SOLUTION: n airplane flies 355 miles to city A. Then, with better winds, it continues on to city B, 448 miles from A, at a speed 14.0 mi./h greater than on the first leg of the trip. The tot

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: n airplane flies 355 miles to city A. Then, with better winds, it continues on to city B, 448 miles from A, at a speed 14.0 mi./h greater than on the first leg of the trip. The tot      Log On

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Question 1105147: n airplane flies 355 miles to city A. Then, with better winds, it continues on to city B, 448 miles from A, at a speed 14.0 mi./h greater than on the first leg of the trip. The total flying time was 5.20 h. Find the speed at which the plane travelled to city A.
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Better winds must mean, "increased wind speed in the traveling direction".
            SPEED       TIME              DISTANCE
To A           r        355/r               355
To B         r+14       448/(r+14)          448
Total                   5.2

highlight_green%28355%2Fr%2B448%2F%28r%2B14%29=5.2%29
Solve this for r.
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Simplification using the algebra steps should lead to something equivalent to 5.2r%5E2-730.2r-4970=0

Discriminant is 636568.04 and sqrt%28636568.04%29=797.85214;
continuing on to using general quadratic formula solution,
r=%28730.2%2B797.85214%29%2F10.4
highlight%28r=146.9%29
and when checked in the original equation might be about 5.19 or 5.2, so this solution works.