document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=877947>Question 877947</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 2x^2+3x-9=0 Complete the Square.\r<BR>\n" );
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document.write( "I know how to do it all the way up till when it should look like this: x^2+ (3x/2)+(9/16)=(81/4)+(9/16). It's adding the fractions to the right that doesn't make sense. I get 90/64, but mymathlab tells me i should have 81/16. **ugh* my mathlab makes everything more complicated by making me square root and square and cube fractions, and not put it in decimal form, so some help would be much obliged. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #529618 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=529618\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Easier would be to factor 2 out first: 2(x^2 + (3/2)x - 9/2)\r<BR>\n" );
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document.write( "Since half of 3/2 is 3/4, we want it to look something like 2(x + 3/4)^2 + constant. 2(x + 3/4)^2 has a constant term of 2*(3/4)^2 = 9/8, and we want 9, so add 63/8.\r<BR>\n" );
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document.write( "Completed square form: 2(x + 3/4)^2 + 63/8 (can be checked via Wolfram Alpha)<BR>\n" );
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