SOLUTION: the length of the rectangle is 5 inches more than the width. the area is 33 square inches. find the length and width. round to the nearest tenth if necessary.

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Question 966893: the length of the rectangle is 5 inches more than the width. the area is 33 square inches. find the length and width. round to the nearest tenth if necessary.
Answer by amarjeeth123(569) About Me  (Show Source):
You can put this solution on YOUR website!
Let the width be x.
Then the length is (x+5).
Area=length*width
x(x+5)=33
x^2+5x-33=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B-33+=+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-33=157.

Discriminant d=157 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+157+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%285%29%2Bsqrt%28+157+%29%29%2F2%5C1+=+3.76498204307083
x%5B2%5D+=+%28-%285%29-sqrt%28+157+%29%29%2F2%5C1+=+-8.76498204307083

Quadratic expression 1x%5E2%2B5x%2B-33 can be factored:
1x%5E2%2B5x%2B-33+=+1%28x-3.76498204307083%29%2A%28x--8.76498204307083%29
Again, the answer is: 3.76498204307083, -8.76498204307083. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-33+%29

x=3.764=3.8
The width is 3.8 units.
The length is 8.8 units.