What is a quadratic equation and how is it structured in Pure Mathematics 1?

A quadratic equation is a second-degree polynomial equation in a single variable x, with the general form ax^2 + bx + c = 0, where a, b, and c are constants, and a ≠ 0. In the Edexcel International A Levels Pure Mathematics 1 curriculum, understanding the structure of quadratic equations is fundamental as it forms the basis for solving and analyzing these equations.

How do you solve a quadratic equation using the quadratic formula?

To solve a quadratic equation using the quadratic formula, you apply the formula x = (-b ± √(b² - 4ac)) / (2a). This formula provides the solutions (roots) of the quadratic equation ax^2 + bx + c = 0. It's essential to calculate the discriminant (b² - 4ac) first, as it determines the nature of the roots. This method is widely taught in the Edexcel International A Levels Pure Mathematics 1 syllabus.

What is the significance of the discriminant in a quadratic equation?

The discriminant of a quadratic equation, given by b² - 4ac, indicates the nature of the roots. If the discriminant is positive, the equation has two distinct real roots. If it is zero, there is exactly one real root (a repeated root). If the discriminant is negative, the equation has two complex roots. Understanding the discriminant is crucial in the Edexcel International A Levels Pure Mathematics 1 curriculum for analyzing quadratic equations.

How can you factorize a quadratic equation?

To factorize a quadratic equation of the form ax^2 + bx + c = 0, you need to express it as a product of two binomials, (px + q)(rx + s) = 0. This involves finding two numbers that multiply to ac and add to b. Factorization is a key technique in the Edexcel International A Levels Pure Mathematics 1 course, as it simplifies solving quadratic equations and understanding their properties.

What are the applications of quadratic equations in real-world scenarios?

Quadratic equations are used in various real-world applications, such as calculating projectile motion, optimizing areas and volumes, and modeling economic problems. In the Edexcel International A Levels Pure Mathematics 1 curriculum, students learn how to apply quadratic equations to solve practical problems, enhancing their understanding of the relevance and utility of mathematics in everyday life.

Why is "For $ax^{2} + bx + c$ with $a > 1$, completing the square without first factoring out $a$" a common mistake in Quadratics?

Why it happens: Skipping the factor-out step. How to avoid it: Write a(x^2 + (b/a)x) + c first. Complete the square INSIDE. Then redistribute. Source: WMA11/01 examiner reports — flagged in June 2023 Q5 commentary

Why is "Using $\Delta \ge 0$ when the question says 'distinct real roots'" a common mistake in Quadratics?

Why it happens: Confusing 'real' with 'distinct real'. How to avoid it: Distinct real roots → > 0 (strict). 'Real roots' (with repeated allowed) → 0. 'No real roots' → < 0.

Why is "In the quadratic formula, getting $-b$ wrong when $b$ is negative" a common mistake in Quadratics?

Why it happens: Mixing signs of b and -b. How to avoid it: If b = -7, then -b = +7. ALWAYS rewrite: x = -(-7) √()2a = 7 √()2a.

Why is "Failing to reject negative $u$ values when $u = \sqrt{x}$ or $u = e^{x}$" a common mistake in Quadratics?

Why it happens: Forgetting domain restriction. How to avoid it: √(x) 0 always; e^x > 0 always. After solving for u, REJECT any negative value. Only positive u-values give valid x. Source: WMA11/01 examiner reports flag this for hidden-quadratic questions

Why is "Reading vertex of $(x + 3)^{2} - 4$ as $(3, -4)$ instead of $(-3, -4)$" a common mistake in Quadratics?

Why it happens: Forgetting that the x-value flips sign. How to avoid it: Vertex of (x + p)^2 + q is at x = -p (because x + p = 0 x = -p). So (x + 3)^2 has minimum when x = -3.

Pure Mathematics 1Edexcel International A Levels Mathematics (XMA01-YMA01)

Quadratics

Q: Why is "In the quadratic formula, getting $-b$ wrong when $b$ is negative" a common mistake in Quadratics?

Why it happens: Mixing signs of b and -b. How to avoid it: If b = -7, then -b = +7. ALWAYS rewrite: x = -(-7) √()2a = 7 √()2a.

Q: Why is "Failing to reject negative $u$ values when $u = \sqrt{x}$ or $u = e^{x}$" a common mistake in Quadratics?

Why it happens: Forgetting domain restriction. How to avoid it: √(x) 0 always; e^x > 0 always. After solving for u, REJECT any negative value. Only positive u-values give valid x. Source: WMA11/01 examiner reports flag this for hidden-quadratic questions

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Quadratics Study Notes — Edexcel IAL Mathematics P1 (WMA11/01, 2018 spec — 2026 onwards)

Completing the square, the quadratic formula, the discriminant, and hidden quadratics. The single most heavily-examined topic in P1 — every paper has at least one quadratic question.

What you’ll learn

Mapped to the Cambridge IGCSE XMA01-YMA01 syllabus (2026 onwards).

2.1 — Work with quadratic functions and their graphs; the discriminant of a quadratic function, including the conditions for real and repeated roots.
2.2 — Completing the square.
2.3 — Solution of quadratic equations including those reducible to quadratic form (hidden quadratics).
2.4 — Solve quadratic equations using factorisation, completing the square, and the quadratic formula.

Three ways to solve a quadratic

Factorise → formula → completing the square. Pick the right tool.

Method 1: Factorisation. Quickest when it works.

$x^{2} - 5x + 6 = 0 \Rightarrow (x - 2)(x - 3) = 0 \Rightarrow x = 2 \text{ or } x = 3.$

For non-monic ( $a > 1$ ), use grouping or trial — e.g. $2x^{2} + 5x - 3 = (2x - 1)(x + 3)$ .

Method 2: Quadratic formula (in the formula booklet).

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

Use when factorisation isn't obvious, or when the question asks for exact (surd) form.

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

$a = 2$ , $b = 5$ , $c = -4$ .
$\Delta = 25 - 4(2)(-4) = 25 + 32 = 57$ .
$x = \dfrac{-5 \pm \sqrt{57}}{4}$ .

Method 3: Completing the square. Use when you need vertex form, exact-form answers when the formula is messy, or when the question explicitly says 'by completing the square'.

When to use	Method
Quick numbers, integer roots	Factorise
Surd-form answers requested	Formula or completing the square
Vertex / sketch question	Completing the square
Discriminant / 'show real roots'	Discriminant only — don't solve

Factorisation: fastest when it works.
Formula: when factorising fails.
Completing the square: vertex problems.
Pick the method the question's wording hints at.

Completing the square — the IAL essential

Half the coefficient of $x$ , square, subtract. Or factor out $a$ first if non-monic.

Monic case ( $a = 1$ ). For $x^{2} + bx + c$ :

Half of $b$ is $b/2$ .
$(x + b/2)^{2}$ expands to $x^{2} + bx + (b/2)^{2}$ .
Subtract $(b/2)^{2}$ to compensate, then add $c$ :

$x^{2} + bx + c = \left(x + \tfrac{b}{2}\right)^{2} - \tfrac{b^{2}}{4} + c.$

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

Half of 6 is 3. $(x + 3)^{2} = x^{2} + 6x + 9$ .
Compensate: $- 9 - 4 = -13$ .
Answer: $(x + 3)^{2} - 13$ .

Non-monic case ( $a \neq 1$ ). FACTOR OUT $a$ first.

$2x^{2} - 12x + 7 = 2(x^{2} - 6x) + 7.$

Complete inside: $x^{2} - 6x = (x - 3)^{2} - 9$ .

Substitute back: $2[(x - 3)^{2} - 9] + 7 = 2(x - 3)^{2} - 18 + 7 = 2(x - 3)^{2} - 11$ .

Vertex from completed-square form.

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

Vertex at $(-p, q)$ .
$a > 0$ : opens up, vertex is minimum.
$a < 0$ : opens down, vertex is maximum.

Example. $y = 2(x - 3)^{2} - 11$ has minimum at $(3, -11)$ .

Why completing the square works. $(x + p)^{2}$ is always $\ge 0$ . So $a(x + p)^{2} + q$ has its smallest value when $(x + p)^{2} = 0$ — i.e. when $x = -p$ . At that point the expression equals $q$ . So vertex is $(-p, q)$ .

Completed-square form y = 2(x − 3)² − 11 reveals the minimum vertex at (3, −11) and opens upwards because a = 2 > 0.

Monic: half the $x$ -coefficient, square, subtract.
Non-monic: factor out $a$ first.
Vertex: $(-p, q)$ for $(x + p)^{2} + q$ .
Sign of $a$ : + = minimum, − = maximum.

The discriminant — three cases

$\Delta = b^{2} - 4ac$ . Sign tells you everything about the roots.

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

$\Delta$	Roots	Geometry
$> 0$	Two distinct real roots	Parabola cuts $x$ -axis at two points
$= 0$	One repeated real root	Parabola touches $x$ -axis (tangent)
$< 0$	No real roots	Parabola does not meet $x$ -axis

Why? The quadratic formula has $\sqrt{\Delta}$ . If $\Delta < 0$ you can't take a real square root. If $\Delta = 0$ the $\pm \sqrt{0}$ collapses to one value. If $\Delta > 0$ you get two distinct values.

Typical question. Find the values of $k$ for which $x^{2} + (k - 3)x + (3 - 2k) = 0$ has two distinct real roots.

Solution:

Discriminant condition: $\Delta > 0$ (strict for 'distinct').
$\Delta = (k - 3)^{2} - 4(1)(3 - 2k)$ .
Expand: $k^{2} - 6k + 9 - 12 + 8k = k^{2} + 2k - 3$ .
Solve: $k^{2} + 2k - 3 > 0 \Rightarrow (k + 3)(k - 1) > 0$ .
Sketch: upward parabola, roots at $-3$ and $1$ .
Above zero for $k < -3$ or $k > 1$ .

Sign of discriminant — wording cheat sheet:

Wording	Condition
"Two distinct real roots"	$\Delta > 0$
"Equal/repeated roots"	$\Delta = 0$
"Real roots" (no further qualifier)	$\Delta \ge 0$
"No real roots"	$\Delta < 0$
"Roots are real and equal"	$\Delta = 0$
"Quadratic touches the $x$ -axis"	$\Delta = 0$

$\Delta = b^{2} - 4ac$ .
Distinct real roots → $\Delta > 0$ .
Repeated root → $\Delta = 0$ .
No real roots → $\Delta < 0$ .
Read the wording carefully — 'distinct' is strict.

Hidden quadratics — substitution unlocks them

$u = \sqrt{x}$ , $u = e^{x}$ , $u = x^{2}$ , $u = \sin x$ — let $u$ work.

Hidden quadratic. An equation that becomes quadratic after a substitution.

Type 1: $u = \sqrt{x}$ .

$x - 5\sqrt{x} + 6 = 0.$

Let $u = \sqrt{x}$ , so $u^{2} = x$ . The equation becomes $u^{2} - 5u + 6 = 0$ . Factorise: $(u - 2)(u - 3) = 0$ .

So $u = 2$ or $u = 3$ . Both positive, so both valid. $\sqrt{x} = 2 \Rightarrow x = 4$ ; $\sqrt{x} = 3 \Rightarrow x = 9$ .

Type 2: $u = x^{2}$ . (A.k.a. a biquadratic.)

$x^{4} - 13x^{2} + 36 = 0.$

Let $u = x^{2}$ : $u^{2} - 13u + 36 = 0 \Rightarrow (u - 4)(u - 9) = 0$ . So $u = 4$ or $9$ .

Solve $x^{2} = 4 \Rightarrow x = \pm 2$ ; $x^{2} = 9 \Rightarrow x = \pm 3$ . Four roots.

Type 3: $u = e^{x}$ .

$e^{2x} - 5e^{x} + 4 = 0 \Rightarrow u^{2} - 5u + 4 = 0 \Rightarrow (u - 1)(u - 4) = 0.$

So $e^{x} = 1$ or $e^{x} = 4$ . Take logs: $x = 0$ or $x = \ln 4$ .

(Note: $e^{x}$ is always positive, so negative $u$ -values would be rejected — but here both are positive.)

Type 4: $u = \sin x$ . (More P2/P3, but worth noting.)

$2\sin^{2}\theta - 3\sin\theta + 1 = 0 \Rightarrow 2u^{2} - 3u + 1 = 0.$

The reject step. ALWAYS check whether each $u$ -value gives a valid $x$ . If $u = \sqrt{x}$ comes out negative, reject (square root can't be negative). If $u = e^{x}$ comes out negative or zero, reject (exponential is always positive). For $u = \sin x$ , $u$ must be in $[-1, 1]$ .

Spot pattern: equation has $x$ and $\sqrt{x}$ , OR $x^{4}$ and $x^{2}$ , OR $e^{2x}$ and $e^{x}$ .
Let $u =$ the smaller power.
Solve the quadratic in $u$ .
Substitute back and REJECT impossible $u$ -values.

How it’s examined

Quadratics is the most-tested topic in WMA11/01 P1. Almost every paper has: one completing-the-square question (4-6 marks), one discriminant question (4-6 marks), and either a hidden-quadratic or a quadratic-formula question. The 'show that' and 'find the set of values of $k$ ' style questions hit A* level — they require crisp algebra AND careful inequality work. Most-dropped marks: wrong sign on vertex, using $\ge$ instead of $>$ for distinct roots, forgetting to reject negative $u$ -values in hidden quadratics. A* candidates should aim for full marks across the quadratics section.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA11/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

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Step-by-step worked examples — Quadratics

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

1Completing the square (4 marks, P1)
Core• Style: WMA11/01 January 2024 Q5• completing the square
Question
Express $f(x) = 2x^{2} - 12x + 7$ in the form $a(x + b)^{2} + c$ , where $a$ , $b$ , $c$ are integers. Hence write down the minimum value of $f(x)$ and the value of $x$ at which it occurs.
Step-by-step solution
1. Step 1
  Factor 2 out of the $x$ -terms:
  $2x^{2} - 12x + 7 = 2(x^{2} - 6x) + 7$
2. Step 2
  Complete the square inside: $x^{2} - 6x = (x - 3)^{2} - 9$ .
3. Step 3
  Substitute back and simplify:
  $2[(x - 3)^{2} - 9] + 7 = 2(x - 3)^{2} - 18 + 7 = 2(x - 3)^{2} - 11$
4. Step 4
  So $a = 2$ , $b = -3$ , $c = -11$ . Minimum value is $-11$ , occurring when $x = 3$ .
Answer
$f(x) = 2(x - 3)^{2} - 11$ . Minimum value $-11$ at $x = 3$ .
Examiner tip
M1 for factoring out 2 (or equivalent); M1 for completing the square inside; A1 for the final form; B1 for the minimum value AND the $x$ -coordinate (both required). Common error: stopping at $2(x - 3)^{2} - 9 + 7$ without simplifying — loses the final A1.
2Discriminant — find the range of values (5 marks, P1)
Extended• Style: WMA11/01 June 2024 Q6• discriminant
Question
The equation $x^{2} + (k - 3)x + (3 - 2k) = 0$ has two distinct real roots. Find the set of possible values of $k$ .
Step-by-step solution
1. Step 1
  Distinct real roots $\Leftrightarrow$ discriminant $> 0$ . Here $a = 1$ , $b = k - 3$ , $c = 3 - 2k$ .
2. Step 2
  Compute discriminant:
  $\Delta = (k - 3)^{2} - 4(1)(3 - 2k) = k^{2} - 6k + 9 - 12 + 8k = k^{2} + 2k - 3$
3. Step 3
  Solve $k^{2} + 2k - 3 > 0$ . Factorise: $(k + 3)(k - 1) > 0$ .
4. Step 4
  Sketch the parabola $y = (k + 3)(k - 1)$ : roots at $k = -3$ and $k = 1$ , opens upwards. Above zero outside the roots.
5. Step 5
  Answer: $k < -3$ or $k > 1$ .
Answer
$k < -3$ or $k > 1$
Examiner tip
M1 for stating discriminant condition $> 0$ ; M1 for correct $a, b, c$ identification and discriminant expression; A1 for simplified discriminant $k^{2} + 2k - 3$ ; M1 for factorising and solving the quadratic inequality; A1 for the correct set notation. Common errors: using $\ge$ (gives equal roots, not distinct); reading the sign of the inequality the wrong way after sketching.
3Solve using the quadratic formula — exact form (3 marks, P1)
Core• Style: WMA11/01 January 2023 Q4• quadratic formula
Question
Solve $2x^{2} + 5x - 4 = 0$ , giving your answers in the form $\dfrac{p \pm \sqrt{q}}{r}$ where $p$ , $q$ , $r$ are integers.
Step-by-step solution
1. Step 1
  Identify $a = 2$ , $b = 5$ , $c = -4$ .
2. Step 2
  Apply formula (given in the formula booklet):
  $x = \dfrac{-5 \pm \sqrt{25 - 4(2)(-4)}}{2(2)} = \dfrac{-5 \pm \sqrt{25 + 32}}{4}$
3. Step 3
  Simplify under the root: $25 + 32 = 57$ . Final answer:
  $x = \dfrac{-5 \pm \sqrt{57}}{4}$
Answer
$x = \dfrac{-5 \pm \sqrt{57}}{4}$ (so $p = -5$ , $q = 57$ , $r = 4$ ).
Examiner tip
M1 for correct substitution into the quadratic formula; A1 for $\sqrt{57}$ ; A1 for the fully correct exact answer. The 'in the form' instruction forbids a decimal answer — students who write $x \approx 0.64, -3.14$ score M1 only.
4Hidden quadratic — substitution (5 marks, P1)
Extended• Style: WMA11/01 June 2023 Q7• hidden quadratic, substitution
Question
Solve $x - 5\sqrt{x} + 6 = 0$ , giving your answers as integers.
Step-by-step solution
1. Step 1
  Let $u = \sqrt{x}$ , so $u^{2} = x$ . The equation becomes:
  $u^{2} - 5u + 6 = 0$
2. Step 2
  Factorise: $(u - 2)(u - 3) = 0$ , so $u = 2$ or $u = 3$ .
3. Step 3
  Substitute back: $\sqrt{x} = 2 \Rightarrow x = 4$ . $\sqrt{x} = 3 \Rightarrow x = 9$ .
4. Step 4
  Check: $4 - 5(2) + 6 = 0$ ✓; $9 - 5(3) + 6 = 0$ ✓.
Answer
$x = 4$ or $x = 9$ .
Examiner tip
M1 for recognising the hidden quadratic and substituting $u = \sqrt{x}$ (or equivalent); A1 for the correct quadratic in $u$ ; M1 for factorising/solving; A1 for both values of $u$ ; A1 for the final $x$ -values. Crucially: both $\sqrt{x} = 2$ and $\sqrt{x} = 3$ give valid solutions because both $u$ -values are positive. If a $u$ -value had been negative, that solution would be rejected (since $\sqrt{x} \ge 0$ ).
5Completing the square to sketch a parabola (6 marks, P1)
Extended• Style: WMA11/01 January 2024 Q8• completing the square, sketch
Question
$f(x) = x^{2} + 4x - 12$ . (a) Express $f(x)$ in completed square form. (b) State the coordinates of the vertex of $y = f(x)$ . (c) Find the $x$ -intercepts of $y = f(x)$ and sketch.
Step-by-step solution
1. Step 1
  (a) Complete the square. Half of $4$ is $2$ , so $(x + 2)^{2} = x^{2} + 4x + 4$ . Subtract $4$ to compensate:
  $f(x) = (x + 2)^{2} - 4 - 12 = (x + 2)^{2} - 16$
2. Step 2
  (b) Vertex of $(x + p)^{2} + q$ is at $(-p, q)$ . So vertex is at $(-2, -16)$ .
3. Step 3
  (c) For $x$ -intercepts, set $f(x) = 0$ : $(x + 2)^{2} = 16 \Rightarrow x + 2 = \pm 4 \Rightarrow x = 2$ or $x = -6$ .
4. Step 4
  $y$ -intercept (for the sketch): $f(0) = -12$ . Sketch a U-shaped parabola through $(-6, 0)$ , $(0, -12)$ , $(2, 0)$ with minimum at $(-2, -16)$ .
Answer
(a) $(x + 2)^{2} - 16$ . (b) Vertex $(-2, -16)$ . (c) $x$ -intercepts $x = -6$ and $x = 2$ .
Examiner tip
Mark scheme: (a) M1 A1 (2). (b) B1 ft (1) — follow-through from (a). (c) M1 for setting $f = 0$ or factorising, A1 A1 for both roots, B1 for a sketch showing minimum, U-shape, and both intercepts (4 marks). A* candidates should label vertex coordinates AND intercepts on the sketch.

Key Formulae — Quadratics

The formulae you need to memorise for quadratics on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Quadratic formula
$x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$
$a, b, c$
coefficients in $ax^{2} + bx + c = 0$ , $a \neq 0$
When to use
When a quadratic does not factorise nicely, or when exact-surd answers are required. IS in the IAL formula booklet, so you don't need to memorise it — but you must apply it accurately.
Example
$2x^{2} + 5x - 4 = 0$ gives $x = \dfrac{-5 \pm \sqrt{57}}{4}$ .
Discriminant
$\Delta = b^{2} - 4ac$
$a, b, c$
coefficients in $ax^{2} + bx + c = 0$
When to use
To determine the nature of the roots of a quadratic. NOT in the formula booklet — but the quadratic formula in the booklet contains it. Memorise the three cases: $\Delta > 0$ two distinct real roots; $\Delta = 0$ one repeated real root; $\Delta < 0$ no real roots.
Example
For 'two distinct real roots', set $\Delta > 0$ and solve for the parameter.
Completing the square (general)
$ax^{2} + bx + c = a\left(x + \dfrac{b}{2a}\right)^{2} + c - \dfrac{b^{2}}{4a}$
$a, b, c$
coefficients
When to use
When the form $(x + p)^{2} + q$ is required — for finding vertex, proving non-negativity, or solving in surd form. NOT in the formula booklet.
Example
$x^{2} + 6x - 4 = (x + 3)^{2} - 13$ — vertex at $(-3, -13)$ .
Vertex (turning point) from completed square
$y = a(x + p)^{2} + q \Rightarrow \text{vertex at } (-p,\; q)$
$a, p, q$
constants ( $a$ determines opening direction)
When to use
Read off the vertex directly from completed-square form. If $a > 0$ it's a minimum; if $a < 0$ it's a maximum.
Example
$y = 2(x - 3)^{2} - 11$ has minimum at $(3, -11)$ .

Key Definitions and Keywords — Quadratics

Definitions to memorise and the exact keywords mark schemes credit for quadratics answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Quadratic equation
Examiner keyword
An equation of the form $ax^{2} + bx + c = 0$ where $a \neq 0$ . A polynomial equation of degree 2.
Discriminant
Examiner keyword
$\Delta = b^{2} - 4ac$ . Controls the number of real roots.
Example
$\Delta > 0$ : two distinct real roots. $\Delta = 0$ : one repeated root. $\Delta < 0$ : no real roots.
Vertex (turning point)
The minimum or maximum point of a parabola. For $y = a(x + p)^{2} + q$ , vertex is at $(-p, q)$ .
Hidden quadratic
An equation that becomes a quadratic after a suitable substitution. Common substitutions: $u = \sqrt{x}$ , $u = e^{x}$ , $u = x^{n}$ , $u = \sin x$ .
Example
$x^{4} - 5x^{2} + 4 = 0 \Rightarrow u^{2} - 5u + 4 = 0$ with $u = x^{2}$ .
Roots of a quadratic
The $x$ -values where the parabola meets the $x$ -axis. Found by setting the quadratic equal to zero and solving.
Related: vertex, discriminant

Common Mistakes and Misconceptions — Quadratics

The traps other students keep falling into on quadratics questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕For $ax^{2} + bx + c$ with $a > 1$ , completing the square without first factoring out $a$
WMA11/01 examiner reports — flagged in June 2023 Q5 commentary
Why it happens
Skipping the factor-out step.
How to avoid it
Write $a(x^{2} + (b/a)x) + c$ first. Complete the square INSIDE. Then redistribute.
✕Using $\Delta \ge 0$ when the question says 'distinct real roots'
Why it happens
Confusing 'real' with 'distinct real'.
How to avoid it
Distinct real roots → $\Delta > 0$ (strict). 'Real roots' (with repeated allowed) → $\Delta \ge 0$ . 'No real roots' → $\Delta < 0$ .
✕In the quadratic formula, getting $-b$ wrong when $b$ is negative
Why it happens
Mixing signs of $b$ and $-b$ .
How to avoid it
If $b = -7$ , then $-b = +7$ . ALWAYS rewrite: $x = \dfrac{-(-7) \pm \sqrt{\ldots}}{2a} = \dfrac{7 \pm \sqrt{\ldots}}{2a}$ .
✕Failing to reject negative $u$ values when $u = \sqrt{x}$ or $u = e^{x}$
WMA11/01 examiner reports flag this for hidden-quadratic questions
Why it happens
Forgetting domain restriction.
How to avoid it
$\sqrt{x} \ge 0$ always; $e^{x} > 0$ always. After solving for $u$ , REJECT any negative value. Only positive $u$ -values give valid $x$ .
✕Reading vertex of $(x + 3)^{2} - 4$ as $(3, -4)$ instead of $(-3, -4)$
Why it happens
Forgetting that the $x$ -value flips sign.
How to avoid it
Vertex of $(x + p)^{2} + q$ is at $x = -p$ (because $x + p = 0 \Rightarrow x = -p$ ). So $(x + 3)^{2}$ has minimum when $x = -3$ .

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Quadratics

Short Study Notes in the form of Slides

Short Study Notes — Quadratics

Detailed Notes

What you’ll learn

How it’s examined

1Completing the square (4 marks, P1)

2Discriminant — find the range of values (5 marks, P1)

3Solve using the quadratic formula — exact form (3 marks, P1)

4Hidden quadratic — substitution (5 marks, P1)

5Completing the square to sketch a parabola (6 marks, P1)

Quadratic formula

Discriminant

Completing the square (general)

Vertex (turning point) from completed square

Quadratic equation

Discriminant

Vertex (turning point)

Hidden quadratic

Roots of a quadratic

✕For ax2+bx+cax^{2} + bx + cax2+bx+c with a>1a > 1a>1, completing the square without first factoring out aaa

✕Using Δ≥0\Delta \ge 0Δ≥0 when the question says 'distinct real roots'

✕In the quadratic formula, getting −b-b−b wrong when bbb is negative

✕Failing to reject negative uuu values when u=xu = \sqrt{x}u=x​ or u=exu = e^{x}u=ex

✕Reading vertex of (x+3)2−4(x + 3)^{2} - 4(x+3)2−4 as (3,−4)(3, -4)(3,−4) instead of (−3,−4)(-3, -4)(−3,−4)

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Quadratics — frequently asked questions

✕For $ax^{2} + bx + c$ with $a > 1$ , completing the square without first factoring out $a$

✕Using $\Delta \ge 0$ when the question says 'distinct real roots'

✕In the quadratic formula, getting $-b$ wrong when $b$ is negative

✕Failing to reject negative $u$ values when $u = \sqrt{x}$ or $u = e^{x}$

✕Reading vertex of $(x + 3)^{2} - 4$ as $(3, -4)$ instead of $(-3, -4)$