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Find the width of a rectangle when its length is greater by twice and 2 feet more than its width and has an area of 40 square feet?
Wiki User
∙ 13y ago
Let the length be (2x+2) and the width be x:
length*width = area
(2x+2)*x = 40 sq ft
2x2+2x = 40
2x2+2x-40 = 0
Solving the above quadratic equation works out as:
x = -5 or x = 4 it must be the latter because dimensions can't be negative.
Therefore: width = 4 feet and length = 10 feet
Check: 10*4 = 40 sq ft
Wiki User
∙ 13y ago
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