SOLUTION: The length of a rectangle is twice it's width. if the width is reduced by 1cm and the length is also reduced by 2cm, the area will be 15cm^2. find the dimension of the original rec

Question 1064531: The length of a rectangle is twice it's width. if the width is reduced by 1cm and the length is also reduced by 2cm, the area will be 15cm^2. find the dimension of the original rectangle.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
he length of a rectangle is twice it's width.
if the width is reduced by 1cm and the length is also reduced by 2cm, the area will be 15cm^2.
find the dimension of the original rectangle.
:
Let L = the original length
let w = the original width
Given that
L = 2w
Then
(L-2) = the new length
(w-1) = the new width
Area equation
(L-2)*(w-1) = 15
Replace L with 2w
(2w-1)*(w-1) = 15
FOIL
2w^2 - 2w - 2w + 2 = 15
2w^2 - 4w + 2 - 15 = 0
2w^2 - 4w - 13 = 0
Use the quadratic formula, a=2; b=-4; c=-13, (this will not factor)
I go a positive solution of w = 3.7386 is the width
therefore: 2 * 3.7386 = 7.4772 is the length
:
:
You can check this for yourself on a calc
(7.4772-2) * (3.7386-1) =
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