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Completing the square
Algebra

YOU ARE LEARNING:

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 x2+4x−3=0 x^2 + 4x - 3 = 0. Can you factorise this normally?

There aren't any factors which add to make 44, and multiply to make −3-3. So we need a new approach! One way we can do this is through completing the square.

Let's have a go! Factorise x2+4x−3=0x^2 + 4x - 3 = 0

1

First, halve the coefficient of xx

42=2\dfrac{4}{2}=2

2

Put this in square brackets along with xx

(x+2)2(x+ 2)^2

3

Expand these brackets to see what you get

(x+2)2=x2+4x+4(x+ 2)^2 = x^2 + 4x + 4

4

Compare this to the original equation

(x2+4x+4)(x^2 + 4x + 4) vs. (x2+4x−3)(x^2 + 4x - 3)

5

What's the difference between x2+4x−3x^2+4x-3 and the original equation?

6

The difference is 7

The difference between 44 and −3-3 is 7.

7

Take the difference away from the brackets

(x+2)2−7(x+ 2)^2 - 7

8

Multiply out of the brackets to check

(x+2)2−7=x2+4x−3(x+ 2)^2 - 7=x^2 + 4x - 3

For x2−8x−6x^2 - 8x - 6 in the form (x+p)2+q(x + p)^2 + q where pp and qq are integers, what will be the value of pp?

For x2−6x+5=(x+p)2−4x^2 - 6x + 5 = (x + p)^2 - 4 what is the value of pp?

For x2+8x+6=(x+4)2+qx^2 + 8x + 6 = (x + 4)^2 + q, what is the value of qq?

Sometimes you will have to complete the square when the x2x^2 coefficient is bigger than 1.

Let's try one! 4x2−16x−8=0 4x^2 - 16x - 8 = 0

1

Divide 4x2−16x−8=04x^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 x2−4x−2=0x^2 - 4x - 2=0.

2

Halve the coefficient of xx

42=2\dfrac{4}{2}=2

3

Put this in square brackets along withx x

(x−2)2(x- 2)^2

4

Expand these brackets to see what you get

(x−2)2=x2−4x+4(x- 2)^2=x^2 - 4x + 4

5

Compare this to the original equation

(x2−4x+4)(x^2 - 4x + 4) vs. (x2−4x−2)(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(x- 2)^2 - 6

What is x2−14x−8x^2 - 14x - 8 once you have completed the square?