document.write( "Question 224569: Please help\r
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\n" ); document.write( "\n" ); document.write( "sq root of 1-3x = x+1\r
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\n" ); document.write( "\n" ); document.write( "I am supposed to solve the radical equation for the given variable.
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Algebra.Com's Answer #167748 by Earlsdon(6294)\"\" \"About 
You can put this solution on YOUR website!
Given:\"sqrt%281-3x%29+=+x%2B1\" solve for x:
\n" ); document.write( "\"sqrt%281-3x%29+=+x%2B1\" Square both sides of the equation.
\n" ); document.write( "\"%28sqrt%281-3x%29%29%5E2+=+%28x%2B1%29%5E2\"
\n" ); document.write( "\"1-3x+=+x%5E2%2B2x%2B1\" Add 3x to both sides.
\n" ); document.write( "\"1-3x%2B3x+=+x%5E2%2B2x%2B3x%2B1\" Simplify.
\n" ); document.write( "\"1+=+x%5E2%2B5x%2B1\" Subtract 1 from both sides.
\n" ); document.write( "\"0+=+x%5E2%2B5x\" Factor an x from the right side.
\n" ); document.write( "\"0+=+x%28x%2B5%29\" Apply the \"zero product\" rule.
\n" ); document.write( "\"x+=+0\" or \"x%2B5+=+0\"
\n" ); document.write( "If \"x%2B5+=+0\" then \"x+=+-5\"
\n" ); document.write( "So the solutions are:
\n" ); document.write( "\"x+=+0\"
\n" ); document.write( "\"x+=+-5\"
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