SOLUTION: Here is another problem that I have posted within the past 2 days that I am having trouble with setting up. If someone could give me a little hint, I would appreciate it. Thanks.

Question 68077: Here is another problem that I have posted within the past 2 days that I am having trouble with setting up. If someone could give me a little hint, I would appreciate it. Thanks.
A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? Give your answer in decimal form, rounded to the nearest thousandth.
Found 2 solutions by checkley71, stanbon:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
(30-2X)(20-2X)=400
600-40X-60X+4X^2=400
4X^2-100X+600-400=0
4X^2-100X+200=0
X^2-25X+50=0
using the quadratic equation we get
x=(-b+-sqrt[b^2-4ac])/2a
x=(25+-sqrt[-25^2-4*1*50])/2*1
x=(25+-sqrt625-200])/2
x=(25+-sqrt425)/2
x=(25+-20.6155)/2
x=(25+20.6155)/2
x=45.6155
x=(25-20.6155)/2
x=4.3845/2
x=2.192 answer for the width of the path.
-------------------------------------------------------------
IF YOU DRAW A DIAGRAM OF THE PROBLEM (WHICH I ALWAYS RECOMMEND) YOU'LL SEE THAT THE RECTANGLE INSIDE ANOTHER RECTANGLE WITH THE OUTSIDE DIMENTIONS OF 20 & 30 FT
YOU'LL SEE THAT THE INNER RECTANGLE LENGTH IS 30 & 20 -2 TIMES THE BORDER (X) THUS THE TERMS 30-2X & 20-2X TO DESCRIBE THE INNER RECTANCLE.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!

A garden area is 30 ft long and 20 ft wide. A path of uniform width is set around the edge. If the remaining garden area is 400 ft2, what is the width of the path? Give your answer in decimal form, rounded to the nearest thousandth.
-------------------------
Draw a rectangle and mark the length as 30 ft and the width as 20 ft.
Draw a path INSIDE the perimeter of the rectangle that is "x" wide.
Indicate that the remaining area of the garden is 400 sq ft.
-----------------------
Try to see the following dimensions:
The length of the "remaining" area is 30-2x ft
The width of the "remaining" area is 20-2x ft
----------------
EQUATION:
(remaining length)(remaining width)= remaining area = 400 sq ft
(30-2x)(20-2x)=400
4(15-x)(10-x)=400
(15-x)(10-x)=100
150-25x+x^2=100
x^2-25x+50=0
x=[25+-sqrt(625-4*50)]/2
x=[25+-sqrt425)]/2
x=[25+-5sqrt17]/2
x=[25+5sqrt17]/2 or x=[25-5sqrt17]/2
x=22.81 ft. or x=2.19 ft.
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x cannot be 22.81 ft as there is not that much remaining width.
Therefore x=2.19 ft (This is the width of the path)
Cheers,
Stan H.

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