A Perfect Square

$x^2+8x+16$ is known as a perfect square because it can be expressed as a single term squared:

$x^2+8x+16=(x+4)^2$

We can confirm this by multiplying out the brackets on the right hand side:

$(x+4)^2=(x+4)(x+4)=x^2+4x+4x+16=x^2+8x+16$

$x^2+8x+17$ is almost a perfect square. It is just one unit more than $(x+4)^2$ . It can be written:

$x^2+8x+17=(x+4)^2+1$

‘Completing the square’ is to assume that your quadratic expression is a perfect square, and then to make an adjustment so that the constant term is correct.

Remember that in general, a perfect square has the form $x^2+bx+\left(\frac{b}{2}\right)^2$ . (Assuming $a=1$ .)

Suppose the following are perfect squares – what is the constant $c$ ?

$x^2+10x+c$
$x^2-8x+c$
$x^2+24x+c$

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Completing the Square when $a\ne0$

When the coefficient of the $x^2$ term $a=1$ we can write the quadratic expression $x^2+bx+c$ in completed square form using this process:

half the coefficient $b$ , and write $\left(x+\frac{b}{2}\right)^2$ ;
subtract the square of $\frac{b}{2}$ , to write $\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2$ ;
Add the original constant and simplify, $\left(x+\frac{b}{2}\right)^2-\left(\frac{b}{2}\right)^2+c$ .

For example: $x^2+8x+17$

half the coefficient $8$ , write $\left(x+4\right)^2$ . This is equal to $x^2+8x+16$ ;
subtract the square of $4$ : write $\left(x+4\right)^2-16$ This is equal to $x^2+8x$ ;
Add the original constant and simpilfy: $\left(x+4\right)^2-16+17=\left(x+4\right)^2+1$ This is equal to $x^2+8x+17$ .

These three steps can all be done in one line:

$x^2+8x+17=(x+4)^2-16+17=(x+4)^2+1$

Example 1

$x^2+10x+1$

Half of $10$ is $5$ ; $5^2=25$ .

$x^2+10x+1=(x+5)^2-25+1=(x+5)^2-24$

For brain muscle memory: ‘half the coefficient $b$ , subtract its square, remember original constant’.

Example 2

$x^2-6x+12$

Half of $-6$ , is $-3$ , the square of $-3$ is $9$ .

$x^2-6x+12=(x-3)^2-9+12=(x-3)^2+3$

To check that the right hand side is equal to the left hand side, lets multiply out and simplify:

$\begin{align*}&(x-3)^2+3\\[10 pt]&=(x-3)(x-3)+3\\[10 pt]&=x^2-6x+9+3\\[10 pt]&=x^2-6x+12 \quad \checkmark\end{align}$

Example 3

‘half the coefficient of $x$ , subtract its square’

$\begin{align*}&x^2+6x-4\\[10 pt]=&(x+3)^2-9-4\\[10 pt]=&(x+3)^2-13\end{align}$

Example 4, odd coefficient of $x$

$\begin{align*}&x^2+7x-1\\[10 pt]=&\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-1\\[10 pt]=&\left(x+\frac{7}{2}\right)^2-\frac{49}{4}-\frac{4}{4}\\[10 pt]=&\left(x+\frac{7}{2}\right)^2-\frac{53}{4}\end{align}$

Practice

Use this applet to develop muscle memory for the complete the square process:

applet link

When the coefficient of $x^2$ is not 1

How we complete the square depends if we are solving an equation or rewriting an expression.

If we have an equation, we can divide both sides by the coefficient $a$ :

$\begin{align*}&3x^2-4x-5=0\\[10 pt]\Rightarrow &x^2-\frac{4}{3}x-\frac{5}{3}=0\end{align}$

(remember that $0\div 3=0$ ). The rest of this work is detailed in Example 4 on another page.

If we have an expression, we need to factor the coefficient of $a$ :

$\begin{align*}&3x^2-4x-5\\[10 pt]=&3\left(x^2-\frac{4}{3}x-\frac{5}{3}\right)\end{align}$

We can then complete the square within the large brackets:

$\begin{align*}&3\left(x^2-\frac{4}{3}x-\frac{5}{3}\right)\\[10 pt]=&3\left(\left(x-\frac{2}{3}\right)^2-\left(\frac{2}{3}\right)^2-\frac{5}{3}\right)\\[10 pt]=&3\left(\left(x-\frac{2}{3}\right)^2-\frac{4}{9}-\frac{15}{9}\right)\\[10 pt]=&3\left(\left(x-\frac{2}{3}\right)^2-\frac{19}{9}\right)\\[10 pt]=&3\left(x-\frac{2}{3}\right)^2-\frac{19}{3}\end{align}$

This algebra is fairly tricksome and can take some practice to master. However, in a pinch, one might wish to use the following:

Using the Vertex to Complete the Square

On another page, we argue that if $y=a(x-h)^2+k$ , the vertex has coordinates $(h,k)$ . We can find the values $h$ and $k$ using any ‘find the vertex’ method.