<PRE><B>Question 80688</B>

distance(d) equals rate(r) times time(t) or d=rt; t=d/r and r=d/t

let r=initial speed

travel time at initial speed =200/r

travel time at 10 mph faster=200/(r+10)

Now we are told that the travel time at 10 mph faster (200/(r+10)) is 1 hour less than the travel time at the initial speed (200/r).  So our equation to solve is:

(200/r)-1=200/(r+10)  multiply both sides by r(r+10) to get rid of fractions

200(r+10)-r(r+10)=200r  get rid of parens

200r+2000-r^2-10r=200r subtract 200r from both sides

200r-200r+2000-r^2-10r=200r-200r  collect like terms

2000-r^2-10r=0  multiply each term by -1

r^2+10r-2000=0  quadratic in standard form and it can be factored.
(Note: If the A coefficient of a quadratic is 1, then the B coefficient will be the sum of the factors of the C coefficient.  In this case, the factors of -2000 that add up to +10 are +50 and -40)

(r+50)(r-40)=0
r=-50 mph-------------------------discount negative speed

and
r=40 mph-------------------------------initial speed

Check

200/40=5 hr
200/50=4 hr
5 hr - 4 hr =1 hr


Hope this helps---ptaylor


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