Quadratic Functions and Equations in One Variable
How to Complete the square
Use completing the square to find the minimum or maximum point of a quadratic, or to solve one that will not factorise neatly.
Before you start
- Expanding a squared bracket like (x + p)²
- Squaring positive and negative numbers
- Solving simple equations of the form (x + p)² = k
When to use it
Use completing the square to find the minimum or maximum point of a quadratic, or to solve one that will not factorise neatly.
The steps
- Make sure the coefficient of x² is 1; if not, divide the whole expression by it.
- Take half of the coefficient of x, then square it.
- Add and subtract that square inside the expression.
- Write the first three terms as a perfect square (x + p)², and simplify the constant.
- Read the turning point (−p, the constant) or solve by making the square equal to the other side.
Worked example
Express x² + 6x + 5 in the form (x + p)² + q and state the minimum point.
- The coefficient of x² is already 1, so no dividing is needed.
- Half of the coefficient of x: half of 6 is 3, and 3² = 9.
- Add and subtract 9: x² + 6x + 9 − 9 + 5.
- Write the first three terms as a square and simplify: (x + 3)² − 9 + 5 = (x + 3)² − 4.
- Here p = 3 and q = −4, so the turning point is (−3, −4); since a > 0 it is a minimum.
A second example, with a twist
The coefficient of x² is not 1, so you divide first, and here you solve an equation instead of just finding the turning point. Solve 2x² − 12x + 10 = 0 by completing the square.
- The coefficient of x² is 2, so divide the whole equation by 2: x² − 6x + 5 = 0.
- Half of the coefficient of x: half of −6 is −3, and (−3)² = 9.
- Add and subtract 9: x² − 6x + 9 − 9 + 5 = 0.
- Write the square and simplify the constant: (x − 3)² − 4 = 0.
- Make the square equal to the other side and solve: (x − 3)² = 4, so x − 3 = ±2, giving x = 5 or x = 1.
Practise this in a KBAT problem
Frequently asked questions
Do I halve the number in front of x or the whole x term?
You halve the coefficient, the plain number multiplying x, not the x itself. In x² + 6x, the coefficient is 6, so you use half of 6, which is 3.
Then you square that 3 to get 9. Keep the sign: with x² − 6x you halve −6 to get −3.
When is the point a minimum and when is it a maximum?
Look at the number in front of x². If it is positive, the parabola opens upward and the turning point is a minimum.
If it is negative, it opens downward and the turning point is a maximum. Completing the square gives the same point either way; only the shape decides which.
Why do I add and subtract the same number?
Adding and then subtracting the same value does not change the expression, so it stays equal. The trick lets you build a perfect square from the first three terms while the subtracted number keeps everything balanced.
It is the same expression, just rewritten in a more useful form.