SWBAT: Write a Quadratic Equation and Solve.

SWBAT: Solve Consecutive Integer Word Problems by
Factoring Quadratic Equations
B
D
-40
-5
8
3
π‘₯ +3
π‘₯= πŸ“
𝒙(𝒙 + πŸ‘) = πŸ’πŸŽ
π’™πŸ +πŸ‘π’™ = πŸ’πŸŽ
π’™πŸ + πŸ‘π’™ βˆ’ πŸ’πŸŽ = 0
𝒙+πŸ– π’™βˆ’πŸ“ =𝟎
𝒙+πŸ– =𝟎
𝒙 = βˆ’πŸ–
Reject
π’™βˆ’πŸ“ =𝟎
𝒙=πŸ“
SWBAT: Solve Consecutive Integer Word Problems by
Factoring Quadratic Equations
Homework
SWBAT:
Equations
Solve Consecutive Integer Word Problems by Factoring Quadratic
SWBAT:
Equations
Solve Consecutive Integer Word Problems by Factoring Quadratic
SWBAT:
Equations
Solve Consecutive Integer Word Problems by Factoring Quadratic
First x
Second x + 1
-56
-7
8
1
π‘­π’Šπ’“π’”π’• × π‘Ίπ’†π’„π’π’π’… = πŸ“πŸ”
𝒙(𝒙 + 𝟏) = πŸ“πŸ”
π’™πŸ +πŸπ’™ = πŸ“πŸ”
π’™πŸ + πŸπ’™ βˆ’ πŸ“πŸ” = 0
𝒙+πŸ– π’™βˆ’πŸ• =𝟎
𝒙+πŸ– =𝟎 π’™βˆ’πŸ• =𝟎
𝒙 = βˆ’πŸ–
𝒙=πŸ•
x = -8
x=7
First
Second x + 1 = -7 x + 1 = 8
SWBAT:
Equations
Solve Consecutive Integer Word Problems by Factoring Quadratic
First x
Second x + 2
-48
-6
8
2
π‘­π’Šπ’“π’”π’• × π‘Ίπ’†π’„π’π’π’… = πŸ’πŸ–
𝒙(𝒙 + 𝟐) = πŸ’πŸ–
π’™πŸ +πŸπ’™ = πŸ’πŸ–
π’™πŸ + πŸπ’™ βˆ’ πŸ’πŸ– = 0
𝒙+πŸ– π’™βˆ’πŸ” =𝟎
𝒙+πŸ– =𝟎 π’™βˆ’πŸ” =𝟎
𝒙 = βˆ’πŸ–
𝒙=πŸ”
x = -8
x=6
First
Second x + 2 = -6 x + 2 = 8
SWBAT:
Equations
Solve Consecutive Integer Word Problems by Factoring Quadratic
First x
Second x + 1
F
S
x=0
x=5
x+1=1
x+1=6
𝒙=𝟎
𝑭 βˆ™ 𝑺 = πŸ‘ π’”π’–π’Ž βˆ’ πŸ‘
𝒙(𝒙 + 𝟏) = πŸ‘(πŸπ’™ + 𝟏) βˆ’ πŸ‘
π’™πŸ + πŸπ’™ = πŸ”π’™ + πŸ‘ βˆ’ πŸ‘
π’™πŸ + πŸπ’™ = πŸ”π’™
βˆ’πŸ”π’™ = βˆ’πŸ”π’™
0
π’™πŸ βˆ’ πŸ“π’™ = 𝟎
-5
0
𝒙 π’™βˆ’πŸ“ =𝟎
-5
π’™βˆ’πŸ“ =𝟎
𝒙=πŸ“
SWBAT: Write a Quadratic Equation and Solve.
First x
Second x + 2
F
S
x = -1
x=3
x+2=1 x+2=5
𝑭 βˆ™ 𝑺 = 𝟐 π’”π’–π’Ž βˆ’ 𝟏
𝒙(𝒙 + 𝟐) = 𝟐(πŸπ’™ + 𝟐) βˆ’ 𝟏
π’™πŸ +πŸπ’™ = πŸ’π’™ + πŸ’ βˆ’ 𝟏
π’™πŸ +πŸπ’™ = πŸ’π’™ + πŸ‘
βˆ’πŸ’π’™ βˆ’ πŸ‘ = βˆ’πŸ’π’™ -3
π’™πŸ βˆ’πŸπ’™ βˆ’ πŸ‘ = 𝟎
𝒙+𝟏 π’™βˆ’πŸ‘ =𝟎
𝒙+𝟏 =𝟎 π’™βˆ’πŸ‘ =𝟎
𝒙 = βˆ’πŸ
𝒙=πŸ‘
-3
-3
1
-2
SWBAT: Write a Quadratic Equation and Solve.
First x
Second x + 1
x=4
F x = -2
S x + 1 = -1 x + 1 = 5
𝑭 βˆ™ 𝑺 = πŸ‘ 𝑺𝒆𝒄𝒐𝒏𝒅 + πŸ“
𝒙 𝒙+𝟏 =πŸ‘ 𝒙 +𝟏 +πŸ“
π’™πŸ + πŸπ’™ = πŸ‘π’™ + πŸ‘ + πŸ“
π’™πŸ +πŸπ’™ = πŸ‘π’™ + πŸ–
βˆ’πŸ‘π’™ βˆ’ πŸ– = βˆ’πŸ‘π’™ βˆ’ πŸ–
π’™πŸ βˆ’πŸπ’™ βˆ’ πŸ– = 𝟎
𝒙+𝟐 π’™βˆ’πŸ’ =𝟎
𝒙+𝟐 =𝟎 π’™βˆ’πŸ’ =𝟎
𝒙 = βˆ’πŸ
𝒙=πŸ’
-8
2 -4
-2
SWBAT: Write a Quadratic Equation and Solve.
Example 5:
Find three consecutive odd integers such that the product of
the first and third is one more than four times the second.
πΉπ‘–π‘Ÿπ‘ π‘‘
π‘₯
π‘†π‘’π‘π‘œπ‘›π‘‘
Third
π‘₯ + 2
π‘₯ + 4
F
S
T
x = -3
x=3
x + 2 = -1
x+2=5
x+4= 1
x+4=7
𝑭 βˆ™ 𝑻 = πŸ’ 𝑺𝒆𝒄𝒐𝒏𝒅 + 𝟏
𝒙 𝒙+πŸ’ =πŸ’ 𝒙 +𝟐 +𝟏
π’™πŸ +πŸ’π’™ = πŸ’π’™ + πŸ– + 𝟏
π’™πŸ +πŸ’π’™ = πŸ’π’™ + πŸ—
βˆ’πŸ’π’™ βˆ’ πŸ— = βˆ’πŸ’π’™ βˆ’πŸ—
π’™πŸ βˆ’πŸ— = 𝟎
𝒙+πŸ‘ π’™βˆ’πŸ‘ =𝟎
𝒙+πŸ‘ =𝟎 π’™βˆ’πŸ‘ =𝟎
𝒙 = βˆ’πŸ‘
𝒙=πŸ‘
π’™πŸ βˆ’
πŸ— = 𝟎
SWBAT: Write a Quadratic Equation and Solve.
Practice: Solve.
Find three consecutive integers such that the product of the
first and second is 2 more than three times the third.
πΉπ‘–π‘Ÿπ‘ π‘‘
π‘†π‘’π‘π‘œπ‘›π‘‘
Third
F
x = -2
S
x + 1 = -1
T
x+2= 0
𝑭 βˆ™ 𝑺 = πŸ‘ π‘»π’‰π’Šπ’“π’… + 𝟐
𝒙 𝒙+𝟏 =πŸ‘ 𝒙 +𝟐 +𝟐
π‘₯ +1
𝟐
𝒙
+ πŸπ’™ = πŸ‘π’™ + πŸ” + 𝟐
π‘₯ +2
𝟐
𝒙
+πŸπ’™ = πŸ‘π’™ + πŸ–
x=4
βˆ’πŸ‘π’™ βˆ’ πŸ– = βˆ’πŸ‘π’™ βˆ’πŸ–
x+1=5
x+2=6
π’™πŸ βˆ’πŸπ’™ βˆ’ πŸ– = 𝟎 2 -8-4
-2
𝒙+𝟐 π’™βˆ’πŸ’ =𝟎
𝒙+𝟐 =𝟎 π’™βˆ’πŸ’ =𝟎
𝒙 = βˆ’πŸ
𝒙=πŸ’
π‘₯
SWBAT: Write a Quadratic Equation and Solve.
Homework Changes
The product of two even consecutive
integers is 120. Find the integers
1.
The product of two consecutive integers is one
less than their sum. Find the integers.
3.
SWBAT: Write a Quadratic Equation and Solve.
Challenge
SWBAT: Write a Quadratic Equation and Solve.
SWBAT: Write a Quadratic Equation and Solve.
Exit Ticket
πŸ‘ π‘­π’Šπ’“π’”π’•
πŸ‘π’™
πŸ‘π’™
πŸπ’™
𝒙
πΉπ‘–π‘Ÿπ‘ π‘‘
π‘₯
π‘†π‘’π‘π‘œπ‘›π‘‘
Third
π‘₯ + 2
π‘₯ + 4
= π‘»π’‰π’Šπ’“π’… + πŸ’πŸ–
= 𝒙 + πŸ’ + πŸ’πŸ–
= 𝒙 + πŸ“πŸ
= πŸ“πŸ
= πŸπŸ”