document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=485509>Question 485509</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The side of one square is 3 cm longer than the side of a second square. If the combined area is 65 cm^2, find the lengths of the sides of both squares. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #332049 by </B><A HREF=/tutors/aboutme.mpl?userid=John10><B>John10(297)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=John10><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=John10><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=332049\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let x be the first square's side then other square's side is (x + 3) cm\r<BR>\n" );
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document.write( "The first square area is x^2\r<BR>\n" );
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document.write( "The second square area is (x + 3)^2\r<BR>\n" );
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document.write( "The sum of the areas is: x^2 + (x + 3)^2 = 65   (***)\r<BR>\n" );
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document.write( "Solve (***)\r<BR>\n" );
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document.write( "x^2 + x^2 + 6x + 9 = 65<BR>\n" );
document.write( "2x^2 + 6x - 56 = 0<BR>\n" );
document.write( "x^2 + 3x - 28 = 0<BR>\n" );
document.write( "(x + 7)(x - 4)= 0<BR>\n" );
document.write( "x = -7 or x = 4<BR>\n" );
document.write( "Since the length can NOT be negative,the first square side is 4 cm\r<BR>\n" );
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document.write( "The second square side is 4 + 3 = 7 cm.\r<BR>\n" );
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document.write( "Hope it will help you! John10:) </SPAN>\n" );
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