% October 30, 2007 % Math 589 \documentclass{beamer} % \usetheme{default} % \usetheme{Boadilla} % \usetheme{Madrid} % \usetheme{Montpellier} % \usetheme{Warsaw} % \usetheme{Copenhagen} % \usetheme{Goettingen} % \usetheme{Hannover} \usetheme{Berkeley} % \usecolortheme{crane} % \beamertemplatesolidbackgroundcolor{craneorange!25} \title{Solving a quadratic equation} \subtitle{a case study} \author{Steven Hurder} \institute[UIC] {University of Illinois at Chicago\\www.math.uic.edu/$\sim$hurder} \date[May 10, 2007] {Math 589 Presentation -- October 30, 2007} \begin{document} \frame{\titlepage} % # 1 \section[Outline]{} \frame{\tableofcontents} \section{The Problem} \frame % # 2 { \frametitle{A pesky problem} Your paycheck has been held up, and they keep asking, \medskip Are you really a mathematician?'' \bigskip \pause How to convince them? \medskip \pause What to do? \medskip \pause And then the idea hits you - you'll show them you can solve a quadratic equation! \medskip \pause If that doesn't convince the admin type, what will? \vfill } \frame % # 2 { \frametitle{Choosing a quadratic equation} Now, it is only a matter to select a quadratic equation which will impress them. \pause \begin{enumerate} \item $x^2 = 0$ \pause (nah, too obvious. it would be shameful if this worked) \pause \item $x^2 -2x +1 = 0$ \pause (more of the same) \pause \item $x^2 -3x -1 = 0$ \pause (sort of fancy... just right!) \end{enumerate} } \section{Picturing the Solution} \frame % 3 { \frametitle{Grab your calculators:} A picture may be worth a thousand words, but is it worth a thousand bucks? \medskip \pause Let's try! If they buy this, we are done. So plot $y = x^2 -3x -1$ \pause \begin{center} \includegraphics[width=0.5\textwidth]{pix/quadratic.pdf} \end{center} \vfill } \frame % # 2 { \frametitle{Not even close...} You want money for your one lousy graph?'' \pause \bigskip Give the solution to 10 decimals, and we'll show you the money!'' \pause \bigskip Oh, for ~ @\#\%\*\&\@ ~ sake!'' } \section{Some Algebra} \frame % 3 { \frametitle{factor, factor, complete...} \begin{eqnarray*} 0 & = & x^2 -3x -1 \\ \pause 0 & = & x^2 -3x + (-3/2)^2 - (3/2)^2 -1 \\ \pause 0 & = & (x - 3/2)^2 -9/4 - 4/4 \\ \pause 0 & = & (x - 3/2)^2 -13/4 \end{eqnarray*} } \frame % 3 { \frametitle{Progress} Now let's solve it: \begin{eqnarray*} 0 = (x - 3/2)^2 -9/4 - 4/4 ~ & \Longrightarrow & ~ (x-3/2)^2 = 13/4 \\ \pause & \Longrightarrow & ~ (x-3/2) = \pm \sqrt{13/4} \\ \pause & \Longrightarrow & ~ x = 3/2\pm \sqrt{13/4} \end{eqnarray*} \pause Think this is enough to get the money? \medskip \pause Not likely... } \frame % 3 { \frametitle{Pay Up!} There are two solutions: \bigskip \pause $x = 3/2 + \sqrt{13/4}$, or $$x = 3.302775637731994646559610633735247973125648286922623106355226528113583474146 505222602309541009245359$$ \bigskip \pause and $x = 3/2 - \sqrt{13/4}$, or $$x = -0.302775637731994646559610633735247973125648286922623106355226528113583474146505222602309541009245359$$ } \section{The Formula} \frame % 3 { \frametitle{Mathematical Proof} The final proof that we are Mathematicians? \bigskip \pause Give them the Magic Formula, $$ax^2 + bx + c = 0 \Longrightarrow x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ \bigskip \pause and tell them to try this first next time... } \end{document}