document.write( "Question 2598: solve the questions give below by factorization
\n" ); document.write( " (1)3x2+x=1 (2)4x2+9=12x (3)5x(x+2)=4(3x=1) (4)2x+7=x2 (5)x2-3x=3x=7 (6)3(4x-1)=x2
\n" ); document.write( "

Algebra.Com's Answer #1114 by longjonsilver(2297)\"\" \"About 
You can put this solution on YOUR website!
by pure factorisation? \r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "1) cannot be factorised easily. You would have to do the \"complete the square\" method.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "2) \"4x%5E2+-+12x+%2B+9+=+0+\"
\n" ); document.write( "(2x - 3)(2x - 3) = 0
\n" ); document.write( "so 2x-3 = 0
\n" ); document.write( "2x = 3
\n" ); document.write( "x = 3/2\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "3) 5x(x+2) = 4(3x+1) --> + or - on the last term, seeing as you have \"=\"?\r
\n" ); document.write( "\n" ); document.write( "\"5x%5E2+%2B+10x+=+12x+%2B+4\"
\n" ); document.write( "\"5x%5E2+-+2x+-+4+=+0\"\r
\n" ); document.write( "\n" ); document.write( "this does not factorise easily. Again, do you want the \"complete the square\" method?\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "4) \"2x%2B7+=+x%5E2\"
\n" ); document.write( "\"x%5E2+-+2x+-+7\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Same as 3).\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "5) \"x%5E2+-+6x+-+7\" Again you have 2 \"=\" signs..which is wrong?\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "6) \"12x-3+=+x%5E2\"
\n" ); document.write( "\"x%5E2+-+12x+%2B+3+=+0\"\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "Again this does not factorise easily.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "jon.
\n" ); document.write( "
\n" ); document.write( "
\n" );