SOLUTION: a car moving at constant speed covers a distanceof 120km.when the car travels 5km/h slower, the journey takes 15 minutes longer calculate the constant speed

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Question 726711: a car moving at constant speed covers a distanceof 120km.when the car travels 5km/h slower, the journey takes 15 minutes longer calculate the constant speed
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
a car moving at constant speed covers a distanceof 120km.when the car travels 5km/h slower, the journey takes 15 minutes longer calculate the constant speed
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Speeds are in km/hr
15 mins = 0.25 hours
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d = r*t
120 = rt --> t = 120/r
120 = (r-5)*(t + 0.25)
Sub for t
120 = (r-5)*(120/r + 0.25)
120 = 120 + 0.25r - 600/r - 1.25
0.25r - 600/r - 1.25 = 0
r - 2400/r - 5 = 0
r%5E2+-+5r+-+2400+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-5x%2B-2400+=+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=%28-5%29%5E2-4%2A1%2A-2400=9625.

Discriminant d=9625 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+9625+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+9625+%29%29%2F2%5C1+=+51.5535421758715
x%5B2%5D+=+%28-%28-5%29-sqrt%28+9625+%29%29%2F2%5C1+=+-46.5535421758715

Quadratic expression 1x%5E2%2B-5x%2B-2400 can be factored:
1x%5E2%2B-5x%2B-2400+=+%28x-51.5535421758715%29%2A%28x--46.5535421758715%29
Again, the answer is: 51.5535421758715, -46.5535421758715. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-5%2Ax%2B-2400+%29

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r = x =~ 51.55 km/hr (the faster speed)
Ignore the negative solution