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Homework Help: Simple Algebra- Mental Block!

  1. Mar 6, 2007 #1

    Gib Z

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    ahh it says to you the template provided below, but theres nothing here...o well

    The Current in a river moves at a speed of 4km/h. A boat travels 32km upriver and 32km downriver in a total time of 6 hours. What is the speed of the boat in still water?

    Relevant Equations:


    I don't know whats wrong! This is meant to be a simple question, maybe my brains farting...

    Let x be the speed in still water.
    On its 32km upriver with the current, its speed is x+4 km/h,
    on its 32km down its speed is x-4 km/h.

    Let the time it took for the way upriver be t_1, and down t_2,

    then on its way up, by s=d/t,

    Solving for the times, and adding them,
    we get [tex]\frac{32}{x+4} + \frac{32}{x-4} = 6[/tex] since t_1 + t_2 must equal 6 hours, as given.

    I multiplyed everything by x^2 - 16, expanded a simplyfied the quadratic equation, i end up with [tex]3x^2 - 32x -48=0[/tex] which i know isnt right >.< Help
  2. jcsd
  3. Mar 6, 2007 #2


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    What's the problem? I get that quadratic too, and it gives me 12 km/h or -4/3 km/h, and I just ignored the negative solution. Btw, I think that going "upriver" usually means travelling against the current, not that it matters here. It would have mattered if he had travelled different distances each way.
  4. Mar 6, 2007 #3


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    I agree with cepheid. Why do you say "which I know isn't right"?
  5. Mar 6, 2007 #4

    Gib Z

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    Ahh I used the quadratic formula, and for some reason I get something else

    [tex]x=\frac{32+\sqrt{32^2+12(48)}}{6}[/tex] which isnt 12..

    EDIT: BTW, Halls, 11,111 posts :D Niice
    Last edited: Mar 6, 2007
  6. Mar 6, 2007 #5


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    ?? My calculator says it is! But what does it know?
  7. Mar 6, 2007 #6
    12 is correct.Try putting [tex]x=12[/tex] in the equation.You should be able to verify.Also you can do this by simple factorization.Can i help?
  8. Mar 6, 2007 #7
    Now,uve got the equation
    [tex]3x^2 - 32x -48=0[/tex]
    This can be written as:
    [tex]3x^2 - 36x +4x -48=0[/tex]


    [tex]3x(x - 12) +4(x -12)=0[/tex]
    So you can group the common terms and write
    [tex](3x+4)(x - 12) =0[/tex]

    As speed cant be negative,so we get [tex]x=12[/tex]
  9. Mar 7, 2007 #8

    Gib Z

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    Ahh ok thanks guy, i think I was just having calculator issues...thanks for the help! You can close this thread is you want.
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