# Length and width

The width of a rectangle is 2ft less than its length. The area is 8ft^2. Find the length and width.

I can't seem to come up with anything that equals 64 sq ft.

Do I start with area 64sqft.=W x L-2.

I feel like I should be coming up with something around 7 times 9 but I can't seem to come up with 64, ideas?

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1. In the first statement you say area is 8 ft^2
Then you say it is 64 ft^2

If it is eight, it is obvious by inspection
2*4 = 8

If it is 64, we better try to do it right
W * (W+2) = 64
W^2 + 2 W - 64 = 0
W = -2/2 +/- (1/2)sqrt (4 + 4*64)
W = -1 +/- (1/2) sqrt ( 4*65)
W = -1 +/- sqrt(65)
W = -1 - sqrt 65 forget this solution, no negative width
W = 8.06 - 1 = 7.06
W + 2 = 9.06
so indeed very close to 7 and 9

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2. x(x+2) = 64

If you are studying quadratic equations, you should be able to solve this one.

It comes out as a radical
(7.0622 and 9.0622)

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3. You are right. I should sleep and try again later. I am starting to make silly mistakes.

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