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complete-the-square.tex

#### 42 complete-the-square.tex View File

 @ -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}