document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=203173>Question 203173</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solving Quadratics\r<BR>\n" );
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document.write( "2x2 - 4x + 1 = 0<BR>\n" );
document.write( "  \r<BR>\n" );
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document.write( " 8.      3x2 + 2x = 7<BR>\n" );
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document.write( " 9.      x2 + 8x + 9 = -7<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #153288 by </B><A HREF=/tutors/aboutme.mpl?userid=jejess><B>jejess(7)</B></A><img src=\"/images/stars/stars-05.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=jejess><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=153288\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is all about factoring <BR>\n" );
document.write( " for the first one we have 2x^2 -4x +1 =0<BR>\n" );
document.write( " You need to find two number that multiply 2 but add to negative 4<BR>\n" );
document.write( "-2, -2 \r<BR>\n" );
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document.write( "now the question will read like this <BR>\n" );
document.write( "2x^2-2x-2x+1=0<BR>\n" );
document.write( "To factor this one we group the first two terms together and the last two terms together. <BR>\n" );
document.write( "(2x^2-2x)(-2x+1)=0<BR>\n" );
document.write( "Now we factor out what we can from the barkets<BR>\n" );
document.write( "In the first one (2x^2-2x) What is common about the two terms? 2 and x<BR>\n" );
document.write( "so we remove them <BR>\n" );
document.write( "2x(x-1) <BR>\n" );
document.write( "now do the same for the second set of barkets (-2x+1)<BR>\n" );
document.write( "-1 is common <BR>\n" );
document.write( "-(2x-1)<BR>\n" );
document.write( "Now the 2x-1 is common so we rewrite the question as this <BR>\n" );
document.write( "(2x-1)(x-1)=0<BR>\n" );
document.write( "Solve for x by splitting the barkets up <BR>\n" );
document.write( "2x-1 = 0   OR    x-1=0<BR>\n" );
document.write( "slove by rearranging <BR>\n" );
document.write( "2x-1=0               or x-1=0<BR>\n" );
document.write( "2x=1                    x=1<BR>\n" );
document.write( "x=1/2\r<BR>\n" );
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document.write( "If you do not understand factoring and rearraging please say so in next question.\r<BR>\n" );
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document.write( "I will answer the third one and you can try the second your own if you don't understand it please re-post with where you seem to go wrong. ie; factoring.\r<BR>\n" );
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document.write( "The next one is a little tricky because the question is equal to something (-7)<BR>\n" );
document.write( "We must re-write the question then to make it equal zero in order to factor it. <BR>\n" );
document.write( "x^2+8x+9=-7<BR>\n" );
document.write( "now will read this <BR>\n" );
document.write( "x^2+8x+9+7=0  REMEMBER when you move a number or term it changes signs a  postive become a negative and vis versa. <BR>\n" );
document.write( "Collect like terms; so you add 9 and 7 to get 16  <BR>\n" );
document.write( "x^2+8x+16=0<BR>\n" );
document.write( "Now you need to find two numbers that add to 8 but mult. to 16<BR>\n" );
document.write( "4 and 4 and you split the middle term (8x) into these two numbers so the question reads like this now.<BR>\n" );
document.write( "x^2+4x+4x+16=0 <BR>\n" );
document.write( "(x^2 +4x)(4x+16)=0<BR>\n" );
document.write( "x(x+4)4(x+4)=0<BR>\n" );
document.write( "(x+4)(x+4)=0<BR>\n" );
document.write( "or  (x+4)^2<BR>\n" );
document.write( "x must be = to -4<BR>\n" );
document.write( "x+4=0<BR>\n" );
document.write( "x=-4<BR>\n" );
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