SOLUTION: The width of a rectangle is three more than twice its length. The area of the rectangle is 44cm^2. Find the dimensions of the rectangle. Show work please

Algebra ->  Rectangles -> SOLUTION: The width of a rectangle is three more than twice its length. The area of the rectangle is 44cm^2. Find the dimensions of the rectangle. Show work please      Log On


   

Question 282188: The width of a rectangle is three more than twice its length. The area of the rectangle is 44cm^2. Find the dimensions of the rectangle. Show work please
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
w=2l+3
a=lw=44
44=l%282l%2B3%29
44=2l%5E2%2B3l
0=2l%5E2%2B3l-44l=1.31 w=2(1.31)+3=5.62
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation al%5E2%2Bbl%2Bc=0 (in our case 2l%5E2%2B31l%2B-44+=+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=%2831%29%5E2-4%2A2%2A-44=1313.

Discriminant d=1313 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-31%2B-sqrt%28+1313+%29%29%2F2%5Ca.

l%5B1%5D+=+%28-%2831%29%2Bsqrt%28+1313+%29%29%2F2%5C2+=+1.30883546599672
l%5B2%5D+=+%28-%2831%29-sqrt%28+1313+%29%29%2F2%5C2+=+-16.8088354659967

Quadratic expression 2l%5E2%2B31l%2B-44 can be factored:
2l%5E2%2B31l%2B-44+=+2%28l-1.30883546599672%29%2A%28l--16.8088354659967%29
Again, the answer is: 1.30883546599672, -16.8088354659967. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B31%2Ax%2B-44+%29