document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=41227>Question 41227</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of an airport is 1 mile more than twice its width. The area is 21 square miles. Identify the length and width of the airport.<BR>\n" );
document.write( "Thank you </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #26586 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=26586\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let w be the width.  Then the length is 2w + 1.<BR>\n" );
document.write( "Area  = length x width so we have<BR>\n" );
document.write( "A = LW<BR>\n" );
document.write( "21 = (2w + 1)w<BR>\n" );
document.write( "21 = 2w^2 + w<BR>\n" );
document.write( "2w^2 + w - 21 = 0<BR>\n" );
document.write( "(2w + 7)(w - 3) = 0<BR>\n" );
document.write( "w = -7/2 or w = 3<BR>\n" );
document.write( "but since the width can't be negative,<BR>\n" );
document.write( "w = 3 miles<BR>\n" );
document.write( "and the length is 7 miles </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );