Updates and typofixes

complete-the-square.tex

@@ -82,7 +82,7 @@ For example, the equation $3x^2+2x=5$ is a quadratic equation because it is writ
 
 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 $3x^2+0x+3=0$.)
 
-\paragraph{Perfect Square Quadratic Equations} The simplest quadratic equation to solve is the type where both sides are a perfect square. A \textbf{perfect square} is an integer whose square root is also an integer, such as 9 or 25, because $\sqrt{9}=3$ and $\sqrt{25}=5$. This term can also apply to a term who coefficient is a perfect square, such as $16x^2$ and $9x^2$ because $\sqrt{16x^2}=4x$ and $\sqrt{9x^2}=3x$.
+\paragraph{Perfect Square Quadratic Equations} The simplest quadratic equation to solve is the type where both sides are a perfect square. A \textbf{perfect square} is an integer whose square root is also an integer, such as 9 or 25, because $\sqrt{9}=3$ and $\sqrt{25}=5$. This term can also apply to a term whose coefficient is a perfect square, such as $16x^2$ and $9x^2$ because $\sqrt{16x^2}=4x$ and $\sqrt{9x^2}=3x$.
 
 These are easy to solve because you can solve them by taking the square root of both sides:
 
@@ -128,7 +128,7 @@ x &= -2
 
 Using either of these values for $x$ in equation \ref{eq2} will result in one of the factors being equal to zero, meaning both sides of the equation will be zero.
 
-\section{The Quadratic Equation}
+\section*{The Quadratic Equation}
 
 For equations that are not easily factored, a general solution called ``the 	quadratic equation'' can be used to solve any quadratic.
 
@@ -142,7 +142,7 @@ In reality, the quadratic equation is a generalized form of the solving techniqu
 
 We will demonstrate how to obtain the quadratic equation in section ``\nameref{expqe}'' on page~\pageref{expqe}.
 
-\section{Completing the Square}
+\section*{Completing the Square}
 
 The general procedure for completing the square is to first make the left side of the equation into a perfect square (the right side will just be a number), and then solving it as we did in \ref{eq1}. This is not a difficult process, but most students tend to get stuck on the first step---making the left side into a perfect square. 
 
@@ -159,9 +159,9 @@ Here is the general procedure:
 \item Solve for $x$.
 \end{enumerate}
 
-\paragraph{Note:} Our goal is to make the \textit{left} side of the equation into a square. However, the right side will often not be a perfect square, as in equation set~(\ref{eq1}) where the right side was $36$, which is simply $6^2$. This means that we will have a $\sqrt{~~}$ sign in our solution.
+\paragraph{Note:} Our goal is to make the \textit{left} side of the equation into a square. However, the right side will often not be a perfect square, as in equation set~(\ref{eq1}) where the right side was $36$, which is simply $6^2$. In such cases, we will have a $\sqrt{~~}$ sign in our solution, which is perfectly acceptable.
 
-\subsection{Example \#1}
+\subsection*{Example \#1}\label{ex01}
 
 For our first example, we'll use an equation in which $a$ is already 1, so we can skip the first step. Our equation is \[x^2+4x+1=0\]
 
@@ -198,9 +198,9 @@ x &= -2\pm\sqrt{3} && && &&\text{Step 6}
 
 \hrule
 
-\subsection{Example \#2}
+\subsection*{Example \#2}\label{ex02}
 
-For our second example, we will look at an equation where $a$ is not equal to 1. Our equation is \[2x^2+5x-3=0\]
+For our second example, we will look at an equation where $a$ is \textit{not} equal to 1. Our equation is: \[2x^2+5x-3=0\]
 
 Step 1: The value for $a$ is 2, so we begin by dividing each term by 2: \[\frac{2}{2}x^2+\frac{5}{2}x-\frac{3}{2}=\frac{0}{2}\] \[x^2+\frac{5}{2}x-\frac{3}{2}=0\]
 
@@ -221,7 +221,7 @@ Step 5: Take the square root of both sides, remembering that because the value o
 \[\sqrt{\biggl(x+\frac{5}{4}\biggl)^2}=\pm\sqrt{\frac{49}{16}}\]
 \[x+\frac{5}{4}=\pm\frac{7}{4}\]
 
-Step 6: all that is left is to solve for $x$:
+Step 6: All that is left is to solve for $x$:
 \[x+\frac{5}{4}-\frac{5}{4}=-\frac{5}{4}\pm\frac{7}{4}\]
 \[x=-\frac{5}{4}\pm\frac{7}{4}\]
 \[x=\frac{-5\pm7}{4}\]
@@ -230,23 +230,23 @@ Step 6: all that is left is to solve for $x$:
 
 
 
-\subsection{Example \#3}
+\subsection*{Example \#3}\label{ex03}
 
 For our third example, we will look at an equation where the right side is not a perfect square.
 
-\subsection{Example \#4}
+\subsection*{Example \#4}\label{ex04}
 
 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.
 
-\section{More Worked Examples}
+\section*{More Worked Examples}
 
-\subsection{Example \#5.1}
+\subsection*{Example \#5}\label{ex05}
 
-\subsection{Example \#5.2}
+\subsection*{Example \#6}\label{ex06}
 
-\section{An Explanation of the Quadratic Equation}\label{expqe}
+\section*{An Explanation of the Quadratic Equation}\label{expqe}
 
-Figuring out the quadratic equation from completing the square is not difficult once you are familiar with this problem solving technique. Most people get hung up doing the math on the coefficients. 
+Figuring out the quadratic equation by completing the square from the standard form (i.e., $ax^2+bx+c=0$) is not difficult once you are familiar with this problem solving technique. Most people get hung up doing the math on the coefficients. 
 
 \begin{equation}\label{quadsolv}
 \setlength{\jot}{10pt} % Add space between each equation
@@ -267,9 +267,9 @@ x &= \frac{-b\pm\sqrt{b^2-4ac}}{2a} && && && \text{Step 6b}
 \end{aligned}
 \end{equation}
 
-Remember we said that most people get hung up doing the math on the coefficients? This is why we divided Step 3 into six substeps, and Steps 5 and 6 were both divided into two substeps. 
+Remember we said that most people get hung up doing the math on the coefficients? This is why we divided Step 3 into six substeps, and Steps 5 and 6 were both divided into two substeps. This suggests one trouble-shooting method if you are having difficulty solving a problem: work out each step on paper and do every single bit of arithmetic on paper, not in your head. Often, you'll find that you can complete the square without any problems, but have made a mistake in the arithmetic when working on the right side of the equation.
 
-In Step 3, we basically left the right side alone after Step 3a. This suggests a trouble-shooting method if you can't work out a problem: only work one side of the equation at a time. 
+Also note that in Step 3, we basically left the left side of the equation alone after Step 3a. This suggests yet another trouble-shooting method if you can't work out a problem: only work one side of the equation at a time. Make sure you have that portion of the problem worked out correctly before attempting the rest of it.
 
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