@ -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:

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:

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: