Why do we complete the square?

It rewrites a quadratic so its turning point can be read off directly, and it solves equations that do not factorise. The quadratic formula itself comes from completing the square.

How do I find the turning point from the completed square?

For a(x − p)² + q, the turning point is (p, q). Watch the sign: (x + 3)² − 4 has its turning point at (−3, −4).

What if the coefficient of x² is negative?

Take out −1 (or the negative number) from the x terms first, complete the square inside the bracket, then multiply back out carefully.

How to Complete the Square (Even When a Is Not 1)

Short answer

How do you complete the square?

Halve the coefficient of x, square it, then add and subtract that number: x² + 6x + 5 = (x + 3)² − 9 + 5 = (x + 3)² − 4. If the x² coefficient is not 1, take it out as a factor of the x terms first.

Completing the square rewrites $a x^{2} + b x + c$ as $a (x - p)^{2} + q$ . In that form the turning point is simply $(p, q)$ .

When a = 1

Take $x^{2} + 6 x + 5$ .

Halve the coefficient of $x$ : half of 6 is 3.
Write $(x + 3)^{2}$ . Expanding it gives $x^{2} + 6 x + 9$ , which is 9 too many.
Subtract that 9 and keep the constant:

x^{2} + 6 x + 5 = (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4

The turning point is $(- 3, - 4)$ .

When a is not 1

Take $2 x^{2} - 8 x + 3$ . Factor the 2 out of the $x$ terms only, and leave the constant outside.

2 x^{2} - 8 x + 3 = 2 (x^{2} - 4 x) + 3

Complete the square inside the bracket, then multiply the correction by 2:

= 2 [(x - 2)^{2} - 4] + 3 = 2 (x - 2)^{2} - 8 + 3 = 2 (x - 2)^{2} - 5

Another one, with $a = 3$ :

3 x^{2} + 12 x - 1 = 3 (x^{2} + 4 x) - 1 = 3 [(x + 2)^{2} - 4] - 1 = 3 (x + 2)^{2} - 13

When a is negative

Take $- x^{2} + 4 x + 1$ . Factor out $- 1$ :

- x^{2} + 4 x + 1 = - (x^{2} - 4 x) + 1 = - [(x - 2)^{2} - 4] + 1 = - (x - 2)^{2} + 5

The curve opens downwards, so $(2, 5)$ is a maximum.

Try it

Predict the completed square first, then check the working.

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Mistakes that cost marks

Forgetting to multiply the correction by $a$ . In $2 [(x - 2)^{2} - 4]$ the $- 4$ becomes $- 8$ .
Reading the turning point with the wrong sign. $(x + 3)^{2}$ means $p = - 3$ .
Factoring $a$ out of the constant too. Only the $x^{2}$ and $x$ terms go in the bracket.

For exam questions that use this, see 4024 quadratic equations practice.

How to complete the square