SOLUTION: Hello, I am having a really hard time with these problems. What is the proper way to set up this equation? The length of a rectangle is 5 feet more than twice its width. If it

Question 133786: Hello,
I am having a really hard time with these problems. What is the proper way to set up this equation?
The length of a rectangle is 5 feet more than twice its width. If its area is 52 in�, find the dimensions of the length and width.
Any help would be greatly appreciated.
Thank you,
Mrs. Gibson
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let w=width
Then length=2w+5
Area of a rectangle =L*W, so:
w(2w+5)=52 (w(2w+5)=L*W and we are told that this equals 52 sq in. OK)
2w^2+5w=52 subtract 52 from each side
2w^2+5w-52=0 quadratic in standard form
It's now set up ready to be solved --you can use the quadratic formula
You will come out with a positive and a negative value for w--neglect the negative value ---lengths and widths in this problem are positive

Hope this helps--ptaylor
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