To complete the square for ax² + bx + c: halve the coefficient of x, write it inside a squared bracket (x + b/2)², then subtract the square of that number, and add c. The result a(x + h)² + k gives the minimum (or maximum) point at (−h, k), the line of symmetry at x = −h, and lets you solve the quadratic by isolating x. This form is called vertex form.
The standard form ax² + bx + c tells you almost nothing useful at a glance. The completed square form a(x + h)² + k immediately reveals three things examiners love to ask about:
Minimum / maximum point
(−h, k)
Minimum if a > 0, maximum if a < 0
Line of symmetry
x = −h
The vertical axis of the parabola
Minimum / maximum value
y = k
The y-coordinate of the turning point
Notice the negative sign: if the completed square form is (x + 3)² − 5, then h = 3, so the line of symmetry is x = −3 and the minimum point is (−3, −5). Getting the sign wrong on the x-coordinate of the turning point is the single most common error in this topic.
Start with the simpler case where the leading coefficient is already 1.
Example 1 — express x² + 6x + 2 in completed square form
6 ÷ 2 = 3. Write the squared bracket: (x + 3)².(x + 3)² actually gives: x² + 6x + 9. Our expression has +2 at the end, not +9. So we've introduced an extra +9 that must be corrected.(x + 3)² − 9 + 2.(x + 3)² − 7.(−3, −7), line of symmetry x = −3, minimum value y = −7.When a ≠ 1, factorise the coefficient out of the x² and x terms first before completing the square inside the bracket.
Example 2 — express 2x² + 12x + 5 in completed square form
2(x² + 6x) + 5.6 ÷ 2 = 3. Write (x + 3)² and subtract 9 inside the bracket: 2(x² + 6x + 9 − 9) + 5.2(x + 3)² + 2(−9) + 5.2(x + 3)² − 18 + 5 = 2(x + 3)² − 13.(−3, −13), line of symmetry x = −3, minimum value y = −13. Since a = 2 > 0, it's a minimum (U-shaped).The most common mistake with a ≠ 1
When the −9 comes out of the bracket, students forget to multiply it by the coefficient outside. The bracket was 2(... − 9), so the −9 exits as 2 × (−9) = −18, not just −9. Every term inside the bracket is multiplied by 2 — including the correction term.
Once in vertex form, you can solve a(x + h)² + k = 0 without factorising or using the quadratic formula.
Example 3 — solve x² + 6x + 2 = 0, leaving answers in surd form
(x + 3)² − 7 = 0.(x + 3)² = 7.x + 3 = ±√7.x = −3 + √7 or x = −3 − √7.x ≈ −0.354 or x ≈ −5.646 (3 s.f.).When the question says "leave in surd form" or "give exact answers", completing the square is the intended method — not the quadratic formula. Surds appear naturally from the ±√ step and require no further simplification.
When the coefficient of x² is negative, the parabola is ∩-shaped, and the turning point is a maximum, not a minimum.
Example 4 — express −x² + 4x + 1 in completed square form
−(x² − 4x) + 1.(x − 2)², subtract the square inside: −(x² − 4x + 4 − 4) + 1.−(x − 2)² + (−1)(−4) + 1 = −(x − 2)² + 4 + 1.−(x − 2)² + 5.(2, 5). Line of symmetry: x = 2. Maximum value: y = 5. Note: h = −2, so x = −(−2) = 2. Sign reversal applies here too.Can I use the quadratic formula instead of completing the square?
For finding roots only, yes. But the quadratic formula does not give you the turning point or line of symmetry directly. If the question asks for the minimum point, line of symmetry, or says "express in the form a(x + h)² + k", you must complete the square — the quadratic formula will not earn the marks. See three ways to solve a quadratic for when each method is the right choice.
The x-coordinate of the minimum point is −h, but my completed square has (x + 3)². Doesn't that make h = 3, so x = 3?
This is the most common sign error. The bracket (x + 3)² equals zero when x = −3, not +3. The minimum occurs at the x-value that makes the squared term zero, because that's where the expression is smallest. So minimum point x-coordinate = −3, not +3. Always reverse the sign of whatever number is inside the bracket.
How do I check my completed square answer?
Expand your answer back to standard form. If you started with x² + 6x + 2 and got (x + 3)² − 7, expand: x² + 6x + 9 − 7 = x² + 6x + 2. Matches — correct. This check takes 20 seconds and catches sign errors before they cost marks.
Does completing the square appear in O-Level Paper 1 or Paper 2?
Both, but in different forms. Paper 1 typically asks a direct "express in the form" or "state the minimum point" question worth 2–3 marks. Paper 2 usually embeds it inside a longer graph question where you also need to sketch the curve, find intercepts, and describe transformations — a 6–8 mark question where completing the square is the foundation for everything that follows.
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