SOLUTION: Hank travels home to Smithville every weekend from Ottawa.The distance each way is 800km. Anxious to return home, he decides to increase his speed by 20km/h and realizes that he ar

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Question 123257: Hank travels home to Smithville every weekend from Ottawa.The distance each way is 800km. Anxious to return home, he decides to increase his speed by 20km/h and realizes that he arrives home two hours earlier than the time it took him to travel from home to Ottawa. How long does the trip to Ottawa usually take?
Answer by rajagopalan(174) About Me  (Show Source):
You can put this solution on YOUR website!
Hank travels home to Smithville every weekend from Ottawa.The distance each way is 800km. Anxious to return home, he decides to increase his speed by 20km/h and realizes that he arrives home two hours earlier than the time it took him to travel from home to Ottawa. How long does the trip to Ottawa usually take?
Let the normal speed=x kmph
Time for onward trip =dist/speed=800/x .. .. .. .. ..A
Time for return trip=dist/speed=800/(x+20).. .. .. ..B
Difference in times =A-B
gives us:
(800/x)-(800/(x+20))=2
taking x(x+20) as LCM, we get the quadratic 2x^2+40x-16000=0
which boils down to x^2+40x-8000
solution is:
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B20x%2B-8000+=+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=%2820%29%5E2-4%2A1%2A-8000=32400.

Discriminant d=32400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+32400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2820%29%2Bsqrt%28+32400+%29%29%2F2%5C1+=+80
x%5B2%5D+=+%28-%2820%29-sqrt%28+32400+%29%29%2F2%5C1+=+-100

Quadratic expression 1x%5E2%2B20x%2B-8000 can be factored:
1x%5E2%2B20x%2B-8000+=+%28x-80%29%2A%28x--100%29
Again, the answer is: 80, -100. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B20%2Ax%2B-8000+%29


we get x=80
at speed of 80 time=800/80=10hours
at speed 80+20 time=80/100=8 hours
then 10-8 hours=2 hours
okay