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Added explanation of the quadratic equation

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Kenneth John Odle 2023-07-17 13:25:16 -04:00
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complete-the-square.tex
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@ -7,8 +7,12 @@
\usepackage{float} \usepackage{float}
\raggedbottom \raggedbottom
% Control spacing of list environments
\usepackage{enumitem} \usepackage{enumitem}
% Refer to sections by title
\usepackage{nameref}
% Where are our images? % Where are our images?
\graphicspath{{images/}} \graphicspath{{images/}}
@ -134,6 +138,8 @@ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
In reality, the quadratic equation is a generalized form of the solving technique called ``completing the square''. Using the quadratic equation is generally much easier (it can be programmed into some calculators and spreadsheets, for instance), but completing the square is used in certain calculus problems and for graphing some functions. In reality, the quadratic equation is a generalized form of the solving technique called ``completing the square''. Using the quadratic equation is generally much easier (it can be programmed into some calculators and spreadsheets, for instance), but completing the square is used in certain calculus problems and for graphing some functions.
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. 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.
@ -145,7 +151,7 @@ Here is the general procedure:
\item Move the constant term to the right side of the equation. \item Move the constant term to the right side of the equation.
\item Divide $b$ by 2, square it, and add it to both sides of equation. \item Divide $b$ by 2, square it, and add it to both sides of equation.
(\textbf{Note:} If dividing by 2 results in a fraction, leave it as a square. That is, leave it as $(\frac{5}{2})^2$ rather than doing the math and arriving at $\frac{25}{4}$. This will make step four easier.) \textbf{Note:} If dividing by 2 results in a fraction, leave it as a square \textit{on the left side}. That is, leave it as $(\frac{5}{2})^2$ rather than doing the math and arriving at $\frac{25}{4}$. This will make step four easier. (However, you will need to do the math on the right side.)
\item Write the left side as a perfect square. \item Write the left side as a perfect square.
\item Take the square root of both sides. \item Take the square root of both sides.
\item Solve for $x$. \item Solve for $x$.
@ -185,7 +191,6 @@ x^2+4x &= 3 && && &&\text{Step 3}\\
\sqrt{(x+2)^2} &= \pm\sqrt{3} && && && \text{Step 5}\\ \sqrt{(x+2)^2} &= \pm\sqrt{3} && && && \text{Step 5}\\
x+2 &= \pm\sqrt{3} && && && \text{Step 5}\\ x+2 &= \pm\sqrt{3} && && && \text{Step 5}\\
x &= -2\pm\sqrt{3} && && &&\text{Step 6} x &= -2\pm\sqrt{3} && && &&\text{Step 6}
%\end{split}
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -209,7 +214,32 @@ For our fourth example, we will look at an equation where neither $a$ is equal t
\subsection{Example \#5.2} \subsection{Example \#5.2}
\section{An Explanation of the Quadratic Equation} \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.
\begin{equation}\label{quadsolv}
\setlength{\jot}{10pt} % Add space between each equation
\begin{aligned}
ax^2+bx+c &= 0 && && &&\text{General Form}\\
x^2+\frac{b}{a}+\frac{c}{a} &= 0 && && &&\text{Step 1} \\
x^2+\frac{b}{a} &= -\frac{c}{a} && && &&\text{Step 2}\\
x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2 &= -\frac{c}{a}+\left(\frac{b}{2a}\right)^2 && && &&\text{Step 3a}\\
x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2 &= -\frac{c}{a}+\frac{b^2}{4a^2} && && && \text{Step 3b}\\
x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2 &= -\biggl(\frac{c}{a}\biggl)\left(\frac{4a}{4a}\right)+\frac{b^2}{4a^2} && && && \text{Step 3c}\\
x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2 &= -\frac{4ac}{4a^2}+\frac{b^2}{4a^2} && && && \text{Step 3d} \\
x^2+\frac{b}{a}+\left(\frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2} && && && \text{Step 3e} \\
\left(x+\frac{b}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2} && && && \text{Step 4} \\
\sqrt{\left(x+\frac{b}{2a}\right)^2} &= \pm\sqrt{\frac{b^2-4ac}{4a^2}} && && && \text{Step 5a} \\
x+\frac{b}{2a} &= \frac{\pm\sqrt{b^2-4ac}}{2a} && && && \text{Step 5b} \\
x+\frac{b}{2a}-\frac{b}{2a} &= \frac{\pm\sqrt{b^2-4ac}}{2a}-\frac{b}{2a} && && && \text{Step 6a} \\
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.
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.
%\pagestyle{lastpage} %\pagestyle{lastpage}
% Remove the header from the last page; comment this out if the document ends on an odd-numbered page % Remove the header from the last page; comment this out if the document ends on an odd-numbered page