document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=588089>Question 588089</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Problem:  Find the inverse of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+y+=+x%5E2+%2B+sqrt+%28x%29+ ALIGN=MIDDLE  ALT=\"+y+=+x%5E2+%2B+sqrt+%28x%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "I've fought the good fight, working on this on and off for two weeks.\r<BR>\n" );
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document.write( "For x^2 + x^1/2 = y, the equivalent is x = x^4 - 2x^2y + y^4.  I cannot figure out what algebraic manipulative trick(s) to use to separate x, and render it as ONE X term in terms of y. [x = f(y)]  Do you have a bag of such applicable tricks?\r<BR>\n" );
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document.write( "Since I can't manage that, here's another approach:\r<BR>\n" );
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document.write( "x^1/2 = y - x^2   and -(x^1/2) = X^2 - y.\r<BR>\n" );
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document.write( "While x^1/2 is not = -(x^1/2), (x^1/2)^2 = x and [-(x^1/2)]^2 = x.\r<BR>\n" );
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document.write( "Therefore:  x = (y - x^2)^2  and x = (x^2-y)^2.\r<BR>\n" );
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document.write( "Now take the square root of each of the above:\r<BR>\n" );
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document.write( "x^1/2 = y - x^2        and x^1/2 = x^2 - y. \r<BR>\n" );
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document.write( "(Of course, the sq rts can also have a \"-\" sign.)  \r<BR>\n" );
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document.write( "Therefore:  y - x^2 = x^2 - y.\r<BR>\n" );
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document.write( "Is my logic correct so far?\r<BR>\n" );
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document.write( "(HOWEVER, in the expression above, let x = 1, and y>1.  \r<BR>\n" );
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document.write( "If y = 4, 4 - 1 NOT = 1 - 4.)\r<BR>\n" );
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document.write( "  \r<BR>\n" );
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document.write( "In any event, if my logic holds up, it seems that I can write:\r<BR>\n" );
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document.write( "-2x^2 = -2y            AND       2x^2 = 2y.\r<BR>\n" );
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document.write( "Each of these is a quadratic.  So, it seems that I should be able to use\r<BR>\n" );
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document.write( "ax^2 + bx + c = (d)y\r<BR>\n" );
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document.write( "inputing known values of x and y  [x = 1, y = 2;   x = 4, y = 18], and come up with a specific quadratic equation of some form of:\r<BR>\n" );
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document.write( " x = -b/2a +/- [(sqrt b^2 + y - 4ac)/2a]\r<BR>\n" );
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document.write( "I'll leave it at that.  Thanks so much.  Cheers!<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 #374374 by </B><A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><B>ankor@dixie-net.com(22740)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ankor@dixie-net.com><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=ankor@dixie-net.com><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=374374\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Problem: Find the inverse of <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=y+=+x%5E2+%2B+sqrt%28x%29 ALIGN=MIDDLE  ALT=\"y+=+x%5E2+%2B+sqrt%28x%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "You seem to have exhausted the usual methods, so I tried this approach<BR>\n" );
document.write( "I don't think there is an inverse, anyway<BR>\n" );
document.write( "The table for the original equation<BR>\n" );
document.write( " x | y<BR>\n" );
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document.write( " 1 | 2<BR>\n" );
document.write( " 4 |18<BR>\n" );
document.write( " 9 |84<BR>\n" );
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document.write( "Swapping x & y<BR>\n" );
document.write( " x | y<BR>\n" );
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document.write( " 2 | 1<BR>\n" );
document.write( "18 | 4<BR>\n" );
document.write( "84 | 9<BR>\n" );
document.write( "After some tedious math, solving for a, b, c in the y = ax^2+bx+c, I came up with an even more tedious equation<BR>\n" );
document.write( "y = -.005452x^2 + .4952x + .5759<BR>\n" );
document.write( "A graph kind of shows an inverse, but negative value probably make it invalid<BR>\n" );
document.write( "plus the quadratic equation will curve back towards 0<BR>\n" );
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document.write( "After I finished this I thought it was kind of silly, and was not going to bother sending it, but you can just toss in the usual place of rejected stuff. <BR>\n" );
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