Ch 9: Quadratic EquationsG) Quadratic Word Problems
Objective:
To solve word problems using various methods for solving quadratic equations.
DefinitionsProjectile Motion: h = at2 + vt + s
The path of an object that is thrown, shot, or dropped.h =height, t = time, v =velocity, s = initial height
Area of a Rectangle: Length width
Perimeter of a Rectangle: 2Length + 2width
Product: multiplication
Sum: addition
Difference: subtraction
Less than: subtraction & switch order −x yless than
x
y
time (in seconds)
hei
ght
(in
feet
)
Example 1: Projectile Motion
The height of a model rocket that is fired into the air can be represented by the equation: h = -16t2 + 64ta)What will be its maximum height? b)How long will it stay in the air?
Vertex: x = -b2a
= -(64)2(-16)
= -64 -32
= 2
y = -16(2)2 + 64(2) = 64
x y
Left
Vertex
Right
2 64
0 0
4 0
10 2
0 30
40
50 6
0 70
80
90 1
00
1 2 3 4 5 6 7 8 9 10 11
Find the vertex Find the time (t) when h = 0
Vertex
(2 seconds, 64 ft)
(4 seconds, 0 ft)
Example 2: Projectile Motion
An object is launched at 19.6 m/s from a 58.8 meter-tall platform. When does the object strike the ground?Note: h = -4.9t2 + 19.6t + 58.8 Find the time (t) when h = 0
t
h
time (in seconds)
hei
ght
(in
met
ers)
10 2
0 30
40
50 6
0 70
80
90 1
00
1 2 3 4 5 6 7 8 9 10 11
(6 seconds, 0 ft)
Solve by Graphing or Use the Quadratic Formula
€
−b ± b2 − 4ac
2a
€
t =−19.6 ± (19.6)2 − 4(−4.9)(58.8)
2(−4.9)
€
t =−19.6 ± 38
−9.8
t = -2 or t = 6
Not Possible!Time can’t be negative
t
h
time (in seconds)
hei
ght
(in
feet
)10
20
30 4
0 50
60
70 8
0 90
100
1 2 3 4 5 6 7 8 9 10 11
ClassworkA model rocket is shot into the air and its path is approximated by the equation: h = -5t2 + 30ta)When will it reach its highest point? b)When will the rocket hit the ground?
2)
t
h
time (in seconds)
hei
ght
(in
met
ers)
10 2
0 30
40
50 6
0 70
80
90 1
00
1 2 3 4 5 6 7 8 9 10 11
1) An object is launched at 64 ft/s from a platform 80 ft high. Note: h= -16t2 + 64t + 80 a)What will be the objects maximum height? b)When will it attain this height?
(2 seconds, 144 ft)Vertex
(3 seconds, 45 ft)
(6 seconds, 0 ft)
Example 1: IntegersFind two numbers whose product is 65 and difference is 8.
Let x = one of the numbers Let y = the other number
Equation 1: Product
Equation 2: Differencex y = 65x – y = 8 x = y + 8
y = 65Substitute xSolve for x
(y + 8)
Simplify y2 + 8y = 65 y2 + 8y − 65 = 0
Solve
€
y =−8 ± (8)2 − 4(1)(−65)
2(1)
€
=−8 ± 324
2= 5 or -13
Plug in and solve for x x = 65y5x = 13
x = 65y-13x = -5
Solution 5 & 13 and -13 and -5
Example 2: IntegersFind two numbers whose product is 640 and difference is 12.
Let x = one of the numbers Let y = the other number
Equation 1: Product
Equation 2: Differencex y = 640x – y = 12 x = y + 12
y = 640Substitute xSolve for x
(y + 12)
Simplify y2 + 12y = 640 y2 + 12y − 640 = 0
Solve
€
y =−12 ± (12)2 − 4(1)(−640)
2(1)
€
=−12 ± 2704
2= 20 or -32
Plug in and solve for x x = 640y20x = 32
x = 640y-32x = -20
Solution 20 & 32 and -32 and -20
Find two numbers whose product is 36 and difference is 5.
4)3) Find two numbers whose product is 48 and difference is 8.
9 & 4 or
-4 and -9
12 & 4 or
-4 and -12
x y = 36x – y = 5
x y = 48x – y = 8
y2 + 5y − 36 = 0 y2 + 8y − 48 = 0
Classwork
Example 1: Dimensions
You have 70 ft of material to fence in a rectangular garden that has an area of 150 ft2. What will be the dimensions of the fence? Let L = length Let w = widthEquation 1: AreaEquation 2: Perimeter
L w = 1502L + 2w = 70 L = 35 − w
w = 150Substitute LSolve for L
(35 − w)
Simplify -w2 + 35w = 150 -w2 + 35w − 150 = 0
Solve
€
w =−35 ± (35)2 − 4(−1)(−150)
2(−1)
€
=−35 ± 625
−2= 30 or 5
Plug in and solve for L L = 150w30L = 5
Solution 5 ft x 30 ft
Example 2: Dimensions
The length of a rectangular garden is 143 ft less than the perimeter. The area of the rectangle is 2420 ft2. What are the dimensions of the rectangle?
Let L = length Let w = width Equation 1: AreaEquation 2: Perimeter
L w = 24202L + 2w = L = 143 − 2w
w = 2420Substitute LSolve for L
(143 − 2w)
Simplify -2w2 + 143w = 2420 -2w2 + 143w − 2420 = 0
Solve
€
w =−143 ± (143)2 − 4(−2)(−2420)
2(−2)
€
=−143 ± 1089
−4= 44 or 27.5
Plug in and solve for L L = 2420w44L = 55
Solution 55 ft x 44 ft
L = P − 143
L + 143
L + 143 = P
or L = 2420w27.5L = 88
27.5 ft x 88 ft
P
The perimeter of a rectangle is 52 ft and its area is 168 ft2. What are the dimensions of the rectangle?
6)5) The width of a rectangle is 46 ft less than 2 times its length. The area of the rectangle is 8580 ft2. What are the dimensions of the rectangle?
12 ft x 14 ft or
11 ft x 15 ft
110 ft x 78 ft or
156 ft x 55 ft 311
L w = 1682L + 2w = 52
L w = 8580L (2L – 46) = 8580
-w2 + 26w − 168 = 0 2L2 − 46L − 8580 = 0
Classwork