# Simple linear and quadratic inequalities

> Edexcel International GCSE Further Pure Mathematics · 4PM1
> Source: https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s3-simple-linear-and-quadratic-inequalities/

This guide teaches you to solve linear and quadratic inequalities as required for Edexcel IGCSE Further Pure Math (4PM1), with step-by-step worked examples and exam-focused tips.

**Prerequisites:** [Rearranging linear equations](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s1-rearranging-equations/); [Factorising quadratic expressions](https://www.owlsprep.com/study/edexcel-igcse-further-pure-math-s2-quadratic-factorisation/)

## Learning objectives

- Solve linear inequalities of the form ax + b > cx + d with correct inequality sign preservation
- Rearrange quadratic inequalities to standard form and find critical values via factorisation
- Use sign diagrams or parabola sketches to select the correct solution region for quadratic inequalities
- Present solution sets using standard inequality notation for exams

## Solving Simple Linear Inequalities

**Linear inequality** — An inequality that can be written in the form $ax + b \bowtie cx + d$, where $\bowtie$ is $<, >, \leq$ or $\geq$, and the highest power of $x$ is 1.

To solve linear inequalities, follow the same steps as rearranging linear equations, with one key rule: **reverse the inequality sign if you multiply or divide both sides by a negative number.**

**Worked example:** Solve the inequality $3x + 7 \geq 19 - 2x$, giving your answer in the form $x \bowtie k$.

1. Add $2x$ to both sides to collect $x$ terms on the left: $5x + 7 \geq 19$
2. Subtract 7 from both sides: $5x \geq 12$
3. Divide by 5 (positive, no sign flip required): $x \geq \frac{12}{5} = 2.4$

**Worked example:** Solve the inequality $11 - 4x < 3$.

1. Subtract 11 from both sides: $-4x < -8$
2. Divide by -4 (negative, flip sign): $x > 2$

> **Exam tip:** Always double-check if you multiplied/divided by a negative number at the final step — forgetting to flip the inequality sign is the most common linear inequality mistake.

*Calculator:* allowed

## Rearranging Quadratic Inequalities & Finding Critical Values

Quadratic inequalities are rearranged first to get all terms on one side, with 0 on the other, and a positive leading coefficient to simplify sign checking. Critical values are the roots of the corresponding quadratic equation.

**Standard form of quadratic inequality** — A quadratic inequality written as $ax^2 + bx + c \bowtie 0$, where $a > 0$, and $\bowtie$ is $<, >, \leq$ or $\geq$.

**Worked example:** Rearrange $2x^2 + 3x - 5 < x^2 + 2x + 1$ into standard form and find its critical values.

1. Subtract the right-hand side from both sides: $x^2 + x - 6 < 0$
2. Factorise the quadratic expression: $(x + 3)(x - 2) = 0$
3. Solve for $x$ to get critical values: $x = -3$ and $x = 2$

> **tip**
>
> If the leading coefficient is negative after rearrangement, multiply the entire inequality by -1 and flip the inequality sign immediately to make $a$ positive — this simplifies finding the solution region later.

*Calculator:* allowed

## Selecting Solution Regions for Quadratic Inequalities

Once you have critical values, you can use either a sign diagram or a sketch of the quadratic parabola to find which regions satisfy the inequality. For a positive leading coefficient, the parabola opens upwards, so it is below the x-axis between the roots, and above the x-axis outside the roots.

**Comparing methods**

Two valid exam-accepted methods for finding solution regions:

- **Sign Diagram** — Mark critical values on a number line, test a value in each interval to see if the quadratic is positive or negative, then select intervals matching the inequality sign.
  - Pros: Fast for simple quadratics; No graph drawing required
  - Cons: Requires manual testing of values; Easy to mix up signs if you skip steps

- **Parabola Sketch** — Sketch the U-shaped parabola (positive leading coefficient) crossing the x-axis at critical values, then read off regions where the parabola is above or below the x-axis as required.
  - Pros: Visual, lower risk of sign error; Works for all quadratic inequality types
  - Cons: Requires basic sketching skills; Takes slightly longer

**Worked example:** Solve $x^2 + x - 6 < 0$ using the sign diagram method.

1. Critical values are $x=-3$ and $x=2$, splitting the number line into 3 intervals: $x < -3$, $-3 < x < 2$, $x > 2$
2. Test $x=-4$: $(-4+3)(-4-2) = 6 > 0$ (does not match < 0)
3. Test $x=0$: $(0+3)(0-2) = -6 < 0$ (matches inequality)
4. Test $x=3$: $(3+3)(3-2) = 6 > 0$ (does not match < 0)
5. Final solution: $-3 < x < 2$

> **Exam tip:** Always check if the inequality is strict ($<, >$) or includes equality ($\leq, \geq$) to know if you should include critical values in your solution set.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Forgetting to flip the inequality sign when dividing/multiplying by a negative number
  - Why it fails: Inequalities reverse when multiplied by a negative number because the order of values on the number line flips (e.g. $3 < 5$ becomes $-3 > -5$)
  - Correct: Immediately reverse the inequality direction any time you multiply or divide both sides by a negative value, and double-check this step before submitting your answer.
- **Wrong:** Picking the wrong region for quadratic inequalities (e.g. outside roots for $<0$ with positive leading coefficient)
  - Why it fails: Confusion over parabola shape: positive leading coefficient gives a U-shape, so it is below the x-axis between roots, not outside
  - Correct: Memorise that for $a>0$: $<$ gives between roots, $>$ gives outside, or draw a quick parabola sketch to confirm the region.
- **Wrong:** Forgetting to rearrange all terms to one side before finding critical values for quadratic inequalities
  - Why it fails: If terms are on both sides, you cannot correctly factorise to find the roots of the quadratic expression, leading to incorrect critical values
  - Correct: Always rearrange the inequality to standard form $ax^2 + bx + c \bowtie 0$ with $a>0$ before factorising or finding critical values.
- **Wrong:** Including critical values for strict inequalities ($<, >$)
  - Why it fails: Strict inequalities mean the expression cannot equal zero, so the roots (where the expression equals zero) are not part of the solution set
  - Correct: Use $<$ or $>$ for strict inequalities (open intervals) and $\leq$ or $\geq$ when equality is allowed (closed intervals).

## Cheatsheet

| Inequality Type | Standard Form ($a>0$) | Solution Region |
| --- | --- | --- |
| Linear | $ax + b > cx + d$ | Rearrange, flip sign if multiplying/dividing by negative |
| Quadratic $< / \leq$ | $ax^2 + bx + c < 0$ | Single interval: $x_1 < x < x_2$ (or $\leq$) |
| Quadratic $> / \geq$ | $ax^2 + bx + c > 0$ | Two intervals: $x < x_1$ or $x > x_2$ (or $\leq/\geq$) |

## What's next

Now that you can solve simple linear and quadratic inequalities, you are ready to move on to more advanced inequality topics in Edexcel IGCSE Further Pure Math. Next, you will learn to graph linear inequalities in two variables for region and linear programming problems, and apply inequality solving to intersection questions in coordinate geometry and roots of quadratics problems. Mastering this foundational topic will also support your work on function domains and optimisation problems in later units. Make sure to practice past paper questions to reinforce your understanding and avoid common mistakes under exam conditions.

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