What is 'completing the square' in the context of solving quadratic equations?

Completing the square is a method used in Mathematics, particularly in the Edexcel IGCSE curriculum, to solve quadratic equations. It involves rewriting a quadratic equation in the form of (x + p)^2 = q, which makes it easier to solve for x. This technique is part of the Equations, Formulae, and Identities topic and helps in understanding the properties of quadratic functions.

How do you complete the square for the quadratic equation x^2 + 6x + 5?

To complete the square for the equation x^2 + 6x + 5, you first focus on the x terms: x^2 + 6x. Take half of the coefficient of x, which is 6, giving you 3, and square it to get 9. Add and subtract 9 inside the equation: x^2 + 6x + 9 - 9 + 5. This simplifies to (x + 3)^2 - 4. Thus, the equation x^2 + 6x + 5 can be rewritten as (x + 3)^2 - 4.

Why is completing the square useful in Mathematics?

Completing the square is useful because it transforms quadratic equations into a form that is easier to solve and analyze. It is particularly helpful in finding the vertex of a parabola, solving quadratic equations, and deriving the quadratic formula. This method is a key component of the Edexcel IGCSE Mathematics curriculum, aiding students in understanding the geometric interpretation of quadratic functions.

Can completing the square be used to derive the quadratic formula?

Yes, completing the square can be used to derive the quadratic formula. By applying the method to the general quadratic equation ax^2 + bx + c = 0, you can transform it into a completed square form and solve for x. This process leads to the quadratic formula x = (-b ± √(b^2 - 4ac)) / (2a), which is a fundamental tool in solving quadratic equations in the Edexcel IGCSE Mathematics syllabus.

What are the steps to complete the square for a quadratic equation with a leading coefficient other than 1?

To complete the square for a quadratic equation with a leading coefficient other than 1, such as ax^2 + bx + c, follow these steps: 1) Divide the entire equation by a to make the coefficient of x^2 equal to 1. 2) Focus on the x terms, x^2 + (b/a)x. 3) Take half of the coefficient of x, square it, and add and subtract this square inside the equation. 4) Rewrite the equation in the form (x + p)^2 = q. This method is crucial for solving complex quadratic equations in the Edexcel IGCSE Mathematics curriculum.

Why is "Forgetting to subtract the extra term" a common mistake in Completing the square?

Why it happens: Hurried. How to avoid it: (x + 3)^2 = x^2 + 6x + 9, NOT x^2 + 6x. So when you write (x + 3)^2 for x^2 + 6x + ..., you must SUBTRACT 9 to compensate.

Why is "Not factoring out the leading coefficient when $a > 1$" a common mistake in Completing the square?

Why it happens: Skipping a step. How to avoid it: 2x^2 - 12x + 5 → factor 2: 2(x^2 - 6x) + 5. Complete inside, then re-multiply. Source: 4MA1/1H May/Jun 2024 — examiner report Q18

Why is "Reading vertex as $(p, q)$ instead of $(-p, q)$" a common mistake in Completing the square?

Why it happens: Form has +p, but vertex flips sign. How to avoid it: (x + p)^2 + q → minimum at x = -p (because that's when squared term is zero).

Edexcel IGCSE Mathematics (4MA1)

Completing the square

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Short Notes - Completing the square

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Detailed Study Notes

Detailed notes on Equations, Formulae and Identities for Edexcel IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Completing the Square Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Convert $ax^{2} + bx + c$ to $a(x + p)^{2} + q$ . Useful for solving, finding turning points, and proof. Standard A* technique.

At a glance

Form: $(x + p)^{2} + q$ (vertex form).
$p = b/2$ when leading coef is $1$ .
$q = c - b^{2}/4$ for monic.
Non-monic: factor out leading coef first.
Turning point: $(x = -p, y = q)$ .
Used for: solve, vertex, proof.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

2.7 — Complete the square for monic and non-monic quadratics.
2.7 — Use completing the square to solve equations.
2.7 — Find the turning point of a parabola.

Completing the square — leading coefficient $1$

Half the coefficient. Square it. Subtract.

Form: $x^{2} + bx + c$ .

Method:

Take half the coefficient of $x$ : $p = b/2$ .
Write $(x + p)^{2}$ , which expands to $x^{2} + bx + p^{2}$ .
Subtract $p^{2}$ to compensate, then add $c$ :

$x^{2} + bx + c = (x + b/2)^{2} - (b/2)^{2} + c$

Example. $x^{2} + 6x - 4$ .

$p = 6/2 = 3$ .
$(x + 3)^{2}$ expands to $x^{2} + 6x + 9$ . So we need to subtract $9$ .
$x^{2} + 6x - 4 = (x + 3)^{2} - 9 - 4 = (x + 3)^{2} - 13$ .

More examples:

Original	Completed
$x^{2} + 4x - 3$	$(x + 2)^{2} - 7$
$x^{2} - 8x + 5$	$(x - 4)^{2} - 11$
$x^{2} + 5x + 2$	$(x + 5/2)^{2} - 17/4$
$x^{2} - 10x$	$(x - 5)^{2} - 25$

Building the square (x + 3)² adds a 3 × 3 corner of area 9 — the amount you subtract to compensate.

Worked qualitative. Why does $(x + p)^{2}$ work?

$(x + p)^{2} = x^{2} + 2px + p^{2}$ .
The middle term is $2p$ , so to match $bx$ we need $p = b/2$ .
The 'extra' is $p^{2}$ , which we subtract to keep the value the same.

Edexcel tip. Memorise: half-the-coefficient is your best friend. Mark schemes credit each step.

Half coefficient of $x$ gives $p$ .
$(x + p)^{2}$ expands to $x^{2} + 2px + p^{2}$ .
Subtract $p^{2}$ to compensate.
Add original $c$ .

Non-monic — leading coefficient > 1

Factor out the leading coefficient first. Complete inside. Re-distribute.

Form: $ax^{2} + bx + c$ where $a > 1$ .

Method:

Factor $a$ out of the $x$ -terms: $a(x^{2} + (b/a)x) + c$ .
Complete the square INSIDE the bracket.
Re-distribute (multiply $a$ through).

Example. $2x^{2} - 12x + 5$ .

Factor: $2(x^{2} - 6x) + 5$ .
Inside: $x^{2} - 6x = (x - 3)^{2} - 9$ .
Substitute: $2[(x - 3)^{2} - 9] + 5 = 2(x - 3)^{2} - 18 + 5 = 2(x - 3)^{2} - 13$ .

More examples:

Original	Completed
$3x^{2} + 12x + 5$	$3(x + 2)^{2} - 7$
$4x^{2} - 8x + 1$	$4(x - 1)^{2} - 3$
$-x^{2} + 6x - 5$	$-(x - 3)^{2} + 4$

Edexcel tip. ALWAYS factor the leading coefficient OUT before completing. The factored form makes the inner step monic.

Factor leading coefficient first.
Complete inside.
Re-distribute.
Final form: $a(x + p)^{2} + q$ .

Why complete the square?

Three uses: solving, turning point, proof.

Use 1: Solving quadratics.

Once in form $a(x + p)^{2} + q = 0$ :

Move $q$ : $a(x + p)^{2} = -q$ .
Divide by $a$ : $(x + p)^{2} = -q/a$ .
Take root: $x + p = \pm \sqrt{-q/a}$ .
Solve for $x$ .

Useful when the equation doesn't factor or when surd-form answers are required.

Use 2: Finding the turning point of a parabola.

For $y = a(x + p)^{2} + q$ :

Turning point: $(x = -p, y = q)$ .
If $a > 0$ : parabola opens UP, turning point is MINIMUM.
If $a < 0$ : parabola opens DOWN, turning point is MAXIMUM.

Example. $y = x^{2} - 4x + 7$ . Complete: $y = (x - 2)^{2} + 3$ . Min at $(2, 3)$ .

In vertex form a(x + p)² + q, the turning point sits at (−p, q) — here a minimum at (2, 3).

Use 3: Showing positive/negative definiteness.

If completing gives $(x + p)^{2} + q$ with $q > 0$ , the expression is ALWAYS positive (since squares are non-negative).

Example. Show $x^{2} + 6x + 13 > 0$ for all $x$ .

$(x + 3)^{2} - 9 + 13 = (x + 3)^{2} + 4$ .
$(x + 3)^{2} \ge 0$ , so $+ 4 > 0$ . Done.

Worked qualitative. Why doesn't $x^{2} + 6x + 13 = 0$ have real solutions?

Completed form: $(x + 3)^{2} + 4 = 0$ .
$(x + 3)^{2}$ is non-negative, so adding 4 always gives ≥ 4. Never zero.
No real $x$ satisfies the equation.
Confirms what the discriminant would show: $36 - 52 = -16 < 0$ .

Edexcel tip. When asked to find the minimum or maximum, complete the square — DON'T differentiate (that's A-level).

Solve: when no nice factoring.
Vertex: read off minimum/maximum.
Proof: show positivity.
Avoid differentiation at IGCSE.

Quick recap

Form: $(x + p)^{2} + q$ where $p = b/2$ .
Non-monic: factor out leading coef first.
Used for solving, vertex, proof.
Vertex: $(-p, q)$ .
Discriminant relation: $q < 0 \Leftrightarrow$ real solutions.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$x^{2} + bx + c = (x + b/2)^{2} + c - b^{2}/4$
Vertex: $(-p, q)$ .
Non-monic: factor $a$ out first.
Use for: solving, turning point, proof.

How it’s examined

Completing the square appears Paper 1H + 2H (4-6 marks). Often combined with finding turning points or solving in surd form. Examiner reports flag (1) forgetting to compensate for the extra term, (2) not factoring out for non-monic, (3) wrong sign on vertex.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H May/Jun 2024. Last reviewed 2026-05-07.

Take this whole topic with you

Step-by-step worked examples — Completing the square

Step-by-step solutions to past-paper-style questions on completing the square, written exactly the way a tutor would explain them at the board.

1Complete the square (a = 1)
Higher• completing the square
Question
Complete the square: $x^{2} + 6x - 4$ .
Step-by-step solution
1. Step 1
  Take HALF the coefficient of $x$ : $6/2 = 3$ .
2. Step 2
  Write $(x + 3)^{2}$ and check: $(x + 3)^{2} = x^{2} + 6x + 9$ .
3. Step 3
  Subtract the extra constant ( $9$ ) and add the original ( $-4$ ):
  $x^{2} + 6x - 4 = (x + 3)^{2} - 9 - 4 = (x + 3)^{2} - 13$
Answer
$(x + 3)^{2} - 13$
Examiner tip
Half-the-coefficient is the magic number. The 'square' part contains $x + (\text{half coef})$ . Then adjust the constant.
2Complete the square (a > 1)
Higher• Adapted from 4MA1/1H May/Jun 2024 Q18• completing the square
Question
Complete the square: $2x^{2} - 12x + 5$ .
Step-by-step solution
1. Step 1
  Factor out $2$ from the $x$ -terms:
  $2x^{2} - 12x + 5 = 2(x^{2} - 6x) + 5$
2. Step 2
  Inside, complete the square: $x^{2} - 6x = (x - 3)^{2} - 9$ .
3. Step 3
  Substitute back:
  $2(x^{2} - 6x) + 5 = 2[(x - 3)^{2} - 9] + 5$
4. Step 4
  Expand: $2(x - 3)^{2} - 18 + 5 = 2(x - 3)^{2} - 13$ .
Answer
$2(x - 3)^{2} - 13$
Examiner tip
Factor out the leading coefficient FIRST. Complete inside. Then re-distribute. Mark scheme awards M1 for factoring out, M1 for the inner completion, A1 for the answer.
3Solve a quadratic using completing the square
Higher• completing the square, solve
Question
Solve $x^{2} + 8x + 5 = 0$ using completing the square.
Step-by-step solution
1. Step 1
  Complete the square: $x^{2} + 8x + 5 = (x + 4)^{2} - 16 + 5 = (x + 4)^{2} - 11$ .
2. Step 2
  Set equal to zero:
  $(x + 4)^{2} - 11 = 0$
3. Step 3
  Rearrange: $(x + 4)^{2} = 11$ . Take square root: $x + 4 = \pm \sqrt{11}$ .
4. Step 4
  Solve: $x = -4 \pm \sqrt{11}$ .
Answer
$x = -4 \pm \sqrt{11}$
Examiner tip
Completing the square is the WORKING that derives the quadratic formula. Useful when surd-form solutions are required.
4Find the turning point
Higher• turning point
Question
Find the coordinates of the minimum point of $y = x^{2} - 4x + 7$ .
Step-by-step solution
1. Step 1
  Complete the square: $x^{2} - 4x + 7 = (x - 2)^{2} + 3$ .
2. Step 2
  Form $(x + p)^{2} + q$ has minimum value $q$ at $x = -p$ .
3. Step 3
  Here $p = -2$ , so minimum at $x = 2$ . Minimum value: $y = 3$ .
Answer
Minimum at $(2, 3)$
Examiner tip
Turning point: $(x = -p, y = q)$ for form $(x + p)^{2} + q$ . Up-opening (coef of $x^{2}$ positive) → minimum.

Key Formulae — Completing the square

The formulae you need to memorise for completing the square on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Completing the square (monic)
$x^{2} + bx + c = \left(x + \dfrac{b}{2}\right)^{2} + c - \dfrac{b^{2}}{4}$
$b/2$
half the coefficient of $x$
When to use
When converting $x^{2} + bx + c$ to vertex form.
Example
$x^{2} + 6x - 4 = (x + 3)^{2} - 13$ .

Key Definitions and Keywords — Completing the square

Definitions to memorise and the exact keywords mark schemes credit for completing the square answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Vertex form (or completed square form)
Examiner keyword
A quadratic written as $a(x + p)^{2} + q$ . Easy to identify the vertex (turning point) and the minimum/maximum value.
Example
$y = (x + 3)^{2} - 13$ has vertex at $(-3, -13)$ .
Turning point (vertex)
The minimum or maximum point of a parabola. Coordinates: $(-p, q)$ for $(x + p)^{2} + q$ .

Common Mistakes and Misconceptions — Completing the square

The traps other students keep falling into on completing the square questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to subtract the extra term
Why it happens
Hurried.
How to avoid it
$(x + 3)^{2} = x^{2} + 6x + 9$ , NOT $x^{2} + 6x$ . So when you write $(x + 3)^{2}$ for $x^{2} + 6x + ...$ , you must SUBTRACT $9$ to compensate.
✕Not factoring out the leading coefficient when $a > 1$
4MA1/1H May/Jun 2024 — examiner report Q18
Why it happens
Skipping a step.
How to avoid it
$2x^{2} - 12x + 5$ → factor $2$ : $2(x^{2} - 6x) + 5$ . Complete inside, then re-multiply.
✕Reading vertex as $(p, q)$ instead of $(-p, q)$
Why it happens
Form has $+p$ , but vertex flips sign.
How to avoid it
$(x + p)^{2} + q$ → minimum at $x = -p$ (because that's when squared term is zero).

Completing the square — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Original

Completed

x^{2} + 4x - 3

(x + 2)^{2} - 7

x^{2} - 8x + 5

(x - 4)^{2} - 11

x^{2} + 5x + 2

(x + 5/2)^{2} - 17/4

x^{2} - 10x

(x - 5)^{2} - 25

Original

Completed

3x^{2} + 12x + 5

3(x + 2)^{2} - 7

4x^{2} - 8x + 1

4(x - 1)^{2} - 3

-x^{2} + 6x - 5

-(x - 3)^{2} + 4

Use 1: Solving quadratics.

Once in form $a(x + p)^{2} + q = 0$ :

Move $q$ : $a(x + p)^{2} = -q$ .
Divide by $a$ : $(x + p)^{2} = -q/a$ .
Take root: $x + p = \pm \sqrt{-q/a}$ .
Solve for $x$ .

Useful when the equation doesn't factor or when surd-form answers are required.

Use 2: Finding the turning point of a parabola.

For $y = a(x + p)^{2} + q$ :

Turning point: $(x = -p, y = q)$ .
If $a > 0$ : parabola opens UP, turning point is MINIMUM.
If $a < 0$ : parabola opens DOWN, turning point is MAXIMUM.

Example. $y = x^{2} - 4x + 7$ . Complete: $y = (x - 2)^{2} + 3$ . Min at $(2, 3)$ .

In vertex form a(x + p)² + q, the turning point sits at (−p, q) — here a minimum at (2, 3).

Use 3: Showing positive/negative definiteness.

If completing gives $(x + p)^{2} + q$ with $q > 0$ , the expression is ALWAYS positive (since squares are non-negative).

Example. Show $x^{2} + 6x + 13 > 0$ for all $x$ .

$(x + 3)^{2} - 9 + 13 = (x + 3)^{2} + 4$ .
$(x + 3)^{2} \ge 0$ , so $+ 4 > 0$ . Done.

Worked qualitative. Why doesn't $x^{2} + 6x + 13 = 0$ have real solutions?

Completed form: $(x + 3)^{2} + 4 = 0$ .
$(x + 3)^{2}$ is non-negative, so adding 4 always gives ≥ 4. Never zero.
No real $x$ satisfies the equation.
Confirms what the discriminant would show: $36 - 52 = -16 < 0$ .

Edexcel tip. When asked to find the minimum or maximum, complete the square — DON'T differentiate (that's A-level).

Completing the square

Short Notes - Completing the square

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Complete the square (a = 1)

2Complete the square (a > 1)

3Solve a quadratic using completing the square

4Find the turning point

Completing the square (monic)

Vertex form (or completed square form)

Turning point (vertex)

✕Forgetting to subtract the extra term

✕Not factoring out the leading coefficient when a>1a > 1a>1

✕Reading vertex as (p,q)(p, q)(p,q) instead of (−p,q)(-p, q)(−p,q)

Completing the square — frequently asked questions

Completing the square

Short Notes - Completing the square

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Complete the square (a = 1)

2Complete the square (a > 1)

3Solve a quadratic using completing the square

4Find the turning point

Completing the square (monic)

Vertex form (or completed square form)

Turning point (vertex)

✕Forgetting to subtract the extra term

✕Not factoring out the leading coefficient when a>1a > 1a>1

✕Reading vertex as (p,q)(p, q)(p,q) instead of (−p,q)(-p, q)(−p,q)

Completing the square — frequently asked questions

✕Not factoring out the leading coefficient when $a > 1$

✕Reading vertex as $(p, q)$ instead of $(-p, q)$

✕Not factoring out the leading coefficient when $a > 1$

✕Reading vertex as $(p, q)$ instead of $(-p, q)$