SOLUTION: A boat travels 2 km upstream and 2 km back. The time for the round trip is 4 hrs. The speed of the stream is 5 ​km/hr. What is the speed of the boat in still​ water?

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Question 1158426: A boat travels 2 km upstream and 2 km back. The time for the round trip is 4 hrs. The speed of the stream is 5 ​km/hr. What is the speed of the boat in still​ water?
Found 2 solutions by Shin123, ikleyn:
Answer by Shin123(626) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say that the boat can travel x km/hr on still water. It can travel (x-5) km/hr upstream and (x+5) km/hr downstream. 2%2F%28x-5%29%2B2%2F%28x%2B5%29=4. 2%28x%2B5%29%2B2%28x-5%29=4%28x-5%29%28x%2B5%29. 2x%2B10%2B2x-10=4x%5E2-100. 4x%5E2-4x-100=0 x%5E2-x-25=0
Solved by pluggable solver: COMPLETING THE SQUARE solver for quadratics
Read this lesson on completing the square by prince_abubu, if you do not know how to complete the square.
Let's convert 1x%5E2%2B-1x%2B-25=0 to standard form by dividing both sides by 1:
We have: 1x%5E2%2B-1x%2B-25=0. What we want to do now is to change this equation to a complete square %28x%2Bsomenumber%29%5E2+%2B+othernumber. How can we find out values of somenumber and othernumber that would make it work?
Look at %28x%2Bsomenumber%29%5E2: %28x%2Bsomenumber%29%5E2+=+x%5E2%2B2%2Asomenumber%2Ax+%2B+somenumber%5E2. Since the coefficient in our equation 1x%5E2%2Bhighlight_red%28+-1%29+%2A+x%2B-25=0 that goes in front of x is -1, we know that -1=2*somenumber, or somenumber+=+-1%2F2. So, we know that our equation can be rewritten as %28x%2B-1%2F2%29%5E2+%2B+othernumber, and we do not yet know the other number.
We are almost there. Finding the other number is simply a matter of not making too many mistakes. We need to find 'other number' such that %28x%2B-1%2F2%29%5E2+%2B+othernumber is equivalent to our original equation 1x%5E2%2B-1x%2Bhighlight_green%28+-25+%29=0.


The highlighted red part must be equal to -25 (highlighted green part).

-1%5E2%2F4+%2B+othernumber+=+-25, or othernumber+=+-25--1%5E2%2F4+=+-25.25.
So, the equation converts to %28x%2B-1%2F2%29%5E2+%2B+-25.25+=+0, or %28x%2B-1%2F2%29%5E2+=+25.25.

Our equation converted to a square %28x%2B-1%2F2%29%5E2, equated to a number (25.25).

Since the right part 25.25 is greater than zero, there are two solutions:


, or

system%28+%28x%2B-1%2F2%29+=+5.02493781056044%2C+%28x%2B-1%2F2%29+=+-5.02493781056044+%29
system%28+x%2B-1%2F2+=+5.02493781056044%2C+x%2B-1%2F2+=+-5.02493781056044+%29
system%28+x+=+5.02493781056044--1%2F2%2C+x+=+-5.02493781056044--1%2F2+%29

system%28+x+=+5.52493781056044%2C+x+=+-4.52493781056044+%29
Answer: x=5.52493781056044, -4.52493781056044.

The boat can travel about 5.525 km/hr.

Answer by ikleyn(52803) About Me  (Show Source):
You can put this solution on YOUR website!
.

Let x be the spreed of the boat in still water, in kilometres per hour..

Then the speed downstream is (x+5) kilometres per hour, and the time traveled downstream is  2%2F%28x%2B5%29 hours.

The speed upstream is (x-5) miles per hour, and the time traveled upstream is  2%2F%28x-5%29 hours.


The total time equation is

   2%2F%28x%2B5%29 + 2%2F%28x-5%29 = 4


Cancel the factor 2 in both sides


   1%2F%28x%2B5%29 + 1%2F%28x-5%29 = 2


Multiply both sides by (x+5)*(x-5); then simplify


    (x-5) + (x+5) = 2*(x^2-25)

    2x            = 2x^2 - 50

    2x^2 - 2x - 50 = 0

    x^2  -  x  - 25 = 0

     x%5B1%2C2%5D = %281+%2B-+sqrt%281+%2B+4%2A25%29%29%2F2 = %281+%2B-+sqrt%28101%29%29%2F2 = %281+%2B-+10.05%29%2F2.


Only positive root  x = %281+%2B+10.05%29%2F2 = 5.525 is the solution.


ANSWER.  The speed of the boat in still water is 5.525 kilometres per hour.

CHECK.   2%2F%285.525%2B5%29 + 2%2F%285.525-5%29 = 4 hours.   ! Correct !

Solved.

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It is a typical and standard Upstream and Downstream round trip word problem.

You can find many similar fully solved problems on upstream and downstream round trips with detailed solutions in lessons
    - Wind and Current problems
    - More problems on upstream and downstream round trips
    - Wind and Current problems solvable by quadratic equations
    - Unpowered raft floating downstream along a river
    - Selected problems from the archive on the boat floating Upstream and Downstream
in this site, where you will find other similar solved problems with detailed explanations.

Read them attentively and learn how to solve this type of problems once and for all.

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the section "Word problems",  the topic "Travel and Distance problems".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.