SOLUTION: "On a recent trip, a trucker traveled 300 mi at a constant rate. Because of improved road conditions, the trucker then increased the speed by 10 mph. An additional 240 mi was trave

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Question 785103: "On a recent trip, a trucker traveled 300 mi at a constant rate. Because of improved road conditions, the trucker then increased the speed by 10 mph. An additional 240 mi was traveled at the increased rate. The entire trip took 16 h. Find the rate of the trucker for the first 300 mi."
Answer by happyhaps(2) About Me  (Show Source):
You can put this solution on YOUR website!
let the rate with which trucker traveled the first 300 mi be s mph.
Now, the Time = Distance+%2F+speed
time taken for 300mi + time taken for 240 mi = 16 hours
300%2Fs + 240%2F%28s%2B10%29 = 16
which becomes
16s%5E2 -380s -3000 = 0
Solving this quadratic equation
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 16x%5E2%2B-380x%2B-3000+=+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=%28-380%29%5E2-4%2A16%2A-3000=336400.

Discriminant d=336400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--380%2B-sqrt%28+336400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-380%29%2Bsqrt%28+336400+%29%29%2F2%5C16+=+30
x%5B2%5D+=+%28-%28-380%29-sqrt%28+336400+%29%29%2F2%5C16+=+-6.25

Quadratic expression 16x%5E2%2B-380x%2B-3000 can be factored:
16x%5E2%2B-380x%2B-3000+=+16%28x-30%29%2A%28x--6.25%29
Again, the answer is: 30, -6.25. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+16%2Ax%5E2%2B-380%2Ax%2B-3000+%29

Neglecting the negative solution we get s = 30mph.