The problem: five question types, one systematic toolkit
Equations, inequalities and graphs questions on the IGCSE Additional Mathematics paper test whether you can move fluently between algebraic manipulation and graphical reasoning. The challenge is not that any single technique is difficult in isolation. It's that the exam expects you to recognise which technique fits which problem shape, execute it cleanly, and interpret the result correctly.
There are five core problem types in this topic area: cubic inequalities solved graphically, modulus equations with one or two modulus signs, modulus inequalities, substitution to reduce a higher-degree equation to a quadratic, and sketching cubic polynomials along with their modulus transformations. Each one has a predictable structure, and each one rewards a disciplined, step-by-step approach over guesswork.
Cubic inequalities: read the graph, not the algebra
A cubic inequality asks you to find the values of x for which a cubic expression is greater than, less than, or equal to some constant. The standard exam form looks like this: given f(x) = (x - a)(x - b)(x - c), solve f(x) > d.
The method relies entirely on the graph. Here is the systematic process.
- Sketch the cubic. Since f(x) is given as a product of three linear factors, the x-intercepts are x = a, x = b, and x = c. Plot these on the x-axis. The y-intercept is f(0) = (-a)(-b)(-c). If the coefficient of x^3 is positive, the curve runs from bottom-left to top-right.
- Draw the horizontal line y = d. This is the threshold you're comparing against.
- Find the intersection points. Set (x - a)(x - b)(x - c) = d and solve the resulting equation. You may need to expand and use the factor theorem, or the question may give you the roots.
- Read the inequality from the graph. For f(x) > d, identify the x-intervals where the cubic curve lies above the line y = d. For f(x) < d, look where the curve sits below the line.
Worked example 1
Solve (x - 1)(x - 3)(x + 2) > 0.
Step 1: The x-intercepts are x = 1, x = 3, and x = -2. The coefficient of x^3 is positive (expanding gives x^3 - ...), so the curve rises from bottom-left to top-right.
Step 2: Here d = 0, so the threshold line is the x-axis itself.
Step 3: The intersections with the x-axis are already known: x = -2, x = 1, x = 3.
Step 4: Reading from the sketch, the cubic is above the x-axis when -2 < x < 1 and when x > 3.
Answer: -2 < x < 1 or x > 3.
Modulus equations: a decision tree for three forms
Modulus equations appear in three progressively complex forms on the IGCSE Additional Mathematics paper. Each form has a specific solving strategy, and choosing the wrong one wastes time or produces invalid solutions.
| Form | Strategy | Validity check needed? |
|---|---|---|
| |ax + b| = c | Split into ax + b = c and ax + b = -c | Only check c >= 0 (if c < 0, no solution) |
| |ax + b| = cx + d | Split into ax + b = cx + d and ax + b = -(cx + d) | Yes: substitute each solution back into original, verify both sides match |
| |ax + b| = |cx + d| | Square both sides: (ax + b)^2 = (cx + d)^2, or split into two cases | Squaring always valid; if splitting, check both solutions |
Worked example 2
Solve |2x - 5| = x + 1.
This is Form 2 (|ax + b| = cx + d), so we split into two cases and must check validity.
Case 1: 2x - 5 = x + 1. Solving: 2x - x = 1 + 5, so x = 6. Check: |2(6) - 5| = |7| = 7, and 6 + 1 = 7. Valid.
Case 2: 2x - 5 = -(x + 1). Expanding: 2x - 5 = -x - 1. Solving: 3x = 4, so x = 4/3. Check: |2(4/3) - 5| = |8/3 - 15/3| = |-7/3| = 7/3, and 4/3 + 1 = 7/3. Valid.
Answer: x = 6 or x = 4/3.
Worked example 3
Solve |3x + 1| = |x - 5|.
This is Form 3. Squaring both sides: (3x + 1)^2 = (x - 5)^2.
Expanding: 9x^2 + 6x + 1 = x^2 - 10x + 25.
Rearranging: 8x^2 + 16x - 24 = 0. Dividing by 8: x^2 + 2x - 3 = 0.
Factorising: (x + 3)(x - 1) = 0, giving x = -3 or x = 1.
Both solutions are valid since squaring preserves all genuine solutions for this form.
Modulus inequalities: combine algebra with graphical reasoning
Modulus inequalities extend the equation techniques but require you to identify regions rather than individual points. Two approaches work reliably.
Approach A (algebraic): Solve the corresponding equation to find boundary points, then test intervals between those points to determine which satisfy the inequality.
Approach B (graphical): Sketch both sides of the inequality on the same axes. The solution is the set of x-values where the correct graph is higher (for >) or lower (for <).
Worked example 4
Solve |2x - 3| < 5.
Step 1: Solve the equation |2x - 3| = 5. This gives 2x - 3 = 5 (so x = 4) and 2x - 3 = -5 (so x = -1).
Step 2: The boundary points are x = -1 and x = 4. For a "less than" inequality with a single modulus against a positive constant, the solution lies between the boundary points.
Answer: -1 < x < 4.
Worked example 5
Solve 2|x + 1| >= |3x - 2|.
Step 1: Square both sides (both sides are non-negative, so this is valid). 4(x + 1)^2 >= (3x - 2)^2.
Step 2: Expand. 4(x^2 + 2x + 1) >= 9x^2 - 12x + 4. So 4x^2 + 8x + 4 >= 9x^2 - 12x + 4.
Step 3: Rearrange. 0 >= 5x^2 - 20x. So 0 >= 5x(x - 4), meaning 5x(x - 4) <= 0.
Step 4: Solve the quadratic inequality. The roots are x = 0 and x = 4. Since the coefficient of x^2 is positive, the parabola opens upward, and 5x(x - 4) <= 0 when 0 <= x <= 4.
Answer: 0 <= x <= 4.
- Key principle: squaring both sides of a modulus inequality is valid when both sides are guaranteed non-negative, which is always the case when both sides are modulus expressions or products of a positive constant and a modulus
- After squaring: the problem reduces to a standard polynomial inequality that you solve by finding roots and testing intervals
- Always express the answer in interval notation or as a compound inequality rather than listing individual values
Substitution to form a quadratic
Some equations look intimidating at first glance but collapse into a standard quadratic once you spot the right substitution. The technique works whenever the equation contains a term and its square.
The recognition pattern is straightforward: if an equation contains both g(x) and [g(x)]^2 for some expression g(x), let u = g(x) to produce a quadratic in u.
Worked example 6
Solve x^4 - 13x^2 + 36 = 0.
Step 1: Notice that x^4 = (x^2)^2. Let u = x^2.
Step 2: The equation becomes u^2 - 13u + 36 = 0.
Step 3: Factorise: (u - 4)(u - 9) = 0, so u = 4 or u = 9.
Step 4: Substitute back. x^2 = 4 gives x = 2 or x = -2. x^2 = 9 gives x = 3 or x = -3.
Answer: x = -3, -2, 2, or 3.
Worked example 7
Solve 2^(2x) - 5(2^x) + 4 = 0.
Step 1: Notice that 2^(2x) = (2^x)^2. Let u = 2^x.
Step 2: The equation becomes u^2 - 5u + 4 = 0.
Step 3: Factorise: (u - 1)(u - 4) = 0, so u = 1 or u = 4.
Step 4: Substitute back. 2^x = 1 gives x = 0. 2^x = 4 gives x = 2.
Answer: x = 0 or x = 2.
Sketching cubic polynomials and their modulus
When a cubic is given in factored form as f(x) = (x - a)(x - b)(x - c), the sketch requires four pieces of information extracted directly from the expression.
| Feature | How to find it | Example: f(x) = (x + 1)(x - 2)(x - 4) |
|---|---|---|
| x-intercepts | Set each factor to zero | x = -1, x = 2, x = 4 |
| y-intercept | Evaluate f(0) | (0 + 1)(0 - 2)(0 - 4) = 1 x (-2) x (-4) = 8 |
| End behaviour | Check the sign of the leading coefficient | Positive x^3 term: curve goes from bottom-left to top-right |
| General shape | Positive leading coefficient: N-shape (down-up-down-up through the three roots) | Falls through x = -1, rises through x = 2, falls then rises through x = 4 |
The shape between the roots follows a predictable pattern. Between the first and second root, the curve dips below (or above) the x-axis. Between the second and third root, it returns. Mark the turning points approximately and connect with a smooth curve.
Sketching |f(x)|
To sketch y = |f(x)| from y = f(x), apply one rule: any portion of the curve that lies below the x-axis gets reflected above it. The portions already above the x-axis remain unchanged. Graphically, you fold the negative parts upward.
For the example f(x) = (x + 1)(x - 2)(x - 4):
- Sketch the original cubic as described above
- Identify the sections below the x-axis (between x = -1 and x = 2 in this case, since the curve dips below there... but actually for a positive cubic, it's negative for x < -1, positive for -1 < x < 2... you need to check signs carefully)
- Reflect those sections in the x-axis
- The result touches the x-axis at the same roots but never goes below it
Let's verify the signs. For x < -1, say x = -2: (-2 + 1)(-2 - 2)(-2 - 4) = (-1)(-4)(-6) = -24. Negative. For -1 < x < 2, say x = 0: (1)(-2)(-4) = 8. Positive. For 2 < x < 4, say x = 3: (4)(1)(-1) = -4. Negative. For x > 4, say x = 5: (6)(3)(1) = 18. Positive.
So the portions below the x-axis are x < -1 and 2 < x < 4. These get reflected upward in the modulus graph.
Self-check questions
Test your command of each technique with these problems. Work through each one fully before checking your reasoning.
- Solve (x + 1)(x - 2)(x - 5) < 0. Hint: sketch the cubic first, then read off where it lies below the x-axis.
- Solve |4x - 3| = 2x + 7. Hint: this is Form 2. Don't forget the validity check.
- Solve |x + 2| = |2x - 1|. Hint: squaring both sides avoids case-checking entirely.
- Solve |3x - 1| > 5. Hint: the solution set consists of two separate intervals, not one.
- Solve 3^(2x) - 10(3^x) + 9 = 0. Hint: let u = 3^x.
- Sketch y = (x + 2)(x)(x - 3) and y = |(x + 2)(x)(x - 3)| on the same axes, clearly labelling intercepts.
For each question, the method selection follows the decision patterns outlined above. Identify the form, apply the matching strategy, and verify your answer either by substitution or by checking the graph. Precision in these five techniques covers the full range of what the IGCSE Additional Mathematics paper asks from this topic.
A methodical guide to solving cubic inequalities, modulus equations, modulus inequalities, substitution-based equations, and sketching cubic graphs for IGCSE Additional Mathematics 0606, with step-by-step worked examples, a decision-tree approach to choosing methods, and self-check questions to test exam readiness.
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