# 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

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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?
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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 (in our case ) has the following solutons: For these solutions to exist, the discriminant should not be a negative number. First, we need to compute the discriminant : . Discriminant d=1936 is greater than zero. That means that there are two solutions: . Quadratic expression can be factored: Again, the answer is: 12, -10. Here's your graph:

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