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)\"\" \"About 
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\"x=sqrt%283x-12%29%2B4\"\r
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\n" ); document.write( "\n" ); document.write( "Isolate the radical:
\n" ); document.write( "\"x-4=sqrt%283x-12%29\"\r
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\n" ); document.write( "\n" ); document.write( "Square both sides:
\n" ); document.write( "\"x%5E2-8x%2B16=3x-12\"\r
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\n" ); document.write( "\n" ); document.write( "Put the quadratic in standard form:
\n" ); document.write( "\"x%5E2-8x-3x%2B16%2B12=0\"
\n" ); document.write( "\"x%5E2-11x%2B28=0\"\r
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\n" ); document.write( "\n" ); document.write( "Since \"-7%2A-4=28\" and \"-7-4=-11\", this quadratic factors, so:
\n" ); document.write( "\"%28x-7%29%28x-4%29=0\"\r
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\n" ); document.write( "\n" ); document.write( "Therefore, by the zero product rule:
\n" ); document.write( "\"x-7=0\" or \"x-4=0\"
\n" ); document.write( "\"x=7\" or \"x=4\"\r
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\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
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\n" ); document.write( "\n" ); document.write( "Is \"7=sqrt%283%2A7-12%29%2B4\" a true statement? \"sqrt%2821-12%29%2B4=sqrt%289%29%2B4=3%2B4=7\" Checks, the equation is true when \"x=7\", so \"x=7\" is a valid solution.\r
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\n" ); document.write( "\n" ); document.write( "Is \"4=sqrt%283%2A4-12%29%2B4\" a true statement? \"sqrt%2812-12%29%2B4=sqrt%280%29%2B4=0%2B4=0\" Checks, the equation is true when \"x=4\", so \"x=4\" is a valid solution.\r
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\n" ); document.write( "\n" ); document.write( "No extraneous roots were introduced, so both solutions are valid.
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