document.write( "Question 731318: 2 square root of 2x+4 end of square root then minus 5=11
\n" ); document.write( "I dont know what to do with the 2 in the front and where to start. thank you! \n" ); document.write( "

Algebra.Com's Answer #447088 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

You can put this solution on YOUR website!
2 square root of 2x+4 end of square root then minus 5=11
\n" ); document.write( " $\"2%2Asqrt%282x%2B4%29+-+5\"$ = 11
\n" ); document.write( "Add 5 to both sides
\n" ); document.write( " $\"2%2Asqrt%282x%2B4%29\"$ = 16
\n" ); document.write( "Divide both sides by 2
\n" ); document.write( " $\"sqrt%282x%2B4%29+\"$ = 8
\n" ); document.write( "square both sides
\n" ); document.write( "2x + 4 = 64
\n" ); document.write( "subtract 4 from both sides
\n" ); document.write( "2x = 60
\n" ); document.write( "divide both sides by 2
\n" ); document.write( "x = 30
\n" ); document.write( ";
\n" ); document.write( ":
\n" ); document.write( "Check this on a calc: enter $\"2%2Asqrt%282%2830%29%2B4%29\"$ - 5, results is 11 \n" ); document.write( "

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