Completing the Square · Study Notes · Edexcel IGCSE 4MA1 Higher

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 $1$

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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

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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?

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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.

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
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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
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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
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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
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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.
5Least value of a completed square [1 mark]
Getting started• Paper 1H/2H style — 1 mark• completing the square, least value
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Question
Write down the least value of $(2x - 1)^{2} + 6$ [1]
Step-by-step solution
1. Step 1
  The squared term $(2x - 1)^{2}$ can never be negative. Its smallest value is $0$ , when $x = \dfrac{1}{2}$ .
2. Step 2
  B1 So the smallest value of the whole expression is $0 + 6$ .
  $0 + 6 = 6$
Answer
$6$
Examiner tip
Examiner tip: The question asks for the least value of the expression, not the value of $x$ where it occurs. Read the question again before writing your answer.

Common mistake: Giving $\dfrac{1}{2}$ (the value of $x$ ) or $-1$ (the number in the bracket). Neither is the least value.
6Completing the square with no constant term [2 marks]
Getting started• Paper 1H/2H style — 2 marks• completing the square, fractions
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Question
Write $x^{2} - 9x$ in the form $(x - p)^{2} - q$ , where $p$ and $q$ are fractions. [2]
Step-by-step solution
1. Step 1
  M1 Half of $9$ is $\dfrac{9}{2}$ , so the bracket is $\left(x - \dfrac{9}{2}\right)^{2}$ . Expanding it gives an extra $\dfrac{81}{4}$ , which must be taken away.
  $x^{2} - 9x = \left(x - \frac{9}{2}\right)^{2} - \left(\frac{9}{2}\right)^{2}$
2. Step 2
  A1 Square the fraction.
  $\left(x - \frac{9}{2}\right)^{2} - \frac{81}{4}$
Answer
$\left(x - \dfrac{9}{2}\right)^{2} - \dfrac{81}{4}$ , so $p = \dfrac{9}{2}$ and $q = \dfrac{81}{4}$
Examiner tip
Examiner tip: With no constant term the method is exactly the same. There is simply nothing to add at the end.

Common mistake: Writing $\left(\dfrac{9}{2}\right)^{2} = \dfrac{81}{2}$ , which squares only the numerator. The denominator must be squared too.
7Leading coefficient 2 with a fractional bracket [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• completing the square, a(x + b)² + c
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Question
Express $2x^{2} + 6x - 1$ in the form $a(x + b)^{2} + c$ , where $a$ , $b$ and $c$ are fractions or integers. [3]
Step-by-step solution
1. Step 1
  M1 Take out the factor $2$ from the first two terms.
  $2x^{2} + 6x - 1 = 2(x^{2} + 3x) - 1$
2. Step 2
  M1 Complete the square inside the bracket. Half of $3$ is $\dfrac{3}{2}$ .
  $2\left[\left(x + \frac{3}{2}\right)^{2} - \frac{9}{4}\right] - 1$
3. Step 3
  A1 Multiply $-\dfrac{9}{4}$ by $2$ and collect the constants.
  $2\left(x + \frac{3}{2}\right)^{2} - \frac{9}{2} - 1 = 2\left(x + \frac{3}{2}\right)^{2} - \frac{11}{2}$
Answer
$2\left(x + \dfrac{3}{2}\right)^{2} - \dfrac{11}{2}$
Examiner tip
Examiner tip: A fraction inside the bracket is normal when the coefficient of $x$ divided by the coefficient of $x^{2}$ is odd (here $6 \div 2 = 3$ ). Decimals are also accepted: $2(x + 1.5)^{2} - 5.5$ .

Common mistake: Forgetting to multiply the $-\dfrac{9}{4}$ by $2$ , which gives $-\dfrac{13}{4}$ instead of $-\dfrac{11}{2}$ . Expanding your answer is a quick check.
8Form a quadratic from a rectangle, then solve exactly [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• completing the square, solving quadratics, context
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Question
A rectangle has length $(x + 6)$ cm and width $x$ cm. The area of the rectangle is $11$ cm².

(a) Show that $x^{2} + 6x - 11 = 0$ [1]

(b) By completing the square, find the exact value of $x$ . Give your answer in the form $p + q\sqrt{5}$ , where $p$ and $q$ are integers. Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  B1 (a) Area = length × width, so $x(x + 6) = 11$ . Expand and subtract $11$ from both sides.
  $x^{2} + 6x = 11 \quad \Rightarrow \quad x^{2} + 6x - 11 = 0$
2. Step 2
  M1 (b) Complete the square and rearrange.
  $(x + 3)^{2} - 9 - 11 = 0 \quad \Rightarrow \quad (x + 3)^{2} = 20$
3. Step 3
  M1 Take the square root of both sides, with $\pm$ , and simplify $\sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5}$ .
  $x + 3 = \pm 2\sqrt{5} \quad \Rightarrow \quad x = -3 + 2\sqrt{5} \ \text{ or } \ x = -3 - 2\sqrt{5}$
4. Step 4
  A1 $-3 - 2\sqrt{5}$ is negative (about $-7.47$ ), and a width cannot be negative, so reject it. The other value, about $1.47$ , is positive.
  $x = -3 + 2\sqrt{5}$
Answer
(b) $x = -3 + 2\sqrt{5}$ (the negative solution is rejected because a width cannot be negative)
Examiner tip
Examiner tip: Find both solutions first, then use the context to reject one, and say why. The final A1 is for $-3 + 2\sqrt{5}$ only.

Common mistake: Leaving the answer as $-3 + \sqrt{20}$ , which is not in the form asked, or giving both values as the answer. Giving the decimal $1.47$ from a calculator scores 0, because clear algebraic working and an exact answer are required.
9Negative x² coefficient: the greatest value [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• completing the square, negative coefficient, greatest value
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Question
(a) Write $8 - 6x - x^{2}$ in the form $a - (x + b)^{2}$ , where $a$ and $b$ are integers. [3]

(b) Write down the greatest value of $8 - 6x - x^{2}$ [1]
Step-by-step solution
1. Step 1
  M1 (a) Take out $-1$ from the $x^{2}$ and $x$ terms. Both signs inside change.
  $8 - 6x - x^{2} = -(x^{2} + 6x) + 8$
2. Step 2
  M1 Complete the square inside the bracket.
  $-\left[(x + 3)^{2} - 9\right] + 8$
3. Step 3
  A1 Remove the square bracket: $-(-9) = +9$ .
  $-(x + 3)^{2} + 9 + 8 = 17 - (x + 3)^{2}$
4. Step 4
  B1 (b) $(x + 3)^{2} \geqslant 0$ and it is subtracted, so the expression is greatest when $(x + 3)^{2} = 0$ .
  $17 - 0 = 17$
Answer
(a) $17 - (x + 3)^{2}$ (b) $17$
Examiner tip
Examiner tip: When the square is subtracted, the expression has a greatest value, not a least value. It is largest when the squared term is $0$ .

Common mistake: Writing $-(x^{2} - 6x) + 8$ . Taking out $-1$ changes the sign of every term inside, so $-6x$ becomes $+6x$ in the bracket. Another common error is $-(x + 3)^{2} - 9 + 8$ , which loses the bracket and gives $-1$ instead of $17$ .
10Completing the square as (ax + b)² + c, then solve [5 marks]
Stretch• Paper 1H/2H style — 5 marks• completing the square, solving quadratics, surds
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Question
(a) Express $9x^{2} - 12x + 7$ in the form $(ax + b)^{2} + c$ , where $a$ , $b$ and $c$ are integers and $a > 0$ . [3]

(b) Hence solve $9x^{2} - 12x + 7 = 8$ . Give your answers in the form $\dfrac{p \pm \sqrt{q}}{r}$ , where $p$ , $q$ and $r$ are integers. [2]
Step-by-step solution
1. Step 1
  M1 (a) $9x^{2} = (3x)^{2}$ , so $a = 3$ . The middle term of $(3x + b)^{2}$ is $6bx$ .
  $(3x + b)^{2} = 9x^{2} + 6bx + b^{2}$
2. Step 2
  M1 Compare the $x$ terms: $6b = -12$ , so $b = -2$ and $b^{2} = 4$ .
  $(3x - 2)^{2} = 9x^{2} - 12x + 4$
3. Step 3
  A1 Add the $3$ still needed to reach $7$ .
  $9x^{2} - 12x + 7 = (3x - 2)^{2} + 3$
4. Step 4
  M1 (b) Use (a): subtract $3$ from both sides, then take the square root with $\pm$ .
  $(3x - 2)^{2} + 3 = 8 \quad \Rightarrow \quad (3x - 2)^{2} = 5 \quad \Rightarrow \quad 3x - 2 = \pm\sqrt{5}$
5. Step 5
  A1 Add $2$ , then divide the WHOLE right-hand side by $3$ .
  $3x = 2 \pm \sqrt{5} \quad \Rightarrow \quad x = \frac{2 \pm \sqrt{5}}{3}$
Answer
(a) $(3x - 2)^{2} + 3$ (b) $x = \dfrac{2 \pm \sqrt{5}}{3}$
Examiner tip
Examiner tip: The form $(ax + b)^{2} + c$ does not take a factor out. Match $a^{2}$ to the $x^{2}$ coefficient, then use $2ab$ to find $b$ . Check (a) by expanding: $(3x - 2)^{2} + 3 = 9x^{2} - 12x + 4 + 3$ .

Common mistake: In (b), dividing only one term by $3$ and writing $x = \dfrac{2}{3} \pm \sqrt{5}$ . Both the $2$ and the $\sqrt{5}$ must be divided by $3$ . Another slip is forgetting the $\pm$ and giving only one solution.

Question 1
Paper 1H/2H style — 1 mark1 mark
Write down the greatest value of $9 - 4(x - 1)^{2}$ [1]
Model answer
$4(x - 1)^{2} \geqslant 0$ , so the greatest value is when $x = 1$ .

Answer: $9$
Why this scores
Mark scheme:
- B1 for $9$ cao
Why this scores full marks: The squared term is subtracted, so the expression is largest when that term is $0$ . The answers $1$ (the value of $x$ ) and $5$ (from $9 - 4$ ) are the two usual wrong answers.
Question 2
Paper 1H/2H style — 2 marks2 marks
Write $x^{2} - 6ax + 10a^{2}$ in the form $(x - p)^{2} + q$ , giving $p$ and $q$ in terms of $a$ . [2]
Model answer
$x^{2} - 6ax + 10a^{2} = (x - 3a)^{2} - 9a^{2} + 10a^{2}$

$= (x - 3a)^{2} + a^{2}$

Answer: $(x - 3a)^{2} + a^{2}$ , so $p = 3a$ and $q = a^{2}$
Why this scores
Mark scheme:
- M1 for $(x - 3a)^{2}$ seen, or $p = 3a$
- A1 for $(x - 3a)^{2} + a^{2}$ oe
Why this scores full marks: The script treats $a$ exactly like a number: half of $-6a$ is $-3a$ , and $(3a)^{2} = 9a^{2}$ , not $3a^{2}$ . The $-9a^{2}$ is written down before it is combined with $10a^{2}$ , so the method is clear. Expanding gives $x^{2} - 6ax + 9a^{2} + a^{2}$ , which checks the answer.
Question 3
Paper 1H/2H style — 2 marks2 marks
$x^{2} - 4x - 41 \equiv (x - 2)^{2} - 45$

Hence solve $x^{2} - 4x - 41 = 0$ . Give your answers in the form $a \pm b\sqrt{5}$ , where $a$ and $b$ are integers. [2]
Model answer
$(x - 2)^{2} = 45$

$x - 2 = \pm\sqrt{45} = \pm 3\sqrt{5}$

$x = 2 \pm 3\sqrt{5}$
Why this scores
Mark scheme:
- M1 for $x - 2 = \pm\sqrt{45}$ oe
- A1 for $2 \pm 3\sqrt{5}$ cao
Why this scores full marks: The completed square is given, so only two steps are needed. The script keeps the $\pm$ , and it simplifies $\sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5}$ to match the form asked for.
Question 4
Paper 1H/2H style — 3 marks3 marks
By completing the square, show that $4x^{2} - 8x + 1$ can never be equal to $-4$ [3]
Model answer
$4x^{2} - 8x + 1 = 4(x^{2} - 2x) + 1$

$= 4\left[(x - 1)^{2} - 1\right] + 1$

$= 4(x - 1)^{2} - 4 + 1 = 4(x - 1)^{2} - 3$

$(x - 1)^{2} \geqslant 0$ for all $x$ , so $4(x - 1)^{2} - 3 \geqslant -3$ .

The least value of $4x^{2} - 8x + 1$ is $-3$ , and $-4 < -3$ , so $4x^{2} - 8x + 1$ can never equal $-4$ .
Why this scores
Mark scheme:
- M1 for $4(x^{2} - 2x) + 1$ or $4\left[(x - 1)^{2} - 1\right] + 1$ oe
- A1 for $4(x - 1)^{2} - 3$
- A1 for a complete conclusion: $(x - 1)^{2} \geqslant 0$ so the expression is $\geqslant -3$ , and $-4$ is less than $-3$
Why this scores full marks: The square bracket makes sure the $-1$ is multiplied by $4$ ; without it the result would be $4(x - 1)^{2}$ with no constant, and the argument collapses. The conclusion states why the least value is $-3$ (a square is never negative) and compares it with $-4$ . An equivalent ending is $4(x - 1)^{2} = -1$ , which is impossible because a square cannot be negative.
Question 5
Paper 1H/2H style — 3 marks3 marks
$3x^{2} + px + q \equiv 3(x + 2)^{2} - 11$ for all values of $x$ .

Find the value of $p$ and the value of $q$ . [3]
Model answer
$3(x + 2)^{2} - 11 = 3(x^{2} + 4x + 4) - 11$

$= 3x^{2} + 12x + 12 - 11 = 3x^{2} + 12x + 1$

$p = 12$ , $q = 1$
Why this scores
Mark scheme:
- M1 for $3(x^{2} + 4x + 4) - 11$ or $3x^{2} + 12x + 12 - 11$ oe
- A1 for $p = 12$
- A1 for $q = 1$
Why this scores full marks: The script expands the right-hand side in full, multiplying every term in the bracket by $3$ , and then compares coefficients. The usual error is $3(x + 2)^{2} = 3x^{2} + 12x + 4$ , which applies the $3$ to only some of the terms.
Question 6
Paper 1H/2H style — 4 marks4 marks
The least value of $x^{2} - 6x + c$ is $8$ .

(a) Find the value of $c$ . [2]

(b) Solve $x^{2} - 6x + c = 12$ [2]
Model answer
(a)

$x^{2} - 6x + c = (x - 3)^{2} - 9 + c$

Least value is $c - 9 = 8$ , so $c = 17$

(b)

$(x - 3)^{2} + 8 = 12$

$(x - 3)^{2} = 4$

$x - 3 = \pm 2$

$x = 5$ or $x = 1$
Why this scores
Mark scheme:
- M1 for $(x - 3)^{2} - 9 + c$ oe
- A1 for $c = 17$
- M1 for $(x - 3)^{2} = 4$ ft their $c$ , or $x^{2} - 6x + 5 = 0$ factorised correctly oe
- A1 for $1$ and $5$ (both required)
Why this scores full marks: Completing the square gives the least value, $c - 9$ , directly. In (b) the script takes both square roots, $+2$ and $-2$ . Giving only $x = 5$ loses the final mark.
Question 7
Paper 1H/2H style — 4 marks4 marks
(a) Express $2 - 8x - 4x^{2}$ in the form $p - q(x + r)^{2}$ , where $p$ , $q$ and $r$ are integers. [3]

(b) Write down the maximum value of $2 - 8x - 4x^{2}$ [1]
Model answer
(a)

$2 - 8x - 4x^{2} = -4(x^{2} + 2x) + 2$

$= -4\left[(x + 1)^{2} - 1\right] + 2$

$= -4(x + 1)^{2} + 4 + 2 = 6 - 4(x + 1)^{2}$

(b) $6$
Why this scores
Mark scheme:
- M1 for $-4(x^{2} + 2x)$ oe
- M1 for $-4\left[(x + 1)^{2} - 1\right]$ oe
- A1 for $6 - 4(x + 1)^{2}$ cao
- B1 for $6$ ft their $p$
Why this scores full marks: The script takes out $-4$ first, so the bracket starts with $+x^{2}$ . It keeps the square bracket so that $-4 \times (-1) = +4$ . Losing that bracket gives $-4(x + 1)^{2} - 1 + 2 = 1 - 4(x + 1)^{2}$ . The June 2024 examiner report names missing brackets, and not taking out the negative factor first, as the errors on this type of question.
Question 8
Paper 1H/2H style — 5 marks5 marks
The curve with equation $y = 5 + px - 2x^{2}$ has a maximum point at $(q, 13)$ , where $p$ and $q$ are constants.

Find all the possible pairs of values of $p$ and $q$ . Show clear algebraic working. [5]
Model answer
The coefficient of $x^{2}$ is $-2$ and the maximum point is $(q, 13)$ , so

$5 + px - 2x^{2} \equiv 13 - 2(x - q)^{2}$

$13 - 2(x - q)^{2} = 13 - 2x^{2} + 4qx - 2q^{2}$

Constants: $13 - 2q^{2} = 5$ , so $q^{2} = 4$ and $q = 2$ or $q = -2$

$x$ terms: $p = 4q$

$p = 8$ , $q = 2$ or $p = -8$ , $q = -2$
Why this scores
Mark scheme:
- B1 for $13 - 2(x - q)^{2}$ oe, or completing the square to get $5 + \dfrac{p^{2}}{8} - 2\left(x - \dfrac{p}{4}\right)^{2}$ oe
- M1 for expanding to $13 - 2x^{2} + 4qx - 2q^{2}$ oe
- M1 for $13 - 2q^{2} = 5$ or $5 + \dfrac{p^{2}}{8} = 13$ oe
- A1 for $q = 2$ and $q = -2$ (or $p = 8$ and $p = -8$ )
- A1 for both pairs correctly matched: $p = 8$ with $q = 2$ , and $p = -8$ with $q = -2$
Why this scores full marks: The script writes the curve in completed-square form straight from the maximum point, keeping the $-2$ outside the bracket, and then compares the constant terms and the $x$ terms. It takes both square roots of $q^{2} = 4$ and pairs each $q$ with its own $p$ . Check: $5 + 8x - 2x^{2} = 13 - 2(x - 2)^{2}$ and $5 - 8x - 2x^{2} = 13 - 2(x + 2)^{2}$ . Giving only $p = 8$ , $q = 2$ loses the final mark.
Question 9
Paper 1H/2H style — 5 marks5 marks
(a) Prove that $(x + 4)(x - 2) + 10$ is positive for all values of $x$ . [3]

(b) Write down the least value of $(x + 4)(x - 2) + 10$ and the value of $x$ at which it occurs. [2]
Model answer
(a)

$(x + 4)(x - 2) + 10 = x^{2} + 2x - 8 + 10 = x^{2} + 2x + 2$

$= (x + 1)^{2} - 1 + 2 = (x + 1)^{2} + 1$

$(x + 1)^{2} \geqslant 0$ for all $x$ , so $(x + 1)^{2} + 1 \geqslant 1 > 0$ . The expression is positive for all values of $x$ .

(b) Least value $1$ , when $x = -1$
Why this scores
Mark scheme:
- M1 for expanding to $x^{2} + 2x + 2$
- M1 for $(x + 1)^{2} + 1$
- A1 for a complete argument: a square is $\geqslant 0$ , so the expression is $\geqslant 1$ and hence positive
- B1 for least value $1$
- B1 for $x = -1$
Why this scores full marks: The script expands first, which is needed before the square can be completed. It then states both facts the proof depends on: the square is never negative, and adding $1$ makes the total at least $1$ . Trying a few values of $x$ is not a proof and scores 0.
Question 10
Paper 1H/2H style — 6 marks6 marks
A shop sells a product at a price of £ $x$ . The weekly profit, £ $P$ , is given by

$P = 60x - 3x^{2} - 200$

(a) Express $60x - 3x^{2} - 200$ in the form $a - b(x - c)^{2}$ , where $a$ , $b$ and $c$ are integers. [3]

(b) Write down the maximum weekly profit. [1]

(c) Using your answer to (a), find the two prices that give a weekly profit of £ $25$ . Show clear algebraic working. [2]
Model answer
(a)

$60x - 3x^{2} - 200 = -3(x^{2} - 20x) - 200$

$= -3\left[(x - 10)^{2} - 100\right] - 200$

$= -3(x - 10)^{2} + 300 - 200 = 100 - 3(x - 10)^{2}$

(b) £ $100$

(c)

$100 - 3(x - 10)^{2} = 25$

$3(x - 10)^{2} = 75 \quad \Rightarrow \quad (x - 10)^{2} = 25$

$x - 10 = \pm 5$

£ $5$ and £ $15$
Why this scores
Mark scheme:
- M1 for $-3(x^{2} - 20x)$ oe
- M1 for $-3\left[(x - 10)^{2} - 100\right]$ oe
- A1 for $100 - 3(x - 10)^{2}$ cao
- B1 for £ $100$ ft their $a$
- M1 for $(x - 10)^{2} = 25$ ft their (a)
- A1 for £ $5$ and £ $15$ (both required)
Why this scores full marks: The script takes out $-3$ with a bracket and uses a square bracket, so $-3 \times (-100) = +300$ is correct. In (b), the profit is greatest when the squared term is $0$ , so the maximum is £ $100$ at a price of £ $10$ . In (c), both square roots give the two prices. Trial and improvement would score 0, because clear algebraic working is required.

4MA1 Completing the Square

At a glance

What you’ll learn

Completing the square — leading coefficient $1$

Non-monic — leading coefficient > 1

Why complete the square?

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Complete the square (a = 1)

2Complete the square (a > 1)

3Solve a quadratic using completing the square

4Find the turning point

5Least value of a completed square [1 mark]

6Completing the square with no constant term [2 marks]

7Leading coefficient 2 with a fractional bracket [3 marks]

8Form a quadratic from a rectangle, then solve exactly [4 marks]

9Negative x² coefficient: the greatest value [4 marks]

10Completing the square as (ax + b)² + c, then solve [5 marks]

Model AnswersFull mark-scheme-style answers

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

4MA1 Completing the Square

At a glance

What you’ll learn

Completing the square — leading coefficient 111

Non-monic — leading coefficient > 1

Why complete the square?

Quick recap

Memorise this

How it’s examined

Worked ExamplesStep-by-step solutions to past-paper-style questions

1Complete the square (a = 1)

2Complete the square (a > 1)

3Solve a quadratic using completing the square

4Find the turning point

5Least value of a completed square [1 mark]

6Completing the square with no constant term [2 marks]

7Leading coefficient 2 with a fractional bracket [3 marks]

8Form a quadratic from a rectangle, then solve exactly [4 marks]

9Negative x² coefficient: the greatest value [4 marks]

10Completing the square as (ax + b)² + c, then solve [5 marks]

Model AnswersFull mark-scheme-style answers

Completing the square — leading coefficient $1$