document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=228412>Question 228412</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 20x^(2/3) - 6x^(1/3) - 2 = 0\r<BR>\n" );
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document.write( "I have tried using my book's suggestion to solve this problem, which is to use a variable, in this case, for (x^1/3) = m. What winds up happening is that I wind up with a factorization problem that doesn't make sense (unfortunately, the text book shows us only the SIMPLEST form of a certain type of problem, leaving you on your own when it comes time to solve something a little more challenging). I am having difficulty grasping the steps and proper execution of this problem and would greatly appreciate your help. Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #169562 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=169562\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Your equation to solve is:\r<BR>\n" );
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document.write( "20x^(2/3) - 6x^(1/3) - 2 = 0 \r<BR>\n" );
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document.write( "If you let y = x^(1/3), then y^2 = x^(1/3)^2 = x^(2/3)\r<BR>\n" );
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document.write( "Your equation would become:\r<BR>\n" );
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document.write( "20y^2 - 6y - 2 = 0\r<BR>\n" );
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document.write( "This might be able to be factored to something like:\r<BR>\n" );
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document.write( "(5y + 1) * (4y - 2)\r<BR>\n" );
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document.write( "To test this out, do the multiplications of the factors as shown below:\r<BR>\n" );
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document.write( "-2 * 1 = -2<BR>\n" );
document.write( "-2 * 5y = -10y<BR>\n" );
document.write( "4y * 1 = 4y<BR>\n" );
document.write( "4y * 5y = 20y\r<BR>\n" );
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document.write( "Add these together and combine like terms and you get:\r<BR>\n" );
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document.write( "20y - 10y + 4y - 2 becomes:<BR>\n" );
document.write( "20y - 6y - 2\r<BR>\n" );
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document.write( "So you have found the factors of:\r<BR>\n" );
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document.write( "(5y + 1) * (4y - 2) = 0\r<BR>\n" );
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document.write( "This results in:\r<BR>\n" );
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document.write( "5y + 1 = 0 which results in y = -.2<BR>\n" );
document.write( "4y - 2 = 0 which results in y = .5\r<BR>\n" );
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document.write( "Now you had made y = x^(1/3) so this equation becomes:\r<BR>\n" );
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document.write( "x^(1/3) = -.2<BR>\n" );
document.write( "and:<BR>\n" );
document.write( "x^(1/3) = .5\r<BR>\n" );
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document.write( "x^(2/3) = x^(1/3)^2 = -.2^ = .04<BR>\n" );
document.write( "and:<BR>\n" );
document.write( "x^(2/3) = x^(1/3)^2 = .5^ = .25\r<BR>\n" );
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document.write( "We hold these values to test against the original equation.\r<BR>\n" );
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document.write( "First values are x^(1/3) = -.2 and x^(2/3) = .04\r<BR>\n" );
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document.write( "Plug into the original equation of:\r<BR>\n" );
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document.write( "20x^(2/3) - 6x^(1/3) - 2 = 0 to get:\r<BR>\n" );
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document.write( "20*.04 - 6*(-.2) - 2 = 0<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( ".8 + 1.2 - 2 = 0<BR>\n" );
document.write( "combine like terms to get:<BR>\n" );
document.write( "0 = 0 which is true so it looks like our first values for x^(1/3) and x^(2/3) are good.\r<BR>\n" );
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document.write( "Our second values are x^(1/3) = .5 and x^(2/3) = .25\r<BR>\n" );
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document.write( "Plug into original equation of :\r<BR>\n" );
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document.write( "20x^(2/3) - 6x^(1/3) - 2 = 0 to get:\r<BR>\n" );
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document.write( "20*.25 - 6*.5 - 2 = 0<BR>\n" );
document.write( "simplify to get:<BR>\n" );
document.write( "5 - 3 - 2 = 0<BR>\n" );
document.write( "combine like terms to get:<BR>\n" );
document.write( "0 = 0\r<BR>\n" );
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document.write( "Looks like our second value for x^(1/3) and x^(2/3) are also good.\r<BR>\n" );
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document.write( "Your answer are that:\r<BR>\n" );
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document.write( "x^(1/3) = -.2<BR>\n" );
document.write( "and:<BR>\n" );
document.write( "x^(1/3) = .5\r<BR>\n" );
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document.write( "To solve for x, you would need to cube both sides of this equation to get:\r<BR>\n" );
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document.write( "x = (-.2)^3 = -.0008<BR>\n" );
document.write( "and:<BR>\n" );
document.write( "x = (.5)^3 = .125\r<BR>\n" );
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document.write( "You can use your calculator to prove that these values are good.\r<BR>\n" );
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document.write( "Your original equation is:\r<BR>\n" );
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document.write( "20x^(2/3) - 6x^(1/3) - 2 = 0 \r<BR>\n" );
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document.write( "If x = -.008, then this equation becomes:\r<BR>\n" );
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document.write( "20*(-.008)^(2/3) - 6*(-.008)^(1/3) - 2 = 0\r<BR>\n" );
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document.write( "This  becomes:\r<BR>\n" );
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document.write( "20* ((-.008)^2)^(1/3) - 6*(-.008)^(1/3) - 2 = 0\r<BR>\n" );
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document.write( "-.008^2 = .000064^(1/3) = .04<BR>\n" );
document.write( "-.008^(1/3) = -.2\r<BR>\n" );
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document.write( "Equation becomes 20 * .04 - 6 * (-.2) - 2 = 0\r<BR>\n" );
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document.write( "This becomes .8 - (-1.2) - 2 = 0\r<BR>\n" );
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document.write( "This becomes .8 + 1.2 - 2 = 0\r<BR>\n" );
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document.write( "This becomes 2 - 2 = 0 which is true so the first value for x is good.\r<BR>\n" );
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document.write( "I'll leave the confirmation of the second value of x to you.\r<BR>\n" );
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document.write( "You're right.\r<BR>\n" );
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document.write( "It was a bear of a problem to solve.\r<BR>\n" );
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document.write( "The key was understanding that y = x^(1/3) and that y^2 equal x^(2/3).\r<BR>\n" );
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document.write( "This was one of the properties of exponents.\r<BR>\n" );
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document.write( "(x^a)^2 = x^a * x^a = x^(a+a) = x^(2a)\r<BR>\n" );
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document.write( "Even knowing this, the problem was difficult.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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