Quadratic Equations · Study Notes · Edexcel IGCSE 4MA1 Higher

Solve $ax^{2} + bx + c = 0$ via factoring, completing the square, or the quadratic formula. Featured every Higher Tier paper.

At a glance

Standard form: $ax^{2} + bx + c = 0$ .
Three methods: factor, complete the square, formula.
Quadratic formula: $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ .
Discriminant: $b^{2} - 4ac$ .
Two solutions typically (or one repeated, or none real).
Always reject contextually invalid answers.

What you’ll learn

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

2.7 — Solve quadratic equations by factorising.
2.7 — Solve quadratic equations using the quadratic formula.
2.7 — Use the discriminant to determine the nature of roots.
2.7 — Set up quadratics from real-world problems.

The three methods

▼

Factor (when easy), formula (always works), completing the square (sometimes).

1. Factorising.

Use when the quadratic factors nicely.

Standard form: $ax^{2} + bx + c = 0$ → $(\square x + \square)(\square x + \square) = 0$ .

Each factor zero gives a solution.

Example. $x^{2} - 7x + 12 = 0$ → $(x - 3)(x - 4) = 0$ → $x = 3$ or $x = 4$ .

2. Quadratic formula.

Always works. Memorise:

$x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$

Example. $2x^{2} + 5x - 7 = 0$ .

$a=2, b=5, c=-7$ .
Discriminant: $25 + 56 = 81$ .
$x = \dfrac{-5 \pm 9}{4} = 1$ or $-3.5$ .

3. Completing the square.

Used for proof and finding turning points. Covered in dedicated subtopic.

Choosing the method:

Situation	Method
Factors nicely	Factorise
Asked for surd answers	Formula
'Show that' / 'Prove that'	Often completing the square
Don't know if factors	Try factor first, then formula

Worked qualitative. When does the formula give you 'no real solutions'?

When the discriminant is negative: $b^{2} - 4ac < 0$ .
The square root of a negative isn't real.
This means the quadratic doesn't cross the $x$ -axis.

Edexcel tip. Quote the formula explicitly. Mark schemes award M1 for the formula, M1 for substitution, A1 for both correct answers.

Factor: when it factors nicely.
Formula: always works.
Memorise the formula — not on Edexcel formula sheet.
Always state BOTH solutions.

The discriminant — what kind of solutions?

▼

$\Delta = b^{2} - 4ac$ . Tells you the nature of the roots.

Definition. Discriminant $\Delta = b^{2} - 4ac$ .

Discriminant	Roots
$\Delta > 0$	Two distinct real roots
$\Delta = 0$	One repeated real root
$\Delta < 0$	No real roots

The roots of a quadratic are where its parabola meets the x-axis — twice, once, or never.

Why? From the quadratic formula:

$\Delta > 0$ : $\sqrt{\Delta}$ is real and non-zero. $x = (-b \pm$ real $)/2a$ → two distinct.
$\Delta = 0$ : $\sqrt{\Delta} = 0$ . $x = -b/2a$ → one repeated.
$\Delta < 0$ : $\sqrt{\Delta}$ not real. No real solutions.

Worked example. Show that $x^{2} - 4x + 5 = 0$ has no real solutions.

$\Delta = 16 - 20 = -4 < 0$ .
No real roots.

Edexcel applications. Sometimes Edexcel asks 'find values of $k$ for which the equation has real roots'.

Example. For what values of $k$ does $x^{2} - 4x + k = 0$ have real roots?

Need $\Delta \ge 0$ .
$\Delta = 16 - 4k$ .
$16 - 4k \ge 0$ → $k \le 4$ .

Worked qualitative. Why is $\Delta = 0$ called 'one repeated root'?

The formula gives $x = (-b \pm 0)/2a = -b/2a$ .
'Plus zero' and 'minus zero' both give the same answer.
One value, but it counts as a 'double root'.

Edexcel tip. Always state the discriminant and its sign. 'Since $\Delta = -4 < 0$ , no real roots' earns the mark.

$\Delta = b^{2} - 4ac$ .
$\Delta > 0$ : 2 distinct real.
$\Delta = 0$ : 1 repeated.
$\Delta < 0$ : no real.
Used to find ranges of $k$ .

Quadratic word problems

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Set up the quadratic, solve, check context.

Common contexts:

Areas (length × width).
Volumes (squared or cubed dimensions).
Numbers (consecutive products).
Projectile motion (height vs time).

Example. Rectangle: length $(x + 3)$ , width $(x - 1)$ . Area $= 32$ .

$(x + 3)(x - 1) = 32$ .
$x^{2} + 2x - 3 = 32$ .
$x^{2} + 2x - 35 = 0$ .
Factor: $(x + 7)(x - 5) = 0$ .
$x = -7$ or $x = 5$ .
Length must be positive: $x = 5$ .

Always reject invalid solutions.

Negative lengths.
Fractional people.
Times before zero.

Worked qualitative. A ball is thrown upward. Its height (m) at time $t$ (s) is $h = -5t^{2} + 20t$ . When does it hit the ground?

$h = 0$ : $-5t^{2} + 20t = 0$ .
Factor: $-5t(t - 4) = 0$ .
$t = 0$ or $t = 4$ .
$t = 0$ is when it was thrown. $t = 4$ is when it lands.

The two solutions of h = 0 are the launch and landing times where the parabola meets the t-axis.

Edexcel tip. When setting up a quadratic from a word problem, always rearrange to the form $ax^{2} + bx + c = 0$ before solving. Mark schemes credit this step.

Set up equation from problem.
Rearrange to $ax^{2} + bx + c = 0$ .
Solve and check context.
Reject invalid solutions.

Quick recap

Standard: $ax^{2} + bx + c = 0$ .
Methods: factor, formula, completing the square.
Formula: $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ .
Discriminant: $\Delta = b^{2} - 4ac$ .
Two solutions; reject invalid.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Quadratic formula — NOT on Edexcel formula sheet.
Discriminant: $b^{2} - 4ac$ .
Three methods: factor, formula, complete square.
Always state BOTH solutions.
Reject contextually invalid.

How it’s examined

Quadratic equations on EVERY Higher Tier paper (4-8 marks). Often combined with word problems, areas, formulae. Examiner reports flag (1) only stating one solution, (2) not quoting the formula, (3) including invalid solutions in word problems.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/2H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

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

1Solve by factorising
Higher• quadratic, factorise
▼
Question
Solve $x^{2} - 7x + 12 = 0$ .
Step-by-step solution
1. Step 1
  Find two numbers that multiply to $12$ and add to $-7$ .
2. Step 2
  Pair: $(-3, -4)$ . Both negative, multiply to $12$ , sum to $-7$ . ✓
3. Step 3
  Factorise: $(x - 3)(x - 4) = 0$ .
4. Step 4
  Each factor zero: $x = 3$ or $x = 4$ .
Answer
$x = 3$ or $x = 4$
2Solve using the quadratic formula
Higher• Adapted from 4MA1/2H Jan 2024 Q15• quadratic formula
▼
Question
Solve $2x^{2} + 5x - 7 = 0$ . Give your answers to $2$ d.p.
Step-by-step solution
1. Step 1
  Identify $a = 2$ , $b = 5$ , $c = -7$ .
2. Step 2
  Apply the formula:
  $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
3. Step 3
  Discriminant: $b^{2} - 4ac = 25 - 4(2)(-7) = 25 + 56 = 81$ . $\sqrt{81} = 9$ .
4. Step 4
  Compute: $x = \dfrac{-5 \pm 9}{4}$ .
5. Step 5
  Two solutions: $x = \dfrac{4}{4} = 1$ or $x = \dfrac{-14}{4} = -3.5$ .
Answer
$x = 1$ or $x = -3.5$
Examiner tip
ALWAYS quote the quadratic formula explicitly. The Edexcel formula sheet does NOT include it — you must memorise.
3Solve by factorising (a > 1)
Higher• quadratic, non-monic
▼
Question
Solve $3x^{2} + 11x + 6 = 0$ .
Step-by-step solution
1. Step 1
  AC method: $a \times c = 18$ . Need pair multiplying to $18$ , adding to $11$ .
2. Step 2
  Pair: $(2, 9)$ . Sum $11$ . ✓
3. Step 3
  Split middle term:
  $3x^{2} + 2x + 9x + 6 = 0$
4. Step 4
  Factor by grouping: $x(3x + 2) + 3(3x + 2) = (3x + 2)(x + 3) = 0$ .
5. Step 5
  Solutions: $x = -\dfrac{2}{3}$ or $x = -3$ .
Answer
$x = -\dfrac{2}{3}$ or $x = -3$
4Quadratic that doesn't factor neatly
Higher• quadratic formula
▼
Question
Solve $x^{2} - 4x - 2 = 0$ . Give your answer in surd form.
Step-by-step solution
1. Step 1
  $a = 1$ , $b = -4$ , $c = -2$ . Discriminant: $16 + 8 = 24$ .
2. Step 2
  $\sqrt{24} = 2\sqrt{6}$ .
3. Step 3
  Apply formula:
  $x = \frac{4 \pm 2\sqrt{6}}{2} = 2 \pm \sqrt{6}$
Answer
$x = 2 + \sqrt{6}$ or $x = 2 - \sqrt{6}$
Examiner tip
When asked for SURD FORM, simplify the radical and don't approximate. $\sqrt{24} = 2\sqrt{6}$ , then divide.
5Quadratic from a word problem
Higher• Adapted from 4MA1/2H May/Jun 2024 Q16• quadratic, word problem
▼
Question
A rectangle has length $(x + 3)$ cm and width $(x - 1)$ cm. The area is $32$ cm². Find $x$ .
Step-by-step solution
1. Step 1
  Area = length × width: $(x + 3)(x - 1) = 32$ .
2. Step 2
  Expand: $x^{2} + 2x - 3 = 32$ .
3. Step 3
  Rearrange: $x^{2} + 2x - 35 = 0$ .
4. Step 4
  Factor: pair multiplying to $-35$ , summing to $2$ : $(7, -5)$ . So $(x + 7)(x - 5) = 0$ .
5. Step 5
  Solutions: $x = -7$ or $x = 5$ . Length must be positive, so $x = 5$ .
Answer
$x = 5$
Examiner tip
After solving, REJECT solutions that don't make physical sense. Length $> 0$ , so $x > 1$ here.
6A quadratic with one repeated solution [2 marks]
Getting started• Paper 1H/2H style — 2 marks• quadratic equations, factorising
▼
Question
Solve $x^{2} - 14x + 49 = 0$ . Show clear algebraic working. [2]
Step-by-step solution
1. Step 1
  M1 Look for two numbers that multiply to $49$ and add to $-14$ : they are $-7$ and $-7$ .
  $(x - 7)(x - 7) = 0 \quad \Rightarrow \quad (x - 7)^{2} = 0$
2. Step 2
  A1 Both brackets give the same value, so there is only one solution.
  $x = 7$
Answer
$x = 7$
Examiner tip
Examiner tip: The expression is a perfect square, $(x - 7)^{2}$ . The graph of $y = x^{2} - 14x + 49$ just touches the $x$ -axis at $x = 7$ , which is why there is only one solution.

Common mistake: Writing $x = 7$ or $x = -7$ . Only the square is zero here, and $(-7 - 7)^{2} = 196$ , not $0$ . The extra wrong value loses the A mark.
7Factorising when a = 4 [3 marks]
Building confidence• Paper 1H/2H style — 3 marks• quadratic equations, factorising, rearranging
▼
Question
Solve $4x^{2} = 4x + 15$ . Show clear algebraic working. [3]
Step-by-step solution
1. Step 1
  M1 Rearrange so that the right-hand side is $0$ .
  $4x^{2} - 4x - 15 = 0$
2. Step 2
  M1 Factorise. $ac = -60$ , and $-10 \times 6 = -60$ with $-10 + 6 = -4$ .
  $4x^{2} - 10x + 6x - 15 = 2x(2x - 5) + 3(2x - 5) = (2x - 5)(2x + 3) = 0$
3. Step 3
  A1 Solve each bracket.
  $x = \dfrac{5}{2} \quad \text{or} \quad x = -\dfrac{3}{2}$
Answer
$x = \dfrac{5}{2}$ or $x = -\dfrac{3}{2}$ (or $2.5$ and $-1.5$ )
Examiner tip
Examiner tip: When $a = 4$ , try $2x$ in both brackets as well as $4x$ and $x$ . Splitting the middle term with $ac = -60$ finds the right pair for you.

Common mistake: Moving $4x$ across without changing its sign, which gives $4x^{2} + 4x - 15 = 0$ . That factorises to $(2x + 5)(2x - 3)$ and gives $x = -2.5$ and $1.5$ , the right numbers with the wrong signs. That loses the first M1 and the A1.
8Sum of squares of consecutive integers [4 marks]
Building confidence• Paper 1H/2H style — 4 marks• quadratic equations, forming equations
▼
Question
The sum of the squares of two consecutive integers is $145$ .

Use algebra to find the two possible pairs of integers. [4]
Step-by-step solution
1. Step 1
  M1 Call the integers $n$ and $n + 1$ and form an equation.
  $n^{2} + (n + 1)^{2} = 145$
2. Step 2
  A1 Expand $(n + 1)^{2} = n^{2} + 2n + 1$ , collect terms and divide by $2$ .
  $2n^{2} + 2n - 144 = 0 \quad \Rightarrow \quad n^{2} + n - 72 = 0$
3. Step 3
  M1 Factorise.
  $(n + 9)(n - 8) = 0 \quad \Rightarrow \quad n = -9 \text{ or } n = 8$
4. Step 4
  A1 Both values are integers, so both give a valid pair. Neither is rejected.
  $8 \text{ and } 9, \qquad -9 \text{ and } -8$
Answer
$8$ and $9$ , or $-9$ and $-8$
Examiner tip
Examiner tip: Only reject a root when the context rules it out. Integers can be negative, and $(-9)^{2} + (-8)^{2} = 81 + 64 = 145$ .

Common mistake: Writing $(n + 1)^{2} = n^{2} + 1$ and losing the $2n$ term. The other common loss is giving only $8$ and $9$ when the question asks for both possible pairs.
9Exact solutions in the form (a ± √b)/c [4 marks]
Stretch• Paper 1H/2H style — 4 marks• quadratic formula, surds, exact form
▼
Question
Solve $3x^{2} + 2x - 2 = 0$ . Give your solutions in the form $\dfrac{a \pm \sqrt{b}}{c}$ , where $a$ , $b$ and $c$ are integers. Show clear algebraic working. [4]
Step-by-step solution
1. Step 1
  M1 Substitute $a = 3$ , $b = 2$ , $c = -2$ into the quadratic formula.
  $x = \dfrac{-2 \pm \sqrt{2^{2} - 4 \times 3 \times (-2)}}{2 \times 3}$
2. Step 2
  A1 Simplify: $4 + 24 = 28$ .
  $x = \dfrac{-2 \pm \sqrt{28}}{6}$
3. Step 3
  M1 Simplify the surd.
  $\sqrt{28} = \sqrt{4 \times 7} = 2\sqrt{7} \quad \Rightarrow \quad x = \dfrac{-2 \pm 2\sqrt{7}}{6}$
4. Step 4
  A1 Every term has a factor of $2$ , so cancel it.
  $x = \dfrac{-1 \pm \sqrt{7}}{3}$
Answer
$x = \dfrac{-1 + \sqrt{7}}{3}$ or $x = \dfrac{-1 - \sqrt{7}}{3}$
Examiner tip
Examiner tip: Before you cancel, check that the factor divides every term: $-2$ , $2\sqrt{7}$ and $6$ all divide by $2$ . A quick calculator check: both $0.5486\ldots$ and $-1.2153\ldots$ make $3x^{2} + 2x - 2$ equal $0$ .

Common mistake: Cancelling only part of the fraction, for example $\dfrac{-1 \pm 2\sqrt{7}}{3}$ . This is the same error as dividing only one term of a split fraction, and it loses the final A mark.
10A path of equal width around a garden [5 marks]
Stretch• Paper 1H/2H style — 5 marks• quadratic formula, forming equations, area
▼
Question
A rectangular garden is $12$ m long and $8$ m wide.

A path of width $x$ metres is built all the way round the outside of the garden.

The total area of the garden and the path is $160$ m².

Work out the value of $x$ . Give your answer correct to $3$ significant figures. Show clear algebraic working. [5]
Step-by-step solution
1. Step 1
  M1 The path adds $x$ to each end, so the outer rectangle is $(12 + 2x)$ m by $(8 + 2x)$ m.
  $(12 + 2x)(8 + 2x) = 160$
2. Step 2
  M1 Expand and rearrange.
  $96 + 24x + 16x + 4x^{2} = 160 \quad \Rightarrow \quad 4x^{2} + 40x - 64 = 0$
3. Step 3
  A1 Divide by $4$ .
  $x^{2} + 10x - 16 = 0$
4. Step 4
  M1 This does not factorise, so use the quadratic formula.
  $x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \times 1 \times (-16)}}{2} = \dfrac{-10 \pm \sqrt{164}}{2}$
5. Step 5
  A1 $x = 1.4031\ldots$ or $x = -11.403\ldots$ . A width cannot be negative, so reject the second value.
  $x = 1.40$
Answer
$x = 1.40$
Examiner tip
Examiner tip: Check: $(12 + 2.806)(8 + 2.806) = 14.806 \times 10.806 = 160.0$ to 4 significant figures.

Common mistake: Using $(12 + x)(8 + x)$ , which adds the path to one side only. Another common loss is writing $1.4$ : the question asks for 3 significant figures, so the answer is $1.40$ .

Question 1
Paper 1H/2H style — 1 mark1 mark
Show that $x = -2$ is a solution of $3x^{2} + 5x - 2 = 0$ . [1]
Model answer
$3(-2)^{2} + 5(-2) - 2 = 3 \times 4 - 10 - 2 = 12 - 12 = 0$

So $x = -2$ is a solution.
Why this scores
Mark scheme:
- B1 for $12 - 10 - 2 = 0$ shown, with a conclusion
Why this scores full marks: The substitution is written with brackets round $-2$ , so $(-2)^{2} = 4$ and not $-4$ . The candidate shows every value and finishes by saying what the result means.
Question 2
Paper 1H/2H style — 2 marks2 marks
Solve $24 + 5x - x^{2} = 0$ . Show clear algebraic working. [2]
Model answer
Multiply every term by $-1$ :

$x^{2} - 5x - 24 = 0$

$(x - 8)(x + 3) = 0$

$x = 8$ or $x = -3$
Why this scores
Mark scheme:
- M1 for $(x - 8)(x + 3)$ or $(8 - x)(3 + x)$ oe, or a correct substitution into the quadratic formula
- A1 for $8$ and $-3$
Why this scores full marks: The candidate makes the $x^{2}$ term positive first by multiplying every term by $-1$ , so all three signs change together. Weaker scripts change only some of the signs, or factorise the original order carelessly and give $-8$ and $3$ . Check: $24 + 40 - 64 = 0$ and $24 - 15 - 9 = 0$ .
Question 3
Paper 1H/2H style — 2 marks2 marks
(a) Factorise $x^{2} - 16x + 63$ [1]

(b) Hence solve $x^{2} - 16x + 63 = 0$ [1]
Model answer
(a) $(x - 7)(x - 9)$

(b) $x = 7$ or $x = 9$
Why this scores
Mark scheme:
- B1 for $(x - 7)(x - 9)$ (a)
- B1 for $7$ and $9$ , ft their factors (b)
Why this scores full marks: "Hence" means use part (a). The two brackets are set to $0$ and the solutions follow at once. Both numbers are negative in the brackets because they multiply to $+63$ and add to $-16$ .
Question 4
Paper 1H/2H style — 3 marks3 marks
Solve $(x + 3)^{2} - 2(x + 3) - 15 = 0$ . Show clear algebraic working. [3]
Model answer
Let $y = x + 3$ :

$y^{2} - 2y - 15 = 0$

$(y - 5)(y + 3) = 0$

$y = 5$ or $y = -3$

$x + 3 = 5$ or $x + 3 = -3$

$x = 2$ or $x = -6$
Why this scores
Mark scheme:
- M1 for $y^{2} - 2y - 15 = 0$ oe, or for expanding to $x^{2} + 4x - 12 = 0$
- M1 for $(y - 5)(y + 3)$ or $(x + 6)(x - 2)$ , or a correct substitution into the quadratic formula
- A1 for $2$ and $-6$
Why this scores full marks: The candidate spots that $(x + 3)$ appears twice and replaces it with one letter, which gives a quadratic with small numbers. The last step turns the values of $y$ back into values of $x$ ; stopping at $5$ and $-3$ is the usual way to lose the A mark. Expanding everything first is equally valid, provided $-2(x + 3)$ becomes $-2x - 6$ .
Question 5
Paper 1H/2H style — 3 marks3 marks
Solve $x + \dfrac{6}{x} = 5$ . Show clear algebraic working. [3]
Model answer
Multiply every term by $x$ :

$x^{2} + 6 = 5x$

$x^{2} - 5x + 6 = 0$

$(x - 2)(x - 3) = 0$

$x = 2$ or $x = 3$
Why this scores
Mark scheme:
- M1 for $x^{2} + 6 = 5x$ oe
- M1 for $(x - 2)(x - 3)$ or a correct substitution into the quadratic formula
- A1 for $2$ and $3$
Why this scores full marks: All three terms are multiplied by $x$ , including the $5$ on the right. Weaker scripts write $x^{2} + 6 = 5$ and get a different equation. Both answers can be checked quickly: $2 + 3 = 5$ and $3 + 2 = 5$ .
Question 6
Paper 1H/2H style — 4 marks4 marks
Solve $(2x + 1)(2x + 3) = 3x + 6$ . Give your solutions correct to $3$ significant figures. Show your working clearly. [4]
Model answer
$4x^{2} + 8x + 3 = 3x + 6$

$4x^{2} + 5x - 3 = 0$

$x = \frac{-5 \pm \sqrt{5^{2} - 4 \times 4 \times (-3)}}{2 \times 4} = \frac{-5 \pm \sqrt{73}}{8}$

$x = 0.4430\ldots$ or $x = -1.6930\ldots$

$x = 0.443$ or $x = -1.69$
Why this scores
Mark scheme:
- M1 for $4x^{2} + 5x - 3 = 0$ oe
- M1 for a correct substitution into the quadratic formula
- A1 for $\dfrac{-5 \pm \sqrt{73}}{8}$ oe
- A1 for $0.443$ and $-1.69$ , awrt
Why this scores full marks: The expansion has all four terms ( $4x^{2} + 6x + 2x + 3$ ) before anything is moved. The substitution is shown in full, and the answers are rounded only at the end, to the accuracy the question asks for.
Question 7
Paper 1H/2H style — 4 marks4 marks
The curve $y = x^{2} + 4x - 1$ crosses the $x$ -axis at the points $A$ and $B$ .

Find the exact length of $AB$ . Show your working clearly. [4]
Model answer
At $A$ and $B$ , $y = 0$ : $x^{2} + 4x - 1 = 0$

$x = \frac{-4 \pm \sqrt{4^{2} - 4 \times 1 \times (-1)}}{2} = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2\sqrt{5}}{2} = -2 \pm \sqrt{5}$

$AB = (-2 + \sqrt{5}) - (-2 - \sqrt{5}) = 2\sqrt{5}$

Length $AB = 2\sqrt{5}$
Why this scores
Mark scheme:
- M1 for $x^{2} + 4x - 1 = 0$
- M1 for a correct substitution into the quadratic formula
- A1 for $-2 \pm \sqrt{5}$ oe
- A1 for $2\sqrt{5}$ oe, e.g. $\sqrt{20}$
Why this scores full marks: "Exact" means surd form, so the candidate never converts to decimals. The length is the larger $x$ -value minus the smaller one, and the $-2$ terms cancel. A decimal answer of $4.47$ would lose the final A mark.
Question 8
Paper 1H/2H style — 5 marks5 marks
Sara cycles $60$ km at an average speed of $x$ km/h.

On the return journey her average speed is $5$ km/h faster, and the journey takes $1$ hour less.

Work out the value of $x$ . Show clear algebraic working. [5]
Model answer
Time there $= \dfrac{60}{x}$ hours; time back $= \dfrac{60}{x + 5}$ hours.

$\frac{60}{x} - \frac{60}{x + 5} = 1$

$60(x + 5) - 60x = x(x + 5)$

$300 = x^{2} + 5x$

$x^{2} + 5x - 300 = 0$

$(x + 20)(x - 15) = 0$

$x = 15$ (a speed cannot be $-20$ )
Why this scores
Mark scheme:
- M1 for $\dfrac{60}{x}$ and $\dfrac{60}{x + 5}$
- M1 for $\dfrac{60}{x} - \dfrac{60}{x + 5} = 1$ oe, with fractions cleared
- A1 for $x^{2} + 5x - 300 = 0$
- M1 for $(x + 20)(x - 15)$ or a correct substitution into the quadratic formula
- A1 for $15$ only
Why this scores full marks: The slower journey takes longer, so the candidate subtracts the return time from the outward time. After clearing the fractions with brackets, $60x - 60x$ cancels and leaves a tidy quadratic. The candidate checks: $4$ hours at $15$ km/h and $3$ hours at $20$ km/h.
Question 9
Paper 1H/2H style — 5 marks5 marks
A polygon with $n$ sides has $d$ diagonals, where

$d = \frac{n(n - 3)}{2}$

(a) A polygon has $54$ diagonals. Work out the number of sides it has. [3]

(b) Show that no polygon has exactly $30$ diagonals. [2]
Model answer
(a) $n(n - 3) = 108$

$n^{2} - 3n - 108 = 0$

$(n - 12)(n + 9) = 0$

$n = 12$ ( $n$ cannot be $-9$ )

(b) $n(n - 3) = 60 \Rightarrow n^{2} - 3n - 60 = 0$

$n = \frac{3 \pm \sqrt{9 + 240}}{2} = \frac{3 \pm \sqrt{249}}{2}$

$n = 9.389\ldots$ or $n = -6.389\ldots$

Neither value is a positive whole number, so no polygon has exactly $30$ diagonals.
Why this scores
Mark scheme:
- M1 for $n^{2} - 3n - 108 = 0$ oe (a)
- M1 for $(n - 12)(n + 9)$ or a correct substitution into the quadratic formula (a)
- A1 for $12$ cao (a)
- M1 for $n^{2} - 3n - 60 = 0$ and a correct method to solve it (b)
- A1 for $9.39$ (or $\dfrac{3 + \sqrt{249}}{2}$ ) and a conclusion that $n$ is not an integer (b)
Why this scores full marks: Part (b) is a "show that", so the candidate solves the equation and then says clearly why the result proves the statement. Showing that $249$ is not a square number would also be a complete reason.
Question 10
Paper 1H/2H style — 6 marks6 marks
A trapezium has parallel sides of length $(x + 2)$ cm and $3x$ cm.

The perpendicular distance between the parallel sides is $(x - 1)$ cm.

The area of the trapezium is $44$ cm².

(a) Show that $2x^{2} - x - 45 = 0$ [3]

(b) Work out the length of the longest parallel side. Show clear algebraic working. [3]
Model answer
(a) $\frac{1}{2}\left[(x + 2) + 3x\right](x - 1) = 44$

$\dfrac{1}{2}(4x + 2)(x - 1) = 44$

$(2x + 1)(x - 1) = 44$

$2x^{2} - x - 1 = 44$

$2x^{2} - x - 45 = 0$

(b) $(2x + 9)(x - 5) = 0$

$x = 5$ ( $x = -4.5$ would make $x - 1$ negative)

Longest parallel side $= 3 \times 5 = 15$ cm
Why this scores
Mark scheme:
- M1 for $\dfrac{1}{2}(x + 2 + 3x)(x - 1)$ oe (a)
- M1 for $(2x + 1)(x - 1) = 44$ or $4x^{2} - 2x - 2 = 88$ oe (a)
- A1 for a correct, complete argument reaching $2x^{2} - x - 45 = 0$ (a)
- M1 for $(2x + 9)(x - 5)$ or a correct substitution into the quadratic formula (b)
- A1 for $x = 5$ (b)
- A1 for $15$ cao (b)
Why this scores full marks: In (a) every line follows from the one before, and nothing is skipped on the way to the printed equation. In (b) the negative root is rejected with a reason, and the candidate answers the question asked: the longest side is $3x = 15$ cm, not $x + 2 = 7$ cm. Check: $\dfrac{1}{2}(7 + 15) \times 4 = 44$ .

4MA1 Quadratic Equations

At a glance

What you’ll learn

The three methods

The discriminant — what kind of solutions?

Quadratic word problems

Quick recap

Memorise this

How it’s examined

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

1Solve by factorising

2Solve using the quadratic formula

3Solve by factorising (a > 1)

4Quadratic that doesn't factor neatly

5Quadratic from a word problem

6A quadratic with one repeated solution [2 marks]

7Factorising when a = 4 [3 marks]

8Sum of squares of consecutive integers [4 marks]

9Exact solutions in the form (a ± √b)/c [4 marks]

10A path of equal width around a garden [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