Completing the square writes ax² + bx + c in the form a(x + p)² + q. From that form you can read off the turning point, state the range, find exact roots, prove that an expression is always positive, and build a quadratic from its vertex. At A-Level you must handle a ≠ 1, including negative values of a, by factoring a out of the x terms first.
Completing the square
- x² + bx + c = (x + b/2)² − (b/2)² + c
- ax² + bx + c: factor a out of the x² and x terms first
- a(x − h)² + k has turning point (h, k)
- a > 0: minimum, range y ≥ k. a < 0: maximum, range y ≤ k
Completing the square when a = 1
Halve the coefficient of x, square the bracket, then subtract the square you have added and include the constant: x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8. This works for odd b too, which simply gives a fraction.
The number inside the bracket is half of b, not b. That is the most common slip.
When a is not 1, including negative a
Factor a out of the x² and x terms only, not the constant. Complete the square inside the bracket, then multiply back out and collect the constants. With a negative a, factoring it out flips every sign inside: −x² + 6x − 1 = −(x² − 6x) − 1 = −[(x − 3)² − 9] − 1 = −(x − 3)² + 8.
Turning points, maximum and minimum
The completed form a(x − h)² + k gives the turning point (h, k) directly. If a > 0 the curve is U-shaped, the point is a minimum, and the range is y ≥ k. If a < 0 the curve is an upside-down U, the point is a maximum, and the range is y ≤ k.
Watch the sign in the bracket: (x − 3)² means h = 3, not −3.
Exact roots and “for all x” proofs
Setting the completed form equal to zero gives exact roots: (x + 3)² = 7 means x = −3 ± √7. Completing the square also proves inequalities that hold for every x. A squared bracket is never negative, so (x − h)² + k ≥ k. To show an expression is always positive, complete the square and point to the square being at least zero.
Building a quadratic from its turning point
Reverse the process: a turning point (h, k) gives y = a(x − h)² + k. One more point on the curve fixes a; substitute it and solve. A maximum needs a < 0 and a minimum needs a > 0, which is a useful check.
Worked examples
Example 1. Write x² + 6x + 1 in the form (x + p)² + q.
- Half of 6 is 3, so start with (x + 3)².
- (x + 3)² = x² + 6x + 9, which is 9 too many, so subtract 9 and add the 1.
Answer: (x + 3)² − 8.
Marks: M1 for (x + 3)², A1 for (x + 3)² − 8.
Example 2. Write 2x² + 8x + 3 in the form a(x + p)² + q.
- Factor 2 from the x terms: 2(x² + 4x) + 3.
- Complete the square inside: x² + 4x = (x + 2)² − 4, so 2[(x + 2)² − 4] + 3.
- Multiply out: 2(x + 2)² − 8 + 3.
Answer: 2(x + 2)² − 5.
Marks: M1 for factoring 2, M1 for completing inside, A1 for the answer.
Example 3. By completing the square, find the minimum point of y = x² − 4x + 7.
- x² − 4x + 7 = (x − 2)² − 4 + 7 = (x − 2)² + 3.
- The turning point is (2, 3), and a = 1 > 0, so it is a minimum.
Answer: Minimum point (2, 3).
Marks: M1 for the completed form, A1 for (2, 3).
Example 4. Solve x² + 6x + 2 = 0 by completing the square.
- (x + 3)² − 9 + 2 = 0, so (x + 3)² = 7.
- x + 3 = ±√7.
Answer: x = −3 ± √7.
Marks: M1 for (x + 3)² = 7, A1 for x = −3 ± √7.
Example 5. A curve has turning point (2, −3) and passes through (0, 5). Find its equation.
- Vertex form: y = a(x − 2)² − 3.
- Use (0, 5): 4a − 3 = 5, so a = 2.
- Expand: y = 2(x − 2)² − 3 = 2x² − 8x + 5.
Answer: y = 2x² − 8x + 5.
Marks: M1 for the vertex form, M1 for finding a, A1 for the equation.
Common mistakes
- Using b instead of b/2 in the bracket. x² + 6x needs (x + 3)², not (x + 6)².
- Factoring a out of the constant. Take a out of the x² and x terms only.
- Sign errors with negative a. Factoring out −1 changes every sign inside the bracket.
- Reading the turning point's sign wrongly. (x − 2)² + 3 has turning point (2, 3).
- Forgetting ± when square-rooting. (x + 3)² = 7 gives two roots, −3 + √7 and −3 − √7.
How completing the square is examined at A-Level
Completing the square is core Pure 1 content (WMA11) in Edexcel International A-Level, and part of first-year Pure in Edexcel A-Level (9MA0). Questions ask you to:
- write a quadratic in the form a(x + p)² + q, often with a negative or non-unit a
- state the turning point and sketch the curve
- find the range of a quadratic function
- find a quadratic from its maximum or minimum point and one other point
Questions in these formats, with mark schemes, are among the Pure 1 quadratics exam questions. The quadratics worksheet and the algebra revision engine give extra practice.
Completing the square is used again for the equation of a circle and for the range of a function. For the GCSE version, see completing the square at GCSE.
Frequently asked questions
What is completing the square used for?
Finding the turning point and range of a quadratic, solving it exactly, proving it is always positive or negative, and rewriting circle equations.
How do you complete the square when a is negative?
Factor the negative out of the x² and x terms, complete the square inside the bracket, then multiply back out, taking care with every sign.
How do you find the turning point from a completed square?
For a(x − h)² + k the turning point is (h, k): a minimum if a > 0 and a maximum if a < 0.