Algebra

The width of a rectangle is 2 less than twice its length. If the area of the rectangle is 82 cm^2, what is the length of the diagonal? Note: Your answer must be a number. It may not contain any arithmetic operations.

w(w+20) = 8000
w^2+20w = 8000
w^2+20w-8000 = 0
(w-80)(w+100) = 0

w-80 = 0
w=80

or

w+100 = 0
w = -100

I think that x=80 makes more sense. However what would the length be if the width is equal to 80 in this problem ? please help Im stuck and am not sure what the final answer for this question would be.

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  1. Your equation doesn't even match your given data

    length --- x
    width ---- 2x - 2

    area = x(2x-2) = 82

    2x^2 - 2x - 82 = 0
    x^2 - x - 41 = 0
    use the formula ...
    x = (1 ± √165)/2
    = appr 6.92 or some negative

    length = 6.92 , length = 11.85
    check: 6.92x11.85 = appr 82

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    posted by Reiny

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