SOLUTION: Please help!! I have tried to come up with a solution...but I am majorly stuck and frustrated!! If I've made an error, I can't see it. And with what I do have, I have no idea what

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: Please help!! I have tried to come up with a solution...but I am majorly stuck and frustrated!! If I've made an error, I can't see it. And with what I do have, I have no idea what       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 212720This question is from textbook Introductory Algebra
: Please help!! I have tried to come up with a solution...but I am majorly stuck and frustrated!! If I've made an error, I can't see it. And with what I do have, I have no idea what the next step should be. ~ Thanks! Here's the problem and my attempt:
Boat Speed. 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?
t(1) = 45/r-3 t(2) = 45/r+3 t(1) + t(2) = 8, where r is the rate of the boat in still water
45/(r-3)+ 45/(r+3)=8 The LCD is (r-3)(r+3)
((r-3) (r+3) (45/(r-3)) + ((r-3) (r+3) (45/(r+3)) = 8 (r-3)(r+3)
45 (r + 3) + 45 (r - 3) = 8 (r^2 – 9)
45r + 135 + 45r – 135 = 8r^2 – 72
90r = 8r^2 – 72
-90r -90r
0 = 8r2 – 90r – 72
Now What? I know the next step is to factor but I can’t determine a factorization in the form of (x + ?)(x - ?)…the best I can do is…
0 = 2 (r^2 – 45r – 32)
And IF this is the correct factoring answer…I have NO CLUE what to do next! PLEASE HELP!!
 This question is from textbook Introductory Algebra

Found 2 solutions by stanbon, nerdybill:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
8r^2 -90r-72 = 0
2(4r^2-45r-36) = 0
2(r-12)(4r+3) = 0
Positive solution:
r = 12
=============
Cheers,
Stan H.

Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Boat Speed. 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?
.
Applying the distance formula:
d = rt)
Solving for t:
t = d/r
.
Let r = speed of boat in still water
then
"time going upstream" + "time going downstream" = 8
.
45/(r-3) + 45/(r+3) = 8
Now, multiplying both sides by (r-3)(r+3):
45(r+3) + 45(r-3) = 8(r-3)(r+3)
Expanding:
45r+135 + 45r-135 = 8(r^2-9)
45r+45r = 8r^2-72
90r = 8r^2-72
0 = 8r^2-90r-72 (Exactly what you had)
Solving for r using the quadratic formula yields:
r = {12, -0.75}
We can toss out the negative answer leaving:
r = 12 mph
.
Details of quadratic to follow:
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ar%5E2%2Bbr%2Bc=0 (in our case 8r%5E2%2B-90r%2B-72+=+0) has the following solutons:

r%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=%28-90%29%5E2-4%2A8%2A-72=10404.

Discriminant d=10404 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--90%2B-sqrt%28+10404+%29%29%2F2%5Ca.

r%5B1%5D+=+%28-%28-90%29%2Bsqrt%28+10404+%29%29%2F2%5C8+=+12
r%5B2%5D+=+%28-%28-90%29-sqrt%28+10404+%29%29%2F2%5C8+=+-0.75

Quadratic expression 8r%5E2%2B-90r%2B-72 can be factored:
8r%5E2%2B-90r%2B-72+=+8%28r-12%29%2A%28r--0.75%29
Again, the answer is: 12, -0.75. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+8%2Ax%5E2%2B-90%2Ax%2B-72+%29