document.write( "Question 46370: Solve and check solution\r
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\n" ); document.write( "\n" ); document.write( "sqrt (2y + 7) + 4 = y\r
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\n" ); document.write( "\n" ); document.write( " Thank You!!!
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Algebra.Com's Answer #30759 by Earlsdon(6294)\"\" \"About 
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Solve and check:
\n" ); document.write( "\"sqrt%282y%2B7%29%2B4+=+y\" Subtract 4 from both sides.
\n" ); document.write( "\"sqrt%282y%2B7%29+=+y-4\" Now square both sides.(Caution! Squaring may introduce an extraneous root)
\n" ); document.write( "\"2y%2B7+=+y%5E2-8y%2B16\" Collect like-terms.
\n" ); document.write( "\"y%5E2-10y%2B9+=+0\" Solve by factoring.
\n" ); document.write( "\"%28y-9%29%28y-1%29+=+0\" Apply the zero products principle.
\n" ); document.write( "\"y-9+=+0\" and/or \"y-1+=+0\"\r
\n" ); document.write( "\n" ); document.write( "The roots are(?):
\n" ); document.write( "y = 9
\n" ); document.write( "y = 1\r
\n" ); document.write( "\n" ); document.write( "Check y=9
\n" ); document.write( "\"sqrt%282%289%29%2B7%29%2B4+=+sqrt%2825%29%2B4\" = \"5%2B4+=+9\" Checks! or \"-5%2B4+=+-1\"
\n" ); document.write( "The last solution is an extraneous solution introduced by squaring. Remember that\"sqrt%2825%29+=+%2B5\" or \"-5\"
\n" ); document.write( "Check y=1
\n" ); document.write( "\"sqrt%282%281%29%2B7%29%2B4+=+sqrt%289%29%2B4\" = \"3%2B4+=+7\" or \"-3%2B4+=+1\" The first solution doesn't check. Extraneous solution was introduced when squaring during the solving process.
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