# Completing the Square

### HT Goodwill

• Introduction

A brief discussion about what completing the square is and what we use it for. Focuses on using the technique for other reasons than solving equations.

• Background Math

Development of the patterns we use for completing the square. This part is important since it introduces the notation we will be using!

• Introductory Examples

The first three examples step through the process very slowly with a great deal of explanation for each step. These are intended for people who are unfamiliar with the process and (perhaps) need to take the algebra slowly.

Note that the three examples increase in difficulty.

• Example 1: How to Complete the Square
• Example 2: A Little More Complicated Example
• Example 3: Dealing With Leading Coefficients
• Fast Example

Example 4 is similar in difficulty to Example 3, but worked through quickly and with less explanation.

• The Quick Method

With a solid grounding in the ideas and methods of completing the square, this discussion shows how we can cut out most of the background steps and basically just do things in one step (at least for the easy ones).

• Quick Method Examples

Demonstrating the Quick Method.

• Example 5: Simple example with a leading coefficient of 1.
• Example 6: Quick method when the leading coefficient is not 1.
• Solving Equations by Completing the Square

An example and discussion of solving a quadratic equation by Completing the Square

• Conclusion

### Introduction

What Is Completing the Square?

Completing the Square is a technique in algebra that allows us to rewrite a quadratic expression that is in standard form in vertex form. That is

$% ax^2 + bx + c\quad \underset{\mbox{\scriptsize the Square}}{\xrightarrow{\mbox{\scriptsize Complete}}}\quad a(x – h)^2 + k$

so that $$ax^2 + bx + c = a(x-h)^2 + k$$.

Why Do We Complete the Square?

It should be noted that many times, students think (are taught?) that completing the square is just a way to solve a quadratic equation that has the form:

$% ax^2 + bx + c = 0.$

However, in other areas of mathematics, we sometimes need to express a quadratic in the vertex form. The following are two examples of this.

Please note, you don’t have to understand these examples! That’s not the point of them!
\begin{itemize}
\item \textit{Integral Calculus:}
\begin{center}
\begin{tikzpicture}
\node [align=center] at (-1,0.75) {\scriptsize Cannot\-2mm]\scriptsize Integrate}; \node [align=center] at (1,0.75) {\scriptsize Can Now\\[-2mm] \scriptsize Integrate}; \node at (0,0) {$$\int \frac{dx}{x^2 + 4x + 5} = \int \frac{dx}{(x+2)^2 + 1}$$}; \draw [-latex] (-1,-0.3) arc [start angle=-180, end angle=0, x radius=1cm, y radius=0.5cm]; \node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator}; \node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com}; \end{tikzpicture} \end{center} \item \textit{Differential Equations (Laplace Transforms):} \begin{center} \begin{tikzpicture} % \draw[help lines,step=5mm] (-2,-1) grid (2.5,1); \node [align = center] at (-1.25,1) {\scriptsize Cannot Find\\[-2mm] \scriptsize Transform}; \node [align=center] at (1.25,1) {\scriptsize Can Now Find\\[-2mm] \scriptsize Transform}; \node at (0,0) {$$\mathcal{L}^{-1}\left\{\frac{5}{s^2 + 4s + 5}\right\} = \mathcal{L}^{-1}\left\{\frac{5}{(s+2)^2 + 1}\right\}$$}; \draw [-latex] (-1.25,-0.3) arc [start angle=-180, end angle=0, x radius=1.5cm, y radius=0.5cm]; \node at (0,-1) [anchor=north,inner sep=0pt,align=center] {\scriptsize Complete the Square\\[-1mm] \scriptsize in Denominator}; \node at (3,-2) [anchor=east] {\tiny www.mathtutoringacademy.com}; \end{tikzpicture} \end{center} \end{itemize} The two examples above are \textit{only} to show you that completing the square is used for \textit{other reasons} than to solve equations.\vskip 1.5cm \textbf{\large Background (Leading up to \textit{How To\ldots})} \vskip 2mm Suppose we have a perfect square binomial that we expand using FOIL: \begin{align*} (x + B)^2 & = (x + B)(x + B)\\ & = x^2 + Bx + Bx + B^2\\ & = x^2 + 2Bx + B^2 \end{align*} If we reverse this string of equations, we see that any quadratic that has this pattern: $$x^2 + 2Bx + B^2$$ is a perfect square and we can factor it as $$(x + B)^2$$. \[% \underbrace{x^2 + 2Bx + B^2}_{\mbox{\parbox{3cm}{\centering \scriptsize The Perfect-Square\\Pattern\ldots}}} = \underbrace{(x+B)^2}_{\mbox{\parbox{3cm}{\centering\scriptsize \ldots always factors\\this way.}}}\vskip 2mm

To Complete the Square, we adjust a quadratic expression so that it exhibits the perfect-square pattern. Here is an overview of the rest of the article so you can focus on what you need most.\vskip 1.5cm

\textbf{\large Introductory Examples}\vskip 2mm

\textbf{Example 1 How To Complete the Square}\vskip 2mm

Complete the square on $$x^2 + 14x + 40$$.
\vskip 5mm

\textbf{Solution}\vskip 2mm

Match up the quadratic expression with the perfect square pattern, starting with the $$x^2$$ term.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\end{align*}
\vskip 5mm

Step 1: Matching The Quadratic Terms\vskip 2mm

The quadratic terms match since they have the same coefficient.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\end{align*}
\vskip 5mm

Step 2: Matching the Linear Terms\vskip 2mm

The \textit{value} of $$2B$$ is assigned be us and is whatever we need it to be so we can match the perfect-square pattern. In this example, they match, if we assign $${\color{blue} 2B = 14}$$.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\end{align*}
\vskip 5mm

Step 3: Matching the Constant Term\vskip 2mm

In order to match the perfect square pattern, the constant term has to be equal to $$B^2$$. Since we defined $$2B = 14$$ in Step 2, we know that $$B = 7$$ which implies $${\color{red} B^2 = 7^2 = 49}$$.
\vskip 2mm

To adjust our constant term, we add zero (which won’t change the value of anything) and choose to rewrite the zero as $$B^2 – B^2$$.

\begin{align*}
x^2 + 14x + 40 & = x^2 + 14x + {\color{red} 0} + 40 &&\mbox{Adding }{\color{red} 0}.\\
& = x^2 + 14x + {\color{red} (49 – 49)} + 40 &&\mbox{Since } B^2 – B^2 = {\color{red} 49-49 = 0}\\
& = x^2 + 14x + 49 – 49 + 40\\
& = (x^2 + 14x + 49) – 49 + 40&&\mbox{Regrouping}
\end{align*}

Matching up to the pattern:

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic:} & \quad (x^2 + 14x + 49) – 49 + 40\\
\end{align*}

Since the first three terms (the ones in the parentheses) match the perfect-square pattern, we know those three terms will factor into $$(x+B)^2$$.

\begin{align*}
{\color{blue}\overbrace{(x^2 + 14x + 49)}^{\mbox{Perfect Square}}} – 49 + 40
& = {\color{blue}\overbrace{(x+7)^2}^{\mbox{Factored}}} {\color{red}- 49 + 40}\\
& = (x+7)^2 {\color{red}- 9}
\end{align*}

\textbf{Answer:} $$x^2 + 14x + 40 = (x+7)^2 – 9$$
\vskip 1.5cm

\textbf{Example 2 (A \textit{Little} More Complicated)}\vskip 2mm
Complete the square on $$x^2 – 9x + 3$$.
\vskip 5mm

\textbf{Solution} As before, we match up our quadratic with the perfect-square pattern. We’ll do this one a bit more quickly.\vskip 2mm

Step 1: Quadratic and Linear Terms

Both quadratic terms have a coefficient of $$1$$ so they already match. The linear terms will match if we assign $${\color{blue}2B = -9}$$. So without any real work or effort we match the first two terms already.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + {\color{blue} 2B}x + B^2\\
\end{align*}

Step 2: Constant Term

Since we assigned $${\color{blue} 2B = -9}$$, we divide by 2 to get $$B = -\frac 9 2$$. And we know our constant term needs to be
$% {\color{red}B^2 = \left(-\frac 9 2\right)^2 = \frac{81} 4},$

and we introduce it the same way we did before, by adding zero.

\begin{align*}
x^2 – 9x + 3
& = x^2 – 9x + {\color{red} 0} + 3 && \mbox{Adding }{\color{red} 0}\\
& = x^2 – 9x + {\color{red}\left(\frac{81} 4 – \frac{81} 4\right)} + 3 && \mbox{Since } B^2 – B^2 = {\color{red} \frac{81} 4-\frac{81} 4 = 0}\\
& = x^2 – 9x + \frac{81} 4 – \frac{81} 4 + 3\\
& = \left(x^2 – 9x + \frac{81} 4\right) – \frac{81} 4 + 3 && \mbox{Regrouping}
\end{align*}

Making sure we match the pattern, we have

\begin{align*}
\mbox{Our Quadratic:} & \quad \left(x^2 + 14x + \frac{81} 4\right) – \frac{81} 4 + 40\\
\end{align*}

Since the three terms in the parentheses match the perfect-square pattern, we know those three terms factor as a perfect square.

\begin{align*}
{\color{blue}\overbrace{\left(x^2 – 9x + \frac{81} 4\right)}^{\mbox{Perfect Square}}} {\color{red}- \frac{81} 4 + 3}
& = {\color{blue} \overbrace{\left(x – \frac 9 2\right)^2}^{\mbox{Factored}}} {\color{red} – \frac{81} 4 + \frac{12} 4}\\
& = \left(x – \frac 9 2\right)^2 {\color{red} – \frac{69} 4}
\end{align*}
\vskip 5mm

\textbf{Answer:} $$x^2 – 9x + 3 = \left(x – \frac 9 2\right)^2 – \frac{69} 4$$
\vskip 1.5cm

\textbf{Example 3: Dealing With Leading Coefficients}\vskip 2mm

Complete the square on $$2x^2 + 12x – 5$$.
\vskip 5mm

\textbf{Solution} In order to complete the square, we first need to reduce the leading coefficient to 1.
\vskip 2mm

Step 1: Factor out the leading coefficient out of the first two terms.

$% {\color{blue} 2}x^2 + 12x – 5 = {\color{blue} 2}[x^2 + 6x] – 5$
\vskip 5mm

Step 2: Complete the Square \textit{Inside the Braces}\vskip 2mm

\qquad Linear Term: \textit{Inside} the braces, we set $${\color{blue} 2B = 6}$$.\vskip 2mm

A quick check and we see that we are matching the perfect-square pattern \textit{inside the braces}.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad\,\,\,\, x^2 + {\color{blue}2B}x + B^2\\
\end{align*}

Let’s move on to the constant term.\vskip 2mm

\qquad Constant Term: \textit{Inside} the braces we have $${\color{blue} 2B = 6}$$ which means $$B = 3$$ and so $${\color{red} B^2 = 3^2 = 9}$$. As before, adjust our constant term by adding {\color{red} 0}, but now it is \textit{inside the braces}.

\begin{align*}
2[x^2 + {\color{blue} 6}x] – 5
& = 2[x^2 + {\color{blue} 6}x + {\color{red} 0}] – 5 && \mbox{Adding } {\color{red} 0} \mbox{ \textit{inside} the braces}\\
& = 2[x^2 + {\color{blue} 6}x + {\color{red} (9 – 9)}] – 5 && \mbox{Since } B^2 – B^2 = {\color{red} 9 – 9 = 0}\\
& = 2[x^2 + 6x + 9 – 9] – 5\\
& = 2[(x^2 + 6x + 9) – 9] – 5 && \mbox{Regrouping \textit{inside} the braces}
\end{align*}
\vskip 2mm

Still keeping our attention focused \textit{inside} the braces, we check to see if we match the perfect-square pattern.

\begin{align*}
\mbox{Perfect-Square Pattern:} & \quad x^2 + 2Bx + B^2\\
\mbox{Our Quadratic (Inside Braces):} & \quad 2[(x^2 + 6x + 9) – 9] – 5\\
\end{align*}

Since the three terms inside the parentheses match the pattern, we know those three terms form a perfect square, so they will easily factor into $$(x + B)^2$$.

\begin{align*}
2\big[{\color{blue}\overbrace{\left(x^2 + 6x + 9\right)}^{\mbox{Perfect Square}}} – 9\big] – 5
& = 2\big[{\color{blue}\overbrace{\left(x + 3\right)^2}^{\mbox{Factored}}} – 9\big] – 5
\end{align*}
\vskip 5mm

Step 3: Adjusting the Final Form\vskip 2mm

Start by multiplying the leading coefficient \textit{through the square braces} (NOT the parentheses!), and simplify.

\begin{align*}%
{\color{blue} 2}\big[\left(x + 3\right)^2 – 9\big] – 5
& = \big[{\color{blue} 2}(x+3)^2 – {\color{blue} (2)} 9\big] – 5 && \mbox{Multiply through braces}\\
& = \big[2(x + 3)^2 – 18\big] – 5\\
& = 2(x + 3)^2 – 18 – 5 && \mbox{Braces no longer needed}\\
& = 2(x+3)^2 – 23 && \mbox{Combine the constants}
\end{align*}
\vskip 5mm

\textbf{Answer:} $$2x^2 + 12x – 5 = 2(x+3)^2 – 23$$
\vskip 1.5cm

\textbf{\large Quick Example}\vskip 2mm

\textbf{Example 4: A Quick Example}\vskip 2mm

Complete the Square on $$3x^2 – 4x + 8$$.\vskip 5mm

\textbf{Solution:}\vskip 2mm

\begin{align*}%
3x^2 – 4x + 8 & = 3\left[x^2 – \frac 4 3 x\right] + 8 && \mbox{Factor out leading coefficient}\6pt] & = 3\left[x^2 {\color{blue}- \frac 4 3} x\right] + 8 && \mbox{Identify } {\color{blue} 2B = -\frac 4 3} \longrightarrow {\color{red} B^2 = \left(-\frac 2 3\right)^2 = \frac 4 9}\\[6pt] & = 3\left[x^2 {\color{blue}- \frac 4 3} x + {\color{red} \left(\frac 4 9 – \frac 4 9\right)}\right] + 8 && \mbox{Add } {\color{red} 0}\\[6pt] & = 3\left[\left(x^2 {\color{blue}- \frac 4 3} x + {\color{red}\frac 4 9}\right)\,\,{\color{red}- \frac 4 9}\right] + 8 && \mbox{Regrouping}\\[6pt] & = 3\left[\left(x – \frac 2 3\right)^2 – \frac 4 9\right] + 8 && \mbox{Factor}\\[6pt] & = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + 8 && \mbox{Distribute}\\[6pt] & = 3\left(x – \frac 2 3\right)^2 – \frac 4 3 + \frac{24} 3\\[6pt] & = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3 && \mbox{Combine like terms} \end{align*} \vskip 5mm \textbf{Answer:} $$3x^2 – 4x + 8 = 3\left(x – \frac 2 3\right)^2 + \frac{20} 3$$ \vskip 1.5cm \textbf{\large The Quick Method}\vskip 2mm The method shown in the examples can be made more efficient if we recognize that the pattern is always the same. For a simple quadratic with a leading coefficient of $$1$$, the completed square form looks like this: \begin{center} \begin{tikzpicture} % Help Lines % \draw [step=0.25cm,help lines] (-2,-1) grid (2,1); % \draw (-2,-1) grid (2,1); % Equation \node {% $$% x^2 + {\color{blue} 2B}x + c = (x + {\color{blue} B})^2 – {\color{blue} B^2} + c$$}; % Arrows \node (linear) [inner sep=0pt] at (-1.625,-0.15){}; \node (BLow) [inner sep=0pt] at (0.75, -0.15){}; \node (BUp) [inner sep=0pt] at (0.75, 0.2){}; \node (B2) [inner sep=0pt] at (1.75,0.2){}; \draw [-latex] (linear.south) to [out=270,in=270] (BLow.south); \draw [-latex] (BUp.north) to [out=90,in=90] (B2.north); % Labels \node at (-0.375, -0.625) {$$\div 2$$}; \node at (1.5, 0.75) {\scriptsize Minus the Square}; \node at (3,-1) [anchor=east] {\tiny www.mathtutoringacademy.com}; \end{tikzpicture} \end{center} Inside the final parentheses we always end up with $$x + B$$, where $$B$$ is half of the coefficient of the original $$x$$ term.\vskip 2mm Next, we subtract $$B^2$$ \textit{outside} the parentheses. Let’s try it with one of our previous examples to see it in action.\vskip 5mm \textbf{Example 5: Simple Quick Method}\vskip 2mm Use the quick version of Completing the Square on $$x^2 + 14x + 40$$. \begin{center} \begin{tikzpicture} % Help Lines % \draw [step=0.25cm,help lines] (0,-1) grid (5,1); % \draw (0,-1) grid (5,1); % Equation \node at (1.2,-0.25){% $$% x^2 + {\color{blue} 14}x + 40 = (x + {\color{blue} 7})^2 \underbrace{- {\color{blue} 7^2} + 40}_{\mbox{\scriptsize Add Together}} = (x + 7)^2 – 9$$}; % Arrows \node (linear) [inner sep=0pt] at (-1.625,-0.15){}; \node (BLow) [inner sep=0pt] at (0.75, -0.15){}; \node (BUp) [inner sep=0pt] at (0.75, 0.2){}; \node (B2) [inner sep=0pt] at (1.75,0.2){}; \draw [-latex] (linear.south) to [out=270,in=270] (BLow.south); \draw [-latex] (BUp.north) to [out=90,in=90] (B2.north); \draw [-latex] (2.85,-0.575) .. controls (3,-0.625) and (5,-0.75)..(5,-0.25); % Labels \node at (-0.375, -0.625) {$$\div 2$$}; \node at (1.5, 0.75) {\scriptsize Minus the Square}; \node at (5,-1) [anchor=east] {\tiny www.mathtutoringacademy.com}; \end{tikzpicture} \end{center} \vskip 5mm \textbf{Answer:} $$x^2 + 14x + 40 = (x + 7)^2 – 9$$ Note: This is the same answer we got in Example 1.\vskip 1.5cm \textbf{Example 6: Quick Method with Leading Coefficient}\vskip 2mm The quick version can also be used when the leading coefficient isn’t $$1$$. Like before, we just factor out the leading coefficient first, and then work inside the braces.\vskip 2mm Let’s complete the square on $$5x^2 + 80x + 13$$. \noindent\textbf{Solution}\vskip 2mm Step 1) Factor out the leading coefficient. \[% 5x^2 + 80x + 13 = 5[x^2 + 16x] + 13

Step 2) Now use the quick method to complete the square on the \textit{inside} of the braces.

\begin{center}
\begin{tikzpicture}
% Help Lines
% \draw [step=0.25cm,help lines] (-2,-2) grid (4,2);
% \draw (-2,-1) grid (4,1);
% \draw [red, step = 2] (0,-1) grid (3,1);
% Equation
\node at (1.2,0){%
\begin{minipage}{\textwidth}
\begin{align*}%
5[x^2 + {\color{blue} 16}x] + 13
& = 5[x^2 + {\color{blue} 16}x] + 13\5mm] & = 5[(x + {\color{blue} 8})^2 – {\color{blue} 64}] + 13\\[6mm] & = 5(x + {\color{blue} 8})^2 – 320 + 13\\ & = 5(x + {\color{blue} 8})^2 – 307% \end{align*}% \end{minipage} }; % Arrows \node (linear) [inner sep=0pt] at (2.375,0.95){}; \node (BLow) [inner sep=0pt] at (2.23, -0.05){}; \node (BUp) [inner sep=0pt] at (2.23, 0.25){}; \node (B2) [inner sep=0pt] at (3.23,-0.05){}; \draw [-latex] (linear.south) to [out=270,in=90] (BUp.south); \draw [-latex] (BLow.south) to [out=270,in=270] (B2.south); % Labels \node at (2.625, 0.625) {\scriptsize $$\div 2$$}; \node at (3, -0.55) {\scriptsize Minus the Square}; \node at (4.5,-2) [anchor=east] {\tiny www.mathtutoringacademy.com}; \end{tikzpicture} \end{center} \vskip 5mm \textbf{Answer:} $$5x^2 + 80x + 13 = 5(x + 8)^2 – 307$$ \vskip 1.5cm \textbf{\large Solving Equations}\vskip 2mm When solving a quadratic equation, I find the technique of completing the square is not very efficient, except in the simple cases where the leading coefficient is $$1$$ and the linear coefficient is even.\vskip 2mm But that doesn’t mean we shouldn’t know \textit{how} to solve more complicated quadratic equations with this technique.\vskip 5mm \textbf{Two Approaches}\vskip 2mm When solving an equation through completing the square, there are two basic approaches: (1) Complete the square first, \textit{then} solve the equation, or (2) Solve the equation \textit{as} you complete the square.\vskip 5mm \textbf{Example 7: Solving a Quadratic Equation by Completing the Square}\vskip 2mm Use Completing the Square to solve $$4x^2 + 3x – 10 = 0$$.\vskip 5mm \textbf{Solution (Method 1)}\vskip 2mm For this approach, we’ll first complete the square, then solve the equation.\vskip 2mm Step 1) Complete the Square on $$4x^2 + 3x – 10$$.\vskip 2mm Using the techniques we’ve discussed above, we get the following. \[% 4x^2 + 3x – 10 = 4\left(x + \frac 3 8\right)^2 – \frac{169}{16}
\vskip 5mm

Step 2) Solve the equation.

\begin{align*}
4x^2 + 3x – 10 & = 0\6pt] 4\left(x + \frac 3 8\right)^2 – \frac{169}{16} & = 0 && \mbox{Complete the Square}\\[6pt] 4\left(x + \frac 3 8\right)^2 & = \frac{169}{16} && \mbox{Add the constant}\\[6pt] \left(x + \frac 3 8\right)^2 & = \frac{169}{64} && \mbox{Divide by $$4$$}\\[6pt] x + \frac 3 8 & = \pm\sqrt{\frac{169}{64}} && \mbox{Take the square root}\\[6pt] x + \frac 3 8 & = \pm\frac{13} 8 && \mbox{Simplify}\\[6pt] x & = – \frac 3 8 \pm\frac{13} 8 && \mbox{Subtract} \end{align*} The two solutions are \[% \begin{array}{rl} x & = \displaystyle – \frac 3 8 + \frac {13} 8 = \frac{10} 8 = \frac 5 4\\[12pt] x & = \displaystyle – \frac 3 8 – \frac{13} 8 = – \frac{16} 8 = – 2 \end{array}
\vskip 5mm

\textbf{Solution (Method 2)}\vskip 2mm

This time, we’ll complete the square as we solve the equation. It should be noted that this approach is what many students think \textit{is} Completing the Square.\vskip 2mm

\begin{align*}
4x^2 + 3x – 10 & = 0\6pt] 4x^2 + 3x & = 10 && \mbox{Add the constant}\\[6pt] x^2 + \frac 3 4 x & = \frac{10} 4 && \mbox{Divide by 4}\\[6pt] x^2 + \frac 3 4 x + \frac 9 {64} & = \frac{10} 4 + \frac 9 {64} && \mbox{Add $$B^2$$}\\[6pt] \left(x + \frac 3 8\right)^2 & = \frac{160}{64} + \frac 9 {64} && \mbox{Factor the left side}\\[6pt] x + \frac 3 8 & = \pm \sqrt{\frac{169}{64}} && \mbox{Take square root}\\[6pt] x & = -\frac 3 8 \pm \frac{13} 8 && \mbox{Subtract} \end{align*} Again, we see the two answers are $$x =-\frac 3 8 + \frac{13} 8 = \frac 5 4$$ and $$x = -\frac 3 8 – \frac{13} 8 = -2$$. \vskip 5mm \textbf{Answer:} $$x = \frac 5 4$$ or $$x = -2$$ \vskip 1.5cm \textbf{Conclusion}\vskip 2mm Admittedly, most high school student will Complete the Square in the context of solving quadratic equations. However, it is important that they understand that the technique is more than just an equation-solving technique. Outside of high school math, much of its use is in changing the form of a quadratic function so that it can be used in higher mathematical techniques. ## Find the Vertex by Using the Quadratic Formula The Quadratic Formula is primarily used to identify the roots ($$x$$-intercepts) of a quadratic function. What many people don’t know is that you can also easily find the vertex of the function by simply looking at the Quadratic Formula! ## Graphing Quadratic Functions When graphing quadratic functions by hand, we will often use the quadratic formula to obtain the roots (or $$x$$-intercepts) of the function. Then separately, we will work on finding the vertex of the function since this point defines the lowest (or highest) point on the graph, as well as the line of symmetry. For the general quadratic function, $$f(x) = ax^2 + bx + c$$, the quadratic formula identifies the roots at: \[% x = \frac{-b \pm\sqrt{b^2 – 4ac}}{2a}
while the the vertex is found at $$\left(-\frac b {2a}, f\left(-\frac b {2a}\right)\right)$$.

The $$x$$-value of the vertex, $$x = -\frac b {2a}$$, is often referred to in algebra textbooks as The Vertex Formula.

## Finding the Vertex in the Quadratic Formula

Many people notice (or are taught) that the Vertex Formula can be found nestled in the Quadratic Formula. Specifically,

$% \mbox{Quadratic Formula: } x = \frac{\color{blue}{-b} \pm\sqrt{b^2 – 4ac}}{\color{blue}{2a}}\longrightarrow \color{blue}{\frac{-b}{2a}} = x\mbox{-value of the vertex.}$

Less well known is the fact that the $$y$$-coordinate of the vertex can also be obtained easily from the quadratic formula. The vertex of a quadratic function can be written as

$% \mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right)$
where $$D$$ is the discriminant of the quadratic function. Recall that the discriminant of a quadratic function is $$\color{red}{D = b^2 – 4ac}$$ and it appears under the radical in the quadratic formula:

$% \mbox{Quadratic Formula: } x = \frac{-b \pm\sqrt{\color{red}{b^2 – 4ac}}}{2a} = \frac{-b \pm\sqrt{\color{red} D}}{2a}.$

Also, notice that the denominator of the $$x$$-coordinate and $$y$$-coordinate of the vertex are very similar.  The $$y$$-coordinate’s denominator is just twice as big as the $$x$$-coordinate’s!

Now, let’s try out this idea and find the vertex of a quadratic from just the quadratic formula.

### Example 1:

Suppose $$f(x) = 5x^2 + 4x – 3$$. Use the quadratic formula to identify the entire vertex of the function, and then verify the answer using the standard method.

Solution: The quadratic formula for this function is
$% x = \frac{-4 \pm \sqrt{4^2 – 4(5)(-3)}}{2(5)} = \frac{-4\pm \sqrt{76}}{10}.$

The discriminant for this function is $$\color{red}{D = 76}$$. So, the vertex should be at

$% \mbox{Vertex: } \left(\color{blue}{-\frac b {2a}}, \color{red}{-\frac D {4a}}\right) = \left(-\frac 4 {10}, -\frac{\color{red}{76}}{20}\right) = \left(-\frac 2 5, -\frac{19} 5\right).$

Let’s double check our $$y$$-coordinate by using the standard method (plugging $$x = -\frac b {2a} = -\frac 2 5$$ into the original function).

\begin{align*}
f\left(\color{blue}{-\frac 2 5}\right)
& = 5\left(\color{blue}{-\frac 2 5}\right)^2 + 4\left(\color{blue}{-\frac 2 5}\right) – 3 && \mbox{Plugging in the $$x$$-value.}\6pt] & = 5\left(\frac 4 {25}\right) – \frac 8 5 – \frac 3 1 && \mbox{Simplifying.}\\[6pt] & = \frac 4 5 – \frac 8 5 – \frac{15} 5\\[6pt] & = \frac{-19} 5 && \mbox{$$y$$-value of the vertex.} \end{align*} ### Example 2 Let’s try another example. Suppose $$f(x) = -2x^2 +12x -25$$. Identify the vertex using the quadratic formula, and verify the answer with the standard method. Solution: The quadratic formula for this function is \[% x = \frac{-12 \pm\sqrt{(-12)^2-4(-2)(-25)}}{2(-2)} = \frac{\color{blue}{-12} \pm\sqrt{\color{red}{-56}}}{-4}

The vertex should be $$\displaystyle \left(\frac{\color{blue}{-12}}{-4}, -\frac{\color{red}{-56}}{-8}\right) = \left(3, -7\right)$$.  Let’s verify the result. Using the standard method, we evaluate $$f(3)$$.

\begin{align*}
f(\color{blue} 3) & = -2(\color{blue} 3)^2 + 12(\color{blue} 3) – 25\6pt] & = -2(9) + 36 – 25\\ & = -18 + 11\\ & = -7 \end{align*} Again, we see the method works. However, the savvy math student will immediately ask, “Does it work every time? Or was there something special’ or unusual’ about these particular quadratic functions?” ## Does it Always Work? This is an important question. In order to claim something is valid in mathematics, we have to do more than look at a few examples. We have to prove its validity. So, without further ado, here is the proof of this technique. Given: The $$x$$-coordinate of the vertex of a quadratic function is $$x = -\frac b {2a}$$. Show: The $$y$$-coordinate of the vertex of a quadratic function is $$y = -\frac D {2a}$$, where $$D$$ is the discriminant of the function. Proof: Simply evaluate $$f(x) = ax^2 + bx + c$$ at $$x = -\frac b {2a}$$. \begin{align*} f\left(\color{blue}{-\frac b {2a}}\right) & = a\left(\color{blue}{-\frac b {2a}}\right)^2 + b\left(\color{blue}{-\frac b {2a}}\right) + c\\[6pt] & = a\left(\frac{b^2}{4a^2}\right) -\frac{b^2}{2a} + c && \mbox{Simplify a little.}\\[6pt] & = \frac{b^2}{4a} -\frac{b^2}{2a} + \frac c 1 && \mbox{Cancel the common factor.}\\[6pt] & = \frac{b^2}{4a} -\frac{2b^2}{4a} + \frac{4ac}{4a} && \mbox{Common denominator.}\\[6pt] & = \frac{-b^2 + 4ac}{4a} && \mbox{Combine numerators.}\\[6pt] & = -\frac{\color{red}{b^2 – 4ac}}{4a} && \mbox{Factor out a negative 1.}\\[6pt] & = -\frac{\color{red}{D}}{4a} \end{align*} Well, how about that. It seems like any time we evaluate a quadratic at $$x = -\frac b {2a}$$, we will end up with $$-\frac D {4a}$$. ## So What? So, what does this do for us in practical terms? Suppose we were asked to graph (by hand) the function $$f(x) = 3x^2 -6x + 8$$. A good first step is to find the roots, and a good method for that is to use the quadratic formula. \begin{align*} x & = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\\[6pt] & = \frac{6 \pm \sqrt{(-6)^2 – 4(3)(8)}}{2(3)}\\[6pt] & = \frac{6 \pm \sqrt{-60}} 6 \end{align*} Oops. This quadratic function doesn’t have real-valued $$x$$-intercepts (Yes, it has complex roots, but that won’t help us graph, will it?). But the function has a graph, we just have to find it. If only we had the vertex and one other point! Then we could graph the quadratic! But wait! We already have written out the quadratic formula, so that should tell us the vertex: \[% \mbox{Vertex: } \left(-\frac b {2a}, -\frac D {4a}\right) = \left(-\frac 6 6, -\frac{-60}{12}\right) = (1, 5).

Now, together with the $$y$$-intercept at $$(0,8)$$, we can easily plot the function.

## Conclusion

In conclusion, the more we understand about the tools we have at our disposal, the easier it becomes to work through the mathematics that crop up in school and careers. And the more we learn, the more we become intrigued by the idea of other possibilities.  I mean, doesn’t this insight into the quadratic formula just beg the question, “What else can we learn from the Quadratic Formula? Or the discriminant?”

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