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
ο Radical expressions contain a radical sign π ( ).
β’ Match the name with the correct part of
the radical expression:
radical
π
π
radicand
index
β’ Square Root when π is 2: 2 ( ) β ( )
3
β’ Cube Root when π is 3:
( ) β¦.and so on.
π
οDepending on the _____,
( ) can either be
index
an even root or an odd root.
Radical Expressions and Functions
οDefinition: The ________
Principal Square Root (p693)
π = ____
π ,
If π2 = π, then (___)
where π and π 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
οΆ β (π) is βthe negative ofβ square root π.
(Not the same as square root of negative π.)
β’ For example:
β 9 is read as βthe negative ofβ 9 , so
β 9= β 9
Since 9 is a real number,
=β 3
= β3
P
then β 9 is also.
Radical Expressions and Functions
οΆ β¦but, what about (βπ), 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 β3
3 2 = ____.
β9 Is it?
?
= β3 only if β3 2 = ____.
β9 Is it?
β9 is not a real number, so in generalβ¦
β9 = 3
Radical Expressions and Functions
β’ Practice: Evaluate the following.
121 = 11
β 81 = β9
β4 πππ π ππππ ππππππ
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:
βWhy is β25 not a real number?β
οIn 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 π π₯ = π₯ .
Radical Expressions and Functions
ο Complete the table to plot the graph of π π₯ = π₯.
πβ₯π
π π = π
(π, π)
0
π 0 = 0=0
(0,0)
1
π 1 = 1=1
(1,1)
π 4 = 4=2
(4,2)
β
5
2
3
4
ββ
6
7
8
9
β
ββ
5
π 9 = 9=3
(9,3)
9
Radical Expressions and Functions
ο The domain of a square root function π π₯ = ( )
radicand be nonnegative.
requires that the ________
(Nonnegative means β₯ 0.)
ο Practice: To find domain of π π₯ = 6π₯ β 18
1) set the radicand β₯ 0
6π₯ β 18 β₯ 0
2) solve for π₯
6π₯ β₯ 18
π₯β₯3
ο The domain written in Interval Notation is [3, β).
P
Radical Expressions and Functions
ο Definition: The Cube Root of a number
3
π 3 = ____
π
π = π, means that (___)
real numbers have cube roots.
β’ All ____
3
3
3
since if 5 3 = 125.
125 = 5
β64 = β4 since if β4 3 = β64.
8
β
27
=
2
β
3
since if
ο The domain of π π₯ =
3
2 3
β
3
=
8
β .
27
real numbers.
π₯ is all ____
Radical Expressions and Functions
οSuggested 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
β¦
β¦
β¦
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)
ο Properties(π, π are real and π, π are rational exponents):
1)
2)
3)
ππ
β
ππ
π π+π
=_____________
Multiply like bases: keep base, add powers,
then simplify
ππ
πβπ
π
=
_____________
ππ
π πβπ
π π π = _____________
Divide like bases: keep base, subtract powers,
then simplify
ππ π π
= _____________
Product to power: raise each factor to power,
then simplify
4)
ππ
5)
π π
π
6) π
π
βπ
π
ππ
ππ
= _____________
1
=
π
ππ
_____________
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
π
power π , then simplify. *π π must be nonzero.
Rational Exponents
fraction form.
β’ Rational exponents: exponents in _______
4
5
1
2
2
3
Example: 9 , (β32) , (16π₯ 3 )
ο Definition:
1
π
π =
π
π
Radical notation
Rational Exponent
notation
β’ If index is even, radicand must be β₯ 0.
81 = 81P
(β25) = β25
P
O
β’ If index is odd, radicand can be any real number.
0 = 0P
32 = 32P
(β128) = β128P
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
π
π =
1
4
81 =
4
π
81 =
P
4
3β3β3β3 =3
1
3
3
β64 =
1
5
5
7π₯ π¦ =
(β64) =
(7π₯ 4 π¦)
π
=
4
3
P
(β4) β (β4) β (β4) = β4
P
5
7βπ₯βπ₯βπ₯βπ₯βπ¦
Rational Exponents
β’ Practice: Use the previous definition to only
rewrite in rational exponent notation.
π
4
5
3
π=π
1
π
21 = 21
π€3
6
1
4
1
3 5
=
π€
6
β5π₯π¦ =
P
P
β5π₯π¦
1
3
P
Rational Exponents
β’ Numerator of rational exponent can be > 1.
4
5
3
2
Example: 9 , (β32) ,
π
π
οDefinition: If π is real and
reduced, then
ο§
π
π
π =
π
π
ο§ andt π =
π
π
π
ππ
π
π
π
2
(16π₯ 3 )7
is positive and
This form will be easier to
work with.
Rational Exponents
β’ Practice: Evaluate using definitions only.
π
π
π =
No calculator.
π
π
π
ππ
3
5
and π =
Exp. Prop. #6
2
125
3
4
=
π
π
P�
= )3 P
32
= β¦?
= β27 P
3
5
3
3
3
β32768
(β32)
= β2
(β32)
= β8=
β32 =
55
(β32) =
β3
π
2
33
β81 = β
ππ 2
33
(β32)
πππ
πππ
3
4
(11 ) ==
81
1
3
π
π
π
2
1π ( πππππ ππ
=β 3
3
Rational Exponents
β’ Practice: Simplify using properties.
Property
1
2
3
2
Expo.
Prop. #1
6 β6 =
Expo.
Prop. #2
7
86
5
86
Expo.
Prop. #6
3
1
1 33
+ )
( 2+
2
6 22
7
1 35
β )
( 2+
2
=8
1
β4
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
π
(
)
ππ
=
= 6(2) = 36
π
π
1
6
(2
)
3 6) 3
( (5)
1 25π₯
5
π₯
1 1
2 1
3
3β 3
= (5
5) 1( π₯5)633 β 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 β 25 = 100 = 10
4 β 25 = 2
β5
P
= 10
3
b)
3
3
8 β 27 =
3
216 = 6
3
8 β 27 = 2
β3
Since 4 β 25 = 4 β 25 and 8 β 27 =
Even and Odd Roots seem to follow the
same type of rule for multiplication.
3
3
P
=6
8 β 27,
Multiplying and Simplifying
Radical Expressions
ο Definition: Product Rule for Radicals
If
π
π and
π
π are real numbers,
π
π
π
then
πβ π β πβπ.
(*index π must be the same.)
Practice: Multiply and simplify.
3
5 β 11
=
3
5 β 11
=
3
55
a)
3
P
b)
π₯+3β π₯β3
=
(π₯ + 3) β (π₯ β 3)
=
π₯2
β9
P
c)
=
=
7
3π₯ 2
7
7
3π₯ 2 β 18π₯ 3
7
54π₯ 5
β 18π₯ 3
P
Multiplying and Simplifying
Radical Expressions
β’ Question?
Can the radicals below be multiplied directly?
3
6β 7
No, the index π must be the same.
β’ Question?
Is π₯ 2 β 9 the same as π₯ 2 β 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 β 40 ? 3 5 β 16
3
3
3
4
β
20
?
8 β 10
?
80
3
3
80 = 8 β 10
1) Factor radicand.
P
β¦make sure one factor of 80 is the
largest perfect ππππ.
2) Apply Product Rule.
3) Simplify.
=
3
3
8 β 10
3
= 2 β 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 β 2
1) Factor radicand.
β¦make sure one factor is the
largest perfect 5π‘β power of 64.
2) Apply Product Rule.
3) Simplify.
=
5
5
32 β 2
5
=2β 2
5
=2 2
P
Multiplying and Simplifying
Radical Expressions
β’ Simplify the Radical Expression by Factoring
and the Product Rule:
500π₯π¦ 2
1) Factor radicand.
500π₯π¦ 2 = 100 β 5 β π₯ β π¦ 2
=
100 β π¦ 2 β 5 β π₯
2) Apply Product Rule.
=
100π¦ 2 β 5π₯
3) Simplify.
= 10 π¦ β 5π₯
p.696: Simplifying π2
= 10 π¦ 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₯ = Canβt simplify, index not the same
3
3
Example: 6 7π₯ + 5 2π₯ = Canβt 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 β 4 π₯ + 1 + 7 π₯ + 1 = (6 β 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 β 6 27 = 5 4 β 3 β 6 9 β 3
factor
factor = 5 β 2 3 β 6 β 3 β 3
= 10 3 β 18 3 = β8 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
ο Definition: Quotient Rule for Radicals
If
π
π and
then
π
π
π
π
π are real numbers and π β 0,
β
π
π
π
π
.
Read the βGreat Question!β 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
=
25β2
81
=
5 2
9
P
b)
3
=
27π₯ 8
π¦ 12
3
27βπ₯ 8
3
π¦ 12
3
=
c)
3π₯
π₯2 π₯ 2
3
π¦4
P
500π₯ 3
20π₯ β1
=
500π₯ 3
20π₯ β1
=
25π₯π₯33β(β1)
= 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:
ο±Properties
of Exponents and Radicals,
P
ο±Factoring
skills, Product Rule, Quotient Rule
P
ο±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₯ β 8 = π₯ β 2,
3
2π₯ β 1 + 5 = 0
Radical Equations
β’ Think about what you already know about
β¦toβ¦..hmmm,
get π₯ = 16,
the
value
of
π₯
in
the
following
equation?
How
the
β¦so
this
how I
π = to
π
β¦but
ifcan
Iissquare
I need
definition
of
can
solve
get
onlysquare
if π = 4,β¦
π 2 β¦.
the
right to
side,
π₯ = 4
πβ¦
π₯ π=
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 πth power and simplify,
3) solve, then check solution(s).
οΆ(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₯ β 3 3= 3
6π₯ β 3 = 27
6π₯ = 30
π₯=5
3
P
Check:
Radical Equations
Practice: Solve.
Check:
Step 1)
2π₯ β 1 β 4 = 3
2π₯ β 1 2= 7 2 Step 2)
2π₯ β 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 β 1 β 4 = 3
a)
P
P
50 β 1 β 4 = 3
49 β 4 = 3
7β4=3
P
Check:
Radical Equations
Practice: Solve.
Check:
Step 1)
π₯β1+7 =2
20
26 β 1 + 7 = 2
π₯ β 1 2= β5 2 Step 2)
25 + 7 = 2
Step 3)
π₯ β 1 = 25
5+7=2
π₯ = 26
This is considered an βextraneous solutionβ.(See p745)
Your turn.
b)
7π₯ + 8 + 15 = 9
Did you figure out when it
would not have a solution?
7π₯ + 8 2 = β6 2
7π₯ + 8 = 36
7π₯ = 28
Check:
π₯=4
βextraneous solutionβ
a)
O
O
O
Radical Equations
Practice: Solve.
a)
6π₯ + 7 β 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 β 2π₯ + 4
0 = π₯ 2 β 2π₯ β 3 Factor
0 = (π₯ β 3)(π₯ + 1)
π₯β3 =0
π₯=3
π₯+1 =0
π₯ = β1
P
P
Donβt forget to check for βextraneous solutionsβ.
Radical Equations
Practice: Solve.
b)
6π₯ + 2 β 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
Donβt forget to check for βextraneous solutionsβ.
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
ο± what a Complex Number is,
ο± the components of a Complex Number,
ο± and how to operate with Complex Numbers.
β’ To prepare, we revisit β4 in Section 10.1.
Was β4 a real number? Yes or No
radicand must
Why? For even roots, the ________
be non-negative.
Complex Numbers
ο± Complex numbers: A number of the form
π + ππ
imaginary part
real part
ο§ where π and π are real number coefficients,
ο§ and π is the imaginary unit.
P
P
ο Definition: π = β1
π 2 = β1
π 2 = β1
So what is
β4 ?
2
Product Rule
β4 = 4 β β1 = 4 β β1
= 2π
imaginary
The square root of a negative number is ________.
οΆ You will see this again in College Algebra.(learn it well)
Complex Numbers
Practice: Write as a multiple of π.
a)
β81
= 81 β β1
= 81 β β1
= 9π
P
b)
β26
c)
β45
= 26 β β1
= 2 β 13 β β1
= 45 β β1
= 9 β 5 β β1
= 26 β β1
= 9 β 5 β β1
= 26 π
P
=3 5π
P = 3π
5
Your turn.
d)
β16
= 4π
P
e)
β35
= 35 π
P
f)
β28
=2 7π
P = 2π
7
Complex Numbers
ο± Operations with Complex numbers.
ο§ To add or subtract, combine only like terms.
ο§ To multiply, use Distributive Property with π 2 = β1.
ο§ To divide, use a Conjugate to rationalize denominator.
ο The conjugate of 2 β 5π is _______.
2 + 5π
β4 β 9π
ο The conjugate of β4 + 9π is ________.
β3π
ο The conjugate of 3π is _____.
οΆ All important in College Algebra-Chapter 5.
Complex Numbers
Practice: Perform the indicated operations.
Write the result in the form π + ππ.
Combine like terms.
a)
2 β π + (β6 + 3π)
Real parts
b)
Imaginary parts
(2) + (β6) (βπ) + (3π)
(β4)
(2π)
+
β4 + 2π
P
Your turn.
c) 9 β 5π + (β2 + 8π)
7 + 3π
P
β1 + 4π β (5 + 8π)
β1 + 4π + (β5 β 8π)
Real parts
Imaginary parts
(β1) + (β5) (4π) + (β8π)
(β6)
+
(β4π)
β6 β 4π
P
d)
β3 + 7π β (6 + 2π)
β9 + 5π
P
Complex Numbers
Practice: Perform the indicated operations.
Write the result in the form π + ππ.
a) 5π(β6 + 3π)
b)
Distribute.
Distribute(FOIL).
β5 β4π +10π +8π 2
β5 + 6π + 8π 2
β5 + 6π + 8(β1)
β5 + 6π β 8
β13 + 6π
β30π +15π 2
β30π + 15(β1)
β15 β 30π
P
Your turn.
c) 2π(4 + 3π)
β6 + 8π
β1 + 2π (5 + 4π)
P
P
d)
2 β 8π (β1 + 6π)
46 + 20π
P
Complex Numbers
Practice: Divide and simplify to form π + ππ.
a)
2+π β
5β3π5+3π
Rationalize denominator
5+3π
by multiplying conjugate.
=
10 + 6π + 5π + 3π 2
25 + 15π β 15π β 9π 2
=
10+11π+3(β1)
25β9(β1)
=
10+11πβ3
25+9
=
=
7+11π
34
7
11
+ π
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
οSquare Root Property: suppose π’ is an
algebraic expression and π is a nonzero real
number,
If π’2 = π,
then π’ = π or π’ = β π .
(This is the same as π’ = ± π.)
MATH 0322 Intermediate Algebra
Unit 3
Quadratic Formula
Section: 11.2
Quadratic Formula
ο Quadratic Formula: a formula used to solve
Quadratic Equations in the form ππ₯ 2 + ππ₯ + π = 0.
β(π)± (π)2 β4(π)(π)
π₯=
2(π)
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’ π 2 β 4ππ is called the discriminant. (Read Table 11.2 page 793)
οΆ if π 2 β 4ππ = 0, one repeated real solution exists.
(1 π-intercept)
οΆ if π 2 β 4ππ > 0, two distinct real solutions exist.
(2 π-intercepts)
(When are they rational? When are they irrational?)
οΆ if π 2 β 4ππ < 0, two distinct non-real solutions exist π ± ππ. (no π-intercepts)
Quadratic Formula
οMethods to solve Quadratic Equations(p795)
Form of
Quadratic Equations
ππ₯ 2 + ππ₯ + π = 0
ππ₯ 2 + ππ₯ + π can be factored
easily.
ππ₯ 2 + π = 0
Equation has no π₯-term.
(π = 0)
π’2 = π
π’ is first-degree polynomial.
ππ₯ 2 + ππ₯ + π = 0
ππ₯ 2 + ππ₯ + π canβt 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 β 5 = 0
(π₯ β 3)2 β6 = 1
π₯ 2 β 2π₯ β 1 = 0
Quadratic Formula
ο± Practice: Solve. 3π₯ 2 + π₯ β 2 = 0
ο§ Standard Form 3π₯ 2 + π₯ β 2 = 0
β(π) ± (π)2 β4 π π
ο§ Quadratic Formula π₯ =
2(π)
β( 1 ) ± ( 1 )2 β4 3 β2
π₯=
1 5
2
2( 3 )
β + =
6 6
3
β1 ± 5
1 5
β1 ± 25
=
= β ±
=
1 5
6
6 6
6
β β = β1
6 6
οSolutions are rational, so Factoring Method more efficient.
3π₯ 2 + π₯ β 2 = 0
3π₯ β 2 π₯ + 1 = 0
3π₯ β 2 = 0
π₯+1=0
2
π₯ = β1
π₯=
P
P
3
P
Quadratic Formula
ο± Practice: Solve. 2π₯ 2 = 6π₯ β 1
ο§ Standard Form 2π₯ 2 β 6π₯ + 1 = 0
ο§ Quadratic Formula
β(π) ± (π)2 β4 π π
π₯=
2(π)
β(β6) ± ( β6)2 β4 2 1
π₯=
2( 2 )
6 ± 2 7 36 12 7 3
6 ± 28
6 ± 41β 7
7
=
= ±
=
=
= ±
24
24
4
4
4
2
2
3
7 3
7
π₯= +
,,, β
2
2
2
2
P
οΆ Know what the discriminant says about the solutions?
Quadratic Formula
ο± Practice: Solve. 4π₯ 2 β 8π₯ = π₯ 2 β 7
ο§ Standard Form 3π₯ 2 β 8π₯ + 7 = 0
ο§ Quadratic Formula
β(π) ± (π)2 β4 π π
π₯=
2(π)
β(β8) ± ( β8)2 β4 3 7
π₯=
2( 3 )
8 ± β20 8 ± β41β 5 8 ± 2π 5 48 12π 5 4
5
=
=
=
= ±
= ±π
3
3
6
6
6
6
6
3
3
4
5 4
5
π₯ = +π
,,, β π
3
3
3
3
P
οΆ Know what the discriminant says about the solutions?
Quadratic Formula
ο± Practice: Suppose a projectile follows a parabolic
trajectory given by the function
π π₯ = β0.0064π₯ 2 + 2π₯ + 3
where π₯ is horizontal distance traveled in meters
and π(π₯) is the height along the path.
Find π₯ when the projectile strikes the ground.
(Round to nearest whole number.)
οΆ Hint: When it strikes the ground, the height π(π₯) is zero.
Quadratic Formula
ο± Solution:
π π₯ = β0.0064π₯ 2 + 2π₯ + 3
0 = β0.0064π₯ 2 + 2π₯ + 3
β( 2 ) ± ( 2 )2 β4 β0.0064
π₯=
2(β0.0064 )
Factoring Method?
Square Root Method?
Quadratic Formula?
3
β2 ± 4.0768
π₯=
β0.0128
β2
4.0768
π₯=
+
β0.0128 β0.0128
π₯ = 156.25 + (β157.74)
π₯ = β1.49
?
β2
4.0768
π₯=
β
β0.0128 β0.0128
π₯ = 156.25 β (β157.74)
P
π₯ = 313.99 = 314π
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:
π π₯ = ππ₯ 2 + ππ₯ + π, π β 0
Axis of symmetry π₯ = β.
parabola
ο± Graph is a ___________.
up
ο± If π > π, graph opens ______.
Vertex is minimum
(β, π) if π > 0.
ο± If π < π, graph opens ______.
down
(β, π) Vertex is maximum
if π < 0.
ο± Point (π, π) is the Vertex.
ο± y-intercept:
x-intercepts:
Compute value π π .
π 0 = π(0)2 +π 0 + π
π 0 =π
Set ππ₯ 2 + ππ₯ + π = 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
οMethods used to solve Quadratic Equations:
(Know when one is more efficient to use. Table11.3 p795)
β’ Factoring Method to solve ππ₯ 2 + ππ₯ + π = 0.
(when ππ₯ 2 + ππ₯ + π is factorable)
β’ Square Root Method to solve ππ₯ 2 + π = 0.
(when π = 0)
β’ Quadratic Formula to solve ππ₯ 2 + ππ₯ + π = 0.
(when ππ₯ 2 + ππ₯ + π cannot be factored or
is too difficult to factor.)