SOLUTION: The area of a rectangular wall hanging in Elizabeth's room is 240 square inches. The width of the hanging is four inches less than twice the length. What are the dimensions of th

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: The area of a rectangular wall hanging in Elizabeth's room is 240 square inches. The width of the hanging is four inches less than twice the length. What are the dimensions of th      Log On

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Question 193383: The area of a rectangular wall hanging in Elizabeth's room is 240 square inches. The width of the hanging is four inches less than twice the length. What are the dimensions of the wall hanging?
Answer by yuendatlo(23) About Me  (Show Source):
You can put this solution on YOUR website!
Let width=w, and length=l
w=2l-4
w*l=240
replace w by 2l-4
(2l-4)*l=240
2l^2-4l=240
2l^2-4l-240=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation al%5E2%2Bbl%2Bc=0 (in our case 2l%5E2%2B-4l%2B-240+=+0) has the following solutons:

l%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-4%29%5E2-4%2A2%2A-240=1936.

Discriminant d=1936 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--4%2B-sqrt%28+1936+%29%29%2F2%5Ca.

l%5B1%5D+=+%28-%28-4%29%2Bsqrt%28+1936+%29%29%2F2%5C2+=+12
l%5B2%5D+=+%28-%28-4%29-sqrt%28+1936+%29%29%2F2%5C2+=+-10

Quadratic expression 2l%5E2%2B-4l%2B-240 can be factored:
2l%5E2%2B-4l%2B-240+=+2%28l-12%29%2A%28l--10%29
Again, the answer is: 12, -10. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-4%2Ax%2B-240+%29

so l can be 12 because the length cannot be negative
replace l with 12
w=2*12-4=24-4=20
Checking:20*12=240