document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=130272>Question 130272</A>This question is from textbook <A HREF=http://www.amazon.com/exec/obidos/ASIN/1559533404/algebrahomeworkh>Discovering Algebra</A><BR>\n" );
document.write( ": <I><SPAN STYLE=\"word-break: break-all;\"> I am supposed to \"Solve each quadratic equation by completing the square. Leave your answer in radical form.\" I'm not really sure what the radical form is. I don't even know if this question falls under the section of quadratic equation or radicals. My textbook says I need to complete the square or something. The equation is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-4x-8=0 ALIGN=MIDDLE  ALT=\"x%5E2-4x-8=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >. I tried to solve it symbollically (I think..) and I: added the 8 to the zero, square-rooted the whole thing and somehow I ended up with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-4x=_%2B4 ALIGN=MIDDLE  ALT=\"x-4x=_%2B4\"   DISABLED_style= \"max-width:90%; height:auto;\" >. I'm not sure if I'm done or not... Please help!  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #95167 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=95167\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let's start from where you are with this right now and work backwards.\r<BR>\n" );
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document.write( "You said that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28x%5E2-4x%29=sqrt%288%29 ALIGN=MIDDLE  ALT=\"sqrt%28x%5E2-4x%29=sqrt%288%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and therefore <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-4x=4 ALIGN=MIDDLE  ALT=\"x-4x=4\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-4x=-4 ALIGN=MIDDLE  ALT=\"x-4x=-4\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "If <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%288%29=4 ALIGN=MIDDLE  ALT=\"sqrt%288%29=4\"   DISABLED_style= \"max-width:90%; height:auto;\" > then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=8+=+4+%2A+4 ALIGN=MIDDLE  ALT=\"8+=+4+%2A+4\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  We know that isn't true, so the right side of your result is certainly incorrect.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-4x=-3x ALIGN=MIDDLE  ALT=\"x-4x=-3x\"   DISABLED_style= \"max-width:90%; height:auto;\" > and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-3x%29%5E2=9x ALIGN=MIDDLE  ALT=\"%28-3x%29%5E2=9x\"   DISABLED_style= \"max-width:90%; height:auto;\" >, and this is only true for two values of x (specifically 1 and -1).  Therefore, the right side of your result is also incorrect in general.\r<BR>\n" );
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document.write( "So let's go back one more step, the one before you took the square root of both sides.  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-4x=8 ALIGN=MIDDLE  ALT=\"x%5E2-4x=8\"   DISABLED_style= \"max-width:90%; height:auto;\" > is a good start.  But what you need to do now is 'complete the square'  That means that you are going to add a constant to both sides of the equation that makes that left side a perfect square.  The process for this is actually simpler than it sounds.\r<BR>\n" );
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document.write( "Step 1:  Divide both sides of your equation by the coefficient on the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2 ALIGN=MIDDLE  ALT=\"x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > term.  This is very easy for your problem because the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2 ALIGN=MIDDLE  ALT=\"x%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > term coefficient is 1, and you don't have to change anything.\r<BR>\n" );
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document.write( "Step 2:  Take the coefficient on the <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x ALIGN=MIDDLE  ALT=\"x\"   DISABLED_style= \"max-width:90%; height:auto;\" > term and divide it by 2.  Your problem, the coefficient is -4, divided by 2 is -2.\r<BR>\n" );
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document.write( "Step 3:  Square the result of step 2.  For your problem, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28-2%29%5E2=4 ALIGN=MIDDLE  ALT=\"%28-2%29%5E2=4\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
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document.write( "Step 4:  Add the result of step 3 to both sides of your equation.  Your problem:  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-4x%2B4=12 ALIGN=MIDDLE  ALT=\"x%5E2-4x%2B4=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Step 5:  Now that the left side of the equation is a perfect square, you can take the square root of both sides.  First note that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-4x%2B2=%28x-2%29%5E2 ALIGN=MIDDLE  ALT=\"x%5E2-4x%2B2=%28x-2%29%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" > (You can use FOIL on <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-2%29%28x-2%29 ALIGN=MIDDLE  ALT=\"%28x-2%29%28x-2%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > to verify that fact)\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28x-2%29%5E2=12 ALIGN=MIDDLE  ALT=\"%28x-2%29%5E2=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >, so:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-2=sqrt%2812%29 ALIGN=MIDDLE  ALT=\"x-2=sqrt%2812%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x-2=-sqrt%2812%29 ALIGN=MIDDLE  ALT=\"x-2=-sqrt%2812%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=2%2B-sqrt%2812%29 ALIGN=MIDDLE  ALT=\"x=2%2B-sqrt%2812%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "That is a representation of the solution set, in radical form (a radical is the square root sign).  However, there is one more simplification step that you can take.  Remember that <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%28ab%29=sqrt%28a%29%2Asqrt%28b%29 ALIGN=MIDDLE  ALT=\"sqrt%28ab%29=sqrt%28a%29%2Asqrt%28b%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  12 = 4 * 3, so <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sqrt%2812%29=sqrt%284%29sqrt%283%29=2%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"sqrt%2812%29=sqrt%284%29sqrt%283%29=2%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  That means your final answer is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=2%2B2%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"x=2%2B2%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > or <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x=2-2%2Asqrt%283%29 ALIGN=MIDDLE  ALT=\"x=2-2%2Asqrt%283%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Check the answer.  If we did the work correctly, we should be able to substitute either of these values for x into the original equation and have it be a true statement.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282%2B2%2Asqrt%283%29%29%5E2-4%282%2B2%2Asqrt%283%29%29-8=0 ALIGN=MIDDLE  ALT=\"%282%2B2%2Asqrt%283%29%29%5E2-4%282%2B2%2Asqrt%283%29%29-8=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%284%2B8%2Asqrt%283%29%2B12%29-%288%2B8%2Asqrt%283%29%29-8=0 ALIGN=MIDDLE  ALT=\"%284%2B8%2Asqrt%283%29%2B12%29-%288%2B8%2Asqrt%283%29%29-8=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=16%2B8%2Asqrt%283%29-16-8%2Asqrt%283%29=0 ALIGN=MIDDLE  ALT=\"16%2B8%2Asqrt%283%29-16-8%2Asqrt%283%29=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=0=0 ALIGN=MIDDLE  ALT=\"0=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "So, the first answer works.  I'll leave it to you to prove the second one.  Or just trust me. </SPAN>\n" );
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