document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1023131>Question 1023131</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Explain how to solve X=(y+2)2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #638660 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=638660\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i presume you mean exponentiation, which is modeled using the exponentiation symbol of ^.\r<BR>\n" );
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document.write( "based on that, x = (y+2)2 would be modeled as x = (y+2)^2.\r<BR>\n" );
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document.write( "assuming that's what you meant, then you would take the square root of both sides of the equation to get plus or minus sqrt(x) = y+2.\r<BR>\n" );
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document.write( "you would then subtract 2 from both sides of the equation to get plus or minus sqrt(x) - 2 = y.\r<BR>\n" );
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document.write( "this can also be shown as y = - 2 plus or minus sqrt(x).\r<BR>\n" );
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document.write( "to confirm that this is the equivalent of the original equation, then do the following:\r<BR>\n" );
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document.write( "start with y = -2 plus or minus sqrt(x).\r<BR>\n" );
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document.write( "add 2 to both sides of this equation to get y + 2 = plus or minus sqrt(xj).\r<BR>\n" );
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document.write( "square both sides of this equation to get (y+2)^2 = (sqrt(x))^2 or (-sqrt(x))^2.\r<BR>\n" );
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document.write( "in either case, you get (y+2)^2 = x.\r<BR>\n" );
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document.write( "when you deal with square roots, the following rules apply.\r<BR>\n" );
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document.write( "if the problem states that y^2 = 4, then the solution is y = plus or minus 2.\r<BR>\n" );
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document.write( "is the problem states that y = sqrt(4), then the solution is y = 2, not plus or minus 2.\r<BR>\n" );
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document.write( "it's a subtle difference that can be difficult to understand.\r<BR>\n" );
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document.write( "here's a reference that discusses it.\r<BR>\n" );
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document.write( "<a  rel=nofollow HREF= \"https://www.mathsisfun.com/algebra/square-root.html\" target = \"_blank\">https://www.mathsisfun.com/algebra/square-root.html</a>\r<BR>\n" );
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