document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=50101>Question 50101</A>: <I><SPAN STYLE=\"word-break: break-all;\"> (3/2m+4)=(1/m+2)-2\r<BR>\n" );
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document.write( "Thankyou. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #33284 by </B><A HREF=/tutors/aboutme.mpl?userid=Roric><B>Roric(1)</B></A><img src=\"/images/stars/iconNewId_16x16.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Roric><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=33284\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> For this question, we need to solve the equation for the value of m. To do this, we must get m alone. The parenthesis in the equation are irrelavent and confusing, so lets get rid of them. (3/2)m + 4 = 1/m + 2 - 2\r<BR>\n" );
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document.write( "The +2-2 cancels out, leaving us with (3/2)m + 4 = 1/m. Next we multiply the entire equation by m in order to get rid of the m in the denominator. <BR>\n" );
document.write( "m((3/2)m + 4) = m(1/m) -->(3/2)m^2 + 4m = 1 --> (3/2)m^2 + 4m - 1 = 0.<BR>\n" );
document.write( "We can now use the quadratic equation to solve for m.\r<BR>\n" );
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document.write( " <TR><TD  ALIGN=MIDDLE>Solved by <A  rel=nofollow HREF=http://wiki.algebra.com/index.php/Tutoring_-_pluggable_solvers>pluggable</A> solver: <A  rel=nofollow HREF=quadratic.solver>SOLVE quadratic equation with variable</A></TD></TR>\n" );
document.write( "<TR><TD  ALIGN=LEFT>Quadratic equation <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=am%5E2%2Bbm%2Bc=0 ALIGN=MIDDLE  ALT=\"am%5E2%2Bbm%2Bc=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (in our case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.5m%5E2%2B4m%2B-1+=+0 ALIGN=MIDDLE  ALT=\"1.5m%5E2%2B4m%2B-1+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" >) has the following solutons:<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"m%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  <BR>\n" );
document.write( "  For these solutions to exist, the <U>discriminant</U> <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" > should not be a negative number.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  First, we need to compute the discriminant <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac ALIGN=MIDDLE  ALT=\"b%5E2-4ac\"   DISABLED_style= \"max-width:90%; height:auto;\" >: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%5E2-4ac=%284%29%5E2-4%2A1.5%2A-1=22 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%284%29%5E2-4%2A1.5%2A-1=22\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Discriminant d=22 is greater than zero. That means that there are two solutions: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+x%5B12%5D+=+%28-4%2B-sqrt%28+22+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28-4%2B-sqrt%28+22+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5B1%5D+=+%28-%284%29%2Bsqrt%28+22+%29%29%2F2%5C1.5+=+0.23013858660781 ALIGN=MIDDLE  ALT=\"m%5B1%5D+=+%28-%284%29%2Bsqrt%28+22+%29%29%2F2%5C1.5+=+0.23013858660781\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=m%5B2%5D+=+%28-%284%29-sqrt%28+22+%29%29%2F2%5C1.5+=+-2.89680525327448 ALIGN=MIDDLE  ALT=\"m%5B2%5D+=+%28-%284%29-sqrt%28+22+%29%29%2F2%5C1.5+=+-2.89680525327448\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.5m%5E2%2B4m%2B-1 ALIGN=MIDDLE  ALT=\"1.5m%5E2%2B4m%2B-1\"   DISABLED_style= \"max-width:90%; height:auto;\" > can be factored:<BR>\n" );
document.write( "  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1.5m%5E2%2B4m%2B-1+=+1.5%28m-0.23013858660781%29%2A%28m--2.89680525327448%29 ALIGN=MIDDLE  ALT=\"1.5m%5E2%2B4m%2B-1+=+1.5%28m-0.23013858660781%29%2A%28m--2.89680525327448%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 0.23013858660781, -2.89680525327448.\n" );
document.write( "Here's your graph:<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1.5%2Ax%5E2%2B4%2Ax%2B-1+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1.5%2Ax%5E2%2B4%2Ax%2B-1+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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