college algebra exam 1 material. special binomial products to memorize when a binomial is squared,...

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


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



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


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


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



• 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



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


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


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


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)


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


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”?


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?


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



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



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




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


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



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



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



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




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