A Complete Guide to Quadratic Inequalities in IGCSE Maths

What Are Quadratic Inequalities?

A quadratic inequality involves a quadratic expression and an inequality sign. Instead of finding exact solutions as you would with a quadratic equation, you find a range of values that satisfy the inequality. Examples include x² − 5x + 6 > 0, 2x² + 3x − 2 ≤ 0, and x² < 9.

These appear on the Extended tier of IGCSE Maths and require a solid understanding of both quadratic factorisation and graphical interpretation.

The Graphical Method: Your Best Approach

While algebraic methods exist, the graphical approach is the most reliable and least error-prone method for solving quadratic inequalities at IGCSE level. The steps are:

• Step 1: Rearrange so one side is zero (e.g., x² − 5x + 6 > 0)
• Step 2: Solve the corresponding equation (x² − 5x + 6 = 0) to find the critical values
• Step 3: Sketch the quadratic graph, marking the roots on the x-axis
• Step 4: Determine which part of the graph satisfies the inequality

For a positive quadratic (U-shaped parabola), the curve is above the x-axis outside the roots and below the x-axis between the roots. For a negative quadratic (inverted U-shape), the opposite is true.

Worked Example 1: Greater Than Zero

Solve x² − 5x + 6 > 0.

Step 1: Already in the right form.

Step 2: Factorise: (x − 2)(x − 3) = 0, so x = 2 and x = 3.

Step 3: Sketch a U-shaped parabola crossing the x-axis at 2 and 3.

Step 4: We need the curve to be above the x-axis (> 0). Looking at the sketch, the curve is above the x-axis when x < 2 or x > 3.

Answer: x < 2 or x > 3.

Worked Example 2: Less Than or Equal to Zero

Solve 2x² + 3x − 2 ≤ 0.

Step 1: Already in the right form.

Step 2: Factorise: (2x − 1)(x + 2) = 0, so x = 1/2 and x = −2.

Step 3: Sketch a U-shaped parabola crossing the x-axis at −2 and 1/2.

Step 4: We need the curve to be on or below the x-axis (≤ 0). The curve is at or below zero between the roots.

Answer: −2 ≤ x ≤ 1/2.

Worked Example 3: Negative Leading Coefficient

Solve −x² + 4x − 3 > 0.

Step 1: You could multiply through by −1 (remembering to flip the inequality): x² − 4x + 3 < 0. Or work with the original.

Let us use the original. Step 2: −x² + 4x − 3 = 0, multiply by −1: x² − 4x + 3 = 0, giving (x − 1)(x − 3) = 0, so x = 1 and x = 3.

Step 3: The original has a negative x² coefficient, so the parabola is inverted (∩-shaped), crossing the x-axis at 1 and 3.

Step 4: For > 0, we need the curve above the x-axis. For an inverted parabola, this is between the roots.

Answer: 1 < x < 3.

Writing the Solution Correctly

Cambridge expects solutions in one of these formats:

• Two separate inequalities joined by “or”: x < 2 or x > 5
• A compound inequality: 2 ≤ x ≤ 5
• Set notation: {x : x < 2} ∪ {x : x > 5}

Important: for “greater than zero” with a positive quadratic, the solution is always two separate regions (x < smaller root or x > larger root). Never write this as “smaller root < x < larger root” because that represents the values between the roots, which is the opposite of what you want.

For “less than zero” with a positive quadratic, the solution is always one region between the roots: smaller root < x < larger root.

Special Cases

Some quadratic inequalities have special solutions:

• x² ≥ 0: True for all real numbers (a perfect square is always non-negative)
• x² < 0: No solutions (a square cannot be negative)
• x² > 0: True for all x except x = 0
• (x − 3)² ≤ 0: Only true when x = 3 (equal to zero, never negative)
• (x − 3)² < 0: No solutions

These cases do not require graphing but do require clear thinking about what squares can and cannot equal.

Linking to the Discriminant

The discriminant b² − 4ac tells you about the roots of the quadratic, which in turn affects the inequality solution:

• If discriminant > 0: two distinct roots, standard approach applies
• If discriminant = 0: one repeated root, the graph touches the x-axis at one point
• If discriminant < 0: no real roots, the parabola does not cross the x-axis. A positive quadratic is always above the axis (always > 0), and a negative quadratic is always below (always < 0).

Quadratic Inequalities in Context

In exam questions, quadratic inequalities sometimes arise from context problems. For example, “a ball is thrown upward and its height h metres after t seconds is given by h = 20t − 5t². Find the times for which the ball is above 15 metres.”

This becomes 20t − 5t² > 15, or −5t² + 20t − 15 > 0, which simplifies to t² − 4t + 3 < 0 (after dividing by −5 and flipping the inequality).

Factorising: (t − 1)(t − 3) < 0, giving 1 < t < 3.

The ball is above 15 metres between 1 and 3 seconds.

Common Mistakes

• Flipping the inequality when dividing by a negative. This is crucial. If you multiply or divide both sides by a negative number, the inequality sign reverses.
• Writing the wrong type of answer. For x² − 4 > 0, the answer is x < −2 or x > 2, not −2 < x < 2.
• Not sketching the graph. Even a rough sketch prevents errors by making the solution visual.
• Forgetting strict vs non-strict inequalities. For ≤ and ≥, use ≤ and ≥ in the answer. For < and >, use strict inequalities.
• Errors in factorising. If you cannot factorise, use the quadratic formula to find the critical values.

Practice Recommendations

Start with simple inequalities like x² < 16 and x² ≥ 25. Progress to factorisable quadratics, then to those requiring the quadratic formula. Always sketch the graph and clearly indicate the solution regions.

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