document.write( "Question 633773: Solve
\n" ); document.write( " √(x+1)+5=x+4
\n" ); document.write( "

Algebra.Com's Answer #399148 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
\r
\n" ); document.write( "\n" ); document.write( "Solve
\n" ); document.write( " √(x+1)+5=x+4\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"sqrt%28x+%2B+1%29+%2B+5+=+x+%2B+4\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"sqrt%28x+%2B+1%29+=+x+%2B+4+-+5\" --- Subtracting 5\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"sqrt%28x+%2B+1%29+=+x+-+1\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"%28sqrt%28x+%2B+1%29%29%5E2+=+%28x+-+1%29%5E2\" ------ Squaring both sides of equation\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x+%2B+1+=+x%5E2+-+2x+%2B+1\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x%5E2+-+2x+-+x+%2B+1+-+1+=+0\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x%5E2+-+3x+=+0\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "x(x - 3) = 0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "x = 0 or x - 3 = 0\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"x+=+0\" or \"x+=+3\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Howewer, when substituting 0 for x in original equation, x = 0 proves to be an ENTRANEOUS solution, so the only solution is \"highlight_green%28x+=+3%29\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Send comments and “thank-yous” to “D” at MathMadEzy@aol.com
\n" ); document.write( "
\n" );