What is a quadratic equation and how is it different from a linear equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and a ≠ 0. It differs from a linear equation, which is of the form ax + b = 0, because it includes a squared term (x^2), resulting in a parabolic graph rather than a straight line. Understanding this distinction is crucial for Cambridge IGCSE Mathematics students.

How do you solve quadratic equations using the quadratic formula?

To solve a quadratic equation using the quadratic formula, apply the formula x = (-b ± √(b^2 - 4ac)) / (2a). This formula provides the solutions for x by substituting the coefficients a, b, and c from the equation ax^2 + bx + c = 0. This method is particularly useful when the equation does not factor easily, a common scenario in Cambridge IGCSE Algebra problems.

What are the methods to solve quadratic inequalities?

Quadratic inequalities can be solved using methods such as graphing, test points, or algebraic manipulation. First, solve the corresponding quadratic equation to find critical points. Then, determine the intervals where the inequality holds true by testing points from each interval. This approach helps Cambridge IGCSE students understand the solution set of inequalities like ax^2 + bx + c > 0 or ax^2 + bx + c < 0.

How can you factorize a quadratic equation?

To factorize a quadratic equation, express it in the form ax^2 + bx + c = 0 and find two numbers that multiply to ac and add to b. Rewrite the middle term using these numbers and factor by grouping. This method is often emphasized in the Cambridge IGCSE curriculum as it provides a straightforward way to find the roots of the equation when they are rational.

What is the significance of the discriminant in quadratic equations?

The discriminant, given by b^2 - 4ac in the quadratic formula, determines the nature of the roots of a quadratic equation. If the discriminant is positive, the equation has two distinct real roots; if zero, it has one real root (a repeated root); and if negative, it has two complex roots. Understanding the discriminant is crucial for Cambridge IGCSE students to predict the type of solutions without solving the equation.

Why is "Sign error inside the quadratic formula" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students forget that -4ac flips sign when c is already negative. How to avoid it: Substitute with brackets: √(b^2) - 4(a)(c) and evaluate carefully. Source: 0580/42 — examiner reports, recurring

Why is "Reporting only one root from a factorised quadratic" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students stop after solving one factor. How to avoid it: If (x - p)(x - q) = 0, BOTH x = p and x = q are solutions. Always state both.

Why is "Picking the wrong region for a quadratic inequality" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students factor correctly but pick "x > 5 or x 0" is above = outside the roots.

Why is "Adding $(\frac{b}{2})$ instead of $(\frac{b}{2})^{2}$" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Forgetting the squaring step in completing the square. How to avoid it: Always square the half-coefficient: b/2 → (b/2)^2.

AlgebraCambridge IGCSE Mathematics (0580)

Quadratic Equations and Inequalities

Q: How can you factorize a quadratic equation?

To factorize a quadratic equation, express it in the form ax^2 + bx + c = 0 and find two numbers that multiply to ac and add to b. Rewrite the middle term using these numbers and factor by grouping. This method is often emphasized in the Cambridge IGCSE curriculum as it provides a straightforward way to find the roots of the equation when they are rational.

Q: What is the significance of the discriminant in quadratic equations?

The discriminant, given by b^2 - 4ac in the quadratic formula, determines the nature of the roots of a quadratic equation. If the discriminant is positive, the equation has two distinct real roots; if zero, it has one real root (a repeated root); and if negative, it has two complex roots. Understanding the discriminant is crucial for Cambridge IGCSE students to predict the type of solutions without solving the equation.

Q: Why is "Sign error inside the quadratic formula" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students forget that -4ac flips sign when c is already negative. How to avoid it: Substitute with brackets: √(b^2) - 4(a)(c) and evaluate carefully. Source: 0580/42 — examiner reports, recurring

Q: Why is "Reporting only one root from a factorised quadratic" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students stop after solving one factor. How to avoid it: If (x - p)(x - q) = 0, BOTH x = p and x = q are solutions. Always state both.

Q: Why is "Picking the wrong region for a quadratic inequality" a common mistake in Quadratic Equations and Inequalities?

Why it happens: Students factor correctly but pick "x > 5 or x 0" is above = outside the roots.

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Quadratic Equations and Inequalities — Cambridge IGCSE 0580 Maths Extended (2026)

Solve $ax^{2} + bx + c = 0$ by factorising, completing the square, or the quadratic formula. Read off solutions, check the discriminant, and turn quadratic inequalities into a sign analysis.

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E2.9 — Solve quadratic equations by factorisation, by completing the square, and by use of the quadratic formula.
E2.10 — Solve quadratic inequalities; sketch quadratic curves to interpret solutions.

Solve by factorising

Set the equation to zero, factorise, then read off the roots from the brackets.

Method.

Move everything to one side so RHS = 0.
Factorise the quadratic.
Apply the zero-product rule: if $(x - p)(x - q) = 0$ then $x = p$ or $x = q$ .

Worked. Solve $x^{2} - 7x + 12 = 0$ .

Factorise: two numbers that add to $-7$ and multiply to $12$ → $-3, -4$ .
$(x - 3)(x - 4) = 0$ .
$x = 3$ or $x = 4$ .

Worked (non-monic). Solve $2x^{2} + x - 6 = 0$ .

AC method: $a c = -12$ . Add to $1$ , multiply to $-12$ : $4$ and $-3$ .
Split: $2x^{2} + 4x - 3x - 6 = 0$ .
Group: $2x(x + 2) - 3(x + 2) = (x + 2)(2x - 3) = 0$ .
$x = -2$ or $x = \dfrac{3}{2}$ .

Move everything to one side first.
Factorise (HCF, DOTS, monic, AC method).
Zero-product: each bracket gives a root.
Always state both roots.

The quadratic formula

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

When factorising fails (or is tedious), use: $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.$

Identify $a$ , $b$ , $c$ from $ax^{2} + bx + c = 0$ and substitute.

Worked. Solve $2x^{2} - 5x - 3 = 0$ .

$a = 2$ , $b = -5$ , $c = -3$ .
Discriminant: $b^{2} - 4ac = 25 - 4(2)(-3) = 25 + 24 = 49$ .
$\sqrt{49} = 7$ .
$x = \dfrac{5 \pm 7}{4}$ .
$x = 3$ or $x = -\dfrac{1}{2}$ .

Worked (irrational roots). Solve $x^{2} - 4x + 1 = 0$ .

$a = 1$ , $b = -4$ , $c = 1$ .
Discriminant: $16 - 4 = 12$ .
$x = \dfrac{4 \pm \sqrt{12}}{2} = \dfrac{4 \pm 2\sqrt{3}}{2} = 2 \pm \sqrt{3}$ .

When asked for exact answers, leave the surd form. When asked for decimals, round as instructed (usually 2 d.p.).

$a$ , $b$ , $c$ from $ax^{2} + bx + c = 0$ .
Compute the discriminant first.
$\sqrt{}$ gives two values via $\pm$ .
Simplify any surd factor (pull out perfect squares).

The discriminant — counting roots

$b^{2} - 4ac$ tells you how many real roots there are without solving.

The discriminant is $\Delta = b^{2} - 4ac$ .

$\Delta$	Real roots
$> 0$	TWO distinct real roots
$= 0$	ONE repeated real root
$< 0$	NO real roots (graph doesn't touch the $x$ -axis)

Worked. How many real roots does $x^{2} + 4x + 5 = 0$ have?

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

Worked. Find $k$ such that $x^{2} + kx + 9 = 0$ has exactly one real root.

One root → $\Delta = 0$ .
$k^{2} - 36 = 0$ → $k = \pm 6$ .

$\Delta = b^{2} - 4ac$ .
$\Delta > 0$ : 2 roots. $\Delta = 0$ : 1 (repeated). $\Delta < 0$ : 0.
Use to determine when an equation has solutions, or to find unknown coefficients.

Completing the square

Rewrite $x^{2} + bx + c$ as $(x + b/2)^{2} - (b/2)^{2} + c$ . Useful for finding turning points and proving identities.

Identity. $x^{2} + bx = (x + \tfrac{b}{2})^{2} - (\tfrac{b}{2})^{2}$ .

For a full quadratic with leading coefficient $1$ : $x^{2} + bx + c = (x + b/2)^{2} - (b/2)^{2} + c$ .

Worked. Complete the square for $x^{2} + 6x + 5$ .

$b/2 = 3$ . $(b/2)^{2} = 9$ .
$= (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ .

Use 1: solving. Set to zero. $(x + 3)^{2} - 4 = 0 \Rightarrow (x + 3)^{2} = 4 \Rightarrow x + 3 = \pm 2 \Rightarrow x = -1$ or $-5$ .

Use 2: turning point of $y = x^{2} + 6x + 5$ . From completed-square form, the vertex is at $(-3, -4)$ . Minimum value is $-4$ .

Non-monic. Factor out $a$ first: $2x^{2} + 8x + 5 = 2(x^{2} + 4x) + 5 = 2[(x + 2)^{2} - 4] + 5 = 2(x + 2)^{2} - 8 + 5 = 2(x + 2)^{2} - 3$ .

$x^{2} + bx + c = (x + b/2)^{2} - (b/2)^{2} + c$ .
Half $b$ , square it, subtract.
Vertex of $y = (x + h)^{2} + k$ is at $(-h, k)$ .
For $a \ne 1$ : factor out $a$ first.

Quadratic inequalities

Find the roots, sketch the parabola, then read which side(s) of the curve match.

Method.

Move everything to one side: $ax^{2} + bx + c \lessgtr 0$ .
Factorise (or use the quadratic formula) to find the roots — these are the $x$ -intercepts.
Sketch the parabola: opens up if $a > 0$ , opens down if $a < 0$ .
Read off the regions where the curve sits ABOVE or BELOW the $x$ -axis as required.

Worked. Solve $x^{2} - 5x + 6 < 0$ .

Factorise: $(x - 2)(x - 3)$ . Roots: $x = 2, 3$ .
Parabola opens UP ( $a = 1 > 0$ ).
Curve below $x$ -axis BETWEEN the roots.
Solution: $2 < x < 3$ .

Worked. Solve $x^{2} - 5x + 6 \ge 0$ .

Same parabola, same roots.
Curve above (or on) the $x$ -axis OUTSIDE the roots.
Solution: $x \le 2$ or $x \ge 3$ .

A handy sketch. A parabola that opens up is positive outside the roots, negative between them. Opens down: opposite.

Find the roots first.
Sketch direction (up or down).
Curve below 0: between the roots (for upward parabola).
Curve above 0: outside the roots (for upward parabola).
Inclusive ( $\le, \ge$ ): include the roots in the solution.

How it’s examined

Quadratic equations appear on every Paper 4 — typically 4-6 marks total across factorising, formula, completing the square. Paper 2 has 2-3 mark factorising-only items. Quadratic inequalities show up most years as 3-4 mark questions. Examiner reports flag sign errors in the quadratic formula and incorrect direction of inequality (between vs outside) as the recurring failures.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E2.9-2.10); 0580/22 May/Jun 2024 — Q15 (factorise + solve); 0580/42 Oct/Nov 2024 — Q15 (formula with surd answer); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

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Step-by-step worked examples — Quadratic Equations and Inequalities

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

1Solve a quadratic by factorising
Extended• factorising
Question
Solve $x^{2} - 5x - 6 = 0$ .
Step-by-step solution
1. Step 1
  Find two numbers whose product is $-6$ and sum is $-5$ : $-6$ and $1$ .
  $(x - 6)(x + 1) = 0$
2. Step 2
  Set each factor equal to zero.
  $x - 6 = 0\ \text{or}\ x + 1 = 0$
3. Step 3
  Solve each.
  $x = 6\ \text{or}\ x = -1$
Answer
$x = 6$ or $x = -1$
2Solve using the quadratic formula
Extended• Adapted from 0580/42 May/Jun 2024 Q12• quadratic formula
Question
Solve $2x^{2} + 5x - 4 = 0$ , giving your answer to $2$ decimal places.
Step-by-step solution
1. Step 1
  Identify $a = 2$ , $b = 5$ , $c = -4$ .
2. Step 2
  Substitute into $x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$ .
  $x = \dfrac{-5 \pm \sqrt{25 - 4(2)(-4)}}{2(2)}$
3. Step 3
  Simplify the discriminant.
  $= \dfrac{-5 \pm \sqrt{57}}{4}$
4. Step 4
  Compute both roots to $2$ d.p.
  $x \approx 0.64\ \text{or}\ x \approx -3.14$
Answer
$x \approx 0.64$ or $x \approx -3.14$
Examiner tip
Be careful with the negative inside $b^{2} - 4ac$ . With $c = -4$ , $-4ac$ becomes $+32$ , not $-32$ .
3Solve by completing the square
Extended• completing the square
Question
Solve $x^{2} + 6x - 5 = 0$ by completing the square. Give your answers in surd form.
Step-by-step solution
1. Step 1
  Move constant to the right.
  $x^{2} + 6x = 5$
2. Step 2
  Add $(\frac{6}{2})^{2} = 9$ to both sides.
  $x^{2} + 6x + 9 = 14$
3. Step 3
  Factor the perfect square.
  $(x + 3)^{2} = 14$
4. Step 4
  Take square roots.
  $x + 3 = \pm \sqrt{14}$
5. Step 5
  Solve.
  $x = -3 \pm \sqrt{14}$
Answer
$x = -3 + \sqrt{14}$ or $x = -3 - \sqrt{14}$
4Solve a quadratic inequality
Extended• Adapted from 0580/42 Oct/Nov 2024 Q14• inequality
Question
Solve $x^{2} - 3x - 10 < 0$ .
Step-by-step solution
1. Step 1
  Factorise.
  $(x - 5)(x + 2) < 0$
2. Step 2
  Roots are at $x = -2$ and $x = 5$ .
3. Step 3
  The parabola opens upwards (coefficient of $x^{2}$ is positive). It's BELOW zero between the roots.
  $-2 < x < 5$
Answer
$-2 < x < 5$
Examiner tip
Sketch the parabola: positive coefficient → smile shape → negative between the roots, positive outside. " $<$ " picks the negative region.
5Translate a word problem to a quadratic
Extended• word problem
Question
The product of two consecutive positive integers is $156$ . Find the integers.
Step-by-step solution
1. Step 1
  Let the first integer be $x$ . The next is $x + 1$ .
  $x(x + 1) = 156$
2. Step 2
  Expand and rearrange.
  $x^{2} + x - 156 = 0$
3. Step 3
  Factorise: numbers $13$ and $-12$ work ( $13 \times -12 = -156$ , $13 + (-12) = 1$ ).
  $(x - 12)(x + 13) = 0$
4. Step 4
  Reject $x = -13$ (must be positive). $x = 12$ , so the integers are $12$ and $13$ .
Answer
$12$ and $13$

Key Formulae — Quadratic Equations and Inequalities

The formulae you need to memorise for quadratic equations and inequalities on the Cambridge IGCSE 0580 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 $ax^{2} + bx + c = 0$ with $a \neq 0$
When to use
When the quadratic doesn't factorise nicely or the question asks for decimals/surds.
Discriminant
$\Delta = b^{2} - 4ac$
When to use
Determines the nature of the roots: $\Delta > 0$ → two real roots; $\Delta = 0$ → repeated root; $\Delta < 0$ → no real roots.
Completing the square
$x^{2} + bx = (x + \tfrac{b}{2})^{2} - (\tfrac{b}{2})^{2}$
When to use
When asked to solve, find min/max, or sketch a quadratic — completing the square reveals the vertex and exact-value roots.

Key Definitions and Keywords — Quadratic Equations and Inequalities

Definitions to memorise and the exact keywords mark schemes credit for quadratic equations and inequalities answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Quadratic equation
Examiner keyword
An equation of the form $ax^{2} + bx + c = 0$ with $a \neq 0$ .
Roots / solutions
Examiner keyword
The values of $x$ that satisfy the equation. A quadratic has at most two real roots.
Discriminant
Examiner keyword
$b^{2} - 4ac$ — its sign tells you how many real roots the quadratic has.
Perfect-square trinomial
An expression of the form $(x + k)^{2} = x^{2} + 2kx + k^{2}$ .

Common Mistakes and Misconceptions — Quadratic Equations and Inequalities

The traps other students keep falling into on quadratic equations and inequalities questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Sign error inside the quadratic formula
0580/42 — examiner reports, recurring
Why it happens
Students forget that $-4ac$ flips sign when $c$ is already negative.
How to avoid it
Substitute with brackets: $\sqrt{b^{2} - 4(a)(c)}$ and evaluate carefully.
✕Reporting only one root from a factorised quadratic
Why it happens
Students stop after solving one factor.
How to avoid it
If $(x - p)(x - q) = 0$ , BOTH $x = p$ and $x = q$ are solutions. Always state both.
✕Picking the wrong region for a quadratic inequality
Why it happens
Students factor correctly but pick " $x > 5$ or $x < -2$ " when the question asks for $< 0$ .
How to avoid it
Sketch the parabola. " $< 0$ " is below the x-axis = between roots; " $> 0$ " is above = outside the roots.
✕Adding $(\frac{b}{2})$ instead of $(\frac{b}{2})^{2}$
Why it happens
Forgetting the squaring step in completing the square.
How to avoid it
Always square the half-coefficient: $b/2$ → $(b/2)^{2}$ .

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Short walkthrough of the concepts students most often get stuck on.

Quadratic Equations and Inequalities — frequently asked questions

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

\Delta

Real roots

> 0

TWO distinct real roots

= 0

ONE repeated real root

< 0

NO real roots (graph doesn't touch the

x

-axis)

Identity. $x^{2} + bx = (x + \tfrac{b}{2})^{2} - (\tfrac{b}{2})^{2}$ .

For a full quadratic with leading coefficient $1$ : $x^{2} + bx + c = (x + b/2)^{2} - (b/2)^{2} + c$ .

Worked. Complete the square for $x^{2} + 6x + 5$ .

$b/2 = 3$ . $(b/2)^{2} = 9$ .
$= (x + 3)^{2} - 9 + 5 = (x + 3)^{2} - 4$ .

Use 1: solving. Set to zero. $(x + 3)^{2} - 4 = 0 \Rightarrow (x + 3)^{2} = 4 \Rightarrow x + 3 = \pm 2 \Rightarrow x = -1$ or $-5$ .

Use 2: turning point of $y = x^{2} + 6x + 5$ . From completed-square form, the vertex is at $(-3, -4)$ . Minimum value is $-4$ .

Non-monic. Factor out $a$ first: $2x^{2} + 8x + 5 = 2(x^{2} + 4x) + 5 = 2[(x + 2)^{2} - 4] + 5 = 2(x + 2)^{2} - 8 + 5 = 2(x + 2)^{2} - 3$ .

Quadratic Equations and Inequalities

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Solve a quadratic by factorising

2Solve using the quadratic formula

3Solve by completing the square

4Solve a quadratic inequality

5Translate a word problem to a quadratic

Quadratic formula

Discriminant

Completing the square

Quadratic equation

Roots / solutions

Discriminant

Perfect-square trinomial

✕Sign error inside the quadratic formula

✕Reporting only one root from a factorised quadratic

✕Picking the wrong region for a quadratic inequality

✕Adding (b2)(\frac{b}{2})(2b​) instead of (b2)2(\frac{b}{2})^{2}(2b​)2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Quadratic Equations and Inequalities — frequently asked questions

Quadratic Equations and Inequalities

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Solve a quadratic by factorising

2Solve using the quadratic formula

3Solve by completing the square

4Solve a quadratic inequality

5Translate a word problem to a quadratic

Quadratic formula

Discriminant

Completing the square

Quadratic equation

Roots / solutions

Discriminant

Perfect-square trinomial

✕Sign error inside the quadratic formula

✕Reporting only one root from a factorised quadratic

✕Picking the wrong region for a quadratic inequality

✕Adding (b2)(\frac{b}{2})(2b​) instead of (b2)2(\frac{b}{2})^{2}(2b​)2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Quadratic Equations and Inequalities — frequently asked questions

✕Adding $(\frac{b}{2})$ instead of $(\frac{b}{2})^{2}$

✕Adding $(\frac{b}{2})$ instead of $(\frac{b}{2})^{2}$