Question 816082: he area of a rectangle is 27yd^2 , and the length of the rectangle is 3yd less than double the width. find the dimensions of the rectangle.
Found 2 solutions by TimothyLamb, algebrahouse.com:
Answer by TimothyLamb(4379)   (Show Source):
You can put this solution on YOUR website! a = Lw = 27
L = 2w - 3
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Lw = 27
(2w - 3)w = 27
2w^2 - 3w - 27 = 0
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the above quadratic equation is in standard form, with a=2, b=-3, and c=-27
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to solve the quadratic equation, plug this:
2 -3 -27
into this: https://sooeet.com/math/quadratic-equation-solver.php
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the roots of the quadratic are:
4.5
-3
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the negative root doesn't make sense for a length, so use the positive root
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w = 4.5
L = 6
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Answer by algebrahouse.com(1659)  (Show Source):
You can put this solution on YOUR website! x = width
2x - 3 = length {length is 3 less than double the width}
Area of a rectangle = width x length
x(2x - 3) = 27 {substituted width, length, and area into area formula}
2x² - 3x = 27 {used distributive property}
2x² - 3x - 27 = 0 {subtracted 27 from each side}
2x² - 9x + 6x - 27 = 0 {split -3x into two terms whose coefficients multiply to get -54 and add to get -3}
x(2x - 9) + 3(2x - 9) = 0 {factored x out of first two terms and 3 out of last two terms}
(2x - 9)(x + 3) = 0 {factored (2x - 9) out of the two terms}
2x - 9 = 0 or x + 3 = 0 {set each factor equal to 0}
2x = 9 or x = -3 {added 9 in first equation, subtracted 3 in second equation}
x = 4.5 {divided first equation by 2, the width cannot be negative, so -3 is out}
2x - 3 = 6 {substituted 4.5, in for x, into 2x - 3}
width is 4.5 yds
length is 6 yds
For more help from me, visit: www.algebrahouse.com
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