SOLUTION: the width of a rectangle is 5cm greater than its length. if the area of the reactangle is 124cm^2, what are the approximate dimensions of the rectangle?

Algebra ->  Rectangles -> SOLUTION: the width of a rectangle is 5cm greater than its length. if the area of the reactangle is 124cm^2, what are the approximate dimensions of the rectangle?      Log On


   

Question 860161: the width of a rectangle is 5cm greater than its length. if the area of the reactangle is 124cm^2, what are the approximate dimensions of the rectangle?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

Let x represent the length.  The width is (x+5)

x(x+5) = 124cm^2

x^2  + 5x = 124

x^2 + 5x - 124 = 0

 x = `8.9, the length.  The width ` 13.9

 


 
Solved by pluggable solver: SOLVE quadratic equation with variable


Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B5x%2B-124+=+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-124=521.

  

  Discriminant d=521 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-5%2B-sqrt%28+521+%29%29%2F2%5Ca.

  

    x%5B1%5D+=+%28-%285%29%2Bsqrt%28+521+%29%29%2F2%5C1+=+8.91271221051333

    x%5B2%5D+=+%28-%285%29-sqrt%28+521+%29%29%2F2%5C1+=+-13.9127122105133

    

    Quadratic expression 1x%5E2%2B5x%2B-124 can be factored:

  1x%5E2%2B5x%2B-124+=+1%28x-8.91271221051333%29%2A%28x--13.9127122105133%29

  Again, the answer is: 8.91271221051333, -13.9127122105133.
Here's your graph:

graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B5%2Ax%2B-124+%29