faculty.ivytech.edufaculty.ivytech.edu/~rwulf/MA 035 notes actual.docx · Web viewMATH 035 STUDY GUIDE NOTES WORKSHEETS Notes contain some copyrighted material from Lial, Hornsby,

MATH 035

STUDY GUIDE

NOTES

WORKSHEETS

Notes contain some copyrighted material from Lial, Hornsby, McGinnis Introductory and Intermediate algebra Published by Pearson Education

1

Math 035 Survival GuideNeed a 70% for Math 035 – Must attend Final – Time and Attitude are the Keys

Commit to Checking off everything on this list and you will succeedHOMEWORK – 100%

All problems copied from the book Show all work needed to complete the problem and clearly indicate your answer Correct problems in class during homework time at the start of each class

Copying from the solutions manual or another student IS NOT completing the homework-no credit. No credit for answers without required work!

TUTORIALS – 100% Every problem must be done correctly at least once (Attempted is not the same as done!) ON TIME on due date. Problems must be 100% before exams!

The tutorials have guided solutions, videos, help me solve this, you can e-mail your instructor, and get help from the LRC tutors-There is no excuse to not finish tutorials!

ATTENDANCE – 100% Attend EVERY class. DO NOT MISS! Contact Instructor as soon as possible when emergencies arrive You may be able to make-up missed days by watching pencasts

TEST/QUIZ/FINAL – how to study/prepare Attend class that day Tutorials must be completed before exams Complete chapter review and study it Study and understand the directions for every type of problem, know when it is finished Study/Rework quizzes, homework, and tutorial problems “Practice Test” – complete and grade yourself Attend review sessions Ask tutors any questions you are not sure about

Make-ups & Retests Retest for scores below 85% - below 70% is failing and should be retested

Regular tests can be retaken to improve scores and/or for skill mastery (except the last test)Retests must be approved by an LRC tutor – a blue sheet is required

Allowed 1 make-up test to be done in the assessment centerMake-up test should be finished within one week of the original test day

USE ALL AVAILABLE RESOURCES

2

LRC tutors – Tutorials – Study Guides – Review Sessions - Teachers

Syllabus and Survival Guide Worksheet NAME:_______________________

1. What is your instructor’s name (first and last)?

2. Where can you find free tutoring for this course?

3. What supplies are needed for this course?

4. What is the lowest grade that you can receive in this course and still progress to the next course?

5. What procedures must you follow when doing homework?

6. Why is attendance important?

7. When must you complete the tutorials for each chapter?

What happens if the tutorials are not done before the test?

8. What does it mean for the tutorials to be complete before each test?

9. What is the procedure for making up a test?

10. What is the process for taking a Retest?

11. What is the difference between a make-up test and a Retest?

12. List three things that may be considered cheating in math class.

13. Summarize you teacher’s cell phone policy.

14. Where do you take make-up tests and Retest? What will you be required to show there?

15. How many late homeworks can you have?

16. What is the final exam policy?

3

MA 035 8.1: Review of Linear Equations/InequalitiesGoals: Review how to solve linear equations in one variable.

Recall that to solve a linear equation we follow these steps:

1. Do all distributive property2. Combine like terms without crossing the equal sign3. Combine like terms by adding the opposite to both sides of the equation

a. Move the variables first, by moving the smallest one to the largestb. Move the numbers to the ‘other’ side of the equal sign

4. Multiply or divide to get a coefficient of 1 on the variable5. Check your answer!

Practice: Solve and check1) 3p + 2p + 1 = -24

2) 4x + 8x = 17x – 9 – 1

3) 5p + 4(3 – 2p) = 2 + p – 10

4) -2 + 3(x + 4) = 8x

4

Clearing Fractions and Decimals: For equations involving fractions, we are allowed to multiply both sides of the equation by the Least Common Denominator (LCD) of the fractions to obtain integer coefficients. Similarly, we can multiply equations involving decimals by a power of 10 in order to obtain integer coefficients.

5)

x4−2 x

3=5

6)

x+76

+ 2 x−82

=−4

7) 0.04x + 0.06(20 – x) = 0.05(50)

8) 0.09k + 0.13(k + 300) = 61

Special Cases. What kind of solutions to the following have?

9) 5(x + 2) – 2(x + 1) = 3x + 1

10) 6(2y + 4) = 4(3y + 5) + 45

Review MA 035 Ch 5: PolynomialsGoals: Perfom operations on polynomials.

To Add polynomials, just combine like terms, by adding their coefficients.1) (2x2 + 5x – 2) + (3x2 – x + 5)

2) Add 5z2 – 10z – 12 3z 2 + 7z – 9

To Subtract, you must distribute the negative and then add.3) (14y3 – 6y2 + 2y – 5) – (2y3 – 7y2 – 4y + 6)

4) Subtract 3mn + 2m – 4n -mn + 4m + n

When multiplying all monomials, just use the rules of exponents.1 ) (8k3 y )(9 ky3 )(−k2 y2)

When multiplying a monomial by a polynomial (other than another monomial), distribute.

2) −7 rx (5r−2rx+7 x 2)

When multiplying two binomials, distribute both terms, the acronym FOIL may be used.3) (4 a−5 )(3 a+7 )

Another Method for multiplying is to work vertically.

4)

(3 k−4 m)¿(k+2m)

6

Multiplying a sum and a difference of two binomials:

( x+ y )( x− y )=x2− y2

1) (4−5u )(4+5u)

2) (8 x2+ y )(8 x 2− y )

To Square a Binomial: 1.) Square the first term

( x+ y )2=x2+2 xy+ y2

( x− y )2=x2−2xy+ y2 2.) Double the Product of 1st term and 2nd term

3.) Square the second term

3) (3 p−4 )2

4) ( y+6 m)2

Caution: ( x+ y )n≠xn+ y n

*Using order of operations on polynomials

5) (4−x )2+3 x ( x−4 )

6) (4 x+1)( x−2)( x+1)

7) (2 x+5 y )(2 x−5 y )−( x2+3 xy+4 y2)+4 x (−2 y )

7

MA 035 Review: Integer Exponents Goals: Use and understand product and power rules for exponents.

Exponential Expression: an is an exponential expression, a is the base, n is the exponent and this is shorthand for a times itself n times.

Practice:a) 25 b) (-3)3

c) (-4)2 d) -42

e) ( -½ )3 f) (-3x)2

Product RuleWhat is the shortcut to find 25·23 ?

Product Rule

Practice:a) 53·57 b) (-3)2(-3)6 c) m8·m6

d) x3·x e) 23·32 f) m2 p3

g) z2·z5·z3 h) 62 + 65

i) 3x2·5x3 j) (-5p4)(9p5)

k) (-3x2y3)(7xy4)

8

Power RuleWhat is the shortcut? (34)2

Power Rule Note: xm + xn ≠ x(m+n)

Practice:a) (52)3 b) (r5)4 c) (a7)8

Examples (“special cases of power rule”)a) (xy)2 = x2y2 b) (3x5)3 = 33x15 c)Practice:a) (uv)3 b) (6p7)2 c) (-2y3)2 d)

Combining Rules Part 1:a) (3x)3(3x)5 b) (-11r)2(-11r)4 c) (x3)2(x5)3

d) (2x2y3z2)3(x4yz2)3 e) (-w4x)2(-w2x3)5

Goals: Use and understand zero and negative exponents and quotient rule for exponents.

Zero ExponentsWhat does it mean to take something to the zero power? Think about the example below.

42·40

9

( xy )

4

= x4

y4

( 23 )

4

Zero Exponent Rule

Practice: a) 60 b) -60

c) (-6)0 d) –(-6)0

e) -60 + 60 f) 01 + 10

g) (x2·x7)0 h) x0·x2

Negative ExponentsWhat does it mean to have a negative exponent, use the example below to find out.

42·4-2

Negative Exponent Rule

Examples: a) 3-2 b) -3-2 c) (-3)-2

Note: Negative exponents and Negative numbers are two different things! i.e -3 ≠ ⅓

Practice:a) 6-1 b) (5z)-3

c) 5z-3 d) 4-1 – 2-1

10

What if the negative exponent is already in the denominator?

e) f)

Quotient RuleWhat is the shortcut?

Quotient Rule

Practice:a) b)

c) d)

11

3−2

2−31

2−3

a8

a3

m8

m1337

35

75 x2

73 x55−6

5−8

Practice: Use rules of exponents to simplifya) x-4·x-6·x8 b)

c) d)

e)

a−12 b−10 c3

a3 b−5 c3f) ( 4 xy−1 z3)2

g)

(−3 x−2 yz5 x−5 )

4

12

(m2 n )−2

m−3n

( 3 x2

y )2

( 4 x3

y−2 )−1

( 2 yx3 )

2

( 4 yx )

−1

Exponent Rules and Examples

A Zero Exponent Rule

a0=1Examples:1) 120=1

2) ( xyz)0=1

3) 4 wrv0=4 wr (1)=4 wr

4) 60+4=1+4=5

5)

6 x0

h0 =6(1)

1=6

B Negative Exponent Rules

a−n= 1an 1

a−n =an ( ab )

−n=( b

a )n

Examples:

1)(−3)−4= 1

(−3)4= 1

81

2)

2 a−2

b−3 c=2 b3

a2c Notice: Only factors with negative exponents were 'moved'.

3)( 2

x )−3

=( x2 )

3= x3

8

4)4−2m−3 n5 p= n5 p

42 m3=n5 p

16 m3

Caution: 4−2

is not a negative number. A negative exponent does not necessarily mean the value is negative. Think reciprocal rather than negative.

C Product Rule

aman=am+n (bases must be the same)

Examples:1) 4 r2r−5r 8=4 r5

2)( x2 y5 z−4 )( xy 3 z2)=x3 y8 z−2= x3 y8

z2

13

3)3−2 a2b−1 a−4 b−10 a=3−2a−1 b−11= 1

32 ab11= 1

9ab11

D Quotient Rule

am

an =am−n

(bases must be the same)Examples:

1)

45

48=4−3= 143=

164

2)

h3 k 4

h−2 k 6=h5 k−2=h5

k2

3)( 2−1 x4 y

x−2 y3 )( 23 y4

x−6 y5 )=22 x4 y5

x−8 y8 =22 x12 y−3= 4 x12

y3

Notice: Product rule was used across first, then quotient rule.

E Power Rules

(am )n=amnPower to a Power: Multiply exponents

(ab )n=an bn or

( ab )

n

=an

bn

Every factor is raised to exponent outside parentheses.

Examples:

1)( s4)−2

(r3 )2=s−8r 6=r6

s8

2)( m3

b−5 )2

(b2 m2)−1= m6

b−10⋅1

b2 m2=m6

m2b−8 =m4 b8

When simplifying, product and quotient rules were used.

3)(5 ab2c−6)2=52 a2b4c−12=25 a2 b4

c12

4)( x−5 y2

x12 z−1 )−3

= x15 y−6

x−36 z3 = x15 x36

y6 z3 = x51

y6 z3

When simplifying, negative exponent and product rules were used.

14

These rules only involve multiplication, division, and powers; not addition or subtraction.am+an≠am+n am−an≠am−n

Worksheet: Simplifying Exponential Expressions

Name______________________________

Simplify each exponential expression using the rules of exponents.

1)(−2 x0 y6 z

y 4 )3

2)( 3 x−4 y

x−3 y2 )−2

3) ( x−2 y−3 )−2(3 x−1 y−2 )34)

(−5ab3 )−2

(−2 a8b−3 )3

5)

a−12 b−10 c3

a3 b−5 c36) (6 x−3 y−5 x0 )(−2x3 y−6 y 0)2

7)( 3 ab−2 c3

4 a0 b4 )2

(5a3 b−2 )3

8) (3 a−2 b)(−2a−1 b−4 )−2 (4−1 a5 b )−1

15

9)(−2 x5 y−3 z−2

3 xy−2 z7 )( 6 x−3 yz5

−4 x4 y2 z−1 )3

10) −(−3 x2 )−4 (6 y−4 x0 y−5 )3 (4 y5 )−1

Math 035 Notes 6.1: Factoring (GCF’s and Grouping)Goals: Find and factor out GCFs, Factor by grouping.

Finding GCF’sAn integer that is a factor of two or more integers is called a common factor. We will be interested in find the largest or Greatest Common Factor of the numbers in our polynomials.

Example: 18 and 24 are both evenly divisible by 6, and 6 is the largest number that will evenly divide into both so 6 is their GCF.

For larger numbers the GCF may not be so obvious. In these cases we can look at the prime factorization of each number to determine what factors should be in the GCF.

Example: 72 = 2*2*2*3*3120 = 2*2*2*3*5432 = 2*2*2*2*3*3*3

All the numbers have factors of both 2 and 3. All of the numbers contain three factors of 2 and each contains 1 factor of 3. Therefore the GCF will also contain those factors: GCF = 2*2*2*3 = 24. Note each of the given numbers can in fact be written as a factor of 24.

Practice: Find the GCF of :32

48

160

16

What is the GCF of :6

12

13 Now let’s add in variables too. What is the GCF?21m7

-18m6

45m8

Find the GCF of:x4y2

x7y5

x3y7

y15

Find the GCF of:-a2b

-ab2

Factoring out GCF’sExample: Factor 5y2 + 10y The GCF is 5y, to write the polynomial as factors we are really dividing out the common factor and rewriting the polynomial using the

distributive property. 5 y ( 5 y2+10 y

5 y )=5 y ( y+2 )

Practice: Always look for ‘Taking out the Greatest Common Factor (GCF)’ FIRST!

17

1) 32 m2+24 m=

2) 16 t 3+24 t2−40 t

3) 2 x5+2 x

4) 5 m4 x+15 m5 x3−20 m4 x8=

5) 12 a2b2+15 ab+6 a3−24 ab2=

In the following polynomial, factor out both a positive GCF and a negative GCF.

6) −6 k 3+3k2−9k=

Sometimes the GCF may be a ‘grouping’.

7) r ( a+3)+4( a+3) (a + 3) is common to both, this will be one factor

= (a+3 )(r+4 ) What was in front of the (a + 3) will be the other.

8) 2 k (k−1)+5( k−1 )

9) x2( x+7 )+9( x+7)

Now we will add a step. You will need to group these 4 term polynomials into 2 pairs. Factor the GCF out of each pair and repeat the steps from 7-9.Example: pq + 5q + 2p + 10

pq + 5q + 2p + 1018

q(p + 5) + 2(p + 5)(p + 5)(q + 2)

Grouping (by pairs) method:

10) ax−ay+2 bx−2by=

11) 2 ab2−8b2+a−4

12) 10 p2+15 p−12 p−18=

13) Sometimes a rearrangement may be necessary4 y−xz+xy−4 z=

14) 6 m2 n+36m+12 mn+18 m2=

15) !the parenthesis must match!

24 xy−42 x−20 y−35=

19

16) 6 ab3−24 b3+3 ab−12 b

Math 035 Lesson 6.2: Factoring TrinomialsGoals: Factor trinomials by product/sum methods.

In this section, we will be writing trinomials as the product of binomial factors:6 x2−7x−20=(2 x−5 )(3 x+4 )2 x2+3 x+1=(2 x+1 )( x+1 )x2−5 x+6=( x−2 )( x−3 ) Those trinomials that cannot be factored are PRIME.

SHORT PRODUCT/SUM METHOD: x2+bx+c Try to find a pair of numbers that have a product of c and a sum of b .

Example:

x2+9 x+20 , the numbers 4 and 5 have a product of 20 and sum of 9, use them to split 9xx2+4x +5x+20 , now use the grouping method from 6 . 1x ( x+4 )+5( x+4 )=( x+4 )( x+5)

Practice:

20

p2+6 p+5

b2−7 b+10

m2−m−6

Product = Sum =First*Last Middle



Using this method will also cause any ‘extra’ letters to fall into place

21

r2+2rs−8 s2

x2 y2+9 xy+20

y2−8 y+6

5 x4−5 x3−100 x2

−3 z4+15 z3−18 z2






Math 035 Lesson 6.3: Factoring TrinomialsGoals: Factor trinomials by product/sum methods.

22

If you used a ‘shortcut’ in section 6.2 it does not work for trinomials that do not start with a 1!!!

Example: 3 y2+7 y+2 , We are looking for two numbers with a product of 3*2 = 6 and sum of 7 . 1 and 6 work3 y2+ y+6 y+2 , Use 1 and 6 to split 7y into two parts, then use grouping to factory (3 y+1 )+2(3 y+1 )(3 y+1)( y+2)

Product =-12 Sum =-11Practice: First*Last Middle

3 y2−11 y−4

6 k 2−19 k+10

12 y2−5 yz−2 z2

7 p2 r2+15 pr+2

48 x2−64 x−35





23

When the leading coefficient is negative, it is helpful but not necessary to factor out a -1 first.

−2 x2−5 x+3

Remember to always look for GCF first!

30 y5−55 y4−50 y3

Higher Powers:y4+ y2−6

6 r8−13 r 4+5

36 m10+18 m5 n4−4 n8

24

MA 035 Notes 6.5: Special Cases of FactoringGoals: Factor using special case patterns.

Difference of Squares:

1) 9 a2−16 b2=

2) 242 y3−128 y=

3) 81 m4−16=

4) 64 y2−49 z2=

5) x16−1=

6) 4 y6−25 r8=

Perfect Square Trinomial:

11) z2+12 z+36=

12) 9 a2+48 ab+64 b2=

25

13) 49 y2 z2−14 yz+1=

14) 25 y4−30 y2 z3+9 z6=

15) Is 9 x2−51 x+16 a perfect square? Can it be factored?

Difference/Sum of Cubes:

7) x3−1000=

8) 27 u3−64=

9) 8 k3+125 y6=

10) 2 x8+54 x2=

26

Quick Check For Factoring Polynomials

Number Of TermsCHECK FOR Two Three Four or

MoreGreatest Common Factor X X X

Difference of Squares

a2−b2=(a+b)( a−b )X

Sum of Cubes

a3+b3=(a+b )(a2−ab+b2)X

Difference of Cubes

a3−b3=( a−b )(a2+ab+b2 )X

Perfect Square Trinomial

a2+2 ab+b2=(a+b)2

a2−2ab+b2=( a−b )2

X

Product Sum Method(‘Grouping Number’ Method) or Reverse

FOIL

X

Grouping by Pairs XGrouping by Cubes or Squares X

Also check your factoring by multiplying back out, and check that you have factored completely.

27

Factoring Decision Tree

Factoring Polynomial Worksheet (or Group Project)

Name_______________________________

Factor each polynomial completely. Write ‘prime’ if the polynomial cannot be factored.

1) 2 x2−18

2) 6 x2−7 x−5

3) 27 x2+36 xy+12 y2

4) 2 x2−100 x+98

28

Check GCF

How Many Terms are there?

2 terms

Check Difference of Squares

Check Sum/Difference

of Cubes

3 terms

Check Perfect Square Trinomial

Use Product/Sum Method

4 terms

Check Factor by grouping pairs

Check Factor by grouping 3 and 1

5) 5 ax+5 ay−20 x−20 y

6) x2+7 x−12

7) x2−10 x−96

8) 1−16 x4

9) 8w3+awx+4 aw2+2w2 x

10) 4 a4+20 a2 b4+25 b8

11) 20 y2−13 y−72

12) a9−27

13) 25 x2+49

29

14) 91+20 x+x2

15) 12 w4+38 w3u−154 w2u2

16) 225 a2 c2−400 b4 c2−36 a2 d2+64 b4 d2

17) x7−16 x3−x4+16

18) 100 a3−9 ab2

19) 14 b2−3 bk−2 k2

20) x5+27 x2−4 x3−108

30

Math 035 Lesson 6.7: Solving Equations by Factoring

Goals: Use zero-factor property and factoring to solve polynomials.

Zero-Factor Property: If ab=0 , then a=0 or b=0 .

Solve each equation.

1) x ( x−1)=0 2) ( x−2 )(3 x+1)=0

3) 2 k (3 k+5 )(2 k+1)=0 4) 2(8 a+3)(2 a−1)(a+5 )=0

Quadratic Equation Form: ax 2+bx+c=0

31

1. Write the equation in standard form, ax 2+bx+c=0 .2. Factor the polynomial.3. Apply the zero-factor property by setting each factor equal to zero.

5) 4 k2−49=0 6) p2=12 p

7) 15 m2+7 m=2 8) −x3+x2=−6 x

9) (a+6 )(a−2 )=a−8

10) 30 k2−14 k=8

11) 3 n3+n2=4 n

32

12) 2 r3+5 r2−2 r−5=0

Math 035 Lesson 6.8: Applications of Quadratic EquationsGoals: Solve applications using the 5 step process.

Steps to follow:1. DATA- Make a list of the data that is given.2. VARIABLE – Represent the unknown quantity/quantities with variable(s).3. PLAN – Tell what relation you will use to get an equation.4. EQUATION – Using the plan and the data, write an equation to solve.5. SOLUTION - Solve the equation and put the answer in terms of the question.

Example Problem One.The Monroes want to plant a rectangular garden in their yard. The width of the garden will be 4 ft less than the length, and they want to have an area of 96 ft2. Find the length and width of the garden.

Data: Rectangular garden, width will be 4 ft less than length. Area of 96 ft2

33

Variable: Length = xWidth = x - 4

Plan: A = L * W

Equation: x(x – 4) = 96x2 – 4x = 96x2 – 4x – 96 = 0(x – 12)(x + 8) = 0x – 12 = 0 x + 8 = 0x = 12 x = -8

Solution: Length = 12 ft Check: 12 * 8 = 96 Width = 8 ft

Practice Problem One

The length of a rectangular room is 2 meters more than the width. The area of the floor is 48 meters squared. Find the length and width of the room.

Data:

Variable:

Plan:

Equation:

34

x

x – 4

Solution:

Example Problem TwoThe product of the numbers on two consecutive post-office boxes is 210. Find the box numbers.

Data: Product of two consecutive boxes is 210

Variable: Box 1 = xBox 2 = x + 1

Plan: Box 1 * Box 2 = 210

Equation: x(x + 1) = 210x2 + x = 210x2 + x – 210 = 0(x + 15)(x – 14) = 0x + 15 = 0 x – 14 = 0x = -15 x = 14

Solution: The box numbers are probably positively numbered so we ignore the negative answer in this case. So Box 1 = 14, Box 2 = 15

Practice Problem TwoThe product of the numbers on two consecutive lockers at the health club is 132. Find the locker numbers.Data:

Variable:

Plan:

Equation:

Solution:

Example Problem Three35

The product of two consecutive odd integers is 1 less than five times their sum. Find the integers.

Data: The product of two consecutive odd integers is 1 less than five times their sum

Variable: Number1 = xNumber2 = x + 2

Plan: Number1 * Number2 = 5(Number1 + Number2) – 1

Equation: x(x + 2) = 5(x + x + 2) – 1 x2 + 2x = 5x + 5x + 10 – 1x2 + x = 10x + 9x2 – 8x – 9 = 0(x – 9)(x + 1) = 0x – 9 = 0 x + 1 = 0x = 9 x = -1

Solution: Since both are odd integers, both are correct solutions.Number1 = 9, Number2 = 11 or Number1 = -1, Number2 = 1

Check: 9*11 = 5(9 + 11) – 1 -1*1 = 5(-1 + 1) -199 = 100 – 1 True -1 = 5*0 – 1 True

Practice Problem Three

Find three consecutive odd integers such that the product of the smallest and the largest is 16 more than the middle integer.

Data:

Variable:

Plan:

Equation:

Solution:36

Example Problem Four

Ed and Mark leave their office, with Ed traveling north and Mark traveling east. When Mark is 1 mile farther than Ed from the office, the distance between them is 2 miles more than Ed’s distance from the office. Find the distances each drove, and the distance between them.

Data: Ed traveled north, Mark traveled east. Mark goes 1 mile farther. Distance between is 2 more than Ed’s

Variable: Ed’s Distance = xMark’s Distance = x + 1Distance Between = x + 2

Plan: Use Pythagorean Theorem:Ed2 + Mark2 = Distance2

Equation: x2 + (x + 1)2 = (x + 2)2

x2 + x2 + 2x + 1 = x2 + 4x + 4 Tip: Square binomials correctly!x2 – 2x – 3 = 0(x – 3)(x + 1) = 0x = 3 x = -1

Solution: Ed’s Distance 3 miles, Mark’s 4 miles, Between 5 miles

Practice Problem FourThe hypotenuse of a right triangle is 3 inches longer than the longer leg. The shorter leg is 3 inches shorter than the longer leg. Find the lengths of the sides for this triangle.

Data:

Variable:

Plan:

Equation:

37

x + 2

x +1

x

x

x - 3 x +3

Solution:

Example Problem Five

A tennis player can hit a ball 180 ft per sec. If she hits a ball directly upward, the height h of the ball in feet at time t in seconds is modeled by the quadratic equation:

h = -16t2 + 180t + 6

How long will it take for the ball to reach a height of 206 feet?

Data: h = -16t2 + 180t + 6, height 206 feet

Variable: t = time, h = height

Plan: h = -16t2 + 180t + 6, substitute in 206 for h and solve for t

Equation: 206 = -16t2 + 180t + 6-16t2 + 180t – 200 = 0-4(4t2 – 45t + 50) = 0(4t – 5)(t – 10) = 04t – 5 = 0 t – 10 = 0t = 5/4 t = 10

Solution: The ball reaches a height of 206 feet at both 1.25 and 10 seconds

Practice Problem Five

An object propelled from a height of 96 ft with an initial velocity of 80 ft per sec after t seconds has height: h = -16t2 + 80t + 96

After how many seconds is the height 160 ft?

38

When does the object reach the ground?

39

ASA Math Test Self Assessment

Name:

You should complete this form (front and back) after you have received your graded test from your instructor. Your teacher may add up to 2 points to your test score, if you complete this form to their satisfaction. Form must be completed within one week of the test being returned to your class.

First check your exam preparation. Put a checkmark on what you did to study for this test, and what you plan on doing for the next test.

Method of Preparation This Test Next TestAttended every class session.

Completed all book homework, by copying, working and checking problems.

Worked tutorials ahead of time to study.

Took notes.

Read the textbook.

Asked questions about confusing parts of the assignment.

Visited tutor(s).

Visited instructor’s office hours.

Attended help session, or supplemental instruction.

Watched pencasts, or MyMathLab videos.

Worked practice test.

Worked in a study group.

Used online tutor, or e-mailed instructor.

Estimate the hours, outside of class, that you spent studying in the week before this test:______

Do you feel that you adequately prepared for this exam? If no what changes could you make to your study habits to improve in the future? __________________________________________________

Were there any study methods that you found helpful on this exam that you will try again?__________

___________________________________________________________________________________

over →

40

What type of errors did you make on this exam? Review the different types of common errors below, and then categorize your errors by listing problem numbers from your test in the right column.

Problem # from your exam

1.) Misread directions Skipped instructions or Misread directions

2.) Careless errors Arithmetic errors or sign errors

3.) Concept errors Did not know how to do the problem, or did not understand the underlying principle, or did not know a formula

4.) Application errors Understood the arithmetic/algebra, but could not apply it to this problem. (common on word problems and graph problems)

5.) Test taking errors Ran out of time, missed problems early on the test due to anxiety or rushing, didn’t finish problem(s), changed answer to incorrect answer, turned test in early without checking answers

6.) Study errors Never learned materials missed during an absence, waited until the last minute to finish tutorials, did not complete homeworks or reviews

For more on test taking skills see: “Winning at Math” by Paul Nolting.

What test taking error cost you the most points for this exam? __________________________________

What steps can you take to fix some of your errors on future tests? ______________________________

____________________________________________________________________________________

What is the most challenging topic for you from this test? ______________________________________

What is the easiest topic for you on this exam? ______________________________________________

Additional comments/questions.__________________________________________________________

Math 035 Notes 7.1: Rational ExpressionsRational Expressions

41

Goals: Find domain, reduce, multiply and divide rational expressions.

A rational expression is a quotient of two numbers, variables, or polynomials. Below are examples of rational expressions. You will notice a rational expression is commonly called a fraction.

34

3 x2

4 x+2 6 x2−4 x+1

x 2 x+ y

3 x−4 y 2x2−3 x

4 x2+6 x+11

Recall that the set of x values is called the domain. A zero denominator makes an undefined rational expression. Any value of x that makes the denominator be zero is NOT in the domain. A) What values of x would have to be restricted from the domains of these functions? B) Write the Domain in set notation

1)2 x+1x−5 4)

82 x2−4 x

2)3+2 x

5 5)4 x

x2+2

3)x+6

x2−x−6

Writing Rational Expressions in Lowest Terms

The fraction 2428 is not in lowest terms (reduced). To get it in lowest terms, divide

the numerator and denominator by the greatest common factor (GCF). The key word here is factor. Factors are multiplied together; terms are added/subtracted together. FACTORS CAN BE CANCELLED; TERMS CANNOT BE CANCELLED!

2428

=4∗64∗7

=67

2428

≠48

or 12

by canceling the 2's

Never, Never Cancel Terms! Only Cancel Factors!

So to reduce rational expressions, we must first factor.3 x−125 x−20

=3 (x−4 )5 (x−4 )

= 35 You CAN NOT CANCEL TERMS!

3 x−125x−20

≠3−35−5

To write a rational expression in lowest terms:

42

1. Factor both numerator and denominator.2. Cancel common (or opposite) factors.3. Multiply across and leave in factored form.

Write the following rational expressions in lowest terms.

1)

2 y+4y2−4

=2)

2 y+4y2+4

=

3)

x2+2x−3x2−3 x+2

=4)

8 m2 n6 mn3 =

5)

a2−93 a+9

=

CAUTION:

a−33

≠a or a−1

6)

ac−ad+bc−bdc2−d2

=

7)

3 z2+z18 z+6

=

43

Opposites in a numerator and denominator equals -1.−33

=−1

2 x−2 x

=−1

−(a−b )(a−b )

=−1

Rational expressions of the form a−bb−a

=−1

7)

2(n−4 )( p−1)(4−n)(1+ p)( p+1 )

=

8) x2−11−x

9)

2r 2−r−64−r2 =

Multiplication or Division of Rational Expressions

To perform multiplication or division with rational expression:1. Factor each numerator and denominator.2. For division only, change to multiply by the reciprocal of the divisor.3. Cancel common (or opposite) factors.4. Multiply across and leave in factored form.

Multiply or Divide.

1)

c2+2cc2−4

⋅c2−4 c+4c2−c

=

2)r2−16r+2

⋅ r3

r 2+4 r=

44

3)

4 x+26x

÷15 x+65 x

=

4)

y2+2 y5+ y

÷ 4− y2

3 y−6=

5)

4 a2−492 a2+a−21

⋅3 a2+14 a+156 a2−11a−35

=

6*) ( 2x3+3 x2−2 x

3 x−15÷ 2 x3−x2

x2−3x−10 )⋅ 5 x2−10 x3 x2+12 x+12

45

Math 035 Notes 7.2: Adding and Subtracting Rational ExpressionsGoals: Add and subtract rational expressions with the same denominator. Find Least common denominator. Add and subtract rational expressions with different denominators.

Addition/Subtraction of Rational ExpressionsIn order to add or subtract rational expressions (fractions), there must be a common denominator.

310

+ 510

= 810

or 45

4x+ y

x=4+ y

xSteps to add or subtract rational expressions with common denominator:

1. Combine the numerators terms together and keep the denominator.2. Factor the numerator, if possible.3. Write in lowest terms, if possible, and leave in factored form.

Examples:

1)

2 ab

+ 3 ab

+ 5b=

2)

2 y−1y2+ y−2

− yy2+ y−2

=

3)

3 x2

x2−4−12

x2−4=

4)

3m2 n

− 2 nm2 n

=

If there are not common denominators, a least common denominator must be found and the numerators ‘rewritten’ for that LCD. Essentially, both numerator and denominators are multiplied by the same quantity.

310

+ 415

=3(3)+4(2 )30

=1730

45 x

+ m2 xy

=4 (2 y )+m(5 )10 xy

= 8 y+5 m10 xy

Steps for find the lowest common denominator:1. Factor each denominator.2. Multiply each factor together the GREATEST NUMBER OF TIMES A FACTOR

APPEARS IN A DENOMINATOR.3. Leave in factored form (except for numbers).

1) 72,54

72=23⋅32

54=2⋅33LCD = 23⋅33=8(27 )=216

46

Numbers can be factored using a ‘factor tree’ or repeated division.

2) 75 x,60 x2 75=3⋅52

60=22⋅3⋅5

LCD = 22 (3 )(52 )x2=300 x2

3) 2m2n2,6 m3 n

4) r,

r+6

5) 6 x−18,

4 x2−12 x

6) 2 x2−6 x,2 x2−18

,x2+6 x+9

7) 2 x2+5 x−3,6 x2+x−2

8) x3−x2−x+1,

x2−x−2

47

Steps to add or subtract rational expressions without common denominators:1. Use the previous steps to find the LCD.2. Rewrite the numerators by multiplying by the ‘missing’ factor necessary to make the

LCD.3. Multiply and combine these terms and place over the LCD.4. Factor the numerator, if possible, and write in lowest terms, if possible.5. Leave in factored form.

Examples:1)

3 x2 x−6

+ 4x+2

− −20x2−x−6

Factor each denominator first, if necessary .

2( x−3 ) ( x−3 )( x+2 )

LCD=2( x+2)( x−3 ) Rewrite each numerator by multiplying by the ‘missing’ factor.

¿3 x (x+2 )+4 (2)( x−3 )−(−20 )(2)2( x+2)( x−3 )

=3 x2+6 x+8 x−24+402( x+2 )( x−3 )

=3 x2+14 x+162( x+2 )( x−3)

¿(3 x+8 )( x+2 )2( x+2)( x−3 )

=3 x+82( x−3 )

2)

6y+ 1

4 y=

3)

5x+3

+ x+2x

+ −6x2+3 x

=

4)

2m

− 4m−2

=

48

5)

2r−2

− r+3r−1

=

6)

yy−2

+ 82− y

=These denominators are opposites. Whenever there are

opposite factors in denominators, multiply one of the rational expressions by−1−1 .

This makes the factors be the same.y

y−2+ 8

2− y⋅−1−1

= yy−2

+ −8y−2

= y−8y−2

7)

x+3x2−6 x+9

− x+2x2−9

− 53−x

=

8)

−aa2+3 a−4

− 4 aa2+7 a+12

=

9)4+ 2

u− 3 u

u+5=

10)

2 x+6x2+6 x+9

+ 5 xx2−9

+ 7x−3

=

11)

3 yy2+xy−2 x2

+ 4 y−xy2−x2

=

49

Math 035 Notes 7.3: Complex Fractions

Goals: Simplify complex fractions. Simplify rational expressions with negative exponents

Complex Fractions: expression having a fraction in the numerator, denominator or both.

Examples:

1+ 12

34−2

3

,

4y

6− 3y

,

m2−9m+1m+3m2−1

It is not acceptable to leave fractions in their complex forms. There are two methods for simplifying them.

Method 1 Simplify numerator and denominator. Then use reciprocal to divide

4+1

x

3+2x2

=

4 xx

+1x

3 x2

x2 +2x2

=

4 x+1x3 x2+2x2

=4 x+1x ÷

3 x2+2x2 =

4 x+1x ⋅

x2

3 x2+2=

( 4 x+1) x3 x2+2

=4 x2+x3 x2+2

Method 2 Multiply numerator and denominator by LCD. Simplify if possible.

4+1

x

3+2x2

=(4+1

x) x2

(3+2x2 )x2

=4 x2+1

xx2

3 x2+2x2 x2

=4 x2+x3 x2+2

Practice: 1) 2)

50

a+25 aa−37 a

2x−35x2−9

3) 4)

3+1

z

3−1z

5y

+7

83 y −1

5) 6)

1w

+1w+1

1w −

2w+1

2 p+5

p−1

3 p−2p

Simplify a rational expression with negative exponents.Start by rewriting the expression with positive exponents then use methods above.7) 8)

51

m−1+ p−2

2 m−2−p−1 =

1m

+1p2

2m2 −1

p

b−4

b−5+2

Math 035 Notes 7.4: Solving Rational Equations

Goals: Determine domain of rational expression. Solve rational equations. Graph rational functions.

Any possible solution for a rational equation that ‘makes’ a denominator be zero, must be disregarded. It can not be a solution.

What x values would have to be excluded from the Domain (or disregarded as possible solutions)?

1)

35 x

+ 2x−1

=52)

12 x( x−2 )(3 x+1)

= x+1x (3 x+1 )(x+5 )

To solve a rational equation:1. Factor the denominators and find the LCD.2. Find all values that would make that LCD be zero. These values cannot be solutions.3. Multiply each term of the equation by the LCD.4. Solve the resulting equation.5. Disregard any possible solutions that are values found in step 2.

Solve the following equations.

1)

−320

+ 2x= 5

4 x LCD = x≠¿ ¿

52

2)

3x+1

= 1x−1

− 2x2−1 LCD =

x≠¿ ¿

3)

4x2+x−6

− 1x2−4

= 2x2+5 x+6 LCD =

x≠¿ ¿

4)

2x+3

− 1x−1

= −x2−3 xx2+2 x−3 LCD =

x≠¿ ¿

5)

4 x−12 x+3

=12 x−256 x−2 LCD =

x≠¿ ¿

6)

xx−3

+ 2x+3

=129−x2

LCD =x≠¿ ¿

53

7)

10m2

− 3m

=1LCD =m≠¿ ¿

8)

5 x+14x2−9

=−2 x2−5x+2x2−9

+ 2 x+4x−3 LCD=

x≠¿ ¿

Graph of f(x) = 1/x Graph g(x)=

−2x−3

Asymptote:

Math 035 Notes 7.5 (Part 1)

Solve for the given variable. Start by clearing the denominator using the LCD.

1)

1f= 1

p+ 1

q for p

2)

32 a

+ 5b=1

c for b

54

3)L= nE

R+nr for r

4)S=nd

2(a+L )

for n

Name:_________________________

GROUP WORK OVER RATIONALEXPRESSIONS AND EQUATIONS

Find all values that would make each rational expression be undefined; in other words, the values that x could not equal.

1) r+7r2−25

2) a+7a3−3 a2+2 a

Write each rational expression in lowest terms.

3)−8 x6 y5

24 x 4 y7 4)11r2−22 r3

6−12r

5)r3−m3

r2−m2 6)8 x2+6 x−916 x2−9

55

Perform the indicated operations.

7)9 k−186 k+12

⋅ 3 k+615k−30 8)

4 z+122 z−10

÷ z2−9z2−z−20

9)p+8

p2−16 p+64− 1

p−8 10)m

m−1− 9

1−m

11)

1x

−1x+1

1x +

3x+1

12)b

a2−b2− a

a2−b2

56

13)4 x−94 x2−9

÷4 x2−13 x+94 x2−12 x+9 14)

5+1

y

5−1y

15)2 x−1

x2+x−2− x

x2+ x−2 16)4− y2

8 y⋅16 y2+8 y

2 y2−3 y−2

17)16−r2

r2+2 r−8÷ r2−2r−8

4−r2 18)4 p−3

3 p− p−5

2 p+ 2

p

Solve each rational equation.

19)9x=5−1

x

57

20)2

x−1+ 1

x+1= 5

4

21)3

r+1− 1

r−1= 2

r2−1

22)5−a

a+ 3

4=7

a

23)x+3

x− x+4

x+5=15

x2+5 x

24)x

x+4=2− 4

x+4

58

25)5

x−2+ 2

2−x= 4

x+1

Math 035 Notes 7.5: Applications of Rational Equations

Proportion Problems1) Biologist tagged 600 fish in a lake on the first day of May. One month later they returned and collected a random sample of 100 fish, 3 of which had been previously tagged. Approximately how many fish does the lake support based on this experiment?

Variable: x = total fish in lake

Plan: taggedfish1totalfish1

=taggedfish 2totalfish2

Equation: 600

x= 3

100

Solution: x = 20000 fish

2) The distance between Lafayette and Vincennes is 160 miles, it is represented by 6 inches on my map. If the distance between Lafayette and Fort Wayne is 120 miles, how far apart are they on the map?

59

3) If the two triangles are similar, find the value for x and all missing sides of the triangles.

Motion Problems4) A plane travels 100 miles against the wind in the same time that it takes to travel 120 miles with the wind. The wind speed averages 20 miles per hour. Find the cruising speed of the plane (sometimes called ground speed).

rate time distanceagainst wind 100

with wind 120

5) A canal has a current of 2 miles per hour. Find the speed of Casey’s boat in still water, if the boat travels 11 miles down the canal in the same time that it goes 8 miles up the canal.

rate time distancedown

up

60

A

B C

D

E F

2x + 44x - 57

10.5

x

6) While on vacation, Jim and Annie decided to drive to Houston. During the first part of their trip, which was on the interstate, they averaged 60 miles per hour. When they got to Houston, traffic caused them to slow down considerably and they only averaged 30 miles per hour. The distance in Houston was 100 miles less than the distance on the interstate. What was their TOTAL distance, if they spend 50 minutes more on the interstate than they did in Houston?

rate time distanceinterstate 60

Houston 30

7) Lucy drove 300 miles north from San Antonio, mostly on the interstate. She usually averaged 55 miles per hour, but an accident slowed her speed through Dallas to 15 miles per hour. If her trip took 6 hours total, how many miles did she drive at the reduced speed?

rate time distanceinterstate 55

Dallas 15

Total 6

Job or Work Problems8) Stan needs 45 minutes to do the dishes, while Bobby can do them in 30 minutes. How long will it take them, if they work together?

rate time (together) fractional partStan 1

45x

61

Bobby 130

x

9) Ben and Patty must clean up often after their grandson has made a mess in his playroom. Ben could always do the job in 15 minutes alone while Patty, working alone, could always clean up in 12 minutes. How long will it take them, if they work together?

rate time (together) fractional partBen

Patty

10) Ron can paint his den in 6 hours working alone. If Dee helps him, the job takes 4 hours. Estimate how long it would take Dee alone to paint the den.

rate time partRon

Dee

11) Ann can clean the bathroom alone in 20 minutes. One day her twin sister, Amy, helps Ann and they complete the cleaning in 8 minutes. How long would it take Amy alone to clean the bathroom?

rate time distanceAnn

Amy

62

12) A vat of acid can be filled by an inlet pipe in 10 hours and emptied by an outlet pipe in 25 hours. How long will it take to fill the vat, if both pipes are left open while acid is filling the vat?

rate time partinlet pipe 1

10outlet pipe − 1

25

63


Name:






Took notes.

Read the textbook.


Visited tutor(s).










___________________________________________________________________________________

over →

64












____________________________________________________________________________________




MA 035 Notes 9.1: Radical ExpressionsGoals: Find square roots, higher roots. Use calculator to estimate roots. Graph.

65

We are already familiar with square numbers:12 = 1 22 = 4 32 = 9 42 = 1652 = 25 62 = 36 72 = 49 82 = 6492 = 81 102 = 100 112 = 121 122 = 144

In this chapter we will be interested in the opposite process.

If a2 = 64, then a = If a2 = 81, then a = If a2 = ¼, then a = In each case, to find a we ask ourselves “what times itself will equal the

number?” Because we are finding the number that is at the root of the square, we call

this finding square roots.

Find all square roots of the following numbers:a) 49

b) 169

c) 25/196

d) .49

e) 18 ?

Principal Root:

Negative Root:

Find the following Roots:

Irrational Numbers:

66

√225 −√169

√3625

−√. 81

√−16 √17≈4 . 123 √37≈¿ ¿

Use a calculator to approximate:

Higher Roots:Similar to square roots, there are cubed roots.

We know that 27 = 33 , and we can find the cubed or third root is 3, by asking what times itself 3 times is 27?

So What are the following roots?

Definition of a Root: if and only if

Radical Sign: n: the index‘a’: the radicand

Signs for higher roots:For roots with an even index, we will follow the same convention as for square roots: is the principal, is the negative root, and the nth root of a negative number is not a real number

Find the roots below:

For roots with an odd index, there will be 1 unique root, and we can take the root of negative numbers.

Special Properties of Roots:If the index = the exponent, the following is generally true.

67

3√1254√163√216

bn=an√a=b√

n√ - n√

4√16−√814√−625

3√−27 3√85√−32−3√125195√−1

n√a n=a

*The exception is if a is negative and n is even.

Be very careful when there is a negative radicand.

If the exponent is a multiple of the index, it is easy to simplify using division.

Approximate these examples using your calculator.

Example: The time for one complete swing or rotation of a pendulum is given by the following formula, where t = time in seconds, L = the length of the pendulum in feet, and g = 32 ft/sec2.

Find the time to the nearest tenth of a second for a pendulum 2 feet long.

Graphing: Graph and give the domain and range.

68

√−72 is not real .√(−7 )2 is 7, because √49 is 7Therefore, √ x2=|x| in general for even n, n√an=|a|

a ) 5√510

b ) 3√r9

c ) √m8

√153√104√2

t=2 π √ Lg

f ( x )=3√ xf ( x )=√ x

Graph and give the domain and range.

MA 035 Lesson 9.2: Rational Exponents

69

f ( x )=3√ x−1f ( x )=√ x+2

Goals: Convert between radical and rational exponent notation. Evaluate expressions of the form am/n Apply rules of exponents to rational exponents.

What would be the value of

Definition of rational exponent:

Evaluate Each by first changing to the radical expression:

Expanded Definition of rational exponents: The numerator becomes the powerAnd the denominator becomes the index of the root.

Convert the following to radical notation:

Convert the following to rational exponents:

Evaluate using definition of rational exponents:

Simplify using rules of exponents. Leave in rational exponent form:70

91 /2?

a1/n=n√a if the root is defined .

321/5=−811/4=(−81 )1/4=

641/2=

(127 )1 /3

=

(0 .16 )1/2=

am/n=n√am or ( n√a )m

m3/4=53 /8=

5√ g2=(√7 )3=

a ) 253/2

b ) (−27 )2/3

c ) −163/4

d ) (−81)5/2

Simplify by using rational exponents, write answer as radical:

Simplify by first converting to rational exponents, then using rules of exponents, write answers with rational exponents.

Math 035 Notes 9.3: Simplifying Radical Expressions

71

(21/2 ) ( 21/3 )=

(−64125 )−

23=

a5/3

a7 /3 =

(m1/2 n2 /3 )4=

( x2 y )1/2

x3 /4 y−1/4 =

n3 /4 (2n5 /4−3n1/4 )=

( x−2/3

y−3/4 )4

( x−2/3 y1/4 )−2=

√212

3√49

√√81

3√ x⋅√x3√ t4

5√ t4

6√ y5⋅3√ y2

3√5√√ y

Goals: Use product and quotient rules for radicals. Simplify radicals. Use Pythagorean theorem and distance formulas.

Product Rule for radicals: √a⋅√b=√ab

Use product rule:

1) √5√13 =

2) √7√x =

3)3√2 3√7=

4)6√8 r2 6√2 r3=

5)3√7√5=

Quotient Rule for radicals:

√a√b

=√ ab

Use quotient rule:

1) √10081

=

2)

3√18125

=

3)

4)

3√ x2

r 12=

72

5)

√18√2

=

Use the product rule to SIMPLIFY each radical.

1) √32=√16⋅2=√16⋅¿√2=4√2 or ¿

√32=√2⋅2⋅2⋅2⋅2=2⋅2⋅√2=4 √2

2) √300=

3) √18=

4) √48=

5) Is 3√8=6 √2 ? Which is considered simplified?

6)3√54=

7) 3√48=

8)4√243=

9) √35=

10) √25 p7=

11) √72 x3 y5=

12) √32 a5 b11=73

13)3√ x12 y7 z5=

14)

4√32 x12 y7 z5=

Simplify by getting a smaller index, if possible.

1)12√23=

2)4√49=

3)9√a3 b6c15=

Don’t forget to simplify radical too!

Multiplying radicals with different indexes...a) Convert to exponentsb) Get a common denominator for exponentsc) Convert back to radical form

1) √5 3√4=

2)3√2 4√3=

Formulas with radicals.... Pythagorean Theorem: a2+b2=c2

1) 2)

74

8

14 20

5

Distance between 2 points: d=√( x2−x1 )2+( y2− y1 )

2

3) (2, -1)(5, 3)

4) (-3, 2)(0, -4)

MA 035 Notes 9.4: Adding and Subtracting Radicals

Goals: Simplify radical expressions using addition and subtraction.

Combining radicals with addition and subtraction follows the same rules as combining like terms. The radicals must have exactly the same radicand and index in order to add them.

Combine where possible:

1) 3√5−6√5+12√5−√5=

2) 5 3√12 y+ 3√12 y−14 3√12 y=

You must simplify these radicals first.

3) 2√11−√99+3√44=

4) 5 3√54−6 3√24+ 3√16−8 3√375=

75

5) 7 3√81+3 3√24−11 3√3=

6) 3√20 x−6√45 x+2√80 x=

7) −2 4√32−7 4√162+9 4√2=

8) 4 √20+5√18=

9) 2 xy3√ x2 y+

3√8x5 y 4=

10) √1216

+√4864

11)

3√ 5x6 + 3√ 4

x9

76

12) Find the perimeter of this rectangle.

MA 035 Notes 9.5: Multiplying and Rationalizing RadicalsGoals: Multiply radical expressions. Rationalize denominators. Write quotients in lowest terms.Review: 1) √2∙√3=¿ 2) 4 ∙√5=¿ 3) √2∙√2=¿

4) (a+b )2=¿ 5) (a+b ) (a−b )=¿

Multiplication of radicals:Multiply ‘outside’ coefficients together and ‘inside' radicands together using product rule.

1) (4 √5 )(2√5 )=

2) (2 3√6 )(−3 3√4 )(6 3√5 )=

3) (4 √2 )(8√8)(−2√5 )=

Distributing with radicals:4) √3(2+√2)

5) √2¿)

77

√48

√192

FOIL with radical binomials:1) (3+√5)¿

2) (2+3√3)(4−2√5)=

3) (√6−2√5)2=

4) (2√5−3√3 )(4 √5+2√3 )=

5) (√5+√2)(√5−√2)=

6) (4+3√3)( 4−3√3 )=

7) (3+ 3√5)(3−3√5)

Division of radicals:Reduce ‘outside’ coefficients, then Reduce ‘inside’ radicands using quotient rule.

1)

14√82√2

=2)

15√610

=

78

3)

3√1625 3√54

=

Rationalizing the denominator: So that all answers are in standard form, we can not leave radicals in the denominator. If there is a single radical in the denominator, multiply the top and bottom by the offending radical.

4)

8√7

5) √136

Always simplify AND rationalize. If you simplify radicals first, numbers that reduce at the end of the problem will be smaller.

6)

3√48

7)−√ 8

45

8) √72y

79

9) √200 y6

k 7

Binomial denominators: Must use conjugates to rationalize.This text will not distribute if the numerator is a monomial, but will distribute if the numerator is a binomial.

10)

−42+√5

11)

15√x+√2

=

12)

√2+√5√2−√5

=

Simplifying radical expressions with binomials: If the numerator and denominator have a common factor, then the quotient can be reduced by that factor. But You Must Factor the numerator first!

1)

6+2√54

=2(3+√5 )

4=3+√5

2

2)

24−36√716

=

Still simplify radicals!

80

3)

2√27+48

=

4)

4 x−6√x3

2 x2=

5)

189+√18

=

Simplifying Radical Expressions

There are three rules that you must ALWAYS follow when Radicals are involved:

1.) Simplify all radicals Examples √18=√9 ∙ 2=3√2 or √45 x3=√9 ∙5 ∙ x2∙ x=3 x√5 x

2.) Rationalize Denominators

Examples 8√7

= 8√7

∙ √7√7

=8√77 or 5

1+√3= 5

1+√3∙ 1−√31−√3

=5 (1−√3 )

1−3=

−5 (1−√3)2

3.) Reduce all Fractions

Examples 2√34

=√32

or 4+2√56

=2¿¿

81

MA 035 Notes 9.6: Solving Radical EquationsGoals: Solve radical equations using power rule. Solve radical equations with indexes greater than 2.

Radical equation: An equation that contains one or more radical expressions.

Examples: √ x−4=8 , √5 x+12=3√2 x−1 ,3√6+x=27

Power Rule: If both sides of an equation are raised to the same power, all solutions of the original equation are also solutions of the new equation.Note: This does not say that all solutions of the new equation are solutions of the original, i.e some solutions you get may be extraneous so you must check answers!

Solving an Equation with radicals1.) Isolate the radical2.) Apply the power rule3.) Solve the equation, if there is still a radical repeat steps 1 and 24.) Check the solution to determine if they are extraneous or not

1.) √5 x+1=4

82

2.) √6 k−1=1

3.) √5q−1+3=0

4.) √9a−4=√8a+1

5.) 3√ z−1=2√2 z+2

6.) √5−x−1=x

7.) √1−2 p−p2=p+1

83

Using the power rule, squaring twice

8.) √2x+3+√x+1=1

Higher powers:

9.) 3√2m−1=3√m+13

10.) 3√r+1+1=0

11.) 4√8 z−3+2=0

84

12.) √1+√24−10 x=√3 x+5

13.) Do not actually solve. What would the first step be?

√ x+3=3√5+4 x

Solving Formulas:14.) Solve for k

V=√ 2 km

MA 035 Notes 9.7: Complex NumbersGoals: Simplify numbers of the form √−b . Perform operations on complex numbers.

Imaginary Numbers: i=√−1and i2=−1

For any number b, √−b=i√b

Practice: Simplify using imaginary numbers.

1) √−100=10 i

2) √−37

3) √−49

4) −√−36

5) √−32

85

Note: It is customary to write the i before the radical so that i √2 is not confused

with √2i where i is under the radical.

When finding products like √−4⋅√−9 we must convert to i form first because √−1⋅√−1=i≠√+1=1 Therefore √−4⋅√−9=2i⋅3 i=6i2=−6

Practice:

1) √−16⋅√−25

2) √−2⋅√−10

3) √−15⋅√−2

4)

√−75√−3

5)

√−50√2

Complex Numbers: If a and b are real numbers, then any number of the form: a +bi is called a complex number. a is the real part and b is the imaginary part.

Since i is a square root, all operations follow the same rules used for simplifying radical expressions.Add:1) (4 + 5i) + (7 + 3i)

2) (-2 + i) + (5 – 7i)

Subtract:3) (6 + 5i) – (8 + 11i)

4) [(4 – 2i) – (3 + 5i)] – (2 + i)

Multiply:

86

4) 4i(2 + 3i)

5) (2 + i)(3 + 2i)

6) (4 + 6i)(3 – 2i)

7) (3 + 2i)(3 – 2i)

Divide: For quotients use the ‘complex conjugate’ to clear i from the denominator and write your answer in the form a + bi

8)

2+ i3+2i

=(2+i)(3+2 i )

⋅(3−2i )(3−2i )

=6−4 i+3 i−2 i2

9−4 i2 = 6−i+29+4

=8−i13

= 813

− 113

i

9)

8−4 i1−i

10)

53−2 i

11)

5−ii

Simplifying powers of i: Powers of i are cyclic. Any power of i can be simplified to i,-1, -i, 1.

If you know these first 4 powers of i, you can figure out any power of i.

i1 = ii2 = -1i3 = i2i1=-ii4 = i2i2= 1

Watch what happens as we continue:

i5 = i4i = 1i = ii6 =i4i2 =1(-1) = -1i7 = i4i3=-i

87

+1

+i

-1

-i

i8 = i4i4= 1

The powers of i cycle every fourth power. We can use this to find any power of i.

i14= i12i2 = (i4)3i2=1(-1) = -1

i33 = i32i = 1i = i

i12

i31

i-2

i50

88


Name:






Took notes.

Read the textbook.


Visited tutor(s).










___________________________________________________________________________________

over →

89












____________________________________________________________________________________




MA 035 Notes 10.1: Square root PropertyGoals: Use Square root property. Solve quadratics with non-real roots.

90

Quadratic Equations: Equation of the form ax 2+bx+c=0where a, b, c are real numbers and a≠0.

We have already solved some quadratic equations by factoring, but as we know not all such equations can be factored.

If the quadratic equation is of the form x2=k then it can be solved by taking the square root of

both sides. x=±√k This can be justified using our knowledge of solving by factoring:

x2=kx2−k=0( x−√k )( x+√k )=0x=√k , x=−√k

1.) x2=5 2a.) x

2=64 2b.) x2=12

3.) ( x−5 )2=36 4.) 3 x2−54=0

5.) ( x−4 )2=3 6.)(3 k+1 )2=2

Solve: These equations have imaginary roots.1.) x2+1=0 2.) x

2=−4

3.) (4 m−7 )2=−27 4.) ( t+2)2=−16

91

MA 035 Notes 10.2: Completing the Square Goals: Use the completing the square method to solve quadratics.

Recall from section 10.6, if you have( x+k )2=m

where k and m are numbers, it will be easy to solve. Most problems do not come in this form, but we can ‘complete the square’ to put any quadratic into this form.

Examples of perfect squares: x2 + 4x + 4 = (x + 2)2 y2 – 6y + 9 = (y – 3)2

Now you try to fill in the blanks: 1.) x2 + 2x __ = (x ___)2 2.) u2 – 32u ___ = (u __)2

3.) n2 + 3n __ = (n ___)2 4.) p2 + ⅓p ___ = (p ___)2

1.) Now we will try this on an equation to solve:x2+8x+10=0 This can't be factored, so move the 10 to the other sidex2+8 x+ = -10 What number can we put in the blank spot to make the left side a perfect square?x2+8 x+16=−10+16 since 8 is the middle number if we take 8/2=4 and square it . So add 16 to both( x+4 )2=6 So we completed the square, now use square root property to solve itx+4=±√6x=−4±√6

2.) x2−2 x−10=0 3.) k

2+5 k−1=0

4.) 2 x2−4 x=5 5.) 8 x2−4 x=2

92

Now we can generalize this idea for any quadratic equation ax 2+bx+c=0

ax2+bx+c=0 First divide by a coefficient

x2+ba

x+ca

=0 Now get the number part(c/a ) on the other side

x2+ba

x =−ca

to complete the square take b/a divide by 2 and square

x2+ba

x+b2

4 a2 =−ca

+b2

4 a2 Add the fractions on the right, and rewrite left as a square

( x+b2a

)2=b2−4 ac4 a2 Now apply the square root property and solve

x+b2a

=±√b2−4 ac4 a2

x=-b±√b2−4 ac2 a

Now you have the quadratic formula

MA 035 Notes 10.3: Quadratic FormulaGoals: Solve quadratic equations by using quadratic formula. Use the discriminant to determine the number and type of solutions.

Quadratic formula x=−b±√b2−4 ac

2 a Where ax 2+bx+c=0

1) x2+20 x+100=0

2) 15 x2+11 x=14

3) 6 x2−5 x−4=0

93

4) 3 q2−9 q+4=0

5) 2 k2+19=14 k

6) (n+1 )(n−7 )=1

7) m2=4 m−5

The discriminant is the radical part of the quadratic formula, b2−4 ac . It will determine how many and what types of solutions a quadratic equation will have.

Value of Discriminant Number of Solutions Types of Solutions0 1 Rational

Positive perfect square 2 RationalPositive non-perfect square 2 Irrational

Negative 2 ImaginaryUse the discriminant to describe how many and what types of solutions (roots) there will be?

8) 4 x2−28 x+49=0

9) 4 x2=4 x+3

94

10) 18 y2+60 y+82=0MA 035 Notes 10.4: Quadratic Formula and Applications

Goals: Solve equations that must be put in quadratic form by using quadratic formula. Solve application problems.

Clear the denominators by using the LCD, then apply quadratic formula:

1) 4−7

r− 2

r2=0

2)

32 x

− 12x+4

=1

3)

1x2

+1=−1x

Remember that when dealing with rational equations, you must check your solution against the domain.

4) z=√5 z−4

5) 2 x=√ x+1

95

Remember that if you apply the power rule, you must also check your solutions.

SubstitutionsThe following may be solved using substitutions and factoring, or substitutions and the quadratic formula. Let u = the middle variable(s) in the quadratic form, solve the new quadratic for u, then ‘undo’ your substitution and solve for the original variable.

1.) 9 y4−37 y2+4=0

2.) 5(r+3 )2+9(r+3 )=2

3.) 4 m2

3−3 m1

3−1=0

96

ApplicationsJob Problems1) Two pipes can fill a tank together in 5 hours. It takes one of the pipes two hours longer than the other to fill the tank alone. How long does it take each pipe to fill the tank alone?

= time alone for the fast pipe = time alone for the slow pipe

Rate Time PartFast pipe 5

Slow Pipe 5

2) Carl can complete a certain lab test in two hours less time than Jamie can. If they work together, they can finish the job in two hours. How long would it take each of them working alone to complete the job?

x = time alone for Jamiex – 2 = time alone for Carl

Rate Time PartJamie

Carl

97

Motion (d = rt) Problems3) In Indiana, Lafayette and Evansville are 200 miles apart. If Lennie drives his Lexus 25 miles per hour faster than Minnie drives her Mercedes, What is Lennie’s average speed if he travels from Lafayette to Evansville in 1 ⅓ hours less time than Minnie?

Rate Time DistanceLennie 200

Minnie 200

4) Lisa flew her plane for 6 hours at a constant cruising speed. She traveled 810 miles with the wind, then turned around and traveled 720 miles against the wind. The wind speed was a constant 15 miles per hour. Find the constant cruising speed of the plane.

x = cruising speed of the planeRate Time Distance

With wind x + 15 810

Against wind x – 15 720

Time with wind + Time against wind = Total Time

5) In a total of 1 ¾ hours, Ken rows his boat 5 miles upriver and returns back to his starting point. He rows at a constant rate and the current is flowing at 3 miles per hour. What is his ‘rowing speed’ (as if in still water)?

x = rowing speed

98

MA 035 Notes 10.5: Applications of QuadraticsGoals: Solve formulas. Solve application problems involving quadratic equations.

Formulas: Solve the formulas involving squares.

1) A=πr 2 for r 2) D=√kh for h

3) Use the quadratic formula to solve: s=2t2+kt for t

Projectile Problems1) A toy rocket is launched upward from ground level. Its distance from the ground (s) in feet after t seconds is given by the equations=−16 t2+208t . After how many seconds will the rocket be 550 feet above the ground?

550=−16 t2+208 t16 t2−208 t +550=0

99

2) The Toronto Dominion Center is 407 feet high. A ball is projected upward from the top of the Center and its position in feet (s) above the ground after t seconds is given by the equation s=−16 t2+75t +407 . How many seconds have elapsed when the ball reaches a height of 450 feet above the ground?

2a) (Refer to the previous problem.) After how many seconds will the ball hit the ground?

2b) When will the ball reach 500 feet? How can you explain this answer?

100

Pythagorean Theorem3) A ladder is leaning against a house. The distance from the bottom of the ladder to the house is 5 feet. The distance from the top of the ladder to the ground is 14 feet less than double the length of the ladder. How long is the ladder? x = ladder

4) Two cars left an intersection at the same time, one heading due north, and the other due west. Some time later, they were exactly 100 miles apart. The car heading north has gone 20 miles farther than the car heading west. How far has each car traveled?

Area Problems5) A piece of cardboard has a length that is 4 in less than twice the width. A square piece 2 in on each side is cut from every corner so that the piece of cardboard’s sides can be turned up to form a box of volume 256 in3 Find the length and width of the original piece of cardboard.

101

5

x

MA 035 Notes 10.6: Graphing Quadratic Functions

Goals: Graph quadratic functions. Use standard form to predict graphs of quadratic functions.

A Quadratic Function has the form f ( x )=ax 2+bx+c or y=ax2+bx+c .

1) Graph f ( x )=x2 x -3 -2 -1 0 1 2 3

y or x2

The shape above is called a parabola. All graphs of quadratic functions will be parabolas. The point at the bottom is called the vertex. A parabola has symmetry about a vertical axis.

2) Graph F (x )=x2−1 x -3 -2 -1 0 1 2 3x2

x2−1

Notice: The graph has been 'shifted' down 1 unit from the graph of f(x) in example 1.

102

A graph of the form F (x )=ax 2±k has the same shape as the graph of f ( x )=ax 2, but has

been shifted k units down (with the -) or k units up (with the +).

Predict what x2 + 2 would look like.

3) Graph F (x )=( x−1 )2

x -3 -2 -1 0 1 2 3 4

y or ( x−1 )2

This parabola has been 'shifted' one unit to the right.

A graph of the form F (x )=a (x±h)2 has the same shape as the graph of f ( x )=ax 2

, but has been shifted h units right (with the -) or h units left (with the +). Notice: Hop the opposite direction of the sign. HOPPOSITE Rules!

A graph could be shifted both up or down and left or right.

4) Graph F (x )=( x+3 )2+2 x -3 -2 -1 0 1 2 3

ySubtract 3 from the x and add 2 to the y value.

103

5) Graph f ( x )=−1

2x2

x -3 -2 -1 0 1 2 3y

The graph now opens downward instead of upward. The negative sign in the front caused a vertical reflection over the x axis.

A graph of the form F (x )=−ax 2 is a vertical reflection of the graph of f ( x )=ax 2

.

x -3 -2 -1 0 1 2 3

6.) Graph f ( x )=2 x2y

104

Notice that the graph on Example 5 became wider with a coefficient of -1/2 and narrower in Example 6 with a coefficient of 2. How can you generalize this rule?

In general the standard form of a graph of the quadratic function is given by F (x )=a (x−h)2+k (h,k) will be the vertex, and a will determine the direction the parabola opens and how wide it is.

These transformations will work for all functions. Prove that is works for lines by considering the equations: y = x, y = x + 1, y = 2x, y = ½ x,

Ma 035 Notes 10.7: More on Graphing ParabolasGoals: Find vertex of parabola. Graph quadratic functions. Find x-intercepts. Solve max/min.

In the previous section the general the standard form of a graph of the quadratic function was given by F (x )=a (x−h)2+k Where (h,k) was the vertex, and a will determine the direction the parabola opens and how wide it is.

But the standard quadratic equation is given as y=ax2+bx+c

So to make the general quadratic equation easier to graph, we will use completing

the square to rewrite it as F (x )=a (x−h)2+k

Example: Use completing the square to find the vertex of f ( x )=x2−4 x+5x2−4 x+5 This is not an equation, so when completing the square we must x2−4 x +5 keep everything on one side. But we still take the x coefficient x2−4 x+4 -4+5 and divide by 2 then square it to complete the square, but now ( x−2 )2+1 we have to add and subtract the number to maintain equality. In this new format, we can easily see that the vertex is (2, 1)

105

Find the vertex of the following parabolas

1.) g( x )=x2+4 x−9

2.) f ( x )=x2+6 x+7

3.) h( x )=−3 x2+6 x−1

4.) p( x )=2 x2−4 x+1

Alt Method for Finding the Vertex, use a formula.Similar to the way we generalized completing the square to find the quadratic formula, we can generalize completing the square to find a formula for the vertex:

f ( x )=ax 2+bx+c→ f ( x )=a( x2+ba

x )+c→ f ( x )=a( x2+ba

x+b2

4 a2 −b2

4 a2 )+c→

f ( x )=a ( x2+ba

x+b2

4 a2 )+c−b2

4 a2 → f (x )=a (x+b2 a

)2+4 ac−b2

4 a

The most useful way to use this information is to see that the x-part of the vertex will be –b/2a . We can then plug this value back into the given quadratic formula to find the y part of the vertex.

Example:

106

f ( x )=x2+6 x+7

Has x-part of vertex =−62(1)

=−3

now find f (−3) to find the y-partf (−3 )=(−3)2+6(−3)+7=−2So vertex ( -3, -2)

Practice: Find the vertex

1) g( x )=x2−4 x+1

2) f ( x )=4 x2−x+5

Finding the x-intercepts: Another useful piece of information about the parabola would be where it intersects the x-axis. This would be the x-intercepts, to find them set y = 0 and solve for x.

Example:f ( x )=x2−x−6

From the above examples this has vertex (½, -6¼)

We can also see that if

0=x2−x−60=( x−3 )( x+2)x=3 , x=−2 So we have points (3, 0 ), (-2, 0 )

If you can’t solve by factoring, just use the quadratic equation. With just those 3 points and symmetry we can get a fairly good graph.

Practice: Find the x-intercepts.

1) f ( x )=x2+6 x+8

2)g( x )=x2−3 x−3

107

3) h( x )=x2+2 x+3

Recall that the discriminant is the b2 – 4ac part of the quadratic formula. How could the discriminant help us predict the number of x-intercepts for an equation?

How would you find the y intercepts?

Practice graphing some of the Parabolas from this section.

Max and MinThe vertex of an upward facing parabola is the parabola’s lowest points, and the vertex of a downward facing parabola is the highest points. Using this fact we can answer application problems involving maximums and minimums when the relationship can be modeled with a quadratic equation.

Example: A rocket is launched from a 60 foot platform with an initial velocity of 112 feet. It’s height is as a function of time is given by: h( t )=−16 t2+112 t+60

108

What is the maximum height it will reach?

We will actually find the time at which it reaches the max by using the vertex

formula for the first coordinate: x=t=−b

2 a= −112

2(−16 )=3 . 5

So at time 3.5 seconds the rocket is at its max height, which we find by:

h(3 . 5)=−16(3 .5 )2+112 (3. 5 )+60=256 feet

Practice: A farmer has 120 feet of fence to enclose a rectangular area next to a building, using the building as one side of the rectangle. Find the maximum area he can enclose and the width he will use.

Practice: Find a pair of numbers that adds to 80 and whose product is a maximum.

Practice: A charter boat company charges $50 per passenger plus a $5 additional dollars per person fee for any unfilled seats. The boat has a capacity of 100 people, let x represent the number of no-shows.

a) Find the cost function f(x) = cost*number of people = (50 + 5x)(100 – x)

b) What number of no-shows will cost you the most?109

c) What would this cost be?

110

111


Name:






Took notes.

Read the textbook.


Visited tutor(s).










___________________________________________________________________________________

over →

112












____________________________________________________________________________________




Ma 035 Notes: 8.1 InequalitiesGoals: Solve and graph linear inequalities in one variable.

113

Solving linear inequalities follows the same rules as linear equations with one major exception. If you multiply or divide both sides by a negative number, you must reverse the inequality symbol.

Practice: Solve the following inequalities, graph and give answer in interval notation.

graphs intervals1.) x – 5< -10

2.) 3x + 3 ≥ 2x – 1

3.) -m + 6 < 2m

4.) 8k ≤ -40

5.) -7s > -63

6.) 6(x – 1) + 3x ≥ -x – 3(x + 2)

7.)

14(m+3 )+2≤3

4(m+8 )

Three part inequalities follow the same rules but you must do each operation to all three parts8.) 3 < x + 10 ≤ 8

114

9.) 5 ≤ 3x – 4 < 8

10.) 4 ≤ 5 – 9x < 8

11.) 2(2x+1) ≤ 4x+5

MA 035 Notes 8.2: Compound InequalitiesGoals: Distinguish between AND and OR. Solve and graph compound inequalities.

115

The word AND in math means both criteria must be true at the same time, this will be the intersection of the graphs. The word OR means either criteria is true, this will be the union of the graphs.

Use the Venn diagram to answer the questions:

Who is in Math 035 AND Eng 111?Which students are in Math 035 OR Eng 111?

First solve each inequality, then decide if the solution is an intersection or a union.

x + 2 < 5 AND x – 3 ≥ -4

Copyright © 2010 Pearson Education, Inc. All rights reserved.

3x + 7 ≤ 1 OR -2x < -10 x ≤ -2 OR x > 5

Practice: Solve, Graph and give interval solution.1.) x + 3 < 1 and x – 4 ≥ -12

116

John Joe Jesse James

Jenny

JeffJuan

Math 035 Eng 111

My Advisees

2.) x + 2 > 3 or 2x + 1 < -3

3.) x – 1 > 2 or 3x + 5 ≤ 2x + 6

4.) 3x – 2 ≤ 13 and x + 5 ≤ 7

5.) x < 3 and x ≥ 6

6.) 3x ≤ x + 12 or 3x – 8 ≥ 10

MA 035 Notes 8.3: Absolute Value

Goals: Solve equations and inequalities involving absolute value.

117

Absolute Value: |x| is the distance from x to 0.

If |x| = 5, then x = 5 or x = -5.

Copyright © 2010 Pearson Education, Inc. All rights reserved.

Because absolute value represents the distance from zero, |x| > 5 will be all numbers that are more than 5 units from 0.Therefore, |x| > 5 has solution: x > 5 OR x < -5. Which is (-∞, -5) U (5, ∞)

For |x| < 5, the solution will be all numbers that are less than 5 units from zero, which will be all numbers between -5 and 5, which is -5 < x < 5 x > -5 AND x < 5

Note: We can check the solutions by plugging any value in the solution set into the original to see if it does make the equation / inequality true.

Based on the above examples, all equations / inequalities that involve absolute value will require us to solve 2 separate equations / inequalities.

Solving Absolute Value Equations and InequalitiesLet k be a positive real number, and p and q be real numbers.

Case 1: To solve |ax + b| = k, solve ax + b = k OR ax + b = -kThis case will usually result in an answer of the form {p, q}.

Case 2: To solve |ax + b| > k, solve ax + b > k OR ax + b < -kThis case will result in a solution set of the form (-∞, p) U (q, ∞)

Case 3: To solve |ax + b| < k, solve -k < ax + b < k This case will result in an answer of the form (p, q)Practice: Solve the following equations/inequalities, graph and give the set/interval solution.1) |x| = 7 2) |x + 3| = 5

118

3) |3x – 4| = 11 4) |1 - ¾x| = 7

5) |x| > 7 6) |x + 3| > 5

7) |3x – 4| ≥ 11 8) |5 – x| > 8

9) |x| < 7 10) |x + 3| < 5

11) |3x – 4 | ≤ 11 12) |2s – 6| < 6

Note: The absolute value must be isolated in order to solve13) |x + 3| + 5 = 12 14) |5x + 2| - 9 = -7

15) |x + 2| - 3 ≥ 2 16) |3x + 2| + 4 < 15

Solving an equation with two absolute valuesTo solve an equation of the form |ax + b|= |cx + d| solve the compound equation :

ax + b = cx + d OR ax + b = -(cx + d)119

Example: Solve |3x + 1| = |2x + 4|3x + 1 = 2x + 4 OR 3x + 1 = -(2x + 4) x + 1 =4 3x + 1 = -2x – 4 x = 3 5x = -5 → x = -1

Solution: {-1, 3}Check: |3(-1) + 1| = |2(-1) + 4| →|-2| = |2| or |3(3) + 1| = |2(3) + 4| → |10| = |10| Yes!

Practice:1) |k – 1| = |5k + 7| 2) |z + 6| = |2z – 3|

Special Cases with Absolute ValueNote: Absolute value can never be negative, and is equal to zero only when the expression in the absolute value is equal to zero.

Examples: |2x + 7| = -3 Impossible! Therefore, No Solution

|4z – 10| < -13 Impossible! Therefore, No Solution

|5w – 3| = 0 Only If 5w – 3 = 0. Therefore w = 3/5

|3d – 2 | > -1 This will always be true! Therefore, All Real Numbers

|x + 2| ≤ 0 Need only solve for = 0. Therefore x = -2

Practice:1) |5x – 2| ≥ 0 2) |8n + 4| ≤ -4

3) |3r + 6| = 0 4) |4 + 9y| + 5 ≥ 4

5) |5n – 2| + 2 < 1 6) |3w + 9| + 2 = 2

Review of Linear EquationsStandard Form:

Slope:120

Point-Slope Form:

Slope Intercept Form:

Example A: Find an equation in standard form for a line with Slope -3 and through (10, 8)

Example B: Find an equation in slope-intercept form with Slope 2/5 and through (3, -4)

Example C: Find an equation in standard form for a line through (-2, 6) and (1, 4)

Example D:What is the slope and y-intercept of the line with the equation: 2x – 5y = 1

Example E: Find an equation in slope-intercept form through (-8, 3) and perpendicular to 2x – 3y = 10

Example F: Find an equation in Standard form through (2, 1) and parallel to 4x – 3y = 16

Horizontal lines:

Vertical lines:

Example G: Find the equation of a horizontal line through the point (-2, 6)

Example H: Find the equation of a vertical line though the point (-2, 6)Review of Graphing y=mx+bTo quickly graph a line with this form, we first plot the y-intercept which is the point (0,b) Then we use that point and the slope m to find another point. Example: Graph

121

y=35

x−2

First graph the y-intercept which is the point (0, -2)

Now use the y-intercept as a starting point and the rise/run definition of slope to find more points.

Practice: Use slope-intercept to graphy=−2x+6 3x – 5y = 25

y = 5 Graph a line with = ½ through ( 2, -3)

Math 035 Notes 3.5: Linear Inequalities in Two Variables

Goals: Graph linear inequalities in two variables, graph the intersection and union of two linear inequalities.

122

Linear Inequality: Ax + By < C A, B, C are real numbers and A and B not both zero

Recall graphing a linear inequality in 1 variable: 2x + 1 ≤ 3 so x ≤ 1 First we would mark the boundary with a ] then we showed the less than portion with shading. Similarly, we will use shading when dealing with (x, y) graphs. We will be shading in all the points (x, y) that make the inequality true. 0 1 2

Graph: x + y ≤4First put the inequality in slope-intercept form by solving for y: y ≤- x + 4 Note that part of this inequality is the equality y = -x + 4. This that is part of the solution, it will be our boundary. Because y tells us the vertical portion of our graph, the values where y ≤ -x + 4 will be all points below or < the boundary line.

Graph: x – 3y < 4First put in Slope-Intercept form. Caution! If you divide by a negative you reverse the inequality: -3y < -x + 4 so y > ⅓x – 4/

3 In this problem the borderline is not part of the answer, so we use a dotted line instead.

123

Hint: you can also check your answer by testing a point in the shaded areaGraph: x + 4y ≥ -3 Graph: 3x – y < 3

Graph: x ≥ 1 Graph: y > 2

Graph: x – y > 2 Graph: x – 3y < 4

124

Math 035 Notes 3.6: Introduction to FunctionsGoals: Define relations and Functions, find Domain and Range, use vertical line test, use function notation.

Relations and FunctionsDefinition: A relation is a set of ordered pairs. Relations can be represented by lists (tables, sets, mappings), graphs, or equations of two variables.

All of these are relations.

x y {(0 ,−4 ) ,(9,9 ), ( 23

,−8 ) ,(3,9)}2 -3

0 5½ -2

y=2 x3+3 x2−5 x

Definition: A function is a relation where each x value is paired with one y value. This means that for each x input, there can be only one y output. The x value is called the independent variable and the y value is called the dependent variable.

A) Determine if these relations are functions.

1) {(2,3) ,(4,8 ) ,(−1,6 ) ,(0 ,−2)}

2)

3) x 4 2 5 6 y 3 1 1 7

125

3-24

-2

8

-4-39

12310

B) To determine if an equation is a function, First write it in the form y = a quantity with the variable x. Consider what happens for each input x, is there only 1 output y ?

Are these equations functions?

1) 4 x+2 y=12 2) x2− y=0

3) y2=x 4) y=√x−3

5) y=±3x 6)y=3 x−5

2 x

C If a graph passes the ‘vertical line test’, it is a function.Vertical Line Test: If every vertical line intersects the graph of a relation only once, then the relation is a function.

Are these graphs functions?

1) 2) 3)

4) 5)

126

D Definition: The domain of a function is the set of x values. The range of a function is the set of y values.

Find the domain and range of each function.1) {(−3,0 ) ,(−2,5 ) ,(6,0 ),(2,3 )}

2)

Find Domain only

3)y=3

x 3b.) y= x+2

x2−4

4) y=√x 4b) y=√x+2

E Function Notation replaces the variable y with f(x) We read this as f of x, it does not mean f times x, rather it gives our function a name and acts as a template for the function. We can use almost any letter besides f, but f, g, h are the most common.

Given the following function, find the values.

1)f ( x )=5−3 x

2 f (7) means use the given template and replace the x's with a 7 and evaluate,

therefore f (7 )=5−3(7 )2

=5−212

=−162

=−8 , so f (7 )=−8

f (−3 )=f (0)=

127

(-3,-3)(8,-3)

(0,10)

2)g( x )=5 x2−1

g(−1)=g(2 )=g(u )g( x+1)

3) Suppose f ( x )is a function defined by this set: {(2,3 ),(5 ,−1) ,(6,0 ),(−2,9 )}

Find:

f (5)=f (6 )=

4) Suppose g( x )is a function defined by this mapping:

Find:

g(3 )=g( 4 )=

Write these equations using function notation, then evaluate f(2).

1) 3 x−4 y=18 2) 2 y=3 x2−2

Many real life applications can be put into function form.

If r represents the length of the femur, the height of a man is determined by the function h(r )=69 . 09+2 .24 r . (Both r and h(r) are in centimeters.) Find the height of the man, if his femur bone measures 57 cm.

Johnny gets paid $10 per hour, plus a $1 safety bonus. So his pay is a function of how many hours he works p ( x )=10 x+1 What is p(15) and what does it represent?

128

234

16-1

MA 035 Notes 4.1: Solving Systems of Linear Equations(Graphing)Goals: Determine it a point is a solution of a set of equations. Solve systems by graphing.

System of Linear Equations: A set of two or more linear equations containing the same variables.

Solution of a system of linear equations: An ordered pair that makes BOTH equations true at the same time.

Is the ordered pair (3, -12) a solution of the system:

Is the ordered pair (2, 5) a solution of the system:

Solve the system by graphing:

129

2 x+ y=−6x+3 y=2

3 x−2 y=−45 x+ y=15

x+ y=52 x− y=4

How Many points might a set of lines have in Common?




130

x−3 y=−24 x+ y=5

x+ y=84 x+4 y=32

If two lines have the same slope, but a different intercept, how many solutions will there be?

If two lines have the same slope and intercept, how many solutions are there?

If two lines have different slopes, how many solutions are there?

Practice:Without actually graphing, determine how many solutions the set has:

See figure 1: The blue line represents the Sales in millions of digital cameras since 2000. The red line represents the sales of conventional cameras in millions since 2000. Where do these lines intersect and what does this ordered pair represent?

131

2 x− y=46 x−3 y=18

2 x−3 y=53 y=2 x−7

2 x+ y=4x+2 y=9

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

MA 035 Notes 4.2: Solving Systems of Linear Equations (Substitution)Goals: Solve systems of equations by substitution. Solving by graphing may not be accurate, and can be time consuming. Another method to solve systems of equations is substitution.

Solving by SubstitutionExample

This system is a good candidate for substitution, we can substitute the y = 5x from the second equation into the first equation, in orderTo get a linear equation in one variable to solve.

Now substitute the x-value into any equation to get the corresponding y-value

Practice Problems: Solve the systems by substitution and check.1)

132

2 x− y=6y=5 x

2 x−5 x=6→−3 x=6→ x=−2

y=5 (−2 )=−10Solution: (-2, -10)

5 x−3 y=−6x=2− y

2)

3)

4)

5)

6)

133

4 x+ y=52 x−3 y=13

3 x+2 y=13 x−4 y=−11

x=5−2 y2 x+4 y=6

2 x=6−2 yx+ y=3

0 .08 x−0 .01 y=1 .30 .22 x+0 .15 y=8 . 9

7)

MA 035 Notes 4.3: Solving Systems of Linear Equations (Elimination)Goals: Solve systems of equations by elimination.

Substitution is a good method, because it always works, but it can also become cumbersome if one of the variables is not easy to solve for. A more sophisticated method for solving systems of equations is the Elimination method.

Elimination Method:Consider the system

Since the left side equals the right side of each equation, the left sides can be ‘added’ and the right sides can be ‘added’ and a new equation is formed. When this is done, the ‘y’ terms are eliminated. ‘Plug’ this value for x in either of the original equations and the matching y value is found.

The solution is (2,1).

Practice:1)

134

12

x+14

y=14

15

x−110

y=310

2 x+3 y=73 x−3 y=3

5 x=10x=2

2(2)+3 y=74+ y=73 y=3→ y=1

−2 x+3 y=−102 x+2 y=5

This method is called elimination because when we added the equations, one of the variables was eliminated. The elimination happened because the coefficients were opposite. If the coefficients are not opposite, we must make them opposite by multiplying the equations to form equivalent equations.Example

It does not matter which variable you eliminate. I chose x.2 and 3 have an LCM of 6, so I multiply the first equation by 3And the second equation by -2, to eliminate x.

Practice: Solve by elimination method1)

*2)

3)Hint: you are going to multiply anyway, why not clear the fractions?

135

2 x+3 y=193 x−7 y=−6

6 x+9 y=57−6 x+14 y=12

3(2 x+3 y=19)→−2(3 x−7 y=−6)→

23 y=69y=3→ 2 x+3 (3)=19

2 x=10x=5

(5,3)12 x+10 y=78 x−15 y=10

12 x+10 y=78 x−15 y=10

56

x−13

y=23

12

x+34

y=134

4) Hint: you need to have like terms lined up!

5)

MA 035 Notes 4.4: Applications of Systems of Equations

Goals: Solve application problems using two equations and two unknowns.

For these application problems: (1) Define two variables.(2) Write two equations to relate the information and variables.(3) Use the elimination or substitution methods.

Example: A rectangular soccer field has a perimeter of 360 yards. Its length is 20 yards more than its width. What are the dimensions of the field?

Variables: W = width, L = length

Plan: P = 2L + 2W and Length is 20 yards more than width

Equations: 2L + 2W = 360 L = W + 20Solution: by substitution method: 2(W + 20) + 2W = 3602W + 40 + 2W = 360 4W = 320 Solution: Width 80 yards, Length 100

yards W = 80L = 80 + 20 = 100

136

2 x+ y=6−8 x=4 y−24

x=4−3 y2 x+6 y=−3

Practice: 1) Tennis courts are 42 feet longer than they are wide, and have perimeters of 228 ft. What are the dimensions of a standard Tennis court?

2) During the summer my family visited a waterpark. For our family 3 kids and 2 adult tickets cost $105.05, for our friends 2 kids tickets and 1 adult ticket cost $58.12. What were we charged for each kind of ticket?K = price per kids ticketA = price per adult ticket

3) A grocer has some $4 per pound coffee and some $8 per pound coffee, which she will mix to make 40 pounds of coffee selling for $5.60 per pound. How many pounds of each kind of coffee should she use?

Number of Pounds

unit price/pound final worth

cheaper coffee x 4expensive coffee y 8final mix 40 5.6

4) During the 2009 MLB season, the Padres played 162 games. They lost 2 more games than they won. What was their record that year?

137

5) See the drawing. Find the measures of angles x and y.*There isn’t a problem like this in our book, but why not try this one anyway?

65

x

y 2x+10

6) How many liters each of 15% acid and 33% acid should be mixed to get 40 liters of 21% acid?

+ =

7) My brother’s lives 200 miles away. If we leave our houses at the same time and I drive 20 mph faster than he does, it takes us 2 hours to meet at a State Park located in between our cities. How fast were we driving?

Rate Time DistanceMe

Brother

138

8) In his motorboat, Nigel travels upstream at top speed to his favorite spot, a distance of 36 miles, in two hours. Returning downstream, still at top speed, the trip only takes 1.5 hours. Find the top speed of the boat (as if it was in still water) and the speed of the current.

Rate Time Distanceupstream

downstream

139


Name:






Took notes.

Read the textbook.


Visited tutor(s).










___________________________________________________________________________________

140

over →












____________________________________________________________________________________




141

MA 035 Notes 5.6: Dividing a Polynomial by a Monomial

Goals: Divide polynomials by monomials.

Note that when we add fractions with a common denominator, we combine the numerator and keep that denominator:

Using this same fact backwards, we could take the fraction on the right and break it into two fractions with the same denominator. When we apply this to polynomials, the resulting fractions can then be reduced by using the quotient rule

Examples:

Note on the last term of this example, there are more a’s in the bottom!

Practice:

142

ac+b

c=a+b

c

12 x2+6 x6 x

=12 x2

6 x+ 6x

6x=2 x+1

16 a5−12a4+8a2

4a3 =16 a5

4a3 −12 a4

4 a3 + 8a2

4a3 =4 a2−3a+ 2a

6 p4+18 p7

3 p2

20 x4−25 x3+5 x5 x2

45 x4 y3+30 x3 y2−60 x2 y−15 x2 y

MA 035 Notes Section 5.7: Dividing a Polynomial by a PolynomialGoals: Divide a polynomial using long division. Apply division to geometry problems.

First let us review how to do long division with numbers:

23 will not go into 5, but it will go into 52, 2 times, subtract 46 2×23 = 46 from 52, 61 23 will only go into 61 2 times, now subtract 46 from 61.

46 61 46

15 23 will not divide into 15, so we will write it as a remainder, over the divisor of 23

Answer:

Now let’s try the same thing with polynomials:

Start by looking at p, it will go into p3 , p2 times, giving:

Now we will subtract.-4p2 – 5p + 9 And continue the division process.

p goes into -4p2 , -4p times

-4p2 – 5p + 9 Now subtract, remember to change the signs.143

23 2|521

23 22|521

22 1523

p3−2 p2−5 p+9p+2

=p+2|p3−2 p2−5 p+9

p+2 p2 |p3−2 p2−5 p+9

p3+2 p2

p+2 p2−4 p |p3−2 p2−5 p+9

p3+2 p2

-4p 2 -8p 3p + 9 continues --->

p goes into 3p +9 3 times

-4p2 – 5p + 9 -4p 2 -8p Subtracting leaves 3, which p does not divide into

3p + 9 So we will write 3 as a remainder 3p + 6

3Answer:

Practice: 1)

2)

3)

4) If a rectangle has area x3 + 4x2 +8x + 8, and Width x +2, What is its Length?

144

p+2 p2−4 p+3 |p3−2 p2−5 p+9

p3+2 p2

p2−4 p+3+ 3p+2

x2−x−6x+2

x3−1x−1

3 k3+9k−14k−2

Documents

faculty.ivytech.edufaculty.ivytech.edu/~rwulf/MA 035 notes actual.docx · Web viewMATH 035 STUDY GUIDE NOTES WORKSHEETS Notes contain some copyrighted material from Lial, Hornsby,