SOLUTION: rectangle's length is 5 ft more than its width. if the width is not changed & the length is increased by 3ft, the new area will be 128 ft^2. What are the dimensions of the original

Algebra ->  Rectangles -> SOLUTION: rectangle's length is 5 ft more than its width. if the width is not changed & the length is increased by 3ft, the new area will be 128 ft^2. What are the dimensions of the original      Log On


   

Question 552083: rectangle's length is 5 ft more than its width. if the width is not changed & the length is increased by 3ft, the new area will be 128 ft^2. What are the dimensions of the original rectangle?
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
l=5+w
w=1-5
%28l%2B3%29%28l-5%29=128 2nd rectangle dimensions: 16x8
l%5E2-2l-15=128
l%5E2-2l-143=0
l=13
13-5=8
w=8
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation al%5E2%2Bbl%2Bc=0 (in our case 1l%5E2%2B-2l%2B-143+=+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=%28-2%29%5E2-4%2A1%2A-143=576.

Discriminant d=576 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--2%2B-sqrt%28+576+%29%29%2F2%5Ca.

l%5B1%5D+=+%28-%28-2%29%2Bsqrt%28+576+%29%29%2F2%5C1+=+13
l%5B2%5D+=+%28-%28-2%29-sqrt%28+576+%29%29%2F2%5C1+=+-11

Quadratic expression 1l%5E2%2B-2l%2B-143 can be factored:
1l%5E2%2B-2l%2B-143+=+1%28l-13%29%2A%28l--11%29
Again, the answer is: 13, -11. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-2%2Ax%2B-143+%29