document.write( "Question 138197This question is from textbook Intermediate Algebra
\n" ); document.write( ": how to solve the square root of y+3 = y-3 \n" ); document.write( "
Algebra.Com's Answer #100911 by solver91311(24713)\"\" \"About 
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\"sqrt%28y%2B3%29=y-3\"\r
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\n" ); document.write( "\n" ); document.write( "Square both sides (since we are squaring the equation, we have to remember to check for extraneous roots at the end)
\n" ); document.write( "\"y%2B3=%28y-3%29%5E2\"
\n" ); document.write( "\"y%2B3=y%5E2-6y%2B9\"
\n" ); document.write( "\"y%5E2-6y-y%2B9-3=0\"
\n" ); document.write( "\"y%5E2-7y%2B6=0\"\r
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\n" ); document.write( "\n" ); document.write( "The quadratic factors because \"-1%2A-6=6\" and \"-1%2B%28-6%29=-7\"\r
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\n" ); document.write( "\n" ); document.write( "\"%28y-6%29%28y-1%29=0\"\r
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\n" ); document.write( "\n" ); document.write( "Using the zero product rule:\r
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\n" ); document.write( "\n" ); document.write( "\"y-6=0\" => \"y=6\"\r
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\n" ); document.write( "\n" ); document.write( "or\r
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\n" ); document.write( "\n" ); document.write( "\"y-1=0\" => \"y=1\"\r
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\n" ); document.write( "\n" ); document.write( "Check for extraneous roots:
\n" ); document.write( "\"sqrt%286%2B3%29=6-3\"
\n" ); document.write( "\"sqrt%289%29=3\" True\r
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\n" ); document.write( "\n" ); document.write( "\"sqrt%281%2B3%29=1-3\"
\n" ); document.write( "\"sqrt%284%29%3C%3E-2\" (radical sign means the positive square root by convention)
\n" ); document.write( "So \"1\" is an extraneous root.\r
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\n" ); document.write( "\n" ); document.write( "The solution set is {6}
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