Boat Speed Word Problem - Help

[mcruz65]

The current in a stream moves at a speed of 3 mph. A boat travels 45 mi upstream and 45 mi downstream in a total time of 8 hr. What is the speed of the boat in still water?

----------- distance--- r ----t
Upstream ---- 45 mi -- r-3---t1
Downstream - 45 mi ---r+3---t2
Total Time = ----------------8 hr.

t1=45/r-3, t2=45/r+3, t1 +t2 = 8

45/r-3 + 45/r+3 = 8

Multiply both sides of the equation with (r-3)(r+3)

45(r+3)+45(r-3)=8(r-3)(r+3)
45r + 135+45r-135=8r^2-72
90r=8r^2-72
90r-90r=8r^2-90r-72
0=8r^2-90r-72
2(4r^2-45r-36)=0
(-r+12)(-4r-3)=0
-r+12=0, -4r-3=0
-r=-12, -4r=3
(-1)-r= -12(-1)
r=12

-4r=3
-4r/-4=3/-4
r= -3/4

So the original question was the speed of the boat in still water.
1) Did I do the computations correctly?
2) When I plug in the answers into the equation, it comes out correctly. So the speed of the boat in still water is ??

 

[JuicyBurger]

What I want to know is how you factored that without the quadratic equation!

r = 12.

Plug them both back into the equations for time and figure out which makes sense. Using r = -3/4 will give you negative time.

Hope this helps!

 

[mcruz65]

Okay so when I got to 8r^2-90r-72=0. This is when I use the quadratic equation? I just want to know the proper steps on completing this exercise.

0=8r^2-90r-72

(-b±?(b^2-4ac))/2a

(-(-90)±?((-90)^2-4(8)(-72) ))/2(8)

(90±?(8100-4(-576) ))/16

(90±?(8100+2304))/16

(90±?10404)/16

(90±102)/16

90/16+102/16=5.625+6.375=12

r=12