Download Lecture Notes: Intermediate Algebra Contents Joseph Lee Metropolitan Community College

Lecture Notes: Intermediate Algebra
Joseph Lee
Metropolitan Community College
Contents
1 Linear Equations
1.1 Linear Equations
1.2 Linear Equations,
1.3 Functions . . . .
1.4 Systems of Linear
1.5 Systems of Linear
1.6 Systems of Linear
. . . . . . . . . . . . . .
Continued . . . . . . . .
. . . . . . . . . . . . . .
Equations . . . . . . . .
Equations, Continued .
Equations: Applications
2 Polynomials
2.1 Polynomials . . . . . . . . . .
2.2 Polynomials, Continued . . .
2.3 Factoring . . . . . . . . . . .
2.4 Factoring, Continued . . . . .
2.5 Factoring III . . . . . . . . . .
2.6 Factoring IV: An Application
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5
5
9
13
18
24
30
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32
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53
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54
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85
90
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94
94
99
105
109
109
3 Rational Expressions
3.1 Rational Expressions . . . . . . . . . . . . . . . . . . . .
3.2 Rational Expressions, Continued . . . . . . . . . . . . . .
3.3 Rational Expressions III: Complex Rational Expressions
3.4 Rational Expressions IV . . . . . . . . . . . . . . . . . .
4 Radicals
4.1 Radicals: An Introduction . .
4.2 Radicals: Rational Exponents
4.3 Radicals III . . . . . . . . . .
4.4 Radicals IV . . . . . . . . . .
4.5 Radicals V . . . . . . . . . . .
4.6 Radicals VI . . . . . . . . . .
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5 Quadratic Equations
5.1 The Complex Number System . . . . . . . . . . . . . . . . .
5.2 Quadratic Equations . . . . . . . . . . . . . . . . . . . . . .
5.3 Quadratic Equations: The Grand Finale . . . . . . . . . . .
5.4 Quadratic Equations III . . . . . . . . . . . . . . . . . . . .
5.5 Prelude To College Algebra: Graphing Quadratic Equations
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Preface
I created this set of notes during the winter quarter 2012 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 students1 , 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 2 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 5.23 .
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
3
Al-Khwarizmi “completes the square” geometrically
2
2
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.
2. Expand upon operations involving exponents, polynomials, and the methods of factoring.
3. 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.
3
6. Solve quadratic equations with real and complex solutions.
A test designed to assess the student’s mastery of objective 2, 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 2013 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 2013 Quarter
I made some revisions to a few of my lectures. In particular, I removed some examples from Sections
1.1 and 1.2 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
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
Ax + By = C,
where A, B, and C are real numbers.
Examples
1. 3x + 2y = 6
1
2. y = − x − 4
3
3. 4x = 20 + 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 = − 32 x + 3, determine the slope and the y-intercept. Then graph
the equation.
Solution: Since the equation is in slope-intercept form, we know the slope is m = − 23 and the
y-intercept is (0, 3).
5
.
Example 1.b For the equation y = 14 x − 2, determine the slope and the y-intercept. Then graph
the equation.
Solution: Since the equation is in slope-intercept form, we know the slope is m =
y-intercept is (0, −2).
1
4
and the
.
Example 1.c For the equation 4x − 3y = 9, determine the slope and the y-intercept. Then graph
the equation.
Solution:
4x − 3y = 9
−3y = −4x + 9
4
y = x−3
3
.
Example 1.d For the equation 5x + 2y = 6, determine the slope and the y-intercept. Then graph
the equation.
Solution:
5x + 2y = 6
2y = −5x + 6
5
y =− x+3
2
6
.
Example 2 Find the slope of the line passing through the points (x1 , y1 ) and (x2 , y2 ).
Solution:
m =
=
vertical change
horizontal change
y2 − y1
x2 − x1
.
Example 3.a Find the slope of the line connecting (1, 6) and (4, 2).
Solution:
m=
y2 − y1
x2 − x1
=
2−6
4−1
=
−4
3
=−
4
3
.
Example 3.b Find the slope of the line connecting (−2, −4) and (6, 12).
Solution:
m=
y2 − y1
x2 − x1
=
12 − (−4)
6 − (−2)
=
16
8
=2
.
Example 3.c Find the slope of the line connecting (−3, 2) and (4, 2).
7
Solution:
m=
y2 − y1
x 2 − x1
=
2−2
4 − (−3)
=
0
7
=0
.
Example 3.d Find the slope of the line connecting (3, −2) and (3, −8).
Solution:
m=
y2 − y1
x2 − x 1
=
−8 − (−2)
3−3
=
−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
1.2
Linear Equations, Continued
.
Example 1.a Write the equation of the line, in slope-intercept form, that has a slope of − 23 and a
y-intercept of (0, 4).
Solution:
2
y =− x+4
3
.
Example 1.b Write the equation of the line, in slope-intercept form, that has a slope of 3 and a
y-intercept of 0, − 23 .
Solution:
y = 3x −
3
2
.
Definition: Point-Slope Form An equation of the line with slope m passing through the point
(x1 , y1 ) is given by
y − y1 = m(x − x1 ).
The equation above is said to be in point-slope form.
.
Example 2.a Write the equation of the line, in slope-intercept form, with a slope of − 21 that passes
though the point (−4, 3).
Solution:
y − y1 = m(x − x1 )
1
y − 3 = − (x − (−4))
2
1
y − 3 = − (x + 4)
2
1
y−3= − x−2
2
1
y = − x+1
2
.
Example 2.b Write the equation of the line, in slope-intercept form, with a slope of −2 that passes
though the point (3, −8).
Solution:
y − y1 = m(x − x1 )
y − (−8) = −2(x − 3)
y + 8 = −2x + 6
y = −2x − 2
9
.
Example 3.a Write the equation of the line, in slope-intercept form, that passes though the points
(2, −3) and (−2, 1).
Solution: First we note the slope of the line is
m=
4
1 − (−3)
=
= −1.
−2 − 2
−4
Then the equation of the line with slope −1 and passing through (2, −3) is
y − (−3) = −(x − 2)
y + 3 = −x + 2
y = −x − 1.
.
Example 3.b Write the equation of the line, in slope-intercept form, that passes though the points
(5, −2) and (−1, 1).
Solution: First we note the slope of the line is
m=
1 − (−2)
3
1
=
=− .
−1 − 5
−6
2
Then the equation of the line with slope − 12 and passing through (5, −2) is
1
y − (−2) = − (x − 5)
2
1
5
y+2= − x+
2
2
1
1
y = − x+ .
2
2
.
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 = 2x − 3
2x − y = 3
2
y =− x+1
3
2x + 3y = 3
10
.
Example 4.a Write the equation of the line, in standard form, with slope
the point (−2, 2).
Solution:
y−2=
1
(x + 2)
3
y−2=
1
2
x+
3
3
y=
1
8
x+
3
3
1
3
that passes though
8
1
− x+y =
3
3
x − 3y = −8.
.
Example 4.b Write the equation of the line, in standard form, that passes though the points (4, 0)
and (−2, 3).
Solution: First we note the slope of the line is
m=
3
1
3−0
=
=− .
−2 − 4
−6
2
Then the equation of the line with slope − 12 and passing through (4, 0) is
1
y − 0 = − (x − 4)
2
1
y = − x+2
2
1
x+y = 2
2
x + 2y = 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
2x + y = 6 and passes though the point (3, 1).
11
Solution: Since our line is parallel to
2x + y = 6
y = −2x + 6,
we know the slope of our line must be −2.
Then the equation of the line with slope −2 and passing through (3, 1) is
y − 1 = −2(x − 3)
y − 1 = −2x + 6
y = −2x + 7.
.
Example 5.b Write the equation of the line, in standard form, that is perpendicular to the line
2x + 3y = 5 and passes though the point (0, 3).
Solution: Since our line is perpendicular to
2x + 3y = 5
y = − 32 x + 35 ,
we know the slope of our line must be 32 .
Then the equation of the line with slope
3
2
and y-intercept (0, 3) is
y=
3
x+3
2
3
− x+y = 3
2
3x − 2y = −6.
12
1.3
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 3 elements.
2. {1, 2, 3, 4} is the set of positive integers less than 5.
3. {−2, 2} is the set of solutions to the equation x2 = 4.
4. ∅ is the set containing no elements. For example, the set of real solutions to the equation
x2 = −4.
.
Example 1.b We may also want to talk about sets with an infinite number of elements.
1. {1, 2, 3, 4, 5, ...} could be the set of positive integers.
2. {3, 5, 7, ...} could be the set of odd numbers greater than 1: {3, 5, 7, 9, 11, 13, ...}
3. On the other hand, {3, 5, 7, ...} could be the set of odd prime numbers: {3, 5, 7, 11, 13, 17, ...}
.
Definition: Set-Builder Notation Set builder notation is a method to describe a set by its
properties.
.
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}
2. {x| x is an odd prime number} = {3, 5, 7, 11, 13, 17, ...}
3. {x| x is an integer & 0 < x < 5} = {1, 2, 3, 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.
13
.
Example 2.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}.
2. {(1, 3), (2, 4), (−1, 1)} is a relation of x and y values that satisfy the equation y = x + 2. The
domain of this relation is {−1, 1, 2}. The range of this relation is {1, 3, 4}.
3. {(3, 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 2.b Determine which of the following relations are functions.
1. {(Joseph, turkey), (Joseph, roast beef), (Michael, ham)}
2. {(1, 3), (2, 4), (−1, 1)}
3. {(3, 5), (4, 5), (5, 5)}.
Solution: 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 3.a The graph of the relation y = 3x − 2 is shown below.
1. Determine the domain of the relation. R (all real numbers)
2. Determine the range of the relation.
R (all real numbers)
3. Determine if the relation is represents a function. Yes, it is a function.
14
.
Example 3.b The graph of the relation x = y 2 − 1 is shown below.
1. Determine the domain of the relation. {x| x ≥ −1}
2. Determine the range of the relation.
R (all real numbers)
3. Determine if the relation is represents a function. No, it is not a function.
.
Example 3.c The graph of the relation y = x2 − 2x − 3 is shown below.
1. Determine the domain of the relation. R (all real numbers)
2. Determine the range of the relation.
{y| y ≥ −4}
3. 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 2x + 3y = 6 in function notation.
15
Solution:
2x + 3y = 6
3y = −2x + 6
2
y = − x+2
3
2
f (x) = − x + 2
3
.
2
Example 4.b Evaluate the function f (x) = − x + 2 for the following values.
3
1. f (−3) = 4
2. f (0) = 2
3. f (1) =
4
3
4. f (3) = 0
.
√
Example 4.c Evaluate the function f (x) = x + 6 for the following values.
1. f (−2) = 2
2. f (19) = 5
3. f (−7) is undefined
.
Example 5.a The graph of the function f is shown below.
Evaluate.
16
1. f (−4) = −3
2. f (−1) = 0
3. f (1) = 2
4. f (4) = 1
.
Example 5.b The graph of the function f is shown below.
Evaluate.
1. f (−5) = 0
2. f (−1) = −2
3. f (2) = 0
17
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 − 2y = 8
4x + y = 5
Solution. First, we will check if (0, −4) is a solution of x − 2y = 8.
x − 2y = 8
?
(0) − 2(−4) = 8
8=8
Thus, (0, −4) is a solution of x − 2y = 8, so now we will check if it is also a solution of 4x + y = 5.
4x + y = 5
?
4(0) + (−4) = 5
−4 6= 5
Thus, (0, −4) is not a solution of this system of equations.
.
Example 1.b Determine if (2, −3) is a solution to the following system of linear equations:
x − 2y = 8
4x + y = 5
Solution. First, we will check if (2, −3) is a solution of x − 2y = 8.
x − 2y = 8
?
(2) − 2(−3) = 8
8=8
Thus, (2, −3) is a solution of x − 2y = 8, so now we will check if it is also a solution of 4x + y = 5.
4x + y = 5
?
4(2) + (−3) = 5
5=5
Thus, (2, −3) is a solution of this system of equations.
.
Example 1.c Graph both equations on the same coordinate plane.
x − 2y = 8
4x + y = 5
18
The line for x − 2y = 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 2.a Determine if (3, 2) is a solution to the following system of linear equations:
2x + 3y = 12
y = − 23 x + 4
Solution. First, we will check if (3, 2) is a solution of 2x + 3y = 12.
2x + 3y = 12
?
2(3) + 3(2) = 12
12 = 12
Thus, (3, 2) is a solution of 2x + 3y = 12, so now we will check if it is also a solution of y = − 32 x + 4.
y = − 23 x + 4
?
(2) = − 23 (3) + 4
2=2
Thus, (3, 2) is a solution of this system of equations.
.
Example 2.b Determine if (9, −2) is a solution to the following system of linear equations:
2x + 3y = 12
y = − 32 x + 4
Solution. First, we will check if (9, −2) is a solution of 2x + 3y = 12.
2x + 3y = 12
?
2(9) + 3(−2) = 12
12 = 12
Thus, (9, −2) is a solution of 2x+3y = 12, so now we will check if it is also a solution of y = − 32 x+4.
y = − 32 x + 4
?
(−2) = − 32 (9) + 4
−2 = −2
Thus, (9, −2) is a solution of this system of equations.
.
Example 2.c Graph both equations on the same coordinate plane.
2x + 3y = 12
y = − 23 x + 4
19
Notice the equations 2x + 3y = 12 and y = − 32 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 2x + 3y = 12.
.
Example 3 Graph both equations on the same coordinate plane.
y = −3x + 4
3x + y = −1
Notice the lines represented by the equations y = −3x + 4 and 3x + 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
2. Substitution
3. Elimination
.
Example 4.a Solve the following system of equations using substitution.
x − 2y = 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 − 2y = 8 and solve for x.
x − 2y = 8
x = 2y + 8
Now, we may substitute x = 2y + 8 into our other equation 4x + y = 5.
4x + y
4(2y + 8) + y
8y + 32 + y
9y
y
20
=5
=5
=5
= −27
= −3
Since y = −3, we may substitute that value back into any equation to solve for x.
x = 2y + 8
x = 2(−3) + 8
x=2
Thus, (2, −3) is the solution of this system of equations.
.
Example 4.b Solve the following system of equations using substitution.
5x + 3y = 6
4x + 2y = 2
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 + 2y = 2 and solve for y.
4x + 2y = 2
2y = −4x + 2
y = −2x + 1
Now, we may substitute y = −2x + 1 into our other equation 5x + 3y = 6.
5x + 3y
5x + 3(−2x + 1)
5x − 6x + 3
−x
x
=6
=6
=6
=3
= −3
Since x = −3, we may substitute that value back into any equation to solve for y.
y = −2x + 1
y = −2(−3) + 1
y =7
Thus, (−3, 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 =3
x+y =7
or
x+y =7
x−y =1
(x + y) + (x − y) = 8
2x = 8
.
Example 5.a Solve the following system of equations using elimination.
x − 2y = 8
4x + y = 5
21
Solution. To solve by elimination, we will eliminate one of our variables. Let’s eliminate y.
x − 2y = 8
4x + y = 5
x − 2y = 8
2(4x + y) = (5)2
x − 2y = 8
8x + 2y = 10
9x = 18
x=2
Since x = 2, we may substitute that value back into any equation to solve for y.
4x + y
4(2) + y
8+y
y
=5
=5
=5
= −3
Thus, (2, −3) is the solution of this system of equations.
.
Example 5.b Solve the following system of equations using elimination.
5x + 3y = 6
4x + 2y = 2
Solution. To solve by elimination, we will eliminate one of our variables. Let’s eliminate y.
5x + 3y = 6
4x + 2y = 2
2(5x + 3y) = (6)2
−3(4x + 2y) = (2)(−3)
10x + 6y = 12
−12x − 6y = −6
−2x = 6
x = −3
Since x = −3, we may substitute that value back into any equation to solve for y.
4x + 2y
4(−3) + 2y
−12 + 2y
2y
y
=2
=2
=2
= 14
=7
Thus, (−3, 7) is the solution of this system of equations.
22
.
Example 6.a Solve the following system of equations using elimination.
4x + 6y = 24
y = − 23 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 = 24
3(y) = − 32 x + 4 3
4x + 6y = 24
3y = −2x + 12
4x + 6y = 24
2x + 3y = 12
4x + 6y = 24
−2(2x + 3y) = (12)(−2)
4x + 6y = 24
−4x − 6y = −24
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 = 24.
.
Example 6.b Solve the following system of equations using elimination.
y = −3x + 4
6x + 2y = −2
Solution. First, we will move our variables over to the left side. Then we may choose a variable
to eliminate.
y = −3x + 4
6x + 2y = −2
3x + y = 4
6x + 2y = −2
−2(3x + y) = (4)(−2)
6x + 2y = −2
−6x − 2y = −8
6x + 2y = −2
0 = −10
Arriving at this contradiction, we may conclude that this system is inconsistent. There is no
solution.
23
1.5
Systems of Linear Equations, Continued
.
Example 1.a Solve the following system of equations.
x+y+z =6
2x − 3y − z = 5
3x + 2y − 2z = −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
2x − 3y − z = 5
3x − 2y = 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
3x + 2y − 2z = −1
2(x + y + z) = (6)2
3x + 2y − 2z = −1
2x + 2y + 2z = 12
3x + 2y − 2z = −1
5x + 4y = 11
We now have two equations with two variables. We may proceed as usual from here.
3x − 2y = 11
5x + 4y = 11
2(3x − 2y) = (11)2
5x + 4y = 11
6x − 4y = 22
5x + 4y = 11
11x = 33
x=3
We substitute x = 3 into an equation containing only x and y.
3x − 2y
3(3) − 2y
9 − 2y
−2y
y
24
= 11
= 11
= 11
=2
= −1
Similarly, we substitute x = 3 and y = −1 into an equation containing x, y, and z.
x+y+z
(3) + (−1) + z
2+z
z
=6
=6
=6
=4
Thus, (3, −1, 4) is the solution to this system of equations.
.
Example 1.b Solve the following system of equations.
3x + 4y = −4
5y + 3z = 1
2x − 5z = 7
Solution. First, we will take two equations and eliminate a variable. Let’s take the last two
equations and eliminate z.
5y + 3z = 1
2x − 5z = 7
5(5y + 3z) = (1)5
3(2x − 5z) = (7)3
25y + 15z = 5
6x − 15z = 21
6x + 25y = 26
Notice we have two equations with two variables.
3x + 4y = −4
6x + 25y = 26
−2(3x + 4y) = (−4)(−2)
6x + 25y = 26
−6x − 8y = 8
6x + 25y = 26
17y = 34
y =2
We may now substitute y = 2 into an equation only containing x and y to solve for x or into an
25
equation containing y and z to solve for z.
3x + 4y
3x + 4(2)
3x + 8
3x
x
= −4
= −4
= −4
= −12
= −4
5y + 3z
5(2) + 3z
10 + 3z
3z
z
=1
=1
=1
= −9
= −3
Thus, (−4, 2, −3) is the solution to the system of equations.
.
Example 1.c Solve the following system of equations.
2x + 3y − z = 7
−3x + 2y − 2z = 7
5x − 4y + 3z = −10
Solution. First, we will take two equations and eliminate a variable. Let’s take the first two
equations and eliminate z.
2x + 3y − z = 7
−3x + 2y − 2z = 7
−2(2x + 3y − z) = (7)(−2)
−3x + 2y − 2z = 7
−4x − 6y + 2z = −14
−3x + 2y − 2z = 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.
3(2x + 3y − z) = (7)3
5x − 4y + 3z = −10
6x + 9y − 3z = 21
5x − 4y + 3z = −10
11x + 5y = 11
26
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
−35x − 20y = −35
44x + 20y = 44
9x = 9
x=1
We substitute x = 1 into an equation containing only x and y.
−7x − 4y
−7(1) − 4y
−7 − 4y
−4y
y
= −7
= −7
= −7
=0
=0
Similarly, we substitute x = 1 and y = 0 into an equation containing x, y, and z.
2x + 3y − z
2(1) + 3(0) − z
2−z
−z
z
=7
=7
=7
=5
= −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 2.a Solve the following system of equations.
x + 2y − z = 3
2x + 3y − 5z = 3
5x + 8y − 11z = 9
Solution. First, we will take two equations and eliminate a variable. Let’s take the first two
27
equations and eliminate x.
x + 2y − z = 3
2x + 3y − 5z = 3
−2(x + 2y − z) = (3)(−2)
2x + 3y − 5z = 3
−2x − 4y + 2z = −6
2x + 3y − 5z = 3
−y − 3z = −3
Now, we want to take two different equations but eliminate the same variable: x. We will take the
first and last equations.
−5(x + 2y − z) = (3)(−5)
5x + 8y − 11z = 9
−5x − 10y + 5z = −15
5x + 8y − 11z = 9
−2y − 6z = −6
We now have two equations with two variables. We may proceed as usual from here.
y − 3z = −3
−2y − 6z = −6
−2(−y − 3z) = (−3)(−2)
−2y − 6z = −6
2y + 6z = 6
−2y − 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 2.b Solve the following system of equations.
3x − y + 2z = 4
x − 5y + 4z = 3
6x − 2y + 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.
28
equations and eliminate x.
3x − y + 2z = 4
6x − 2y + 4z = −8
−2(3x − y + 2z) = (4)(−2)
6x − 2y + 4z = −8
−6x + 2y − 4z = −8
6x − 2y + 4z = −8
0 = −16
Arriving at this contradiction, we conclude that this is an inconsistent system of equations, and
thus it has no solution.
29
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 $3.70. Joseph knows he has 25 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 $3.70, and the other equations should use the fact that Joseph has
25 coins. If x represents the number of quarters Joseph possesses and y represents the number of
dimes Joseph possesses, we may write:
.25x + .10y = 3.70
x + y = 25
Solve the system using substitution or elimination.
.25x + .10y = 3.70
x + y = 25
−10(.25x + .10y) = (3.70)(−10)
x + y = 25
−2.5x − y = −37
x + y = 25
−1.5x = −12
x=8
x + y = 25
8 + y = 25
y = 17
Joseph has 8 quarters and 17 dimes.
.
Example 2 Ten liters of a 12% HCl solution is mixed with a 20% HCl solution to make a mixture
that is 15% HCl. How many liters of the 20% 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 20% 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
.20x + .12(10) = .15y
30
Solve the system using substitution or elimination.
x + 10 = y
.20x + .12(10) = .15y
.20x + .12(10)
.20x + 1.2
.05x
x
= .15(x + 10)
= .15x + 1.5
= 0.3
=6
6 liters or 20% solution were used to make the mixture.
.
Example 3 Against the wind, a plane can fly 2880 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 = 2880
(x + y)(4.5) = 2880
Solve the system using substitution or elimination.
(x − y)5 = 2880
(x + y)(4.5) = 2880
5x − 5y = 2880
4.5x + 4.5y = 2880
9(5x − 5y) = (2880)9
10(4.5x + 4.5y) = (2880)10
45x − 45y = 25920
45x + 45y = 28800
90x = 54720
x = 608
5x − 5y
5(608) − 5y
3040 − 5y
−5y
y
= 2880
= 2880
= 2880
= −160
= 32
The speed of the plane in still air is 608 miles per hour and the speed of the wind is 32 miles per
hour.
31
2
Polynomials
2.1
Polynomials
.
Definition: Polynomial A polynomial is a mathematical expression containing the operations of
addition, subtraction, and multiplication.
Examples
1. 4x5 − 3x2 + 4
2. 3m4 n2 p3
3.
1
3
x−
2
4
4. x − 3y + 2z − 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.
7x4 + 1
Solution: The polynomial has 2 terms, so it is called a binomial.
Term
7x4
1
Coefficient
7
1
32
Degree
4
0
Finally, 7x4 + 1 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.
1
4x3 − 7x2 + x − 2
2
Solution: The polynomial has 4 terms, so it does not have a special name.
Term
4x3
−7x2
1
x
2
−2
Coefficient
4
−7
1
2
−2
Degree
3
2
1
0
1
Finally, 4x3 − 7x2 + x − 2 is a third degree polynomial.
2
.
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.
−3.2x4 y 2 z 3 + 12.3x3 yz 2 − 7.9y 2 z 4
Solution: The polynomial has 3 terms, so it is called a trinomial.
Term
−3.2x4 y 2 z 3
12.3x3 yz 2
−7.9y 2 z 4
Coefficient
−3.2
12.3
−7.9
Degree
9
6
6
Finally, −3.2x4 y 2 z 3 + 12.3x3 yz 2 − 7.9y 2 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 2.a Add the polynomials.
(4x2 − 7x + 12) + (3x2 − 8x − 9)
Solution:
(4x2 − 7x + 12) + (3x2 − 8x − 9) = 7x2 − 15x + 3
.
Example 2.b Subtract the polynomials.
(4x2 − 7x + 12) − (3x2 − 8x − 9)
33
Solution:
(4x2 − 7x + 12) − (3x2 − 8x − 9) = 4x2 − 7x + 12 − 3x2 + 8x + 9
= x2 + x + 21
.
Example 2.c Add the polynomials.
10
2 3
5
7
3 3 1 2
2
x − x − 5x +
+
x − 4x + x +
4
2
3
3
4
2
Solution:
3 3 1 2
10
x − x − 5x +
4
2
3
=
=
+
2 3
5
7
x − 4x2 + x +
3
4
2
20
9 3 1 2 20
x − x − x+
12
2
4
6
+
8 3 8 2 5
21
x − x + x+
12
2
4
6
17 3 9 2 15
41
x − x − x+
12
2
4
6
.
Example 2.d Subtract the polynomials.
7
1
3
x+
−
2x −
4
8
3
Solution:
3
2x −
4
−
1
7
x+
8
3
= 2x −
1
3 7
− x−
4 8
3
=
9
7
4
16
x−
− x−
8
12 8
12
=
13
9
x−
8
12
.
Example 2.e Subtract the polynomials.
(−3.4x3 y 2 + 7.8x2 y + 9.12x − 4.3y) − (2.78x3 y 2 − 2.9x2 y + 4.75y − 5.3)
Solution:
(−3.4x3 y 2 + 7.8x2 y + 9.12x − 4.3y)
−(2.78x3 y 2 − 2.9x2 y + 4.75y − 5.3)
= −3.4x3 y 2 + 7.8x2 y + 9.12x − 4.3y
−2.78x3 y 2 + 2.9x2 y − 4.75y + 5.3
= −6.18x3 y 2 + 10.7x2 y + 9.12x − 9.05y + 5.3
34
2.2
Polynomials, Continued
.
Example 1.a Multiply.
3x2 (4x3 − 7x + 2)
Solution:
3x2 (4x3 − 7x + 2) = 3x2 (4x3 ) + 3x2 (−7x) + 3x2 (2)
= 12x5 − 21x3 + 6x2
.
Example 1.b Multiply.
(2x + 7)(3x − 2)
Solution:
(2x + 7)(3x − 2) = 6x2 − 4x + 21x − 14
= 6x2 + 17x − 14
.
Example 1.c Multiply.
(x3 + 1)(x3 − 8)
Solution:
(x3 + 1)(x3 − 8) = x6 − 8x3 + x3 − 8
= x6 − 7x3 − 8
.
Theorem. Square of a Binomial For any a and b,
(a + b)2 = a2 + 2ab + b2 .
Proof.
(a + b)2 = (a + b)(a + b)
= a2 + ab + ab + b2
= a2 + 2ab + b2
.
Example 2.a Multiply.
(x + 4)2
Solution:
(x + 4)2 = (x)2 + 2(x)(4) + (4)2
= x2 + 8x + 16
.
Example 2.b Multiply.
(x − 4)2
35
Solution:
2
(x − 4)2 = x + (−4)
= (x)2 + 2(x)(−4) + (−4)2
= x2 − 8x + 16
.
Example 2.c Multiply.
(3x2 y + 7z)2
Solution:
(3x2 y + 7z)2 = (3x2 y)2 + 2(3x2 y)(7z) + (7z)2
= 9x4 y 2 + 42x2 yz + 49z 2
.
Example 3.a Multiply.
(a + b + c)2
Solution:
2
(a + b + c)2 = (a + b) + c
= (a + b)2 + 2(a + b)(c) + (c)2
= (a + b)2 + 2(ac + bc) + c2
= (a + b)2 + 2ac + 2bc + c2
= a2 + 2ab + b2 + 2ac + 2bc + c2
.
Example 3.b Multiply.
(3x − 2y − 4z)2
Solution:
2
(3x − 2y − 4z)2 = (3x − 2y) + (−4z)
= (3x − 2y)2 + 2(3x − 2y)(−4z) + (−4z)2
= (3x − 2y)2 + 2(−12xz + 8yz) + 16z 2
= (3x − 2y)2 − 24xz + 16yz + 16z 2
= 9x2 − 12xy + 4z 2 − 24xz + 16yz + 16z 2
.
Example 3.c Multiply.
(x − 2y + 4z + 7)2
36
Solution:
(x − 2y + 4z + 7)2
2
= (x − 2y) + (4z + 7)
= (x − 2y)2 + 2(x − 2y)(4z + 7) + (4z + 7)2
= (x − 2y)2 + 2(4xz + 7x − 8yz − 14y) + (4z + 7)2
= (x − 2y)2 + 8xz + 14x − 16yz − 28y + (4z + 7)2
= x2 − 4xy + 4y 2 + 8xz + 14x − 16yz − 28y + 16z 2 + 56z + 49
.
Theorem. Product of Conjugates For any a and b,
(a + b)(a − b) = a2 − b2 .
Proof.
(a + b)(a − b) = (a + b)(a − b)
= a2 − ab + ab − b2
= a2 − b 2
.
Example 4.a Multiply.
(x + 9)(x − 9)
Solution:
(x + 9)(x − 9) = (x)2 − (9)2
= x2 − 81
.
Example 4.b Multiply.
(3m2 − 8n2 )(3m2 + 8n2 )
Solution:
(3m2 − 8n2 )(3m2 + 8n2 ) = (3m2 )2 − (8n2 )2
= 9m4 − 64n4
.
Example 4.c Multiply.
(6x2 y − 5z 3 )(6x2 y + 5z 3 )
Solution:
(6x2 y − 5z 3 )(6x2 y + 5z 3 ) = (6x2 y)2 − (5z 3 )2
= 36x4 y 2 − 25z 6
.
Example 5.a Multiply.
(x + y + z)(x + y − z)
37
Solution:
(x + y + z)(x + y − z) = (x + y) + z (x + y) − z
= (x + y)2 − (z)2
= x2 + 2xy + y 2 − z 2
.
Example 5.b Multiply.
(x − y + z)(x + y − z)
Solution:
(x − y + z)(x + y − z) = x − (y − z) x + (y − z)
= (x)2 − (y − z)2
= x2 − (y 2 − 2yz + z 2 )
= x2 − y 2 + 2yz − z 2
.
Example 5.c Multiply.
(3x2 − y + z + 1)(3x2 − y − z − 1)
Solution:
(3x2 − y + z + 1)(3x2 − y − z − 1)
= (3x2 − y) + (z + 1) (3x2 − y) − (z + 1)
= (3x2 − y)2 − (z + 1)2
= (9x4 − 6x2 y + y 2 ) − (z 2 + 2z + 1)
= 9x4 − 6x2 y + y 2 − z 2 − 2z − 1
38
2.3
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 3x · 2y = 6xy, 3x and 2y 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 = 3x · 2y and 12x2 = 3x · 4x, 3x is a common factor of 6xy and 12x2 .
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.
15x3 − 25x2 y
Solution: First, we note the greatest common factor of 15x3 and −25x2 y is 5x2 .
15x3 − 25x2 y = 5x2 (3x) + 5x2 (−5y)
= 5x2 (3x − 5y)
.
Example 1.b Factor.
14x2 + 35x − 21
Solution: First, we note the greatest common factor of 14x2 , 35x, and −21 is 7.
14x2 + 35x − 21 = 7(2x2 ) + 7(5x) + 7(−3)
= 7(2x2 + 5x − 3)
.
Example 1.c Factor.
12p2 q 4 r2 − 6p3 q 2 r2 + 20pq 4 r3
Solution: First, we note the greatest common factor of 12p2 q 4 r2 , −6p3 q 2 r2 , and 20pq 4 r3 is 2pq 2 r2 .
12p2 q 4 r2 − 6p3 q 2 r2 + 20pq 4 r3
= 2pq 2 r2 (6pq 2 ) + 2pq 2 r2 (−3p2 ) + 2pq 2 r2 (10q 2 r)
= 2pq 2 r2 (6pq 2 − 3p2 + 10q 2 r)
.
Example 2.a Factor.
4x(3x − 1) − y(3x − 1)
39
Solution: First, we note the greatest common factor of 4x(3x − 1) and −y(3x − 1) is 3x − 1.
4x(3x − 1) − y(3x − 1) = (3x − 1)(4x) + (3x − 1)(−y)
= (3x − 1)(4x − y)
.
Example 2.b Factor.
4x2 (x2 + 2) − 3x(x2 + 2)
Solution: First, we note the greatest common factor of 4x2 (x2 + 2) and −3x(x2 + 2) is x(x2 + 2).
4x2 (x2 + 2) − 3x(x2 + 2) = [x(x2 + 2)](4x) + [x(x2 + 2)](−3)
= x(x2 + 2)(4x − 3)
.
Example 2.c Factor.
(3x − 2)(x + 3) + y(x + 3)
Solution: First, we note the greatest common factor of (3x − 2)(x + 3) and y(x + 3) is x + 3.
(3x − 2)(x + 3) + y(x + 3) = (x + 3)(3x − 2) + (x + 3)(y)
= (x + 3)[(3x − 2) + y]
= (x + 3)(3x − 2 + y)
.
Example 3.a Factor.
12x3 − 9x2 + 8x − 6
Solution:
12x3 − 9x2 + 8x − 6 = (12x3 − 9x2 ) + (8x − 6)
= 3x2 (4x − 3) + 2(4x − 3)
= (4x − 3)(3x2 + 2)
.
Example 3.b Factor.
xy + 3y − 5x − 15
Solution:
xy + 3y − 5x − 15 = (xy + 3y) + (−5x − 15)
= y(x + 3) − 5(x + 3)
= (x + 3)(y − 5)
.
Example 3.c Factor.
7x2 y − 14xy + 28xy − 56y
40
Solution:
7x2 y − 14xy + 28xy − 56y = 7y(x2 − 2x + 4x − 8)
= 7y[(x2 − 2x) + (4x − 8)]
= 7y[x(x − 2) + 4(x − 2)]
= 7y(x − 2)(x + 4)
.
Example 3.d Factor.
8mn − 12m − 12n + 18
Solution:
8mn − 12m − 12n + 18 = 2(4mn − 6m − 6n + 9)
= 2[(4mn − 6m) + (−6n + 9)]
= 2[2m(2n − 3) − 3(2n − 3)]
= 2(2n − 3)(2m − 3)
41
2.4
Factoring, Continued
.
Example 1.a Multiply.
(3x − 7)(2x + 5)
Solution:
(3x − 7)(2x + 5) = 6x2 + 15x − 14x − 35
= 6x2 + x − 35
.
Example 1.b Factor.
6x2 + x − 35
Solution:
6x2 + x − 35 = (3x − 7)(2x + 5)
.
Example 2.a Factor.
2x2 + 3x − 5
Solution:
2x2 + 3x − 5 = (2x
)(x
)
We need factors of 5, and the only factors of 5 are 1 and 5. Observe, our two choices:
(2x 1)(x 5) = 2x2
(2x 5)(x 1) = 2x2
10x x 5
2x 5x 5
The only way to come up with 2x2 + 3x − 5 would be
(2x 5)(x 1) = 2x2 − 2x + 5x 5.
Finally,
(2x + 5)(x − 1) = 2x2 + 3x − 5.
.
Example 2.b Factor.
3x2 + 11x − 20
Solution:
3x2 + 11x − 20 = (3x
We need factors of 20:
)(x
)
20 = 1 · 20
2 · 10
4·5
We have the following choices:
(3x 1)(x 20) =
(3x 20)(x 1) =
(3x 2)(x 10) =
(3x 10)(x 2) =
(3x − 4)(x + 5) =
(3x 5)(x 4) =
3x2 60x x 20
3x2 3x 20x 20
3x2 30x 2x 20
3x2 6x 10x 20
3x2 + 15x − 4x − 20
3x2 12x 5x 20
42
.
Example 2.c Factor.
5x2 − 13x + 6
Solution:
5x2 − 13x + 6 = (5x
We need factors of 6:
)(x
)
6= 1·6
2·3
We have the following choices:
(5x
(5x
(5x
(5x
1)(x
6)(x
2)(x
3)(x
6) =
1) =
3) =
2) =
5x2
5x2
5x2
5x2
30x
5x
15x
10x
x
6x
2x
3x
6
6
6
6
Our choice is:
(5x − 3)(x − 2) = 5x2 − 10x − 3x + 6
.
Example 2.d Factor.
4x2 + 12x − 7
Solution: Note we have two possibilities to begin with:
4x2 + 12x − 7 = (4x
= (2x
)(x
)(2x
)
)
The only factors of 7, however, are 1 and 7.
Thus, we have the following choices:
(4x 1)(x 7) = 4x2
(4x 7)(x 1) = 4x2
(2x 1)(2x 7) = 4x2
28x x 7
4x 7x 7
14x 2x 7
Our choice is:
(2x − 1)(2x + 7) = 4x2 + 14x − 2x − 7
.
Example 2.e Factor.
6x2 − 27x + 12
Solution: We should first observe that we may factor out a common factor.
6x2 − 27x + 12 = 3(2x2 − 9x + 4)
= 3(2x
)(x
)
We need factors of 4:
4= 1·4
2·2
43
We have the following choices:
3(2x 1)(x 4) = 3(2x2
3(2x 4)(x 1) = 3(2x2
3(2x 2)(x 2) = 3(2x2
8x x 4)
2x 4x 4)
4x 2x 4)
Our choice is:
3(2x − 1)(x − 4) = 3(2x2 − 8x − x + 4)
.
Example 2.f Factor.
9m4 n − 15m3 n2 − 6m2 n3
Solution: We should first observe that we may factor out a common factor.
9m4 n − 15m3 n2 − 6m2 n3 = 3m2 n(3m2 − 5mn − 2n2 )
= 3m2 n(3m
)(m
)
The only factors of 2n2 that we need to consider are 2n and n.
We have the following choices:
3m2 n(3m n)(m 2n) = 3m2 n(3m2
3m2 n(3m 2n)(m n) = 3m2 n(3m2
6mn mn 2n2 )
3mn 2mn 2n2 )
Our choice is:
3m2 n(3m + n)(m − 2n) = 3m2 n(3m2 − 6mn + mn − 2n2 )
.
Example 3.a Factor.
(x + 7)2 − 6(x + 7) − 16
Solution: Let u = x + 7.
(x + 7)2 − 6(x + 7) − 16 = u2 − 6u − 16
= (u − 8)(u + 2)
= [(x + 7) − 8][(x + 7) + 2]
= (x − 1)(x + 9)
.
Example 3.b Factor.
4x6 − 4x3 − 15
Solution: Let u = x3 .
4x6 − 4x3 − 15 = 4u2 − 4u − 15
= (2u
)(2u
)
= (2u + 3)(2u − 5)
= (2x3 + 3)(2x3 − 5)
44
.
Example 3.c Factor.
4(2n2 + 1)2 − 7(2n2 + 1) + 3
Solution: Let u = 2n2 + 1.
4(2n2 + 1)2 − 7(2n2 + 1) + 3 = 4u2 − 7u + 3
= (4u
)(u
)
= (4u − 3)(u − 1)
= [4(2n2 + 1) − 3][(2n2 + 1) − 1]
= (8n2 + 4 − 3)(2n2 )
= (8n2 + 1)(2n2 )
45
2.5
Factoring III
.
Factoring III Recall two theorems from multiplying polynomials:
(a + b)2 = a2 + 2ab + b2
(a + b)(a − b) = a2 − b2
.
Example 1.a Factor.
4x2 + 28x + 49
Solution:
4x2 + 28x + 49 = (2x)2 + 2(2x)(7) + (7)2
= (2x + 7)2
.
Example 1.b Factor.
9x2 − 30x + 25
Solution:
9x2 − 30x + 25 = (3x)2 + 2(3x)(−5) + (−5)2
= (3x − 5)2
.
Example 1.c Factor.
100x4 y 2 + 60x2 yz 3 + 9z 6
Solution:
100x4 y 2 + 60x2 yz 3 + 9z 6 = (10x2 y)2 + 2(10x2 y)(3z 3 ) + (3z 3 )2
= (10x2 y + 3z 3 )2
.
Example 1.d Factor.
32x2 − 80xy + 50y 2
Solution:
32x2 − 80xy + 50y 2 = 2(16x2 − 40xy + 25y 2 )
= 2[(4x)2 + 2(4x)(−5y) + (−5y)2 ]
= 2(4x − 5y)2
.
Example 2.a Factor.
x2 − 16
Solution:
x2 − 16 = (x)2 − (4)2
= (x + 4)(x − 4)
46
.
Example 2.b Factor.
4m8 − 9n2
Solution:
4m8 − 9n2 = (2m4 )2 − (3n)2
= (2m4 + 3n)(2m4 − 3n)
.
Example 2.c Factor.
12x3 − 3x
Solution:
12x3 − 3x = 3x(4x2 − 1)
= 3x[(2x)2 − (1)2 ]
= 3x(2x + 1)(2x − 1)
.
Example 2.d Factor.
x4 − 16
Solution:
x4 − 16 = (x2 )2 − (4)2
= (x2 + 4)(x4 − 4)
= (x2 + 4)[(x2 )2 − (2)2 ]
= (x2 + 4)(x + 2)(x − 2)
.
Example 3.a Factor.
(x + y)2 + 2(x + y) + 1
Solution: Let u = x + y.
(x + y)2 + 2(x + y) + 1 = u2 + 2u + 1
= (u + 1)2
= [(x + y) + 1]2
= (x + y + 1)2
.
Example 3.b Factor.
25 − (3z + 4)2
47
Solution: Let u = 3z + 4.
25 − (3z + 4)2 = 25 − u2
= (5 + u)(5 − u)
= [5 + (3z + 4)][5 − (3z + 4)]
= (5 + 3z + 4)(5 − 3z − 4)
= (3z + 9)(−3z + 1)
= 3(z + 3)(−3z + 1)
.
Theorem. Factoring the Sum of Cubes For any a and b,
a3 + b3 = (a + b)(a2 − ab + b2 ).
Proof.
(a + b)(a2 − ab + b2 ) = (a + b)(a2 ) + (a + b)(−ab) + (a + b)(b2 )
= a3 + a2 b − a2 b − ab2 + ab2 + b3
= a3 + b 3
.
Theorem. Factoring the Difference of Cubes For any a and b,
a3 − b3 = (a − b)(a2 + ab + b2 ).
Proof. Left for the student.
.
Example 4.a Factor.
8x3 + 27
Solution:
8x3 + 27 = (2x)3 + (3)3
= (2x + 3)[(2x)2 − (2x)(3) + (3)2 ]
= (2x + 3)(4x2 − 6x + 9)
.
Example 4.b Factor.
1 − 64x3
Solution:
1 − 64x3 = (1)3 − (4x)3
= (1 − 4x)[(1)2 + (1)(4x) + (4x)2 ]
= (1 − 4x)(1 + 4x + 16x2 )
48
.
Example 4.c Factor.
125x6 − 27y 3 z 9
Solution:
125x6 − 27y 3 z 9 = (5x2 )3 + (3yz 3 )3
= (5x2 + 3yz 3 )[(5x2 )2 − (5x2 )(3yz 3 ) + (3yz 3 )2 ]
= (5x2 + 3yz 3 )(25x4 − 15x2 yz 3 + 9z 6 )
.
Example 5.a Factor.
216 − (a + b)3
Solution: Let u = a + b.
216 − (a + b)3 = 216 − u3
= (6)3 − u3
= (6 − u)(36 + 6u + u2 )
= [6 − (a + b)][36 + 6(a + b) + (a + b)2 ]
= (6 − a − b)(36 + 6a + 6b + a2 + 2ab + b2 )
.
Example 5.b Factor.
(3x − 2y)3 + (x + 1)3
Solution: Let u = a + b and let v = x + 1.
(3x − 2y)3 + (x + 1)3
= u3 + v 3
= (u + v)(u2 − uv + v 2 )
= [(3x − 2y) + (x + 1)]
[(3x − 2y)2 − (3x − 2y)(x + 1) + (x + 1)2 ]
= (4x − 2y + 1)
[(9x2 − 12xy + 4y 2 ) − (3x2 + 3x − 2xy − 2y) + (x2 + 2x + 1)]
= (4x − 2y + 1)
(9x2 − 12xy + 4y 2 − 3x2 − 3x + 2xy + 2y + x2 + 2x + 1)
= (4x − 2y + 1)(7x2 − 10xy + 4y 2 − x + 2y + 1)
49
2.6
Factoring IV: An Application
.
Example 1.a Solve.
x2 − 5x + 6 = 0
Solution:
x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0
Thus, by the zero factor property, either
x−2= 0
or
x−3= 0
x=2
x=3
The solution set is {2, 3}.
.
Example 1.b Solve.
3x2 − 10x − 8 = 0
Solution:
3x2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
Thus, by the zero factor property, either
3x + 2 = 0
or
x−4= 0
3x = −2
x=−
The solution set is
x=4
2
3
2
− ,4 .
3
.
Example 1.c Solve.
3x3 − 3x2 − 6x = 0
Solution:
3x3 − 3x2 − 6x = 0
3x(x2 − x − 2) = 0
3x(x + 1)(x − 2) = 0
Thus, by the zero factor property, either
3x = 0
x=0
or
x+1= 0
x = −1
50
or
x−2= 0
x=2
The solution set is {−1, 0, 2}.
.
Example 2.a Solve.
2x2 = 5x + 3
Solution:
2x2 = 5x + 3
2x2 − 5x − 3 = 0
(2x + 1)(x − 3) = 0
Thus, by the zero factor property, either
2x + 1 = 0
x=−
The solution set is
or
x−3= 0
1
2
x=3
1
− ,3 .
2
.
Example 2.b Solve.
(x − 3)(x − 5) = −1
Solution:
(x − 3)(x − 5) = −1
x2 − 8x + 15 = −1
x2 − 8x + 16 = 0
(x − 4)2 = 0
Thus, by the zero factor property,
x−4= 0
x=4
The solution set is {4}.
.
Example 2.c Solve.
3x(x − 2) = 4(x + 1) + 4
Solution:
3x(x − 2) = 4(x + 1) + 4
3x2 − 6x = 4x + 4 + 4
3x2 − 6x = 4x + 8
3x2 − 10x − 8 = 0
(3x + 2)(x − 4) = 0
51
The solution set is
2
− ,4 .
3
.
Pythagorean Theorem If a right triangle has legs of lengths a and b and hypotenuse of length c,
then
a2 + b2 = c2 .
.
Example 3.a
Solution:
x2 + (x + 1)2 =
x + x2 + 2x + 1 =
2x2 + 2x + 1 =
x2 − 2x − 3 =
(x − 3)(x + 1) =
2
(x + 2)2
x2 + 4x + 4
x2 + 4x + 4
0
0
The solution set is {−1, 3}. Since x represents the length of a leg of the triangle, we know x = 3.
Thus, the sides of the triangle are 3, 4, and 5.
.
Example 3.b
Solution:
(x − 8)2 + (x − 1)2 =
x − 16x + 64 + x2 − 2x + 1 =
2x2 − 18x + 65 =
x2 − 18x + 65 =
(x − 5)(x − 13) =
2
x2
x2
x2
0
0
The solution set is {5, 13}. Since x − 8 represents the length of a leg of the triangle, we know x = 13
(otherwise x − 8 = −3). Thus, the sides of the triangle are 5, 12, and 13.
52
3
3.1
Rational Expressions
Rational Expressions
The notes for this section will be presented from the text.
3.2
Rational Expressions, Continued
The notes for this section will be presented from the text.
3.3
Rational Expressions III: Complex Rational Expressions
The notes for this section will be presented from the text.
3.4
Rational Expressions IV
The notes for this section will be presented from the text.
53
4
4.1
Radicals
Radicals: An Introduction
.
Definition: nth Root, Principle nth Root The nth root of a is b if
bn = a.
The principle nth root of a is b if
bn = a
b ≥ 0 if possible.
and
The principle nth root of a is denoted by
√
n
a = b.
If n = 2, the principle square root of a is denoted by
√
a = b.
.
Example 1.a Evaluate.
Solution: Since 92 = 81,
√
81
√
.
Example 1.b Evaluate.
Solution: Since 132 = 169,
81 = 9.
√
√
169
169 = 13.
.
Example 1.c Evaluate.
r
1
25
2
1
1
Solution: Since
= ,
5
25
r
.
Example 1.d Evaluate.
1
1
= .
25
5
√
−36
Solution: Notice (−6)2 = 36. In fact, for any real number x, x2 ≥ 0. Thus,
number.
.
Example 1.e Evaluate.
√
3
64
54
√
−36 is not a real
Solution: Since 43 = 64,
√
3
.
Example 1.f Evaluate.
Solution: Since (−4)3 = 64,
√
3
−64
√
3
−64 = −4.
.
Example 1.g Evaluate.
Solution: Since 24 = 16,
64 = 4.
√
4
16
√
4
.
Example 1.h Evaluate.
16 = 2.
√
4
−16
Solution: Notice (−2)4 = 16. In fact, for any real number x, x4 ≥ 0. Thus,
number.
.
Example 1.i Evaluate.
√
15
Solution: Using a calculator, we can approximate
√
15 ≈ 3.873.
.
Example 1.j Evaluate.
√
3
31
Solution: Using a calculator, we can approximate
√
3
31 ≈ 3.141.
.
Example 2.a Simplify.
Solution:
√
√
.
Example 2.b Simplify.
Solution:
x2 = |x|
√
√
x2
x10
x10 = |x5 |
55
√
4
−16 is not a real
.
Example 2.c Simplify.
Solution:
√
√
16x6 = 4|x3 |
.
Example 2.d Simplify.
Solution:
√
√
√
3
x3
√
3
.
Example 2.f Simplify.
Solution:
x3 = x
√
4
√
4
.
Example 2.g Simplify.
Solution:
x4
x4 = x 2
.
Example 2.e Simplify.
Solution:
16x6
x4
x4 = |x|
√
4
81x8
√
4
81x8 = 3x2
.
Example 3.a Simplify. Assume all variables represent nonnegative values.
√
x2
Solution:
√
x2 = x
.
Example 3.b Simplify. Assume all variables represent nonnegative values.
√
9x6
Solution:
√
9x6 = 3x3
.
Example 3.c Simplify. Assume all variables represent nonnegative values.
p
25x6 y 2
56
Solution:
p
25x6 y 2 = 5x3 y
.
Example 3.d Simplify. Assume all variables represent nonnegative values.
s
4x2
y4
Solution:
s
4x2
2x
=
y4
y2
.
Example 3.e Simplify. Assume all variables represent nonnegative values.
√
3
8x6
Solution:
√
3
8x6 = 2x2
.
Example 3.f Simplify. Assume all variables represent nonnegative values.
p
4
625x8 y 12
Solution:
p
4
625x8 y 12 = 5x2 y 3
.
Example 3.g* Simplify. Assume all variables represent nonnegative values.
√
x2 + 2x + 1
Solution:
p
√
x2 + 2x + 1 = (x + 1)2
=x+1
.
√
Example 4.a Let f (x) = x + 7. Find f (2).
Solution:
f (2) =
=
p
√
(2) + 7
9
=3
.
√
Example 4.b Let f (x) = x + 7. Find f (18).
Solution:
57
f (18) =
=
p
√
(18) + 7
25
=5
.
√
Example 4.c Let f (x) = x + 7. Find f (−7).
Solution:
f (−7) =
=
p
√
(−7) + 7
0
=0
.
√
Example 4.d Let f (x) = x + 7. Find f (−11).
Solution:
p
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.
√
Solution: For f (x) = x + 7 to be defined,
=
x+7 ≥0
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 − 42.
√
Solution: For g(x) = 7x − 42 to be defined,
7x − 42 ≥ 0
7x ≥ 42
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)
=
2x + 5.
√
Solution: For h(x) = 2x + 5 to be defined,
58
2x + 5 ≥ 0
2x ≥ −5
5
x≥− .
2
Thus, h(x) is defined for all real numbers x ≥ − 52 . The domain of h(x) is x|x ≥ − 25 .
59
4.2
Radicals: Rational Exponents
.
Definition: x1/n We define
x1/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.
91/2
Solution:
91/2 =
√
9
=3
.
Example 1.b Simplify.
641/3
Solution:
641/3 =
√
3
64
=4
.
Example 1.c Simplify.
−251/2
Solution:
√
−251/2 = − 25
= −5
.
Example 1.d Simplify.
(−25)1/2
Solution:
(−25)1/2 =
√
−25
Thus, (−25)1/2 is not a real number.
.
Example 1.e Simplify. Assume all variables represent positive values.
(81x2 )1/2
Solution:
(81x2 )1/2 =
√
81x2
= 9x
60
.
Example 1.f Simplify. Assume all variables represent positive values.
4x1/2
Solution:
√
4x1/2 = 4 x
.
Example 1.g Simplify. Assume all variables represent positive values.
(4x)1/2
Solution:
(4x)1/2 =
√
4x
√
=2 x
.
Example 1.h Simplify. Assume all variables represent positive values.
1/2
144x8
169y 6
Solution:
144x8
169y 6
s
1/2
=
=
.
Theorem: xm/n
xm/n =
Proof.
144x8
169y 6
12x4
13y 3
m
√
n
x
1
xm/n = x n ·m
1 m
= xn
√ m
= ( n x)
.
Example 2.a Simplify. Assume all variables represent nonnegative values.
813/4
Solution:
813/4 =
√
3
4
81
= 33
= 27
61
.
Example 2.b Simplify. Assume all variables represent nonnegative values.
642/3
Solution:
√
2
3
64
642/3 =
= 42
= 16
.
Example 2.c Simplify. Assume all variables represent nonnegative values.
323/5
Solution:
√
3
5
32
323/5 =
= 23
=8
.
Example 2.d Simplify. Assume all variables represent nonnegative values.
(8x3 )2/3
Solution:
3 2/3
(8x )
√ 2
3
8x3
=
= (2x)2
= 4x2
.
Theorem 2. xm/n
xm/n =
√
n
Proof.
xm
1
xm/n = xm· n
= (xm )1/n
=
√
n
xm
.
Example 3.a Simplify. Assume all variables represent nonnegative values.
x2/3
62
Solution:
√
3
x2/3 =
x2
.
Example 3.b Simplify. Assume all variables represent nonnegative values.
2
(x2 y 3 ) 11
Solution:
2
p
(x2 y 3 )2
(x2 y 3 ) 11 =
11
=
11
p
x4 y 6
.
Example 3.c Simplify. Assume all variables represent nonnegative values.
(x + 4)3/5
Solution:
(x + 4)3/5 =
p
5
(x + 4)3
.
Definition: x−m
x−m =
1
xm
.
Example 4.a Simplify.
16−1/2
Solution:
16−1/2 =
1
161/2
1
=√
16
=
1
4
.
Example 4.b Simplify.
−49−1/2
Solution:
−49−1/2 = −
1
491/2
1
= −√
49
=−
1
7
.
Example 4.c Simplify.
(−49)−1/2
63
Solution:
(−49)−1/2 =
1
(−49)1/2
1
=√
−49
√
However, −49 is not a real number, so neither is (−49)−1/2 .
.
Example 4.d Simplify.
27−2/3
Solution:
27−2/3 =
1
272/3
1
= √
2
3
27
=
1
32
=
1
9
.
Example 4.e Simplify.
Solution:
1
125
1
125
−2/3
−2/3
= 1252/3
=
√
3
= 52
= 25
.
Example 4.f Simplify.
−64−2/3
64
2
125
Solution:
−64−2/3 = −
1
642/3
1
=− √
2
3
64
=−
1
42
=−
1
16
.
Basic Rules for Exponents
xm · xn = xm+n
xm
= xm−n
n
x
(xm )n = xmn
ax
by
m
=
am x m
bm y m
.
Example 5.a Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
x1/2 · x2/3
Solution:
1
2
3
4
x1/2 · x2/3 = x 2 + 3
= x6+6
= x7/6
.
Example 5.b Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
x3/4
x2/3
Solution:
x3/4
3
− 23
4
=
x
x2/3
9
8
= x 12 − 12
= x1/12
65
.
Example 5.c Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
1/2
x3/4
Solution:
x3/4
1/2
3 1
= x4·2
= x3/8
.
Example 5.d Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
(x2 y 3 )2/3
Solution:
2/3
(x2 y 3 )2/3 = (x2 )
2/3
(y 3 )
= x4/3 y 2
.
Example 5.e Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
3 1/2
x
y4
Solution:
x3
y4
1/2
1/2
=
=
(x3 )
(y 4 )1/2
x3/2
y2
.
Example 5.f Simplify. Write the answer with positive exponents. Assume all variables represent
positive values.
6 1/2
9x
y8
Solution:
9x6
y8
1/2
1/2
=
(9x6 )
(y 8 )1/2
=
91/2 x3
y4
=
3x3
y4
.
Example 6.a Simplify by writing as a root with a smaller index.
√
4
25
66
Solution:
√
4
25 = (25)1/4
1/4
= (52 )
= 51/2
√
=
5
.
Example 6.b Simplify by writing as a root with a smaller index.
√
6
8
Solution:
√
6
8 = (8)1/6
1/6
= (23 )
= 21/2
=
√
2
.
Example 6.c Simplify by writing as a root with a smaller index.
√
8
x6
Solution:
√
8
x6 = (x6 )1/8
= x3/4
=
√
4
x3
.
Example 6.d Simplify by writing as a root with a smaller index.
p
9
x3 y 6
Solution:
p
9
x3 y 6 = (x3 y 6 )1/9
= x1/3 y 2/3
1/3
= (xy 2 )
=
p
3
xy 2
.
Example 7.a Simplify. Write the answer as a radical. Assume all variables represent positive
values.
√ √
x3x
67
Solution:
√ √
x 3 x = x1/2 · x1/3
= x3/6 · x2/6
= x5/6
=
√
6
x5
.
Example 7.b Simplify. Write the answer as a radical. Assume all variables represent positive
values.
√
√
4
5
x3 x2
Solution:
√
√
4
5
x3 x2 = x3/4 · x2/5
= x15/20 · x8/20
= x23/20
=
√
20
x23
.
Example 7.c Simplify. Write the answer as a radical. Assume all variables represent positive
values.
√
3
x2
√
x
Solution:
√
3
x2
x2/3
√ = 1/2
x
x
2
1
4
3
= x3−2
= x6−6
= x1/6
=
√
6
x
.
Example 7.d Simplify. Write the answer as a radical. Assume all variables represent positive
values.
√ √
3
5· 7
68
Solution:
√
5·
√
3
7 = 51/2 · 71/3
= 53/6 · 72/6
= (53 · 72 ) 1/6
= (6125)1/6
=
√
6
6125
.
Example 8.a Simplify. Write the answer as a single radical. Assume all variables represent positive
values.
q
3 √
x
Solution:
p
√ 1/3
√
3
x = ( x)
= x1/2
1/3
= x1/6
=
√
6
x
.
Example 8.b Simplify. Write the answer as a single radical. Assume all variables represent positive
values.
q
3 √
4
x
Solution:
p
√
√ 1/3
3 4
x = ( 4 x)
= x1/4
= x1/12
=
√
12
69
x
1/3
4.3
Radicals III
.
√ √
√
Theorem: n x · n y = n xy For any x ≥ 0 and y ≥ 0,
√
√
√
n
x · n y = n xy.
Proof.
√
n
x·
√
n
y = x1/n y 1/n
= (xy)1/n
√
n
=
.
Example 1.a Simplify.
Solution:
xy
√ √
3 · 27
√
3·
√
27 =
√
81
=9
.
Example 1.b Simplify.
Solution:
√
√
5·
5·
√
√
125
125 =
√
625
= 25
.
Example 1.c Simplify.
Solution:
√
√
3
3
2 · 32
√
3
2·
√
3
32 =
√
3
64
=4
.
Example 1.d Simplify.
Solution:
√
4
4·
√
4
4
√
√
√
4
4 · 4 4 = 4 16
=2
.
Example 1.e Simplify. Assume all variables represent nonnegative values.
p
√
3
5x · 3 4y
70
Solution:
√
√
√
3
5x · 3 4y = 3 20xy
.
Example 1.f Simplify. Assume all variables represent nonnegative values.
√
√
x 3 · x5
Solution:
√
x3 ·
√
x5 =
√
x8
= x4
.
√ √
Corollary: x · x = x For any x ≥ 0,
√
Proof.
√
x·
x·
√
√
x = x.
x=
√
x2
=x
.
Example 2.a Simplify.
Solution:
√
√
7·
7·
√
7
√
√
7 = 72
=7
.
Example 2.b Simplify. Assume all variables represent nonnegative values.
p
p
3x2 y · 3x2 y
Solution:
q
p
p
2
2
3x y · 3x y = (3x2 y)2
= 3x2 y
.
Rules for Simplifying Radicals A radical expression is simplified if:
1. there are no factors inside any radical raised to a power greater than or equal to the index,
2. there are no fractions inside any radicals, and
3. there are no radicals in any denominator of any fraction.
71
.
q
Theorem: n xy =
√
nx
√
ny
For any x ≥ 0 and y ≥ 0,
r
n
√
n
x
x
= √
.
n y
y
Proof.
r
n
x
=
y
1/n
x
y
x1/n
y 1/n
√
n
x
= √
n y
=
.
Example 3.a Simplify.
r
Solution:
r
.
Example 3.b Simplify.
Solution:
5
9
√
5
5
=√
9
9
√
5
=
3
√
14
√
2
√
14 √
√ = 7
2
.
Example 3.c Simplify.
r
3
Solution:
r
3
.
Example 3.d Simplify.
25
27
√
3
25
25
= √
3
27
27
√
3
25
=
3
√
4
48
√
4
3
72
Solution:
√
4
√
48
4
√
=
16
4
3
=2
.
Example 3.e Simplify. Assume all variables represent positive values.
r
9
3
x3
Solution:
r
3
√
3
9
9
√
=
3
3
x
x3
√
3
9
=
x
.
Example 4.a Simplify.
Solution:
√
√
8=
=
8
√
√
23
22 ·
√
2
√
=2 2
Alternatively,
√ √
√
√
8= 4· 2=2 2
.
Example 4.b Simplify.
Solution:
√
48
√
48 =
=
√
√
24 · 3
22 · 22 · 3
=2·2·
√
3
√
=4 3
Alternatively,
.
Example 4.c Simplify.
√
√
√
√
48 = 16 · 3 = 4 3
√
72
73
Solution:
√
√
72 =
√
=
23 · 32
22 · 2 · 32
√
2·3
=2·
√
=6 2
Alternatively,
√
√
√
√
72 = 36 · 2 = 6 2
.
Example 4.d Simplify.
√
56
Solution:
√
56 =
=
√
√
23 · 7
22 · 2 · 7
=2·
√
2·7
√
= 2 14
Alternatively,
.
Example 4.e Simplify.
Solution:
√
56 =
√
4·
√
√
14 = 2 14
√
4 12
√
√
4 12 = 4 22 · 3
=4·2·
√
3
√
=8 3
Alternatively,
.
Example 4.f Simplify.
Solution:
√
√ √
√
4 12 = 4 · 4 · 3 = 8 3
√
−3 54
√
√
−3 54 = −3 2 · 33
√
= −3 2 · 32 · 3
= −3 · 3 ·
√
= −9 6
74
√
2·3
Alternatively,
√ √
√
√
−3 54 = −3 · 9 · 6 = −9 6
.
Example 4.g Simplify.
√
Solution:
252
√
√
252 = 22 · 32 · 7
=2·3·
√
7
√
=6 7
Alternatively,
√
252 =
√
.
Example 4.h Simplify.
36 ·
√
Solution:
√
686 =
√
√
7=6 7
686
√
2 · 73
=
√
2 · 72 · 7
=
√
√
2·7· 7
√
= 7 14
Alternatively,
√
686 =
√
.
Example 4.i Simplify.
49 ·
√
Solution:
√
√
14 = 7 14
720
√
√
720 = 24 · 32 · 5
=
√
22 · 22 · 32 · 5
=2·2·3·
√
5
√
= 12 5
Alternatively,
.
Example 4.j Simplify.
√
720 =
√
144 ·
√
√
3
32
75
√
5 = 12 5
Solution:
√
√
3
32 = 25
√
3
=
23 · 22
=2·
√
3
22
√
=234
Alternatively,
√
3
32 =
√
3
.
Example 4.k Simplify.
8·
√
3
Solution:
√
3
162 =
√
3
√
3
4=2 4
162
√
3
2 · 34
=
√
3
2 · 33 · 3
=
√
√
3
2·3· 33
√
=336
Alternatively,
√
3
162 =
√
3
.
Example 4.l Simplify.
27 ·
√
4
Solution:
√
3
√
3
6=3 6
208
√
√
4
4
208 = 24 · 13
√
= 2 4 13
Alternatively,
√
4
208 =
√
4
16 ·
√
4
√
4
13 = 2 13
.
Example 5.a Simplify. Assume all variables represent nonnegative values.
√
x3
Solution:
√
x3 =
√
x2 · x
√
=x x
.
Example 5.b Simplify. Assume all variables represent nonnegative values.
√
x11
76
Solution:
√
x11 =
√
x10 · x
√
= x5 x
.
Example 5.c Simplify. Assume all variables represent nonnegative values.
√
3
x11
Solution:
√
3
x11 =
√
3
x9 · x2
√
3
= x3 x2
.
Example 5.d Simplify. Assume all variables represent nonnegative values.
p
32x5 y 4
Solution:
p
p
32x5 y 4 = 25 · x5 · y 4
√
= 22 x2 y 2 2x
√
= 4x2 y 2 2x
.
Example 5.e Simplify. Assume all variables represent nonnegative values.
p
3
54x8 y 7
Solution:
p
p
3
54x8 y 7 = 3 2 · 33 · x8 · y 7
= 3x2 y 2
p
3
2x2 y
.
Example 5.f Simplify. Assume all variables represent nonnegative values.
p
4
8x8 y 7
Solution:
p
p
4
8x8 y 7 = 4 23 · x8 · y 7
= x2 y
.
Example 6.a Simplify.
√
12 ·
√
77
p
4
8y 3
15
Solution:
√
√
√
√
12 · 15 = 22 · 3 · 3 · 5
√
=
22 · 32 · 5
√
=6 5
.
Example 6.b Simplify. Assume all variables represent nonnegative values.
p
p
5x3 y 2 · 10xy 3
Solution:
p
p
p
p
5x3 y 2 · 10xy 3 = 5x3 y 2 · 2 · 5 · xy 3
=
p
2 · 52 · x4 · y 5
√
= 5x2 y 2 2y
.
Example 6.c Simplify. Assume all variables represent nonnegative values.
p
p
4
2
2
4x y 2x y 3xy 6xy
Solution:
p
p
√
2
4x y 2x y 3xy 4 6xy = 12x3 y 5 2x2 y · 6xy
2
= 12x3 y 5
p
12x3 y 2
= 12x3 y 5
p
22 · 3 · x3 · y 2
√
= 12x3 y 5 · 2xy 3x
√
= 24x4 y 6 3x
.
Example 6.d Simplify.
Solution:
√
540
√
12
r
√
540
540
√
=
12
12
=
√
45
√
=3 5
.
Example 6.e Simplify. Assume all variables represent positive values.
p
72x5 y 7
p
9x2 y
78
Solution:
s
p
5
7
72x y
72x5 y 7
p
=
9x2 y
9x2 y
=
p
8x3 y 6
√
= 2xy 3 2x
79
4.4
Radicals IV
.
Example 1.a Add.
Solution:
√
√
4 3+7 3
√
√
√
4 3 + 7 3 = 3(4 + 7)
√
= 11 3
.
Example 1.b Subtract.
Solution:
√
√
3
3
2 7−6 7
√
√
√
2 3 7 − 6 3 7 = 3 7(2 − 6)
√
= −4 3 7
.
Example 1.c Subtract. Assume all variables represent nonnegative values.
√
√
4 5x − 3 5x
Solution:
√
√
√
4 5x − 3 5x = 5x(4 − 3)
=
√
5x
.
Example 1.d Subtract. Assume all variables represent nonnegative values.
√
√
2x2 x − 5x2 x
Solution:
√
√
√
2x2 x − 5x2 x = x2 x(2 − 5)
√
= −3x2 x
.
Example 2.a Add.
Solution:
√
18 +
√
32
√
√
√
√
18 + 32 = 9 · 2 + 16 · 2
√
√
=3 2+4 2
√
=7 2
.
Example 2.b Subtract.
√
√
6 20 − 2 80
80
Solution:
√
√
√
√
6 20 − 2 80 = 6 4 · 5 − 2 16 · 5
√
√
=6·2 5−2·4 5
√
√
= 12 5 − 8 5
√
=4 5
.
Example 2.c Subtract.
Solution:
√
3
√
3
√
3
54 − 5 2
√
√
√
54 − 5 3 2 = 3 27 · 2 − 5 3 2
√
√
=332−532
√
= −2 3 2
.
Example 2.d Subtract. Assume all variables represent nonnegative values.
√
√
2 48x5 − 7x 12x3
Solution:
√
√
√
√
2 48x5 − 7x 12x3 = 2 16x4 · 3x − 7x 4x2 · 3x
√
√
= 2 · 4x 3x − 7x · 2x 3x
√
√
= 8x2 3x − 14x2 3x
√
= −6x2 3x
.
Example 2.e Simplify.
Solution:
√
√
√
4
4
2x 16x − 4 4 x + 7 x5
√
√
√
√
√
√
4
4
4
2x 4 16x − 4 4 x + 7 x5 = 2x 24 · x − 4 4 x + 7 x4 · x
√
√
√
= 2x · 2 4 x − 4 4 x + 7 · x 4 x
√
√
√
= 4x 4 x − 4 4 x + 7x 4 x
√
√
= 11x 4 x − 4 4 x
√
= (11x − 4) 4 x
.
Example 3.a Simplify.
√ √
√ 6
5− 3
81
Solution:
√
6
√
√ √ √
√ √
5− 3 = 6· 5− 6· 3
=
=
.
Example 3.b Simplify.
Solution:
√
√
30 −
√
18
√
30 − 3 2
√ √ √
6 5 2 8 − 3 10
√
√
√ √
√
√
√
6 5 2 8 − 3 10 = 6 5 · 2 8 − 6 5 · 3 10
√
√
= 12 40 − 18 50
√
√
= 12 · 2 10 − 18 · 5 2
√
√
= 24 10 − 90 2
.
Example 3.c Simplify.
Solution:
4+
√ √ 4+ 3 6− 7
√ √ √ √ √ √ 3 6 − 7 = (4)(6) + (4) − 7 +
3 (6) +
3 − 7
√
√
√
= 24 − 4 7 + 6 3 − 21
.
Example 3.d Simplify.
Solution:
√
√ √
√ 2 6−3 5 5 6−9 5
√
√ √
√ √
√
√
√
2 6 − 3 5 5 6 − 9 5 = 10 36 − 18 30 − 15 30 + 27 25
√
√
= 10 · 6 − 18 30 − 15 30 + 27 · 5
√
√
= 60 − 18 30 − 15 30 + 135
√
= 195 − 33 30
.
Example 3.e Simplify. Assume all variables represent nonnegative values.
√
√
x+3
x+7
Solution:
√
√
√
√
√
( x + 3) ( x + 7) = x2 + 7 x + 3 x + 21
√
= x + 10 x + 21
82
.
Example 3.f Simplify.
Solution:
√ 2
2+5 3
√ 2
√ √ 2+5 3 = 2+5 3 2+5 3
√
√
√
= 4 + 10 3 + 10 3 + 25 9
√
= 4 + 20 3 + 75
√
= 79 + 20 3
.
Example 3.g Simplify.
Solution:
√
√ 2
3 2−6 7
√
√ 2
√ √
√ √
3 2−6 7 = 3 2−6 7 3 2−6 7
√
√
√
√
= 9 4 − 18 14 − 18 14 + 36 49
√
= 18 − 36 14 + 252
√
= 270 − 36 14
.
Example 3.h Simplify. Assume all variables represent nonnegative values.
2
√
x+4
Solution:
√
√
√
2
( x + 4) = ( x + 4) ( x + 4)
=
√
√
√
x2 + 4 x + 4 x + 16
√
= x + 8 x + 16
.
Example 3.i Simplify.
Solution:
8+
8+
√ √ 3 8− 3
√ √ √
√
√
3 8 − 3 = 64 − 8 3 + 8 3 − 9
= 64 − 3
= 61
.
Example 3.j Simplify.
√
√ √
√ 3 7−4 5 3 7+4 5
83
Solution:
√ √
√ √
√
√
√
√
3 7 − 4 5 3 7 + 4 5 = 9 49 + 12 35 − 12 35 − 16 25
= 63 − 80
= −17
.
Example 4.a Simplify.
Solution:
√
√
8·
√
8·
6−
√
6−
√
5·
√
15
√ √
√
√
5 · 15 = 48 − 75
√
√
=4 3−5 3
√
=− 3
.
Example 4.b Simplify.
Solution:
√
3
1500 √
3
√
+ 108
3
3
√
3
√
√
1500 √
3
3
3
√
+
108
=
500
+
108
3
3
√
√
= 3 125 · 4 + 3 27 · 4
√
√
=534+334
√
=834
84
4.5
Radicals V
.
Example 1.a Simplify.
1
√
2
Solution:
√
1
2
1
√ =√ ·√
2
2
2
√
2
=
2
.
Example 1.b Simplify.
4
√
3
Solution:
√
4
4
3
√ =√ ·√
3
3
3
√
4 3
=
3
.
Example 1.c Simplify.
r
Solution:
r
9
5
√
9
9
=√
5
5
3
=√
5
√
3
5
=√ ·√
5
5
√
3 5
=
5
.
Example 1.d Simplify. Assume all variables represent positive values.
r
2
x
85
Solution:
r
√
2
2
=√
x
x
√ √
2
x
=√ ·√
x
x
√
=
2x
x
.
Example 1.e Simplify.
1
√
3
2
Solution:
√
3
1
22
1
√
√
√
=
·
3
3
3
2
2
22
√
3
4
=
2
.
Example 1.f Simplify.
1
√
3
4
Solution:
1
1
√
= √
3
3
4
22
√
3
1
2
√
= √
·
3
3
2
22
√
3
2
=
2
.
Example 1.g Simplify. Assume all variables represent positive values.
r
3x
3
4y 2
86
Solution:
r
3
√
3
3x
3x
p
=
2
3
4y
4y 2
√
3
3x
= p
3
22 y 2
√
√
3
3
3x
2y
√
= p
·
3
3
2y
22 y 2
√
3
=
6xy
2y
.
Example 1.h Simplify.
1
√
4
8
Solution:
1
1
√
= √
4
4
8
23
√
4
1
2
= √
·√
4
4
2
23
√
4
2
=
2
.
Example 1.i Simplify.
Solution:
.
Example 2.a Multiply.
√
3 8
√
15
√
√
3 8
6 2
√ =√
15
15
√ √
6 2
15
=√ ·√
15
15
√
6 30
=
15
√
2 30
=
5
√
x+
√ √
√ y
x− y
87
Solution:
√
x+
p
√ √
√ √
√
√
y
x − y = x2 − xy + xy − y 2
=x−y
.
Example 2.b Simplify.
3
√
4− 2
Solution:
√
3
3
4+ 2
√ =
√ ·
√
4− 2
4− 2 4+ 2
√
12 + 3 2
=
16 − 2
√
12 + 3 2
=
14
.
Example 2.c Simplify.
√
√
Solution:
√
√
3+
.
Example 2.d Simplify.
Solution:
6
√
6
√
3 + 18
√
√
√
6
3 − 18
√ ·√
√
=√
18
3 + 18
3 − 18
√
√
18 − 108
=
3 − 18
√
√
3 2−6 3
=
−15
√
√ √
√
3 2−2 3
2−2 3
=−
=−
15
5
√
2− 3
√
2+ 3
√
√
√
2− 3 2− 3
2− 3
√ =
√ ·
√
2+ 3
2+ 3 2− 3
√
4−4 3+3
=
4−3
√
=7−4 3
88
.
Example 2.e Simplify. Assume all variables represent positive values.
√
x
√
√
x + 2y
Solution:
√
√
√
x
x
x − 2y
√ =√
√ ·√
√
√
x + 2y
x + 2y
x − 2y
√
√
√
x2 − 2xy
=
x − 2y
√
x − 2xy
=
x − 2y
.
Example 3.a Rationalize the numerator.
Solution:
√
2
3
√
√ √
2
2
2
=
·√
3
3
2
2
= √
3 2
.
Example 3.b Rationalize the numerator.
√
3− x
√
2
Solution:
√
√
√
3− x
3− x 3+ x
√
√
= √
·
3+ x
2
2
9−x
√
= √
3 2 + 2x
89
4.6
Radicals VI
.
Example 1.a Solve.
Solution:
√
x=4
√ 2
( x) = (4)2
x = 16
√
Check: 16 = 4
.
Example 1.b Solve.
Solution:
√
x = 13
√ 2
( x) = (13)2
x = 169
√
Check: 169 = 13
.
Example 1.c Solve.
Solution:
√
x = −5
√ 2
( x) = (−5)2
x = 25
√
Check: 25 6= −5. Thus, x = 25 is an extraneous solution, so there is no solution.
.
Example 2.a Solve.
√
x−3=2
Solution:
2
√
x − 3 = (2)2
x−3 =4
x=7
√
Check: 7 − 3 = 2
.
Example 2.b Solve.
√
4x + 1 = 5
90
Solution:
√
4x + 1
2
= (5)2
4x + 1 = 25
4x = 24
x=6
p
Check: 4(6) + 1 = 5
.
Example 2.c Solve.
√
3
Solution:
x + 8 = −3
3
√
3
x + 8 = (−3)3
x + 8 = −27
x = −35
√
3
Check: −35 + 8 = −3
.
Example 2.d Solve.
√
4
2x + 1 = −2
Solution:
√
4
2x + 1
4
= (−2)4
2x + 1 = 16
2x = 15
x=
15
2
q
Check: 4 2 15
+ 1 6= −2. Thus, x = 15
is an extraneous solution, so there is no solution.
2
2
.
Example 3.a Solve.
√
√
2x + 8 = 3x + 2
Solution:
√
2x + 8
2
=
2
√
3x + 2
2x + 8 = 3x + 2
6=x
p
p
√
Check: 2(6) + 8 = 20 = 3(6) + 2
91
.
Example 3.b Solve.
√
3
Solution:
√
3
7x − 4 =
√
3
1 − 3x
3
3
√
7x − 4 = 3 1 − 3x
7x − 4 = 1 − 3x
10x = 5
x=
q
q
q
Check: 3 7 12 − 4 = 3 − 12 = 3 1 −
.
Example 3.c Solve.
1
2
3
2
x+1=
√
x + 13
Solution:
(x + 1)2 =
√
x + 13
2
x2 + 2x + 1 = x + 13
x2 + x − 12 = 0
(x + 4)(x − 3) = 0
We have solutions of x = −4 and x = 3. Checking those solutions however, we find that x = −4 is
extraneous, as
√
−4 + 1 6= −4 + 13.
Our solution is x = 3.
.
Example 3.d Solve.
√
Solution:
√
√
x2 − x + 3 − 1 = 2x
x2 − x + 3 = 2x + 1
x2 − x + 3
2
= (2x + 1)2
x2 − x + 3 = 4x2 + 4x + 1
0 = 3x2 + 5x − 2
0 = (3x − 1)(x + 2)
We have solutions of x =
extraneous, as
1
3
and x = −2. Checking those solutions however, we find that x = −2 is
p
(−2)2 − (−2) + 3 − 1 6= 2(−2).
92
Our solution is x = 31 .
.
Example 3.e Solve.
Solution:
√
√
x − 1 = 2x + 2
2
√
√
2
( x − 1) =
2x + 2
√
x − 2 x + 1 = 2x + 2
√ 2
(−2 x) = (x + 1)2
4x = x2 + 2x + 1
0 = x2 − 2x + 1
0 = (x − 1)2
p
√
The solution, x = 1, is extraneous, as 1 − 1 = 0 6= 2(1) + 2 = 2. Thus, there is no solution.
93
5
5.1
Quadratic Equations
The Complex Number System
.
Definition: Complex Number We define i to be the number such that
i2 = −1.
We call i the imaginary unit.
The product of a real number b and the imaginary unit i,
bi,
is called an imaginary number.
The sum of any real number a and imaginary number bi,
a + bi,
is called a complex number.
.
Example 1.a Simplify.
Solution:
√
−36
√
√
√
−36 = −1 · 36
= 6i
.
Example 1.b Simplify.
Solution:
√
−12
√
√
√
−12 = −1 · 12
=
√
−1 ·
√
4·3
√
= 2i 3
.
Example 1.c Simplify.
Solution:
√
√
−98 =
=
−98
√
√
−1 · 98
√
√
−1 · 49 · 2
√
= 7i 2
94
.
Example 2.a Add the complex numbers.
(−2 + 3i) + (6 − 9i)
Solution:
(−2 + 3i) + (6 − 9i) = 4 − 6i
.
Example 2.b Subtract the complex numbers.
(5 − 12i) − (8 − 3i)
Solution:
(5 − 12i) − (8 − 3i) = 5 − 12i − 8 + 3i
= −3 − 9i
.
Example 3.a Multiply.
(4i)(5i)
Solution:
(4i)(5i) = 20i2
= −20
.
Example 3.b Multiply.
(−2i)(i)
Solution:
(−2i)(i) = −2i2
=2
.
Example 3.c Multiply.
4i(5 − 2i)
Solution:
4i(5 − 2i) = 20i − 8i2
= 20i + 8
= 8 + 20i
.
Example 3.d Multiply.
(3 + 4i)(5 − i)
Solution:
(3 + 4i)(5 − i) = 15 − 3i + 20i − 4i2
= 15 + 17i + 4
= 19 + 17i
95
.
Example 3.e Multiply.
(1 − 3i)(3 − 2i)
Solution:
(1 − 3i)(3 − 2i) = 3 − 2i − 9i + 6i2
= 3 − 11i − 6
= −3 − 11i
.
Example 3.f Multiply.
(4 + i)2
Solution:
(4 + i)2 = (4 + i)(4 + i)
= 16 + 8i + i2
= 16 + 8i − 1
= 15 + 8i
.
Example 3.g Multiply.
(2 − 5i)2
Solution:
(2 − 5i) = 4 − 20i + 25i2
= 4 − 20i − 25
= −21 − 20i
.
Example 3.h Multiply.
(7 − 2i)(7 + 2i)
Solution:
(7 − 2i)(7 + 2i) = 49 + 14i − 14i − 4i2
= 49 + 4
= 53
.
Example 3.i Multiply.
(3 + 4i)(3 − 4i)
Solution:
(3 + 4i)(3 − 4i) = 9 − 12i + 12i − 16i2
= 9 + 16
= 25
96
.
Example 4.a Divide.
3
4i
Solution:
3
3 i
=
·
4i
4i i
=
3i
4i2
=
3i
−4
3
=− i
4
.
Example 4.b Divide.
Solution:
4 − 5i
3i
4 − 5i i
4 − 5i
=
·
3i
3i
i
=
4i − 5i2
3i2
=
4i + 5
−3
5 4
=− − i
3 3
.
Example 4.c Divide.
7 + 4i
2 − 3i
Solution:
7 + 4i
7 + 4i 2 + 3i
=
·
2 − 3i
2 − 3i 2 + 3i
=
14 + 29i + 12i2
4 − 9i2
=
14 + 29i − 12
4+9
=
2 + 29i
13
=
2
29
+ i
13 13
97
.
Example 4.d Divide.
Solution:
5−i
4 + 5i
5−i
5 − i 4 − 5i
=
·
4 + 5i
4 + 5i 4 − 5i
=
20 − 29i + 5i2
16 − 25i2
=
20 − 29i − 5
16 + 25
=
15 − 29i
41
=
15 29
− i
41 41
98
5.2
Quadratic Equations
.
Definition: Quadratic Equation A quadratic equation is an equation that can be written as
ax2 + bx + c = 0
where a, b, and c are real numbers and a 6= 0.
.
Example 1 Solve.
x2 = 4
Solution:
x2 = 4
x2 − 4 = 0
(x + 2)(x − 2) = 0
Then either:
x+2= 0
x−2= 0
or
x = −2
x=2
.
√
√
Theorem: Square Root Principal If x2 = a, then x = a or x = − a.
Proof.
x2 = a
x2 − a = 0
√
(x +
Then either:
x+
a)(x −
√
a=0
√
a) = 0
or
√
x=− a
x2 = 9
√
√
x2 = ± 9
x = ±3
.
Example 2.b Solve.
x2 = 12
Solution:
√
a=0
x=
.
Example 2.a Solve.
Solution:
x−
√
√
x2 = ± 12
√
x = ±2 3
99
√
a
.
Example 2.c Solve.
x2 = −25
Solution:
√
√
x2 = ± −25
x = ±5i
.
Example 2.d Solve.
x2 − 4 = 32
Solution:
x2 − 4 = 32
x2 = 36
√
√
x2 = ± 36
x = ±6
.
Example 2.e Solve.
3x2 + 4 = 58
Solution:
3x2 + 4 = 58
3x2 = 54
x2 = 18
√
√
x2 = ± 18
√
x = ±3 2
.
Example 2.f Solve.
(x − 3)2 = 4
Solution:
(x − 3)2 = 4
p
√
(x − 3)2 = ± 4
x − 3 = ±2
x =3±2
Thus, x = 5 or x = 1.
100
.
Example 2.g Solve.
(2x − 1)2 = −5
Solution:
(2x − 1)2 = −5
p
√
(2x − 1)2 = ± −5
√
2x − 1 = ±i 5
√
2x = 1 ± i 5
√
√
1−i 5
1+i 5
or x =
.
Thus, x =
2
2
.
Example 3.a Solve.
√
1±i 5
x=
2
x2 + 6x + 9 = 49
Solution:
x2 + 6x + 9 = 49
(x + 3)2 = 49
p
√
(x + 3)2 = ± 49
x + 3 = ±7
x = −3 ± 7
Thus, x = 4 or x = −10.
.
Example 3.b Solve.
x2 − 10x + 25 = 10
Solution:
x2 − 10x + 25 = 10
(x − 5)2 = 10
p
√
(x − 5)2 = ± 10
√
x − 5 = ± 10
x =5±
Thus, x = 5 +
√
10 or x = 5 −
√
10.
101
√
10
.
Example 3.c Solve.
4x2 + 12x + 9 = −1
Solution:
4x2 + 12x + 9 = −1
(2x + 3)2 = −1
p
√
(2x + 3)2 = ± −1
2x + 3 = ±i
2x = −3 ± i
x=
−3 ± i
2
3 1
3 1
Thus, x = − + i or x = − − i.
2 2
2 2
.
Example 4.a Solve.
x2 + 8x + 2 = −5
Solution:
x2 + 8x + 2 = −5
x2 + 8x
= −7
x2 + 8x + 16 = −7 + 16
(x + 4)2 = 9
p
√
(x + 4)2 = ± 9
x + 4 = ±3
x = −4 ± 3
Thus, x = −1 or x = −7.
.
Example 4.b Solve.
x2 − 6x + 1 = 0
102
Solution:
x2 − 6x + 1 = 0
x2 − 6x
= −1
x2 − 6x + 9 = −1 + 9
(x − 3)2 = 8
p
√
(x − 3)2 = ± 8
√
x − 3 = ±2 2
√
x =3±2 2
.
Example 4.c Solve.
x2 + x − 2 = 5
Solution:
x2 + x
=7
x2 + x +
q
1
4
=7+
1
4
x+
1 2
2
=
x+
1 2
2
q
= ± 29
4
x+
29
4
√
1
2
=±
x=
29
2
√
−1± 29
2
.
Example 4.d Solve.
x2 − 5x +
Solution:
= − 41
4
x2 − 5x
x2 − 5x +
q
41
=0
4
25
4
= − 41
+
4
x−
5 2
2
= −4
x−
5 2
2
√
= ± −4
x−
5
2
= ±2i
x=
103
5
2
± 2i
25
4
.
Example 5.a Solve.
9x2 + 12x − 5 = 0
Solution:
9x2 + 12x − 5 = 0
9x2 + 12x
=5
9x2 + 12x
9
=
5
9
4
5
x2 + x
=
3
9
2
2
5
4
2
2
= +
x2 + x +
3
3
9
3
.
Example 5.a (Continued)
4
4
5 4
x2 + x + = +
3
9
9 9
2
2
x+
=1
3
s
2
√
2
=± 1
x+
3
x+
2
= ±1
3
2
x =− ±1
3
Thus, x = 13 or x = − 35 .
.
Example 5.b Solve.
4x2 − 20x + 33 = 0
104
Solution:
4x2 − 20x + 33 = 0
4x2 − 20x
= −33
4x2 − 20x
4
=
−33
4
x2 − 5x
=
−33
4
2
2
−33
5
5
=
+
x − 5x +
2
4
2
2
.
Example 5.b (Continued)
25
−33 25
=
+
4
4
4
2
5
= −2
x−
2
x2 − 5x +
s
5
x−
2
2
x−
√
= ± −2
√
5
= ±i 2
2
x=
5.3
√
5
±i 2
2
Quadratic Equations: The Grand Finale
.
Definition: Quadratic Equation A quadratic equation is an equation that can be written as
ax2 + bx + c = 0
where a, b, and c are real numbers and a 6= 0.
.
Theorem: Quadratic Formula For any equation ax2 + bx + c = 0, where a, b, and c are real
numbers, and a 6= 0,
√
−b ± b2 − 4ac
.
x=
2a
105
Proof.
ax2 + bx + c = 0
ax2 + bx
= −c
ax2 + bx
a
=
−c
a
b
c
x2 + x
=−
a
a
2
2
b
b
c
b
2
=− +
x + x+
a
2a
a
2a
b
b2
b2
c
x2 + x + 2 = − + 2
a
4a
a 4a
2
4ac
b2
b
=− 2 + 2
x+
2a
4a
4a
2
b2
4ac
b
= 2− 2
x+
2a
4a
4a
2
b2 − 4ac
b
=
x+
2a
4a2
s
r
2
b
b2 − 4ac
x+
=±
2a
4a2
√
b
b2 − 4ac
x+
=± √
2a
4a2
√
b
b2 − 4ac
x+
=±
2a
2a
√
b
b2 − 4ac
x=− ±
2a
2a
√
−b ± b2 − 4ac
x=
2a
.
Example 1.a Solve.
x2 + 2x − 8 = 0
106
Solution:
√
b2 − 4ac
2a
p
(−2) ± 22 − 4(1)(−8)
=
2(1)
x=
=
=
−b ±
−2 ±
√
4 + 32
2
√
−2 ± 36
−2 ± 6
=
2
2
Thus, x = 2 or x = −4.
.
Example 1.b Solve.
3x2 − 5x − 2 = 0
Solution:
x=
=
−b ±
5±
√
b2 − 4ac
2a
p
(−5)2 − 4(3)(−2)
2(3)
√
25 + 24
=
6
√
5±7
5 ± 49
=
=
6
6
5±
Thus, x = 2 or x = − 13 .
.
Example 1.c Solve.
x2 − 3x − 7 = 0
Solution:
x=
=
−b ±
3±
√
b2 − 4ac
2a
p
(−3)2 − 4(1)(−7)
2(1)
√
9 + 28
=
2
√
3 ± 37
=
2
3±
Thus, x =
3+
√
√
37
3 − 37
or x =
.
2
2
107
.
Example 1.d Solve.
2x2 + x + 1 = 0
Solution:
√
b2 − 4ac
2a
p
(−1) ± 12 − 4(2)(1)
=
2(2)
x=
=
−b ±
−1 ±
√
1−8
4
√
√
−1 ± i 7
−1 ± −7
=
=
4
4
√
√
−1 + i 7
−1 − i 7
Thus, x =
or x =
.
4
4
.
Example 1.e Solve.
4x2 = 7x − 3
Solution:
4x2 = 7x − 3
4x2 − 7x + 3 = 0
x=
=
7±
7±
p
(−7)2 − 4(4)(3)
2(4)
√
49 − 48
8
√
7± 1
7±1
=
=
8
8
Thus, x = 1 or x = 34 .
.
Example 1.f Solve.
3x − x2 = 1
Solution:
3x − x2 = 1
−x2 + 3x − 1 = 0
108
x=
(−3) ±
p
32 − 4(−1)(−1)
2(−1)
√
−3 ± 9 − 4
=
−2
√
√
−3 ± 5
3∓ 5
=
=
−2
2
Thus, x =
5.4
√
3− 5
2
or x =
√
3+ 5
.
2
Quadratic Equations III
The notes for this section will be presented from the text.
5.5
Prelude To College Algebra: Graphing Quadratic Equations
The notes for this section will be presented from the text.
109