document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1196478>Question 1196478</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How do you know that the equation has no real roots?\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #829322 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=829322\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Almost certainly the explanation by tutor @ikleyn is the easiest way to show that the given polynomial has no real roots.<br><BR>\n" );
document.write( "But note that Descartes' rule of signs can also be used.<br><BR>\n" );
document.write( "The given polynomial has 0 sign changes, so the number of positive real roots is 0.<br><BR>\n" );
document.write( "The polynomial with x replaced by -x also has 0 sign changes, so the number of negative real roots is also 0.<br><BR>\n" );
document.write( "Along with the fact that x=0 is obviously not a root, that means the number of real roots is 0.<br><BR>\n" );
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