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

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

 @ -20,6 +20,9 @@  margin=0.5in }   % Add an \hrule with space above and below \newcommand\krule[2]{\vspace{#1}\hrule\vspace{#2}}   % Adjust the top and bottom margins % http://kb.mit.edu/confluence/pages/viewpage.action?pageId=3907057 \addtolength{\topmargin}{0.4in} @ -68,13 +71,14 @@ The simplest quadratic equation to solve is the type where both sides are a perf \begin{split} 4x^2 &= 36 \\ \sqrt{4x^2} &= \sqrt{36} \\ \sqrt{4}\times\sqrt{x^2} &= \sqrt{36} \\ 2x &= \pm6 \\ x &= \pm\frac{6}{2} \\ x &= \pm3 \end{split} \end{align}   \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, hence $\pm3$. In story problems, the situation may mean that we can safely ignore one of these values. \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,   @ -109,7 +113,7 @@ Using either of these values for $x$ in equation \ref{eq2} will result in one of   For equations that are not easily factored, a general solution called the quadratic equation'' can be used to solve any quadratic.   For any equation of the form $ax^2+bx+c=0$, the solution can be found by using: For any equation of the form $ax^2+bx+c=0$, the solution (that is, the value for $x$) can be found by using:   \begin{equation}\label{eq5} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} @ -119,7 +123,7 @@ In reality, the quadratic equation is a generalized form of the solving techniqu   \section{Completing the Square}   The general procedure for completing the square is to first make both sides of the equation into a perfect square, 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 equation 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.    Here is the general procedure:   @ -132,17 +136,17 @@ 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}). 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 $6^2$. This means that we will have a $\sqrt{~~}$ sign in our solution.   \subsection{Example \#1}   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 $x^2+4x+1=0$   The second step is the move the constant term to the right side of the equation by subtracting 1 from both sides: $x^2+4x+1-1=0-1$ $x^2+4x=-1$ The second step is to move the constant term to the right side of the equation by subtracting 1 from both sides: $x^2+4x+1-1=0-1$ $x^2+4x=-1$   Our value for $b$ is 4. $4\div2 = 2$ and $2^2=4$, so we will add 4 to both sides of the equation (step three): $x^2+4x+4=-1+4$ $x^2+4x+4=3$   The left side is now a perfect square. Because $x^2+4x+4=(x+2)^2$ we can rewrite it as a perfect square (step four): $(x+2)^2=3$ The left side is now a perfect square, even though it doesn't look like it. Because $x^2+4x+4=(x+2)^2$ we can rewrite it as a perfect square (step four): $(x+2)^2=3$   All that is left to do is to take the square root of both sides (step five): $\sqrt{(x+2)^2}=\pm\sqrt{3}$   @ -150,9 +154,9 @@ which gives us $x+2=\pm\sqrt{3}$   and then solve for $x$ (step six): $x=\pm\sqrt{3} -2$   Conventionally, we would write this as $x=-2+\sqrt{3}$ or $x=-2-\sqrt{3}$. Conventionally, we would write this as $x=-2+\sqrt{3}, -2-\sqrt{3}$.   \bigskip \krule{6pt}{6pt}   \noindent{}Here is the entire sequence all together:   @ -170,6 +174,20 @@ x &= -2\pm\sqrt{3} && && &&\text{Step 6} \end{aligned} \end{equation}   \hrule   \section{Example \#2}   For our second example, we will look at an equation where $a$ is not equal to 1.   \section{Example \#3}   For our third example, we will look at an equation where the right side is not a perfect square.   \section{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   \end{document}
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