document.write( "Question 1122133: The length of a rectangle is 5 inches more than twice the width. If the area is to be at least 33 square inches, what are the possibilities for the width? \n" ); document.write( "
Algebra.Com's Answer #738190 by Boreal(15235)\"\" \"About 
You can put this solution on YOUR website!
width=x
\n" ); document.write( "length=x+5
\n" ); document.write( "x(2x+5)>=33
\n" ); document.write( "2^2+5x-33>=0
\n" ); document.write( "(2x+11)(x-3)>=0
\n" ); document.write( "only positive root is at least x=3 inches width and (2x+5)=11 inches length
\n" ); document.write( "the width has to be at least 3 inches.
\n" ); document.write( "
\n" ); document.write( "
\n" );