document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=223710>Question 223710</A>: <I><SPAN STYLE=\"word-break: break-all;\"> what is the length of a rectangle is twice its width if the width is increased by one meter and the length is diminished by three meters the area will be 12 square meters  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #167206 by </B><A HREF=/tutors/aboutme.mpl?userid=drj><B>drj(1380)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=drj><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=drj><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=167206\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> What is the length of a rectangle is twice its width if the width is increased by one meter and the length is diminished by three meters the area will be 12 square meters.\r<BR>\n" );
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document.write( "Step 1.  Let w be the width\r<BR>\n" );
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document.write( "Step 2.  Let 2w be the length.\r<BR>\n" );
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document.write( "Step 3.  Let w+1 be the width increased by one meter.\r<BR>\n" );
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document.write( "Step 4.  Let 2w-3 be the length diminished by three meters.\r<BR>\n" );
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document.write( "Step 5.  The Area A=(w+1)(2w-3)=12 since the area will be 12 meters with these changed dimensions.\r<BR>\n" );
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document.write( "Step 6.  Solving the equation in Step 5 will lead to a quadratic equation as follows:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2w%5E2-3w%2B2w-3=12 ALIGN=MIDDLE  ALT=\"2w%5E2-3w%2B2w-3=12\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Subtract 12 from both sides of the equation\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2w%5E2-w-3-12=12-12 ALIGN=MIDDLE  ALT=\"2w%5E2-w-3-12=12-12\"   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=2w%5E2-w-15=0 ALIGN=MIDDLE  ALT=\"2w%5E2-w-15=0\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "Step 7.  To solve, use the quadratic formula given as <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;\" > \r<BR>\n" );
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document.write( "where a=2, b=-1, and c=-15.\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=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=2x%5E2%2B-1x%2B-15+=+0 ALIGN=MIDDLE  ALT=\"2x%5E2%2B-1x%2B-15+=+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" );
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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=%28-1%29%5E2-4%2A2%2A-15=121 ALIGN=MIDDLE  ALT=\"b%5E2-4ac=%28-1%29%5E2-4%2A2%2A-15=121\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
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document.write( "  Discriminant d=121 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--1%2B-sqrt%28+121+%29%29%2F2%5Ca ALIGN=MIDDLE  ALT=\"+x%5B12%5D+=+%28--1%2B-sqrt%28+121+%29%29%2F2%5Ca\"   DISABLED_style= \"max-width:90%; height:auto;\" >.<BR>\n" );
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document.write( "    <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+121+%29%29%2F2%5C2+=+3 ALIGN=MIDDLE  ALT=\"x%5B1%5D+=+%28-%28-1%29%2Bsqrt%28+121+%29%29%2F2%5C2+=+3\"   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-%28-1%29-sqrt%28+121+%29%29%2F2%5C2+=+-2.5 ALIGN=MIDDLE  ALT=\"x%5B2%5D+=+%28-%28-1%29-sqrt%28+121+%29%29%2F2%5C2+=+-2.5\"   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=2x%5E2%2B-1x%2B-15 ALIGN=MIDDLE  ALT=\"2x%5E2%2B-1x%2B-15\"   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=2x%5E2%2B-1x%2B-15+=+2%28x-3%29%2A%28x--2.5%29 ALIGN=MIDDLE  ALT=\"2x%5E2%2B-1x%2B-15+=+2%28x-3%29%2A%28x--2.5%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "  Again, the answer is: 3, -2.5.\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+2%2Ax%5E2%2B-1%2Ax%2B-15+%29 ALIGN=MIDDLE  ALT=\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+2%2Ax%5E2%2B-1%2Ax%2B-15+%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ></TD></TR>\n" );
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document.write( "Selecting the positive solution given as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=w=3 ALIGN=MIDDLE  ALT=\"w=3\"   DISABLED_style= \"max-width:90%; height:auto;\" >, then <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=2w=6 ALIGN=MIDDLE  ALT=\"2w=6\"   DISABLED_style= \"max-width:90%; height:auto;\" >.  Check area A with width 3+1=4 and length 2w-3=3.  So the area with these dimensions is 12 as given by the problem statement.\r<BR>\n" );
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document.write( "Step 8.  ANSWER:  The length is 6 meters.\r<BR>\n" );
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document.write( "I hope the above steps were helpful.\r<BR>\n" );
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document.write( "For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.\r<BR>\n" );
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document.write( "Good luck in your studies!\r<BR>\n" );
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document.write( "Respectfully,<BR>\n" );
document.write( "Dr J </SPAN>\n" );
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