Solving Quadratic Inequalities
Master quadratic inequalities for Queensland Year 11 Mathematical Methods (QCAA). A quadratic inequality asks where a parabola sits above or below the x-axis.
You will learn to rearrange against zero, factorise to find the zeros, sketch the parabola and read off the solution, reverse the sign when dividing by a negative, and write answers in interval notation with the right open or closed boundaries.
Every question with a fully worked solution.
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Theory
In Year 11 Mathematical Methods (QCAA, Unit 1), a quadratic inequality such as \(ax^2+bx+c>0\) is solved by factorising to find the zeros, then reading which side of the parabola is above or below the \(x\)-axis. This page shows how to solve product-form, monic and non-monic inequalities and write the answer in inequality or interval form.
A quadratic inequality compares a quadratic with \(0\) using \(>,\ \ge,\ <\) or \(\le\). Solving it means finding all \(x\)-values that make the statement true — usually an interval or a pair of rays, not single points.
The key idea is the sign of the parabola. Factorise to find the zeros \(x_1\lt x_2\); an upward parabola (\(a>0\)) is below the axis between the zeros and above it outside them. A downward parabola is the other way around.
Strict inequalities (\(<,\ >\)) exclude the boundary zeros (open circles); inclusive inequalities (\(\le,\ \ge\)) include them (closed circles).
The sign pattern of \(y=a(x-x_1)(x-x_2)\) with \(x_1\lt x_2\) and \(a\gt0\):
How to solve a quadratic inequality
- Rearrange to compare with \(0\) (all terms on one side), and if the leading coefficient is negative you may multiply by \(-1\) — then flip the inequality sign.
- Factorise to find the zeros, and sketch the parabola the correct way up.
- Read the region where the parabola is on the required side of the axis, and write the answer with strict or inclusive boundaries.
Zeros — set each factor to zero:
| \(x-2=0\) | \(=\) | \(x=2\) |
| \(x+3=0\) | \(=\) | \(x=-3\) |
The parabola opens up (\(a=1>0\)), so it is above the axis outside the zeros.
Read the region where \(y\gt0\) (strict, so exclude the zeros):
| \(x\lt-3\) | \(=\) | \(\text{or}\quad x\gt2\) |
\(x\lt-3\) or \(x\gt2\).
Factorise the left side:
| \(x^2-x-6\) | \(=\) | \((x-3)(x+2)\) |
Zeros:
| \(x\) | \(=\) | \(3\ \text{or}\ -2\) |
The parabola opens up, so it is below the axis between the zeros. Because the inequality is \(\le\), the boundaries are included.
Write the solution:
| \(-2\le x\) | \(\le\) | \(3\) |
\(-2\le x\le3\).
Multiply both sides by \(-1\) and reverse the inequality:
| \(x^2-4x+3\) | \(\le\) | \(0\) |
Factorise:
| \((x-1)(x-3)\) | \(\le\) | \(0\) |
This upward parabola is below (or on) the axis between its zeros \(1\) and \(3\), inclusive.
Write the solution:
| \(1\le x\) | \(\le\) | \(3\) |
\(1\le x\le3\).
Profit positive means \(P>0\). Multiply by \(-1\) and reverse the sign:
| \(-x^2+10x-16\) | \(\gt\) | \(0\) |
| \(x^2-10x+16\) | \(\lt\) | \(0\) |
Factorise — two numbers multiplying to \(16\), adding to \(-10\) are \(-2\) and \(-8\):
| \((x-2)(x-8)\) | \(\lt\) | \(0\) |
The upward parabola is below the axis between its zeros, and the inequality is strict.
Read the region:
| \(2\lt x\) | \(\lt\) | \(8\) |
Profit is positive for \(2\lt x\lt8\) — between \(2000\) and \(8000\) items.
Common pitfalls
Frequently asked questions
How do I solve a quadratic inequality?
Rearrange to compare with \(0\), factorise to find the zeros, sketch the parabola, then read the region where it is on the required side of the \(x\)-axis.
When do I reverse the inequality sign?
Whenever you multiply or divide both sides by a negative number — for example after taking out a factor of \(-1\).
For an upward parabola, where is it negative?
Between its two zeros. It is positive outside them, so \(y\lt0\) gives \(x_1\lt x\lt x_2\).
What is the difference between strict and inclusive?
Strict inequalities \(<,\ >\) exclude the boundary zeros (open circles); inclusive \(\le,\ \ge\) include them (closed circles).
How do I write the answer in interval notation?
A region between zeros is \((x_1,x_2)\) or \([x_1,x_2]\); a pair of rays is \((-\infty,x_1)\cup(x_2,\infty)\), using square brackets when the boundary is included.
Do I use calculus for these?
No. In Year 11 you factorise and read the sign from the parabola sketch; no differentiation is needed.