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