document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=498427>Question 498427</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A large rectangular movie screen in an Imax theater has an area of 8,364 square feet. Find the dimensions of the screen if it is 20 feet longer than it is wide. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #337037 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=337037\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "If a rectangle has an area of <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large A\"> square units and the length, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large l\"> is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large d\"> units longer than the width, <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large w\">, find the dimensions.\r<BR>\n" );
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document.write( "The length can be expressed as <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large l\ =\ w\ +\ d\"> and the width as simply <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large w\">.  The product of the length times the width is the area, so:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ w(w\ +\ d)\ =\ A\">\r<BR>\n" );
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document.write( "A little algebra music there Maestro!\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ w^2\ +\ dw\ -\ A\ =\ 0\">\r<BR>\n" );
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document.write( "And we have a quadratic.  Since the constant term (the <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large A\">) is negative, we are assured that the signs on the binomial factors are opposite  and therefore the roots have opposite signs.  Since we are looking for a positive measure of length we can discard the negative root.  The quadratic formula solution to the general problem is therefore:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ w\ =\ \frac{-d\ +\ \sqrt{d^2\ +\ 4A}}{2}\">\r<BR>\n" );
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document.write( "Just plug in the given difference and the given area and then do the arithmetic.\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
document.write( "My calculator said it, I believe it, that settles it<BR>\n" );
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