SOLUTION: The length of a rectangle is 3 times it's width. Is both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?

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Question 824982: The length of a rectangle is 3 times it's width. Is both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?
Found 2 solutions by jim_thompson5910, TimothyLamb:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The length of a rectangle is 3 times it's width. If both dimensions are increased by 2 centimeters, the area is 32 square centimeters. What was the original length?


Width = x
Length = 3x

"If both dimensions are increased by 2 centimeters, the area is 32 square centimeters" ---> %28x%2B2%29%283x%2B2%29=32

%28x%2B2%29%283x%2B2%29=32

x%283x%2B2%29%2B2%283x%2B2%29=32

3x%5E2%2B2x%2B6x%2B4=32

3x%5E2%2B2x%2B6x%2B4-32=0

3x%5E2%2B8x-28=0

%28x-2%29%283x%2B14%29=0

x-2=0 or 3x%2B14=0

x=2 or 3x=-14

x=2 or x=-14%2F3

Toss the negative solution. So the only solution is x=2

Width = x = 2

Length = 3x = 3*2 = 6

The width is 2 centimeters and the length is 6 centimeters

Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
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x = length
y = width
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x = 3y
(x + 2)(y + 2) = 32
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(x + 2)(y + 2) = 32
(3y + 2)(y + 2) = 32
3yy + 6y + 2y + 4 = 32
3yy + 8y - 28 = 0
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the above quadratic equation is in standard form, with a=3, b=8, and c=-28
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to solve the quadratic equation, by using the quadratic formula, copy and paste this:
3 8 -28
into this solver: https://sooeet.com/math/quadratic-equation-solver.php
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this quadratic has two real roots (the x-intercepts), which are:
y = 2
y = -4.66666667
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the negative root does not fit the problem statement, so use the positive root
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answer:
x = length = 6 cm
y = width = 2 cm
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Solve quadratic equations, quadratic formula:
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