Completing the Square

A quadratic curve is usually written in the form $$ \begin{equation}\begin{aligned} y=ax^2+bx+c\\ \end{aligned}\end{equation} $$

This can be rewritten in the form $a(x+h)^2+k$ in order to extract important information such as:

• The axis of symmetry $x=-h$
• The maximum/minimum value $y=k$ and
• The turning point $(-h,k)$.

Example

Complete the square for the curve $2x^2+5x+6$.

Solution First, we pair off the terms in $x$: $$ \begin{equation}\begin{aligned} (2x^2+5x)+6\\ \end{aligned}\end{equation} $$

Next we factorize out the coefficient of $x^2$: $$ \begin{equation}\begin{aligned} 2(x^2+\frac{5}{2}x)+6\\ \end{aligned}\end{equation} $$

Now we add the square of half of the coefficient of $x$ to the bracket, subtracting the equivalent from the outside of the bracket to balance the change: $$ \begin{equation}\begin{aligned} \textcolor{forestgreen}{2}(x^2+\frac{5}{2}x+\textcolor{crimson}{[\frac{5}{4}]^2})+6\textcolor{forestgreen}{-2}\textcolor{crimson}{[\frac{5}{4}]^2}\\ \end{aligned}\end{equation} $$

Notice that by adding the $[\frac{5}{4}]^2$ to the bracket, we effectively added $2[\frac{5}{4}]^2$ overall to the expression. Hence, we had to subtract $2[\frac{5}{4}]^2$ in order to balance this increase.

We now apply the perfect square factorization to the bracket: $$ \begin{equation}\begin{aligned} 2(x+\frac{5}{4})^2+6-2[\frac{5}{4}]^2\\ \end{aligned}\end{equation} $$

And simplify the terms after the bracket: $$ \begin{equation}\begin{aligned} &2(x+\frac{5}{4})^2+6-2[\frac{25}{16}]\\ &2(x+\frac{5}{4})^2+6-\frac{25}{8}\\ &2(x+\frac{5}{4})^2+\frac{48-25}{8}\\ &2(x+\frac{5}{4})^2-\frac{23}{8}\\ \end{aligned}\end{equation} $$

We thus have the quadratic in the form $a(x+h)^2+k$ where $h=\frac{5}{4}$ and $k=\frac{23}{8}$.

Example 2

Complete the square for the quadratic curve $y=-x^2+7x-3$.

Solution First we pair the terms in $x$: $$ \begin{equation}\begin{aligned} (-x^2+7x)-3\\ \end{aligned}\end{equation} $$

Factorize out the coefficient of $x^2$: $$ \begin{equation}\begin{aligned} -1(x^2-7x)-3\\ \end{aligned}\end{equation} $$

Add the square of half of the coefficient of $x$ to the bracket, subtracting the equivalent from the outside: $$ \begin{equation}\begin{aligned} -1(x^2-7x+[\frac{7}{2}]^2)-3--1[\frac{7}{2}]^2\\ \end{aligned}\end{equation} $$

Notice how we subtract $-1[\frac{7}{2}]^2$ from the outside. It will then add in the next line. Applying the perfect square factorization and simplifying: $$ \begin{equation}\begin{aligned} -1(x^2-\frac{7}{2})^2-3+1[\frac{7}{2}]^2\\ -1(x^2-\frac{7}{2})^2-3+\frac{49}{4}\\ -1(x^2-\frac{7}{2})^2+\frac{-12+49}{4}\\ -1(x^2-\frac{7}{2})^2+\frac{37}{4}\\ \end{aligned}\end{equation} $$

Therefore the quadratic is rewritten as: $$ \begin{equation}\begin{aligned} -1(x^2-\frac{7}{2})^2+\frac{37}{4}\\ \end{aligned}\end{equation} $$ where $h=\frac{-7}{2}$ and $k=\frac{37}{4}$.

The nature of the turning point

It should be noted that the turning point of the curve is given as $(-h,k)$ and its nature is indicated by the sign of the coefficient of $x^2$:

• If $a>0$, the curve is minimum (positive people smile)
• If $a<0$, the curve is maximum (negative people frown)

Notice that with both forms of a quadratic, the coefficient of $x^2$ is easily read: $$ \begin{equation}\begin{aligned} &ax^2+bx+c\\ &a(x+h)^2+k\\ \end{aligned}\end{equation} $$

Thus, we can tell the nature of the turning point of a quadratic curve regardless of whether we have completed the square or not.