document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199265>Question 1199265</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for x.\r<BR>\n" );
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document.write( "2x^(2/3) - 5x^(1/3) - 3 = 0\r<BR>\n" );
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document.write( "Using the Diamond Method, I got the following:\r<BR>\n" );
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document.write( "2x^(2/3) - 6x^(1/3) + x^(1/3) - 3 = 0\r<BR>\n" );
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document.write( "I then applied factoring by group.\r<BR>\n" );
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document.write( "2x^(1/3)[x^(1/3) - 3] \r<BR>\n" );
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document.write( "Setting each factor to 0 and solving for x, I got the following \r<BR>\n" );
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document.write( "values: 0, 27\r<BR>\n" );
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document.write( "According to the textbook, x = 27 is one of the solutions.\r<BR>\n" );
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document.write( "The second value of x should be -1/8. \r<BR>\n" );
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document.write( "I did everything I could think of to find the second value of x but to no avail.\r<BR>\n" );
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document.write( "How do I find x = -1/8?\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 #833059 by </B><A HREF=/tutors/aboutme.mpl?userid=ikleyn><B>ikleyn(52790)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ikleyn><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=ikleyn><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=833059\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .\r<BR>\n" );
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document.write( "Solve for x   &nbsp;&nbsp;&nbsp;&nbsp;2x^(2/3) - 5x^(1/3) - 3 = 0<BR>\n" );
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document.write( "Prepare your equation to factoring using grouping-regrouping &nbsp;(in the same way as you did in your post)\r\n" );
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document.write( "    2x^(2/3) - 6x^(1/3) + x^(1/3) - 3 = 0\r\n" );
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document.write( "Now factor left side step by step\r\n" );
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document.write( "    2x^(1/3)*(x^(1/3)-3) + (x^(1/3) - 3) = \r\n" );
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document.write( "    (x^(1/3)-3)*(2x^(1/3)+1) = 0.\r\n" );
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document.write( "From this factored form you have EITHER first parentheses expression is zero, \r\n" );
document.write( "OR the second parentheses expression is zero.\r\n" );
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document.write( "(a)  Case  x^(1/3)-3 = 0.\r\n" );
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document.write( "     It gives  x^(1/3) = 3,  which implies  x = 3^3 = 27.\r\n" );
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document.write( "(b)  Case  2x^(1/3)+1 = 0.\r\n" );
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document.write( "     It gives  2x^(1/3) + 1  0,  which implies  x^(1/3) = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2F2 ALIGN=MIDDLE  ALT=\"-1%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >,  x = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-1%2F2%29%5E3 ALIGN=MIDDLE  ALT=\"%28-1%2F2%29%5E3\"   DISABLED_style= \"max-width:90%; height:auto;\" > = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2F8 ALIGN=MIDDLE  ALT=\"-1%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r\n" );
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document.write( "Solved.\r<BR>\n" );
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document.write( "The answers/(the solutions) &nbsp;&nbsp;are  &nbsp;&nbsp;x= 27  &nbsp;&nbsp;and/or  &nbsp;&nbsp;x= <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=-1%2F8 ALIGN=MIDDLE  ALT=\"-1%2F8\"   DISABLED_style= \"max-width:90%; height:auto;\" >,  &nbsp;&nbsp;same as in your textbook.\r<BR>\n" );
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document.write( "Victory (!)\r<BR>\n" );
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document.write( "In your post, your very first step was correct, but your factoring after that was wrong.\r<BR>\n" );
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