document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=828560>Question 828560</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The area of a rectangle is 28 sq inches. The length is 8 more than thrice the width. Find the length and the width.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #499429 by </B><A HREF=/tutors/aboutme.mpl?userid=LinnW><B>LinnW(1048)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=LinnW><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=LinnW><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=499429\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> area = L*W<BR>\n" );
document.write( "28=(3W+8)(W)<BR>\n" );
document.write( "28 = 3W^2+8W<BR>\n" );
document.write( "subtract 28 from each side<BR>\n" );
document.write( "0 = 3W^2+8W-28<BR>\n" );
document.write( "The product of the outer coefficients is 3*(-28) = -84<BR>\n" );
document.write( "The product of 14 and -6 is -84 and their sum is 8<BR>\n" );
document.write( "so we can solve by parts<BR>\n" );
document.write( "3W^2 +14W -6W -28 = 0<BR>\n" );
document.write( "W(3W + 14) -2(3W + 14) + 0<BR>\n" );
document.write( "(W - 2)(3W + 14) = 0<BR>\n" );
document.write( "So W = 2 or W = -14/3<BR>\n" );
document.write( "We cannot have a negative length so W = 2<BR>\n" );
document.write( "length = (3W+8) = (3*3 + 8 ) = 17<BR>\n" );
document.write( "width = 2 and length = 17<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );