SOLUTION: Lisa traveled 1040 mi at a certain speed. If the speed had been 13 mph faster, the trip could have been in 4 hr. less time. Find the speed.

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Question 909916: Lisa traveled 1040 mi at a certain speed. If the speed had been 13 mph faster, the trip could have been in 4 hr. less time. Find the speed.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
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1040%2Fr+-+1040%2F%28r%2B13%29+=+4 |Multiplying thru by r(r+13) so as all denominators = 1
1040(r+13) - 1040r = 4r^2 + 52r
4r^2 +52r - 13(1040) = 0
r^2 + 13r - 13(260) = 0
r^2 + 13r - 3380 = 0 (Tossing out the negative solution for unit measure)
r = 52mph. 65mph 52%2B13 to save 4hrs
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B13x%2B-3380+=+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=%2813%29%5E2-4%2A1%2A-3380=13689.

Discriminant d=13689 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-13%2B-sqrt%28+13689+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%2813%29%2Bsqrt%28+13689+%29%29%2F2%5C1+=+52
x%5B2%5D+=+%28-%2813%29-sqrt%28+13689+%29%29%2F2%5C1+=+-65

Quadratic expression 1x%5E2%2B13x%2B-3380 can be factored:
1x%5E2%2B13x%2B-3380+=+1%28x-52%29%2A%28x--65%29
Again, the answer is: 52, -65. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B13%2Ax%2B-3380+%29