# Inequalities

> Solve cubic inequalities, inequalities with x in the denominator and absolute value inequalities, with algebra and with graphs.

From HSC Maths Extension 1 on WhizzNotes, free study notes for NSW students: https://whizznotes.com/mathematics-extension-1/inequalities/. Updated 3 November 2026.
Year 11 and Year 12 boxes go deeper than the Year 10 text around them.

### Story time: The safe speed zone
A car's cruise control must keep the speed within 5 km/h of 60 km/h. So the speed $v$ must satisfy $55 \le v \le 65$. Engineers write this in one short line: $|v - 60| \le 5$.

Real problems are full of rules like "at least", "no more than" and "within". These are **inequalities**. In this chapter you solve harder ones, where $x$ is in a fraction or inside an absolute value.

**In short:** To solve an inequality, find where the expression equals zero, sketch it, and read off where it is above or below the x-axis. With $x$ in the denominator, multiply both sides by the denominator squared, and $|x - a| < k$ means $a - k < x < a + k$.

## Starting point: what you already know

- Solve a linear inequality like an equation, but **flip the sign if you multiply or divide by a negative number**: $-2x < 6$ gives $x > -3$.
- **Quadratic inequalities** (Mathematics Advanced): to solve $x^2 - x - 6 < 0$, factorise to $(x - 3)(x + 2) < 0$, sketch the smile parabola and read where it is below the axis: $-2 < x < 3$.
- $|x|$ is the distance of $x$ from 0. So $|x - a|$ is the distance between $x$ and $a$.

### Key idea: The graph method
To solve $f(x) > 0$, find where $f(x) = 0$, sketch the graph, and read where it is **above** the axis. For $f(x) < 0$, read where it is **below**.

## Cubic inequalities

A cubic like $(x + 1)(x - 2)(x - 4)$ has three roots. A positive cubic (leading coefficient positive) starts low on the left and ends high on the right, crossing the axis at each single root.

### Worked example: Solve (x + 1)(x − 2)(x − 4) ≥ 0
Roots: $-1$, 2, 4. The graph comes up from below, crosses at $-1$, goes back down at 2, and up again at 4.

Above or on the axis: $-1 \le x \le 2$ or $x \ge 4$.

### Exam tip: Sign table
If you prefer, test one number in each gap between the roots. The sign is the same everywhere in a gap.

## Inequalities with x in the denominator

You **cannot** just multiply both sides by $x$, because you do not know if $x$ is positive or negative. Instead, **multiply both sides by the square of the denominator**, which is always positive.

### Worked example: Solve 3/(x − 1) ≤ 1
First, $x \ne 1$. Multiply both sides by $(x - 1)^2$:
$$3(x - 1) \le (x - 1)^2$$
$$0 \le (x - 1)^2 - 3(x - 1) = (x - 1)(x - 4)$$
A smile parabola with roots 1 and 4 is at or above the axis when $x \le 1$ or $x \ge 4$. Remove $x = 1$ (not allowed):
$$x < 1 \quad\text{or}\quad x \ge 4$$
Check $x = 0$: $\dfrac{3}{-1} = -3 \le 1$. True.

### Watch out: Never include the asymptote
The value that makes the denominator zero can never be part of the answer, even if the sign says "or equal to".

## Absolute value inequalities

Read $|x - a|$ as the distance from $x$ to $a$. Then:

- $|x - a| < k$ means $x$ is **within** $k$ of $a$: $a - k < x < a + k$.
- $|x - a| > k$ means $x$ is **further** than $k$ from $a$: $x < a - k$ or $x > a + k$.

*Diagram: The V shaped graph of y equals the absolute value of x minus 1 and the horizontal line y equals 2. They meet at x equals minus 1 and x equals 3. The V is below the line between them, shaded on the x-axis.*
*|x − 1| &lt; 2 where the V is under the line: −1 &lt; x &lt; 3.*

### Worked example: Solve |2x + 3| ≥ 5
Either $2x + 3 \ge 5$, which gives $x \ge 1$, or $2x + 3 \le -5$, which gives $x \le -4$.

So $x \le -4$ or $x \ge 1$.

**Mind map**

```text
Inequalities
  Basics
    Flip sign for negative multiply
    Sketch and read the graph
  Cubic
    Roots, then shape
    Above or below the axis
  x in the denominator
    Multiply by denominator squared
    Exclude the asymptote
  Absolute value
    Distance from a
    Less than k: between
    More than k: outside
```

### Year 11: Both sides with absolute value
Solve $|x - 1| < |x + 3|$. Both sides are never negative, so we can square without changing the inequality:
$$(x - 1)^2 < (x + 3)^2$$
$$x^2 - 2x + 1 < x^2 + 6x + 9 \quad\Rightarrow\quad -8 < 8x \quad\Rightarrow\quad x > -1$$
Meaning: $x$ is closer to 1 than to $-3$, so it is past the midpoint $-1$.

### Year 11: A harder fraction
Solve $\dfrac{x + 2}{x - 3} > 2$. Note $x \ne 3$. Multiply by $(x - 3)^2$:
$$(x + 2)(x - 3) > 2(x - 3)^2$$
$$0 > (x - 3)\big[2(x - 3) - (x + 2)\big] = (x - 3)(x - 8)$$
So $(x - 3)(x - 8) < 0$, giving $3 < x < 8$.

**Repeated roots.** $(x - 2)^2$ is never negative and is zero only at $x = 2$, so it does not change the sign. For $(x - 2)^2(x + 1)(x - 5) \leq 0$, solve $(x + 1)(x - 5) \leq 0$, which gives $-1 \leq x \leq 5$; the point $x = 2$ is already inside.

**The $|x|^2$ trick.** $x^2 = |x|^2$, so $x^2 - 3|x| - 4 < 0$ is $(|x| - 4)(|x| + 1) < 0$. The second bracket is always positive, so $|x| < 4$, that is $-4 < x < 4$.

**Comparing two fractions.** Move everything to one side and combine: $\dfrac{1}{x - 2} > \dfrac{1}{x + 2}$ becomes $\dfrac{4}{(x - 2)(x + 2)} > 0$, which needs $(x - 2)(x + 2) > 0$: $x < -2$ or $x > 2$.

**A leftover factor that never changes sign.** Multiply $\dfrac{4}{x + 1} < 3 - x$ by $(x + 1)^2$: $4(x + 1) < (3 - x)(x + 1)^2$. Take out $(x + 1)$: $(x + 1)\big[(3 - x)(x + 1) - 4\big] > 0$, which is $(x + 1)(-x^2 + 2x - 1) > 0$, that is $-(x + 1)(x - 1)^2 > 0$. Since $(x - 1)^2 \geq 0$, this needs $x + 1 < 0$: $x < -1$. In general, if a quadratic factor has $\Delta \leq 0$, it keeps one sign, so only the other factor decides the answer.

**Positive for every $x$.** $ax^2 + bx + c > 0$ for all $x$ needs $a > 0$ and $\Delta < 0$. For $(k - 2)x^2 + 2x + (k - 2)$: $k > 2$ and $4 - 4(k - 2)^2 < 0$, so $k > 3$.

### Year 12: Inequalities in proofs
In Year 12, some proofs by induction and calculus problems ask you to show one expression is bigger than another for all values of $x$. Moving everything to one side and showing it is a perfect square, which is never negative, is a common way.

## Quick summary

- Find where the expression equals zero, sketch, then read above the axis for $> 0$ or below it for $< 0$.
- With $x$ in the denominator, multiply both sides by the denominator squared and leave out the value that makes it zero.
- $|x - a| < k$ means $a - k < x < a + k$. $|x - a| > k$ means $x < a - k$ or $x > a + k$.

## Quiz

Try these questions from the chapter quiz.

1. Solve $-3x > 12$.
   - A. $x > -4$
   - B. $x < -4$
   - C. $x < 4$
   - D. $x > 4$
   - Answer: B. $x < -4$. Dividing by $-3$ flips the sign.

2. Solve $(x - 1)(x - 5) < 0$.
   - A. $1 < x < 5$
   - B. $x < 1$ or $x > 5$
   - C. $x > 5$
   - D. $-5 < x < -1$
   - Answer: A. $1 < x < 5$. A smile parabola is below the axis between its roots.

3. Solve $x^2 \ge 16$.
   - A. $-4 \le x \le 4$
   - B. $x \le -4$ or $x \ge 4$
   - C. $x \ge 4$
   - D. $x \ge 16$
   - Answer: B. $x \le -4$ or $x \ge 4$. $x^2 - 16 = (x - 4)(x + 4) \ge 0$ outside the roots.

The whole chapter quiz is on the page, for signed-in students: https://whizznotes.com/mathematics-extension-1/inequalities/#quiz