The purpose of this document is to describe how to complete the square---a common method for factoring quadratic equations. It also described how the quadratic formula is derived from the standard form for quadratic equations by completing the square.
\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 $x^2$.
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. 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 description can also apply to a term whose coefficient is a perfect square, such as $16x^2$ and $9x^2$ because $\sqrt{16x^2}=4x$ and $\sqrt{9x^2}=3x$.
\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.)
Using either of these values for $x$ in equation \ref{eq2} will result in one of the factors being equal to zero, meaning both sides of the equation will be zero.
In reality, the quadratic formula is the result of solving the general form quadratic equation $ax^2+bx+c=0$ by completing the square. Using the quadratic formula is generally much easier (it can be programmed into some calculators and spreadsheets, for instance), but completing the square is used in certain calculus problems and for graphing some functions.
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.
\textbf{Note:} If dividing by 2 results in a fraction, leave it as a square \textit{on the left side}. That is, leave it as $(\frac{5}{2})^2$ rather than doing the math and arriving at $\frac{25}{4}$. This will make step four easier. (However, you will need to do the math on the right side.)
\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 $6^2$. In such cases, we will have a $\sqrt{~~}$ sign in our solution, which is perfectly acceptable.
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): \[(x+2)^2=3\]
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:
Figuring out the quadratic formula by completing the square from the standard form (i.e., $ax^2+bx+c=0$) is not difficult once you are familiar with this problem solving technique. Most people get hung up doing the math on the coefficients.
Remember we said that most people get hung up doing the math on the coefficients? This is why we divided Step 3 into six substeps, and Steps 5 and 6 were both divided into two substeps. This suggests one trouble-shooting method if you are having difficulty solving a problem: work out each step on paper and do every single bit of arithmetic on paper, not in your head. Often, you'll find that you can complete the square without any problems, but have made a mistake in the arithmetic when working on the right side of the equation.
Also note that in Step 3, we basically left the left side of the equation alone after Step 3a. This suggests yet another trouble-shooting method if you can't work out a problem: only work one side of the equation at a time. Make sure you have that portion of the problem worked out correctly before attempting the rest of it.