College Algebra Exam 1 Material
Dec 25, 2015
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
223x
254x
29x x12 4216x x40 25
Homework Problems
• 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
(a + b)2 =
(a + b)m is NOT equal to am + bm
(a – b)m is NOT equal to am – bm
22 2 baba
Patterns in Binomials Raised to Whole Number Powers
0ba
1ba
22 121 baba
3ba
2ba
1
ba 11
2baba 22 2 bababa 322223 22 babbaabbaa
3223 1331 babbaa 3ba
3ba
Patterns in Binomials Raised to Whole Number Powers
0ba 1ba
22 121 baba 3ba 2ba
1ba 11
3223 1331 babbaa :exponent with compared termsofNumber more one Always
:exponent with compared termsof Degree exponent as degree same All
:always is first term oft Coefficien 1
term.previous theofnumber
by the divided term,previousfor ""on exponent by multiplied
termprevious oft coefficien always is sother term oft Coefficien
a
each term. up goes and zeroat starts b""on exponent each term,
onedown goes and binomialon exponent at starts a""on Exponent
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:
5ba 51a 2310 baba45 3210 ba 45ab 51b
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
543223455 15101051 babbababaaba
322345 3210321032521 yxyxyxx
532 yx
54 31325 yyx 54322345 243810108072024032 yxyyxyxyxx 532 yx
Homework Problems
• Section:
• Page:
• Problems: Binomial Worksheet
• There is no MyMathLab Homework Assignment that corresponds with these problems
Binomial Expansion WorksheetUse the Binomial Theorem to expand and simplify each of the following:1.
2.
3.
4.
5.
6.
7.
8.
41x 53yx
42yx
432 yx
32yx 32 yx
62x
522 yx
Equation• a statement that two algebraic expressions are equal• Many different types with different names• Examples:
• Many other types of equations – (we will learn names as we go)
xx 6173
xxx 725 2
4552 xx
7254 x
LinearQuadratic
Radical
Value Absolute
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 = 7Is 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 + 5Why 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
• Multiplication Property of EqualityWhen the same non-zero value is multiplied (or divided) on both sides of an equation, the result is an equivalent equation
.
x: toequivalent is 73x 4
x: toequivalent is 23
1x 6
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:
1253 xxx xxx4
37.14
12153 xxx xxx4
37.14
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
• Identify the type of equation:
• Get rid of parentheses:
• Get rid of fractions and decimals by multiplying both sides by LCD:
2
17.
3
22
xxx
linear! isIt
2
17.
3
42 xxx
:is 2 and 10, 3, of LCD 30
2
1307.
3
4230 xxx
1530214060 xxx
Example Continued
• 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 coefficient of variable:
1530214060 xxx
15304039 xx
right on the thoseof ridget you willleft on variableskeep tochoseyou If15409 x
559 x
9
55x
Homework Problems
• Section: 1.1
• Page: 90
• Problems: Odd: 9 – 27
• MyMathLab Homework Assignment 2
Contradiction
Solve: 2x – (x – 3) = x + 7What type of equation?Solve by linear steps: 2x – x + 3 = x + 7 x + 3 = x + 7What’s wrong?This says that 3 added to a number is the same as 7 added to the number - No matter what type of equation, when we reach an obvious impossibility, the equation is a classified as a “contradiction” and has “no solution”
Linear
!IMPOSSIBLESTHAT'
Identity
Solve: x – (2 – 7x) = 2x – 2(1 – 3x)
What type of equation?
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”
Linear
,
Homework Problems
• 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 rectangleP = 2L + 2W Perimeter of rectangleI = PRT Simple InterestD = RT DistanceIn 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:
• 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 variableP = 2L + 2W is solved for P, but can be solved for L or W
B
AAB 2 Isolated! is Variable No
Solving Formulasfor 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”:
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.)
2
2
1gtvts
Solving a Formula for a Specified Variable
Solve for “g”:
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
2
1gtvts
2
Solving a Formula for a Specified Variable
2
2
1gtvts
2
2
122 gtvts
222 gtvts next?What g with termIsolate222 gtvts next?What g oft coefficienby Divide
gt
vts
2
22 ! for solved Now g
:LCDby sidesboth Multiply
Example Two
Solve for y: 323 yyxx
6232 yxyx
xyyx 3622 xyyx 3262
factor! for, solved being able with vari termone than more is When there
yxx 3262
yx
x
32
62
Homework Problems
• 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 Continued4. 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 + 5The 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 xLength of second x + 5Length 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 unknownsWhat other information is given in the problem that has not been used?Total length of pipe is 31 inchesHow 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 questionThis 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 = 4Is 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:P = 2L + 2W
Original New
cm 4 W 2 -OL
1 OW
18 Perimeter
Example Continued
• Make a word list of all unknowns:length of originalwidth of originallength of newwidth 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:
length of original:width of original:length of new:width of new:
• 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 cmUse formula: P = 2L + 2W18 = 2(x + 2) + 2(x + 1)
4xx
2x1x Original New
cm 4 W 2 -OL
1 OW
18 Perimeter
x
4x 2x1x
Example Continued
• Solve equation and answer the original question:18 = 2(x + 2) + 2(x + 1)18 = 2x + 4 + 2x + 218 = 4x + 612 = 4x 3 = xlength of original:x + 4 = 3 + 4 = 7width of original:x = 3
Homework Problems
• 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 DExample: 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 unknown4. Fill out the remainder of that column based on
information given in the problem5. Fill out one more column based on the most specific
information given in the problem6. 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
• 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.
1. Immediately write formula: D = RT as a heading on a chart:2. Make and label one line in the chart for everything “moving”:
D = R TLisa
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
Dionne
x
Example Continued
4. Fill out the remainder of that column based on information given in the problemWhat is the other item in that column?Dionne’s speed.How would we describe it with an algebra description?x + 5
D = R TLisa
Dionne
x
5x
Example Continued
5. Fill out one more column based on the most specific information given in the problemIs 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 TLisa
Dionne
x
5x2
5.2
Example Continued6. 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
Dionne
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
What other information is given in the problem that was not used in making the chart?Dionne’s distance was 40 miles more than Lisa’s distance: 2.5(x + 5) = 2x + 40
x
5x25.2
x 55.2 xx2
Example Continued
2.5(x + 5) = 2x + 40 2.5x + 12.5 = 2x + 4010(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
• 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?5/8Why?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. Immediately write formula: F = RT as a heading on two charts, one labeled “alone” and the other labeled “together”
2. Make and label one line in both charts for everything “working”3. Write “x” in the box for the most basic unknown4. Fill out the remainder of that column based on information given in the
problem5. 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 chartAlways put the same value for R in both the “alone” and “together” chartsUse the formula to fill out the final box in a row when other boxes are known
6. Always write an equation based on the fact that when things are working together, the sum of their fractions, F, must equal 1
7. 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 TogetherF= R T F= R
T
Example Continued2. Make and label one line in both charts for everything “working”
Alone TogetherF= 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 TogetherF= R T F= R T
A
B
x
x
Example Continued4. 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 timeAlone TogetherF= R T F= R T
A
B
5. Fill out other boxes in both charts based on information givenAlways put F = 1 in all boxes in the column in the alone chartUse the formula to fill out the final box in a row when other boxes are knownAlways put the same value for R in both the “alone” and “together” charts
Alone TogetherF= R T F= R T
A
B
xx
xx
59
11
T
FR
5
15
1
9
19
1
RTF
x5
1
x9
1
Example Continued6. Always write an equation based on the fact that when things are
working together, the sum of their fractions, F, must equal 1Looking at the charts below, how would you write an equation that says “the sum of the fractions of their work is one”?
Alone TogetherF= R T F= R
T
A
B
xx
59
11
5
15
1
9
19
1
x5
1
x9
1
19
1
5
1 xx
Example Continued
7. Solve the equation and answer question:
19
1
5
1 xx
1459
1
5
145
xx
4559 xx
4514 x
togetherjob thecomplete tohours 14
33
14
45x
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 it take B to do the job alone?
Alone TogetherF= R T F= R
TA
B
7x
11
x
2
x
1x
14
x
7x
2
x
1 72
x
T
FR
RTF
1714
xx
Unknown?Basic
Column?
of
RemainderTrue? Always n?InformatioOther
:Equation
Example Continued
1714
xx
1714
x
xxx
x714
x21
job. thedo toalone workinghours 21 B"" takeIt would
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. Immediately write formula: IA = (IP)(SA) as a heading on a chart2. Make and label one line in the chart for everything “being mixed”,
and another line for the “mixture”3. Write “x” in the box for the most basic unknown4. Fill out the remainder of that column based on information given
in the problem5. 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
6. 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
7. Solve the equation and answer the question
Solve the mixture problem:
• 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)
10%A
70%A
30%A
x5
5x
10.70.30.
x1.5.3
53. xformula! Use
)5(3.5.31. xx:Equation Write Unknown?Basic
Column?
of
Remaindern?InformatioOther
Example Continued
)5(3.5.31. xx5.13.5.31. xx
5.13.105.31.10 xx15335 xx
x220 x10
needed. is alcohol 10% of liters 10
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 yearsI = ($2,000)(.05)(3.5) = $350
Solving Simple Interest Problems
1. Immediately write formula: I = PRT as a heading on a chart
2. Make and label one line in the chart for each investment
3. Write “x” in the box for the most basic unknown4. Fill out the remainder of that column based on
information given in the problem5. 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
6. Always write an equation based on the fact that the “total interest” is the sum of the individual interests
7. Solve the equation and answer the question
Solve the Simple Interest Problem
• 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?
I = P R T6%Inv
4%Inv
x
2
x
Unkown?Basic
Column?
of
Remainder n?InformatioOther
06.
04.?Investment of Time
1
1
:Formula Use
x06.
204.
x
:Equation Write
4002
04.06. x
x
Example Continued
Solve the equation:
The amount invested at 6% was $5,000 and the amount invested at 4% was half that amount, $2,500.
4002
04.06. x
x
40002.06. xx
40010002.06.100 xx
4000026 xx
400008 x5000x
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: Note: i is not a real numberThis definition is applied in the following way: around.other way not the , by replaced always is 1 i
around.other way not the ,1by replaced always is 2 i
1 and 1 2 ii
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 number4 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:
• Note: A square root of a negative radicand must immediately be changed to an imaginary number before doing any other operations
• Examples:
aia
5
4
82
4i i2
5i
82 ii 4162i 41
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 binomials3. Simplify by changing “i2” to “-1” and combining
like termsExample:(-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:
i
i
23
34
ii 2334
i
i
i
i
23
23
23
34
2
2
49
69812
i
iii
49
69812 ii 13
176 ii
13
17
13
6
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:
answers! possiblefour theseof one ALWAYS is ni
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:
14i ?2i
72i 1 71
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:
33i 162ii i 161i 32ii 1i
Simplifying “in” forNegative 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:
15i 15
1
i
14
1
ii
72
1
ii
71
1
i
i1
i
i
i
1 2i
i 1
i
1
i i
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 = 42x(x – 3) = x – 1
form? standardin theseofany Areform. standardin put be could allbut No,
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?
.
035 2 x
4632 2 x0652 xx
43 xx
missing termdegreeFirst
2exponent with sparenthesein only Variable
Steps in ApplyingSquare 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
035 2 x
35 2 x
5
32 x
5
3x
5
3ix
5
5
5
3ix
5
15ix
solutionsimaginary pure 2 :Note
Second Example of Solving by the Square Root Method
4632 2 x
1032 2 x
53 2 x
53 x
53x
53 x
53x
solutions irrational 2 :Note
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:
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.
352 xx
Solving byZero Factor Method
Put in standard form:
Factor non-zero side:
Apply zero factor principle:
Solve the equations:
352 xx
352 2 xx0352 2 xx
0312 xx
12 x
03 x012 x
2
1x
3x
OR
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)
3. Find “n” by:
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 trinomial6. Solve the resulting equation by means of the “square
root method”
2
termdegree1st oft coefficien 2
1
n
Example
Consider the following equation:
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?
xx 812 2 :Form Standard 0182 2 xx
Square theCompleting
Solve byCompleting the Square Method
Isolate variables on one side:
Divide both sides by coefficient of second degree term:
Calculate “n” by taking ½ times coefficient of first degree term and squaring that:
Add “n” on both sides of equation:
xx 812 2
182 2 xx
2
142 xx
n 442
12
42
1442 xx
2
7442 xx
Example Continued
Factor trinomial as a square of a binomial:
Solve by square root method:
2
7442 xx
2
72 2 x
2
72OR
2
72 xx
2
72 x
2
72 x
2
142 x
2
142 x
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:
.
02 cbxax
cbxax 2
a
cx
a
bx 2
n2
22
42
1
a
b
a
b
a
c
a
b
a
bx
a
bx
2
2
2
22
44
a
c
a
b
a
bx
2
22
42
22
22
4
4
42 a
ac
a
b
a
bx
a
acb
a
bx
2
4
2
2
a
acbbx
2
42
Steps in Using the Quadratic Formula to Solve an Equation
• Write the quadratic equation in standard form:
• Determine the values of “a”, “b”, and “c”
• Plug those values into the quadratic formula:
• Simplify
02 cbxax
a
acbbx
2
42
Solve by Using theQuadratic Formula
Write in standard form:
Find “a”, “b”, and “c”:a = 3, b = -4, and c = -4
Plug these into quadratic formula:
Simplify:
xx 443 2
0443 2 xx
a
acbbx
2
42
32
43444 2 x
6
48164 x
6
644
6
84
26
12x
3
2
6
4x
Homework Problems
• Section: 1.4
• Page: 124
• Problems: Odd: 45 – 57
• MyMathLab Homework Assignment 18
Solving Formulas Using the Quadratic Formula
• If the formula to be solved is “quadratic” in the variable for which you wish to solve:
1. Write the formula in standard form for that variable
2. Identify “a”, “b”, and “c”
3. Plug into quadratic formula
4. Simplify
Example
Solve for t:Is this equation linear or quadratic for t?Quadratic:Put in standard form for t:
Identify a, b and c:
Plug into quadratic formula & simplify:
.
)5.( gtvts
25. gtvts
svtgt 25.0
ga 5. vb sc
a
acbbt
2
42
t
gsgvv
5.2
)(5.42 g
gsvvt
22
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:disc = b2 – 4ac
• 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
a
acbbx
2
42
Examples of Using Discriminate to Determine Nature of Solutions
Disc = b2 – 4ac
5x2 – 3x + 2 = 0Disc = (-3)2 – 4(5)(2) = 9 – 40 = - 31Two non-real complex solutions
3x2 – 4x – 2 = 0Disc = 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!!!