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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

@ -55,7 +55,7 @@
\title{Completing the Square}
\author{Kenneth John Odle}
\date{16 July 2023}
\date{17 July 2023}
\begin{document}
@ -71,18 +71,20 @@ It also provides several worked equations to serve as examples.
\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
When we say ``of the form'' what we mean is that:
\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 $x^2$
\item The letters $a, ~b,$ and $c$ represent constants, and that both $b$ and $c$ can be zero. (The constants are the \textbf{coefficients} of the equation.)
\item There is at least 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 $3x^2+0x+3=0$.)
\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:
\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 $\sqrt{9x^2}=3x$.
These are easy to solve because you can solve them by taking the square root of both sides:
\begin{align}\label{eq1}
\begin{split}
@ -198,7 +200,35 @@ x &= -2\pm\sqrt{3} && && &&\text{Step 6}
\subsection{Example \#2}
For our second example, we will look at an equation where $a$ is not equal to 1.
For our second example, we will look at an equation where $a$ is not equal to 1. Our equation is \[2x^2+5x-3=0\]
Step 1: The value for $a$ is 2, so we begin by dividing each term by 2: \[\frac{2}{2}x^2+\frac{5}{2}x-\frac{3}{2}=\frac{0}{2}\] \[x^2+\frac{5}{2}x-\frac{3}{2}=0\]
Step 2: We now need to move the constant term ($-\frac{3}{2}$) to the right side: \[x^2+\frac{5}{2}x-\frac{3}{2}+\frac{3}{2}=0+\frac{3}{2}\] \[x^2+\frac{5}{2}x=\frac{3}{2}\]
Step 3: Our value for $b$ is $\frac{5}{2}$ so we will divide that by 2, square it, and add it both sides of the equation: \[x^2+\frac{5}{2}x+\biggl(\frac{5}{4}\biggl)^2=\frac{3}{2}+\biggl(\frac{5}{4}\biggl)^2\]
We will leave it as a square term on the left, but let's do the math on the right side of the equation:
\[x^2+\frac{5}{2}x+\biggl(\frac{5}{4}\biggl)^2=\frac{3}{2}+\frac{25}{16}\]
\[x^2+\frac{5}{2}x+\biggl(\frac{5}{4}\biggl)^2=\frac{3}{2}\biggl(\frac{8}{8}\biggl)+\frac{25}{16}\]
\[x^2+\frac{5}{2}x+\biggl(\frac{5}{4}\biggl)^2=\frac{24}{16}+\frac{25}{16}\]
\[x^2+\frac{5}{2}x+\biggl(\frac{5}{4}\biggl)^2=\frac{49}{16}\]
Step 4: We can now rewrite the left side as a perfect square:
\[\biggl(x+\frac{5}{4}\biggl)^2=\frac{49}{16}\]
Step 5: Take the square root of both sides, remembering that because the value on the right is a constant, we need to take into account both negative and positive values:
\[\sqrt{\biggl(x+\frac{5}{4}\biggl)^2}=\pm\sqrt{\frac{49}{16}}\]
\[x+\frac{5}{4}=\pm\frac{7}{4}\]
Step 6: all that is left is to solve for $x$:
\[x+\frac{5}{4}-\frac{5}{4}=-\frac{5}{4}\pm\frac{7}{4}\]
\[x=-\frac{5}{4}\pm\frac{7}{4}\]
\[x=\frac{-5\pm7}{4}\]
\[x=\frac{-5-7}{4}= ~\frac{-12}{4}, ~x=\frac{-2+4}{4}=\frac{2}{4}\]
\[x=-3, ~x=\frac{1}{2}\]
\subsection{Example \#3}

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