Transcript College Algebra - Seminole State College

College Algebra
Exam 1 Material
Special Binomial Products to
Memorize
• When a binomial is squared, the result is always a
“perfect square trinomial”
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
• Both of these can be summarized as a formula:
– Square the first term
– Multiply 2 times first term times second term
– Square the last term
3x  2 
2
4x  5 
2
9x 2  12 x  4
16x 2  40 x  25
Homework Problems
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Section: R.3
Page: 33
Problems: 49 – 52
MyMathLab Homework Assignment 1
Raising a Binomial to a Power
Other Than Two
• You should recall that you CAN NOT distribute
an exponent over addition or subtraction:
We have just seen that:
(a + b)2 is NOT equal to a2 + b2
2
2
2
a

2
ab

b
(a + b) =
(a + b)m is NOT equal to am + bm
(a – b)m is NOT equal to am – bm
Patterns in Binomials Raised to
Whole Number Powers
a  b
0
 1
a  b 
2
a  b 
1a  1b
a  b 
3
a  b 
3
a  b 
a  ba  b2  a  ba2  2ab  b2 
1
3
1a 2  2ab  1b2
a3  2a 2b  ab2  a 2b  2ab2  b3
1a3  3a 2b  3ab2  1b3
Patterns in Binomials Raised to
Whole Number Powers
1
a  b0 
1
1a  1b
a  b 
2
2
2
a  b  1a  2ab  1b
3
2
2
3
3
a  b  1a  3a b  3ab  1b
Number of termscomparedwith exponent: Always one more
Degree of termscomparedwith exponent: All same degree as exponent
Exponenton " a" startsat exponenton binomialand goes down one
each term,exponenton " b" startsat zero and goes up each term.
Coefficient of first termis always: 1
Coefficient of other term
s is always coefficient of previousterm
multipliedby exponenton " a" for previousterm,divided by the
number of thepreviousterm.
Binomial Theorem
• Patterns observed on previous slide are
the basis for the Binomial Theorem that
gives a short cut method for raising any
binomial to any whole number power:
a  b
5
 1a 5  5a 4b  10a 3b 2 10a 2b3  5ab4  1b 5
Using Binomial Theorem
•
To raise any binomial to nth power:
– Write expansion of (a + b)n using patterns
– Use this as a formula for the desired
binomial by substituting for “a” and “b”
– Simplify the result
2x  3 y5
a  b5  1a5  5a4b 10a3b2 10a2b3  5ab4 1b5
5
4
3
2
2
3
 12x  52x  3 y  102x  3 y  102x  3 y  
4
5
 52x 3 y  1 3 y 
2x  3 y5  32x5  240x4 y  720x3 y 2 1080x2 y3  810xy4  243y5
Homework Problems
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Section:
Page:
Problems: Binomial Worksheet
There is no MyMathLab Homework
Assignment that corresponds with these
problems
Binomial Expansion Worksheet
Use the Binomial Theorem to expand and simplify each of the following:
4
1.
x 1
2.
3.
4.
5.
6.
7.
8.
 
x  3y 5
x  2y 4
2x  3 y 4
x  2y 3
2x  y 3
x  26
x
2
y

2 5
Equation
• a statement that two algebraic expressions are equal
• Many different types with different names
• Examples:
3x  7  1  6 x
5x 2  2 x  7 x
2 x  5  5x  4
4x  5  2  7
Linear
Quadratic
Radical
Absolute Value
• Many other types of equations – (we will learn names as
we go)
Solutions to Equations
• Since an equation is a statement, it may
be “true” or “false”
• All values of a variable that make an
equation “true” are called “solutions” of
the equation
• Example consider this statement:
x+3=7
Is there a value of x that makes this true?
“x = 4” is the only solution to this equation
All Types of Equations Classified
in One of Three Categories on
the Basis of Its Solutions:
• Conditional Equation
• Identity
• Contradiction
Conditional Equation
• An equation that is true for only certain
values of the variable, but not for all
• Previous example:
x + 3 = 7 Is true only under the “condition”
that x = 4, and not all values of x make it
true
• Conditional Equation
Identity
• An equation that is true for all possible
values of the variable
• Example:
2(x + 3) = 2x + 6
What values of x make this true?
All values, because this is just a statement
of the distributive property
• Identity
Contradiction
• An equation that is false for every possible
value of the variable
• Example:
x=x+5
Why is it impossible for any value of x to
make this true?
It says that a number is the same as five
added to the number – impossible
• Contradiction
Classifying Equations as
Conditional, Identity, Contradiction
• Classification normally becomes possible
only as an attempt is made to “solve” the
equation
• We will examine classifying equations as
we begin to solve them
Equivalent Equations
• Equations with exactly the same solution sets
• Example:
Why are each of the following equivalent?
2x – 3 = 7
2x = 10
x=5
They all have exactly the same solution set: {5}
Finding Solutions to Equations
• One way to find the solutions to an equation is to
write it as a simpler equivalent equation for
which the solution is obvious
Example which of these equivalent equations
has an obvious solution?
3(x – 5) + 2x = x + 1
x=4
Both have only the solution “4”
obvious for the second equation, but not for first
Procedures that Convert an
Equation to an Equivalent
Equation
• Addition Property of Equality:
When the same value is added (or subtracted) on both
sides of an equation, the result is an equivalent equation
x  3  7 is equivalent to: x   4
• Multiplication Property of Equality
When the same non-zero value is multiplied (or divided)
on both sides of an equation, the result is an equivalent
equation
1
x  2 is equivalent to : x   6
3
.
Linear Equations in One Variable
• Technical Definition: An equation where, after
parentheses are gone, every term is a constant
or a constant times a variable to the first power.
• Shorter Definition: A polynomial equation in one
variable of degree 1.
• Examples:
3x  5  2 x  x  1
3x  15  2 x  x  1
3
4 x   x  1.7   x
4
3
4 x  x  1. 7  x
4
Solving Linear Equations
• Get rid of parentheses
• Get rid of fractions and decimals by multiplying
both sides by LCD
• Collect like terms
• Decide which side will keep variable terms and
get rid of variable terms on other side
• Get rid of non-variable terms on variable side
• Divide both sides by the coefficient of variable
Solve the Equation
2
1

2 x    .7 x  x 
3
2

• Identify the type of equation:
It is linear!
• Get rid of parentheses:
2x 
4
1
 .7 x  x 
3
2
• Get rid of fractions and decimals by multiplying
both sides by LCD: LCD of 3,10,and 2 is : 30
4
1



30 2 x   .7 x   30 x  
3
2



60 x  40  21x  30 x  15
Example Continued
60 x  40  21x  30 x  15
• Collect like terms:
39 x  40  30 x  15
• Decide which side will keep variable terms and
get rid of variable terms on other side:
If youchose to keep variableson left you willget rid of thoseon theright
9 x  40  15
• Get rid of non-variable terms on variable side:
9 x  55
• Divide both sides by coefficient of variable:
55
x
9
Homework Problems
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Section: 1.1
Page: 90
Problems: Odd: 9 – 27
MyMathLab Homework Assignment 2
Contradiction
Solve: 2x – (x – 3) = x + 7
What type of equation? Linear
Solve by linear steps:
2x – x + 3 = x + 7
x+3=x+7
What’s wrong?
This says that 3 added to a number is the same as 7
added to the number - T HATS' IMPOSSIBLE!
No matter what type of equation, when we reach an
obvious impossibility, the equation is a classified as a
“contradiction” and has “no solution”

Identity
Solve: x – (2 – 7x) = 2x – 2(1 – 3x)
What type of equation? Linear
Solve by linear steps:
x – 2 + 7x = 2x – 2 + 6x
8x – 2 = 8x – 2
What looks strange?
Both sides are identical
In any type of equation when this happens we classify
the equation as an “identity” and say that “all real
numbers are solutions”
 , 


Homework Problems
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Section: 1.1
Page: 91
Problems: Odd: 29 – 35
MyMathLab Homework Assignment 3
Formulas
• Any equation in two or more variables can be
called a “formula”
• Familiar Examples:
A = LW
Area of rectangle
P = 2L + 2W
Perimeter of rectangle
I = PRT
Simple Interest
D = RT
Distance
In all these examples each formula has a
variable isolated and we say the formula is
“solved for that variable”
Formulas Continued
• Some formulas may not be solved for a
particular variable:
A
2B  A 
B
No Variable is Isolated!
• In cases like this we need to be able to solve for
a specified variable (A or B)
• In other cases, when an equation is solved for
one variable, we may need to solve it for another
variable
P = 2L + 2W is solved for P, but can be solved
for L or W
Solving Formulas
for a Specified Variable
• When solving for a specified variable, pretend all
other variables are just numbers (their degree is
“zero”)
• Ask yourself “Considering only the variable I
am solving for, what type of equation is
this?”
• If it is “linear” we can solve it using linear
techniques already learned, otherwise we will
have to use techniques appropriate for the type
of equation
Solving a Formula
for a Specified Variable
Solve for “t”:
1 2
s  vt  gt
2
Is this equation “linear in t” ?
No, it is second degree in “t” – not first degree!
Since it’s not linear in t we can’t solve by using
linear equation techniques.
(Later we can solve this for t, but not with linear
techniques.)
Solving a Formula
for a Specified Variable
Solve for “g”:
1 2
s  vt  gt
2
Is this equation “linear in g” ?
Yes, so we can solve like any other linear
equation:
Get rid of parentheses:
(Not necessary for this formula)
Get rid of fractions:
What is LCD?
2
Solving a Formula
for a Specified Variable
1 2
s  vt  gt
2
Multiplyboth sides by LCD :
1 2

2s   2 vt  gt 
2


2s  2vt  gt2
2s  2vt  gt2
2 s  2vt
t
2
g
What next?
Isolatetermwith g
What next? Divide by coefficient of g
Now solvedfor g !
Example Two
Solve for y:
xx  3 y   2 y  3
x 2  3xy  2 y  6
2
x  2 y  6  3xy
x 2  6  2 y  3xy
When thereis morethanone termwith variable being solvedfor,factor!
x  6  2  3xy
x2  6
y
2  3x
2
Homework Problems
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Section: 1.1
Page: 91
Problems: Odd: 39 – 57
MyMathLab Homework Assignment 4
• MyMathLab Homework Quiz 1 will be due
for a grade on the date of our next class
meeting!!!
Linear Applications
• General methods for solving an applied (word)
problem:
1. Read problem carefully taking notes, drawing pictures, thinking
about formulas that apply, making charts, etc.
2. Read problem again to make a “word list” of everything that is
unknown
3. Give a variable name, such as “x” to the “most basic unknown” in
the list (the thing that you know the least about)
4. Give all other unknowns in you word list and algebraic
expression name that includes the variable, “x”
5. Read the problem one last time to determine what information
has been given, or implied by the problem, that has not been
used in giving an algebra name to the unknowns and use this
information to write an equation about the unknowns
6. Solve the equation and answer the original question
Solve the Application Problem
•
A 31 inch pipe needs to be cut into three pieces
in such a way that the second piece is 5 inches
longer than the first piece and the third piece is
twice as long as the second piece. How long
should the third piece be?
1. Read the problem carefully taking notes,
drawing pictures, thinking about formulas that
apply, making charts, etc.
Perhaps draw a picture of a pipe that is labeled
as 31 inches with two cut marks dividing it into
3 pieces labeled first, second and third
1st
2nd
3rd
31
Example Continued
2. Read problem again to make a “word list” of
everything that is unknown
What things are unknown in this problem?
The length of all three pieces (even though the
problem only asked for the length of the third).
Word List of Unknowns:
Length of first
Length of second
Length of third
Example Continued
3. Give a variable name, such as “x” to the “most
basic unknown” in the list (the thing that you
know the least about)
What is the most basic unknown in this list?
Length of first piece is most basic, because
problem describes the second in terms of the
first, and the third in terms of the second, but
says nothing about the first
Give the name “x” to the length of first
Example Continued
4.
Give all other unknowns in the word list an algebraic expression
name that includes the variable, “x”
The second is 5 inches more than the first. How would the length
of the second be named?
x+5
The third is twice as long as the second. How would the length of
the third be named?
2(x + 5)
Word List of Unknowns:
Algebra Names:
Length of first
x
Length of second
x+5
Length of third
2(x + 5)
Example Continued
5.
Read the problem one last time to determine what
information has been given, or implied by the problem,
that has not been used in giving an algebra name to
the unknowns and use this information to write an
equation about the unknowns
What other information is given in the problem that has
not been used?
Total length of pipe is 31 inches
How do we say, by using the algebra names, that the
total length of the three pieces is 31?
x + (x + 5) + 2(x + 5) = 31
Example Continued
6.
Solve the equation and answer the original question
This is a linear equation so solve using the appropriate
steps:
x + (x + 5) + 2(x + 5) = 31
x + x + 5 + 2x + 10 = 31
4x + 15 = 31
4x = 16
x=4
Is this the answer to the original question?
No, this is the length of the first piece.
How do we find the length of the third piece?
The length of the third piece is 2(x + 5):
2(4 + 5) = (2)(9) = 18 inches = length of third piece
Solve this Application Problem
• The length of a rectangle is 4 cm more than its
width. When the length is decreased by 2 and
the width increased by 1, the new rectangle has
a perimeter of 18 cm. What were the
dimensions of the original rectangle?
• Draw of picture of two rectangles and label them
as “original” and “new”. Also write notes about
relationships between the widths and lengths.
Write the formula for perimeter of rectangle:
W  4 cm
OL - 2
P = 2L + 2W
OW  1
Original
New
Perimeter  18
Example Continued
• Make a word list of all unknowns:
length of original
width of original
length of new
width of new
• Give the name “x” to the most basic
unknown:
width of original = x
Example Continued
• Read problem again to give algebra names to all other unknowns:
x4
xOL- 22
length of original: x  4
W  4 cm
width of original:
x
x 1
x
OW  1
x2
length of new:
Original
New
x 1
width of new:
Perimeter  18
• Read problem one more time to determine what other information is
given that has not been used and use it to write an equation:
Perimeter of new rectangle is 18 cm
Use formula: P = 2L + 2W
18 = 2(x + 2) + 2(x + 1)
Example Continued
• Solve equation and answer the original question:
18 = 2(x + 2) + 2(x + 1)
18 = 2x + 4 + 2x + 2
18 = 4x + 6
12 = 4x
3=x
length of original:
x+4=3+4=7
width of original:
x=3
Homework Problems
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Section: 1.2
Page: 101
Problems: Odd: 9 – 17
MyMathLab Homework Assignment 5
Solving Application Problems
with Formulas & Charts
• There are four types of problems that can easily be
solved by means of formulas and charts:
1. Motion problems:
D = RT
(Distance equals Rate multiplied by Time)
2. Work problems: F = RT
(Fraction of job completed equals Rate of work multiplied
by Time worked)
3. Mixture problems:
IA = (IP)(SA)
(Ingredient Amount equals Ingredient Percent multiplied
by Substance Amount)
4. Simple Interest problems:
I = PRT
(Interest equals Principle multiplied by Rate (%)
multiplied by Time (in years or part of a year)
Formula: D = RT
• Given R and T this formula can be used as is to find D
Example: If R = 5 mph and T = 3 hr, what is D?
D = (5)(3) = 15 miles
• Given any two of the three variables in the formula, the
other one can always be found:
• How would you find T if D and R were given?
T=D/R
• How would you find R if D and T were given?
R=D/T
Solving Motion Problems with
Formula and Chart
1. Immediately write formula: D = RT as a heading on a
chart
2. Make and label one line in the chart for everything
“moving”
3. Write “x” in the box for the most basic unknown
4. Fill out the remainder of that column based on
information given in the problem
5. Fill out one more column based on the most specific
information given in the problem
6. Fill out the final column by using the formula at the top
7. Read problem one more time and write an equation
about D, R, or T, based on other information given in the
problem that was not used in completing the chart
8. Solve the equation and answer the original question
Solving a Motion Problem
•
1.
2.
Lisa and Dionne are traveling to a meeting. It takes Lisa 2 hours to
reach the meeting site and 2.5 hours for Dionne, since she lives 40
miles farther away. Dionne travels 5 mph faster than Lisa. Find
their average speeds.
Immediately write formula: D = RT as a heading on a chart:
Make and label one line in the chart for everything “moving”:
D=
R
T
Lisa
Dionne
Example Continued
3. Write “x” in the box for the most basic
unknown:
What is it?
Lisa’s speed, because if we find it, we can
calculate Dionne’s speed by adding 5 mph
D= R
T
Lisa
x
Dionne
Example Continued
4.
Fill out the remainder of that column based on
information given in the problem
What is the other item in that column?
Dionne’s speed.
How would we describe it with an algebra description?
x+5
D=
Lisa
Dionne
R
x
x5
T
Example Continued
5.
Fill out one more column based on the most specific
information given in the problem
Is the most specific information given about how far
each one traveled, or about how much time each one
took?
The time each took: Lisa’s time was 2 hours, and
Dionne’s 2.5 hours
D= R
T
Lisa
x
Dionne
2
x  5 2 .5
Example Continued
6.
Fill out the final column by using the formula at the top
Formula says that D = RT, so the final column is:
D
=
R
T
Lisa
2x
x
2
Dionne
2.5x  5 x  5 2.5
7.
Read problem one more time and write an equation about D, R, or
T, based on other information given in the problem that was not
used in completing the chart
x was not used
What other information is given in the problem that
in making the chart?
Dionne’s distance was 40 miles more than Lisa’s distance:
2.5(x + 5) = 2x + 40
Example Continued
2.5(x + 5) = 2x + 40
2.5x + 12.5 = 2x + 40
10(2.5x + 12.5) = 10(2x + 40)
25x + 125 = 20x + 400
5x + 125 = 400
5x = 275
x = 55 mph (Lisa’s speed)
x + 5 = 60 mph (Dionne’s speed)
Homework Problems
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Section: 1.2
Page: 102
Problems: 19 – 24, 27 – 28
MyMathLab Homework Assignment 6
Solving Work Problems With
Formula: F = RT
• This formula says that the fraction of a job completed
equals the rate of work (portion of the job done per unit
time) multiplied by the amount of time worked.
Example: If the rate tells us that 1/8 of the job is being
done per hour and work is done for 3 hours, then R = 1/8
and T = 3. What is the fraction of the job completed?
F = 3(1/8) = 3/8 (3/8 of the job is completed)
What fraction of the job remains?
5/8
Why?
Because anytime a whole job is done, the fraction
completed will be “1”
Other Forms of Formula:
F = RT
• Given F and R, find T:
T=F/R
• Given F and T, find R:
R=F/T
• General note about formula F = RT when one thing
works alone to complete a job, it’s fraction done is “1”,
but when two or more things work to finish a job, then
the sum of their fractions must be “1”
• Example: If your friend and you work together to finish a
job and you do 2/3 of the job, then your friend must do:
1/3 of the job
Solving Work Problems with
Formula and Chart
1.
2.
3.
4.
5.
6.
7.
Immediately write formula: F = RT as a heading on two charts, one
labeled “alone” and the other labeled “together”
Make and label one line in both charts for everything “working”
Write “x” in the box for the most basic unknown
Fill out the remainder of that column based on information given in the
problem
As you fill out other boxes in both charts based on information given
always:
Always put F = 1 in all boxes in the column in the alone chart
Always put the same value for R in both the “alone” and “together” charts
Use the formula to fill out the final box in a row when other boxes are
known
Always write an equation based on the fact that when things are working
together, the sum of their fractions, F, must equal 1
Solve the equation and answer the original question
Solve the Work Problem:
•
If A, working alone, takes 5 hours to complete a
job, and B, working alone, takes 9 hours to
complete the same job, how long should it take
to do the job if they work together?
1. Immediately write formula: F = RT as a
heading on two charts, one labeled “alone” and
the other labeled “together”
Alone
Together
F= R
T
F= R
T
Example Continued
2.
Make and label one line in both charts for everything “working”
Alone
Together
F=
R
T
F=
R
T
A
B
3.
Put “x” in box for most basic unknown. What is it?
Time working together (same for both)
Alone
Together
F=
R
T
F=
R
A
B
T
x
x
Example Continued
4.
Fill out the remainder of that column based on information given in the problem
Already done in this example since that together A and B worked the same time
Alone
Together
F=
R
T
F=
R
T
A
x
x
B
5.
Fill out other boxes in both charts based on information given
Always put F = 1 in all boxes in the column in the alone chart
Use the formula to fill out the final box in a row when other boxes are known
Always put the same value for R in both the “alone” and “together” charts
Alone
Together
F=
R
T
F=
R
T
1
A
1
1
x
5
5
5
B
1
1
1
x
9
9
9
x
x
5
9
1
1
R
F
T
F  RT
Example Continued
6.
Always write an equation based on the fact that when things are
working together, the sum of their fractions, F, must equal 1
Looking at the charts below, how would you write an equation that
says “the sum of the fractions of their work is one”?
1
1
x  x 1
5
9
A
B
Alone
F=
R
1
1
1
5
1
9
T
5
9
Together
F=
R
1
x
5
1
x
9
1
5
1
9
T
x
x
Example Continued
1
1
x  x 1
5
9
7. Solve the equation and answer question:
1 
1
45 x  x   451
9 
5
9 x  5 x  45
14 x  45
45
3
x
 3 hours to complete the job together
14
14
Solve this problem:
•
When A and B work together they can complete a job
in 7 hours, but A is twice as fast as B. How long would
Basic Unknown?
it take B to do the job alone?
Alone
F=
R
A
B
1
1
Always T rue?
2
x
1
x
F
R
T
Equation:
Together
F=
R
T
14
x
7
x
x
2
x
Remainder
of
Column?
F  RT
14 7
 1
x x
2
x
1
x
T
7
7
Other Informatio n?
Example Continued
14 7
 1
x x
 14 7 
x    x 1
 x x
14  7  x
21  x
It would take" B" 21 hours workingalone to do the job.
Homework Problems
•
•
•
•
Section: 1.2
Page: 103
Problems: 29 – 34
MyMathLab Homework Assignment 7
Solving Mixture Problems With
Formula: IA = (IP) (SA )
• This formula tells us that the amount of an ingredient, IA,
is equal to the percent of the ingredient, IP, multiplied by
the amount of the substance that includes the ingredient,
SA
• Example: If a 20 gallon tank contains 15% gasoline,
what is the amount of gasoline in the tank?
IA = (IP)(SA)
IA = (15%)(20) = (.15)(20) = 3 gallons
• Note: Like all other formulas, this formula can be solved
for any of the variables as necessary
• For mixture problems it is important to realize that “mix”
means “add”
Solving Mixture Problems with
Formula and Chart
1.
2.
3.
4.
5.
6.
7.
Immediately write formula: IA = (IP)(SA) as a heading on a chart
Make and label one line in the chart for everything “being mixed”,
and another line for the “mixture”
Write “x” in the box for the most basic unknown
Fill out the remainder of that column based on information given in
the problem
As you fill out other boxes in the chart based on information given
always use the formula to fill out the final box in a row when other
boxes are known
Always write an equation based on the fact that the sum of the
individual ingredient amounts equals the amount of the ingredient
in the mixture
Solve the equation and answer the question
Solve the mixture problem:
WriteEquation:
•
.1x  3.5  .3( x  5)
Basic Unknown?
A chemist needs a mixture that is 30% alcohol,
but has one bottle labeled 10% alcohol and
another labeled 70%. How much 10% alcohol
should be mixed with 5 liters of 70% to get a
mixture that is 30% alcohol?
IA =
(IP)
(SA)
x
10%A
.1x .10
3.5 .70
70%A
5
30%A .3 x  5 .30
x5


Use formula!
Other Informatio n?
Remainder
of
Column?
Example Continued
.1x  3.5  .3( x  5)
.1x  3.5  .3x  1.5
10.1x  3.5  10.3x  1.5
x  35  3x  15
20  2 x
10  x
10 liters of 10% alcohol is needed.
Homework Problems
•
•
•
•
Section: 1.2
Page: 104
Problems: 35 – 40
MyMathLab Homework Assignment 8
Solving Simple Interest
Problems With Formula: I = PRT
• Formula means that the interest earned from an
investment is equal to the amount of the
investment, P, multiplied by the interest rate, R
(percent), multiplied by the time, T, measured in
years or parts of years
• Example: Calculate the interest earned on
$2,000 invested at 5% interest for 3 years and 6
months:
P = $2,000, R = 5%, T = 3.5 years
I = ($2,000)(.05)(3.5) = $350
Solving Simple Interest
Problems
1.
2.
3.
4.
5.
6.
7.
Immediately write formula: I = PRT as a heading on a
chart
Make and label one line in the chart for each
investment
Write “x” in the box for the most basic unknown
Fill out the remainder of that column based on
information given in the problem
As you fill out other boxes the chart based on
information given always use the formula to fill out the
final box in a row when other boxes are known
Always write an equation based on the fact that the
“total interest” is the sum of the individual interests
Solve the equation and answer the question
Solve the Simple Interest
WriteEquation:
x
Problem
Basic Unkown?
.06 x  .04  400
2
•
A man invests some money at 6%
interest and half that amount at 4%
interest. If his annual income from the
two investments is $400, how much did
he invest at each rate?
Use Formula : I =
P
R
T
6%Inv
.06 x
 x
4%Inv .04 
2
x
.06
1
x
2
.04
1
Remainder
of
Column?
Other Informatio n?
Time of Investment ?
Example Continued
Solve the equation:
x
.06 x  .04  400
2
.06 x  .02 x  400
100.06x  .02x  100400
6 x  2 x  40000
8 x  40000
x  5000
The amount invested at 6% was $5,000 and the
amount invested at 4% was half that amount, $2,500.
Homework Problems
•
•
•
•
Section: 1.2
Page: 105
Problems: 41 – 46
MyMathLab Homework Assignment 9
• MyMathLab Homework Quiz 2 will be due
for a grade on the date of our next class
meeting!!!
Imaginary Unit, i
• Introduction:
In the real number system an equation such as: x2 = - 1
has no solution. Why?
The square of every real number is either 0 or positive.
To solve this equation, a new kind of number, called an
“imaginary unit”, i, has been defined as:
i  1 and i 2  1
Note: i is not a real number
This definition is applied in the following way:
1 is always replacedby i , not theother wayaround.
i 2 is always replacedby 1, not theother wayaround.
Complex Number
• A “complex number” is any number that can be written
in the form:
a + bi where “a” and “b” are real numbers and “i” is the
imaginary unit (This is called “standard form” of a
complex number)
Based on this definition, why is every real number also a
complex number?
Every real number “a” can be written as:
“a + 0i”
Write - 5 in the standard form of a complex number:
- 5 = - 5 + 0i
Complex Number Continued
• Are there some complex numbers that are not real?
Yes, any number of the form “a + bi” where “b” is not
zero.
7 + 3i is a complex number that is not real
• Every complex number that contains “i” is called “a
non-real complex number”
2 - 5i is an “non-real” complex number
4 is a “real” complex number
• Every complex number that contains “i”, but is missing
“a” is called “pure imaginary”
8i is a “pure imaginary” non- real complex number
Square Roots of Negative
Radicands: Imaginary Numbers
• Definition:  a  i a
• Note: A square root of a negative
radicand must immediately be changed
to an imaginary number before doing
any other operations
• Examples:  4  i 4  2i
5  i 5
 2  8  i 2  i 8  i 2 16   14   4
Homework Problems
•
•
•
•
Section: 1.3
Page: 113
Problems: 1 – 16, Odd: 25 – 41
MyMathLab Homework Assignment 10
Addition and Subtraction of
Complex Numbers
1. Pretend that “i” is a variable and that a
complex number is a binomial
2. Add and subtract as you would binomials
Example:
(2 + i) – (-5 + 7i) + (4 – 3i)
2 + i + 5 – 7i + 4 – 3i
11 – 9i
Homework Problems
•
•
•
•
Section: 1.3
Page: 114
Problems: Odd: 43 – 49
MyMathLab Homework Assignment 11
Multiplication of Complex Numbers
1. Pretend that “i” is a variable and that a
complex number is a binomial
2. Multiply as you would binomials
3. Simplify by changing “i2” to “-1” and combining
like terms
Example:
(-4 + 3i)(5 – i)
-20 + 4i + 15i - 3i2
-20 + 4i + 15i + 3
-17 + 19i
Homework Problems
•
•
•
•
Section: 1.3
Page: 114
Problems: Odd: 51 – 67
MyMathLab Homework Assignment 12
Division of Complex Numbers
1. Write division problem in fraction form
2. Multiply fraction by a special “1” where
“1” is the conjugate of the denominator
over itself
3. Simplify and write answer in standard
form: a + bi
Example:
4  3i   3  2i 
4  3i 3  2i 12  8i  9i  6i 2
4  3i




2
3  2i 3  2i
3  2i
9  4i
6 17
6  17 i
12  8i  9i  6
 i


13 13
13
94
Homework Problems
•
•
•
•
Section: 1.3
Page: 115
Problems: 83 – 93
MyMathLab Homework Assignment 13
Simplifying Integer Powers of “i”
• Every integer power of “i” simplifies to one
of four possible values: i, -1, -i, or 1
• When “n” is an integer:
n
i is ALWAYSoneof thesefour possible answers!
Simplifying “in”
for Even Positive Integer “n”
•
Use the following procedure to simplify
any even positive integer power of “i”, in
1. If “n” is even write in = (i2)m for some
integer “m”
2. Change i2 to -1 and simplify
Example:
   i   1
i  i
14
2 ?
2 7
7
 1
Simplifying “in”
for Odd Positive Integer “n”
•
Use the following procedure to simplify
any odd positive integer power of “i”, in
1. Write in = i(i)n-1 (Note: n – 1 will be even)
2. Finish simplifying by using rules for
simplifying even powers of “i”
Example:
 
i  i i  i i
33
32
2 16
 i1  i1  i
16
Simplifying “in” for
Negative Integer “n”
1. First use definition of negative exponent
2. Simplify according to rules for even and odd
exponents as already explained
3. If necessary perform any division (never leave
an “i” in a denominator)
Example:
1
1
1
1
1



i  15  14 
7
7
2
i i
i
i   1
i
i i
i
i
i
1 i

 i

 
2
  1
1
i
i i
15
 
Homework Problems
•
•
•
•
Section: 1.3
Page: 114
Problems: Odd: 69 – 79
MyMathLab Homework Assignment 14
• MyMathLab Homework Quiz 3 will be due
for a grade on the date of our next class
meeting!!!
Quadratic Equations
• Technical Definition: any equation in one variable that
can be written in the form:
ax2 + bx + c = 0 where “a”, “b”, and “c” are real and a ≠ 0
(This form is called the “standard form”)
• Practical Definition: A polynomial equation of degree 2
• Examples:
5x2 + 7 = – 4x
9x2 = 4
2x(x – 3) = x – 1
Are any of thesein standardform?
No, but all could be put in standardform.
Solving Quadratic Equations
• There are four possible methods:
– Square root method
– Zero factor method
– Completing the square method
– Quadratic formula method
• The last two methods will solve any
quadratic equation
• The first two work only in special situations
Square Root Method
• Can be used only when:
– the first degree term is missing,
– or when the variable is found only within
parentheses with an exponent of two on the
parentheses
• Which of these can be solved by this method?
5x 2  3  0
2
x  5x  6  0
2x  3  6  4
xx  3  4
2
First degree termmissing
Variable onlyin parentheses with exponent2
.
Steps in Applying
Square Root Method
1. Write equivalent equations to isolate
either the variable squared, or the
parenthesis squared
2. Square root both sides, being sure to
put a “plus and minus sign” on any real
number that is square rooted (This step
reduces the equation to two linear
equations)
3. Solve the linear equations
Example of Solving by the
Square Root Method
5x 2  3  0
5 x  3
3
2
x 
5
2
3
x 
5
3
x  i
5
3 5
x  i
5 5
i 15
x
5
Note: 2 pure imaginarysolutions
Second Example of Solving by
the Square Root Method
2x  3  6  4
2
2x  3  10
2
x  3
2
5
x 3   5
x  3 5
Note : 2 irrational solutions
x  3 5
x  3 5
Homework Problems
•
•
•
•
Section: 1.4
Page: 124
Problems: Odd: 19 – 29
MyMathLab Homework Assignment 15
Zero Factor Method
• Put equation in standard form (one side
zero other side in descending powers)
• Factor non-zero side
(If it won’t factor this method won’t work!)
• Use zero factor property that says:
ab = 0 if and only if a=0 or b=0
• Set each factor equal to zero
• Solve resulting equations
Example
Consider the following equation:
x2 x  5  3
Is this equation linear or quadratic?
Quadratic! (2nd degree)
Could it be solved by square root method?
No (first degree term is not missing and variable
is not entirely inside parenthesis with a square)
What other method might be used to solve it?
Maybe zero factor method will work.
Solving by
Zero Factor Method
x2 x  5  3
Put in standard form:
2 x 2  5x  3
2 x 2  5x  3  0
Factor non-zero side:
2x 1x  3  0
Apply zero factor principle:
x3 0
2x 1  0
OR
Solve the equations:
x  3
2x  1
1
x
2
Homework Problems
•
•
•
•
Section: 1.4
Page: 124
Problems: 13 – 18
MyMathLab Homework Assignment 16
Completing the Square Method
• The third possible method of solving
quadratic equations will solve every
quadratic equation
• in practice this method is used only
when directions dictate
• This method is essential in developing the
fourth method: Quadratic Formula
Completing the Square Method
1. Isolate variables on one side of equal sign and number
on the other side
2. Divide both sides of equation by coefficient of second
degree term (unless it is already one)
2
3. Find “n” by:
1

n   coefficient of 1st degree term
2

4. Add “n” to both sides of the equation (As a result of
doing this, the trinomial on the left will always factor as
the square of a binomial)
5. Factor the side of the equation containing the trinomial
6. Solve the resulting equation by means of the “square
root method”
Example
Consider the following equation:
2
2
2 x  1  8x Standard Form : 2 x  8x  1  0
Is this equation linear or quadratic?
Quadratic! (2nd degree)
Could it be solved by square root method?
No (first degree term is not missing and variable
is not entirely inside parenthesis with a square)
Could it be solved by zero factor method?
No (non-zero side won’t factor)
What method will work?
CompletingtheSquare
Solve by
Completing the Square Method
2 x 2  1  8x
Isolate variables on one side:
2 x  8x  1
2
Divide both sides by coefficient of second degree term:
1
x  4x  
2
2
Calculate “n” by taking ½ times coefficient of first degree
2
term and squaring that:
1


n    4  4
2

Add “n” on both sides of equation:
1
x  4x  4    4
2
7
2
x  4x  4 
2
2
Example Continued
7
x  4x  4 
2
2
Factor trinomial as a square of a binomial:
7
x  2 
2
2
Solve by square root method:
7
x2  
OR x  2  
2
7
x  2
x  2
2
14
x  2
x  2
2
7
2
7
2
14
2
Homework Problems
•
•
•
•
Section: 1.4
Page: 124
Problems: Odd: 31 – 41
MyMathLab Homework Assignment 17
Quadratic Formula Development
Solve standard form of quadratic equation by
completing the square:
2
2
b
b
c


2
x   2 
ax  bx  c  0
2a  4a a

2
ax  bx  c
2
2
b
b
4ac


b
c
2
x   2  2
x  x
2a 
4a 4a

a
a
2
2
2
b

b
 4ac
b
1 b
x

n     2
2a
2a
 2 a  4a
.
2
2
2
 b  b  4ac
b
b
b
c
x
x2  x  2  2 
2a
a
4a
4a
a
Steps in Using the Quadratic
Formula to Solve an Equation
• Write the quadratic equation in standard
2
form: ax  bx  c  0
• Determine the values of “a”, “b”, and “c”
• Plug those values into the quadratic
formula:
2
 b  b  4ac
x
2a
• Simplify
Solve by Using the
Quadratic Formula
3x 2  4  4 x
Write in standard form:
3x 2  4 x  4  0
Find “a”, “b”, and “c”:
a = 3, b = -4, and c = -4
Plug these into quadratic formula:
 b  b  4ac
x
2a
2
Simplify:
2




  4   4  43 4 
x
23
4  64
4  16  48

x
6
6
48

6
12
x
2
6
4
2
x 
6
3
Homework Problems
•
•
•
•
Section: 1.4
Page: 124
Problems: Odd: 45 – 57
MyMathLab Homework Assignment 18
Solving Formulas Using the
Quadratic Formula
•
1.
2.
3.
4.
If the formula to be solved is
“quadratic” in the variable for which
you wish to solve:
Write the formula in standard form for
that variable
Identify “a”, “b”, and “c”
Plug into quadratic formula
Simplify
Example
Solve for t:
s  t (v  .5gt)
Is this equation linear or quadratic for t?
2
Quadratic:
s  vt  .5gt
Put in standard form for t:
0  .5gt2  vt  s
Identify a, b and c:
a  .5g b  v c  s
Plug into quadratic formula & simplify:
 v  v 2  4.5 g ( s)
.t 
2.5 g 
 b  b 2  4ac
t
2a
 v  v 2  2 gs
t
g
Homework Problems
•
•
•
•
Section: 1.4
Page: 124
Problems: 63 – 70
MyMathLab Homework Assignment 19
“Discriminate” Determines the
Number and Type of Solutions of a
Quadratic Equation: ax2+bx +c = 0
• The “discriminate” of a quadratic equation is the radicand of the
quadratic formula:
 b  b 2  4ac
x
disc = b2 – 4ac
2a
• If disc = 0, then whole quadratic formula becomes
x = -b/2a,
so the equation has one rational solution
• If disc is negative, then solution involves a square root of a
negative radicand with a ± in front, so there will be two non-real
complex solutions
• If disc is positive perfect square, then radical will disappear, but
there is still a ±, so there will be two rational solutions
• If disc is positive but not a perfect square, then radical will remain
with a ±, so there will be two irrational solutions
Examples of Using Discriminate
to Determine Nature of Solutions
Disc = b2 – 4ac
5x2 – 3x + 2 = 0
Disc =
(-3)2 – 4(5)(2) = 9 – 40 = - 31
Two non-real complex solutions
3x2 – 4x – 2 = 0
Disc =
16 – 4(3)(-2) = 16 + 24 = 40 (positive, but not perfect
square)
Two irrational solutions
Homework Problems
•
•
•
•
Section: 1.4
Page: 125
Problems: Odd: 71 – 79
MyMathLab Homework Assignment 21
• MyMathLab Homework Quiz 4 will be due
for a grade on the date of our next class
meeting!!!