document.write( "Question 320117: Solve by taking square root of BOTH sides of the equation.\r
\n" ); document.write( "\n" ); document.write( "1) x^2=16\r
\n" ); document.write( "\n" ); document.write( "2) x^2-36=0\r
\n" ); document.write( "\n" ); document.write( "3) (x-4)^2=25\r
\n" ); document.write( "\n" ); document.write( "4) 3(x+4)^2=12 \n" ); document.write( "

Algebra.Com's Answer #229242 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
1) $\"x%5E2=16+\"$
\n" ); document.write( " $\"x=0+%2B-+4\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "2) $\"+x%5E2-36=0\"$
\n" ); document.write( " $\"x%5E2=36\"$
\n" ); document.write( " $\"x=0+%2B-+6\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "3) $\"%28x-4%29%5E2=25++\"$
\n" ); document.write( " $\"x-4=0+%2B-+5\"$
\n" ); document.write( " $\"x=4+%2B-+5\"$
\n" ); document.write( " $\"x=-1\"$ and $\"+x=9+\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( "4) $\"3%28x%2B4%29%5E2=12\"$ \r
\n" ); document.write( "\n" ); document.write( " $\"%28x%2B4%29%5E2=4\"$
\n" ); document.write( " $\"x%2B4=+0+%2B-+2\"$
\n" ); document.write( " $\"x=-4+%2B-+2\"$
\n" ); document.write( " $\"+x=-6\"$ and $\"x=-2\"$ \n" ); document.write( "

\n" );