document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=38292>Question 38292</A>: <I><SPAN STYLE=\"word-break: break-all;\"> In class, we are learning how to use quadratic techniques to solve polynomial equations.  I understand this, but in the homework there is two problems that have negative problems.  I could not figure out how to do this.  For example, here is one of the problems:\r<BR>\n" );
document.write( "\n" );
document.write( "y^(-1)-8y^(-1/2)+12=0  \r<BR>\n" );
document.write( "\n" );
document.write( "I have tried many ways to do this, but can't figure it out.  Thank you so much.  I look forward to your replay so that I may understand how to do the other problems. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #23800 by </B><A HREF=/tutors/aboutme.mpl?userid=ilana><B>ilana(307)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ilana><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=23800\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Okay, this one is a bit tricky.  Luckily, there's a trick!<BR>\n" );
document.write( "We want it to be of the form Ax^2+Bx+C=0.  So, try letting x^2=y^(-1).  Then x=y^(-1/2).  We are lucky in that this works perfectly- then we get x^2-8x+12=0.<BR>\n" );
document.write( "Now solve for x by factoring to get x=2,6.<BR>\n" );
document.write( "Now, make sure to remember we want y, not x, so if x=y^(-1/2), y=(y(-1/2))^(-2)=x^(-2), so y=1/4 and 1/36.<BR>\n" );
document.write( "You can test these values in the original equation and see they work. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );