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"The length of a rectangle is 3 ft shorter than twice its width. Its area is 54 square feet. Find the length and width, then enter their sum below."
I thought I was getting these types of problems but this one I need help with, thank you!!
I thought I was getting these types of problems but this one I need help with, thank you!!
Answers
Answered by
Anonymous
L= 2 w-3
L w = 54
54/w = 2 w - 3
54 = 2 w^2 - 3 w
2 w^2 - 3 w - 54= 0
w = [ 3 +/- sqrt(9+ 432) ] / 4 = [ 3 +/- 21 ] / 4 so try 6
then L= 12 - 3 = 9
L w = 54
54/w = 2 w - 3
54 = 2 w^2 - 3 w
2 w^2 - 3 w - 54= 0
w = [ 3 +/- sqrt(9+ 432) ] / 4 = [ 3 +/- 21 ] / 4 so try 6
then L= 12 - 3 = 9
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