document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1048804>Question 1048804</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Please help me solve this equation <BR>\n" );
document.write( " Solve the inequality.<BR>\n" );
document.write( "(x - 5)(x^2 + x + 1) > 0 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #664373 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=664373\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> set the equation equal to 0.<BR>\n" );
document.write( "you get (x-5) * (x^2 + x + 1) = 0<BR>\n" );
document.write( "this is true if (x-5) = 0 and/or if (x^2 + x + 1) = 0\r<BR>\n" );
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document.write( "x-5 = 0 when x = 5.\r<BR>\n" );
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document.write( "this means that x-5 is negative when x < 5 and x-5 is positive when x > 5.\r<BR>\n" );
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document.write( "x^2 + x + 1 never equals 0.<BR>\n" );
document.write( "in fact, it is always positive.\r<BR>\n" );
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document.write( "since x^2 + x + 1 = 0 is a quadratic equation in standard form, then it has a max/min point at x = -b/2a.\r<BR>\n" );
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document.write( "since a = 1 and b = 1 and c = 1, that max/min point will be at x = -1/2.\r<BR>\n" );
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document.write( "when x = -1/2, the max/min point will be at y = (-1/2)^2 - 1/2 + 1 which will be at y =  1/4 - 1/2 + 1 which becomes y = 3/4.\r<BR>\n" );
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document.write( "since the coefficient of the x^2 term is positive, then x = -1/2 is a min point and the equation opens up.\r<BR>\n" );
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document.write( "that says that all values of y = x^2 + x + 1 are positive regardless of the value of x.\r<BR>\n" );
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document.write( "not only are they all positive, but they are all greater than or equal to 3/4.\r<BR>\n" );
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document.write( "since x^2 + x + 1 is always positive and since x-5 is positive when x > 5 and negative when x < 5, then:\r<BR>\n" );
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document.write( "(x-5) * (x^2 + x + 1) is positive when x > 5 because positive times positive always gives a positiv<BR>\n" );
document.write( "e result.\r<BR>\n" );
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document.write( "(x-5) * (x^2 + x + 1) is negative when x < 5 because positive times negative always give a negative result.\r<BR>\n" );
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document.write( "your solution is that (x-5) * (x^2 + x + 1) > 0 when x > 5.\r<BR>\n" );
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document.write( "the following graph confirms that this is true.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2016/091901.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "here's the graph of x-5 by itself.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2016/091902.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( "here's the graph of x^2 + x + 1 by itself.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2016/091903.jpg\" alt=\"$$$\" </>\r<BR>\n" );
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document.write( " </SPAN>\n" );
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