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hscscience Maths Ext 1 · Y11
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Module 1 · L2 of 15 ~35 min ⚡ +95 XP available

Linear & Quadratic Inequalities

Equations say two things are equal. Inequalities say one thing is bigger, and that turns out to be far more useful in the real world. Speed limits, dosage thresholds, budget constraints, engineering tolerances: they're all inequalities. In this lesson you'll learn the two core techniques, the sign reversal rule for linear inequalities, and the parabola interval method for quadratics, that are the backbone of every harder inequality type in this module.

Today's hook, Solve $-3x > 12$. Most students write $x > -4$. The correct answer is $x < -4$. One invisible sign flip, four marks lost in the HSC. This lesson makes sure that never happens to you, and shows you a clean visual method so you never need to remember which direction to flip.
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Recall, your gut answer first

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Recall, your gut answer first
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Solve $3x - 5 > 7$. Then answer: what happens to the inequality sign if you multiply or divide both sides by a negative number? Write your reasoning, don't just state the rule.

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Orientation and key terms

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The two moves
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Inequality problems always come down to two moves. For linear inequalities: solve like an equation, but flip the sign when you multiply or divide by a negative. For quadratic inequalities: find the roots, sketch the parabola, and read off the intervals.

The key insight is that multiplying by a negative reverses the order of the number line. If $a > b$ and $c < 0$, then $ac < bc$, the larger number becomes the smaller one when everything flips. For quadratics, the roots divide the number line into sign-constant intervals: between the roots, or outside them, the expression keeps one sign throughout.

The parabola's sign is positive outside its roots and negative between them, giving the solution intervals.
multiply/divide by $c < 0$ $\Rightarrow$ flip $<$ to $>$
Sign reversal rule
Multiplying or dividing an inequality by a negative number reverses the sign. Adding or subtracting never changes the direction.
Parabola interval test
Factorise, find roots, sketch parabola. Between the roots the quadratic has the opposite sign to the leading coefficient; outside the roots it has the same sign.
No-roots shortcut
If $\Delta < 0$ and the leading coefficient is positive, the quadratic is always positive. If negative leading coefficient and $\Delta < 0$, it's always negative. No solving needed.
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What you'll master
Know

Key facts

  • The sign reversal rule: multiply/divide by a negative, flip the inequality sign
  • Interval notation: $[a, b]$ (closed), $(a, b)$ (open), $(-\infty, a)$, $[a, \infty)$
  • For a positive-leading quadratic: negative between roots, positive outside roots
Understand

Concepts

  • Why multiplying by a negative reverses order on the number line
  • Why quadratic inequalities require a parabola sketch or sign table rather than just solving an equation
  • How the discriminant determines whether a quadratic changes sign or is always positive/negative
Can do

Skills

  • Solve linear inequalities and express solutions in interval notation
  • Solve quadratic inequalities using the parabola method or sign table
  • Handle quadratics with no real roots using the discriminant
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Key terms
InequalityA statement that one expression is greater than, less than, or equal to another using $<$, $>$, $\le$, $\ge$.
Interval notationA compact way to write solution sets: $[2, 5)$ means $2 \le x < 5$; $(-\infty, 3]$ means $x \le 3$.
Critical valueA value where the expression equals zero or is undefined, dividing the number line into testable intervals.
Sign reversalWhen multiplying or dividing an inequality by a negative number, the inequality sign flips direction.
Discriminant $\Delta$$\Delta = b^2 - 4ac$. Determines whether $ax^2 + bx + c = 0$ has two real roots ($\Delta > 0$), one ($\Delta = 0$), or none ($\Delta < 0$).
Sign tableA table showing the sign of each factor and their product across each interval separated by critical values.
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Solving linear inequalities

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Solving linear inequalities
core concept

Linear inequalities are solved using the same techniques as linear equations, with one crucial exception:

Sign Reversal Rule: If you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality sign.

All other operations, adding, subtracting, multiplying/dividing by a positive number, leave the direction of the inequality unchanged.

Example: Solve $-2x + 3 > 7$.

  • Subtract 3: $-2x > 4$
  • Divide by $-2$ (reverse the sign): $x < -2$
  • In interval notation: $(-\infty, -2)$

Why does multiplying by a negative flip the sign? Consider $2 < 5$. Multiply both sides by $-1$: $-2$ and $-5$. Now $-2 > -5$ because $-2$ is to the right on the number line. Multiplication by a negative number reflects the number line, reversing all order relationships.

Real-world connection. If a company's profit is modelled by $P = -3x + 60$ (where $x$ is units produced beyond a threshold), the inequality $P > 0$ gives $-3x > -60$, so $x < 20$. Correctly flipping the sign here is the difference between knowing when to produce and when to stop.

Sign reversal rule: multiply/divide by negative flip the inequality sign.; All other operations (add, subtract, multiply/divide by positive) do NOT change the direction.

Pause, copy the sign-reversal rule into your book: only multiplying or dividing both sides by a negative number reverses the inequality direction, all other operations (add, subtract, multiply/divide by a positive) preserve it.

Quick check: Solve $-4x \ge 12$. Which answer is correct?

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Solving quadratic inequalities
core concept

We just saw that multiplying or dividing a linear inequality by a negative number reverses the direction (e.g. $-2x < 6 Rightarrow x > -3$). That raises a question: when the inequality is quadratic, you can't just divide by $x$, how does the sign of a quadratic expression vary, and how do you identify which intervals satisfy it? This card answers it → by finding the roots, sketching the parabola (or sign table), and reading off the required intervals.

To solve a quadratic inequality such as $x^2 - 5x + 6 \le 0$, use the following three-step method:

  1. Find the roots by solving the corresponding equation. For $x^2 - 5x + 6 = 0$: $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$.
  2. Sketch the parabola (or draw a sign table). Since the leading coefficient is positive, the parabola opens upward, it is negative (below the $x$-axis) between the roots and positive outside.
  3. Read off the solution using the inequality sign. For $\le 0$ we want where the parabola is on or below the $x$-axis: $2 \le x \le 3$, i.e., $[2, 3]$.
Key principle: For a positive-leading quadratic $y = a(x - r_1)(x - r_2)$ with $r_1 < r_2$:
$y < 0$ when $r_1 < x < r_2$ (between roots); $y > 0$ when $x < r_1$ or $x > r_2$ (outside roots).

For a negative-leading quadratic, the signs are reversed.

Three-step method: (1) Find roots, (2) Sketch parabola or draw sign table, (3) Read off intervals.; Positive leading coefficient: negative between roots, positive outside.

Pause, copy the three-step quadratic inequality method into your book: (1) find the roots; (2) sketch the parabola or draw a sign table; (3) read off the intervals where the expression has the required sign.

Did you get this? True or false: for $x^2 - 7x + 12 > 0$, the solution is $x < 3$ or $x > 4$.

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Worked examples

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · LINEAR INEQUALITY

Solve $5(x - 2) \le 3x + 4$ and express the answer in interval notation.

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Expand: $5x - 10 \le 3x + 4$. Subtract $3x$: $2x - 10 \le 4$. Add 10: $2x \le 14$.
Expand the bracket first, then collect all $x$-terms on one side. No sign flip needed yet, we have only added and subtracted.
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PROBLEM 2 · QUADRATIC INEQUALITY

Solve $x^2 - x - 12 > 0$ and express the answer in interval notation.

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Factorise: $(x - 4)(x + 3) > 0$. Roots: $x = 4$ and $x = -3$.
Factorising reveals the critical values that divide the number line into sign-constant intervals: $(-\infty, -3)$, $(-3, 4)$, and $(4, \infty)$.
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PROBLEM 3 · NO REAL ROOTS

Solve $x^2 + 2x + 5 > 0$.

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Discriminant: $\Delta = b^2 - 4ac = 4 - 20 = -16 < 0$.
Since $\Delta < 0$, there are no real roots. The parabola $y = x^2 + 2x + 5$ never crosses the $x$-axis.
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Fill the gap: When solving $x^2 - 5x + 6 \le 0$, the roots are $x = 2$ and $x = 3$. The solution is $[$ $, 3]$.

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Misconceptions to fix

Misconceptions to fix · the 3 traps that cost marks

Trap 01
Forgetting to flip the sign
Solving $-2x > 8$ as $x > -4$ is the most common linear inequality error. The moment you divide by $-2$, the sign flips: $x < -4$. Underline the negative coefficient before solving as a visual reminder.
Trap 02
Reading quadratic intervals the wrong way
For $x^2 - 5x + 6 > 0$ (positive leading coefficient), the answer is OUTSIDE the roots: $x < 2$ or $x > 3$. Students often mistakenly take the inside interval. Sketch the parabola, if it opens up and you want $> 0$, you want where it's above the axis.
Trap 03
Using open brackets for $\le$ or $\ge$
If the inequality is $\le 0$ or $\ge 0$, the roots are included, use closed brackets $[\,]$. If the inequality is strict ($< 0$ or $> 0$), the roots are excluded, use open brackets $(\,)$. One bracket wrong costs a mark.

Did you get this? True or false: the solution to $x^2 + 4 > 0$ is all real $x$.

Odd one out: Three of these are solved correctly. Which one contains an error?

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Activities

Activities · practice with the ideas

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Solve $4 - 3x < 10$ and express your answer in interval notation.

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Solve $x^2 - 7x + 10 \le 0$. State your answer in interval notation.

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Solve $2x^2 + 5x - 3 > 0$. State your answer in interval notation.

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Without solving, state whether $x^2 - 2x + 5 > 0$ has solutions for all real $x$, some real $x$, or no real $x$. Justify your answer using the discriminant.

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Verify the sign reversal rule by checking: if $3 < 7$, what happens to the inequality when you multiply both sides by $-2$? Does the result confirm the rule?

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Revisit your thinking

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Revisit your thinking

Earlier you were asked to solve $3x - 5 > 7$, and to explain the sign flip.

Answer: $3x - 5 > 7 \Rightarrow 3x > 12 \Rightarrow x > 4$. In interval notation: $(4, \infty)$. No sign flip here, we divided by positive 3.

The hook question $-3x > 12$: divide by $-3$ (negative!), so flip: $x < -4$, i.e., $(-\infty, -4)$. A student who writes $x > -4$ gets this wrong. The visual check: on a number line, $-3 \times (-5) = 15 > 12$, so $x = -5$ works, confirming $x < -4$.

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Multiple choice

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Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.

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Short answer

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Short answer
ApplyBand 42 marks

Q1. Solve $4 - 3x < 10$ and express your answer in interval notation. (2 marks)

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ApplyBand 42 marks

Q2. Solve $x^2 - 7x + 10 \le 0$. (2 marks)

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AnalyseBand 53 marks

Q3. Solve $2x^2 + 5x - 3 > 0$. (3 marks)

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Comprehensive answers (click to reveal)

Activity answers: 1. $4 - 3x < 10 \Rightarrow -3x < 6 \Rightarrow x > -2$; interval: $(-2, \infty)$. · 2. $(x-2)(x-5) \le 0$; solution $[2, 5]$. · 3. $(2x-1)(x+3) > 0$; roots $x = \frac{1}{2}$, $x = -3$; solution $(-\infty, -3) \cup (\frac{1}{2}, \infty)$. · 4. $\Delta = 4 - 20 = -16 < 0$ and $a = 1 > 0$: always positive, solution is all real $x$. · 5. $3 < 7$; multiply by $-2$: $-6 > -14$. Sign flips from $<$ to $>$. Confirms the rule.

Q1 (2 marks): $4 - 3x < 10 \Rightarrow -3x < 6$ [1 mark]. Divide by $-3$ (flip sign): $x > -2$; interval notation $(-2, \infty)$ [1 mark].

Q2 (2 marks): $(x - 2)(x - 5) = 0 \Rightarrow x = 2, x = 5$ [1 mark]. Upward parabola, $\le 0$ between roots: $2 \le x \le 5$, i.e., $[2, 5]$ [1 mark].

Q3 (3 marks): $2x^2 + 5x - 3 = (2x - 1)(x + 3)$ [1 mark]. Roots $x = \frac{1}{2}$ and $x = -3$ [1 mark]. Positive-leading parabola, $> 0$ outside roots: $x < -3$ or $x > \frac{1}{2}$; interval notation $(-\infty, -3) \cup (\frac{1}{2}, \infty)$ [1 mark].

Module overview · Extension 1 subject page · Checkpoint 1 · Module quiz