Completing the square

In a nutshell

Completing the square involves rearranging the quadratic into a form which can make it easier to solve. It is another way of solving a quadratic equation, instead of factorising or using the quadratic formula. It can be used to derive the quadratic formula. Once a quadratic is in the 'complete the square' form, it also indicates where the vertex of a quadratic is on a graph.

Completing the square

Recall that a quadratic equation can be written in the form

$\boxed{ax^2+bx+c=0}$

where $a$ , $b$ and $c$ are constants.

Note: $a, b$ and $c$ are called coefficients. More specifically, a coefficient is a number before something - so the coefficient of $x$ here is $b$ , the coefficient of $x^2$ is $a$ . You can consider $c$ to be the coefficient of $x^0$ .

To complete the square on a quadratic expression, it should be put into the form

$\boxed{a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a}+c=0}$

Once the quadratic has been put in the completed square form, it can be rearranged to solve for $x$ :

$x=-\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2}{4a^2}-\dfrac{c}{a}}$

This simplifies to

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

which you will recognise as the quadratic formula.

Tip: Rather than remembering a formula to complete the square, you should understand how it is done.

PROCEDURE

$1.$	Write the quadratic equation in the form $ax^2+bx+c=0$ .
$2.$	Factor the $a$ out of the first two terms: $a\left(x^2 +\dfrac{b}{a}x\right)+c=0$ . This is to make the coefficient of $x^2$ equal to $1$ .
$3$ .	Now focus on the $x^2 +\dfrac{b}{a}x$ part. Express this in the form $\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}$ . Note: If you expand this, you will see that these two expressions are equal. All that has been done is that the coefficient of the $x$ (namely $\dfrac{b}{a}$ ) has been halved and put in a pair of brackets with $x$ . These brackets have been squared. But this has a $\left(\dfrac{b}{2a}\right)^2$ term that is not in $x^2 +\dfrac{b}{a}x$ , hence you have to subtract it from $\left(x+\dfrac{b}{2a}\right)^2$ . Note: $\left(\dfrac{b}{2a}\right)^2=\dfrac{b^2}{4a^2}$ .
$4.$	Now you have $a\left(\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2}\right)+c=0$ . You want the $-\dfrac{b^2}{4a^2}$ outside of the brackets. Since it is inside brackets that are being multiplied by $a$ , you must multiply it by $a$ to take it outside the brackets: $-\dfrac{b^2}{4a^2}\times a=-\dfrac{b^2}{4a}$ .
$5.$	Now you have $a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a}+c=0$ . This is the completed the square form.

Example 1

Complete the square on the following quadratic:

$3x^2-12x+5$ 

It may help to identify the coefficients first:

$a=3$ , $b=-12$ and $c=5$ .

To complete the square, start by factoring out the $3$ from the first two terms:

$3(x^2-4x)+5$

Now halve the $-4$ and put it in brackets squared added to $x$ :

$(x-2)^2$

This does not equal $x^2-4x$ , so you must subtract $(-2)^2$ to make them equal:

$(x-2)^2-4$ 

This is now equal to $x^2-4x$ , so it can be substituted into $3(x^2-4x)+5$ :

$3((x-2)^2-4)+5$

Next you want to expand $-4$ out of the brackets. To do so you must multiply it by the $3$ :

$3(x-2)^2-12+5$ 

This can be simplified:

$\underline{3(x-2)^2-7}$ 

This is the correct completed the square form.

Using the completed the square form to solve quadratics

Once in the correct form, it is straightforward to solve the quadratic equation. The important thing to remember is to include the plus or minus ( $\pm$ ) when you take the square root, since both the positive route and the negative route lead to solutions.

Example 2

By completing the square, solve the following quadratic:

$2x^2+12x-3=0$ 

Start by factoring out the $2$ from the first two terms:

$2(x^2-6x)-3=0$

Now re-express the $x^2-6x$ as $(x-3)^2-9$ .

Note: The $-9$ is there because you need to subtract $(-3)^2$ from $(x-3)^2$ to make the two expressions equal.

Now you have

$2((x-3)^2-9)-3=0$ 

Expand the brackets:

$2(x-3)^2-18-3=0$ 

and tidy up:

$2(x-3)^2-21=0$ 

This can be rearranged to be solved:

$2(x-3)^2=21\\\space\\(x-3)^2=\dfrac{21}2\\\space\\x-3=\pm\sqrt{\dfrac{21}2}\\\space\\x=3\pm\sqrt{\dfrac{21}2}$ 

So the solutions are

$x=3+\sqrt{\dfrac{21}2}\approx\underline{6.24}$ and $x=3-\sqrt{\dfrac{21}2}\approx\underline{-0.24}$

Finding the vertex of a quadratic curve

The vertex is also known as the turning point of a quadratic curve. In a positive quadratic, this is the lowest point. In a negative quadratic, this is the highest point. Once in completed the square form

$y=a\left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a}+c$

the turning point has coordinates

$\left(-\dfrac{b}{2a},-\dfrac{b^2}{2a}+c\right)$

In other words, the $x$ -coordinate is the negative of the part inside the brackets with $x$ and the $y$ -coordinate is the bit added on to the bracket.