document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=122433>Question 122433</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The length of a rectangle measures 4 cm less than 4 times the width. If the area is 31.0 cm^2, find the dimensions.  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #89905 by </B><A HREF=/tutors/aboutme.mpl?userid=oscargut><B>oscargut(2103)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=oscargut><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=oscargut><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=89905\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let the length L and the width W then LW=31 (area is 31.0 cm^2)\r<BR>\n" );
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document.write( "and   L=4W-4  (The length of a rectangle measures 4 cm less than 4 times the width)\r<BR>\n" );
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document.write( "then L=4(W-1) then 4(W-1)W=31 then 4W^2-4W-31=0 then W=(4+sqrt(16+16(31)))/8<BR>\n" );
document.write( "or W=(4-sqrt(16+16(31)))/8  then solutions are W=(4+16sqrt(2))/8 or W=(4-16sqrt(2))/8 as the second solution is <0 the answer is W=(4+16sqrt(2))/8 if you simplify \r<BR>\n" );
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document.write( " W = 1/2+2sqrt(2) and<BR>\n" );
document.write( " L = 4(W-1)= 4(-1/2+2sqrt(2)) = -2+8sqrt(2) </SPAN>\n" );
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