document.write( "Question 540815: Find the number that is 3 more than twice the square root. \n" ); document.write( "

Algebra.Com's Answer #353883 by fcabanski(1391) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"x=2sqrt%28x%29%2B3\"$

\n" ); document.write( " $\"x-3=2sqrt%28x%29\"$

\n" ); document.write( "Square both sides

\n" ); document.write( " $\"%28x-3%29%5E2+=+4x\"$

\n" ); document.write( " $\"x%5E2+-+6x+%2B+9+=+4x\"$

\n" ); document.write( " $\"x%5E2+-+10x+%2B+9+=+0\"$

\n" ); document.write( " $\"%28x-9%29%28x-1%29=0\"$

\n" ); document.write( "x-9=0 and x-1=0

\n" ); document.write( "x=9 and x=1

\n" ); document.write( "Try them in the original equation. 9 works. 1 works because recall that $\"sqrt%28x%29\"$ can be positive or negative. If it's negative (-1 * -1 = 1) then the original equation is:
\n" ); document.write( "1 = -2+3 which is correct.
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