SOLUTION: the length of a rectangle is 13 centimeters less than six times its width. it's area is 15 square centimeters. find the dimensions of the rectangle.

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: the length of a rectangle is 13 centimeters less than six times its width. it's area is 15 square centimeters. find the dimensions of the rectangle.      Log On


   

Question 949113: the length of a rectangle is 13 centimeters less than six times its width. it's area is 15 square centimeters. find the dimensions of the rectangle.
Answer by macston(5194) About Me  (Show Source):
You can put this solution on YOUR website!
W=width; L=length=6W-13 cm;A=L*W=15 sq cm
A=L*W
15 sq cm=L*W Substitute for L.
15 sq cm=(6W-13)(W)
15+sq+cm=6W%5E2-13W Subtract 15 sq cm from each side.
0=6W%5E2-13W-15+sq+cm
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation aW%5E2%2BbW%2Bc=0 (in our case 6W%5E2%2B-13W%2B-15+=+0) has the following solutons:

W%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=%28-13%29%5E2-4%2A6%2A-15=529.

Discriminant d=529 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--13%2B-sqrt%28+529+%29%29%2F2%5Ca.

W%5B1%5D+=+%28-%28-13%29%2Bsqrt%28+529+%29%29%2F2%5C6+=+3
W%5B2%5D+=+%28-%28-13%29-sqrt%28+529+%29%29%2F2%5C6+=+-0.833333333333333

Quadratic expression 6W%5E2%2B-13W%2B-15 can be factored:
6W%5E2%2B-13W%2B-15+=+6%28W-3%29%2A%28W--0.833333333333333%29
Again, the answer is: 3, -0.833333333333333. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+6%2Ax%5E2%2B-13%2Ax%2B-15+%29

The answer we want is 3 ANSWER: Width is 3 cm.
L=6W-13=6(3)-13=18-13=5 cm ANSWER: Length is 5 cm.
CHECk:
A=L*W
15 sq cm=5 cm*3 cm
15 sq cm=15 sq cm