Completed Square Form

The completed square form can be used to:

  1. Locate the vertex of a parabola, both x and y coordinates;
  2. Solve a quadratic equation;
  3. Calculate the inverse of a quadratic function – discussed in the grade 12 curriculum.

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?

  1. x^2+10x+c
  2. x^2-8x+c
  3. x^2+24x+c

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.

On a different page, we see that x_{v}=\dfrac{-b}{2a}

Considering our example above, 3x^2-4x-5, x_{v}=\dfrac{-(-4)}{2(3)}=\dfrac{4}{6}=\dfrac{2}{3}. Therefore, h=\dfrac{2}{3}.

To find y_{v} we simply need to substitute x_v to the expression:

    \[3\left(\frac{2}{3}\right)^2-4\left(\frac{2}{3}\right)-5=\frac{4}{3}-\frac{8}{3}-\frac{15}{3}=\frac{-19}{3}\]

Therefore k=\frac{-19}{3}.

After calculating h and k, (a=3), we have

    \[a(x-h)^2+k=3\left(x-\frac{2}{3}\right)^2-\frac{19}{3}\]

which agrees with the algebra above.