document.write( "Question 135978: How to solve x=√3x-12 +4 radical sign over 3x-12
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Algebra.Com's Answer #99612 by solver91311(24713)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Isolate the radical: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Square both sides: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Put the quadratic in standard form: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, by the zero product rule: \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Since we had to square both sides of the equation, there is the possibility that we introduced an extraneous root -- one that is a solution to the derived equation but is NOT a solution to the original equation. Check both answers.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Is \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "No extraneous roots were introduced, so both solutions are valid. \n" ); document.write( " |