An Example of a Short LaTeX Document.

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\begin{document}

\centerline{\LARGE \bf Solving Quadratic Equations}
\smallskip
\centerline{\large Notes by}
\smallskip
\centerline{\Large  Ralph Howard}
\bigskip


It is well known how to solve polynomial equations of the first
degree.  For a first degree equation $ax+b=0$ with $a\neq 0$ the
solution is $x=-b/a$.  We now look at solving $ax^2+bx+c=0$.

\begin{thm}
The equation $ax^2+bx+c=0$ with $a\neq 0$ as the solutions
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
\end{thm}

\begin{proof}
We use the method of completing the square to rewrite $ax^2+bx+c$.
\begin{align*}
ax^2+bx+c&=a\left( x^2 + \frac{b}{a}x+\right)+c \\
  &=a\left( x^2 + \frac{b}{a}x+ \left(\frac{b}{2a}\right)^2
     -\left(\frac{b}{2a}\right)^2 +\right)+c \\
  &=a\left( x+\frac{b}{2a}\right)^2 -  a\left(\frac{b}{2a}\right)^2+c\\
  &= a\left( x+\frac{b}{2a}\right)^2- \frac{b^2-4ac}{4a}.
\end{align*}
Therefore $ax^2+bx+c=0$ can be rewritten as 
$$
a\left( x+\frac{b}{2a}\right)^2- \frac{b^2-4ac}{4a}=0,
$$
which can in turn  be rearranged as
$$
\left( x+\frac{b}{2a}\right)^2= \frac{b^2-4ac}{4a^2}.
$$
Taking square roots gives
$$
x+\frac{b}{2a}= \frac{\pm \sqrt{b^2-4ac}}{2a}
$$
which implies
$$
x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}
$$
as required.
\end{proof}




\end{document}