Completing the Square

How can I rewrite the first two terms of a quadratic expression as the difference of two squares?

Look at the quadratic expression x² + bx + c
The first two terms can be written as the difference of two squares using the following rule

$x^{2} + b x$ is the same as ${(x + p)}^{2} - p^{2}$ where $p$ is half of $b$

Check this is true by expanding the right-hand side
- Is $x^{2} + 2 x$ the same as ${(x + 1)}^{2} - 1^{2}$ ?
  - Yes: (x + 1)(x + 1) - 1² = x² + 2x + 1 - 1 = x² + 2x
This works for negative values of b too
- $x^{2} - 20 x$ can be written as ${(x - 10)}^{2} - {(- 10)}^{2}$ which is ${(x - 10)}^{2} - 100$
- A negative b does not change the sign at the end

How do I complete the square?

Completing the square is a way to rewrite a quadratic expression in a form containing a squared-bracket
To complete the square on x² + 10x + 9
- Use the rule above to replace the first two terms, x² + 10x, with (x + 5)² - 5²
- add 9: (x + 5)² - 5²+ 9
- simplify the numbers: (x + 5)² - 25+ 9
- answer: (x + 5)² - 16

How do I complete the square when there is a coefficient in front of the x² term?

You first need to take $a$ out as a factor of the x² and x terms only
- $a x^{2} + b x + c = a [x^{2} + \frac{b}{a} x] + c$
  - Use square-shaped brackets here to avoid confusion with curly brackets later
Then complete the square on the bit inside the square-brackets: $x^{2} + \frac{b}{a} x$
- This gives $a [{(x + p)}^{2} - p^{2}] + c$
  - where p is half of $\frac{b}{a}$
Finally multiply this expression by the a outside the square-brackets and add the c
- $a {(x + p)}^{2} - a p^{2} + c$
- This looks far more complicated than it is in practice!
  - Usually you are asked to give your final answer in the form $a {(x + p)}^{2} + q$
For quadratics like $- x^{2} + b x + c$ , do the above with a = -1

Completing the square Notes Diagram 2, A Level & AS Level Pure Maths Revision Notes

How do I find the turning point by completing the square?

Completing the square helps us find the turning point on a quadratic graph
- If $y = {(x + p)}^{2} + q$ then the turning point is at $(- p, q)$
  - Notice the negative sign in the x-coordinate
  - This links to transformations of graphs (translating $y = x^{2}$ by p to the left and q up)
- If $y = a {(x + p)}^{2} + q$ then the turning point is still at $(- p, q)$
  - It's at a minimum point if a > 0
  - It's at a maximum point if a < 0
It can also help you create the equation of a quadratic when given the turning point

Completing the square Notes Diagram 3, A Level & AS Level Pure Maths Revision Notes

It can also be used to prove and/or show results using the fact that any "squared term", i.e. the bracket (x ± p)², will always be greater than or equal to 0
- You cannot square a number and get a negative value

Completing the square Notes Diagram 4, A Level & AS Level Pure Maths Revision Notes

Exam Tip

To know if you have completed the square correctly, expand your answer to check.

Worked example

(a)

By completing the square, find the coordinates of the turning point on the graph of

y = x^{2} + 6 x - 11

Find half of +6 (call this p)

p = \frac{6}{2} = 3

Write x² + 6x in the form (x + p)² - p²

x^{2} + 6 x

is the same as

{(x + 3)}^{2} - 3^{2}

Put this result into the equation of the curve

y = {(x + 3)}^{2} - 3^{2} - 11

Simplify the numbers

y = {(x + 3)}^{2} - 20

Use that the turning point of

y = {(x + p)}^{2} + q

is at

(- p, q)

p = 3 and q = -20

turning point at (-3, -20)

(b)

Write

- 3 x^{2} + 12 x + 24

in the form

a {(x + p)}^{2} + q

Factorise -3 out of the first two terms only
Use square-shaped brackets

$- 3 [x^{2} - 4 x] + 24$

Complete the square on the x² - 4x inside the brackets (write in the form (x + p)² - p² where p is half of -4)

$- 3 [{(x - 2)}^{2} - {(- 2)}^{2}] + 24$

Simplify the numbers inside the brackets
(-2)² is 4

$- 3 [{(x - 2)}^{2} - 4] + 24$

Multiply -3 by all the terms inside the square-shaped brackets

$- 3 {(x - 2)}^{2} + 12 + 24$

Simplify the numbers

$- 3 {(x - 2)}^{2} + 36$

This is now in the form a(x + p)² + q where a = -3, p = -2 and q = 36

$- 3 {(x - 2)}^{2} + 36$

Solving by Completing the Square

How do I solve a quadratic equation by completing the square?

To solve x² + bx + c = 0
- replace the first two terms, x² + bx, with (x + p)² - p² where p is half of b
- this is called completing the square
  - x² + bx + c = 0 becomes
    - (x + p)² - p² + c = 0 where p is half of b
- rearrange this equation to make x the subject (using ±√)
For example, solve x² + 10x + 9 = 0 by completing the square
- x² + 10x becomes (x + 5)² - 5²
- so x² + 10x + 9 = 0 becomes (x + 5)² - 5²+ 9 = 0
- make x the subject (using ±√)
  - (x + 5)² - 25+ 9 = 0
  - (x + 5)² = 16
  - x + 5 = ±√16
  - x = ±4 - 5
  - x = -1 or x = -9

If the equation is ax² + bx + c = 0 with a number in front of x², then divide both sides by a first, before completing the square

How does completing the square link to the quadratic formula?

The quadratic formula actually comes from completing the square to solve ax² + bx + c = 0
- a, b and c are left as letters, to be as general as possible
You can see hints of this when you solve quadratics
- For example, solving x² + 10x + 9 = 0
  - by completing the square, (x + 5)² = 16 so x = ± 4 - 5 (from above)
  - by the quadratic formula, $x = \frac{- 10 \pm \sqrt{64}}{2} = - 5 \pm \frac{8}{2}$ = ± 4 - 5 (the same structure)

Exam Tip

When making x the subject to find the solutions at the end, don't expand the squared brackets back out again!
- Remember to use ±√ to get two solutions

Worked example

Solve $2 x^{2} - 8 x - 24 = 0$ by completing the square

Divide both sides by 2 to make the quadratic start with x²

$x^{2} - 4 x - 12 = 0$

Halve the middle number, -4, to get -2
Replace the first two terms, x² - 4x, with (x - 2)² - (-2)²

${(x - 2)}^{2} - {(- 2)}^{2} - 12 = 0$

Simplify the numbers

$\begin{array}{rcl} {(x - 2)}^{2} - 4 - 12 & = & 0 \\ {(x - 2)}^{2} - 16 & = & 0 \end{array}$

Add 16 to both sides

${(x - 2)}^{2} = 16$

Square root both sides
Include the ± sign to get two solutions

$x - 2 = \pm \sqrt{16} = \pm 4$

Add 2 to both sides

$x = \pm 4 + 2$

Work out each solution separately

x = 6 or x = -2

Completing the Square (CIE IGCSE Maths: Extended)

Revision Note