SOLUTION: I need to answer this question using quadratics and I am not sure where to start. Ed needs to enclose a rectangular section of his yard. The area is 35 sq ft and the perimeter is 2

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Question 166189: I need to answer this question using quadratics and I am not sure where to start. Ed needs to enclose a rectangular section of his yard. The area is 35 sq ft and the perimeter is 27 ft. Find the length and the width of the section.
Found 2 solutions by vleith, Mathtut:
Answer by vleith(2983)   (Show Source): You can put this solution on YOUR website!
Perimeter is the length 'around' the rectangular area.
Area is the area of the enclosed rectangle








Now use the quadratic equation to solve
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=169 is greater than zero. That means that there are two solutions: .




Quadratic expression can be factored:

Again, the answer is: 10, 3.5. Here's your graph:
Answer by Mathtut(3670)   (Show Source): You can put this solution on YOUR website!
since A=lw where l is length and w is the width
and P=2(l+w)
we plug in the known values
35=lw
27=2(l+w)--->27/2=l+w---->l=27/2-w---->l=(27-2w)/2
place value of l into 1st equation----->35=((27-2w)/2)(w)----->
factoring (2w-7)(w-10) so w=7/2,10
when w=7/2 l=10 when w=10 l=7/2
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