Completing the square rewrites a quadratic in vertex form — a(x − h)2 + k. This form gives you the vertex, the maximum or minimum value, and lets you solve any quadratic — even ones that won’t factorise.
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What you need to know
The core trick: halve the coefficient of x and square it
For a monic quadratic x2 + bx + c: write as (x + b2)2 − (b2)2 + c
For a non-monic quadratic: factor out a from the x terms first, then complete the square inside
The vertex form a(x − h)2 + k shows the vertex (h, k) directly
The value k is the minimum if a > 0, and the maximum if a < 0
Why Bother?
Completing the square unlocks four useful things:
Finds the maximum or minimum of a quadratic — useful for the range
Gives the vertex instantly when sketching
Lets you solve quadratic equations that don’t factorise nicely
Is the technique used to derive the quadratic formula
The Core Trick
The whole technique relies on one move: halve the coefficient of x, then square it. Why? Because:
(x + b2)2 = x2 + bx + (b2)2
So x2 + bx is hiding inside (x + b2)2 — we just need to subtract the extra (b2)2 back off.
The Halve-and-Square Move
Start
x2 + 6x
→
Halve b
6 ÷ 2 = 3
→
Square it
32 = 9
x2 + 6x = (x + 3)2 − 9
Method 1: Monic Quadratics
For x2 + bx + c (where the coefficient of x2 is 1):
Halve b and write down x + b2 inside a squared bracket: (x + b2)2
Subtract the extra (b2)2 term that the squared bracket created.
Add the original constant c.
Simplify the constants together.
For example, x2 − 4x + 5:
Halve −4 → −2. Square it → 4. So (x − 2)2 = x2 − 4x + 4
Factor out a from the x terms only — leave c outside the bracket. Like this: a(x2 + bax) + c
Complete the square inside the bracket using the monic method.
Multiply through by a to expand the constant that came out.
Add the original c and simplify.
Why leave c outside? Factoring it out would create messy fractions. Keeping it outside means the only fraction you might deal with is ba inside the bracket.
Reading the Vertex Form
Once you’ve completed the square, you have the vertex form. Each part tells you something:
a(x − h)2 + k
a — Stretch
Sign tells you ∪ or ∩
h — x of vertex
Sign flips: (x + 3) means h = −3
k — y of vertex
Min or max value
Finding the Maximum or Minimum
The squared bracket (x − h)2 can never be negative — its smallest value is 0, when x = h. So:
a > 0
Vertex is the minimum. The smallest value is k, reached when x = h.
a < 0
Vertex is the maximum. The largest value is k, reached when x = h.
Watch for sneaky wording: questions don’t always say “complete the square” — they might ask for the maximum profit, minimum cost, or range of a function. If you see a quadratic and a max/min in the same question, completing the square is your tool.
Worked Examples
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Example 1 — Monic with positive middle term
Complete the square on x2 + 6x + 11.
Answer:
Step 1: halve b and square it.b = 6, b/2 = 3, (b/2)² = 9Step 2: write the squared bracket.(x + 3)² = x² + 6x + 9Step 3: subtract the extra 9 and add c = 11.x² + 6x + 11 = (x + 3)² − 9 + 11(x + 3)² + 2Vertex at (−3, 2). Min value is 2.
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Example 2 — Monic with negative middle term
Complete the square on x2 − 8x + 3.
Answer:
Step 1: halve b and square it.b = −8, b/2 = −4, (b/2)² = 16Step 2: write the squared bracket.(x − 4)² = x² − 8x + 16Step 3: subtract 16 and add c = 3.x² − 8x + 3 = (x − 4)² − 16 + 3(x − 4)² − 13Vertex at (4, −13). Min value is −13.
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Example 3 — Non-monic
Complete the square on 3x2 + 12x − 5.
Answer:
Step 1: factor 3 from the x terms only (leave −5 outside).3(x² + 4x) − 5Step 2: complete the square inside the bracket.half of 4 is 2, 2² = 43((x + 2)² − 4) − 5Step 3: multiply through by 3 and combine constants.3(x + 2)² − 12 − 53(x + 2)² − 17Vertex at (−2, −17). Since a = 3 > 0, this is a minimum.
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Example 4 — Non-monic with bigger coefficients
Complete the square on 2x2 + 12x − 3.
Answer:
Step 1: factor 2 from the x terms only.2(x² + 6x) − 3Step 2: complete the square inside.half of 6 is 3, 3² = 92((x + 3)² − 9) − 3Step 3: multiply through by 2.2(x + 3)² − 18 − 32(x + 3)² − 21
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Example 5 — Finding a minimum value
A function is given by f(x) = x2 − 10x + 27. Find the minimum value of f(x) and the value of x for which it occurs.
Answer:
Step 1: complete the square.half of −10 is −5, (−5)² = 25f(x) = (x − 5)² − 25 + 27f(x) = (x − 5)² + 2Step 2: read off the vertex from a(x − h)² + k.h = 5, k = 2Step 3: a = 1 > 0, so vertex is a minimum.Min value is 2, at x = 5The smallest (x − 5)² can be is 0, so f(x) is never less than 2.
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Tips
Mantra: “halve and square” — that’s the entire trick. Halve b, square the result.
For non-monic, leave c outside the bracket. You only factor a from the x terms — never from the constant.
Watch the sign flip when reading vertex coordinates: (x + 3)2 means h = −3, not +3.
Check your work by expanding back. If you complete the square on x2 + 6x + 11 and get (x + 3)2 + 2, expanding should give x2 + 6x + 9 + 2 = x2 + 6x + 11 ✓.
The minimum (or maximum) value is just k — once you have vertex form, you’ve already done the work.
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Common mistakes
Forgetting to subtract the (b/2)2 term. The squared bracket creates an extra term — you must subtract it back. Otherwise you’ve changed the value of the expression.
For non-monic, factoring a out of c as well. Don’t! Factor a only from the x terms. Keep c outside.
For non-monic, forgetting to multiply through by a when expanding the bracket. The −9 inside became −18 after multiplying by 2 — both terms inside get multiplied.
Sign error reading vertex form. (x + 3)2 + 2 has vertex (−3, 2). The h is the value that makes the bracket zero.
Confusing min and max based on the sign of a. Positive a → ∪ shape → vertex is MIN. Negative a → ∩ shape → vertex is MAX.
Stopping too early. Always combine the constants at the end: leaving (x − 4)2 − 16 + 3 instead of (x − 4)2 − 13 loses marks for incomplete simplification.
Final word: Halve, square, subtract, simplify. Once vertex form is in front of you, the vertex, the max/min, and even the roots all fall out without much extra work.
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