SOLUTION: The length of a rectangular field is 15 meters more than its width. If the area is less than 100, what could be the possible dimensions of the field?

Let x be the smaller dimension, in meters (the width).

Then the larger dimension (the length) is (x+15) meters, according to the condition.



They want you determine x in a way to satisfy inequality

    x*(x+15) < 100.


It is equivalent to 

    x^2 + 15x - 100 < 0,


or, in factored form

    (x+20)*(x-5) < 0.


The solution to this inequality is the set

    -20 < x < 5.


But due to the meaning of x, the dimension x must be positive.


Therefore, the final solution is the set of real numbers  0 < x < 5 meters.


ANSWER.  The width must be shorter than 5 meters;  the length is 15 meters greater than the width.