document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=162784>Question 162784</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/9780073533490/algebrahomeworkh>Elementary and Intermediate</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> Winter wheat. While finding the amount of seed needed to plant his three square wheat fields. Hank observed that the side of one field was 1 kilometer longer than the side of the smallest field and that the side of the largest field was 3 kilometers longer than the side of the smallest field. if the total area of the threes fields is 38 square kilometers, then what is the area of each field? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #119994 by </B><A HREF=/tutors/aboutme.mpl?userid=gonzo><B>gonzo(654)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=gonzo><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=119994\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> let s = side of smallest field.<BR>\n" );
document.write( "let m = side of middle field.<BR>\n" );
document.write( "let l = side of largest field.<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "s = s kilometers<BR>\n" );
document.write( "m = s+1 kilometers<BR>\n" );
document.write( "l = s+3 kilometers<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "total area of all 3 fields is 38 square kilometers.<BR>\n" );
document.write( "s^2 + m^2 + l^2 = 38<BR>\n" );
document.write( "since m = s+1 and l = s+3, substitute in equation to get<BR>\n" );
document.write( "s^2 + (s+1)^2 + (s+3)^2 = 38<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "this becomes<BR>\n" );
document.write( "s^2 <BR>\n" );
document.write( "+ s^2 + 2s + 1<BR>\n" );
document.write( "+ s^2 + 6s + 9<BR>\n" );
document.write( "= 38<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "this becomes<BR>\n" );
document.write( "3s^2 + 8s + 10 = 38<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "subtracting 38 from both sides of equation and it becomes<BR>\n" );
document.write( "3s^2 + 8s - 28 = 0<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "factoring this equation yields<BR>\n" );
document.write( "(3s+14) * (s-2) = 0<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "s = 2 was substituted in original equation of s^2 + (s+1)^2 + (s+3)^2 = 38 and is good.<BR>\n" );
document.write( "s = -14/3 was rejected because the side of the field can't be negative.  it was substituted in the original equation and it worked, but it can't be used.<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "smallest field is 2 kilometers per side.<BR>\n" );
document.write( "middle field is 3 kilometers per side.<BR>\n" );
document.write( "largest field is 5 kilometers per side<BR>\n" );
document.write( "2^2 + 3^2 + 5^2 = 4 + 9 + 25 = 38 square kilometers.<BR>\n" );
document.write( "-----<BR>\n" );
document.write( "equation is satisfied.<BR>\n" );
document.write( "area of the smallest field is 4.<BR>\n" );
document.write( "area of the middle field is 9.<BR>\n" );
document.write( "area of the largest field is 25. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );