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# Quadratic Equation - Length, Width / Solution - Explanation

Thu 20 Nov, 2014 01:26 pm
An oblong pond is surrounded by a walkway 4 yards wide. The pond is 15200 square yards and the walkway 2104 square yards. Determine length and width of pond.

Let x = length of pond.

15200 / x = width of pond.

(x + 4) * (15200 / x + 8) = (15200 + 2104) = 17304 **

8x + 121600 /x + 15264 = 17304 **

8x^2 - 2040x = 121600 **

x^2 - 255x = 15200

x = 127 * 5 + - 32 * 5 = 160 or 95 **

length = 160 or 95

width = 15200 / 160 = 95 or 15200 / 95 = 160

** I don't understand these steps. I am not very good at quadratics.
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engineer

1
Thu 20 Nov, 2014 03:38 pm
@Randy Dandy,
You don't understand because the person who gave you this answer is doing so many steps in his head that he is confusing you. Let me try.

Let x = the length of the rectangular pond. (Don't know anyone who would use oblong here.)
Since the area of the pond = length x width = 15,200, the width = 15200 / x
The length of the pond plus the walkway = x + 4 + 4 = x + 8 (the walkway is on each side so I add 4 twice)
The width of the pond plus the walkway = 15200/x + 4 + 4 = 15200/x + 8
The area of the pond and the walkway = 15,200 + 2,104 = 17,304.

The area of the pond and the walkway = length of the pond and the walkway times the width of the pond and the walkway so...

17,304 = (x+8) (15200/x + 8)

That is not the same as what you have below because whoever solved it for you the first time only put a walk on one side of the pond (x+4) instead of both sides (x+4+4 or x+8). Do you understand why the total length is x+8? If not, draw a picture of it.

Now we are going to multiply that out using FOIL (first, outer, inner, last).

17,304 = 15200 + 8x + 8(15200)/x + 64
17,304 = 15264 + 8x + 121600/x
2040 = 8x + 121600/x

Even though the solution you posted has an error in the third line, it is somehow correct at this point. You now multiply both sides by X to get ..

2040x = 8x^2 + 121600

Move all the terms to one side...

0 = 8x^2 - 2040x + 121600

You can solve this several ways. I would just use the quadratic formula if you have been taught that. What your solution below does, is first divide by eight...

0 = x^2 - 255x + 15200

Then factor the problem into

0 = (x-160) (x-95)

So the solutions are x = 160 or x = 95 just like in your solution. Does that help?
Randy Dandy

1
Thu 20 Nov, 2014 05:15 pm
@engineer,
Yes, I understand it now. Also, can you provide some tips/hints for quadratics ?
engineer

1
Thu 20 Nov, 2014 07:55 pm
@Randy Dandy,
Sure, but that is a big field. Is there a specific issue you have? In general, there is no need to try and do ten things at the same time. Just take it step by step like I did in your problem and you can keep everything clear. Remember, you want to end up with an equation that looks like:

ax^2 + bx + c = 0

Always try for this format. Have you learned the quadratic equation yet? That makes things pretty easy once you get the the format above.
Randy Dandy

0
Thu 20 Nov, 2014 10:40 pm
@engineer,
I can't remember the quadratic equation. I have trouble setting up the problems.
engineer

1
Fri 21 Nov, 2014 03:23 pm
@Randy Dandy,
The quadratic equation is your friend. Practice memorizing it until you can do it in your sleep. As for setting up problems, the problems you have posted here are a lot easier if you DRAW A PICTURE. It also helps if you take the words in the problems and translate them into simple equations. If you have a problem that says "The area of a garden is 50 square feet and the length is twice the width" then you should write down:

Area = 50
Length = 2 Width

Then you write down facts you know that will help you. Things like:

Area = Length x Width

Hopefully at this point you can see how to substitute in to get a good equation.

50 (the area) = 2 Width x Width = 2 Width^2
25 = Width^2
5 = Width

The key is to capture the information given to you in the problem and combine it with the information you have in your head.
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