document.write( "Question 125483: How do i solve (3x-5)squard=12 by using the root propert? \n" ); document.write( "
Algebra.Com's Answer #91925 by josmiceli(19441)\"\" \"About 
You can put this solution on YOUR website!
\"%283x+-+5%29%5E2+=+12\"
\n" ); document.write( "Take the square root of both sides
\n" ); document.write( "There are 2 square roots, + and -
\n" ); document.write( "\"%283x+-+5%29+=+sqrt%2812%29\"
\n" ); document.write( "\"3x+-+5+=+2%2Asqrt%283%29\"
\n" ); document.write( "\"x+=+%282%2Asqrt%283%29+%2B+5%29+%2F+3\" answer
\n" ); document.write( "-------------------------------------
\n" ); document.write( "\"-%283x+-+5%29+=+sqrt%2812%29\"
\n" ); document.write( "\"-3x+%2B+5+=+2%2Asqrt%283%29\"
\n" ); document.write( "\"x+=+%285+-+2%2Asqrt%283%29%29+%2F+3\" answer
\n" ); document.write( "-------------------------------------
\n" ); document.write( "check the answers
\n" ); document.write( "\"%283x+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%283%2A%28%282%2Asqrt%283%29+%2B+5%29+%2F+3%29+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%282%2Asqrt%283%29+%2B+5+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%282%2Asqrt%283%29%29%5E2+=+12\"
\n" ); document.write( "\"4%2A3+=+12\"
\n" ); document.write( "\"12+=+12\"
\n" ); document.write( "OK
\n" ); document.write( "--------------------
\n" ); document.write( "\"%283x+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%283%2A%28%285+-+2%2Asqrt%283%29%29+%2F+3%29+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%285+-+2%2Asqrt%283%29+-+5%29%5E2+=+12\"
\n" ); document.write( "\"%28-2%2Asqrt%283%29%29%5E2+=+12\"
\n" ); document.write( "\"4%2A3+=+12\"
\n" ); document.write( "\"12+=+12\"
\n" ); document.write( "OK
\n" ); document.write( "
\n" );