College 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

SORRY I WROTE THE WRONG QUESTION PERVIOUSLY

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. Length = L units.
    Width = (2L-2) units.

    A = L(2L-2) = 82cm^2
    2L^2 - 2L - 82 = 0
    L^2 - L - 41 = 0
    Use Quadratic formula:
    L = 6.92 cm.

    Width = 2L - 2 = 2*6.92 - 2 = 11.84 cm.

    The diagonal divides the rectangle into
    2 congruent rt. triangles.

    D^2 = L^2 + W^2 = 6.92^2 + 11.84^2=188
    Diag. = 13.71 cm.

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    posted by Henry

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