document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1199216>Question 1199216</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for x.\r<BR>\n" );
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document.write( "x^(3/2) - 3x^(1/2) = 0\r<BR>\n" );
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document.write( "sqrt{x^2} - sqrt{3x} = 0\r<BR>\n" );
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document.write( "After squaring both sides, I got x^3 = 3x.\r<BR>\n" );
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document.write( "x^3 - 3x = 0\r<BR>\n" );
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document.write( "x(x^2 - 3) = 0\r<BR>\n" );
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document.write( "x = 0 and x^2 - 3 = 0\r<BR>\n" );
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document.write( "Solving x^2 - 3 = 0 for x, I got sqrt|3}, -sqrt{3}.\r<BR>\n" );
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document.write( "After checking my answers for x, I concludes that the only two solutions<BR>\n" );
document.write( "for x are 0 and sqrt{3}.\r<BR>\n" );
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document.write( "The textbook answers for x are 0 and 3.\r<BR>\n" );
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document.write( "Why am I wrong?\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 #833007 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=833007\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "The <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E%281%2F2%29%5E%22%22 ALIGN=MIDDLE  ALT=\"3x%5E%281%2F2%29%5E%22%22\"   DISABLED_style= \"max-width:90%; height:auto;\" > is not the same as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283x%29%5E%281%2F2%29%5E%22%22 ALIGN=MIDDLE  ALT=\"%283x%29%5E%281%2F2%29%5E%22%22\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "In the first expression, we apply the 1/2 exponent to only the x; while the second expression has the exponent apply to all of 3x.\r<BR>\n" );
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document.write( "Recall that with PEMDAS, the parenthesis is done first, then exponents, then multiplication.<BR>\n" );
document.write( "P = parenthesis<BR>\n" );
document.write( "E = exponents<BR>\n" );
document.write( "M = multiplication<BR>\n" );
document.write( "D = division<BR>\n" );
document.write( "A = addition<BR>\n" );
document.write( "S = subtraction\r<BR>\n" );
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document.write( "The reason I bring this up is that with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283x%29%5E%281%2F2%29%5E%22%22 ALIGN=MIDDLE  ALT=\"%283x%29%5E%281%2F2%29%5E%22%22\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we compute the 3x first in the parenthesis, then apply the exponent later.<BR>\n" );
document.write( "Whereas with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=3x%5E%281%2F2%29%5E%22%22 ALIGN=MIDDLE  ALT=\"3x%5E%281%2F2%29%5E%22%22\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we do the exponent piece <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E%281%2F2%29%5E%22%22 ALIGN=MIDDLE  ALT=\"x%5E%281%2F2%29%5E%22%22\"   DISABLED_style= \"max-width:90%; height:auto;\" > first before multiplying with the 3 out front.\r<BR>\n" );
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document.write( "-----------------------------\r<BR>\n" );
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document.write( "Here is one way to solve for x.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E%283%2F2%29%5E%22%22+-+3x%5E%281%2F2%29%5E%22%22+=+0 ALIGN=MIDDLE  ALT=\"x%5E%283%2F2%29%5E%22%22+-+3x%5E%281%2F2%29%5E%22%22+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x%5E3%29%5E%281%2F2%29%5E%22%22+-+3x%5E%281%2F2%29%5E%22%22+=+0 ALIGN=MIDDLE  ALT=\"%28x%5E3%29%5E%281%2F2%29%5E%22%22+-+3x%5E%281%2F2%29%5E%22%22+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > Rearrange exponents in the first term. Use the rule a^(b*c) = (a^b)^c.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28x%5E3%29+-+3%2Asqrt%28x%29+=+0 ALIGN=MIDDLE  ALT=\"sqrt%28x%5E3%29+-+3%2Asqrt%28x%29+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > Raising to the 1/2 power is the same as a square root.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28x%5E3%29+=+3%2Asqrt%28x%29 ALIGN=MIDDLE  ALT=\"sqrt%28x%5E3%29+=+3%2Asqrt%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28sqrt%28x%5E3%29%29%5E2+=+%283%2Asqrt%28x%29%29%5E2 ALIGN=MIDDLE  ALT=\"%28sqrt%28x%5E3%29%29%5E2+=+%283%2Asqrt%28x%29%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > Square both sides\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+0 ALIGN=MIDDLE  ALT=\"x+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-3=0 ALIGN=MIDDLE  ALT=\"x-3=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%2B3=0 ALIGN=MIDDLE  ALT=\"x%2B3=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+0 ALIGN=MIDDLE  ALT=\"x+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=3 ALIGN=MIDDLE  ALT=\"x=3\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=-3 ALIGN=MIDDLE  ALT=\"x=-3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Those are the potential solutions. <BR>\n" );
document.write( "But we need to check each of them.\r<BR>\n" );
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document.write( "If you were to plug in x = 0 into the original equation, then,<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0+=+0 ALIGN=MIDDLE  ALT=\"0+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "We end up with a true statement, therefore confirming x = 0 is indeed a root.\r<BR>\n" );
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document.write( "Repeat for x = 3<BR>\n" );
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document.write( "We have confirmed x = 3 is also a root.\r<BR>\n" );
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document.write( "Now try x = -3<BR>\n" );
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document.write( "We see that a negative is under the square root, to produce a complex value in the form a+bi, where <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=i+=+sqrt%28-1%29 ALIGN=MIDDLE  ALT=\"i+=+sqrt%28-1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "So we'll ignore x = -3.\r<BR>\n" );
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document.write( "Another approach to take is to use a factoring method similar to what @greenestamps mentioned.<BR>\n" );
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