Download Unit 5 Quadratic Functions

Unit 5
Bellringer 11/12/14
1.
2.
What is a variable?
a letter used to represent a value or unknown
quantity that can change or vary.
What is a coefficient?
a number being multiplied by a variable
3.
What is a term?
a number, a variable, or the product of a number
and variable(s).
How are
quadratic
expressions
and quadratic
equations
alike? How are
they different?
 Algebraic expressions are mathematical statements that
include numbers, operations, and variables to represent a
number or quantity.
 They do NOT have an =
 If you set an algebraic expression equal to something it
becomes an algebraic equation. The left side of the =
has the same value as the right side.
 Linear expressions are expressions where the highest
power of the variable is the first power. Example: 2x+1
Today we are going to work with expressions and equations
to the second power.
 Quadratic expressions are expressions where the highest
power of the variable is the second power.
 Example: 3𝑥 2 − 5𝑥 + 4
 Quadratic Expressions in standard form are written as
𝑎𝑥 2 + 𝑏𝑥 + 𝑐, where x is the variable and a, b, and c are
constants.
 𝑏 and 𝑐 can be any value, but 𝑎 can never equal zero
 Determine whether it is a quadratic expression.
 6(x – 1) – x(3 – 2x) + 12.
6(x – 1) – x(3 – 2x) + 12
Original expression
6x – 6 – x(3 – 2x) + 12
Distribute 6 over x – 1.
6x – 6 – 3x + 2x2 + 12
Distribute –x over 3 – 2x.
3x + 6 + 2x2
2x2 + 3x + 6
Combine like terms: 6x and –3x;
–6 and 12.
Rearrange terms so the powers
are in descending order.
 Determine whether it is a quadratic expression.
 Identify each term, coefficient, and constant.
 Classify the expression as a monomial, binomial, or
trinomial.
1. 2𝑥2 + (3𝑥 + 1) + (3𝑥 + 2)
2. 4𝑥 − (5𝑥 + 8)
3. (𝑥 + 5) (2𝑥 − 7)
Bellringer 11/13/14
1.
Is 3𝑥 − 4 𝑥 + 2 + 𝑥(1 + 6𝑥) a
Quadratic expression?
Yes it is, Simplified: 6𝑥 2 − 8
2.
What are the terms?
The terms are 6𝑥 2 and -8
How are
quadratic
expressions
and quadratic
equations
alike? How are
they different?
 Determine whether it is a quadratic expression.
 Identify each term, coefficient, and constant.
 Classify the expression as a monomial, binomial, or
trinomial.
1. 2𝑥2 + 3𝑥 + 1 + 3𝑥 + 2 = 2𝑥 2 + 6𝑥 + 3, yes it is a
quadratic trinomial
2. 4𝑥 − 5𝑥 + 8 = −x − 8
3. 𝑥 + 5 2𝑥 − 7 = 2𝑥 2 + 3𝑥 − 35
 A factor is one of two or more numbers or expressions that
when multiplied produce a given product.
 Example: 3 and 4 are factors of 12, (3*4=12)
 In practice problem 3 from yesterday
 #3. 𝑥 + 5 2𝑥 − 7 simplified to 2𝑥 2 + 3𝑥 − 35
 𝑥 + 5 𝑎𝑛𝑑 (2𝑥 − 7) are factors of
2𝑥 2 +3𝑥 − 35.
 When presented with a polynomial in factored form, such
as example 3, multiply the factors to see if the polynomial
in standard form is a quadratic.
• Sometimes it is easier to work with the factored form to
determine the values that make an expression negative or
equal to 0.
 The length of each side of a square is increased by 2
centimeters. How does the perimeter change? How does
the area change?
Define what you know, don’t know and what you need to find out.
We know a square has 4 equal sides and we are increasing each side by 2
We don’t know the lengths of the sides of the original square
We need to find the Perimeter(add the lengths of all the sides) and Area
(multiply length times width) and how they change between the original
square and the increased square.
 We don’t know the length of the sides
of the original square so we represent
it with a variable, x
 X=the length of 1 side of a square
 Perimeter of this square is:
 𝑥 + 𝑥 + 𝑥 + 𝑥 = 4𝑥
 Area of this square is:
 𝑥 ∗ 𝑥 = 𝑥2
x
x
x
x
 The increased square’s sides are increased by 2
 Does this mean we add, subtract, multiply, or
divide by 2?
Increase means we add, so our new sides equal x+2
 Perimeter of this square is:
 𝑥 + 2 + 𝑥 + 2 + 𝑥 + 2 + 𝑥 + 2 = 4𝑥 + 8
 Area of this square is:
 𝑥 + 2 ∗ 𝑥 + 2 = 𝑥 2 + 4𝑥 + 4
x+2
x+2
x+2
x+2
x
x
x
x
Perimeter: 4x
Area: 𝑥 2
To see how the Perimeter and
Area change we take the
increased square Perimeter and
Area and subtract the original
square Perimeter and Area
Change in Perimeter: (4x+8)-(4x)=8
Change in Area: ( 𝑥 2 +4𝑥 + 4)-(𝑥 2 )=4x+4
x+2
x+2
x+2
x+2
Perimeter: 4x+8
Area: 𝑥 2 +4𝑥 + 4
 The length of each side of a pentagon is increased by 2
centimeters. All of the pentagon’s 5 sides are equal. How
does the perimeter change?
Define what you know, don’t know and what you need to find out.
Bellringer 11/14/14
1.
Identify the Terms, Coefficients,
2
and Constants in 2𝑥 + 4𝑥
Terms: 2𝑥 2 , and 4𝑥; Coefficients: 2, 4; Constants: 0
2.
What are Factors?
2 or more expressions multiplied to get a certain outcome
3.
Are (𝑥) and (2𝑥 + 4) factors of the
expression in Question 1?
Yes, when you multiply the factors you will get 2𝑥 2 + 4𝑥
How does
changing the
value of a
coefficient,
constant, or
variable in an
expression
change the value
of the expression?
 What do we know, not know, and need to find
 We know:
 It is a triangle
 The sides of the triangle are 8x, 10x, and 11x-2
 X is the length in feet
 We do NOT know:
 What to use for x
 We need to find: …
1) Find
an expression for the perimeter of the park.
 Set up an equation using the given expressions. Then
combine like terms to find an expression to represent the
perimeter, P.
 P = 8x + 10x + (11x – 2)
10x
8x
 P = 8x + 10x + 11x – 2
 P = 29x – 2
11x-2
2) What
long?
is the perimeter of the park if a fence panel is 4 feet
 Substitute 4 for x in the equation for perimeter.
 P = 29(4) – 2 = 116 – 2 = 114
 If a fence panel is 4 feet long, the perimeter of the park is 114 feet.
3) How does the perimeter change if the length of each
fence panel increases to 5 feet?
 Substitute 5 for x in the equation for perimeter and compare the
result to the previous question’s answer.
 P = 29(5) – 2 = 145 – 2 = 143
 If the length of each fence panel increases to 5 feet, the perimeter of
the park increases to 143 feet. Subtracting 114 from 143 indicates this
is an increase of 29 feet.
How do changes to the parts affect the expression?
What did we change?
The variable x
What happened when we changed it?
The Perimeter grew when we went from x=4 to x=5
1. You can use Box method, Distribution, or FOIL method
to multiply the binomials.
2. Make sure you simplify your result by combining any
like terms.
3. Your answer should be…
2𝑥 2 + 3𝑥 − 2
a=2 b=3 c=-2
How do changes to the parts affect the expression?
What can we change?
The variable, x
Now let’s try substituting that x with 2
Now substitute it with -2?
Bellringer 11/17/14
1. Simplify 3(12 – x2) + (3 + x) and put
in standard form
2. Identify the Terms, Coefficients, and
Constants.
3. Identify a, b, and c.
 Remember standard form of a
quadratic expression is ax2 + bx + c,
Are all quadratic
expressions
factorable?
Review
 A factor is one of two or more numbers or expressions that
when multiplied produce a given product.
 Example 1: 3 and 4 are factors of 12, (3*4=12)
 Example 2: 𝑥 + 5 2𝑥 − 7 simplified to 2𝑥 2 + 3𝑥 − 35,
so 𝑥 + 5 𝑎𝑛𝑑 (2𝑥 − 7) are factors of 2𝑥 2 +3𝑥 − 35.
• Sometimes it is easier to work with the factored form to
determine the values that make an expression negative or
equal to 0.
What values of x make (x + 7)(x – 10) equal Zero?
 Remember: 0 times anything is 0
 In order for an expression of factors to have a
product of 0, only one of the factors has to be 0
 What makes (x + 7) = 0?
-7
 What makes (x – 10) = 0?
10
What values of x make the expression (x + 2)(x – 3) positive?
Determine the sign possibilities for each factor.
The expression will be positive when both factors are positive or both
factors are negative.
Determine the values of the variable
that make both factors positive.
Determine the values of the
variable that make both factors
negative.
x + 2 is positive when x > –2.
x + 2 is negative when x < –2.
x – 3 is positive when x > 3.
x – 3 is negative when x < 3.
Both factors are positive when x > –2
and x > 3, or when x > 3.
Both factors are negative when
x < –2 and x < 3, or when x < –2.
What values of x make the
expression (x + 7)(x – 10)
negative?
What values of x make the
expression (x + 7)(x – 10)
positive?
 A negative number is any
 A positive number is any number
number less than 0
 In order to have a negative you
need 1 negative number times
1 positive number
 x+7 is negative when x<-7
 x-10 is negative when x<10
 –7 < x < 10 will result in a
negative expression
greater than than 0
 In order to have a positive you
need 2 negative numbers or 2
positive numbers
 x+7 is positive when x>-7
 x-10 is positive when x>10
 x<-7 or x >10 will result in a
positive expression
Multiply (x + 6)(x +3) and
simplify:
Multiply (x + 7)(x – 10) and
simplify:
 (x + 7)(x – 10)
 (x + 6)(x +3)
 x2 + 7x – 10x – 70
 x2 + 6x+3x + 18
 x2– 3x – 70
 x2 + 9x + 18
Notice: What are the sums of the
constants in the original binomials?
-3
Notice: What are the sums of the
constants in the original binomials?
-70
Notice: What is the product of the 18
constants in the original binomials?
Notice: What is the product of the
constants in the original binomials?
9
Bellringer 11/18/14
1. What are two factors of 18 that add
up to 11?
2. What are two factors of 36 that add
up to -15?
3. What is the greatest common factor
of 36 and 18?
Are all quadratic
expressions
factorable?
Factoring a polynomial
means expressing it as
a product of other
polynomials.
Factoring Method #1
Factoring polynomials with a
common monomial factor
(using GCF).
**Always look for a GCF before
using any other factoring
method.
Steps:
1. Find the greatest common factor
(GCF).
2. Divide the polynomial by the GCF.
The quotient is the other factor.
3. Express the polynomial as the
product of the quotient and the GCF.
Example :
Step 1:
2x  12 x  8 x
3
2
GCF  2x
Step 2: Divide by GCF
(2x  12 x  8 x)  2 x  x  6 x  4
3
2
2
Step 3: Express the polynomial as the product of the
quotient and the GCF.
2x(𝑥 2 − 6𝑥 + 4)
Example :
6c d  12c d  3cd
3
2
2
GCF  3cd
Step 1:
Step 2: Divide by GCF
3
2
2
(6c d  12c d  3cd)  3cd 
2
2c  4cd  1
The answer should look like this:
3
2
2
Ex: 6c d 12c d  3cd
2
 3cd(2c  4cd  1)
Factor these on your
own looking for a GCF.
1. 6x  3x  12x  3 x  2 x
3
2
2
2. 5x  10x  35
2
 x  4
 5 x2  2 x  7
3. 16x y z  8x y z  12xy z
3 4
2
2 3
 4 xy z  4 x y  2 xz  3 yz 
2
2
2
2
3 2
Bellringer 11/19/14
 Factor by finding the GCF
2
1. 15𝑥 + 2𝑥
2. 9𝑥 3 + 18𝑥 2 − 81𝑥
3. 16𝑥 3 𝑦 4 z − 8𝑥 2 𝑦 2 𝑧 3 + 12𝑥𝑦 3 𝑧 2
Are all quadratic
expressions
factorable?
Bellringer 11/20/14
Factor
1. 12a4 - 10ab3 + 18a3
2. 70x2y3 + 56x3 + 49x
3. 10x2y3 + 4xy3 - 4x4
Are all quadratic
expressions
factorable?
Factoring Method #2
Factoring polynomials that are a
difference of squares.
To factor, express each term as a
square of a monomial then apply
2
2
the rule... a  b  (a  b)(a  b)
2
Ex: x 16 
2
2
x 4 
(x  4)(x  4)
Here is another example:
1 2
x  81 
49
2
1 x  92  1 x  91 x  9
7
7

7 
Try these on your own:
1. x  121
2
  x  11 x  11
2. 9y  169x
2
2
  3 y  13 x  3 y  13 x 
3. x  16   x  2  x  2   x  4 
4
Be careful!
2
Hint for #3: remember your rule for
raising a power to a power, and you
may have to factor this one twice!
Bellringer 11/20/14
 Factor
1. v2 – 16
2. 9r2 – 16
3. 75n2 - 3
Are all quadratic
expressions
factorable?
Bellringer 11/21/14
 Have a seat
 Clear your desks of everything
 I mean everything!
How could I have
done better on
the last Test?
Bellringer 12/1/14
1. What is the greatest common
factor of 27 and 18?
9
2. What is the greatest common
factor in 4𝑥 2 − 24𝑥?
 4x
3. Factor 6𝑥 2 − 30𝑥
 6x(x-5)
Are all quadratic
expressions
factorable?
Factoring Technique #3
Factoring By Grouping
for polynomials
with 4 or more terms
Factoring By Grouping
1. Group the first set of terms and last set of
terms with parentheses.
2. Factor out the GCF from each group so
that both sets of parentheses contain
the same factors.
3. Factor out the GCF again
(the GCF is the factor from step 2).
Example 1:
b
3
 3b
Step 1: Group
 b  3b
3
2
2
 4 b  12
 4b  12 
Step 2: Factor out GCF from each
group  b2 b  3  4 b  3


Step 3: Factor out GCF
2
again

 b  3b  4

2 x  16 x  6 x  48
Example 2:
3
2
 2  x  8 x  3 x  24 
3

2
 2  x  8 x    3x  24 
 2 x
3
2
 x  8  3  x  8 
2
 2   x  8   x  3 
2

3
2
1. 15𝑥 − 6𝑥 − 20x + 8
 (3𝑥 2 − 4)(5𝑥 − 2)
3
2
2. 2𝑥 − 14𝑥 − 𝑥 + 7
 (2𝑥 2 − 1)(𝑥 − 7)
3. 18𝑥 3 + 15𝑥 2 − 6𝑥 − 5
 (3𝑥 2 − 1)(6𝑥 + 5)
4. 8𝑥 3 + 5𝑥 2 − 48𝑥 − 30
 (𝑥 2 − 6)(8𝑥 + 5)
Bellringer 12/2/14
Solve by grouping
1. 2𝑥 3 + 12𝑥 2 + 5𝑥 + 30
 (2𝑥 2 + 5)(𝑥 + 6)
2. 8𝑥 3 + 5𝑥 2 − 48𝑥 − 30
 (𝑥 2 − 6)(8𝑥 + 5)
3. 18𝑥 3 + 15𝑥 2 − 6𝑥 − 5
 (3𝑥 2 − 1)(6𝑥 + 5)
Are all quadratic
expressions
factorable?
Factoring Method #4
Factoring a trinomial in the
form:
2
ax  bx  c
FACTOR THE X-BOX WAY
We are going to factor trinomials in Standard form
(ax2 + bx +c) using the X-Box method.
Step 1: Write the polynomial in standard form.
Step 2: Factor all common factors in the trinomial.
(GCF)
Step 3: Use the X method.
Step 4: Write your answer.
Step 5: Check your answer by distributing
FACTOR THE X-BOX WAY
ax2 + bx + c
Sometimes m and n are
easy to find.
When they aren’t it is best to
list out factors of ac and see
which ones add up to b
First and
Last
Coefficients
Product
ac=mn
n
m
b=m+n
Sum
Middle
EXAMPLES
Factor using the x-box method.
-12
6
-2
4
Factors of -12
1 -12
-1 12
2 -6
-2
6
3 -4
-3 4
x2 + 4x – 12
Which Factors add to b?
1+ -12 = -11
-1 + 12 = 11
2 + -6 = -4
-2 + 6 = 4
3 + -4 = -1
-3 + 4 = 1
Solution: x2 + 4x – 12 = (x + 6)(x - 2)
EXAMPLES
CONTINUED
2. x2 - 9x + 20
20
-4
-5
-9
Factors of 20
1 20
2
10
4
5
-1 -20
-2 -10
-4
-5
Which Factors add to b?
1 + 20 = 21
2 + 10 = 12
4+5=9
-1 + -20 = -21
-2 + -10 = -12
-4 + -5 = -9
Solution: x2 - 9x + 20 = (x - 4)(x - 5)
1. x2 – 6x + 5
2. x2 + 6x + 5
3. x2 – 12x + 35
4. 6x2 - 12x – 18
Bellringer 12/3/14
Factor the Trinomials:
1. x2 + 5x + 6
 (x+3)(x+2)
2. x2 + 2x - 8
 (x+4)(x-2)
Are all quadratic
expressions
factorable?
3. 4x3 + 28x2 + 24x
 4x(x+6)(x+1)
Make sure your phone is off or on silent!
If I see it, I take it up for the rest of the day!
FACTOR THE X-BOX WAY
Example: Factor 3x2 -13x -10
(3)(-10)=
(x-15)(x+2)
-30
2
-15
x
-15
x
x2
-15x
+2
2x
-30
-13
Check your work and you get x2 -13x -30
That wasn’t the original problem, …
FACTOR THE X-BOX WAY
Example: Factor
(3)(-10)=
-30
2
-15
-13
3x2
-13x -10
3x2 -13x -10 = (x-5)(3x+2)
The x-box method needs help
when the leading coefficient is not
equal to 1. We must use the fact
that the leading coefficient is 3.
15
2
3x
(𝑥 − )(𝑥 + )
3
3
Now, reduce the fractions, if
possible. The coefficient of x will
be the reduced denominator.
+2
5
2
(𝑥 − )(𝑥 + )
1
3
(𝑥 − 5)(3𝑥 + 2)
x
-5
3x2
-15x
2x
-10
FACTOR THE X-BOX WAY
Another Way to look at it:
ax2 + bx + c = (dx+e)(fx+g)
First and
Last
Coefficients
Product
dx
e
ac=mn
fx
1st
Term
Factor
n
g
Factor
m
Last
term
m
n
Middle
b=m+n
Sum
EXAMPLES
CONTINUED
2x2 - 5x - 7
a)
b)
-14
-7
2
-5
2x
x
+1
-7
2x2 -7x
2x
-7
Solution: 2x2 - 5x – 7 = (2x - 7)(x + 1)
EXAMPLES
CONTINUED
15x2 + 7x - 2
a)
b)
-30
10
-3
7
3x
+2
5x 15x2 10x
-1
-3x
-2
Solution: 15x2 + 7x – 2 = (3x + 2)(5x - 1)
1. 2x2 – 19x + 24
 (x-8)(2x-3)
2. 2x² + 13x + 6
 (x + 6)(2x + 1)
3. 6x2 + x -15
 (3x + 5)(2x – 3)
4.
6x2 + 13x -5
 (2x + 5)(3x – 1)
Bellringer 12/4/14
Factor the Trinomials:
1. x2 - 9x + 18
 (x - 6)(x - 3)
2. 2x2 - 9x - 18
Are all quadratic
expressions
factorable?
 (x - 6)(2x + 3)
3. 9x2 - 4
 (3x - 2)(3x + 2)]
Make sure your phone is off or on silent!
If I see it, I take it up for the rest of the day!
Factoring Technique #5
Factoring a perfect square
trinomial in the form:
a  2ab  b  (a  b)
2
2
a  2ab  b  (a  b)
2
2
2
2
Perfect Square Trinomials can be
factored just like other trinomials,
but if you recognize the perfect
squares pattern, follow the formula!
a  2ab  b  (a  b)
2
2
a  2ab  b  (a  b)
2
2
2
2
2
Ex: x  8x 16
x 
Does the middle
term fit the
pattern, 2ab?
4 
2
2
a
b
2  x  4  8x
Yes, the factors are (a + b)2 :
x  8x  16  x  4
2
2
2
Ex: 4x 12x  9
2x 
Does the middle
term fit the
pattern, 2ab?
3
2
2
a
b
2 2x  3  12x
Yes, the factors are (a - b)2 :
4x  12x  9  2x  3
2
2
1. x2 – 6x + 9
 (x - 3)2
2. 9x2 + 12x + 4
 (3x + 2)2
3. 16x2 – 56x + 49
 (4x - 7)2
4.
4x2 + 20x + 25
Bellringer 12/5/14
Factor the following:
1. - 40x2y - 8y
 - 8y (5x2 + 1)
Are all quadratic
expressions
factorable?
2. 9x2 - 12x + 4
 (3x - 2)2
3. 25n2 - 9
 (5n + 3)(5n - 3)
4. 12p3 + 8p2 + 21p + 14
 (4p2 + 7)(3p + 2)
Make sure your phone is off or on silent!
If I see it, I take it up for the rest of the day!
First group that has every question
answered correctly on everyone’s paper
with work shown wins 5 bonus Points on the
test Tuesday!
2nd group gets 4 points, 3rd gets 3, 4th gets 2,
and 5th gets 1
Bellringer 12/8/14
 Take out your Review sheet that you
should have completed this weekend.
 Make sure your phone is off or on
silent. If I see it, I take it up for the rest
of the day!
Are all quadratic expressions
factorable?
Bellringer 12/9/14
1. What factoring method do you use
when you have 4 terms?
2. What factoring method should we try
to use if there are 2 terms?
3. What factoring methods can we use
when we have 3 terms
 When you are done clear your desks
except for a pencil, calculator, graphic
organizer, and 1 sheet of paper to
show your work if you would like.
 Make sure your phone is off or on
silent!
Factoring a trinomial
using guess and check:
2
ax  bx  c
1. Write two sets of parenthesis, ( )( ).
These will be the factors of the
trinomial.
2. Product of first terms of both binomials
2
must equal first term of the trinomial. ( ax )
Next
Factoring a trinomial
using guess and check :
2
ax  bx  c
3. The product of last terms of both
binomials must equal last term of the
trinomial (c).
4. Think of the FOIL method of
multiplying binomials, the sum of the
outer and the inner products must
equal the middle term (bx).
x  6x  8
2
Example :
x
x
-2
 x
x
Factors of +8:
-4


x x  x
O + I = bx ?
1&8
1x + 8x = 9x
2&4
2x + 4x = 6x
-1 & -8
-1x - 8x = -9x
-2 & -4
-2x - 4x = -6x
2
x  6x  8  (x  2)(x  4)
2
Check your answer by
using FOIL
F
2
O
I
L
(x  2)(x  4)  x  4x  2x  8
2
 x  6x  8
Lets do another example:
2
6x 12x 18
Don’t Forget Method #1.
Always check for GCF before you do anything else.
2
6(x  2x  3)
Find a GCF
6(x  3)(x 1)
Factor trinomial
When a>1 and c<1, there may be
more combinations to try!
Example :
6 x  13x  5
2
Step 1:
2
Find the factors of 6x :
3x  2x
6x  x
Example : 6 x  13 x  5
2
Step 2:
5 -1
-5 1
 -1 5 


 1 -5 
Find the factors of -5:
Order can make
a difference!
Example : 6 x  13 x  5
2
Step 3: Place the factors inside the
parenthesis until O + I = bx.
Try:
F
2
6x  1x  5
O
I
L
6x  30x  x  5
O + I = 30 x - x = 29x
This
doesn’t
work!!
Example : 6 x  13 x  5
2
Switch the order of the second terms
and try again.
6x  5x  1
F
2
O
I
L
6x  6x  5x  5
O + I = -6x + 5x = -x
This
doesn’t
work!!
Try another combination:
Switch to 3x and 2x
(3x 1)(2x  5)
F
2
O
I
L
6x  15x  2x  5
O+I = 15x - 2x = 13x IT WORKS!!
6x  13x  5  (3x 1)(2x  5)
2
Bellringer 12/10/14
1. Factor: 12x3-4x2+18x-6
2. Factor: 9x2-81
3. Factor: 2x2-6x+5
 When you are done clear your desks of
everything except your bellringer sheet!
 Make sure your phone is off or on silent!
Bellringer 12/11/14
 Solve for x
1.
x+8=0
2.
3x - 2 = 0
3.
5x + 6 = - 5 - 3x
 Make sure your phone is off or on silent!
How do the
factors of a
quadratic
functions yield the
zeros for that
function?
Solving Quadratic Equations by Factoring.
A quadratic equation is an equation that can be written in the form
ax2 + bx + c = 0,
where a, b, and c are real numbers, with a ≠ 0.
The form ax2 + bx + c = 0 is the standard form of a quadratic equation. For
example, x 2  5x  6  0, 2 x2  5x  3, and x 2  4 are all quadratic equations, but only
x2 + 5x +6 = 0 is in standard form.
Until now, we have factored expressions, including many quadratic
expressions. In this section we see how we can use factored quadratic
expressions to solve quadratic equations.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
 We use the zero-factor property to solve a
quadratic equation by factoring.
 If a and b are real numbers and if ab = 0,
then a = 0 or
b = 0.
 That is, if the product of two numbers is 0,
then at least one of the numbers must be 0.
One number must, but both may be 0.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 91
Using the Zero-Factor Property
 Solve.
Solution:
 2x  35x  7   0
 3 7
 ,  
 2 5
2 x  3  0 or 5x  7  0
2x  3  3  0  3 5x  7  7  0  7
2 x 3
5 x 7


2
2
5
5
3
7
x
x
2
5
x  2 x  4  0
0, 2
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
x0
or
2x  4  0
2x  4  4  0  4
2 x 4

2
2
x  2
Slide 6.5 - 92
Bellringer 12/12/14
 Solve for x
1.
(x + 5)(x - 4) = 0
2.
(7x - 1)(x+9)= 0
3.
x2 + 3x + 2= 0
 Make sure your phone is off or on silent!
How do the
factors of a
quadratic
functions yield the
zeros for that
function?
Solving Quadratic Equations
Solve.
x2  2 x  8
x2  2x  8  8  8
x2  2x  8  0
 x  4 x  2  0
Solution:
x40
x2  0
or
x44  04 x 22  02
x2
x  4
4, 2
x5  0
x 2  x  30
or
x6  0
x 55  05
x2   x  30  x  30   x  30 x  6  6  0  6
x6
x  5
x 2  x  30  0
 x  6 x  5  0
5,6
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 94
In summary, follow these steps to solve quadratic equations by
factoring.
Step 1: Write the equation in standard form— that is, with
all terms on one side of the equals sign in descending
power of the variable and 0 on the other side.
Step 2: Factor completely.
Step 3: Use the zero-factor property to set each factor with
variable equal to 0, and solve the resulting equations.
Step 4: Check each solution in the original equation.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 95
Solving a Quadratic Equation
with a Common Factor
2
Solve 3m − 9m = 30.
m2  0
m 2  2  0  2
Solution:
m  2
3m2  9m  30  30  30
2
3m  9m  30  0
m5  0
2
3  m  3m  10   0
m55  05
3  m  2 m  5  0
m5
2,5
A common error is to include the common factor 3 as a solution.
Only factors containing variables lead to solutions.
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 96
Solving Quadratic Equations
Solve.
49 x 2  9  0 7 x  3  3  0  3
Solution:
 7 x  3 7 x  3  0
 3 3
 , 
 7 7
x 2  3x
x 2  3x  3 x  3 x
x  x  3  0
7x  3  3  0  3
7x 3

7
7
3
x
7
7 x 3

7
7
x0
x 33  0  3
0,3
x
3
7
x 3
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 97
Solving Quadratic Equations
(cont’d)
x20
x22  02
x  2
Solve.
x  4x  7  2
Solution:
4x  7x  2  2  2
2
4x  7x  2  0
 x  2 4x 1  0
2
1

 2, 
4

4 x 1  0
4 x 1  1  0  1
4x 1

4
4
1
x
4
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 98
Solving Equations with More
than Two Variable Factors

2 x3  50 x  0
Solution: 2 x x 2  25  0
Solve.


2 x  x  5 x  5  0
2x  0
2x 0

2 2
x0
x 55  05
x  5
x 55  05
x 5
0, 5,5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 99
Solving Equations with More
than Two Variable Factors
(cont’d)

Solve.
 2 x  1  2 x 2  7 x  15  0
Solution:  2 x  1 2 x  3 x  5  0
2 x  1 1  0 1
2 x 1

2
2
1
x
2
2x  3  3  0  3
2x 3

2 2
3
x
2
x 55  05
x  5
1 3

5,  , 
2 2

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 100
Solving an Equation Requiring
Multiplication before Factoring

 x  1 2 x  1   x  1
Solve.
2
2 x 2  3x  1  x 2  2 x  1
2
2
2
2
2 x  3x  1   x  2 x  1  x  2 x  1   x  2 x  1
x2  5x  0
x  x  5  0
Solution:
x0
x 55  05
x 5
 0,5
Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Slide 6.5 - 101
Bellringer 12/15/14
 Solve for x
1. 3x + 10 = 2x - 5
2. x2 + 5x – 50 = 0
3. 2x2 – 3 = - 5x
None
Bellringer 12/16/14
 Solve for x
1. x2 + 2x + 10 = 0
None
2. 3x2 + 9x – 12 = 0
3. 2x2 – 5x +3= 0
SOLVING QUADRATIC
EQUATIONS BY THE
QUADRATIC FORMULA
THE QUADRATIC FORMULA
1. When you solve using completing the square
on the general formula ax 2  bx  c  0
you get:
b  b2  4ac
x
2a
2. This is the quadratic formula!
3. Just identify a, b, and c then substitute into
the formula.
WHY USE THE
QUADRATIC FORMULA?
The quadratic formula allows you to solve ANY
quadratic equation, even if you cannot factor
it.
An important piece of the quadratic formula is
what’s under the radical:
b2 – 4ac
This piece is called the discriminant.
The discriminant tells you the number and types of answers
(roots) you will get. The discriminant can be +, –, or 0
which actually tells you a lot! Since the discriminant is
under a radical, think about what it means if you have a
positive or negative number or 0 under the radical.
WHAT THE DISCRIMINANT
TELLS YOU!
Value of the Discriminant
Nature of the Solutions
Negative
2 imaginary solutions
Zero
1 Real Solution
Positive – perfect square
2 Reals- Rational
Positive – non-perfect
square
2 Reals- Irrational
EXAMPLE #1
Find the value of the discriminant and describe the nature of the
roots (real,imaginary, rational, irrational) of each quadratic
equation. Then solve the equation using the quadratic formula)
1.
2 x 2  7 x  11  0
Discriminant =
a=2, b=7, c=-11
b 2  4ac
(7) 2  4(2)( 11)
49  88
Discriminant = 137
Value of discriminant=137
Positive-NON perfect
square
Nature of the Roots –
2 Reals - Irrational
EXAMPLE #1- CONTINUED
Solve using the Quadratic
Formula
a  2, b  7, c  11
2 x 2  7 x  11  0
b  b 2  4ac
2a
7  7 2  4(2)(11)
2(2)
7  137
4
2 Reals - Irrational
Try the following examples. Do your work on your paper and then check
your answers.
1. x  2 x  63  0
2
2. x  8 x  84  0
2
3. x  5 x  24  0
2
4. x  7 x  13  0
2
5. 3 x 2 5 x  6  0
1.  9, 7 
2.(6, 14)
3.  3,8 
 7  i 3 
4. 

2


 5  i 47 
5. 

6


Bellringer 12/17/14
 Solve for x by Completing the
Square
1. x2 + 10x + 16 = 0
 Solve for x by using the Quadratic
Formula
2. 2x2 + x - 1 = 0
None