SOLUTION: thanks for your help. 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 400ft^, what is the width of t

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Question 28978: thanks for your help. 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 400ft^, what is the width of the path?
A=(2)LW
2x30.20=1200
1200-400=800 sq ft left

could the path be 20 ft wide? 20 times 20 times 2=800
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
It sounds like the path is constructed IN and not AROUND the garden, so
it diminishes the amount of gardent area.
Let the uniform width of the path be "x".
If you draw the picture you will see that the remaining garden space is
a rectangle with a width of 30-2x and a length of 20-2x.
EQUATION:
(30-2x)(20-2x)=400
2(15-x)2(10-x)=400
(15-x)(10-x)=100
150-25x+x^2=100
x^2-25x+50=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation (in our case ) has the following solutons:



For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=425 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 22.8077640640442, 2.19223593595585. Here's your graph:

Hope this helps
Cheers,
Stan H.
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