SOLUTION: 1.area of a rectangle is 33 square inches. If the length is 2 more than three times the width, find the length and the width.

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Question 286216: 1.area of a rectangle is 33 square inches. If the length is 2 more than three times the width, find the length and the width.
Answer by mestrydilip(13) About Me  (Show Source):
You can put this solution on YOUR website!
Area = 33 Sq.inches.
Let us assume width to be x inches.
Length = (3x + 2)
x (3x + 2) = 33 Area of Rectangle = Length X Width
3x^2 + 2x = 33
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 3x%5E2%2B2x%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=%282%29%5E2-4%2A3%2A-33=400.

Discriminant d=400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-2%2B-sqrt%28+400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%282%29%2Bsqrt%28+400+%29%29%2F2%5C3+=+3
x%5B2%5D+=+%28-%282%29-sqrt%28+400+%29%29%2F2%5C3+=+-3.66666666666667

Quadratic expression 3x%5E2%2B2x%2B-33 can be factored:
3x%5E2%2B2x%2B-33+=+3%28x-3%29%2A%28x--3.66666666666667%29
Again, the answer is: 3, -3.66666666666667. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+3%2Ax%5E2%2B2%2Ax%2B-33+%29


Hence if x = 3
Length = 11 inches
Width = 3 inches