document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=75079>Question 75079</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle is 1 cm longer than the width.  If the diagonal of the rectangle is 4cm, what are the dimensions (the length and the width) of the rectangle.  \r<BR>\n" );
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document.write( "Please show work. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #53960 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=53960\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> THE DIAGONAL FORMS A TRIANGLE WITH SIDES X, X+1 & 4<BR>\n" );
document.write( "USING THE PATHOGOREAN FORMULA<BR>\n" );
document.write( "A^2+B^2=C^2<BR>\n" );
document.write( "X^2+(X+1)^2=4^2<BR>\n" );
document.write( "X^2+X^2+2X+1=16<BR>\n" );
document.write( "2X^2+2X+1-16=0<BR>\n" );
document.write( "2X^2+2X-15=0<BR>\n" );
document.write( "2X^2+2X-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+-SQRT4+120])/4<BR>\n" );
document.write( "X=(-2+-SQRT124)/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.<BR>\n" );
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