What are the possible dimensions of a 28-square-foot garden if the width is 3 feet shorter than the length?

Question

Are the solutions to the equation reasonable?

Answer 1

Let w = width, so length = w + 3

Area = w(w+3) = 28.

 

This is a quadratic equation: w² + 3w - 28 = 0

(w+7)(w-4) = 0  so w = 4 (the other root, -7, is meaningless as a rectangle's width).  So dimensions are 4 ft by 7 ft.

Answer 2

Area problems are quadratic.  First step is to set up your problem.  I suggest drawing a picture:

 

           ______

   (X-3)|         |

           --------

               X

 

If the area of a rectangle is LxW and equals 28 sq ft, you can set up an equation substituting the "X"s for the L and W:

 

    X(X-3) = A = 28

 

Distribute the X and get X^2 -3X = 28.  Set this equal to zero  -->   X^2 - 3X - 28 = 0.

 

There are several ways to get the answer, but by factoring, what two numbers multiply to -28 and add to -3?

 

  (X-7)(X+4) = 0

 

Your two zeroes are X = 7 and X = -4.  Since negative lengths don't make sense, the only reasonable answer is 7.

 

Plug it back in to the picture above to confirm.