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Completing the Square Method –

GCSEMathsAI Team·8 min read·23 May 2026

Completing the square is a Higher tier technique that rewrites a quadratic expression into a form that reveals the turning point of the parabola. It is also a powerful method for solving quadratic equations, especially when the question demands exact (surd) answers.

What Is Completing the Square?

Completing the square transforms a quadratic expression from the standard form x² + bx + c into the equivalent form (x + p)² + q. In this new form, the turning point (vertex) of the parabola y = x² + bx + c is immediately visible at the coordinates (-p, q).

The technique works by constructing a perfect square from the x² and bx terms, then adjusting the constant to compensate. "Completing" refers to adding the right amount to make a perfect square trinomial, and then subtracting that same amount so the expression stays equivalent.

Once the expression is in completed square form, it can also be used to solve equations. Setting (x + p)² + q = 0 and rearranging gives (x + p)² = -q. If -q is non-negative, taking the square root of both sides (remembering the plus-or-minus) leads to the two solutions.

Key Formulas

x² + bx + c = (x + b/2)² - (b/2)² + c
Turning point of y = (x + p)² + q is (-p, q)

Step-by-Step Method

  1. Start with the quadratic x² + bx + c.
  2. Halve the coefficient of x to get b/2.
  3. Write the perfect square bracket: (x + b/2)².
  4. Expanding this bracket gives x² + bx + (b/2)². Since you only had x² + bx, you have introduced an extra (b/2)².
  5. Subtract (b/2)² and add back the original constant c: the completed form is (x + b/2)² - (b/2)² + c.
  6. Simplify the constant part to get the final form (x + p)² + q.

Worked Example 1 — Foundation Level

Question: Write x² + 6x + 2 in the form (x + p)² + q.

Working:

Step 1 — Halve the coefficient of x: 6 / 2 = 3.

Step 2 — Write the bracket: (x + 3)².

Step 3 — Expanding (x + 3)² gives x² + 6x + 9. The extra amount introduced is 9.

Step 4 — Subtract 9 and add the original constant 2: (x + 3)² - 9 + 2 = (x + 3)² - 7.

Check: (x + 3)² - 7 = x² + 6x + 9 - 7 = x² + 6x + 2. Correct.

Answer: (x + 3)² - 7

Worked Example 2 — Higher Level

Question: Find the turning point of y = x² - 10x + 18.

Working:

Step 1 — Halve the coefficient of x: -10 / 2 = -5.

Step 2 — Write the bracket: (x - 5)².

Step 3 — Expanding gives x² - 10x + 25. Extra amount: 25.

Step 4 — Subtract 25 and add 18: (x - 5)² - 25 + 18 = (x - 5)² - 7.

The turning point is at (-(-5), -7) = (5, -7). Since the coefficient of x² is positive, this is a minimum point.

Answer: Turning point is (5, -7); it is a minimum.

Worked Example 3 — Exam Style

Question: Solve x² + 4x - 3 = 0 by completing the square. Give your answers in surd form. (4 marks)

Working:

Step 1 — Complete the square on x² + 4x - 3.

Halve 4: gives 2. Write (x + 2)². Expanding gives x² + 4x + 4. Extra: 4.

So x² + 4x - 3 = (x + 2)² - 4 - 3 = (x + 2)² - 7.

Step 2 — Set equal to zero: (x + 2)² - 7 = 0.

Step 3 — Rearrange: (x + 2)² = 7.

Step 4 — Square root both sides: x + 2 = ±sqrt(7).

Step 5 — Subtract 2: x = -2 ± sqrt(7).

Check: x = -2 + sqrt(7) is approximately 0.646. (0.646)² + 4(0.646) - 3 = 0.417 + 2.584 - 3 = 0.001, which is approximately 0. Correct.

Answer: x = -2 + sqrt(7) or x = -2 - sqrt(7)

Common Mistakes

  • Forgetting to subtract the square you introduced. When you write (x + 3)², you create an extra +9. You must subtract 9 to keep the expression equivalent to the original.
  • Getting the sign of p wrong. If the coefficient of x is -8, then p = -4, not +4. Always halve including the sign.
  • Confusing the turning point coordinates. In (x + 3)² - 7, the turning point is (-3, -7), not (3, -7). The x-coordinate is the opposite sign of what appears inside the bracket.

Exam Tips

  • Read whether the question asks you to "write in the form (x + p)² + q" or to "solve by completing the square." The first requires only the rewrite; the second requires finding x values.
  • For "find the minimum value" questions, complete the square and state q. The minimum value is q, not the full coordinate pair.
  • Practise with negative coefficients of x, as these are where most students make sign errors.

Practice Questions

Q1 (Higher): Write x² + 8x + 10 in the form (x + p)² + q.

Answer: Halve 8 gives 4. (x + 4)² = x² + 8x + 16. Extra = 16. (x + 4)² - 16 + 10 = (x + 4)² - 6.

Q2 (Higher): Find the minimum value of x² - 12x + 40.

Answer: Halve -12 gives -6. (x - 6)² = x² - 12x + 36. Extra = 36. (x - 6)² - 36 + 40 = (x - 6)² + 4. Minimum value is 4, occurring when x = 6.

Q3 (Higher): Solve x² - 2x - 5 = 0 by completing the square, giving answers in surd form.

Answer: (x - 1)² - 1 - 5 = 0. (x - 1)² = 6. x - 1 = ±sqrt(6). x = 1 + sqrt(6) or x = 1 - sqrt(6).

Practise completing the square method questions with instant feedback — completely free on GCSEMathsAI.

Summary

  • Completing the square rewrites x² + bx + c as (x + b/2)² - (b/2)² + c.
  • The turning point of the parabola y = (x + p)² + q is at (-p, q).
  • For monic quadratics, halve the coefficient of x, square it, then adjust the constant.
  • The method can solve quadratic equations by isolating the squared bracket and taking square roots with plus-or-minus.
  • Always remember that the x-coordinate of the turning point is the opposite sign of what appears in the bracket.

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§Academic References

Further reading from leading academic institutions — free and open-access.

N
QuadraticsNRICH

Quadratic equations and graphs — Cambridge problem sets.

University of Cambridge · Free · Open Access
C
QuadraticsCorbett Maths

Factorising, formula, completing the square — all methods.

Corbett Maths · Free · Open Access
M
Quadratic FunctionsMIT OpenCourseWare

MIT treatment of quadratic functions and their properties.

Massachusetts Institute of Technology · Free · Open Access
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