SOLUTION: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Not sure how to do these word probl

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle. Not sure how to do these word probl      Log On


   

Question 170727This question is from textbook intermediate algebra
: Steve traveled 600 miles at a certain speed. Had he gone 20mph faster, the trip would have taken 1 hour less. Find the speed of his vehicle.
Not sure how to do these word problems and will appreciate any help, so far I have a B in class :) This question is from textbook intermediate algebra

Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
well we know that distance equals rate times time. d=rt
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In our case distance is the same under both scenarios 600 miles.
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what is different is the speeds and thus the times as speed and time are inversely related.
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so in the first trip lets call the rate r and the time t.
in the second scenario the rate would be r+20 and the time is t-1.
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600=rt....................eq 1
600=(r+20)(t-1)...........eq 2
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lets rewrite eq 1 as t=600/r and place that value into eq 2
:
600=%28r%2B20%29%28%28600%2Fr%29-1%29
600=%28r%2B20%29%28%28600-r%29%2Fr%29 multiply by r
600r=%28r%2B20%29%28600-r%29 multiply factors on right side of eq
600r=600r-r%5E2%2B12000-20r combine like terms on one side of equation
r%5E2%2B20r-12000=0
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throw out the negative speedsystem%28r=100mph%2Cr=-120mph%29
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so speed on first trip is 100mph and on second trip is 100%2B20=120mph
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Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ar%5E2%2Bbr%2Bc=0 (in our case 1r%5E2%2B20r%2B-12000+=+0) has the following solutons:

r%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-12000=48400.

Discriminant d=48400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-20%2B-sqrt%28+48400+%29%29%2F2%5Ca.

r%5B1%5D+=+%28-%2820%29%2Bsqrt%28+48400+%29%29%2F2%5C1+=+100
r%5B2%5D+=+%28-%2820%29-sqrt%28+48400+%29%29%2F2%5C1+=+-120

Quadratic expression 1r%5E2%2B20r%2B-12000 can be factored:
1r%5E2%2B20r%2B-12000+=+1%28r-100%29%2A%28r--120%29
Again, the answer is: 100, -120. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B20%2Ax%2B-12000+%29