document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=595164>Question 595164</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What answer do I leave in absolute value when the equation of the other part does not factor. Here's the problem I'm talking about:<BR>\n" );
document.write( "|12x^2-x-2|=1<BR>\n" );
document.write( "One side already factored is <BR>\n" );
document.write( "X=3/2 x=-1<BR>\n" );
document.write( "The other side canbit be factored. Do I have to put 0 or just give the ones I got? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #377021 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=377021\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's start from the beginning. <BR>\n" );
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document.write( "I generally work these absolute value problems as two separate problems. I do this in two steps. First, I assign a plus sign to the entire quantity inside the absolute value signs and work that as one problem without using the absolute value signs. Second, I assign a minus sign to the entire quantity inside the absolute value signs and work that as a second problem without the absolute value signs.<BR>\n" );
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document.write( "How does this system work for this problem?  Here we go ...<BR>\n" );
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document.write( "The quantity inside the absolute value signs is (12x^2 - x - 2). Put a + sign in front of this quantity and it becomes just 12x^2 - x - 2. Now ignoring the absolute value signs of the original problem, the original problem is reduced to solving:<BR>\n" );
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document.write( "12x^2 - x - 2 = 1<BR>\n" );
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document.write( "Subtract 1 from both sides to get rid of the 1 on the right side and you get the standard quadratic form:<BR>\n" );
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document.write( "12x^2 - x - 3 = 0<BR>\n" );
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document.write( "The easiest way to solve for x in this equation is to use the quadratic formula which as you should know by now says that for the standard quadratic form:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=ax%5E2+%2B+bx+%2B+c+=+0 ALIGN=MIDDLE  ALT=\"ax%5E2+%2B+bx+%2B+c+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "the answers for x are given by the form:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" > <BR>\n" );
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document.write( "For this problem, a (the multiplier of the x^2 term is +12, b (the multiplier of the x term) is -1, and c (the constant term) is -3. I'll leave you with the job of substituting these three values into the equation for x, but when you do you should get as the answers for x:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%281%2B-sqrt%28145%29%29%2F24 ALIGN=MIDDLE  ALT=\"x+=+%281%2B-sqrt%28145%29%29%2F24\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "which if you convert to decimals results in the two values for x being x = 0.543399774116 and x = -0.460066440783.<BR>\n" );
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document.write( "That's half the answer to this problem. We got these two values for x by assigning a plus sign to the quantity inside the absolute value signs of the original problem.<BR>\n" );
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document.write( "Next, let's assign a minus sign to the quantity inside the absolute value signs and go through a similar exercise. Begin with the minus sign assigned as follows:<BR>\n" );
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document.write( "-(12x2 - x - 2)<BR>\n" );
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document.write( "Ignore the absolute value signs and set this quantity equal to 1 as the original problem requires. This second equation becomes:<BR>\n" );
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document.write( "-(12x2 - x - 2) = 1<BR>\n" );
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document.write( "You can remove the parentheses on the left side by changing the signs of all the terms inside to get:<BR>\n" );
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document.write( "- 12x^2 + x + 2 = 1<BR>\n" );
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document.write( "Again, get rid of the + 1 on the right side by subtracting 1 from both sides to get the standard quadratic form of:<BR>\n" );
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document.write( "- 12x^2 + x + 1 = 0<BR>\n" );
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document.write( "Apply the quadratic formula as we did in the first part. This time a (the multiplier of the x-squared) is -12, b (the multiplier of the x) is +1, and c (the constant) is +1.<BR>\n" );
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document.write( "Substituting these values into the form:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-b+%2B-+sqrt%28++b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-b+%2B-+sqrt%28++b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "results in the two values:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-1+%2B-+sqrt%28++49+%29%29%2F%28-24%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-1+%2B-+sqrt%28++49+%29%29%2F%28-24%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "which, by taking the square root of 49, reduces to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+%28-1+%2B-+7%29%2F%28-24%29+ ALIGN=MIDDLE  ALT=\"x+=+%28-1+%2B-+7%29%2F%28-24%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "When you work this out, you get the two answers for x as:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+-1%2F4 ALIGN=MIDDLE  ALT=\"x+=+-1%2F4\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+=+1%2F3 ALIGN=MIDDLE  ALT=\"x+=+1%2F3\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "So, in summary, for this problem we end up with four answers for x as follows:  <BR>\n" );
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document.write( "x = 0.543399774116,  -0.460066440783, -1/4, and 1/3<BR>\n" );
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document.write( "I hope this process of working the problem as two different problems, one with a plus sign preceding the entire quantity inside the absolute value signs, and one with a minus sign preceding the entire quantity inside the absolute value signs, is something that you can use to help you get through such problems. I found that it helped me to eliminate some confusion with such problems, and it works because it's just a way of applying the basic definition for absolute value. <BR>\n" );
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