Solve quadratic equations and inequalities
Choose a valid quadratic-solving method, then use roots and the graph's sign to state an inequality solution set.
Cambridge IGCSE Additional Mathematics 0606, 2028–2030 syllabus, Version 1. Paper 1 is non-calculator; Paper 2 requires a scientific calculator. These original lessons provide self-checked topic practice, not a Cambridge grade or a timed paper. Full-paper assessment and human specialist review are still outstanding.
Before you begin
- Cambridge assumes knowledge of IGCSE Mathematics or an equivalent course. Check the domain before using an inverse or logarithm, and retain exact values where the question asks for them.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
When are endpoints included in an inequality solution?
What you will learn
- Solve real-root quadratic equations by factorising, completing the square or using the formula
- Find solution sets for quadratic inequalities
- Write interval endpoints with the correct inclusion
Use a method that preserves every root
Factorising is efficient when integer factors are visible. Completing the square is useful for turning-point information. The quadratic formula works for any quadratic and is supplied in the exam formula list.
Check a root by substitution when possible. If an equation has two roots, keep both unless the domain removes one.
Inequalities depend on the sign between and outside the roots
For a positive-leading quadratic with two simple roots, the graph opens upward: it is non-positive between the roots and non-negative outside. Reverse those sign regions for a negative-leading quadratic.
Use a sketch or sign table, then include each endpoint for ≤ or ≥ and exclude it for < or >.
Pause and explain
Where is (x − 1)(x − 4) negative?
Worked example
Solve x² − 5x + 6 ≤ 0.
Show the worked solution
- Factor: (x − 2)(x − 3) ≤ 0.
- The roots are 2 and 3, and the leading coefficient is positive.
- The upward-opening graph is at or below zero between the roots, including them.
Answer 2 ≤ x ≤ 3.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Solve x² − 7x + 12 = 0.
Give me a hint
Find two numbers whose product is 12 and sum is 7.
Compare my reasoning
- Factor x² − 7x + 12 as (x − 3)(x − 4).
- Set each factor equal to zero.
- Check both values in the original equation.
x = 3 or x = 4.
Look for these in your work
- Factorisation correct
- Both roots found
- Substitution check considered
Solve 2x² − 3x − 2 > 0.
Give me a hint
Factor first, then use the upward-opening sign pattern.
Compare my reasoning
- Factor as (2x + 1)(x − 2) > 0.
- The roots are −1/2 and 2.
- Since the leading coefficient is positive, the expression is positive outside the roots.
x < −1/2 or x > 2.
Look for these in your work
- Roots correct
- Sign region justified
- Strict endpoints excluded
A profit model is P(x) = −x² + 8x − 12, where x is a non-negative production adjustment. For which x is the model profitable?
Give me a hint
Profit is positive between the roots of the downward-opening quadratic.
Compare my reasoning
- Factor P(x) = −(x − 2)(x − 6).
- The roots are x = 2 and x = 6; the parabola opens downward.
- Profit is positive between the roots; zero-profit endpoints are excluded.
2 < x < 6.
Look for these in your work
- Model factored
- Correct sign interval
- Context uses strict profit
Common mistakes
- Dropping one root
- Reversing the interval for a positive-leading quadratic
- Using open endpoints for ≤ or ≥
What can you explain now?
For x² − 9 ≥ 0, where is the expression non-negative?
Compare with the explanation
x ≤ −3 or x ≥ 3
The parabola opens upward and is above the axis outside its roots.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Explain how roots and the leading coefficient determine a quadratic inequality's solution intervals.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.