Question 1160435
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Let w be the width; then the length is 2w+9.<br>
The area (length times width) is greater than or equal to 45 square inches:<br>
{{{w(2w+9)>=45}}}
{{{2w^2+9w-45>=0}}}
{{{(w-3)(2w+15)>= 0}}}
{{{w>=3}}} or {{{w<= -15/2}}}<br>
Clearly the negative solution makes no sense in the problem.  So<br>
ANSWER: The width can be anything 3 inches or greater<br>
NOTE: That result can be obtained in a few seconds with a bit of trial and error and simple arithmetic....<br>