document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=36065>Question 36065</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve for x: sqaure root(3x+10) minus square root(x+3)=-1\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 #22085 by </B><A HREF=/tutors/aboutme.mpl?userid=Prithwis><B>Prithwis(166)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Prithwis><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.nriassociation.com/BeliveITorNOT/mittal70millionHouse.jpg valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=22085\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> sqrt(3x+10) - sqrt(x+3) = -1<BR>\n" );
document.write( "=> sqrt(3x+10) = sqrt(x+3) - 1<BR>\n" );
document.write( "Square both sides:<BR>\n" );
document.write( "(3x+10) = (x+3) + 1 - 2sqrt(x+3)<BR>\n" );
document.write( "=> 2x+6 = -2sqrt(x+3)<BR>\n" );
document.write( "=> x+3 = -sqrt(x+3)<BR>\n" );
document.write( "Square again:<BR>\n" );
document.write( "(x+3)^2 = (x+3)<BR>\n" );
document.write( "=> (x+3)(x+3-1) = 0<BR>\n" );
document.write( "=> (x+3)(x+2) = 0<BR>\n" );
document.write( "=> x = -3 or x= -2<BR>\n" );
document.write( "Answer: x = -3 or x= -2 (you will need to use both the values in the square root to check if the value within sqaure root is >= 0);  In this case they are.\r<BR>\n" );
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