# Inequalities (Edexcel IAL FP2 Further Mathematics)

> Edexcel International A-Level Further Mathematics · Edexcel IAL FMATHS 2018 Spec (Issue 3)
> Source: https://www.owlsprep.com/study/edexcel-ial-further-math-fp2-inequalities/

This guide covers FP2 inequality solving for Edexcel IAL Further Maths, including rational and modulus inequalities, using approved sign-safe methods and exam-valid justification techniques.

**Prerequisites:** [P1-P4 algebra manipulation and modulus functions](https://www.owlsprep.com/study/edexcel-ial-maths-p4-algebra/); [FP1 algebraic manipulation](https://www.owlsprep.com/study/edexcel-ial-further-math-fp1-algebra/)

## Learning objectives

- Solve rational inequalities using sign analysis or squaring of denominators (no multiplication by unknown sign expressions)
- Solve modulus inequalities via case analysis, squaring, or graphical justification
- Correctly exclude undefined values (denominator zeros) from final solution sets
- Justify solutions using sketch graphs where required for exam marks

## Solving Rational Inequalities: Sign Analysis Method

**Rational inequality** — An inequality that contains at least one fraction where the numerator and denominator are polynomials in x, with fractions on one or both sides of the inequality sign.

*Example:* $\frac{1}{x-a} > \frac{x}{x-b}$

The only safe methods for solving rational inequalities are sign analysis of the difference between sides, or multiplying by the square of the denominator. You **must never multiply directly by a denominator of unknown sign**, as this will flip the inequality if the denominator is negative, leading to incorrect solutions. For sign analysis: 1. Rearrange all terms to one side of the inequality, so the other side is 0. 2. Combine terms into a single fully simplified fraction. 3. Find critical values: points where the numerator = 0 (fraction = 0) or denominator = 0 (fraction undefined). 4. Plot critical values on a number line, splitting it into intervals. 5. Test the sign of the fraction in each interval to find which satisfy the inequality. 6. Exclude any values that make the original denominator zero.

**Worked example:** Solve $\frac{1}{x-2} > \frac{x}{x+1}$

1. Step 1: Rearrange all terms to left side, set to > 0:
2. $$\frac{1}{x-2} - \frac{x}{x+1} > 0$$
3. Step 2: Combine into single fraction with common denominator $(x-2)(x+1)$:
4. $$\frac{(x+1) - x(x-2)}{(x-2)(x+1)} > 0$$
5. $$\frac{-x^2 + 3x + 1}{(x-2)(x+1)} > 0$$
6. Multiply numerator and denominator by -1, flipping the inequality sign:
7. $$\frac{x^2 - 3x -1}{(x-2)(x+1)} < 0$$
8. Step 3: Find critical values: numerator roots $x = \frac{3\pm\sqrt{13}}{2} \approx -0.303, 3.303$, denominator undefined at $x=-1, x=2$
9. Step 4: Test intervals between critical values: valid intervals are $(-1, \frac{3-\sqrt{13}}{2}) \cup (2, \frac{3+\sqrt{13}}{2})$
10. Step 5: Exclude undefined points $x=-1, x=2$, which are already excluded from the open intervals.

> **Exam tip:** Always state critical values clearly in your working, as exam markers award partial marks for correct critical values even if your final interval is wrong.

*Calculator:* allowed

## Solving Rational Inequalities: Squaring Denominator Method

An alternative to sign analysis is multiplying both sides of the inequality by the square of the denominator, which is always non-negative, so the inequality sign does not flip. This method is faster for simple inequalities, but you must still explicitly exclude values that make the original denominator zero from your final solution set. Steps: 1. Identify all denominators in the inequality. 2. Multiply both sides of the inequality by the square of each denominator (non-negative, so sign stays the same). 3. Rearrange all terms to one side, factor fully, and solve the resulting polynomial inequality. 4. Exclude any values that make the original denominators zero.

**Worked example:** Solve $\frac{3}{x+4} < 2$ using the squaring method

1. Step 1: The only denominator is $x+4$, square is $(x+4)^2, x \neq -4$
2. Step 2: Multiply both sides by $(x+4)^2$ (positive, inequality sign unchanged):
3. $$3(x+4) < 2(x+4)^2$$
4. Step 3: Rearrange all terms to one side and simplify:
5. $$2(x^2 + 8x +16) - 3x - 12 > 0$$
6. $$2x^2 +13x +20 > 0$$
7. $$(2x +5)(x +4) > 0$$
8. Step 4: Find critical values $x=-4, x=-\frac{5}{2}$, test intervals: valid solutions are $x < -4 \cup x > -\frac{5}{2}$
9. Step 5: Exclude undefined value $x=-4$, which is already excluded from the open intervals.

> **warning**
>
> Never forget to exclude values that make the original denominator zero, even if they appear to satisfy the solved polynomial inequality. These values are undefined in the original problem, so they can never be part of the solution set.

*Calculator:* allowed

## Solving Modulus Inequalities

**Modulus inequality** — An inequality involving the absolute value function $|f(x)|$, which returns the non-negative value of $f(x)$ for all real x.

*Example:* $|x^2 -1| > 2(x+1)$

Three valid methods for solving modulus inequalities are: 1. Case analysis: split into cases where the expression inside the modulus is positive or negative. 2. Squaring both sides: valid only if both sides of the inequality are non-negative. 3. Graphical sketching: sketch both sides of the inequality, find intersection points, and identify intervals where the required side is higher. All methods require you to check your final solutions against the original inequality to eliminate extraneous solutions.

**Worked example:** Solve $|x^2 -1| > 2(x+1)$

1. Step 1: Factor the expression inside the modulus: $|(x-1)(x+1)| > 2(x+1)$
2. Step 2: Case analysis, Case 1: $x^2 -1 \geq 0$ → $x \leq -1$ or $x \geq 1$, so $|x^2-1| = x^2 -1$
3. $$x^2 -1 > 2x + 2 → x^2 - 2x -3 >0 → (x-3)(x+1) >0$$
4. Solution for case 1: $x < -1$ or $x > 3$, which overlaps with case 1 domain to give $x < -1$ or $x>3$
5. Case 2: $x^2 -1 <0$ → $-1 < x <1$, so $|x^2 -1| = 1 -x^2$
6. $$1 -x^2 > 2x + 2 → 0 > x^2 + 2x +1 → 0 > (x+1)^2$$
7. This is never true for real x, so no solutions in case 2.
8. Final solution: $x < -1 \cup x > 3$
9. Verify with graph: sketch $y=|x²-1|$ and $y=2(x+1)$, intersection at x=-1 and x=3, so the modulus curve is above the line when x < -1 or x>3, confirming the solution.

> **Exam tip:** A quick labelled sketch of both sides of the inequality is a valid justification for your solution in the exam, and can help you catch errors in algebraic working quickly.

*Calculator:* allowed

## Exam Justification and Answer Presentation

**Exam command terms**

Common command terms for inequality questions in FP2 exams, with expected responses:

- **Solve** — Find all real values of x that satisfy the inequality, presented as a set or union of intervals, excluding undefined points. *(Write solutions as $x < -2 \cup 1 < x < 4$ rather than listing individual values.)*

- **Justify your answer** — Show all algebraic working, or include a labelled sketch of the functions on either side of the inequality to support your solution.

- **Show that** — Prove the given inequality holds for a specified range of x, using valid algebraic manipulation (no multiplication by unknown sign expressions).

**Check your understanding**

1. Which of the following is a valid first step to solve $\frac{2}{x-3} > x + 1$?

   - Multiply both sides by $x-3$ to get $2 > (x+1)(x-3)$
   - Rearrange to $\frac{2}{x-3} - (x+1) > 0$
   - Multiply both sides by $(x-3)^2$ to get $2(x-3) > (x+1)(x-3)^2$

   *Why:* Option 1 is invalid because $x-3$ could be negative, flipping the inequality sign. Options 2 and 3 are the approved sign analysis and squaring methods respectively.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Multiplying both sides of a rational inequality by a denominator of unknown sign.
  - Why it fails: If the denominator is negative, the inequality sign flips, leading to missing or incorrect solution intervals.
  - Correct: Either rearrange all terms to one side and use sign analysis, or multiply by the square of the denominator (always non-negative, so sign stays the same).
- **Wrong:** Including values that make a denominator zero in the final solution set.
  - Why it fails: These values make the original inequality undefined, so they can never be valid solutions.
  - Correct: Always list critical points where denominators are zero, and explicitly exclude them from your final solution intervals.
- **Wrong:** Squaring both sides of an inequality where one side could be negative.
  - Why it fails: Squaring a negative number makes it positive, leading to extraneous solutions that do not satisfy the original inequality.
  - Correct: Only square both sides if you can confirm both sides are non-negative for all x, otherwise use case analysis or graphical methods.
- **Wrong:** Forgetting to flip the inequality sign when multiplying or dividing both sides by a negative number.
  - Why it fails: Multiplying or dividing by a negative reverses the order of values, so the inequality direction must change to stay valid.
  - Correct: Any time you multiply/divide an inequality by a negative value, immediately flip the direction of the inequality sign.
- **Wrong:** Using open intervals for inequalities with ≥/≤ signs, including endpoints that make the expression valid.
  - Why it fails: Endpoints where the expression is defined and satisfies the equality condition must be included in the solution set for non-strict inequalities.
  - Correct: Use closed square brackets [ ] for endpoints that are valid, and open parentheses ( ) for undefined points or strict inequalities.

## Cheatsheet

| Inequality Type | Approved Method | Key Rule |
| --- | --- | --- |
| Rational inequality | Sign analysis or square of denominator | Never multiply by unknown sign denominator; exclude undefined points |
| Modulus inequality (both sides non-negative) | Square both sides, solve polynomial inequality | Check all solutions against original inequality to remove extraneous values |
| Modulus inequality (one side can be negative) | Case analysis or graphical sketch | Split modulus into positive/negative cases, match solutions to case domain |
| Non-strict inequality (≥/≤) | Any approved method | Include valid endpoints in solution set; exclude undefined points |

## What's next

Now that you have mastered FP2 inequality solving, you can apply these algebraic manipulation skills to other FP2 topics including further calculus and complex number inequalities. These techniques are also frequently tested in combination with series and differential equations questions in later FP2 exam sections. Practicing past paper questions will help you quickly identify the most efficient method for each problem type, and avoid common pitfalls that cost marks in the exam. Make sure you can present your solutions clearly, with justifications where required, to secure full marks for inequality questions.

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