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
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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
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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 inē, 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
