SOLUTION: Miami and Pittsburg are 1000 miles apart. A plane flew into a 50mph headwind from miami to pittsburg. on the return flight the 50mph wind became a tailwind. The lane was in the air

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Question 750065: Miami and Pittsburg are 1000 miles apart. A plane flew into a 50mph headwind from miami to pittsburg. on the return flight the 50mph wind became a tailwind. The lane was in the air a total of 4.5 hrs. for the roundtrip. what would have been the planes average speed without the wind?
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Miami and Pittsburg are 1000 miles apart. A plane flew into a 50mph headwind from miami to pittsburg. on the return flight the 50mph wind became a tailwind. The lane was in the air a total of 4.5 hrs. for the roundtrip. what would have been the planes average speed without the wind?
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p = plane's airspeed
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1000/(p - 50) + 1000/(p + 50) = 4.5
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1000*(2p) = 4.5*(p^2 - 2500)
4000p = 9p^2 - 22500
9p^2 - 4000p - 22500 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=16810000 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 450, -5.55555555555556. Here's your graph:

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Ignore the negative solution,
p = 450 mi/hr
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with no wind, the avg = 450 mi/hr
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