Scholars Gate
Find a TutorOnline Tutoring
GuidesHow It WorksBecome a Tutor
Log inGet Started
All guides
MathematicsGCSE 8 min read

Completing the Square, Step by Step

Step through the algebra one line at a time, then drag the parabola to see exactly what vertex form is telling you.

The ScholarsGate Maths Team·Updated 03 Jul 2026

On this page

  • Why bother?
  • Step by step
  • The "square" you complete
  • What vertex form tells you
  • Worked examples
  • Common mistakes
  • FAQ

Completing the square rewrites a quadratic ax2+bx+cax^2+bx+cax2+bx+c in the form a(x+p)2+qa(x+p)^2+qa(x+p)2+q. That single rearrangement hands you the turning point, the minimum (or maximum) value and the line of symmetry for free — no calculus required. Here’s the method, step by step, with the geometry that gives it its name.

Why bother?

A quadratic in standard form, x2+4x+1x^2+4x+1x2+4x+1, hides its most useful features. In completed-square form, (x+2)2−3(x+2)^2-3(x+2)2−3, everything is on display: the graph is the basic parabola x2x^2x2 shifted 2 left and 3 down, so its lowest point is at (−2,−3)(-2,-3)(−2,−3). Completing the square is also how the quadratic formula itself is derived, and it’s a tool you’ll reuse in integration and coordinate geometry later.

You’re trading a form that’s easy to expand for one that’s easy to read.

Step by step

Watch the manipulation unfold one line at a time for x2+4x+1x^2+4x+1x2+4x+1. The key move is in step 3: halve the coefficient of xxx, then subtract the square of that half to undo what the bracket secretly adds.

InteractiveCompleting the square, line by line
Loading interactive…
Step through the algebra. Each line shows one move and why it keeps the expression equal.
Text description ↓Hide text description ↑

A step-by-step derivation of completing the square for x² + 4x + 1:

  1. x² + 4x + 1 (start)
  2. (x² + 4x) + 1 — group the x terms
  3. (x + 2)² − 4 + 1 — half of 4 is 2; the bracket adds an extra 2² = 4, so subtract it
  4. (x + 2)² − 3 — simplify the constants
The method (when the x² coefficient is 1)
x2+bx+c=(x+b2)2−(b2)2+cx^2+bx+c = \left(x+\tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2 + cx2+bx+c=(x+2b​)2−(2b​)2+c

Halve bbb to build the bracket, then subtract (b/2)2(b/2)^2(b/2)2 to cancel the term the bracket adds by itself.

The “square” you’re completing

The name is literal. Picture x2+4xx^2+4xx2+4x as an x×xx\times xx×x square plus two 2×x2\times x2×x rectangles (splitting the 4x4x4x as 2x+2x2x+2x2x+2x along two sides). Those pieces almost form a bigger square of side x+2x+2x+2 — but a 2×22\times 22×2 corner is missing. Add that 444to “complete” the square into (x+2)2(x+2)^2(x+2)2, and subtract it again to keep the value unchanged. That missing corner is exactly the (b/2)2(b/2)^2(b/2)2 you subtract.

What vertex form tells you

Once a quadratic is a(x+p)2+qa(x+p)^2+qa(x+p)2+q, the graph reads off instantly: the turning point is at (−p, q)(-p,\,q)(−p,q) and the line of symmetry is x=−px=-px=−p. Drag the point on x2+4x+1x^2+4x+1x2+4x+1 below and confirm the lowest point sits at (−2,−3)(-2,-3)(−2,−3), precisely the numbers in (x+2)2−3(x+2)^2-3(x+2)2−3.

InteractiveThe parabola behind the algebra
Loading interactive…
The minimum sits at (−2, −3) — the same −2 and −3 that appear in (x + 2)² − 3.
Text description ↓Hide text description ↑

An interactive parabola of y = x² + 4x + 1. Its vertex (minimum) is at (−2, −3) and its line of symmetry is x = −2, matching the completed-square form (x + 2)² − 3.

Worked examples

1Worked example — a negative constant

Complete the square for x2−6x+5x^2 - 6x + 5x2−6x+5. Half of −6-6−6 is −3-3−3, and (−3)2=9(-3)^2 = 9(−3)2=9:

x2−6x+5=(x−3)2−9+5=(x−3)2−4.x^2-6x+5 = (x-3)^2 - 9 + 5 = (x-3)^2 - 4.x2−6x+5=(x−3)2−9+5=(x−3)2−4.

Turning point (3,−4)(3,-4)(3,−4); minimum value −4-4−4.

2Worked example — when a ≠ 1

For 2x2+12x+72x^2+12x+72x2+12x+7, factor the 222 out of the xxx terms first:

2(x2+6x)+7=2((x+3)2−9)+7=2(x+3)2−11.2\big(x^2+6x\big)+7 = 2\big((x+3)^2-9\big)+7 = 2(x+3)^2 - 11.2(x2+6x)+7=2((x+3)2−9)+7=2(x+3)2−11.

Turning point (−3,−11)(-3,-11)(−3,−11). Remember to multiply the −9-9−9 back by the 222.

Common mistakes

Forgetting to subtract (b/2)²

(x+2)2(x+2)^2(x+2)2 expands to x2+4x+4x^2+4x+4x2+4x+4 — it carries an extra +4+4+4. If you don’t subtract it, your expression is no longer equal to the original. Always compensate.

Not multiplying the correction back by a

When a≠1a\neq 1a=1 and you’ve factored aaa out, the −(b/2)2-(b/2)^2−(b/2)2 is inside the bracket. Bringing it outside means multiplying by aaa (as in Example 2, where −9-9−9 became −18-18−18).

Your turn

Write x2+8x+3x^2 + 8x + 3x2+8x+3 in completed-square form and state its minimum point.

Show the answer ↓Hide the answer ↑

Half of 8 is 4, 42=164^2=1642=16: (x+4)2−16+3=(x+4)2−13(x+4)^2 - 16 + 3 = (x+4)^2 - 13(x+4)2−16+3=(x+4)2−13. Minimum at (−4,−13)(-4,-13)(−4,−13).

Key takeaways
  • Completing the square turns ax2+bx+cax^2+bx+cax2+bx+c into a(x+p)2+qa(x+p)^2+qa(x+p)2+q.
  • When a=1a=1a=1: halve bbb for the bracket, subtract (b/2)2(b/2)^2(b/2)2 to compensate.
  • Vertex form reveals the turning point (−p,q)(-p,q)(−p,q) and line of symmetry x=−px=-px=−p at a glance.
  • When a≠1a\neq1a=1, factor aaa from the xxx-terms first and multiply the correction back.

Frequently asked questions

What does completing the square mean?+
It means rewriting a quadratic ax² + bx + c in the form a(x + p)² + q. This "completed square" form makes the turning point, minimum or maximum value, and line of symmetry immediately readable.
How do you complete the square when the coefficient of x² is 1?+
Halve the coefficient of x to get p, then subtract p² to compensate: x² + bx + c = (x + b/2)² − (b/2)² + c. The bracket squared reproduces the bx term, and you correct the extra p² you introduced.
What is the turning point from completed square form?+
For a(x + p)² + q, the turning point is at (−p, q). If a > 0 it is a minimum; if a < 0 it is a maximum. The line of symmetry is x = −p.
Why is completing the square useful?+
It lets you find the exact vertex without calculus, solve any quadratic (it is how the quadratic formula is derived), sketch parabolas quickly, and integrate certain functions in later study.
SM

The ScholarsGate Maths Team

Oxbridge & Russell Group maths tutors

Written and reviewed by ScholarsGate tutors who teach A-Level and undergraduate mathematics. Every explainer is checked for accuracy against the AQA, Edexcel and OCR specifications.

Keep exploring

GCSE Maths tutorsThe chain rule, made intuitiveSine & cosine from the unit circleAll Mathematics tutors

Want a tutor to walk you through it?

Book a DBS-checked GCSE Mathematics tutor for a 1-on-1 lesson — online or in person.

Find a GCSE Mathematics tutor
Scholars GateTutoring & Learning

ScholarsGate (Scholars Gate) is a premium UK tutoring marketplace connecting ambitious students with expert tutors across every subject, level, and admissions pathway.

hello@scholarsgate.co.uk
+44 7544 736438
Bath, United Kingdom

Platform

  • Find a Tutor
  • Online Tutoring
  • How It Works
  • Become a Tutor
  • Interactive Guides
  • Blog
  • FAQ

Subjects

  • Mathematics & Computing
  • Natural Sciences
  • Humanities
  • Languages
  • All Subjects

Admissions

  • Medicine & UCAT
  • Maths (TMUA, STEP)
  • Law & LNAT
  • Statements & Interviews
  • All Admissions
4.9/5 average tutor rating
All tutors DBS verified
Secure payments via Stripe
500+ expert tutors

© 2026 ScholarsGate Ltd. All rights reserved.

Privacy PolicyTerms of ServiceCookie PolicySafeguarding