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Kenneth John Odle 11 months ago
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      complete-the-square.tex

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complete-the-square.tex

@ -7,6 +7,8 @@
\usepackage{float}
\raggedbottom
\usepackage{enumitem}
% Where are our images?
\graphicspath{{images/}}
@ -63,9 +65,20 @@ It also provides several worked equations to serve as examples.
\section{Quadratic Equations}
\textbf{Quadratic equations} are equations of the form $ax^2+bx+c=0$.
\textbf{Quadratic equations} are equations of the form $ax^2+bx+c=0$, or that can be written in that form.
When we say ``of the form'' what we mean is
\begin{itemize}[noitemsep]
\item The letters $a, ~b,$ and $c$ represent constants, and that both $b$ and $c$ can be zero.
\item There is one term that contains the variable $x^2$
\end{itemize}
For example, the equation $3x^2+2x=5$ is a quadratic equation because it is written in the form $ax^2+bx+c=0$.
However, the equation $5=2-3x^2$ is also a quadratic equation because it can be rewritten as $3x^2+3=0$. In this case, the value for $b$ is zero. (This could also be written as $3x^2+0x+3=0$.)
The simplest quadratic equation to solve is the type where both sides are a perfect square, because you can solve them by taking the square root of both sides:
\paragraph{Perfect Square Quadratic Equations} The simplest quadratic equation to solve is the type where both sides are a perfect square, because you can solve them by taking the square root of both sides:
\begin{align}\label{eq1}
\begin{split}
@ -80,7 +93,7 @@ x &= \pm3
\paragraph{Note:} Because we are taking the square root of a constant, we must include both the positive and negative values of the square root as the solution, hence $\pm3$. Substituting either $3$ or $-3$ into $x$ in the original equation results in a value of 36. (In story problems, the situation may mean that we can safely ignore one of these values.)
\paragraph{Factorable equations} The other type of quadratic equation is one that can easily be solved by factoring. For example,
\paragraph{Factorable Quadratic Equations} The other type of quadratic equation is one that can easily be solved by factoring. For example,
\begin{align}\label{eq2}
\begin{split}
@ -156,7 +169,7 @@ and then solve for $x$ (step six): \[x=\pm\sqrt{3} -2\]
Conventionally, we would write this as $x=-2+\sqrt{3}, -2-\sqrt{3}$.
\krule{6pt}{6pt}
\newpage
\noindent{}Here is the entire sequence all together:
@ -176,18 +189,27 @@ x &= -2\pm\sqrt{3} && && &&\text{Step 6}
\hrule
\section{Example \#2}
\subsection{Example \#2}
For our second example, we will look at an equation where $a$ is not equal to 1.
\section{Example \#3}
\subsection{Example \#3}
For our third example, we will look at an equation where the right side is not a perfect square.
\section{Example \#4}
\subsection{Example \#4}
For our fourth example, we will look at an equation where neither $a$ is equal to 1 nor the right side is a perfect square.
\pagestyle{lastpage}% Remove the header from the last page; comment this out if the document ends on an odd-numbered page
\section{More Worked Examples}
\subsection{Example \#5.1}
\subsection{Example \#5.2}
\section{An Explanation of the Quadratic Equation}
%\pagestyle{lastpage}
% Remove the header from the last page; comment this out if the document ends on an odd-numbered page
\end{document}
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