% 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}