@ -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: