SOLUTION: If the area of a rectangle is 88yd^2, what are the dimensions if, the length of a rectangle is 5yd more than twice its width?
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Question 433325: If the area of a rectangle is 88yd^2, what are the dimensions if, the length of a rectangle is 5yd more than twice its width?
Answer by algebrahouse.com(1659) (Show Source): You can put this solution on YOUR website!
"If the area of a rectangle is 88yd^2, what are the dimensions if, the length of a rectangle is 5yd more than twice its width?"
x = width
2x + 5 = length {length is five more than twice width}
Area of a rectangle is length x width
x(2x + 5) = 88 {area is length x width}
2x² + 5x = 88 {used distributive property}
2x² + 5x - 88 = 0 {subtracted 88 from both sides}
(2x - 11)(x + 8) = 0 {factored into two binomials}
2x - 11 = 0 or x + 8 = 0 {set each factor equal to 0}
2x = 11 or x = -8 {added 11 and subtracted 8, respectively}
x = 11/2 or x = -8 {divided left equation by 2}
x = 11/2 {width cannot be negative}
2x + 5 = 16 {substituted 11/2, in for x, into 2x + 5}
width = 5 1/2 yds
length = 16 yds
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