SOLUTION: the width of a rectangle is 6 feet less than twice its length. If its area is 108 square feet, find the dimensions of the rectangle
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Question 925982: the width of a rectangle is 6 feet less than twice its length. If its area is 108 square feet, find the dimensions of the rectangle Found 2 solutions by TimothyLamb, srinivas.g:Answer by TimothyLamb(4379) (Show Source):
You can put this solution on YOUR website! w = 2L - 6
Lw = 108
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L(2L - 6) = 108
2LL - 6L - 108 = 0
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the above quadratic equation is in standard form, with a=2, b=-6 and c=-108
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
2 -6 -108
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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the quadratic has two real roots at:
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L = 9
L = -6
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the negative root doesn't fit the problem statement, so use the positive root:
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L = 9
w = 2*9 - 6
w = 12
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answer:
L = 9 feet
w = 12 feet
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You can put this solution on YOUR website! let length of rectangle be l
width = 2*length -6
width =2l-6
Area of rectangle = 108 square feet
Area = length* width
108 = l*(2l-6)
108 = l*(2l)+l*(-6)
move 108 to the right
divide each term with 2
from above equation , the possibilities are
l-9= 0 or l+6 = 0
l=9 or l=-6
since l is length , it cannot be negative
hence l is 9 foot
width = 2*length -6
= 2*9-6
=18-6
=12
width = 12 feet
length = 9 feet
