document.write( "Question 5224: Find real solutions.\r
\n" ); document.write( "\n" ); document.write( "\"x%2B8%2Asqrt%28x%29=0\"\r
\n" ); document.write( "\n" ); document.write( "Thanks Caroline
\n" ); document.write( "

Algebra.Com's Answer #2652 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
you can work this out from this starting point if you are OK with working with powers. If not, a very acceptable solution is to say the following:\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Let y = \"sqrt%28x%29\" --> so \"y%5E2+=+x\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Therefore the equation then becomes \"y%5E2+%2B+8y+=+0\" which can be factorised to y(y+8) = 0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "So either y=0 or y+8 = 0
\n" ); document.write( "So, y=0 or y = -8. \r
\n" ); document.write( "\n" ); document.write( "y is a square root, which cannot be negative, so -8 is not a possible solution. \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Hence, \"sqrt%28x%29+=+0\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Squaring sqrt(x) = 0 gives x=0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "jon.
\n" ); document.write( "
\n" ); document.write( "
\n" );