document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=39425>Question 39425</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am having trouble solving absolute value inequalities which involves exponents.\r<BR>\n" );
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document.write( "1. |x - 1| = |x to the second power - 2x +1|\r<BR>\n" );
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document.write( "2. |x to the second power + x - 2| < x + 3<BR>\n" );
document.write( "Your help will be  appreciated. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #25006 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><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=25006\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi,\r<BR>\n" );
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document.write( "There are quite a few ways to solve these, and I don't know which you prefer, so we'll do the first one geometrically, and the second one with a bit of algebra.\r<BR>\n" );
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document.write( "You may not know this but the 'absolute value' operator as you call it, is actually a distance operator. The way to think about it is like this\r<BR>\n" );
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document.write( "|A-B|=D(A, B)\r<BR>\n" );
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document.write( "The distance from A to B is |A-B|. For example |5-3|=|3-5|=2, if you draw a number line and examine the distance from 3 to 5, or from 5 to 3 you will see that it is 2.\r<BR>\n" );
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document.write( "The equation |x|=1, can be written as |x-0|=1 so it really says the distance between x and 0 is 1. There are two points on the number line that are a distance of 1 from 0, and they are 1, and -1. Which as we both know are the solutions to |x|=1.\r<BR>\n" );
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document.write( "Anyway, what we are dealing with is the equation |x-1|=|x^2-2x+1|. We can factorise x^2-2x+1 as (x-1)(x-1). So we have |x-1|=|(x-1)^2| or (because abs can move through products) |x-1|=|x-1|^2.\r<BR>\n" );
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document.write( "As I said above |x-1| is the distance between x and 1. So let's call it d. We have the equation d=d^2. The solutions to this are clearly d=0 and d=1. The only point that is a distance of 0 from 1 is one itsself, so x=1 is a solution. The points which are a distance of 1 from 1 are 0, and 2. So the solutions to this are x=0, 1, and 2.\r<BR>\n" );
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document.write( "Hopefully I didn't lose you in all of that. If it helps, try drawing number lines.\r<BR>\n" );
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document.write( "Now let's get into some real algebra and solve the inequality.\r<BR>\n" );
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document.write( "If we were give something like |f|=g, then we know that we can write that as two equations f=g, and -f=g. The same is true with inequalities, so <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?|f|<g\"> produces two inequalities <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?f<g\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-f<g.\">\r<BR>\n" );
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document.write( "Doing the same with this question gives the two inequalities\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?x^2+x-2<x+3\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-x^2-x+2<x+3\">\r<BR>\n" );
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document.write( "Now, quadratic inequalities are a bit of a pain, because the sign changes if we multiply/divide by a negative number. To get past this we rewrite the inequality and consider sign changes.\r<BR>\n" );
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document.write( "The first inequality can be written as\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?x^2 - 5 < 0\">, so has sign changes at <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?x=\pm\sqrt{5}\"> We have to consider the truthfulness of the inequality ecah side of the sign changes. If we take a number way bigger than <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\sqrt{5}\"> say for example x=10, we can see that <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?10^2-5<0\"> is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?95<0\"> which is false. Similarly if we choose a number way below <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-\sqrt{5}\"> say -10, we can again see that the inequality doesn't hold. Choosing a number in between <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-\sqrt{5}\"> and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\sqrt{5}\"> like 0 we can see that the inequality does hold <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?0-5<0\">. So <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-\sqrt{5} < x < \sqrt{5}\">.\r<BR>\n" );
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document.write( "We also need to check the other inequality. Rearraging that gives\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?x^2+2x+1>0\">\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?(x+1)^2>0\">\r<BR>\n" );
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document.write( "This is always true for any (real) value of x, so doesn't matter.\r<BR>\n" );
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document.write( "So we finally have <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?-\sqrt{5} < x < \sqrt{5}\"> or more simply <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?|x|<\sqrt{5}\">\r<BR>\n" );
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document.write( "Again, I hope I didn't lose you in all of that. The only hard part there is solving quadratic inequalities, which you should have learnt before attempting such inequalities so you should be ok.\r<BR>\n" );
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document.write( "If you want any more help with this or just a clarification, please write back.\r<BR>\n" );
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document.write( "Hope that helps,<BR>\n" );
document.write( "Kev </SPAN>\n" );
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