SOLUTION: The area of a rectangle is 54 yd^2 , and the length of the rectangle is 3 yd more than twice the width. Find the dimensions of the rectangle.

Algebra ->  Rectangles -> SOLUTION: The area of a rectangle is 54 yd^2 , and the length of the rectangle is 3 yd more than twice the width. Find the dimensions of the rectangle.       Log On


   



Question 928565: The area of a rectangle is 54 yd^2 , and the length of the rectangle is 3 yd more than twice the width. Find the dimensions of the rectangle.
Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
Lw = 54
L = 2w + 3
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(2w + 3)w = 54
2ww + 3w - 54 = 0
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the above quadratic equation is in standard form, with a=2, b=3 and c=-54
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
2 3 -54
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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w = 4.5
w = -6
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the negative root doesn't fit the problem statement, so use the positive root:
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w = 4.5
L = 2w + 3
L = 2*4.5 + 3
L = 12
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answer:
w = 4.5 yd
L = 12 yd
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