document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=91977>Question 91977</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am having a problem with this word problem, I am not good at them and need some help.\r<BR>\n" );
document.write( "\n" );
document.write( "Geometry: The length of a rectangle is 1 cm longer than its width. If the diagonial of the rectangle is 4cm what are the dimensions of the rectangle?\r<BR>\n" );
document.write( "\n" );
document.write( "Thanks a million </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #66891 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=66891\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let w=width, so w+1=length ... the length, width, and diagonal form a right triangle with the diagonal as the hypotenuse\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "by Pythagoras, w^2+(w+1)^2=4^2 ... 2w^2+2w+1=16 ... 2w^2+2w-15=0 ... <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=w=%28-2+%2B-+sqrt%284-%284%29%282%29%28-15%29%29%29%2F%282%282%29%29=%28-2+%2B-+sqrt%28124%29%29%2F4=%28-1+%2B-+sqrt%2831%29%29%2F2 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "so w=2.28 and w=-3.28 ... the negative value is not realistic ... length=3.28 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );