SOLUTION: The width of a rectangular painting is 2 inches less than the length. The area is 120 inches squared. Find the length and the width.

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Question 666142: The width of a rectangular painting is 2 inches less than the length. The area is 120 inches squared. Find the length and the width.
Found 2 solutions by swincher4391, Shana-D77:
Answer by swincher4391(1107) (Show Source):
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w = l-2
l * l-2 = 120
l^2 - 2l = 120
l^2 - 2l -120 = 0
(l-12)(l+10) = 0
l = 12 l =-10
Length can't be negative so l = 12.
Then w = 12 -2 = 10
12 x 10.

Answer by Shana-D77(132) (Show Source):
You can put this solution on YOUR website!
The width of a rectangular painting (w) is (=) 2 inches less than ( - 2) the length (L). The area(A) is 120 inches squared. Find the length and the width.

Take out the words, leave the translation:
w = L - 2
A = 120
We also know how to find Area:
A = (L)(w)

Now we need to get creative:
A = (L)(w)
120 = (L )(L - 2) (This is the hardest, and most important, step to know)
120 = L^2 - 2L (multiplied L^2 by L - 2)
L^2 - 2L - 120 = 0 (Get everything to one side before factoring)
Q: What to numbers multiply to get -120 and add to get -2)
A: -12*10
So let's factor:
(L - 12)(L + 10) = 0
So, either L - 12 = 0 or L + 10 = 0
If L + 10 = 0 then L = -10. You can't have a negative length so that makes no sense.
If L - 12 = 0 then L = 12. This makes sense and therefore is the rectangle's length.
If the Length is 12, and the width is "2 less than", then the width is 10.
Length = 12
width = 10