document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=55652>Question 55652</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The sum of the reciprocals of two consecutive odd integers is 8/15. Find the integers. Thank you. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #37760 by </B><A HREF=/tutors/aboutme.mpl?userid=aaaaaaaa><B>aaaaaaaa(138)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=aaaaaaaa><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=37760\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> reciprocal of x = the number which multiplied by x gives 1.\r<BR>\n" );
document.write( "\n" );
document.write( "Therefore, we can express the reciprocal of x as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Fx ALIGN=MIDDLE  ALT=\"1%2Fx\"   DISABLED_style= \"max-width:90%; height:auto;\" >.\r<BR>\n" );
document.write( "\n" );
document.write( "So now, let's assign the variables to the values:\r<BR>\n" );
document.write( "\n" );
document.write( "First number = 2x+1<BR>\n" );
document.write( "Second number = 2x+3<BR>\n" );
document.write( "Reciprocal of first = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%282x%2B1%29 ALIGN=MIDDLE  ALT=\"1%2F%282x%2B1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Reciprocal of second = <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%282x%2B3%29 ALIGN=MIDDLE  ALT=\"1%2F%282x%2B3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
document.write( "\n" );
document.write( "The assugnment of the two numbers requires ome explanation. As the problem requires the two numbers to be odd, we use a little trick (that 2x is always even), to determine that 2x+1 and 2x+3 will be always odd.\r<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2F%282x%2B1%29+%2B+1%2F%282x%2B3%29+=+8%2F15 ALIGN=MIDDLE  ALT=\"1%2F%282x%2B1%29+%2B+1%2F%282x%2B3%29+=+8%2F15\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Take the lcm of the denominators (it's <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15%282x%2B1%29%282x%2B3%29 ALIGN=MIDDLE  ALT=\"15%282x%2B1%29%282x%2B3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >):<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15%282x%2B3%29+%2B+15%282x%2B1%29+=+8%282x%2B3%29%282x%2B1%29 ALIGN=MIDDLE  ALT=\"15%282x%2B3%29+%2B+15%282x%2B1%29+=+8%282x%2B3%29%282x%2B1%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=30x+%2B+45+%2B+30x+%2B+15+=+32x%5E2+%2B+64x+%2B+24 ALIGN=MIDDLE  ALT=\"30x+%2B+45+%2B+30x+%2B+15+=+32x%5E2+%2B+64x+%2B+24\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=60x+%2B+60+=+32x%5E2+%2B+64x+%2B+24 ALIGN=MIDDLE  ALT=\"60x+%2B+60+=+32x%5E2+%2B+64x+%2B+24\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=32x%5E2+%2B+4x+-+36+=+0 ALIGN=MIDDLE  ALT=\"32x%5E2+%2B+4x+-+36+=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<TABLE  BORDER=1>\n" );
document.write( "\n" );
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=ax%5E2%2Bbx%2Bc=0 ALIGN=MIDDLE  ALT=\"ax%5E2%2Bbx%2Bc=0\"   DISABLED_style= \"max-width:90%; height:auto;\" > (in our case <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=32x%5E2%2B4x%2B-36+=+0 ALIGN=MIDDLE  ALT=\"32x%5E2%2B4x%2B-36+=+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=x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"x%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%2A32%2A-36=4624 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%284%29%5E2-4%2A32%2A-36=4624\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
document.write( "  <BR>\n" );
document.write( "  Discriminant d=4624 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+4624+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28-4%2B-sqrt%28+4624+%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=x%5B1%5D+=+%28-%284%29%2Bsqrt%28+4624+%29%29%2F2%5C32+=+1 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%284%29%2Bsqrt%28+4624+%29%29%2F2%5C32+=+1\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B2%5D+=+%28-%284%29-sqrt%28+4624+%29%29%2F2%5C32+=+-1.125 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%284%29-sqrt%28+4624+%29%29%2F2%5C32+=+-1.125\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "    <BR>\n" );
document.write( "    Quadratic expression <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=32x%5E2%2B4x%2B-36 ALIGN=MIDDLE  ALT=\"32x%5E2%2B4x%2B-36\"   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=32x%5E2%2B4x%2B-36+=+32%28x-1%29%2A%28x--1.125%29 ALIGN=MIDDLE  ALT=\"32x%5E2%2B4x%2B-36+=+32%28x-1%29%2A%28x--1.125%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 1, -1.125.\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+32%2Ax%5E2%2B4%2Ax%2B-36+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+32%2Ax%5E2%2B4%2Ax%2B-36+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
document.write( "</TABLE><BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Since 1.125 is not an integer, x = 1, first = 3 and second = 5. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );