word problems with rational equations

[tarheel2006]

I just need help with the following word problems: Diagonal brace. The width of a rectangular gate is 2 meters
(m) larger than its height. The diagonal brace measures
?6m. Find the width and height. I know that A=lw using the quadractic formula
so I guess the first part is x(x+2)+ ?6m=0
=x^2+2x+ ?6m=0

Debbie traveled by boat 5 miles upstream to
fish in her favorite spot. Because of the 4-mph current, it
took her 20 minutes longer to get there than to return. How
fast will her boat go in still water?
D=rt right?
t=20
r=4
And that's all I have, if i could just get help with the set up then i will be fine

 

[Loren]

Did you draw a sketch of the gate with a diagonal brace? Did you label the diagonal brace "?6"? Did you label the height x and the width x+2? Did you then consider the Pythagorean Theorem? After doing that, please explain why you chose to use the formula for the area of a rectangle.

On the second one---The distance upstream is the same as the distance downstream. That can be the basis for an equation, i.e., distance upstream = distance downstream. Notice that the rate is in miles per hour but the time of travel is in minutes. Those need to be in the same units. If you call the rate of the boat in calm water "x" then the rate going downstream is x+4. I'll leave the rest to you.

 

[Denis]

I think you're not ready for these...what grade are you in?

Anyhow, here's hints:

1st one: you have a right triangle with sides x, x+2 and sqrt(6)

2nd one: upstream: (t + 1/3)(b - 4) = 5 ; returning: t(b + 4) = 5
b = boat speed; 1/3 = 20 minutes

 

[tarheel2006]

the book labeled the picture for me and i chose the quadractic formula b/c that is the section we are working on. I was thinkin gwe had to solve for a=lw>>>a=x(x+2)>>x^2+2x=0 and then i used the quad. fourmala then having to use a=1/2bh after getting the answer from the the first part.

but iended up with a -2 and that would not work l=-2 w=(-2+2)

 

[Denis]

the book labeled the picture for me and i chose the quadractic formula b/c that is the section we are working on. I was thinkin gwe had to solve for a=lw>>>a=x(x+2)>>x^2+2x=0 and then i used the quad. fourmala then having to use a=1/2bh after getting the answer from the the first part.
but iended up with a -2 and that would not work l=-2 w=(-2+2)

OK, area = lw, or x(x+2), BUT that's NOT required here; you're not making use of the diagonal = sqrt(6);
PLUS if you have a = x(x + 2), then x^2 + 2x = a, NOT 0.

I told you you had a right triangle with sides x and x+2, and hypotenuse of sqrt(6):
so x^2 + (x+2)^2 = [sqrt(6)]^2
x^2 + x^2+4x+4 = 6

Can you finish that?
If you can't, or don't follow, you need to talk to your teacher.

 

[tarheel2006]

Yes thank you.