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Kenneth John Odle 2023-07-16 15:32:26 -04:00
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complete-the-square.tex
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@ -89,8 +89,8 @@ Because any number times zero is equal to zero, we can conclude that either $x-3
\begin{align}\label{eq3} \begin{align}\label{eq3}
\begin{split} \begin{split}
x-3 &0 \\ x-3 & =0 \\
x=3 x &= 3
\end{split} \end{split}
\end{align} \end{align}
@ -109,7 +109,7 @@ 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 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 $ax^s+bx+c=0$, the solution can be found by using: For any equation of the form $ax^2+bx+c=0$, the solution can be found by using:
\begin{equation} \begin{equation}
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}
@ -129,25 +129,26 @@ Here is the general procedure:
\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.
\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 $x$.
\end{enumerate} \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 \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}). This means that we will have a $\sqrt{~~}$ sign in our solution.
\subsection{Example \#1} \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 \[x^2+4x+1=0\] 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: \[x^2+4x=-1\] The second step is the move the constant term to the right side of the equation by subtracting 1 from both sides: \[x^2+4x+1-1=0-1\] \[x^2+4x=-1\]
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): \[x^2+4x=3\] 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 $x^2+4x=(x+2)^2$ we can rewrite it as a perfect square (step four): \[(x+2)^2=3\] The left side is now a perfect square. 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): \[\sqrt{(x+2)^2}=\pm\sqrt{3}\] All that is left to do is to take the square root of both sides (step five): \[\sqrt{(x+2)^2}=\pm\sqrt{3}\]
which gives us \[x+2=\pm\sqrt{3}\] which gives us \[x+2=\pm\sqrt{3}\]
and then solve for $x$: \[x=\pm\sqrt{3} -2\] 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 $x=-2+\sqrt{3}$ or $x=-2-\sqrt{3}$.