SOLUTION: the length of a rectangle is 5 meters less than twice the width, if the area of the rectangle is 493 sq meters what are the demensions

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: the length of a rectangle is 5 meters less than twice the width, if the area of the rectangle is 493 sq meters what are the demensions      Log On


   

Question 439283: the length of a rectangle is 5 meters less than twice the width, if the area of the rectangle is 493 sq meters what are the demensions
Answer by jorel1380(3719) About Me  (Show Source):
You can put this solution on YOUR website!
L=2W-5
LxW=Area
Wx(2W-5)=493
2W2-5W-493=0
(2W+29)(W-17)=0
W=-29/2 or 17
Throwing out the negative answer, we get length=29 m., width=17 m..
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 2x%5E2%2B-5x%2B-493+=+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=%28-5%29%5E2-4%2A2%2A-493=3969.

Discriminant d=3969 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--5%2B-sqrt%28+3969+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-5%29%2Bsqrt%28+3969+%29%29%2F2%5C2+=+17
x%5B2%5D+=+%28-%28-5%29-sqrt%28+3969+%29%29%2F2%5C2+=+-14.5

Quadratic expression 2x%5E2%2B-5x%2B-493 can be factored:
2x%5E2%2B-5x%2B-493+=+2%28x-17%29%2A%28x--14.5%29
Again, the answer is: 17, -14.5. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-5%2Ax%2B-493+%29