SOLUTION: The length of a rectangle is 2 ft longer than the width. If the area is 16ft^2, then what are the length and width?

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Question 81911: The length of a rectangle is 2 ft longer than the width. If the area is 16ft^2, then what are the length and width?
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
L=W+2
AREA=L*W
16=(W+2)W
16=W^2+2W
W^2+2W-16=0
USING THE QUADRATIC EQUATION x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
WE GET:
W=(-2+-SQRT[2^2-4*1*-16])/2*1
W=(-2+-SQRT[4+64])/2
W=(-2+-SQRT68)/2
W=(-2+-8.246)/2
W=(-2+8.246)/2
W=6.246/2
W=3.123 ANSWER FOR THE WIDTH.
L=3.123+2
L=5.123 ANSWER FOR THE LENGTH.
PROOF
3.123*5.123=16
16=16