@ -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{split}

x-3 &0 \\

x=3

x-3 & =0 \\

x&= 3

\end{split}

\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 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}

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 Write the left side as a perfect square.

\item Take the square root of both sides.

\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 \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}

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): \[\sqrt{(x+2)^2}=\pm\sqrt{3}\]

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 $x=-2-\sqrt{3}$.