document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=277181>Question 277181</A>: <I><SPAN STYLE=\"word-break: break-all;\"> This is an application problem from a chapter on the quadratic formula.<BR>\n" );
document.write( "The diagonal of a square is 2 m longer than a side.  Find the length of a side.<BR>\n" );
document.write( "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 #201848 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=201848\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Garamond\" size=\"+2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Let <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> represent the measure of one of the sides.  Then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ +\ 2\"> is the measure of the diagonal which is the hypotenuse of an isosceles right triangle.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "There are two ways to go about this one.  The 'charge straight at it' way uses the Pythagorean Theorem.\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ (x\ +\ 2)^2\ =\ x^2\ +\ x^2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ +\ 4x\ +\ 4\ =\ 2x^2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x^2\ -\ 4x\ -\ 4\ =\ 0\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Using the Quadratic Formula:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ = \frac{-(-4)\ \pm\ \sqrt{(-4)^2\ -\ 4(1)(-4)}}{2(1)}\ =\ \frac{4\ \pm\ \sqrt{32}}{2}\ =\ \frac{4\ \pm\ 4\sqrt{2}}{2}\ =\ 2\ \pm\ 2\sqrt{2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Exclude the negative root because we are looking for a positive measure of length and we are left with:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 2\ +\ 2\sqrt{2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "+++++++++++++++++++++++++++++++++++++++++++++++++\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "A slightly simpler method requires that you recall that an isosceles right triangle has three sides in proportion:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ 1\,:\,1\,:\,\sqrt{2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Hence, if <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\"> is the measure of one side, then <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\sqrt{2}\"> is the measure of the hypotenuse.  But we are given that the measure of the hypotenuse is <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\ +\ 2\">, therefore:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\sqrt{2}\ =\ x\ +\ 2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Solving for <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large x\">:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\sqrt{2}\ -\ x\ =\ 2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\left(\sqrt{2}\ -\ 1\right)\ =\ 2\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \frac{2}{\sqrt{2}\ -\ 1}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "Finally, using the conjugate of the denominator to rationalize the denominator:\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ \left(\frac{2}{\sqrt{2}\ -\ 1}\right)\left(\frac{\sqrt{2}\ +\ 1}{\sqrt{2}\ +\ 1}\right)\ =\ 2\ +\ 2\sqrt{2}\">\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
document.write( "(Verification of that last step is left as an exercise for the student)\r<BR>\n" );
document.write( "<BR>\n" );
document.write( "\n" );
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( "</font> </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );