SOLUTION: the length of a rectangle is 5 feet more than it's width. the area of the rectangle is 176 sq ft. find the length and the width of the rectangle

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Question 179357: the length of a rectangle is 5 feet more than it's width. the area of the rectangle is 176 sq ft. find the length and the width of the rectangle
Answer by Mathtut(3670) About Me  (Show Source):
You can put this solution on YOUR website!
the length of a rectangle is 5 feet more than it's width. the area of the rectangle is 176 sq ft. find the length and the width of the rectangle
:
A=L*W......eq 1
L=W+5......eq 2
A=176......eq 3
:
take L's value of W+5 from eq 2 and A's value of 176 and plug them into eq 1
:
176=(W+5)W
:
176=W%5E2%2B5W
:
W%5E2%2B5W-176=0
:
W=11 and -16...throw out the negative value
:
highlight%28W=11%29width
:
highlight%28L=11%2B5%29length
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B-176+=+0) has the following solutons:

x%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=%285%29%5E2-4%2A1%2A-176=729.

Discriminant d=729 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+729+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+729+%29%29%2F2%5C1+=+11
x%5B2%5D+=+%28-%285%29-sqrt%28+729+%29%29%2F2%5C1+=+-16

Quadratic expression 1x%5E2%2B5x%2B-176 can be factored:
1x%5E2%2B5x%2B-176+=+%28x-11%29%2A%28x--16%29
Again, the answer is: 11, -16. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-176+%29