SOLUTION: A car travels 380 miles averaging a certain speed. If the car had gone 5 mph faster the trip would have take 1 hour less. What was the average speed?

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Question 1005932: A car travels 380 miles averaging a certain speed. If the car had gone 5 mph faster the trip would have take 1 hour less. What was the average speed?
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
A car travels 380 miles averaging a certain speed. If the car had gone 5 mph faster the trip would have take 1 hour less. What was the average speed?
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d = r*t
t = d/r
t = 380/r
t-1 = 380/(r+5)
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380/r - 1 = 380/(r+5)
(380 - r)/r = 380/(r+5)
(380 - r)*(r+5) = 380r
(r-380)*(r+5) = -380r
r^2 + 5r - 1900 = 0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B-1900+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%285%29%5E2-4%2A1%2A-1900=7625.

Discriminant d=7625 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+7625+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+7625+%29%29%2F2%5C1+=+41.1606229914325
x%5B2%5D+=+%28-%285%29-sqrt%28+7625+%29%29%2F2%5C1+=+-46.1606229914325

Quadratic expression 1x%5E2%2B5x%2B-1900 can be factored:
1x%5E2%2B5x%2B-1900+=+%28x-41.1606229914325%29%2A%28x--46.1606229914325%29
Again, the answer is: 41.1606229914325, -46.1606229914325. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-1900+%29

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r =~ 41.16 mi/hr (Ignore the negative value)