document.write( "Question 705205: Solve the equation. Identify any extraneous solutions. a=√(-2a)\r
\n" ); document.write( "\n" ); document.write( "A. –2 is a solution of the original equation. 0 is an extraneous solution. \r
\n" ); document.write( "\n" ); document.write( "B. 0 and –2 are solutions of the original equation. \r
\n" ); document.write( "\n" ); document.write( "C. 0 is a solution of the original equation. –2 is an extraneous solution. \r
\n" ); document.write( "\n" ); document.write( "D. 2 is a solution of the original equation. 0 is an extraneous solution.
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Algebra.Com's Answer #434468 by KMST(5328)\"\" \"About 
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The answer is
\n" ); document.write( "\"highlight%28C%29\". 0 is a solution of the original equation. –2 is an extraneous solution.
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\n" ); document.write( "\"a=sqrt%28-2a%29\"
\n" ); document.write( "If you square both sides, you get the equation
\n" ); document.write( "\"a%5E2=-2a\"
\n" ); document.write( "That equation has all the solutions that \"a=sqrt%28-2a%29\" has,
\n" ); document.write( "and then some.
\n" ); document.write( "The solutions for \"a%5E2=-2a\" are \"a=0\" and \"a=-2\".
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\n" ); document.write( "When we try them on \"a=sqrt%28-2a%29\",
\n" ); document.write( "for \"a=0\" \"sqrt%28-2a%29=0\".
\n" ); document.write( "Substituting into the equation \"a=sqrt%28-2a%29\",
\n" ); document.write( "that give \"0=0\" which verifies \"a=sqrt%28-2a%29\",
\n" ); document.write( "so \"a=0\" is a solution of the original equation.
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\n" ); document.write( "However, \"a=-2\" makes
\n" ); document.write( "\"sqrt%28-2a%29=sqrt%28-2%28-2%29%29\" --> \"sqrt%28-2a%29=sqrt%284%29\" --> \"sqrt%28-2a%29=2\",
\n" ); document.write( "so substituting \"a=-2\" into the equation \"a=sqrt%28-2a%29\"
\n" ); document.write( "gives \"-2=2\", which is not true.
\n" ); document.write( "So \"a=-2\" is an extraneous solution.
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