Calculating Boat Speed in Still Water: Solving a Simple Algebraic Equation

[Gib Z]

ahh it says to you the template provided below, but there's 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:
$$Speed=\frac{Distance}{Time}$$

Attempt:

I don't know what's 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,
$$x+4=\frac{32}{t_1}$$
Down:$$x-4=\frac{32}{t_2}$$

Solving for the times, and adding them,
we get $$\frac{32}{x+4} + \frac{32}{x-4} = 6$$ 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 $$3x^2 - 32x -48=0$$ which i know isn't right >.< Help

 

[cepheid]

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 traveling against the current, not that it matters here. It would have mattered if he had traveled different distances each way.

 

[HallsofIvy]

I agree with cepheid. Why do you say "which I know isn't right"?

 

[Gib Z]

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

$$x=\frac{32+\sqrt{32^2+12(48)}}{6}$$ which isn't 12..

EDIT: BTW, Halls, 11,111 posts :D Niice

 

[HallsofIvy]

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

$$x=\frac{32+\sqrt{32^2+12(48)}}{6}$$ which isn't 12..

?? My calculator says it is! But what does it know?

 

[anantchowdhary]

12 is correct.Try putting $$x=12$$ in the equation.You should be able to verify.Also you can do this by simple factorization.Can i help?

 

[anantchowdhary]

Now,uve got the equation
$$3x^2 - 32x -48=0$$
This can be written as:
$$3x^2 - 36x +4x -48=0$$

Also

$$3x(x - 12) +4(x -12)=0$$
So you can group the common terms and write
$$(3x+4)(x - 12) =0$$

As speed can't be negative,so we get $$x=12$$

 

[Gib Z]

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