IGCSE Additional Mathematics 0606 · Topic 3

IGCSE Additional Mathematics: Equations, Inequalities and Graphs Practice Questions

A quadratic inequality is solved by finding the critical values and then deciding, from a sketch, whether the solution lies between them or outside them. A modulus inequality of the form less than a value produces a single interval.

Cambridge IGCSE Additional Mathematics (0606) · Topic 3: Equations, Inequalities and Graphs

Topic 3 of Cambridge IGCSE Additional Mathematics 0606 punishes algebraic manipulation without a sketch. Deciding between an inside interval and an outside pair is the whole question, and a rough parabola settles it in seconds. The questions below make that step explicit.

What you need to know for Equations, Inequalities and Graphs

IGCSE Additional Mathematics Equations, Inequalities and Graphs questions and answers

4 exam-style questions written to the 0606 syllabus. Try each one on paper first, then open the worked answer to check your method against the marks.

Question 1[4 marks]

Solve the inequality the modulus of (x minus 3) is less than 5.

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Answer: minus 2 is less than x, which is less than 8
  1. The modulus of an expression being less than 5 means the expression lies between minus 5 and 5.
  2. So minus 5 is less than x minus 3, and x minus 3 is less than 5.
  3. Add 3 throughout: minus 2 is less than x, and x is less than 8.
  4. The solution is a single interval, from minus 2 to 8, excluding both endpoints.
How the marks are awarded. 1 mark for setting up the double inequality. 1 mark for adding 3 throughout. 1 mark for the lower bound. 1 mark for the upper bound written as a single interval.
Where students lose the mark. Writing the answer as two separate regions. The less than form always gives one interval. Only the greater than form gives two.
Question 2[4 marks]

Solve the inequality x2 minus 5x + 6 is greater than 0.

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Answer: x is less than 2, or x is greater than 3
  1. Find the critical values by solving x2 minus 5x + 6 = 0.
  2. Factorising gives (x minus 2)(x minus 3) = 0, so x = 2 or x = 3.
  3. The coefficient of x2 is positive, so the parabola opens upwards and lies above the x-axis outside the roots.
  4. The solution is therefore x is less than 2, or x is greater than 3.
How the marks are awarded. 1 mark for factorising. 1 mark for both critical values. 1 mark for identifying that the curve is above the axis outside the roots. 1 mark for the solution written as two regions.
Where students lose the mark. Giving 2 less than x less than 3. That is where the expression is negative, which answers the opposite inequality.
Question 3[4 marks]

Solve the inequality x2 minus 2x minus 8 is less than or equal to 0.

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Answer: minus 2 is less than or equal to x, which is less than or equal to 4
  1. Factorise: x2 minus 2x minus 8 = (x + 2)(x minus 4).
  2. The critical values are x = minus 2 and x = 4.
  3. The parabola opens upwards, so it lies below the x-axis between the roots.
  4. Since the inequality includes equality, the endpoints are included: minus 2 is less than or equal to x, which is less than or equal to 4.
How the marks are awarded. 1 mark for factorising. 1 mark for both critical values. 1 mark for recognising the solution lies between the roots. 1 mark for including both endpoints.
Where students lose the mark. Using strict inequalities. The original includes equality, so the roots themselves are solutions and must be included.
Question 4[3 marks]

Describe the graph of y equals the modulus of (2x minus 4), stating the coordinates of its vertex.

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Answer: A V shape with vertex at (2, 0)
  1. The graph of y = 2x minus 4 is a straight line crossing the x-axis where 2x minus 4 = 0, so at x = 2.
  2. Taking the modulus reflects every part of the line that lies below the x-axis upwards.
  3. The result is a V shape with its vertex at the point (2, 0), where the original line met the axis.
  4. The right branch has gradient 2 and the left branch has gradient minus 2.
How the marks are awarded. 1 mark for identifying a V shape. 1 mark for the vertex at (2, 0). 1 mark for describing the reflection of the negative part above the axis, or for stating the two gradients.
Where students lose the mark. Drawing a curve. The modulus of a linear function is two straight line segments meeting at a sharp point, not a smooth curve.

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Equations, Inequalities and Graphs FAQs

How do I solve a quadratic inequality?

Rearrange so one side is zero, factorise to find the critical values, then sketch the parabola. For an upward parabola the expression is negative between the roots and positive outside them, which tells you which region satisfies the inequality.

What is the difference between a modulus less than and greater than inequality?

The modulus of an expression being less than k means the expression lies between minus k and k, giving one interval. The modulus being greater than k means the expression is above k or below minus k, giving two separate regions.

Why can't I divide an inequality by x?

The sign of x is unknown, and dividing by a negative reverses the inequality. Dividing by a variable can therefore produce a wrong answer or lose solutions. Rearrange to compare with zero and factorise instead.

What does the graph of a modulus function look like?

Any part of the original graph lying below the x-axis is reflected upwards. For a linear function this gives a V shape with a sharp vertex where the original crossed the axis, and the two branches have gradients equal in size but opposite in sign.

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Written to the published Cambridge IGCSE Additional Mathematics (0606) syllabus. Check your school entry code and syllabus year, because Core and Extended candidates are assessed on different content. Last reviewed 2026-08-12.