MATH 0322 Intermediate Algebra Unit 2

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Transcript MATH 0322 Intermediate Algebra Unit 2

MATH 0322 Intermediate Algebra
Unit 3
Radical Expressions
and Functions
Section: 10.1
Radical Expressions and Functions
οƒ˜ Radical expressions contain a radical sign 𝑛 ( ).
β€’ Match the name with the correct part of
the radical expression:
radical
𝑛
π‘Ž
radicand
index
β€’ Square Root when 𝑛 is 2: 2 ( ) β†’ ( )
3
β€’ Cube Root when 𝑛 is 3:
( ) ….and so on.
𝑛
οƒ˜Depending on the _____,
( ) can either be
index
an even root or an odd root.
Radical Expressions and Functions
οƒ˜Definition: The ________
Principal Square Root (p693)
π‘Ž = ____
𝑏 ,
If 𝑏2 = π‘Ž, then (___)
where π‘Ž and 𝑏 are nonnegative real numbers.
β€’ For example:
6 2 = ____
36 inside the square root.
36 = 6 because (___)
1
9
=
1
3
because
1 2
3
=
1
____
9
inside the square root.
Radical Expressions and Functions
 βˆ’ (π‘Ž) is β€œthe negative of” square root π‘Ž.
(Not the same as square root of negative π‘Ž.)
β€’ For example:
βˆ’ 9 is read as β€œthe negative of” 9 , so
βˆ’ 9= βˆ’ 9
Since 9 is a real number,
=βˆ’ 3
= βˆ’3
P
then βˆ’ 9 is also.
Radical Expressions and Functions
 …but, what about (βˆ’π‘Ž), not a real number
is it a real number also?
β€’ Try to simplify using the definition of the
Principal Square Root.
?
O
O
only if βˆ’3
3 2 = ____.
βˆ’9 Is it?
?
= βˆ’3 only if βˆ’3 2 = ____.
βˆ’9 Is it?
βˆ’9 is not a real number, so in general…
βˆ’9 = 3
Radical Expressions and Functions
β€’ Practice: Evaluate the following.
121 = 11
βˆ’ 81 = βˆ’9
βˆ’4 𝒏𝒐𝒕 𝒂 𝒓𝒆𝒂𝒍 π’π’–π’Žπ’ƒπ’†π’“
49
100
=
7
10
64 + 36 = 8 + 6 = 14
9 + 16 = 25 = 5
Radical Expressions and Functions
β€’ Use your notes to explain the following:
β€œWhy is βˆ’25 not a real number?”
οƒ˜In general, to produce real numbers,
even roots can only operate on
radicand values that are:
P greater than or equal to zero?
A.
B.
less than or equal to zero?
β€’ This also applies to the function 𝑓 π‘₯ = π‘₯ .
Radical Expressions and Functions
οƒ˜ Complete the table to plot the graph of 𝑓 π‘₯ = π‘₯.
𝒙β‰₯𝟎
𝒇 𝒙 = 𝒙
(𝒙, π’š)
0
𝑓 0 = 0=0
(0,0)
1
𝑓 1 = 1=1
(1,1)
𝑓 4 = 4=2
(4,2)
∞
5
2
3
4
βˆ’βˆž
6
7
8
9
∞
βˆ’βˆž
5
𝑓 9 = 9=3
(9,3)
9
Radical Expressions and Functions
οƒ˜ The domain of a square root function 𝑓 π‘₯ = ( )
radicand be nonnegative.
requires that the ________
(Nonnegative means β‰₯ 0.)
οƒ˜ Practice: To find domain of 𝑓 π‘₯ = 6π‘₯ βˆ’ 18
1) set the radicand β‰₯ 0
6π‘₯ βˆ’ 18 β‰₯ 0
2) solve for π‘₯
6π‘₯ β‰₯ 18
π‘₯β‰₯3
οƒ˜ The domain written in Interval Notation is [3, ∞).
P
Radical Expressions and Functions
οƒ˜ Definition: The Cube Root of a number
3
𝑏 3 = ____
π‘Ž
π‘Ž = 𝑏, means that (___)
real numbers have cube roots.
β€’ All ____
3
3
3
since if 5 3 = 125.
125 = 5
βˆ’64 = βˆ’4 since if βˆ’4 3 = βˆ’64.
8
βˆ’
27
=
2
βˆ’
3
since if
οƒ˜ The domain of 𝑓 π‘₯ =
3
2 3
βˆ’
3
=
8
βˆ’ .
27
real numbers.
π‘₯ is all ____
Radical Expressions and Functions
οƒ˜Suggested Practice: Form a table of squares
and cubes, then review for quicker recall of
square roots and cube roots in later sections.
π‘₯
π‘₯2
π‘₯3
0
0
0
1
1
1
2
4
8
3
9
27
4
16
64
5
25
125
…
…
…
MATH 0322 Intermediate Algebra
Unit 3
Rational Exponents
Section: 10.2
Rational Exponents
Complete the Exponent Properties below
for your notes and Formula Sheet.(p.709)
οƒ˜ Properties(π‘Ž, 𝑏 are real and π‘š, 𝑛 are rational exponents):
1)
2)
3)
π‘π‘š
βˆ™
𝑏𝑛
𝑏 π‘š+𝑛
=_____________
Multiply like bases: keep base, add powers,
then simplify
π‘π‘š
π‘šβˆ’π‘›
𝑏
=
_____________
𝑏𝑛
𝑏 π‘šβˆ™π‘›
𝑏 π‘š 𝑛 = _____________
Divide like bases: keep base, subtract powers,
then simplify
π‘Žπ‘› 𝑏 𝑛
= _____________
Product to power: raise each factor to power,
then simplify
4)
π‘Žπ‘
5)
π‘Ž 𝑛
𝑏
6) π‘Ž
π‘š
βˆ’π‘›
𝑛
π‘Žπ‘›
𝑏𝑛
= _____________
1
=
π‘š
π‘Žπ‘›
_____________
Power to power: keep base, multiply powers,
then simplify
Quotient to power: raise numerator and
denominator to power, then simplify
Negative power: take reciprocal of base, apply
π’Ž
π’Ž
power 𝒏 , then simplify. *𝒂 𝒏 must be nonzero.
Rational Exponents
fraction form.
β€’ Rational exponents: exponents in _______
4
5
1
2
2
3
Example: 9 , (βˆ’32) , (16π‘₯ 3 )
οƒ˜ Definition:
1
𝑛
π‘Ž =
𝑛
π‘Ž
Radical notation
Rational Exponent
notation
β€’ If index is even, radicand must be β‰₯ 0.
81 = 81P
(βˆ’25) = βˆ’25
P
O
β€’ If index is odd, radicand can be any real number.
0 = 0P
32 = 32P
(βˆ’128) = βˆ’128P
1
6
6
1
3
3
0 =
0
1
4
4
1
5
5
1
2
1
7
7
Rational Exponents
β€’ Practice: Use the previous definition to rewrite
in radical notation, then simplify.
1
𝑛
π‘Ž =
1
4
81 =
4
π‘Ž
81 =
P
4
3βˆ™3βˆ™3βˆ™3 =3
1
3
3
βˆ’64 =
1
5
5
7π‘₯ 𝑦 =
(βˆ’64) =
(7π‘₯ 4 𝑦)
𝑛
=
4
3
P
(βˆ’4) βˆ™ (βˆ’4) βˆ™ (βˆ’4) = βˆ’4
P
5
7βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘₯βˆ™π‘¦
Rational Exponents
β€’ Practice: Use the previous definition to only
rewrite in rational exponent notation.
𝑛
4
5
3
π‘Ž=π‘Ž
1
𝑛
21 = 21
𝑀3
6
1
4
1
3 5
=
𝑀
6
βˆ’5π‘₯𝑦 =
P
P
βˆ’5π‘₯𝑦
1
3
P
Rational Exponents
β€’ Numerator of rational exponent can be > 1.
4
5
3
2
Example: 9 , (βˆ’32) ,
π‘š
𝑛
οƒ˜Definition: If π‘Ž is real and
reduced, then

π‘š
𝑛
π‘Ž =
π‘š
𝑛
 andt π‘Ž =
𝑛
𝑛
π‘Ž
π‘Žπ‘š
π‘š
π‘š
𝑛
2
(16π‘₯ 3 )7
is positive and
This form will be easier to
work with.
Rational Exponents
β€’ Practice: Evaluate using definitions only.
π‘š
𝑛
π‘Ž =
No calculator.
𝑛
π‘š
𝑛
π‘Žπ‘š
3
5
and π‘Ž =
Exp. Prop. #6
2
125
3
4
=
π‘Ž
π‘š
P…?
= )3 P
32
= …?
= βˆ’27 P
3
5
3
3
3
βˆ’32768
(βˆ’32)
= βˆ’2
(βˆ’32)
= βˆ’8=
βˆ’32 =
55
(βˆ’32) =
βˆ’3
𝑛
2
33
βˆ’81 = βˆ’
𝟏𝟏 2
33
(βˆ’32)
πŸπŸπŸ“
πŸπŸπŸ“
3
4
(11 ) ==
81
1
3
𝟏
𝟏
𝟏
2
1πŸ“ ( πŸπŸ“πŸ”πŸπŸ“ πŸπŸ“
=βˆ’ 3
3
Rational Exponents
β€’ Practice: Simplify using properties.
Property
1
2
3
2
Expo.
Prop. #1
6 βˆ™6 =
Expo.
Prop. #2
7
86
5
86
Expo.
Prop. #6
3
1
1 33
+ )
( 2+
2
6 22
7
1 35
βˆ’ )
( 2+
2
=8
1
βˆ’4
16
6
6
=
P
=8
= 8=2P
1 = P
12
(2)
=8
6
1 1
43
= ( )4 =
16
125π‘₯ 6 =
14
6( 2 )
1
(125π‘₯
125π‘₯6 ) 3
11
(2)
3
3
𝟏
(
)
πŸπŸ”
=
= 6(2) = 36
𝟏
𝟐
1
6
(2
)
3 6) 3
( (5)
1 25π‘₯
5
π‘₯
1 1
2 1
3
3βˆ™ 3
= (5
5) 1( π‘₯5)633 βˆ™ 31
= 5π‘₯ 2
P
Rational Exponents
Practice and Complete
HW10.2
MATH 0322 Intermediate Algebra
Unit 3
Multiplying and Simplifying
Radical Expressions
Section: 10.3
Multiplying and Simplifying
Radical Expressions
β€’ To multiply and simplify Radical Expressions,
you must learn to use:
1) Factoring
2) Product Rule for Radicals
β€’ Compare the following:
a)
4 βˆ™ 25 = 100 = 10
4 βˆ™ 25 = 2
βˆ™5
P
= 10
3
b)
3
3
8 βˆ™ 27 =
3
216 = 6
3
8 βˆ™ 27 = 2
βˆ™3
Since 4 βˆ™ 25 = 4 βˆ™ 25 and 8 βˆ™ 27 =
Even and Odd Roots seem to follow the
same type of rule for multiplication.
3
3
P
=6
8 βˆ™ 27,
Multiplying and Simplifying
Radical Expressions
οƒ˜ Definition: Product Rule for Radicals
If
𝑛
π‘Ž and
𝑛
𝑏 are real numbers,
𝑛
𝑛
𝑛
then
π‘Žβˆ™ 𝑏 ⇔ π‘Žβˆ™π‘.
(*index 𝑛 must be the same.)
Practice: Multiply and simplify.
3
5 βˆ™ 11
=
3
5 βˆ™ 11
=
3
55
a)
3
P
b)
π‘₯+3βˆ™ π‘₯βˆ’3
=
(π‘₯ + 3) βˆ™ (π‘₯ βˆ’ 3)
=
π‘₯2
βˆ’9
P
c)
=
=
7
3π‘₯ 2
7
7
3π‘₯ 2 βˆ™ 18π‘₯ 3
7
54π‘₯ 5
βˆ™ 18π‘₯ 3
P
Multiplying and Simplifying
Radical Expressions
β€’ Question?
Can the radicals below be multiplied directly?
3
6βˆ™ 7
No, the index 𝑛 must be the same.
β€’ Question?
Is π‘₯ 2 βˆ’ 9 the same as π‘₯ 2 βˆ’ 9?
Let π‘₯ = 5 to find out.(read p716)
Multiplying and Simplifying
Radical Expressions
β€’ Simplify the Radical Expression by Factoring
and the Product Rule: ? 3 2 βˆ™ 40 ? 3 5 βˆ™ 16
3
3
3
4
βˆ™
20
?
8 βˆ™ 10
?
80
3
3
80 = 8 βˆ™ 10
1) Factor radicand.
P
…make sure one factor of 80 is the
largest perfect 𝒄𝒖𝒃𝒆.
2) Apply Product Rule.
3) Simplify.
=
3
3
8 βˆ™ 10
3
= 2 βˆ™ 10
3
= 2 10
P
Multiplying and Simplifying
Radical Expressions
β€’ Simplify the Radical Expression by Factoring
and the Product Rule:
5
64
5
5
64 = 32 βˆ™ 2
1) Factor radicand.
…make sure one factor is the
largest perfect 5π‘‘β„Ž power of 64.
2) Apply Product Rule.
3) Simplify.
=
5
5
32 βˆ™ 2
5
=2βˆ™ 2
5
=2 2
P
Multiplying and Simplifying
Radical Expressions
β€’ Simplify the Radical Expression by Factoring
and the Product Rule:
500π‘₯𝑦 2
1) Factor radicand.
500π‘₯𝑦 2 = 100 βˆ™ 5 βˆ™ π‘₯ βˆ™ 𝑦 2
=
100 βˆ™ 𝑦 2 βˆ™ 5 βˆ™ π‘₯
2) Apply Product Rule.
=
100𝑦 2 βˆ™ 5π‘₯
3) Simplify.
= 10 𝑦 βˆ™ 5π‘₯
p.696: Simplifying π‘Ž2
= 10 𝑦 5π‘₯
P
Rational Exponents
Practice and Complete
HW10.3
MATH 0322 Intermediate Algebra
Unit 3
Adding, Subtracting, Dividing
Radical Expressions
Section: 10.4
Adding, Subtracting, Dividing
Radical Expressions
β€’ When adding or subtracting variable expressions,
like terms can be combined.
only ___
β€’ In this section, like radical terms have:
index
1) the same _____
radicand
2) and the same ________.
β€’ When combining like radical terms,
only the __________
coefficients are added or subtracted.
3
3
3
3
P
Example: 6 2π‘₯ + 5 2π‘₯ = 6 + 5 2π‘₯ = 11 2π‘₯
3
Example: 6 2π‘₯ + 5 2π‘₯ = Can’t simplify, index not the same
3
3
Example: 6 7π‘₯ + 5 2π‘₯ = Can’t simplify, radicand not the same
Adding, Subtracting, Dividing
Radical Expressions
Practice: Simplify by combining like radical terms.
3
3
3
3
P
a)
9 5+ 5 = 9+1
b)
6 π‘₯ + 1 βˆ’ 4 π‘₯ + 1 + 7 π‘₯ + 1 = (6 βˆ’ 4 + 7) π‘₯ + 1
5 = 10 5
=9 π‘₯+1
c)
=
7
3π‘₯ 2
7
3π‘₯ 2
7
18π‘₯ 3
7
3π‘₯ 2
7
3π‘₯ 2
+2
+4
= (1 + 4)
P
7
+ 4 3π‘₯ 2
7
+ 2 18π‘₯ 3
+2
7
18π‘₯ 3
=5
7
3π‘₯ 2
7
+ 2 18π‘₯ 3
P
Adding, Subtracting, Dividing
Radical Expressions
Practice: Simplify by combining like radical terms,
if possible.
a)
b)
5 12 βˆ’ 6 27 = 5 4 βˆ™ 3 βˆ’ 6 9 βˆ™ 3
factor
factor = 5 βˆ™ 2 3 βˆ’ 6 βˆ™ 3 βˆ™ 3
= 10 3 βˆ’ 18 3 = βˆ’8 3
3
5
7 2+9 2
Cannot be simplified.
P
P
Adding, Subtracting, Dividing
Radical Expressions
β€’ Compare the following:
a)
36
4
36
4
= 9 =3
=
6
2
P
=3
b)
3
3
3
64
=
8
3
64
4
2
8
=
8 =2
=2
What do you notice?
Looks like a Rule for Dividing Radicals!
P
Adding, Subtracting, Dividing
Radical Expressions
οƒ˜ Definition: Quotient Rule for Radicals
If
𝑛
π‘Ž and
then
𝑛
π‘Ž
𝑏
𝑛
𝑏 are real numbers and 𝑏 β‰  0,
⇔
𝑛
π‘Ž
𝑛
𝑏
.
Read the β€œGreat Question!” on page 727.
Adding, Subtracting, Dividing
Radical Expressions
Practice: Simplify using the Quotient Rule.
(Assume all variables represent positive real numbers.)
50
81
a)
Not square
Square
Factor
=
50
81
=
25βˆ™2
81
=
5 2
9
P
b)
3
=
27π‘₯ 8
𝑦 12
3
27βˆ™π‘₯ 8
3
𝑦 12
3
=
c)
3π‘₯
π‘₯2 π‘₯ 2
3
𝑦4
P
500π‘₯ 3
20π‘₯ βˆ’1
=
500π‘₯ 3
20π‘₯ βˆ’1
=
25π‘₯π‘₯33βˆ’(βˆ’1)
= 25π‘₯ 4
= 5π‘₯ 2
P
Rational Exponents
Practice and Complete
HW10.4
MATH 0322 Intermediate Algebra
Unit 3
Radical Equations
Section: 10.6
Radical Equations
β€’ In this section, students will be asked to solve
Radical Equations by using:
Properties
of Exponents and Radicals,
P
Factoring
skills, Product Rule, Quotient Rule
P
and
P learned strategies for simplifying radical expressions.
β€’ Radical Equation: an equation containing a
radicand of a radical expression.
variable in the ________
Examples:
π‘₯ + 3 = 6,
π‘₯ βˆ’ 8 = π‘₯ βˆ’ 2,
3
2π‘₯ βˆ’ 1 + 5 = 0
Radical Equations
β€’ Think about what you already know about
…to…..hmmm,
get π‘₯ = 16,
the
value
of
π‘₯
in
the
following
equation?
How
the
…so
this
how I
π‘Ž = to
𝑏
…but
ifcan
Iissquare
I need
definition
of
can
solve
get
onlysquare
if π‘Ž = 4,…
𝑏 2 ….
the
right to
side,
π‘₯ = 4
𝑏…
π‘₯ π‘Ž=
I have
to=16
square
help
solve
for π‘₯?
Fan-tastic!
the
left side.
π‘₯ 2= 4 2
π‘₯ = 16
π‘₯ has to be 16,
because 16 = 4.
Radical Equations
β€’ To solve a Radical Equation:
1) isolate the radical,
2) raise both sides to the 𝑛th power and simplify,
3) solve, then check solution(s).
(If equation contains more than one radical,
repeat Steps 1 & 2 until all radicals are eliminated.)
Radical Equations
Practice: Solve.
a)
Check:
3π‘₯ + 4 2 = 8
3π‘₯ + 4 = 64
3π‘₯ = 60
π‘₯ = 20
2
Step 1)
Step 2)
Step 3)
3(20)
20 + 4 = 8
60 + 4 = 8
P 64 = 8
P
Your turn.
3
b)
6π‘₯ βˆ’ 3 3= 3
6π‘₯ βˆ’ 3 = 27
6π‘₯ = 30
π‘₯=5
3
P
Check:
Radical Equations
Practice: Solve.
Check:
Step 1)
2π‘₯ βˆ’ 1 βˆ’ 4 = 3
2π‘₯ βˆ’ 1 2= 7 2 Step 2)
2π‘₯ βˆ’ 1 = 49 Step 3)
2π‘₯ = 50
π‘₯ = 25
Your turn.
b)
6π‘₯ + 3 + 15 = 24
6π‘₯ + 3 2= 9 2
6π‘₯ + 3 = 81
6π‘₯ = 78
π‘₯ = 13
2 20
25 βˆ’ 1 βˆ’ 4 = 3
a)
P
P
50 βˆ’ 1 βˆ’ 4 = 3
49 βˆ’ 4 = 3
7βˆ’4=3
P
Check:
Radical Equations
Practice: Solve.
Check:
Step 1)
π‘₯βˆ’1+7 =2
20
26 βˆ’ 1 + 7 = 2
π‘₯ βˆ’ 1 2= βˆ’5 2 Step 2)
25 + 7 = 2
Step 3)
π‘₯ βˆ’ 1 = 25
5+7=2
π‘₯ = 26
This is considered an β€œextraneous solution”.(See p745)
Your turn.
b)
7π‘₯ + 8 + 15 = 9
Did you figure out when it
would not have a solution?
7π‘₯ + 8 2 = βˆ’6 2
7π‘₯ + 8 = 36
7π‘₯ = 28
Check:
π‘₯=4
β€œextraneous solution”
a)
O
O
O
Radical Equations
Practice: Solve.
a)
6π‘₯ + 7 βˆ’ π‘₯ = 2
6π‘₯ + 7 2 = π‘₯ + 2 2
6π‘₯ + 7 = π‘₯ 2 + 4π‘₯ + 4
Step 1)
Step 2)
Step 3) Need Quadratic
Equation in Standard Form.
7 = π‘₯ 2 βˆ’ 2π‘₯ + 4
0 = π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 3 Factor
0 = (π‘₯ βˆ’ 3)(π‘₯ + 1)
π‘₯βˆ’3 =0
π‘₯=3
π‘₯+1 =0
π‘₯ = βˆ’1
P
P
Don’t forget to check for β€œextraneous solutions”.
Radical Equations
Practice: Solve.
b)
6π‘₯ + 2 βˆ’ 5π‘₯ + 3 = 0
Step 1)
6π‘₯ + 2 2 =
Step 2)
5π‘₯ + 3
6π‘₯ + 2 = 5π‘₯ + 3
π‘₯+2=3
P
2
Step 3)
π‘₯=1
Don’t forget to check for β€œextraneous solutions”.
Radical Equations
Complete and
Practice HW 10.6.
MATH 0322 Intermediate Algebra
Unit 3
Complex Numbers
Section: 10.7
Complex Numbers
β€’ In this section, students complete Chapter 10
by learning:
P
P
P
 what a Complex Number is,
 the components of a Complex Number,
 and how to operate with Complex Numbers.
β€’ To prepare, we revisit βˆ’4 in Section 10.1.
Was βˆ’4 a real number? Yes or No
radicand must
Why? For even roots, the ________
be non-negative.
Complex Numbers
 Complex numbers: A number of the form
π‘Ž + 𝑏𝑖
imaginary part
real part
 where π‘Ž and 𝑏 are real number coefficients,
 and 𝑖 is the imaginary unit.
P
P
οƒ˜ Definition: 𝑖 = βˆ’1
𝑖 2 = βˆ’1
𝑖 2 = βˆ’1
So what is
βˆ’4 ?
2
Product Rule
βˆ’4 = 4 βˆ™ βˆ’1 = 4 βˆ™ βˆ’1
= 2𝑖
imaginary
The square root of a negative number is ________.
 You will see this again in College Algebra.(learn it well)
Complex Numbers
Practice: Write as a multiple of 𝑖.
a)
βˆ’81
= 81 βˆ™ βˆ’1
= 81 βˆ™ βˆ’1
= 9𝑖
P
b)
βˆ’26
c)
βˆ’45
= 26 βˆ™ βˆ’1
= 2 βˆ™ 13 βˆ™ βˆ’1
= 45 βˆ™ βˆ’1
= 9 βˆ™ 5 βˆ™ βˆ’1
= 26 βˆ™ βˆ’1
= 9 βˆ™ 5 βˆ™ βˆ’1
= 26 𝑖
P
=3 5𝑖
P = 3𝑖
5
Your turn.
d)
βˆ’16
= 4𝑖
P
e)
βˆ’35
= 35 𝑖
P
f)
βˆ’28
=2 7𝑖
P = 2𝑖
7
Complex Numbers
 Operations with Complex numbers.
 To add or subtract, combine only like terms.
 To multiply, use Distributive Property with 𝑖 2 = βˆ’1.
 To divide, use a Conjugate to rationalize denominator.
οƒ˜ The conjugate of 2 βˆ’ 5𝑖 is _______.
2 + 5𝑖
βˆ’4 βˆ’ 9𝑖
οƒ˜ The conjugate of βˆ’4 + 9𝑖 is ________.
βˆ’3𝑖
οƒ˜ The conjugate of 3𝑖 is _____.
 All important in College Algebra-Chapter 5.
Complex Numbers
Practice: Perform the indicated operations.
Write the result in the form π‘Ž + 𝑏𝑖.
Combine like terms.
a)
2 βˆ’ 𝑖 + (βˆ’6 + 3𝑖)
Real parts
b)
Imaginary parts
(2) + (βˆ’6) (βˆ’π‘–) + (3𝑖)
(βˆ’4)
(2𝑖)
+
βˆ’4 + 2𝑖
P
Your turn.
c) 9 βˆ’ 5𝑖 + (βˆ’2 + 8𝑖)
7 + 3𝑖
P
βˆ’1 + 4𝑖 βˆ’ (5 + 8𝑖)
βˆ’1 + 4𝑖 + (βˆ’5 βˆ’ 8𝑖)
Real parts
Imaginary parts
(βˆ’1) + (βˆ’5) (4𝑖) + (βˆ’8𝑖)
(βˆ’6)
+
(βˆ’4𝑖)
βˆ’6 βˆ’ 4𝑖
P
d)
βˆ’3 + 7𝑖 βˆ’ (6 + 2𝑖)
βˆ’9 + 5𝑖
P
Complex Numbers
Practice: Perform the indicated operations.
Write the result in the form π‘Ž + 𝑏𝑖.
a) 5𝑖(βˆ’6 + 3𝑖)
b)
Distribute.
Distribute(FOIL).
βˆ’5 βˆ’4𝑖 +10𝑖 +8𝑖 2
βˆ’5 + 6𝑖 + 8𝑖 2
βˆ’5 + 6𝑖 + 8(βˆ’1)
βˆ’5 + 6𝑖 βˆ’ 8
βˆ’13 + 6𝑖
βˆ’30𝑖 +15𝑖 2
βˆ’30𝑖 + 15(βˆ’1)
βˆ’15 βˆ’ 30𝑖
P
Your turn.
c) 2𝑖(4 + 3𝑖)
βˆ’6 + 8𝑖
βˆ’1 + 2𝑖 (5 + 4𝑖)
P
P
d)
2 βˆ’ 8𝑖 (βˆ’1 + 6𝑖)
46 + 20𝑖
P
Complex Numbers
Practice: Divide and simplify to form π‘Ž + 𝑏𝑖.
a)
2+𝑖 βˆ™
5βˆ’3𝑖5+3𝑖
Rationalize denominator
5+3𝑖
by multiplying conjugate.
=
10 + 6𝑖 + 5𝑖 + 3𝑖 2
25 + 15𝑖 βˆ’ 15𝑖 βˆ’ 9𝑖 2
=
10+11𝑖+3(βˆ’1)
25βˆ’9(βˆ’1)
=
10+11π‘–βˆ’3
25+9
=
=
7+11𝑖
34
7
11
+ 𝑖
34
34
P
Radical Equations
Complete and
Practice HW 10.7
MATH 0322 Intermediate Algebra
Unit 3
Solving Quadratic Equations
(Square Root Property)
Section: 11.1
Square Root Property
οƒ˜Square Root Property: suppose 𝑒 is an
algebraic expression and 𝑑 is a nonzero real
number,
If 𝑒2 = 𝑑,
then 𝑒 = 𝑑 or 𝑒 = βˆ’ 𝑑 .
(This is the same as 𝑒 = ± 𝑑.)
MATH 0322 Intermediate Algebra
Unit 3
Quadratic Formula
Section: 11.2
Quadratic Formula
οƒ˜ Quadratic Formula: a formula used to solve
Quadratic Equations in the form π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0.
βˆ’(𝑏)± (𝑏)2 βˆ’4(π‘Ž)(𝑐)
π‘₯=
2(π‘Ž)
β€’ Derived using the Complete the Square method.(p789)
β€’ ± indicates equation has two solutions,
which may be distinct(different) or the same.
β€’ Solutions to a Quadratic Equation are π‘₯-intercepts on the graph.
β€’ 𝑏 2 βˆ’ 4π‘Žπ‘ is called the discriminant. (Read Table 11.2 page 793)
 if 𝑏 2 βˆ’ 4π‘Žπ‘ = 0, one repeated real solution exists.
(1 𝒙-intercept)
 if 𝑏 2 βˆ’ 4π‘Žπ‘ > 0, two distinct real solutions exist.
(2 𝒙-intercepts)
(When are they rational? When are they irrational?)
 if 𝑏 2 βˆ’ 4π‘Žπ‘ < 0, two distinct non-real solutions exist π‘Ž ± 𝑏𝑖. (no 𝒙-intercepts)
Quadratic Formula
οƒ˜Methods to solve Quadratic Equations(p795)
Form of
Quadratic Equations
π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0
π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 can be factored
easily.
π‘Žπ‘₯ 2 + 𝑐 = 0
Equation has no π‘₯-term.
(𝑏 = 0)
𝑒2 = 𝑑
𝑒 is first-degree polynomial.
π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0
π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 can’t be factored
or factoring is too difficult.
Most Efficient Method
Factoring
Square Root
Property
Square Root
Property
Quadratic
Formula
Example
2π‘₯ 2 + 7π‘₯ + 3 = 0
9π‘₯ 2 βˆ’ 5 = 0
(π‘₯ βˆ’ 3)2 βˆ’6 = 1
π‘₯ 2 βˆ’ 2π‘₯ βˆ’ 1 = 0
Quadratic Formula
 Practice: Solve. 3π‘₯ 2 + π‘₯ βˆ’ 2 = 0
 Standard Form 3π‘₯ 2 + π‘₯ βˆ’ 2 = 0
βˆ’(𝑏) ± (𝑏)2 βˆ’4 π‘Ž 𝑐
 Quadratic Formula π‘₯ =
2(π‘Ž)
βˆ’( 1 ) ± ( 1 )2 βˆ’4 3 βˆ’2
π‘₯=
1 5
2
2( 3 )
βˆ’ + =
6 6
3
βˆ’1 ± 5
1 5
βˆ’1 ± 25
=
= βˆ’ ±
=
1 5
6
6 6
6
βˆ’ βˆ’ = βˆ’1
6 6
οƒ˜Solutions are rational, so Factoring Method more efficient.
3π‘₯ 2 + π‘₯ βˆ’ 2 = 0
3π‘₯ βˆ’ 2 π‘₯ + 1 = 0
3π‘₯ βˆ’ 2 = 0
π‘₯+1=0
2
π‘₯ = βˆ’1
π‘₯=
P
P
3
P
Quadratic Formula
 Practice: Solve. 2π‘₯ 2 = 6π‘₯ βˆ’ 1
 Standard Form 2π‘₯ 2 βˆ’ 6π‘₯ + 1 = 0
 Quadratic Formula
βˆ’(𝑏) ± (𝑏)2 βˆ’4 π‘Ž 𝑐
π‘₯=
2(π‘Ž)
βˆ’(βˆ’6) ± ( βˆ’6)2 βˆ’4 2 1
π‘₯=
2( 2 )
6 ± 2 7 36 12 7 3
6 ± 28
6 ± 41βˆ™ 7
7
=
= ±
=
=
= ±
24
24
4
4
4
2
2
3
7 3
7
π‘₯= +
,,, βˆ’
2
2
2
2
P
 Know what the discriminant says about the solutions?
Quadratic Formula
 Practice: Solve. 4π‘₯ 2 βˆ’ 8π‘₯ = π‘₯ 2 βˆ’ 7
 Standard Form 3π‘₯ 2 βˆ’ 8π‘₯ + 7 = 0
 Quadratic Formula
βˆ’(𝑏) ± (𝑏)2 βˆ’4 π‘Ž 𝑐
π‘₯=
2(π‘Ž)
βˆ’(βˆ’8) ± ( βˆ’8)2 βˆ’4 3 7
π‘₯=
2( 3 )
8 ± βˆ’20 8 ± βˆ’41βˆ™ 5 8 ± 2𝑖 5 48 12𝑖 5 4
5
=
=
=
= ±
= ±π‘–
3
3
6
6
6
6
6
3
3
4
5 4
5
π‘₯ = +𝑖
,,, βˆ’ 𝑖
3
3
3
3
P
 Know what the discriminant says about the solutions?
Quadratic Formula
 Practice: Suppose a projectile follows a parabolic
trajectory given by the function
𝑓 π‘₯ = βˆ’0.0064π‘₯ 2 + 2π‘₯ + 3
where π‘₯ is horizontal distance traveled in meters
and 𝑓(π‘₯) is the height along the path.
Find π‘₯ when the projectile strikes the ground.
(Round to nearest whole number.)
 Hint: When it strikes the ground, the height 𝑓(π‘₯) is zero.
Quadratic Formula
 Solution:
𝑓 π‘₯ = βˆ’0.0064π‘₯ 2 + 2π‘₯ + 3
0 = βˆ’0.0064π‘₯ 2 + 2π‘₯ + 3
βˆ’( 2 ) ± ( 2 )2 βˆ’4 βˆ’0.0064
π‘₯=
2(βˆ’0.0064 )
Factoring Method?
Square Root Method?
Quadratic Formula?
3
βˆ’2 ± 4.0768
π‘₯=
βˆ’0.0128
βˆ’2
4.0768
π‘₯=
+
βˆ’0.0128 βˆ’0.0128
π‘₯ = 156.25 + (βˆ’157.74)
π‘₯ = βˆ’1.49
?
βˆ’2
4.0768
π‘₯=
βˆ’
βˆ’0.0128 βˆ’0.0128
π‘₯ = 156.25 βˆ’ (βˆ’157.74)
P
π‘₯ = 313.99 = 314π‘š
Quadratic Formula
Complete and
Practice HW 11.2
MATH 0322 Intermediate Algebra
Unit 3
Quadratic Vocabulary
Quadratic Functions and Their Graphs
β€’ General Form of Quadratic Function:
𝑓 π‘₯ = π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐, π‘Ž β‰  0
Axis of symmetry π‘₯ = β„Ž.
parabola
 Graph is a ___________.
up
 If 𝒂 > 𝟎, graph opens ______.
Vertex is minimum
(β„Ž, π‘˜) if π‘Ž > 0.
 If 𝒂 < 𝟎, graph opens ______.
down
(β„Ž, π‘˜) Vertex is maximum
if π‘Ž < 0.
 Point (𝒉, π’Œ) is the Vertex.
 y-intercept:
x-intercepts:
Compute value 𝒇 𝟎 .
𝑓 0 = π‘Ž(0)2 +𝑏 0 + 𝑐
𝑓 0 =𝒄
Set π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0 and
solve for π‘₯ by: 1) Factoring
2) Square Root Property
3) Quadratic Formula
MATH 0322 Intermediate Algebra
Unit 3
Complete Assignments
Practice for Unit 3 Exam.
Quadratic Formula
οƒ˜Methods used to solve Quadratic Equations:
(Know when one is more efficient to use. Table11.3 p795)
β€’ Factoring Method to solve π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0.
(when π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 is factorable)
β€’ Square Root Method to solve π‘Žπ‘₯ 2 + 𝑐 = 0.
(when 𝑏 = 0)
β€’ Quadratic Formula to solve π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 = 0.
(when π‘Žπ‘₯ 2 + 𝑏π‘₯ + 𝑐 cannot be factored or
is too difficult to factor.)