. Graphing equations is covered with special attention to finding the slope of a line. Point-slope form is used to write the equation of a line. Readers of other texts might prefer to only use only slope-intercept form, but I choose to stress point-slope form in all of my arguments. Functions are introduced in this course, and while I cover them thoroughly, the student should note these ideas will be strongly reinforced if he or she continues on to College Algebra. The second half of this unit deals with solving systems of linear equations. These last sections are omitted from my notes and will be presented from the text. The second unit covers polynomials with an emphasis on factoring. The first section on definitions and the operations of addition and subtraction should be review. The second section on multiplication is also review, but I emphasize what I consider are the more difficult ideas. Again, 1 hopefully not due to my presentation The Compendious Book on Calculation by Completion and Balancing Al-Khwarizmi completes the square geometrically

3 the student might not enjoy my presentation at this point, but I feel the approach I present is the one intended by the course. The remainder of the unit covers factoring in all its glory. The third unit covers rational expressions. As with systems of equations, I have omitted these sections from the notes, and they will be covered from the text. The fourth unit covers radicals. It starts with a basic introduction to the definitions. Rules for radicals are developed using rational exponents, which are defined in the second section. Sections three through five focus on simplifying radicals following three basic rules. The last section of the unit covers solving equations with radicals. The last unit brings the course to what I call The Grand Finale. The genius of the course is that the student has learned all the tools to prove the famous quadratic formula. I feel that many teachers, if not all, present the formula without proof. I think this causes me deep sadness in fact. Having learned some really great algebra completing the square, adding rational expressions, rules of radicals, factoring trinomials it would be a complete waste not to use all these tools to prove the quadratic formula. I feel this last unit completely justifies learning all the algebra throughout the course. This fifth section introduces the complex number system. Quadratic equations are solved by completing the square and the quadratic formula. To The Student I hope you are successful in this course. It will require a lot of work, and more for some than others. You should work on homework every night. Read your textbook, in addition to my notes. When you get frustrated, take a break, but come back to it the next day. Come to my office hours or go to the math center for help. You should complete every homework problem. I am glad you are taking my course, but you have to decide how important this class is to you and what your priorities are. If you decide you have more important things to do in life than doing every single homework problem, I will not fault you for it. I have no children or other commitments, so it s easy for me to spend all day doing math. I realize this may not be the case for you. However, you might find this course does require the kind of dedication I have outlined above. Grading Policy As an assessment tool, quizzes and tests are used to determine a student s mastery of required course objectives. The six course objectives, listed in the syllabus, are as follows: 1. Graph linear functions and other basic functions: define a function and its notation.. Expand upon operations involving exponents, polynomials, and the methods of factoring.. Solve systems of equations and apply them to solving application problems. 4. Simplify rational expressions and solve rational equations. 5. Simplify radical expressions, solve radical equations, define rational exponents, (manipulate and convert from exponential to radical notation and visa/versa), and perform operations with complex numbers.

4 6. Solve quadratic equations with real and complex solutions. A test designed to assess the student s mastery of objective, for example, should indicate the student s proficiency at factoring. An 80% on a quiz or test should indicate that the student is able to correctly factor 80% of the polynomials he or she is asked to factor not that the student has mastered 80% of the steps needed to factor a polynomial. Thus, in most cases, partial credit for incorrect answers will not be given. Preface to the Spring 01 Quarter Since I first prepared these notes, my appreciation for this course has in no way been diminished. Admittedly, I am very evangelical about this course. Intermediate algebra is an amazing course, and all students should experience the ideas embedded in this course. In my opinion, it is the incorrect view of this course that it is designed to program the students with algorithms needed to answer questions the instructor poses. Instead, this course is designed to teach the student fundamental concepts of algebra. These lectures are designed with that intent. Critics would argue there are easier ways to teach a stucent how to answer specific exercises presented throughout the course. I remain convinced, however, that the rationale for the course is not simply to memorize a set of procedures for various mathematical questions, but instead to understand the concepts underlying these procedures. The examples presented in these lectures are therefore intended as a means to discuss these ideas, and not an end in of themselves. Additionally, in many cases, a greater understanding of the underlying concept will improve a student s proficiency in completing a certain problem. Preface to the Fall 01 Quarter I made some revisions to a few of my lectures. In particular, I removed some examples from Sections 1.1 and 1. used to develop the ideas of slope-intercept and point-slope form. Instead, I simply give definitions and proceed straight into examples. I think students will find this more straight forward approach easier to follow in class. 4

5 1 Linear Equations 1.1 Linear Equations Definition: Linear Equation in Two Variables A linear equation in two variables is any equation that can be written in the form where A, B, and C are real numbers. Examples 1. x + y = 6 Ax + By = C,. y = 1 x 4. 4x = 0 + 5y Definition: Slope The slope, m, of a line is the ratio of the vertical change to the horizontal change. vertical change m = horizontal change Definition: y-intercept The y-intercept of a line is the point where the line intersects the y-axis. Definition: Slope-Intercept Form The equation of a line is in slope-intercept form if it is written as y = mx + b, where m is the slope of the line and (0, b) is the y-intercept. Example 1.a For the equation y = x +, determine the slope and the y-intercept. Then graph the equation. Since the equation is in slope-intercept form, we know the slope is m = y-intercept is (0, ). and the 5

6 Example 1.b For the equation y = 1 x, determine the slope and the y-intercept. Then graph 4 the equation. Since the equation is in slope-intercept form, we know the slope is m = 1 4 y-intercept is (0, ). and the Example 1.c For the equation 4x y = 9, determine the slope and the y-intercept. Then graph the equation. 4x y = 9 y = 4x + 9 y = 4 x Example 1.d For the equation 5x + y = 6, determine the slope and the y-intercept. Then graph the equation. 5x + y = 6 y = 5x + 6 y = 5 x + 6

7 Example Find the slope of the line passing through the points (x 1, y 1 ) and (x, y ). m = vertical change horizontal change = y y 1 x x 1 Example.a Find the slope of the line connecting (1, 6) and (4, ). m = y y 1 x x 1 = = 4 = 4 Example.b Find the slope of the line connecting (, 4) and (6, 1). m = y y 1 x x 1 = 1 ( 4) 6 ( ) = 16 8 = Example.c Find the slope of the line connecting (, ) and (4, ). 7

8 m = y y 1 x x 1 = 4 ( ) = 0 7 Example.d Find the slope of the line connecting (, ) and (, 8). = 0 m = y y 1 x x 1 = 8 ( ) = 6 0 However, 6 divided by 0 is undefined. (In other words, the ratio of the vertical change to the horizontal change does not exist.) 8

9 1. Linear Equations, Continued Example 1.a Write the equation of the line, in slope-intercept form, that has a slope of and a y-intercept of (0, 4). y = x + 4 Example 1.b Write the equation of the line, in slope-intercept form, that has a slope of and a y-intercept of ( 0, ). y = x Definition: Point-Slope Form An equation of the line with slope m passing through the point (x 1, y 1 ) is given by y y 1 = m(x x 1 ). The equation above is said to be in point-slope form. Example.a Write the equation of the line, in slope-intercept form, with a slope of 1 that passes though the point ( 4, ). y y 1 = m(x x 1 ) y = 1 (x ( 4)) y = 1 (x + 4) y = 1 x y = 1 x + 1 Example.b Write the equation of the line, in slope-intercept form, with a slope of that passes though the point (, 8). y y 1 = m(x x 1 ) y ( 8) = (x ) y + 8 = x + 6 y = x 9

10 Example.a Write the equation of the line, in slope-intercept form, that passes though the points (, ) and (, 1). First we note the slope of the line is m = 1 ( ) = 4 4 = 1. Then the equation of the line with slope 1 and passing through (, ) is y ( ) = (x ) y + = x + y = x 1. Example.b Write the equation of the line, in slope-intercept form, that passes though the points (5, ) and ( 1, 1). First we note the slope of the line is m = 1 ( ) 1 5 = 6 = 1. Then the equation of the line with slope 1 and passing through (5, ) is y ( ) = 1 (x 5) y + = 1 x + 5 y = 1 x + 1. Definition: Standard Form Recall our definition of a linear equation in two variables is any equation that can be written in the form Ax + By = C, where A, B, and C are real numbers. The linear equation is said to be in standard form if it is written in this manner where A, B, and C are integers and A > 0. Slope-Intercept Form Standard Form y = x x y = y = x + 1 x + y = 10

11 Example 4.a Write the equation of the line, in standard form, with slope 1 the point (, ). that passes though y = 1 (x + ) y = 1 x + y = 1 x x + y = 8 x y = 8. Example 4.b Write the equation of the line, in standard form, that passes though the points (4, 0) and (, ). First we note the slope of the line is m = 0 4 = 6 = 1. Then the equation of the line with slope 1 and passing through (4, 0) is y 0 = 1 (x 4) y = 1 x + 1 x + y = x + y = 4. Definition: Parallel, Perpendicular Two lines are parallel if they have the same slope. Two lines are perpendicular if the slopes are opposite reciprocals (if their product is 1). Moreover: Any two horizontal lines are parallel. Any two vertical lines are parallel. A horizontal line and a vertical line are perpendicular. Example 5.a Write the equation of the line, in slope-intercept form, that is parallel to the line x + y = 6 and passes though the point (, 1). 11

12 Since our line is parallel to x + y = 6 y = x + 6, we know the slope of our line must be. Then the equation of the line with slope and passing through (, 1) is y 1 = (x ) y 1 = x + 6 y = x + 7. Example 5.b Write the equation of the line, in standard form, that is perpendicular to the line x + y = 5 and passes though the point (0, ). Since our line is perpendicular to we know the slope of our line must be. Then the equation of the line with slope x + y = 5 y = x + 5, and y-intercept (0, ) is y = x + x + y = x y = 6. 1

13 1. Functions Definition: Set A set is a collection. Members of the collection are called elements. Example 1.a A set is a collection. Members of the collection are called elements. 1. {black, white, pink} is the set of Joseph s three favorite colors. This set has elements.. {1,,, 4} is the set of positive integers less than 5.. {, } is the set of solutions to the equation x = is the set containing no elements. For example, the set of real solutions to the equation x = 4. Example 1.b We may also want to talk about sets with an infinite number of elements. 1. {1,,, 4, 5,...} could be the set of positive integers.. {, 5, 7,...} could be the set of odd numbers greater than 1: {, 5, 7, 9, 11, 1,...}. On the other hand, {, 5, 7,...} could be the set of odd prime numbers: {, 5, 7, 11, 1, 17,...} Definition: properties. Set-Builder Notation Set builder notation is a method to describe a set by its Example 1.c The following sets are written using set-builder notation. 1. {x x is one of Joseph s three favorite colors} = {black, white, pink}. {x x is an odd prime number} = {, 5, 7, 11, 1, 17,...}. {x x is an integer & 0 < x < 5} = {1,,, 4} Definition: Relation A relation is a correspondence between two sets. Elements of the first set are called the domain. Elements of the second set are called the range. Relations are often expressed as sets of ordered pairs. 1

14 Example.a A relation is a correspondence between two sets. Elements of the first set are called the domain. Elements of the second set are called the range. 1. {(Joseph, turkey), (Joseph, roast beef), (Michael, ham)} is a relation between math instructors and sandwiches they enjoy. The domain of this relation is {Joseph, Michael}. The range of this relation is {turkey, roast beef, ham}.. {(1, ), (, 4), ( 1, 1)} is a relation of x and y values that satisfy the equation y = x +. The domain of this relation is { 1, 1, }. The range of this relation is {1,, 4}.. {(, 5), (4, 5), (5, 5)} is a relation of x and y values that satisfy the equation y = 5. Definition: Function A function is a specific type of a relation where each element in the the domain corresponds to exactly one element in the range. Example.b Determine which of the following relations are functions. 1. {(Joseph, turkey), (Joseph, roast beef), (Michael, ham)}. {(1, ), (, 4), ( 1, 1)}. {(, 5), (4, 5), (5, 5)}. The first relation is not a function, as Joseph is assigned to two elements in the range, turkey and roast beef. The last two relations are functions as each element in the domain is assigned to exactly one element in the range. Example.a The graph of the relation y = x is shown below. 1. Determine the domain of the relation. R (all real numbers). Determine the range of the relation. R (all real numbers). Determine if the relation is represents a function. Yes, it is a function. 14

15 Example.b The graph of the relation x = y 1 is shown below. 1. Determine the domain of the relation. {x x 1}. Determine the range of the relation. R (all real numbers). Determine if the relation is represents a function. No, it is not a function. Example.c The graph of the relation y = x x is shown below. 1. Determine the domain of the relation. R (all real numbers). Determine the range of the relation. {y y 4}. Determine if the relation is represents a function. Yes, it is a function. Definition: Vertical Line Test If any vertical line intersects the graph of a relation more than once, the relation is not a function. Definition: Function Notation If a relation of x and y is a function, than you may solve for y and replace y with f(x). This notation is called function notation. Example 4.a Write the function x + y = 6 in function notation. 15

16 x + y = 6 y = x + 6 y = x + f(x) = x + Example 4.b Evaluate the function f(x) = x + for the following values. 1. f( ) = 4. f(0) =. f(1) = 4 4. f() = 0 Example 4.c Evaluate the function f(x) = x + 6 for the following values. 1. f( ) =. f(19) = 5. f( 7) is undefined Example 5.a The graph of the function f is shown below. Evaluate. 16

17 1. f( 4) =. f( 1) = 0. f(1) = 4. f(4) = 1 Example 5.b The graph of the function f is shown below. Evaluate. 1. f( 5) = 0. f( 1) =. f() = 0 17

18 1.4 Systems of Linear Equations Definition: System of Equations A system of equations is a group of two or more equations. Example 1.a Determine if (0, 4) is a solution to the following system of linear equations: x y = 8 4x + y = 5 Solution. First, we will check if (0, 4) is a solution of x y = 8. x y = 8 (0) ( 4)? = 8 8 = 8 Thus, (0, 4) is a solution of x y = 8, so now we will check if it is also a solution of 4x + y = 5. 4x + y = 5 4(0) + ( 4)? = Thus, (0, 4) is not a solution of this system of equations. Example 1.b Determine if (, ) is a solution to the following system of linear equations: x y = 8 4x + y = 5 Solution. First, we will check if (, ) is a solution of x y = 8. x y = 8 () ( )? = 8 8 = 8 Thus, (, ) is a solution of x y = 8, so now we will check if it is also a solution of 4x + y = 5. 4x + y = 5 4() + ( )? = 5 5 = 5 Thus, (, ) is a solution of this system of equations. Example 1.c Graph both equations on the same coordinate plane. x y = 8 4x + y = 5 18

19 The line for x y = 8 represents all the points that satisfy that equation. The line for 4x + y = 5 represents all the points that satisfy that equation. Therefore, any point where the two lines intersect represents a point that will satisfy both equations our solution. Example.a Determine if (, ) is a solution to the following system of linear equations: x + y = 1 y = x + 4 Solution. First, we will check if (, ) is a solution of x + y = 1. x + y = 1 () + ()? = 1 1 = 1 Thus, (, ) is a solution of x + y = 1, so now we will check if it is also a solution of y = x + 4. y = x + 4 ()? = () + 4 = Thus, (, ) is a solution of this system of equations. Example.b Determine if (9, ) is a solution to the following system of linear equations: x + y = 1 y = x + 4 Solution. First, we will check if (9, ) is a solution of x + y = 1. x + y = 1 (9) + ( )? = 1 1 = 1 Thus, (9, ) is a solution of x+y = 1, so now we will check if it is also a solution of y = x+4. y = x + 4 ( )? = (9) + 4 = Thus, (9, ) is a solution of this system of equations. Example.c Graph both equations on the same coordinate plane. x + y = 1 y = x

20 Notice the equations x + y = 1 and y = x + 4 represent the same line. The lines represent all the points that satisfy the equation therefore, any point on the lines satisfies both equations, i.e. the system. Thus, this system of linear equations has infinitely many solution. In particular, the solution is any point on the line x + y = 1. Example Graph both equations on the same coordinate plane. y = x + 4 x + y = 1 Notice the lines represented by the equations y = x + 4 and x + y = 1 are parallel meaning the lines will never intersect. Since points of intersection are solutions to the system, this system of linear equations has no solution. Definition: Dependent, Independent, Inconsistent Systems A system of equations is called inconsistent if it has no solutions. A system of equations is called dependent if it has infinitely many solutions. A system of equations is called independent if it has a single solution. Methods for Solving Systems of Linear Equations There are three methods we will discuss for solving systems of linear equations. 1. Graphing. Substitution. Elimination Example 4.a Solve the following system of equations using substitution. x y = 8 4x + y = 5 Solution. To solve by substitution, we will take one of our equations and solve for one of our variables. Let s take our first equation x y = 8 and solve for x. x y = 8 x = y + 8 Now, we may substitute x = y + 8 into our other equation 4x + y = 5. 4x + y = 5 4(y + 8) + y = 5 8y + + y = 5 9y = 7 y = 0

21 Since y =, we may substitute that value back into any equation to solve for x. x = y + 8 x = ( ) + 8 x = Thus, (, ) is the solution of this system of equations. Example 4.b Solve the following system of equations using substitution. 5x + y = 6 4x + y = Solution. To solve by substitution, we will take one of our equations and solve for one of our variables. Let s take our second equation 4x + y = and solve for y. 4x + y = y = 4x + y = x + 1 Now, we may substitute y = x + 1 into our other equation 5x + y = 6. 5x + y = 6 5x + ( x + 1) = 6 5x 6x + = 6 x = x = Since x =, we may substitute that value back into any equation to solve for y. y = x + 1 y = ( ) + 1 y = 7 Thus, (, 7) is the solution of this system of equations. Observation: Operations on Equations Observe that equations may be added together if we wanted... x = 4 y = or x + y = 7 x + y = 7 x y = 1 (x + y) + (x y) = 8 x = 8 Example 5.a Solve the following system of equations using elimination. x y = 8 4x + y = 5 1

22 Solution. To solve by elimination, we will eliminate one of our variables. Let s eliminate y. x y = 8 4x + y = 5 x y = 8 (4x + y) = (5) x y = 8 8x + y = 10 9x = 18 x = Since x =, we may substitute that value back into any equation to solve for y. 4x + y = 5 4() + y = y = 5 y = Thus, (, ) is the solution of this system of equations. Example 5.b Solve the following system of equations using elimination. 5x + y = 6 4x + y = Solution. To solve by elimination, we will eliminate one of our variables. Let s eliminate y. 5x + y = 6 4x + y = (5x + y) = (6) (4x + y) = ()( ) 10x + 6y = 1 1x 6y = 6 x = 6 x = Since x =, we may substitute that value back into any equation to solve for y. 4x + y = 4( ) + y = 1 + y = y = 14 y = 7 Thus, (, 7) is the solution of this system of equations.

23 Example 6.a Solve the following system of equations using elimination. 4x + 6y = 4 y = x + 4 Solution. First, let s get rid of the fraction. Then we will move our variables over to the left side. Then we may choose a variable to eliminate. 4x + 6y = 4 (y) = ( x + 4) 4x + 6y = 4 y = x + 1 4x + 6y = 4 x + y = 1 4x + 6y = 4 (x + y) = (1)( ) 4x + 6y = 4 4x 6y = 4 0 = 0 Arriving at this identity, we may conclude that this system is dependent. There are infinitely many solutions, or more specifically, the solution is any point on the line 4x + 6y = 4. Example 6.b Solve the following system of equations using elimination. y = x + 4 6x + y = Solution. First, we will move our variables over to the left side. Then we may choose a variable to eliminate. y = x + 4 6x + y = x + y = 4 6x + y = (x + y) = (4)( ) 6x + y = 6x y = 8 6x + y = 0 = 10 Arriving at this contradiction, we may conclude that this system is inconsistent. solution. There is no

24 1.5 Systems of Linear Equations, Continued Example 1.a Solve the following system of equations. x + y + z = 6 x y z = 5 x + y z = 1 Solution. First, we will take two equations and eliminate a variable. Let s take the first two equations and eliminate z. x + y + z = 6 x y z = 5 x y = 11 Now, we want to take two different equations but eliminate the same variable: z. We will take the first and last equations. x + y + z = 6 x + y z = 1 (x + y + z) = (6) x + y z = 1 x + y + z = 1 x + y z = 1 5x + 4y = 11 We now have two equations with two variables. We may proceed as usual from here. x y = 11 5x + 4y = 11 (x y) = (11) 5x + 4y = 11 6x 4y = 5x + 4y = 11 11x = x = We substitute x = into an equation containing only x and y. x y = 11 () y = 11 9 y = 11 y = y = 1 4

25 Similarly, we substitute x = and y = 1 into an equation containing x, y, and z. x + y + z = 6 () + ( 1) + z = 6 + z = 6 z = 4 Thus, (, 1, 4) is the solution to this system of equations. Example 1.b Solve the following system of equations. x + 4y = 4 5y + z = 1 x 5z = 7 Solution. First, we will take two equations and eliminate a variable. Let s take the last two equations and eliminate z. 5y + z = 1 x 5z = 7 5(5y + z) = (1)5 (x 5z) = (7) 5y + 15z = 5 6x 15z = 1 6x + 5y = 6 Notice we have two equations with two variables. x + 4y = 4 6x + 5y = 6 (x + 4y) = ( 4)( ) 6x + 5y = 6 6x 8y = 8 6x + 5y = 6 17y = 4 y = We may now substitute y = into an equation only containing x and y to solve for x or into an 5

26 equation containing y and z to solve for z. x + 4y = 4 x + 4() = 4 x + 8 = 4 x = 1 x = 4 5y + z = 1 5() + z = z = 1 z = 9 z = Thus, ( 4,, ) is the solution to the system of equations. Example 1.c Solve the following system of equations. x + y z = 7 x + y z = 7 5x 4y + z = 10 Solution. First, we will take two equations and eliminate a variable. Let s take the first two equations and eliminate z. x + y z = 7 x + y z = 7 (x + y z) = (7)( ) x + y z = 7 4x 6y + z = 14 x + y z = 7 7x 4y = 7 Now, we want to take two different equations but eliminate the same variable: z. We will take the first and last equations. (x + y z) = (7) 5x 4y + z = 10 6x + 9y z = 1 5x 4y + z = 10 11x + 5y = 11 6

27 We now have two equations with two variables. We may proceed as usual from here. 7x 4y = 7 11x + 5y = 11 5( 7x 4y) = ( 7)5 4(11x + 5y) = (11)4 5x 0y = 5 44x + 0y = 44 9x = 9 x = 1 We substitute x = 1 into an equation containing only x and y. 7x 4y = 7 7(1) 4y = 7 7 4y = 7 4y = 0 y = 0 Similarly, we substitute x = 1 and y = 0 into an equation containing x, y, and z. x + y z = 7 (1) + (0) z = 7 z = 7 z = 5 z = 5 Thus, (1, 0, 5) is the solution to this system of equations. Suggestion If you come across a system of equations not written with the variables on the left side of the equal sign and the constants on the right side, you may perform basic algebraic manipulation to arrange the system in this more aesthetically pleasing manner. Example.a Solve the following system of equations. x + y z = x + y 5z = 5x + 8y 11z = 9 Solution. First, we will take two equations and eliminate a variable. Let s take the first two 7

28 equations and eliminate x. x + y z = x + y 5z = (x + y z) = ()( ) x + y 5z = x 4y + z = 6 x + y 5z = y z = Now, we want to take two different equations but eliminate the same variable: x. We will take the first and last equations. 5(x + y z) = ()( 5) 5x + 8y 11z = 9 5x 10y + 5z = 15 5x + 8y 11z = 9 y 6z = 6 We now have two equations with two variables. We may proceed as usual from here. y z = y 6z = 6 ( y z) = ( )( ) y 6z = 6 y + 6z = 6 y 6z = 6 0 = 0 Arriving at this identity, we conclude that this is a dependent system of equations, and thus it has infinitely many solutions. Unlike in the previous section, we will not describe these solutions. 4 Example.b Solve the following system of equations. x y + z = 4 x 5y + 4z = 6x y + 4z = 8 Solution. First, we will take two equations and eliminate a variable. Let s take the first and last 4 If you are interested in these solutions, please do sign up for my college algebra course. 8

29 equations and eliminate x. x y + z = 4 6x y + 4z = 8 (x y + z) = (4)( ) 6x y + 4z = 8 6x + y 4z = 8 6x y + 4z = 8 0 = 16 Arriving at this contradiction, we conclude that this is an inconsistent system of equations, and thus it has no solution. 9

30 1.6 Systems of Linear Equations: Applications Example 1 Joseph has a collection of quarters and dimes. A friend counts his change and determines he has $.70. Joseph knows he has 5 coins. How many quarters and dimes does Joseph have? Solution. Our goal is to express this situation as a system of two equations. One equation should use the fact that the coins total $.70, and the other equations should use the fact that Joseph has 5 coins. If x represents the number of quarters Joseph possesses and y represents the number of dimes Joseph possesses, we may write:.5x +.10y =.70 x + y = 5 Solve the system using substitution or elimination. Joseph has 8 quarters and 17 dimes..5x +.10y =.70 x + y = 5 10(.5x +.10y) = (.70)( 10) x + y = 5.5x y = 7 x + y = 5 1.5x = 1 x = 8 x + y = y = 5 y = 17 Example Ten liters of a 1% HCl solution is mixed with a 0% HCl solution to make a mixture that is 15% HCl. How many liters of the 0% HCl solution were used to make the mixture? Solution. Our goal is to express this situation as a system of two equations. Let x represents the number of liters of 0% solution used, and let y represents the number of liters in the 15% mixture. One equation should represent the total number of liters, and the other equation should address the HCl in the mixture. x + 10 = y.0x +.1(10) =.15y 0

31 Solve the system using substitution or elimination. x + 10 = y.0x +.1(10) =.15y.0x +.1(10) =.15(x + 10).0x + 1. =.15x x = 0. x = 6 6 liters or 0% solution were used to make the mixture. Example Against the wind, a plane can fly 880 miles in 5 hours. Flying with the same wind at its tail, it only takes 4.5 hours. Determine the speed of the wind and the speed of the plane in still air. Solution. Our goal is to express this situation as a system of two equations. Let x represents the speed of the plane in still air, and let y represents the speed of the wind. One equation should represent the plane flying against the wind, and the other equation should represent the plane flying with the wind. Keep in mind the distance formula: rate time = distance (x y)5 = 880 (x + y)(4.5) = 880 Solve the system using substitution or elimination. (x y)5 = 880 (x + y)(4.5) = 880 5x 5y = x + 4.5y = 880 9(5x 5y) = (880)9 10(4.5x + 4.5y) = (880)10 45x 45y = x + 45y = x = 5470 x = 608 5x 5y = 880 5(608) 5y = y = 880 5y = 160 y = The speed of the plane in still air is 608 miles per hour and the speed of the wind is miles per hour. 1

32 Polynomials.1 Polynomials Definition: Polynomial A polynomial is a mathematical expression containing the operations of addition, subtraction, and multiplication. Examples 1. 4x 5 x + 4. m 4 n p. 1 x 4 4. x y + z 8 Definition: Terms, Monomial, Binomial, Trinomial The parts of a polynomial that are added or subtracted together are called terms. A polynomial with exactly one term is called a monomial. A polynomial with exactly two terms is called a binomial. A polynomial with exactly three terms is called a trinomial. Definition: Coefficient, Degree of a Term, Degree of a Polynomial The coefficient is the numerical part of the term without the variable factors. The degree of a term is the number of variable factors in the term. The degree of a polynomial is equal to the highest degree term it contains. Example 1.a Determine how many terms the polynomial contains and give its specific name, if it exists. For each term, determine its coefficient and its degree. Finally, determine the degree of the polynomial. 7x The polynomial has terms, so it is called a binomial. Term Coefficient Degree 7x

33 Finally, 7x is a fourth degree polynomial. Example 1.b Determine how many terms the polynomial contains and give its specific name, if it exists. For each term, determine its coefficient and its degree. Finally, determine the degree of the polynomial. 4x 7x + 1 x The polynomial has 4 terms, so it does not have a special name. Term Coefficient Degree 4x 4 7x Finally, 4x 7x + 1 x is a third degree polynomial. Example 1.c Determine how many terms the polynomial contains and give its specific name, if it exists. For each term, determine its coefficient and its degree. Finally, determine the degree of the polynomial..x 4 y z + 1.x yz 7.9y z 4 The polynomial has terms, so it is called a trinomial. Term Coefficient Degree.x 4 y z. 9 1.x yz y z Finally,.x 4 y z + 1.x yz 7.9y z 4 is a ninth degree polynomial. Definition: Descending Order A polynomial is written in descending order if the terms are written from highest degree to lowest degree. If two terms have the same degree, then they are written in alphabetical order. Observe that each polynomial in Example 1 was written in descending order. Example.a Add the polynomials. (4x 7x + 1) + (x 8x 9) (4x 7x + 1) + (x 8x 9) = 7x 15x + Example.b Subtract the polynomials. (4x 7x + 1) (x 8x 9)

34 (4x 7x + 1) (x 8x 9) = 4x 7x + 1 x + 8x + 9 = x + x + 1 Example.c Add the polynomials. ( 4 x 1 x 5x + 10 ) ( + x 4x x + 7 ) ( 4 x 1 x 5x + 10 ) ( + x 4x x + 7 ) = ( 9 1 x 1 x 0 4 x + 0 ) ( x 8 x x + 1 ) 6 = 17 1 x 9 x 15 4 x Example.d Subtract the polynomials. ( x ) ( x + 1 ) ( x ) ( x + 1 ) = x x 1 = 16 8 x x 4 1 Example.e Subtract the polynomials. = 9 8 x 1 1 (.4x y + 7.8x y + 9.1x 4.y) (.78x y.9x y y 5.) (.4x y + 7.8x y + 9.1x 4.y) (.78x y.9x y y 5.) =.4x y + 7.8x y + 9.1x 4.y.78x y +.9x y 4.75y + 5. = 6.18x y x y + 9.1x 9.05y

35 . Polynomials, Continued Example 1.a Multiply. x (4x 7x + ) x (4x 7x + ) = x (4x ) + x ( 7x) + x () Example 1.b Multiply. Example 1.c Multiply. = 1x 5 1x + 6x (x + 7)(x ) (x + 7)(x ) = 6x 4x + 1x 14 = 6x + 17x 14 (x + 1)(x 8) (x + 1)(x 8) = x 6 8x + x 8 Theorem. Square of a Binomial For any a and b, = x 6 7x 8 (a + b) = a + ab + b. Proof. (a + b) = (a + b)(a + b) = a + ab + ab + b Example.a Multiply. Example.b Multiply. = a + ab + b (x + 4) (x + 4) = (x) + (x)(4) + (4) = x + 8x + 16 (x 4) 5

36 (x 4) = [ x + ( 4) ] = (x) + (x)( 4) + ( 4) Example.c Multiply. Example.a Multiply. = x 8x + 16 (x y + 7z) (x y + 7z) = (x y) + (x y)(7z) + (7z) = 9x 4 y + 4x yz + 49z (a + b + c) (a + b + c) = [ (a + b) + c ] Example.b Multiply. = (a + b) + (a + b)(c) + (c) = (a + b) + (ac + bc) + c = (a + b) + ac + bc + c = a + ab + b + ac + bc + c (x y 4z) (x y 4z) = [ (x y) + ( 4z) ] = (x y) + (x y)( 4z) + ( 4z) = (x y) + ( 1xz + 8yz) + 16z = (x y) 4xz + 16yz + 16z Example.c Multiply. = 9x 1xy + 4z 4xz + 16yz + 16z (x y + 4z + 7) 6

37 (x y + 4z + 7) = [ (x y) + (4z + 7) ] = (x y) + (x y)(4z + 7) + (4z + 7) = (x y) + (4xz + 7x 8yz 14y) + (4z + 7) = (x y) + 8xz + 14x 16yz 8y + (4z + 7) = x 4xy + 4y + 8xz + 14x 16yz 8y + 16z + 56z + 49 Theorem. Product of Conjugates For any a and b, (a + b)(a b) = a b. Proof. (a + b)(a b) = (a + b)(a b) = a ab + ab b Example 4.a Multiply. Example 4.b Multiply. Example 4.c Multiply. Example 5.a Multiply. = a b (x + 9)(x 9) (x + 9)(x 9) = (x) (9) = x 81 (m 8n )(m + 8n ) (m 8n )(m + 8n ) = (m ) (8n ) = 9m 4 64n 4 (6x y 5z )(6x y + 5z ) (6x y 5z )(6x y + 5z ) = (6x y) (5z ) = 6x 4 y 5z 6 (x + y + z)(x + y z) 7

38 (x + y + z)(x + y z) = ( (x + y) + z )( (x + y) z ) Example 5.b Multiply. Example 5.c Multiply. = (x + y) (z) = x + xy + y z (x y + z)(x + y z) (x y + z)(x + y z) = ( x (y z) )( x + (y z) ) = (x) (y z) = x (y yz + z ) = x y + yz z (x y + z + 1)(x y z 1) (x y + z + 1)(x y z 1) = ( (x y) + (z + 1) )( (x y) (z + 1) ) = (x y) (z + 1) = (9x 4 6x y + y ) (z + z + 1) = 9x 4 6x y + y z z 1 8

39 . Factoring Definition: Factor, Greatest Common Factor Let a, b, c, m, and n be polynomials. If a b = m, then a and b are called factors of m. Example. Since x y = 6xy, x and y are factors of 6xy. If m = a b and n = a c, then a is called a common factor of m and n. Example. Since 6xy = x y and 1x = x 4x, x is a common factor of 6xy and 1x. Observation: For any m and n, 1 is a common factor. If m = a b and n = a c, and the only common factor between b and c is 1, then a is called the greatest common factor of m and n. Example 1.a Factor. 15x 5x y First, we note the greatest common factor of 15x and 5x y is 5x. 15x 5x y = 5x (x) + 5x ( 5y) = 5x (x 5y) Example 1.b Factor. 14x + 5x 1 First, we note the greatest common factor of 14x, 5x, and 1 is 7. 14x + 5x 1 = 7(x ) + 7(5x) + 7( ) = 7(x + 5x ) Example 1.c Factor. 1p q 4 r 6p q r + 0pq 4 r First, we note the greatest common factor of 1p q 4 r, 6p q r, and 0pq 4 r is pq r. 1p q 4 r 6p q r + 0pq 4 r = pq r (6pq ) + pq r ( p ) + pq r (10q r) = pq r (6pq p + 10q r) Example.a Factor. 4x(x 1) y(x 1) 9

40 First, we note the greatest common factor of 4x(x 1) and y(x 1) is x 1. Example.b Factor. 4x(x 1) y(x 1) = (x 1)(4x) + (x 1)( y) = (x 1)(4x y) 4x (x + ) x(x + ) First, we note the greatest common factor of 4x (x + ) and x(x + ) is x(x + ). Example.c Factor. 4x (x + ) x(x + ) = [x(x + )](4x) + [x(x + )]( ) = x(x + )(4x ) (x )(x + ) + y(x + ) First, we note the greatest common factor of (x )(x + ) and y(x + ) is x +. (x )(x + ) + y(x + ) = (x + )(x ) + (x + )(y) = (x + )[(x ) + y] = (x + )(x + y) Example.a Factor. 1x 9x + 8x 6 1x 9x + 8x 6 = (1x 9x ) + (8x 6) = x (4x ) + (4x ) Example.b Factor. = (4x )(x + ) xy + y 5x 15 xy + y 5x 15 = (xy + y) + ( 5x 15) = y(x + ) 5(x + ) Example.c Factor. = (x + )(y 5) 7x y 14xy + 8xy 56y 40

41 7x y 14xy + 8xy 56y = 7y(x x + 4x 8) Example.d Factor. = 7y[(x x) + (4x 8)] = 7y[x(x ) + 4(x )] = 7y(x )(x + 4) 8mn 1m 1n mn 1m 1n + 18 = (4mn 6m 6n + 9) = [(4mn 6m) + ( 6n + 9)] = [m(n ) (n )] = (n )(m ) 41

42 .4 Factoring, Continued Example 1.a Multiply. (x 7)(x + 5) (x 7)(x + 5) = 6x + 15x 14x 5 Example 1.b Factor. Example.a Factor. = 6x + x 5 6x + x 5 6x + x 5 = (x 7)(x + 5) x + x 5 x + x 5 = (x )(x ) We need factors of 5, and the only factors of 5 are 1 and 5. Observe, our two choices: (x 1)(x 5) = x 10x x 5 (x 5)(x 1) = x x 5x 5 The only way to come up with x + x 5 would be Finally, Example.b Factor. We need factors of 0: We have the following choices: (x 5)(x 1) = x x + 5x 5. (x + 5)(x 1) = x + x 5. x + 11x 0 x + 11x 0 = (x )(x ) 0 = (x 1)(x 0) = x 60x x 0 (x 0)(x 1) = x x 0x 0 (x )(x 10) = x 0x x 0 (x 10)(x ) = x 6x 10x 0 (x 4)(x + 5) = x + 15x 4x 0 (x 5)(x 4) = x 1x 5x 0 4

43 Example.c Factor. We need factors of 6: We have the following choices: 5x 1x + 6 5x 1x + 6 = (5x )(x ) 6 = 1 6 (5x 1)(x 6) = 5x 0x x 6 (5x 6)(x 1) = 5x 5x 6x 6 (5x )(x ) = 5x 15x x 6 (5x )(x ) = 5x 10x x 6 Our choice is: (5x )(x ) = 5x 10x x + 6 Example.d Factor. 4x + 1x 7 Note we have two possibilities to begin with: The only factors of 7, however, are 1 and 7. Thus, we have the following choices: 4x + 1x 7 = (4x )(x ) = (x )(x ) (4x 1)(x 7) = 4x 8x x 7 (4x 7)(x 1) = 4x 4x 7x 7 (x 1)(x 7) = 4x 14x x 7 Our choice is: (x 1)(x + 7) = 4x + 14x x 7 Example.e Factor. 6x 7x + 1 We should first observe that we may factor out a common factor. 6x 7x + 1 = (x 9x + 4) = (x )(x ) We need factors of 4: 4 = 1 4 4

44 We have the following choices: Our choice is: Example.f Factor. (x 1)(x 4) = (x 8x x 4) (x 4)(x 1) = (x x 4x 4) (x )(x ) = (x 4x x 4) (x 1)(x 4) = (x 8x x + 4) 9m 4 n 15m n 6m n We should first observe that we may factor out a common factor. 9m 4 n 15m n 6m n = m n(m 5mn n ) = m n(m )(m ) The only factors of n that we need to consider are n and n. We have the following choices: Our choice is: Example.a Factor. Let u = x + 7. m n(m n)(m n) = m n(m 6mn mn n ) m n(m n)(m n) = m n(m mn mn n ) m n(m + n)(m n) = m n(m 6mn + mn n ) (x + 7) 6(x + 7) 16 (x + 7) 6(x + 7) 16 = u 6u 16 = (u 8)(u + ) = [(x + 7) 8][(x + 7) + ] = (x 1)(x + 9) Example.b Factor. Let u = x. 4x 6 4x 15 4x 6 4x 15 = 4u 4u 15 = (u )(u ) = (u + )(u 5) = (x + )(x 5) 44

45 Example.c Factor. Let u = n (n + 1) 7(n + 1) + 4(n + 1) 7(n + 1) + = 4u 7u + = (4u )(u ) = (4u )(u 1) = [4(n + 1) ][(n + 1) 1] = (8n + 4 )(n ) = (8n + 1)(n ) 45

46 .5 Factoring III Factoring III Recall two theorems from multiplying polynomials: (a + b) = a + ab + b (a + b)(a b) = a b Example 1.a Factor. Example 1.b Factor. 4x + 8x x + 8x + 49 = (x) + (x)(7) + (7) = (x + 7) 9x 0x + 5 9x 0x + 5 = (x) + (x)( 5) + ( 5) Example 1.c Factor. Example 1.d Factor. = (x 5) 100x 4 y + 60x yz + 9z 6 100x 4 y + 60x yz + 9z 6 = (10x y) + (10x y)(z ) + (z ) = (10x y + z ) x 80xy + 50y x 80xy + 50y = (16x 40xy + 5y ) = [(4x) + (4x)( 5y) + ( 5y) ] Example.a Factor. = (4x 5y) x 16 x 16 = (x) (4) = (x + 4)(x 4) 46

47 Example.b Factor. Example.c Factor. 4m 8 9n 4m 8 9n = (m 4 ) (n) = (m 4 + n)(m 4 n) 1x x 1x x = x(4x 1) = x[(x) (1) ] Example.d Factor. = x(x + 1)(x 1) x 4 16 x 4 16 = (x ) (4) = (x + 4)(x 4 4) = (x + 4)[(x ) () ] Example.a Factor. Let u = x + y. = (x + 4)(x + )(x ) (x + y) + (x + y) + 1 (x + y) + (x + y) + 1 = u + u + 1 = (u + 1) = [(x + y) + 1] = (x + y + 1) Example.b Factor. 5 (z + 4) 47

48 Let u = z (z + 4) = 5 u = (5 + u)(5 u) = [5 + (z + 4)][5 (z + 4)] = (5 + z + 4)(5 z 4) = (z + 9)( z + 1) = (z + )( z + 1) Theorem. Factoring the Sum of Cubes For any a and b, a + b = (a + b)(a ab + b ). Proof. (a + b)(a ab + b ) = (a + b)(a ) + (a + b)( ab) + (a + b)(b ) = a + a b a b ab + ab + b = a + b Theorem. Factoring the Difference of Cubes For any a and b, a b = (a b)(a + ab + b ). Proof. Left for the student. Example 4.a Factor. 8x + 7 8x + 7 = (x) + () = (x + )[(x) (x)() + () ] = (x + )(4x 6x + 9) Example 4.b Factor. 1 64x 1 64x = (1) (4x) = (1 4x)[(1) + (1)(4x) + (4x) ] = (1 4x)(1 + 4x + 16x ) 48

49 Example 4.c Factor. 15x 6 7y z 9 15x 6 7y z 9 = (5x ) + (yz ) = (5x + yz )[(5x ) (5x )(yz ) + (yz ) ] Example 5.a Factor. Let u = a + b. = (5x + yz )(5x 4 15x yz + 9z 6 ) 16 (a + b) Example 5.b Factor. 16 (a + b) = 16 u = (6) u Let u = a + b and let v = x + 1. = (6 u)(6 + 6u + u ) = [6 (a + b)][6 + 6(a + b) + (a + b) ] = (6 a b)(6 + 6a + 6b + a + ab + b ) (x y) + (x + 1) (x y) + (x + 1) = u + v = (u + v)(u uv + v ) = [(x y) + (x + 1)] [(x y) (x y)(x + 1) + (x + 1) ] = (4x y + 1) [(9x 1xy + 4y ) (x + x xy y) + (x + x + 1)] = (4x y + 1) (9x 1xy + 4y x x + xy + y + x + x + 1) = (4x y + 1)(7x 10xy + 4y x + y + 1) 49

50 .6 Factoring IV: An Application Example 1.a Solve. x 5x + 6 = 0 x 5x + 6 = 0 (x )(x ) = 0 Thus, by the zero factor property, either x = 0 or x = 0 x = x = The solution set is {, }. Example 1.b Solve. x 10x 8 = 0 x 10x 8 = 0 (x + )(x 4) = 0 Thus, by the zero factor property, either x + = 0 or x 4 = 0 x = x = 4 The solution set is {, 4 }. x = Example 1.c Solve. x x 6x = 0 x x 6x = 0 x(x x ) = 0 x(x + 1)(x ) = 0 Thus, by the zero factor property, either x = 0 or x + 1 = 0 or x = 0 x = 0 x = 1 x = 50

51 The solution set is { 1, 0, }. Example.a Solve. x = 5x + x = 5x + x 5x = 0 (x + 1)(x ) = 0 Thus, by the zero factor property, either x + 1 = 0 or x = 0 x = 1 x = The solution set is { 1, }. Example.b Solve. (x )(x 5) = 1 (x )(x 5) = 1 x 8x + 15 = 1 x 8x + 16 = 0 (x 4) = 0 Thus, by the zero factor property, The solution set is {4}. Example.c Solve. x 4 = 0 x = 4 x(x ) = 4(x + 1) + 4 x(x ) = 4(x + 1) + 4 x 6x = 4x x 6x = 4x + 8 x 10x 8 = 0 (x + )(x 4) = 0 51

52 The solution set is { }, 4. Pythagorean Theorem If a right triangle has legs of lengths a and b and hypotenuse of length c, then a + b = c. Example.a x + (x + 1) = (x + ) x + x + x + 1 = x + 4x + 4 x + x + 1 = x + 4x + 4 x x = 0 (x )(x + 1) = 0 The solution set is { 1, }. Since x represents the length of a leg of the triangle, we know x =. Thus, the sides of the triangle are, 4, and 5. Example.b (x 8) + (x 1) = x x 16x x x + 1 = x x 18x + 65 = x x 18x + 65 = 0 (x 5)(x 1) = 0 The solution set is {5, 1}. Since x 8 represents the length of a leg of the triangle, we know x = 1 (otherwise x 8 = ). Thus, the sides of the triangle are 5, 1, and 1. 5

53 Rational Expressions.1 Rational Expressions The notes for this section will be presented from the text.. Rational Expressions, Continued The notes for this section will be presented from the text.. Rational Expressions III: Complex Rational Expressions The notes for this section will be presented from the text..4 Rational Expressions IV The notes for this section will be presented from the text. 5

54 4 Radicals 4.1 Radicals: An Introduction Definition: nth Root, Principle nth Root The nth root of a is b if The principle nth root of a is b if The principle nth root of a is denoted by b n = a. b n = a and b 0 if possible. n a = b. If n =, the principle square root of a is denoted by a = b. Example 1.a Evaluate. 81 Since 9 = 81, 81 = 9. Example 1.b Evaluate. 169 Since 1 = 169, 169 = 1. Example 1.c Evaluate. 1 5 ( ) 1 Since = 1 5 5, 1 5 = 1 5. Example 1.d Evaluate. 6 Notice ( 6) = 6. In fact, for any real number x, x 0. Thus, 6 is not a real number. Example 1.e Evaluate

55 Since 4 = 64, Example 1.f Evaluate. Since ( 4) = 64, Example 1.g Evaluate. Since 4 = 16, Example 1.h Evaluate. 64 = = = Notice ( ) 4 = 16. In fact, for any real number x, x 4 0. Thus, number. Example 1.i Evaluate is not a real Using a calculator, we can approximate Example 1.j Evaluate. 1 Using a calculator, we can approximate Example.a Simplify. x x = x Example.b Simplify. x 10 x 10 = x 5 55

56 Example.c Simplify. 16x 6 16x6 = 4 x Example.d Simplify. x 4 x4 = x Example.e Simplify. Example.f Simplify. Example.g Simplify. x x = x 4 x 4 4 x4 = x 4 81x x8 = x Example.a Simplify. Assume all variables represent nonnegative values. x x = x Example.b Simplify. Assume all variables represent nonnegative values. 9x 6 9x6 = x Example.c Simplify. Assume all variables represent nonnegative values. 5x6 y 56

57 5x6 y = 5x y Example.d Simplify. Assume all variables represent nonnegative values. 4x 4x y 4 y 4 = x y Example.e Simplify. Assume all variables represent nonnegative values. 8x 6 8x6 = x Example.f Simplify. Assume all variables represent nonnegative values. 4 65x8 y x8 y 1 = 5x y Example.g* Simplify. Assume all variables represent nonnegative values. x + x + 1 x + x + 1 = (x + 1) Example 4.a Let f(x) = x + 7. Find f(). Example 4.b Let f(x) = x + 7. Find f(18). = x + 1 f() = () + 7 = 9 = 57

58 Example 4.c Let f(x) = x + 7. Find f( 7). f(18) = (18) + 7 = 5 = 5 f( 7) = ( 7) + 7 = 0 = 0 Example 4.d Let f(x) = x + 7. Find f( 11). f( 11) = ( 11) + 7 = 4 However, 4 is not a real number, so therefore f( 11) is undefined. Example 5.a State the domain of f(x) = x + 7. For f(x) = x + 7 to be defined, x x 7. Thus, f(x) is defined for all real numbers x 7. The domain of f(x) is {x x 7}. Example 5.b State the domain of g(x) = 7x 4. For g(x) = 7x 4 to be defined, 7x 4 0 7x 4 x 6. Thus, g(x) is defined for all real numbers x 6. The domain of g(x) is {x x 6}. Example 5.c State the domain of h(x) = x + 5. For h(x) = x + 5 to be defined, 58

59 x x 5 x 5. Thus, h(x) is defined for all real numbers x 5. The domain of h(x) is { x x 5 }. 59

60 4. Radicals: Rational Exponents Definition: x 1/n We define x 1/n = n x. There are many reasons for this definition that have to do with our general intuition on how exponents should behave. We will develop these connections throughout the rest of the section and chapter. For now, we can just accept this definition as the one that works. Example 1.a Simplify. 9 1/ 9 1/ = 9 = Example 1.b Simplify. 64 1/ 64 1/ = 64 = 4 Example 1.c Simplify. 5 1/ 5 1/ = 5 = 5 Example 1.d Simplify. ( 5) 1/ ( 5) 1/ = 5 Thus, ( 5) 1/ is not a real number. Example 1.e Simplify. Assume all variables represent positive values. (81x ) 1/ (81x ) 1/ = 81x = 9x 60

61 Example 1.f Simplify. Assume all variables represent positive values. 4x 1/ 4x 1/ = 4 x Example 1.g Simplify. Assume all variables represent positive values. (4x) 1/ (4x) 1/ = 4x = x Example 1.h Simplify. Assume all variables represent positive values. ( ) 144x 8 1/ 169y 6 ( ) 144x 8 1/ 144x 8 = 169y 6 169y 6 Theorem: x m/n Proof. = 1x4 1y x m/n = ( n x ) m x m/n = x 1 n m = ( ) m x 1 n = ( n x) m Example.a Simplify. Assume all variables represent nonnegative values. 81 /4 81 /4 = ( 4 81 ) = = 7 61