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Simple Algebra- Mental Block!

  • Thread starter Gib Z
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  • #1
Gib Z
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ahh it says to you the template provided below, but theres nothing here...o well

Question:
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:
[tex]Speed=\frac{Distance}{Time}[/tex]

Attempt:

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,
[tex]x+4=\frac{32}{t_1}[/tex]
Down:[tex]x-4=\frac{32}{t_2}[/tex]

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
 

Answers and Replies

  • #2
cepheid
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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.
 
  • #3
HallsofIvy
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I agree with cepheid. Why do you say "which I know isn't right"?
 
  • #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:
  • #5
HallsofIvy
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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..
?? My calculator says it is! But what does it know?
 
  • #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?
 
  • #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]

Also

[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]
 
  • #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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