document.write( "Question 62209: sqrt2x+4=Sqrt(x+3)+1 \n" ); document.write( "

Algebra.Com's Answer #42969 by joyofmath(189) $\"\"$ $\"About$

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$\"sqrt%282x%2B4%29=sqrt%28x%2B3%29%2B1\"$ \r
\n" ); document.write( "\n" ); document.write( "Square both sides: $\"2x%2B4+=+%28x%2B3%29%2B2sqrt%28x%2B3%29%2B1\"$ .
\n" ); document.write( "Subtract x+4 from both sides: $\"x+=+2sqrt%28x%2B3%29\"$ .
\n" ); document.write( "Square both sides: $\"x%5E2+=+4%28x%2B3%29\"$ .
\n" ); document.write( "Turn the equation into a quadratic: $\"x%5E2-4x-12+=+0\"$ .
\n" ); document.write( "This factors: $\"%28x-6%29%28x%2B2%29=+0\"$ .
\n" ); document.write( "So, x=6 and x=-2 are possible solutions.\r
\n" ); document.write( "\n" ); document.write( "Let's check the possible solutions:\r
\n" ); document.write( "\n" ); document.write( "When x=6, $\"sqrt%282x%2B4%29=sqrt%2816%29+=+4\"$ and $\"sqrt%28x%2B3%29%2B1+=+sqrt%289%29%2B1+=+4\"$ .
\n" ); document.write( "When x=-2, $\"sqrt%282x%2B4%29=sqrt%280%29+=+0\"$ and $\"sqrt%28x%2B3%29%2B1+=+sqrt%281%29%2B1+=+2\"$ so x= -2 is not an answer.\r
\n" ); document.write( "\n" ); document.write( "So, the only answer is x=6. \n" ); document.write( "

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