document.write( "Question 139453: 3 + sqrt 3x + 1 = x \n" ); document.write( "
Algebra.Com's Answer #101660 by checkley77(12844)\"\" \"About 
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3+sqrt3x+1=x
\n" ); document.write( "sqrt3x=x-4
\n" ); document.write( "now square both sides:
\n" ); document.write( "3x=(x-4)^2
\n" ); document.write( "3x=x^2-8x+16
\n" ); document.write( "x^2-8x-3x+16=0
\n" ); document.write( "x^2-11x+16=0
\n" ); document.write( "usingthe quasratig equation:\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" we get.
\n" ); document.write( "x=(11+-sqrt[-11^2-4*1*16])/2*1
\n" ); document.write( "x=(11+-sqrt[121-64])/2
\n" ); document.write( "x=(11+-sqrt57)/2
\n" ); document.write( "x=(11+-7.55)/2
\n" ); document.write( "x=11+7.55)/2
\n" ); document.write( "x=18.55/2
\n" ); document.write( "x=9.275 answer.
\n" ); document.write( "x=(11-7.55)/2
\n" ); document.write( "x=3.45/2
\n" ); document.write( "x=1.725 answer.
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