Question 570372: the length of a rectangle is 4 meters less than twice the width. If the area of the rectangle is 286 square meters, find the dimensions.
Answer by mathsmiles(68)   (Show Source):
You can put this solution on YOUR website! Length of rectangle is 4 meters less than twice the width
L = 2W - 4
L x W = 286
We need to find L and W. Let's use substitution since we already have L in the form we need for that method.
Substituting 2W - 4 for L in the area equation:
(2W - 4) x W = 286
Multiplying out ...
2W^2 - 4W = 286
Subtracting 286 from each side to get this in standard format:
2W^2 - 4w - 286 = 0
(2W _ _) (W _ _) = 0
We know the operators are different (one neg and one pos), need to find the terms and which operator goes where. 286 makes this a bit too interesting imho.
Finding factors of 286:
I used this website: http://www.calculatorsoup.com/calculators/math/factors.php
2, 11, 13, 22, 26, 143
I'm going to guess here and try 11 and 26. (Call it age or experience :-)
(2W - 26)(W + 11) = 0
Using foil to check: 2W^2 + 22W - 26W -286 = 2W^2 - 4W - 286 = 0. We have a winner!
(2W - 26) = 0
2W = 26
W = 13
AND
(W + 11) = 0
W = -11
Since rectangles have only positive sides in our dimension of reality, we have to use W=13 as the only possible answer. In that case,
L = 2W - 4
L = 2(13) - 4
L = 26 - 4
L = 22
For this rectangle, W = 13 meters, L = 22 meters
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