document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=71642>Question 71642</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please help,thank you.\r<BR>\n" );
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document.write( "    The length of a rectangle is 1 cm longer than it's width. If the diagonal of the rectangle is 4cm,what are the dimensions(the length and width)of the rectangle? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #51237 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley75><B>checkley75(3666)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley75><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=51237\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> LENGTH=W+1 & WIDTH=W <BR>\n" );
document.write( "THUS THE DIAGONAL OR HYPOTENUSE (SQUARED 4^2)=L^2+W^2<BR>\n" );
document.write( "L^2+W^2=4^2<BR>\n" );
document.write( "(W+1)^2+W^2=16<BR>\n" );
document.write( "W^2+2W+1+W^2=16<BR>\n" );
document.write( "2W^2+2W+1-16=0<BR>\n" );
document.write( "2W^2+2W-15=0<BR>\n" );
document.write( "USING THE QUADRATIC EQUATION WE GET<BR>\n" );
document.write( "X=(-2+-SQRT[2^2-4*2*-15])/2*2<BR>\n" );
document.write( "X=(-2+-SQRT[4+120])/4<BR>\n" );
document.write( "X=(-2+-SQRT[124])/4<BR>\n" );
document.write( "X=(-2+-11.13)/4<BR>\n" );
document.write( "X=(-2+11.13)/4<BR>\n" );
document.write( "X=9.13/4<BR>\n" );
document.write( "X=2.28 ANSWER<BR>\n" );
document.write( "X=(-2-11.13)/4<BR>\n" );
document.write( "X=-13.13/4<BR>\n" );
document.write( "X=-3.28 ANSWER\r<BR>\n" );
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