Ch 9: Quadratic Equations
G) Quadratic Word Problems
Objective:
To solve word problems using various methods
for solving quadratic equations.
Definitions
Projectile 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: 2Length + 2width
Product: multiplication
Sum: addition
Difference: subtraction
Less than: subtraction & switch order
less
than
x − y
Example 1: Projectile Motion
The height of a model rocket that is fired into the air can be
represented by the equation: h = -16t2 + 64t
a) What will be its maximum height? Find the vertex
b) How long will it stay in the air? Find the time (t) when h = 0
y
Left
Vertex
Right
x y
0 0
2 64
4 0
10 20 30 40 50 60 70 80 90 100
2a 2(-16) -32
y = -16(2)2 + 64(2) = 64
height (in feet)
Vertex: x = -b = -(64) = -64 = 2
Vertex
(2 seconds, 64 ft)
(4 seconds, 0 ft)
1 2 3 4 5 6 7 8 9 10 11
time (in seconds)
x
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?
Find the time (t) when h = 0
Note: h = -4.9t2 + 19.6t + 58.8
Solve by Graphing
10 20 30 40 50 60 70 80 90 100
height (in meters)
h
or
Use the Quadratic Formula
b  b 2  4ac
2a
19.6  (19.6) 2  4(4.9)(58.8)
t
2(4.9)
(6 seconds, 0 ft)
t  19.6  38
Not Possible!
9.8
Time can’t be negative
t
1 2 3 4 5 6 7 8 9 
10 11
t = -2 or t = 6
time (in seconds)
Classwork
1) A model rocket is shot into the 2) An object is launched at 64 ft/s
(3 seconds, 45 ft)
(6 seconds, 0 ft)
1 2 3 4 5 6 7 8 9 10 11
time (in seconds)
t
10 20 30 40 50 60 70 80 90 100
Vertex
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?
h
height (in feet)
10 20 30 40 50 60 70 80 90 100
height (in meters)
air and its path is approximated
by the equation: h = -5t2 + 30t
a) When will it reach its
highest point?
b) When will the rocket hit
the ground?
h
(2 seconds, 144 ft)
1 2 3 4 5 6 7 8 9 10 11
time (in seconds)
t
Example 1: Integers
Find two numbers whose product is 65 and difference is 8.
Let x = one of the numbers
Let y = the other number
x  y = 65
Equation 2: Difference x – y = 8
Equation 1: Product
(y + 8)
x  y = 65
Simplify y2 + 8y = 65
Solve for x
x=y+8
Substitute
Solve
8  (8)2  4(1)(65)
y
2(1)
8  324

2
= 5 or -13
x  5y = 65
x -13
y = 65
x = -5
 x = 13
Solution 5 & 13 and -13 and -5
Plug in and solve for x

y2 + 8y − 65 = 0
Example 2: Integers
Find two numbers whose product is 640 and difference is 12.
Let x = one of the numbers
Let y = the other number
x  y = 640
Equation 2: Difference x – y = 12 Solve for x x = y + 12
Equation 1: Product
x  y = 640
Substitute (y + 12)
Simplify y2 + 12y = 640
y2 + 12y − 640 = 0
12  2704
12  (12)2  4(1)(640)

= 20 or -32
Solve y 
2
2(1)
x  20
y = 640
x -32
y = 640
x = -20
 x = 32
20 & 32 and -32 and -20
Solution
Plug in and solve for x

Classwork
3) Find two numbers whose
product is 36 and difference is 5.
x  y = 36
x–y=5
y2
+ 5y − 36 = 0
9&4
or
-4 and -9
4) Find two numbers whose
product is 48 and difference is 8.
x  y = 48
x–y=8
y2 + 8y − 48 = 0
12 & 4
or
-4 and -12
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 = width
L  w = 150
Equation 1: Area
= 35 − w
Equation 2: Perimeter 2L + 2w = 70Solve for L
L
Substitute (35 − w)
L  w = 150
Simplify -w2 + 35w = 150
-w2 + 35w − 150 = 0
35  (35) 2  4(1)(150) 35  625

Solve w 
2
2(1)
L 30
w = 150
L=5

5 ft x 30 ft
Solution
Plug in and solve for L

= 30 or 5
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?
L = P − 143
Let L = length
Let w = width
L + 143 = P
L  w = 2420
Equation 1: Area
L + 143 L = 143 − 2w
Equation 2: Perimeter 2L + 2w = P
Substitute (143 − 2w)
L  w = 2420
Simplify -2w2 + 143w = 2420
Solve for L
-2w2 + 143w − 2420 = 0
143 (143) 2  4(2)(2420) 143  1089

= 44 or 27.5
Solvew 
4
2(2)
L 44
w = 2420 or L 27.5
w = 2420
L = 55
L = 88
27.5 ft x 88 ft
Solution 55ft x 44 ft
Plug in and solve for L
Classwork
5) The perimeter of a rectangle is 6) The width of a rectangle is 46 ft
52 ft and its area is 168 ft2.
What are the dimensions of the
rectangle?
L  w = 168
2L + 2w = 52
-w2
+ 26w − 168 = 0
12 ft x 14 ft
or
3
11 ft x 1511ft
less than 2 times its length. The
area of the rectangle is 8580 ft2.
What are the dimensions of the
rectangle?
L  w = 8580
L  (2L – 46) = 8580
2L2 − 46L − 8580 = 0
110 ft x 78 ft
or
156 ft x 55 ft