SOLUTION: Practical Application of Quadratic Equations 1.A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How w

Question 192566: Practical Application of Quadratic Equations
1.A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 192 square feet?

2.A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate?

3.Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle.

4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still water?

Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
1.A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width is to be built around the garden. How wide can the path be if there is enough gravel for 192 square feet?
:
Let x = the width of the path
then
(2x+15) by (2x+11) = overall area of the garden and the path
and FOIL this
Overall Area = 4x^2 + 52x + 165
:
Garden area: 15 * 11 = 165
:
Overall area - garden area = path area (given as 192 sq/ft)
(4x^2 + 52x + 165) - 165 = 192
4x^2 + 52x + 165 - 165 - 192 = 0
4x^2 + 52x - 192 = 0
Simplify, divide by 4
x^2 + 13x - 48
Factors to:
(x + 16)(x - 3) = 0
Positive solution
x = 3 ft is the width of the garden path
Check
(21*17) - (15*11) =
357 - 165 = 192
:
:
2.A business invests $8,000 in a savings account for two years. At the beginning of the second year, an additional $2,500 is invested. At the end of the second year, the account balance is $11,445. What was the annual interest rate?
:
Let r = interest in decimal form
:
(x+1)(8000(x+1) + 2500) = 11445
(x+1)(8000x + 8000 + 2500) = 11445
(x+1)(8000x + 10500) = 11445
FOIL
8000x^2 + 18500x + 10500 = 11445
8000x^2 + 18500x + 10500 - 11445 = 0
8000x^2 + 18500x - 945
Simplify divide by 5
1600x^2 + 3700x - 189 = 0
Solve this using the quadratic formula,
:
:
3.Steve traveled 150 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 2 hours less. Find the speed of his vehicle.
:
Let s = "that certain speed"
then
(s+20) = the faster speed
:
Write a time equation: time = dist/speed
Actual time = faster speed time + 2 hrs
= + 2
Solve this equation for s
;
:
4. The Hudson River flows at a rate of 5 miles per hour. A patrol boat travels 40 miles upriver, and returns in a total time of 6 hours. What is the speed of the boat in still water?
:
Let s = speed in still water
then
(s+5) = speed down stream
(s-5) = speed up stream
:
time downstream + time upstream = 6 hrs
+ = 6
Solve for s
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