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Review Topics(Ch R & 1 in College Algebra Book)• Exponents & Radical Expressions (P. 21-25 and P. 72-77)
• Complex Numbers (P. 109 – 114)
• Factoring (p. 49 – 55)
• Quadratic Equations (P. 97 – 105)
• Rational Expressions (P. 61 – 69)
• Rational Equations & Clearing Fractions (P. 88 – 91)
• Radical Equations (P. 118 – 123)
• 1.5: Solving Inequalities
• 1.6: Equations and Inequalities involving absolute value
Review of Exponents82 =8 • 8 = 64 24 = 2 • 2 • 2 • 2 = 16
x2 = x • x x4 = x • x • x • x Base = x Base = xExponent = 2 Exponent = 4
Exponents of 1 Zero ExponentsAnything to the 1 power is itself Anything to the zero power = 1
51 = 5 x1 = x (xy)1 = xy 50 = 1 x0 = 1 (xy)0 = 1
Negative Exponents
5-2 = 1/(52) = 1/25 x-2 = 1/(x2) xy-3 = x/(y3) (xy)-3 = 1/(xy)3 = 1/(x3y3)
a-n = 1/an 1/a-n = an a-n/a-m = am/an
Powers with Base 10100 = 1101 = 10102 = 100103 = 1000104 = 10000
The exponent is the same as the The exponent is the same as the numbernumber of 0’s after the 1. of digits after the decimal where 1 is placed
100 = 110-1 = 1/101 = 1/10 = .110-2 = 1/102 = 1/100 = .0110-3 = 1/103 = 1/1000 = .00110-4 = 1/104 = 1/10000 = .0001
Scientific Notation uses the concept of powers with base 10.
Scientific Notation is of the form: __. ______ x 10(** Note: Only 1 digit to the left of the decimal)
You can change standard numbers to scientific notationYou can change scientific notation numbers to standard numbers
Scientific NotationScientific Notation uses the concept of powers with base 10.
Scientific Notation is of the form: __. ______ x 10(** Note: Only 1 digit to the left of the decimal)
-25 321
Changing a number from scientific notation to standard formStep 1: Write the number down without the 10n part.Step 2: Find the decimal pointStep 3: Move the decimal point n places in the ‘number-line’ direction of the sign of the exponent.Step 4: Fillin any ‘empty moving spaces’ with 0.
Changing a number from standard form to scientific notationStep1: Locate the decimal point.Step 2: Move the decimal point so there is 1 digit to the left of the decimal.Step 3: Write new number adding a x 10n where n is the # of digits moved left adding a x10-n where n is the #digits moved right
5.321
.05321
.0 5 3 2 1= 5.321 x 10-2
Raising Quotients to Powers
a n
b = an
bna -n b
= a-n
b-n= bn
an= b n
a
Examples: 3 2 32 94 42 16= =
2x 3 (2x)3 8x3
y y3 y3= =
2x -3 (2x)-3 1 y3 y3
y y-3 y-3(2x)3 (2x)3 8x3= = = =
Product Ruleam • an = a(m+n)
x3 • x5 = xxx • xxxxx = x8
x-3 • x5 = xxxxx = x2 = x2
xxx 1
x4 y3 x-3 y6 = xxxx•yyy•yyyyyy = xy9 xxx
3x2 y4 x-5 • 7x = 3xxyyyy • 7x = 21x-2 y4 = 21y4
xxxxx x2
Quotient Rule
am = a(m-n)
an
43 = 4 • 4 • 4 = 41 = 4 43 = 64 = 8 = 442 4 • 4 42 16 2
x5 = xxxxx = x3 x5 = x(5-2) = x3
x2 xx x2
15x2y3 = 15 xx yyy = 3y2 15x2y3 = 3 • x -2 • y2 = 3y2 5x4y 5 xxxx y x2 5x4y x2
3a-2 b5 = 3 bbbbb bbb = b8 3a-2 b5 = a(-2-4)b(5-(-3)) = a-6 b8 = b8
9a4b-3 9aaaa aa 3a6 9a4b-3 3 3 3a6
Powers to Powers
(am)n = amn
(a2)3 a2 • a2 • a2 = aa aa aa = a6
(24)-2 = 1 = 1 = 1 = 1/256 (24)2 24 • 24 16 • 16 28 256
(x3)-2 = x –6 = x 10 = x4
(x -5)2 x –10 x 6
(24)-2 = 2-8 = 1 = 1
Products to Powers
(ab)n = anbn
(6y)2 = 62y2 = 36y2
(2a2b-3)2 = 22a4b-6 = 4a4 = a4(ab3)3 4a3b9 4a3b9b6 b15
What about this problem?
5.2 x 1014 = 5.2/3.8 x 109 1.37 x 109
3.8 x 105
Do you know how to do exponents on the calculator?
Square Roots & Cube Roots
A number b is a square root of a number a if b2 = a
25 = 5 since 52 = 25
Notice that 25 breaks down into 5 • 5So, 25 = 5 • 5
See a ‘group of 2’ -> bring it outside theradical (square root sign).
Example: 200 = 2 • 100 = 2 • 10 • 10 = 10 2
A number b is a cube root of a number a if b3 = a
8 = 2 since 23 = 8
Notice that 8 breaks down into 2 • 2 • 2 So, 8 = 2 • 2 • 2
See a ‘group of 3’ –> bring it outsidethe radical (the cube root sign)
Example: 200 = 2 • 100 = 2 • 10 • 10 = 2 • 5 • 2 • 5 • 2
= 2 • 2 • 2 • 5 • 5 = 2 25
3
3
3 3
3
3
3
3
Note: -25 is not a real number since nonumber multiplied by itself will be negative
Note: -8 IS a real number (-2) since-2 • -2 • -2 = -8
3
Nth Root ‘Sign’ Examples
16
-16
= 4 or -4
not a real number
-164
not a real number
Even radicals of negative numbersAre not real numbers.
-325
= -2 Odd radicals of negative numbersHave 1 negative root.
325
= 2 Odd radicals of positive numbersHave 1 positive root.
Even radicals of positive numbersHave 2 roots. The principal rootIs positive.
Exponent Rules( )x x
x x x
x
xx
m n mn
m n m n
m
nm n
x
xx
x x
mm
m m
0
1
1
1
/
(XY)m = xmym
XY
m
=Xm
Ym
Examples to Work through
3 34
4
3
8
12
81
27
yx
Product Rule and Quotient Rule Example
4/1
4/34/5
8
88
Some Rules for Simplifying Radical Expressions
nmn m
nn
nnn
aa
aa
abba
/
/1
Example Set 1
300
162
75
55
33
x
y
y
x
Example Set 2
4 4
3
512
54
16
x
Example Set 3
55
56
5
6
33
27
8
9
4
84
1255
r
t
r
t
tt
Operations on Radical Expressions
•Addition and Subtraction (Combining LIKE Terms)
•Multiplication and Division
• Rationalizing the Denominator
Radical Operations with Numbers
333 210545162
2423
Radical Operations with Variables
zzz
yxxy
xx
48312332
3
2
27
8
4 54 5
3
3
Multiplying Radicals (FOIL works with Radicals Too!)
)8)(9(
)32)(32(
xx
yxyx
Rationalizing the Denominator
• Remove all radicals from the denominator
3
2
1
y
xy
Rationalizing Continued…
• Multiply by the conjugate
23
3
23
1
Complex Numbers
REAL NUMBERS Imaginary Numbers
IrrationalNumbers
, 8, -13
Rational Numbers(1/2 –7/11, 7/9, .33
Integers(-2, -1, 0, 1, 2, 3...)
Whole Numbers(0,1,2,3,4...)
Natural Numbers(1,2,3,4...)
Complex Numbers(a + bi)
Real Numbersa + bi with b = 0
Imaginary Numbersa + bi with b 0
i = -1 where
i2= -1
IrrationalNumbers
Rational Numbers
Integers
Whole Numbers
Natural Numbers
Simplifying Complex NumbersA complex number is simplified if it is in standard form:
a + bi
Addition & Subtraction)Ex1: (5 – 11i) + (7 + 4i) = 12 – 7i
Ex2: (-5 + 7i) – (-11 – 6i) = -5 + 7i +11 + 6i = 6 + 13i
Multiplication)Ex3: 4i(3 – 5i) = 12i –20i2 = 12i –20(-1) = 12i +20 = 20 + 12i
Ex4: (7 – 3i) (-2 – 5i) [Use FOIL] -14 –35i +6i +15i2
-14 –29i +15(-1) -14 –29i –15 -29 –29i
Complex ConjugatesThe complex conjugate of (a + bi) is (a – bi)The complex conjugate of (a – bi) is (a + bi)
(a + bi) (a – bi) = a2 + b2
Division7 + 4i2 – 5i
2 + 5i 14 + 35i + 8i + 20i2 14 + 43i +20(-1)2 + 5i 4 + 10i –10i – 25i2 4 –25(-1)
14 + 43i –20 -6 + 43i -6 434 + 25 29 29 29
= =
= + i=
Square Root of a Negative Number
25 4 = 100 = 10
-25 -4 = (-1)(25) (-1)(4)
= (i2)(25) (i2)(4) = i 25 i 4 = (5i) (2i) = 10i2 = 10(-1) = -10
Optional Step
Practice – Square Root of Negatives
i 1
12
16
4
Practice – Simplify Imaginary Numbers
i2 =
i3 =
i4 =
i5 =
i6 =
-1
-i
1
i
-1
i0 = 1i1 = i
Another way to calculate in
Divide n by 4. If the remainder is rthen in = ir
Example:i11 = __________
11/4 = 2 remainder 3
So, i11 = i3 = -i
Practice – Simplify More Imaginary Numbers
203
100
26
15
i
i
i
i
Practice – Addition/Subtraction
)7()93(
)7()93(
ii
ii 10 +8i
-4 +10i
Practice – Complex Conjugates
• Find complex conjugate.
i
i
43
25
3i =>
-4i =>
Practice Division w/Complex Conjugates
i
i
4
47
4__2i
=
Adding & Subtracting Polynomials
Combine Like Terms
(2x2 –3x +7) + (3x2 + 4x – 2) = 5x2 + x + 5
(5x2 –6x + 1) – (-5x2 + 3x – 5) = (5x2 –6x + 1) + (5x2 - 3x + 5) = 10x2 – 9x + 6
Types of Polynomialsf(x) = 3 Degree 0 Constant Functionf(x) = 5x –3 Degree 1 Linear f(x) = x2 –2x –1 Degree 2 Quadraticf(x) = 3x3 + 2x2 – 6 Degree 3 Cubic
Multiplication of Polynomials
Step 1: Using the distributive property, multiply every term in the 1st polynomial by every term in the 2nd polynomial
Step 2: Combine Like TermsStep 3: Place in Decreasing Order of Exponent
4x2 (2x3 + 10x2 – 2x – 5) = 8x5 + 40x4 –8x3 –20x2
(x + 5) (2x3 + 10x2 – 2x – 5) = 2x4 + 10x3 – 2x2 – 5x + 10x3 + 50x2 – 10x – 25
= 2x4 + 20x3 + 48x2 –15x -25
Binomial Multiplication with FOIL
(2x + 3) (x - 7)
F. O. I. L.(First) (Outside) (Inside) (Last)
(2x)(x) (2x)(-7) (3)(x) (3)(-7)
2x2 -14x 3x -21
2x2 -14x + 3x -21
2x2 - 11x -21
Division by a Monomial3x2 + x 5x3 – 15x2
x 15x
4x2 + 8x – 12 5x2y + 10xy2
4x2 5xy
15A2 – 8A2 + 12 12A5 – 8A2 + 12 4A 4A
Review: Factoring Polynomials
To factor a polynomial, follow a similar process.
Factor: 3x4 – 9x3 +12x2
3x2 (x2 – 3x + 4)
To factor a number such as 10, find out
‘what times what’ = 10
10 = 5(2)
Another Example:Factor 2x(x + 1) + 3 (x + 1)
(x + 1)(2x + 3)
Solving Polynomial Equations By Factoring
Solve the Equation: 2x2 + x = 0
Step 1: Factor x (2x + 1) = 0
Step 2: Zero Product x = 0 or 2x + 1 = 0
Step 3: Solve for X x = 0 or x = - ½
Zero Product Property : If AB = 0 then A = 0 or B = 0
Question: Why are there 2 values for x???
Factoring Trinomials
To factor a trinomial means to find 2 binomials whose productgives you the trinomial back again.
Consider the expression: x2 – 7x + 10
(x – 5) (x – 2)The factored form is:
Using FOIL, you can multiply the 2 binomials andsee that the product gives you the original trinomial expression.
How to find the factors of a trinomial:
Step 1: Write down 2 parentheses pairs.Step 2: Do the FIRSTSStep3 : Do the SIGNSStep4: Generate factor pairs for LASTSStep5: Use trial and error and check with FOIL
Practice
Factor:
1. y2 + 7y –30 4. –15a2 –70a + 120
2. 10x2 +3x –18 5. 3m4 + 6m3 –27m2
3. 8k2 + 34k +35 6. x2 + 10x + 25
Special Types of FactoringSquare Minus a Square
A2 – B2 = (A + B) (A – B)
Cube minus Cube and Cube plus a Cube
(A3 – B3) = (A – B) (A2 + AB + B2)
(A3 + B3) = (A + B) (A2 - AB + B2)
Perfect Squares
A2 + 2AB + B2 = (A + B)2
A2 – 2AB + B2 = (A – B)2
Quadratic Equations
General Form of Quadratic Equation
ax2 + bx + c = 0 a, b, c are real numbers & a 0
A quadratic Equation: x2 – 7x + 10 = 0 a = _____ b = _____ c = ______
Methods & Tools for Solving Quadratic Equations1. Factor 2. Apply zero product principle (If AB = 0 then A = 0 or B = 0)3. Square root method4. Completing the Square5. Quadratic Formula
Example1: Example 2:x2 – 7x + 10 = 0 4x2 – 2x = 0(x – 5) (x – 2) = 0 2x (2x –1) = 0x – 5 = 0 or x – 2 = 0 2x=0 or 2x-1=0 + 5 + 5 + 2 + 2 2 2 +1 +1
2x=1x = 5 or x = 2 x = 0 or x=1/2
1 -7 10
Square Root Method
If u2 = d then u = d or u = - d. If u2 = d then u = + d
Solving a Quadratic Equation with the Square Root MethodExample 1: Example 2:4x2 = 20 (x – 2)2 = 64 4
x – 2 = +6 x2 = 5 + 2 + 2
x = + 5 x = 2 + 6
So, x = 5 or - 5 So, x = 2 + 6 or 2 - 6
Completing the Square
If x2 + bx is a binomial then by adding b 2 which is the square of half 2
the coefficient of x, a perfect square trinomial results:
x2 + bx + b 2 = x + b 2
2 2
Solving a quadratic equation with ‘completing the square’ method.
Example: Step1: Isolate the Binomialx2 - 6x + 2 = 0 -2 -2 Step 2: Find ½ the coefficient of x (-3 )x2 - 6x = -2 and square it (9) & add to both sides.x2 - 6x + 9 = -2 + 9(x – 3)2 = 7x – 3 = + 7
x = (3 + 7 ) or (3 - 7 )
Note: If the coefficient of x2 is not 1 you must divide by the coefficient of x2 beforecompleting the square. ex: 3x2 – 2x –4 = 0(Must divide by 3 before taking ½ coefficient of x)
Step 3: Apply square root method
(Example 1)
(Completing the Square – Example 2)
2x2 +4x – 1 = 0
2x2 +4x – 1 = 0 (x + 1) (x + 1) = 3/2 2 2 2 2 (x + 1)2 = 3/2x2 +2x – 1/2 = 0 (x2 +2x ) = ½ √(x + 1)2 = √3/2(x2 +2x + 1 ) = 1/2 + 1 x + 1 = +/- √6/2
x = √6/2 – 1 or - √6/2 - 1
Step 1: Check the coefficient of the x2 term. If 1 goto step 2 If not 1, divide both sides by the coefficient of the x2 term.
Step 2: Calculate the value of : (b/2)2 [In this example: (2/2)2 = (1)2 = 1]
Step 3: Isolate the binomial by grouping the x2 and x term together, then add (b/2)2 to both sides of he equation.
Step 4: Factor & apply square root method
Quadratic FormulaGeneral Form of Quadratic Equation: ax2 + bx + c = 0
Quadratic Formula: x = -b + b2 – 4ac discriminant: b2 – 4ac 2a if 0, one real solution if >0, two unequal real solutions if <0, imaginary solutionsSolving a quadratic equation with the ‘Quadratic Formula’
2x2 – 6x + 1= 0 a = ______ b = ______ c = _______
x = - (-6) + (-6)2 – 4(2)(1) 2(2)
= 6 + 36 –8 4
= 6 + 28 = 6 + 27 = 2 (3 + 7 ) = (3 + 7 ) 4 4 4 2
2 -6 1
Solving Higher Degree Equations
x3 = 4x
x3 - 4x = 0x (x2 – 4) = 0x (x – 2)(x + 2) = 0
x = 0 x – 2 = 0 x + 2 = 0
x = 2 x = -2
2x3 + 2x2 - 12x = 0
2x (x2 + x – 6) = 0
2x (x + 3) (x – 2) = 0
2x = 0 or x + 3 = 0 or x – 2 = 0
x = 0 or x = -3 or x = 2
Solving By Grouping
x3 – 5x2 – x + 5 = 0
(x3 – 5x2) + (-x + 5) = 0
x2 (x – 5) – 1 (x – 5) = 0
(x – 5)(x2 – 1) = 0
(x – 5)(x – 1) (x + 1) = 0
x – 5 = 0 or x - 1 = 0 or x + 1 = 0
x = 5 or x = 1 or x = -1
Rational Expressions
Rational Expression – an expression in which a polynomial is divided by another nonzero polynomial.
Examples of rational expressions
4 x 2x 2x – 5 x – 5
Domain = {x | x 0} Domain = {x | x 5/2} Domain = {x | x 5}
Multiplication and Division of Rational Expressions
A • C = A 9x = 3B • C B 3x2 x
5y – 10 = 5 (y – 2) = 5 = 110y - 20 10 (y – 2) 10 2
2z2 – 3z – 9 = (2z + 3) (z – 3) = 2z + 3z2 + 2z – 15 (z + 5) (z – 3) z + 5
A2 – B2 = (A + B)(A – B) = (A – B)A + B (A + B)
Negation/Multiplying by –1
-y – 24y + 8
- = y + 2 4y + 8 OR -y - 2
-4y - 8
Examples
x3 – x x + 1x – 1 x
•
(x3 – x) (x + 1) x(x – 1)=
x (x2 – 1)(x + 1) x(x – 1)
=
= x (x + 1) (x – 1)(x + 1) x(x – 1)
= (x + 1)(x + 1) = (x + 1)2
x2 – 25 x2 –10x + 25x2 + 5x + 4 2x2 + 8x
=x2 – 25 2x2 + 8xx2 + 5x + 4 x2 –10x + 25
•
=(x + 5) (x – 5) • 2x(x + 4)(x + 4)(x + 1) • (x – 5) (x – 5)
=2x (x + 5)(x + 1)(x – 5)
Check Your Understanding
Simplify:
x2 –6x –7 x2 -1
Simplify:
1 3x - 2 x2 + x - 6
(x + 1) (x –7)(x + 1) (x – 1)
(x – 7)(x – 1)
1 x2 + x - 6x – 2 3•
1 (x + 3) (x – 2)x – 2 3
•
(x + 3) 3
Addition of Rational ExpressionsAdding rational expressions is like adding fractions
With LIKE denominators:
1 + 2 = 3 8 8 8
x + 3x - 1 = 4x - 1 x + 2 x + 2 x + 2
x + 2 (2 + x) (2 + x)3x2 + 4x - 4 3x2 + 4x -4 (3x2 + 4x – 4) (3x -2)(x + 2)
= =
= 1 (3x – 2)
Adding with UN-Like Denominators
3 + 14 8
(3) (2) + 18 8
6 + 18 8
7 8
1 + 2x2 – 9 x + 3
1 + 2(x + 3)(x – 3) (x + 3)
1 + 2 (x – 3)(x + 3)(x – 3) (x + 3)(x – 3)
1 + 2(x – 3) 1 + 2x – 6 2x - 5(x + 3) (x – 3) (x + 3) (x – 3) (x + 3) (x – 3)
= =
Subtraction of Rational Expressions
2x - x + 1x2 – 1 x2 - 1
To subtract rational expressions:Step 1: Get a Common DenominatorStep 2: Combine Fractions DISTRIBUTING the ‘negative sign’
BE CAREFUL!!
=2x – (x + 1)x2 -1
= x – 1(x + 1)(x –1)
= 1(x + 1)
= 2x – x - 1x2 -1
Check Your Understanding
Simplify:
b b-12b - 4 b-2-
b b-12(b – 2) b-2
-
b -b+12(b – 2) b-2
+
b2(b – 2)
2(-b+1)2(b – 2)
+
b –2b+22(b – 2)
-b + 22(b – 2)= =
-1(b – 2)2(b – 2)
= -12
Complex Fractions
A complex fraction is a rational expression that contains fractions in its numerator, denominator, or both.
Examples:
15
47
xx2 – 16
1x - 4
1x
+ 2x2
3x
- 1x2
7/20 xx + 4
x + 23x - 1
Rational Equations
3x = 3 x + 1 = 3 6 = x2x – 1 x – 2 x - 2 x + 1
(2x – 1)
3x = 3(2x – 1)3x = 6x – 3-3x = -3
x = 1
(x - 2)
x + 1 = 3
x = 2
(x + 1)
6 = x (x + 1)
6 = x2 + x
x2 + x – 6 = 0
(x + 3 ) (x - 2 ) = 0
x = -3 or x = 2Careful! – What doYou notice about the answer?
Rational Equations Cont…To solve a rational equation:
Step 1: Factor all polynomialsStep 2: Find the common denominatorStep 3: Multiply all terms by the common denominatorStep 4: Solve
x + 1 - x – 1 = 1 2x 4x 3
(12x)
= 6 (x + 1) -3(x – 1) = 4x6x + 6 –3x + 3 = 4x
3x + 9 = 4x -3x -3x 9 = x
Other Rational Equation Examples
3 + 5 = 12x – 2 x + 2 x2 - 4
3 + 5 = 12x – 2 x + 2 (x + 2) (x – 2)
(x + 2)(x – 2)
3(x + 2) + 5(x – 2) = 12
3x + 6 + 5x – 10 = 12
8x – 4 = 12 + 4 + 4
8x = 16
x = 2
1 + 1 = 3x x2 4
(4x2)
4x + 4 = 3x2
3x2 - 4x - 4 = 0
(3x + 2) (x – 2) = 0
3x + 2 = 0 or x – 2 = 0
3x = -2 or x = 2
x = -2/3 or x = 2
Check Your Understanding
Simplify:x 1x2 – 1 x2 – 1
1 3x – 2 x
1 1 2x(x – 1) x2 – 1 x(x + 1)
Solve6 1x 2
3 22x – 1 x + 1
2 3 xx – 1 x + 2 x2 + x - 2
+
-
+ -
- = 1
=
+ =
1x - 1
2(x – 3)x(x – 2)
3x(x – 1)(x + 1)
4
5
-1/4
1 = 1 + 1F p q
Solve for p:Try this one:
Solving Radical Equations
25)63( 2 xX2 = 64
10003 x
1000)4( 3 x
#1
#2
#3
#4
Radical Equations Continued…
Example 1:
x + 26 – 11x = 4
26 – 11x = 4 - x
(26 – 11x)2 = (4 – x)2
26 – 11x = (4-x) (4-x)
26 - 11x = 16 –4x –4x +x2
26 –11x = 16 –8x + x2
-26 +11x -26 +11x0 = x2 + 3x -100 = (x - 2) (x + 5) x – 2 = 0 or x + 5 = 0 x = 2 x = -5
Example 2:
3x + 1 – x + 4 = 1
3x + 1 = x + 4 + 1
(3x + 1)2 = (x + 4 + 1)2
3x + 1 = (x + 4 + 1) (x + 4 + 1)
3x + 1 = x + 4 + x + 4 + x + 4 + 13x + 1 = x + 4 + 2x + 4 + 13x + 1 = x + 5 + 2x + 4 -x -5 -x -5 2x - 4 = 2x + 4 (2x - 4)2 = (2x + 4)2
4x2 –16x +16 = 4(x+4) 4x2 –20x = 0 4x(x –5) = 0, so…4x = 0 or x – 5 = 0 x = 0 or x = 5
4x+16
1.5 Inequality Set & Interval Notation
Set Builder Notation{1,5,6} { } {6}
{x | x > -4} {x | x < 2} {x | -2 < x < 7}x such that x such that x is less x such that x is greaterx is greater than –4 than or equal to 2 than –2 and less than or equal to 7
Interval (-4, ) (-, 2] (-2, 7]Notation
Graph-4 2 -2 0 7
Question: How would you write the set of all real numbers? (-, ) or R
Inequality Example
Statement Reason7x + 15 > 13x + 51 [Given]
-6x + 15 > 51 [-13x]
-6x > 36 [-15]
x < 6 [Divide by –6, so must ‘flip’ the inequality sign
Set Notation: {x | x < 6}
Interval Notation: (-, 6]
Graph:
6
Compound Inequality
-3 < 2x + 1 < 3 Set Notation: {x | -2 < x < 1}
-1 -1 -1-4 < 2x < 2 Interval Notation: (-2, 1]
2 2 2Graph:
-2 < x < 1 -2 0 1
Set Operations and Compound Inequalities
Union () – “OR”A B = {x | x A or x B}
-4x + 1 9 or 5X+ 3 12 X -2 or X -3
Intersection () – “AND”A B = {x | x A and x B}
X+ 1 9 and X – 2 3
X 8 and X 5
Set Notation: {x | X 8 and X 5}
Interval Notation: (- , 8] [5, )
0 5 8
[ ]
Set Notation: {x | X -2 or X -3}
Interval Notation: (- , -2] (- , -3]
-2
1.6 Absolute Value Inequality
| 2x + 3| > 5
2x + 3 > 5 or -(2x + 3) > 5
2x > 2 -2x - 3 > 5
x > 1 -2x > 8 -2 -2
x < -4
Set Notation: {x | x < -4 or x > 1}
Interval Notation: (-, -4] or [1, )
Graph: -4 0 1
Absolute Value Equations
| 2x – 3| = 11
2x – 3 = 11 or -(2x – 3) = 11
2x = 14 -2x + 3 = 11
x = 7 -2x = 8
x = -4