Albert Teen

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Completing the Square

# Completing the Square

### Completing the Square

Completing the square is another option when factorising normally isn't possible.

Quadratic equations can't always be solved by factorising normally. In some cases, we need to factorise in a way that leaves a remainder, then take this away at the end. This is a method called 'completing the square'.

Have a look at $x^2 + 4x - 3 = 0$. Can you factorise this normally?

There aren't any factors which add to make $4$, and multiply to make $-3$. So we need a new approach! One way we can do this is through completing the square.

Let's have a go! Factorise $x^2 + 4x - 3 = 0$

1

First, halve the coefficient of $x$

$\dfrac{4}{2}=2$

2

Put this in square brackets along with $x$

$(x+ 2)^2$

3

Expand these brackets to see what you get

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

4

Compare this to the original equation

$(x^2 + 4x + 4)$ vs. $(x^2 + 4x - 3)$

5

What's the difference between $x^2+4x-3$ and the original equation?

6

The difference is 7

The difference between $4$ and $-3$ is 7.

7

Take the difference away from the brackets

$(x+ 2)^2 - 7$

8

Multiply out of the brackets to check

$(x+ 2)^2 - 7=x^2 + 4x - 3$

For $x^2 - 8x - 6$ in the form $(x + p)^2 + q$ where $p$ and $q$ are integers, what will be the value of $p$?

For $x^2 - 6x + 5 = (x + p)^2 - 4$ what is the value of $p$?

For $x^2 + 8x + 6 = (x + 4)^2 + q$, what is the value of $q$?

Sometimes you will have to complete the square when the $x^2$ coefficient is bigger than 1.

Let's try one! $4x^2 - 16x - 8 = 0$

1

Divide $4x^2 - 16x - 8 = 0$ by 4

Notice that there is a common factor of 4 in each term. Therefore, we can simplify the equation to $x^2 - 4x - 2=0$.

2

Halve the coefficient of $x$

$\dfrac{4}{2}=2$

3

Put this in square brackets along with$x$

$(x- 2)^2$

4

Expand these brackets to see what you get

$(x- 2)^2=x^2 - 4x + 4$

5

Compare this to the original equation

$(x^2 - 4x + 4)$ vs. $(x^2 - 4x - 2)$

6

What's the difference?

The difference is 6

7

Take the difference away from the brackets

$(x- 2)^2 - 6$

What is $x^2 - 14x - 8$ once you have completed the square?