SOLUTION: Please check my work, thank you. Practical Application of Quadratic Equations 1. A rectangular garden has dimensions of 15 feet by 11 feet. A gravel path of uniform width

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Question 193387: Please check my work, thank you.
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?
Ag = area of garden = 15 * 11 = 165 sq feet
Ap = area of gravel path = (15 + 2x)(11 + 2x) - 165
Ap = 165 + 52x + 4x^2 - 165
Ap = 4x^2 + 5x
where
x = width of the gravel path around the garden
and since the available gravel is 192 sq ft, then the above equation becomes
4x2 + 5x = 192
and rewriting,
4x2 + 5x - 192 = 0
Using the quadratic formula,
x = 6.33 feet
ANSWER: The width of the gravel path should be about 6' 4".

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?
the annual interest rate = r
$8,000 in a savings account for two years.--> 8000*(1+r)^2
beginning of the second year, an additional $2,500 is invested-->2500*(100+r)
8000*(1+r)^2 + 2500*(1+r) = 11,445
suppose 100+r = A
8000*A^2 + 2500*A = 11,445
8000*A^2 + 2500*A - 11,445 = 0
A = 1.05
r = 0.05 or 5 % Ans.
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 R = rate of Steve
150/R – 150/(R + 20) = 2
150(R+ 20) – 150(R) = 2(R)(R + 20)
150R + 3000 – 150R = 2Rē + 40R
0 = 2Rē + 40R – 3000
Rē + 20R – 1500 = 0
(R + 50)(R – 30) = 0
(R + 50) = 0
R = –50 Discard, negative rate
(R – 30) = 0
R = 30 mph (speed of 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?
Let R = rate of patrol boat
40/(R – 5) + 40(R + 5) = 6
40(R + 5) + 40(R + 5) = 6(R + 5)(R – 5)
40R – 200 + 40R + 200 = 6Rē – 150
0 = 6Rē – 80R – 150
3Rē – 40R – 75 = 0
(3R + 5)(R – 15) = 0
(3R + 5) = 0
3R = –5
R = –5/3 Discard, negative rate
(R – 15) = 0
R = 15 mph (speed of boat in still water)

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