document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=729758>Question 729758</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find Three consecutive odd integers such that twice the product of the second and third is 1 more than the opposite of the first. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #806414 by </B><A HREF=/tutors/aboutme.mpl?userid=CubeyThePenguin><B>CubeyThePenguin(3113)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CubeyThePenguin><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=CubeyThePenguin><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=806414\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> consecutive odd integers: (x-2), x, (x+2)\r<BR>\n" );
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document.write( "2(x)(x+2) = 1 - (x-2)<BR>\n" );
document.write( "2x^2 + 4x = -x + 3<BR>\n" );
document.write( "2x^2 + 5x - 3 = 0<BR>\n" );
document.write( "(x + 3)(2x - 1) = 0<BR>\n" );
document.write( "x = -3, x = 1/2\r<BR>\n" );
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document.write( "x is an integer, so the integers are -5, -3, -1. </SPAN>\n" );
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