unit 4 lecture 31 applications of the quadratic formula
DESCRIPTION
Unit 4 Lecture 31 Applications of the Quadratic Formula. Applications of the Quadratic Formula. Objectives. Formulate a quadratic equation from a problem situation Solve a quadratic equation by using the quadratic formula. - PowerPoint PPT PresentationTRANSCRIPT
Lecture 31Lecture 31 11
Unit 4 Lecture 31Unit 4 Lecture 31Applications of the Quadratic Applications of the Quadratic
FormulaFormula
Applications of the Quadratic Applications of the Quadratic FormulaFormula
Lecture 31Lecture 31 22
ObjectivesObjectives
• Formulate a quadratic equation from a Formulate a quadratic equation from a problem situationproblem situation
• Solve a quadratic equation by using the Solve a quadratic equation by using the quadratic formulaquadratic formula
Lecture 31Lecture 31 33
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
Lecture 31Lecture 31 44
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 1600
Lecture 31Lecture 31 55
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
Lecture 31Lecture 31 66
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
00 = -16(t = -16(t22 – 100) – 100)
Lecture 31Lecture 31 77
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
00 = -16(t = -16(t22 – 100) – 100)
00 = -16(t – 10)(t + 10) = -16(t – 10)(t + 10)
Lecture 31Lecture 31 88
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
00 = -16(t = -16(t22 – 100) – 100)
(t – 10)=0 or (t + 10)=0(t – 10)=0 or (t + 10)=0
00 = -16(t – 10)(t + 10) = -16(t – 10)(t + 10)
Lecture 31Lecture 31 99
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
00 = -16(t = -16(t22 – 100) – 100)
(t – 10)=0 or (t + 10)=0(t – 10)=0 or (t + 10)=0
00 = -16(t – 10)(t + 10) = -16(t – 10)(t + 10)
t = 10 or t = 10 or
t =-10t =-10
Lecture 31Lecture 31 1010
You drop a rock from the top of the 1600 feet tall You drop a rock from the top of the 1600 feet tall Rears Tower. The height (in feet) of the rock from Rears Tower. The height (in feet) of the rock from the ground is given by the equation:the ground is given by the equation: h = -16t h = -16t22 + 1600 + 1600h is in feet and t is in seconds. How long until the h is in feet and t is in seconds. How long until the rock hits the ground?rock hits the ground?
hh = -16t = -16t22 + 1600 + 160000 = -16t = -16t22 + 1600 + 1600
00 = -16(t = -16(t22 – 100) – 100)
(t – 10)=0 or (t + 10)=0(t – 10)=0 or (t + 10)=0
00 = -16(t – 10)(t + 10) = -16(t – 10)(t + 10)
t = 10 or t = 10 or
t =-10t =-10t = 10t = 10 is is
the only the only
reasonable reasonable
answer.answer.
Lecture 31Lecture 31 1111
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
Lecture 31Lecture 31 1212
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
P = ( 280x – .4xP = ( 280x – .4x22) – () – (5000 + .6x5000 + .6x22))
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
Lecture 31Lecture 31 1313
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
P = ( 280x – .4xP = ( 280x – .4x22) – () – (5000 + .6x5000 + .6x22))
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
P = 280x – .4xP = 280x – .4x22– 5000 - .6x– 5000 - .6x22
Lecture 31Lecture 31 1414
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
P =P = -1x -1x22
P = ( 280x – .4xP = ( 280x – .4x22) – () – (5000 + .6x5000 + .6x22))
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
P = 280x – .4xP = 280x – .4x22– 5000 - .6x– 5000 - .6x22
Lecture 31Lecture 31 1515
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
P =P = -1x -1x2 2 + 280x+ 280x
P = ( 280x – .4xP = ( 280x – .4x22) – () – (5000 + .6x5000 + .6x22))
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
P = 280x – .4xP = 280x – .4x22– 5000 - .6x– 5000 - .6x22
Lecture 31Lecture 31 1616
The equation for profit is, The equation for profit is, Profit = Profit = RevenueRevenue - - CostCost
Revenue:Revenue: R = 280x -.4xR = 280x -.4x22
P =P = -1x -1x2 2 + 280x+ 280x - 5000- 5000
P = ( 280x – .4xP = ( 280x – .4x22) – () – (5000 + .6x5000 + .6x22))
Cost:Cost: C = 5000 + .6xC = 5000 + .6x22
P = 280x – .4xP = 280x – .4x22– 5000 - .6x– 5000 - .6x22
Lecture 31Lecture 31 1717
How many items must be made and sold How many items must be made and sold to generate a$439 profit.to generate a$439 profit.
P =P = -1x -1x2 2 + 280x+ 280x - 5000- 5000
Lecture 31Lecture 31 1818
How many items must be made and sold How many items must be made and sold to generate a$439 profit.to generate a$439 profit.
P =P = -1x -1x2 2 + 280x+ 280x - 5000- 5000
439 =439 = -1x -1x2 2 + 280x+ 280x - 5000- 5000
Lecture 31Lecture 31 1919
How many items must be made and sold How many items must be made and sold to generate a$439 profit.to generate a$439 profit.
P =P = -1x -1x2 2 + 280x+ 280x - 5000- 5000
439 =439 = -1x -1x2 2 + 280x+ 280x - 5000- 5000 Subtract 439 from Subtract 439 from both sidesboth sides
Lecture 31Lecture 31 2020
How many items must be made and sold How many items must be made and sold to generate a$439 profit.to generate a$439 profit.
P =P = -1x -1x2 2 + 280x+ 280x - 5000- 5000
439 =439 = -1x -1x2 2 + 280x+ 280x - 5000- 5000
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
Subtract 439 from Subtract 439 from both sidesboth sides
Lecture 31Lecture 31 2121
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
Lecture 31Lecture 31 2222
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
Lecture 31Lecture 31 2323
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
a = -1a = -1
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
Lecture 31Lecture 31 2424
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
a = -1a = -1 b = 280b = 280
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
Lecture 31Lecture 31 2525
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
a = -1a = -1 b = 280b = 280 c = -5439c = -5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
Lecture 31Lecture 31 2626
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
a = -1a = -1 b = 280b = 280 c = -5439c = -5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 1)
( 1)
280 ( 54392 )80x
Lecture 31Lecture 31 2727
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
a = -1a = -1 b = 280b = 280 c = -5439c = -5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 1)
( 1)
280 ( 54392 )80x
280 78400 21756
2x
Lecture 31Lecture 31 2828
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 1)
( 1)
280 ( 54392 )80x
280 78400 21756
2x
280 56644
2x
Lecture 31Lecture 31 2929
To find the square root of a number, use key in row 6 column 1.
Use of the calculator to Use of the calculator to evaluate a square root.evaluate a square root.
is keyed in as is keyed in as 2nd2nd, ,, , 56644 56644, , ENTERENTER
56644
Lecture 31Lecture 31 3030
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 1)
( 1)
280 ( 54392 )80x
280 78400 21756
2x
280 56644
2x
280 238
2x
Lecture 31Lecture 31 3131
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439280 238
2x
280 238 42
21280 2382 2
2x
Lecture 31Lecture 31 3232
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439280 238
2x
280 238 42
21280 238 2 2
2 280 238 518259
2 2
x
Lecture 31Lecture 31 3333
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -1x -1x2 2 + 280x+ 280x - 5439- 5439280 238
2x
280 238 42
21280 238 2 2
2 280 238 518259
2 2
x
Two solutions to make P = $439: Two solutions to make P = $439: x = 21 and x =259 itemsx = 21 and x =259 items
Lecture 31Lecture 31 3434
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
Lecture 31Lecture 31 3535
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55
Lecture 31Lecture 31 3636
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30
Lecture 31Lecture 31 3737
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
Lecture 31Lecture 31 3838
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010
Lecture 31Lecture 31 3939
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20
Lecture 31Lecture 31 4040
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
Lecture 31Lecture 31 4141
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515
Lecture 31Lecture 31 4242
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515 40-2(40-2(1515)=10)=10
Lecture 31Lecture 31 4343
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515 40-2(40-2(1515)=10)=10 15(10)=15015(10)=150
Lecture 31Lecture 31 4444
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515 40-2(40-2(1515)=10)=10 15(10)=15015(10)=150
XX
Lecture 31Lecture 31 4545
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515 40-2(40-2(1515)=10)=10 15(10)=15015(10)=150
XX 40-2(40-2(XX))
Lecture 31Lecture 31 4646
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Make a table and find the formula rectangle. Make a table and find the formula for the area of the rectangle.for the area of the rectangle. WallWall
L
WWWidthWidth LengthLength AreaArea
55 40-2(40-2(55)=30)=30 5(30)=1505(30)=150
1010 40-2(40-2(1010)=20)=20 10(20)=20010(20)=200
1515 40-2(40-2(1515)=10)=10 15(10)=15015(10)=150
XX 40-2(40-2(XX)) X(40-2X)=X(40-2X)=
40X-2X40X-2X22
Lecture 31Lecture 31 4747
A =A = 40X-2X40X-2X22
65 =65 = -2x -2x2 2 + 40x+ 40x
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Find the dimensions of the rectangle rectangle. Find the dimensions of the rectangle if the area is 65 square inches.if the area is 65 square inches.
WallWall
L
WW
Subtract 65 from Subtract 65 from both sidesboth sides
Lecture 31Lecture 31 4848
A =A = 40X-2X40X-2X22
65 =65 = -2x -2x2 2 + 40x+ 40x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Enclose a rectangle with a 40 inch string and Enclose a rectangle with a 40 inch string and use a wall of the room for one side of the use a wall of the room for one side of the rectangle. Find the dimensions of the rectangle rectangle. Find the dimensions of the rectangle if the area is 65 square inches.if the area is 65 square inches.
WallWall
L
WW
Subtract 65 from Subtract 65 from both sidesboth sides
Lecture 31Lecture 31 4949
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5050
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5151
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -2a = -2
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5252
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -2a = -2 b = 40b = 40
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5353
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -2a = -2 b = 40b = 40 c = -65c = -65
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5454
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -2a = -2 b = 40b = 40 c = -65c = -65
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 2)
( 2)
40 4 650 ( )x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5555
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -2a = -2 b = 40b = 40 c = -65c = -65
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 4
2
( 2)
( 2)
40 4 650 ( )x
40 1600 520
4x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5656
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
40 1600 520
4x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
40 1080
4x
Lecture 31Lecture 31 5757
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
40 1600 520
4x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
40 1080
4x
40 32.86
4x
Lecture 31Lecture 31 5858
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
40 32.86 7.141.840 32.86
4 44
x
40 32.86
4x
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65
Lecture 31Lecture 31 5959
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65280 238
2x
40 32.86 7.14
1.840 32.86 4 4
4 40 32.86 72.8618.2
4 4
x
Lecture 31Lecture 31 6060
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -2x -2x2 2 + 40x+ 40x - 65- 65280 238
2x
40 32.86 7.14
1.840 32.86 4 4
4 40 32.86 72.8618.2
4 4
x
Two solutions to make A = 65 sq in: Two solutions to make A = 65 sq in: x = 1.8 and x =18.2 inchesx = 1.8 and x =18.2 inches
Lecture 31Lecture 31 6161
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
H
Lecture 31Lecture 31 6262
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00
22
44
66
88
1010
bb
Lecture 31Lecture 31 6363
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10
22
44
66
88
1010
bb
Lecture 31Lecture 31 6464
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22
44
66
88
1010
bb
Lecture 31Lecture 31 6565
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8
44
66
88
1010
bb
Lecture 31Lecture 31 6666
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44
66
88
1010
bb
Lecture 31Lecture 31 6767
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6
66
88
1010
bb
Lecture 31Lecture 31 6868
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66
88
1010
bb
Lecture 31Lecture 31 6969
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4
88
1010
bb
Lecture 31Lecture 31 7070
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88
1010
bb
Lecture 31Lecture 31 7171
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2
1010
bb
Lecture 31Lecture 31 7272
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2 (.5)(8)(2)=8 sq cm(.5)(8)(2)=8 sq cm
1010
bb
Lecture 31Lecture 31 7373
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2 (.5)(8)(2)=8 sq cm(.5)(8)(2)=8 sq cm
1010 10-(10-(1010)=0)=0
bb
Lecture 31Lecture 31 7474
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2 (.5)(8)(2)=8 sq cm(.5)(8)(2)=8 sq cm
1010 10-(10-(1010)=0)=0 (.5)(10)(0)=0 sq cm(.5)(10)(0)=0 sq cm
bb
Lecture 31Lecture 31 7575
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2 (.5)(8)(2)=8 sq cm(.5)(8)(2)=8 sq cm
1010 10-(10-(1010)=0)=0 (.5)(10)(0)=0 sq cm(.5)(10)(0)=0 sq cm
bb 10-(10-(bb))
Lecture 31Lecture 31 7676
A 10 cm stick is broken into two pieces. One is placed A 10 cm stick is broken into two pieces. One is placed at a right angle to form an upside down “T” shape. By at a right angle to form an upside down “T” shape. By attaching wires from the ends of the base to the end of attaching wires from the ends of the base to the end of the upright piece, a framework for a sail will be formed.the upright piece, a framework for a sail will be formed.
Base
HBaseBase HeightHeight Area (A=.5bh)Area (A=.5bh)
00 10 -(10 -(00)=10)=10 (.5)(0)(10)=0 sq cm(.5)(0)(10)=0 sq cm
22 10 -(10 -(22)=8)=8 (.5)(2)(8)= 8 sq cm(.5)(2)(8)= 8 sq cm
44 10-(10-(44)=6)=6 (.5)(4)(6)=12 sq cm(.5)(4)(6)=12 sq cm
66 10-(10-(66)=4)=4 (.5)(6)(4)=12 sq cm(.5)(6)(4)=12 sq cm
88 10-(10-(88)=2)=2 (.5)(8)(2)=8 sq cm(.5)(8)(2)=8 sq cm
1010 10-(10-(1010)=0)=0 (.5)(10)(0)=0 sq cm(.5)(10)(0)=0 sq cm
bb 10-(10-(bb)) (.5)(b)(10-b)= 5b-.5b(.5)(b)(10-b)= 5b-.5b22
Lecture 31Lecture 31 7777
What should the base of the sail be if the area What should the base of the sail be if the area must be 10 sq cm? must be 10 sq cm?
Base
H
A = 5b-.5bA = 5b-.5b22
Lecture 31Lecture 31 7878
What should the base of the sail be if the area What should the base of the sail be if the area must be 10 sq cm? must be 10 sq cm?
Base
H
A = 5b-.5bA = 5b-.5b22
10 = 5b-.5b10 = 5b-.5b22
Lecture 31Lecture 31 7979
What should the base of the sail be if the area What should the base of the sail be if the area must be 10 sq cm? must be 10 sq cm?
Base
H
A = 5b-.5bA = 5b-.5b22
10 = 5b-.5b10 = 5b-.5b22 Subtract 10 from Subtract 10 from both sidesboth sides
Lecture 31Lecture 31 8080
What should the base of the sail be if the area What should the base of the sail be if the area must be 10 sq cm? must be 10 sq cm?
Base
H
A = 5b -.5bA = 5b -.5b22
10 = 5b -.5b10 = 5b -.5b22
0 = -.5b0 = -.5b2 2 + 5b -10+ 5b -10
Subtract 10 from Subtract 10 from both sidesboth sides
Lecture 31Lecture 31 8181
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8282
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8383
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -.5a = -.5
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8484
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -.5a = -.5 b = 5b = 5
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8585
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -.5a = -.5 b = 5b = 5 c = -10c = -10
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8686
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -.5a = -.5 b = 5b = 5 c = -10c = -10
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 ( .5 ()
( .5)
4 0)5
2
5 1x
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8787
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
a = -.5a = -.5 b = 5b = 5 c = -10c = -10
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
2 ( .5 ()
( .5)
4 0)5
2
5 1x
5 25 20
1x
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8888
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
5 25 20
1x
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
Lecture 31Lecture 31 8989
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
5 25 20
1x
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
5 5
1x
Lecture 31Lecture 31 9090
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = a axx22 ++ b bx +x + c c2 4
2x
cb b a
a
5 25 20
1x
0 =0 = -.5 -.5bb22 + 5+ 5b b - 10- 10
5 2.24
1x
5 5
1x
Lecture 31Lecture 31 9191
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
5 2.24 2.762.765 2.24
1 11
x
0 =0 = -.5x -.5x2 2 + 5x+ 5x - 10- 10
5 2.24
1x
Lecture 31Lecture 31 9292
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
5 2.24 2.762.765 2.24
1 11
x
0 =0 = -.5x -.5x2 2 + 5x+ 5x - 10- 10
5 2.24
1x
Lecture 31Lecture 31 9393
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -.5x -.5x2 2 + 5x+ 5x - 10- 10
5 2.24 2.762.8
5 2.24 1 11 5 2.24 7.24
7.21 1
x
5 2.24
1x
Lecture 31Lecture 31 9494
Use the Quadratic Formula to solve:Use the Quadratic Formula to solve:
0 =0 = -.5x -.5x2 2 + 5x+ 5x - 10- 10
5 2.24 2.762.8
5 2.24 1 11 5 2.24 7.24
7.21 1
x
5 2.24
1x
Two solutions to make A = 10 sq in: Two solutions to make A = 10 sq in: x = 2.8 and x =7.2 inchesx = 2.8 and x =7.2 inches
Lecture 31Lecture 31 9595
Lecture 31Lecture 31 9696
Lecture 31Lecture 31 9797