Solve quadratic inequalities by factorising and using a sketch or number line, and express the solution set correctly

What this dot point is asking

SEAB wants you to solve a quadratic inequality such as $x^2 - x - 6 < 0$ and write the answer as a range (or two ranges) of values of $x$. The reliable method is: get everything on one side so you compare with zero, factorise to find the critical values (where the quadratic equals zero), then use a quick sketch of the parabola, or a number line, to read off where the curve is above or below the axis. The key idea is that a parabola changes sign only at its roots.

The answer

Step one: compare with zero

An inequality is easiest to handle when one side is $0$. Rearrange first, for example $x^2 \ge 4$ becomes $x^2 - 4 \ge 0$. Never divide both sides by a variable, because you do not know its sign.

Step two: factorise to find the critical values

Factorise the quadratic and set each factor to zero. These critical values are the only places the expression can change between positive and negative. For $(x - 3)(x + 2)$ the critical values are $x = 3$ and $x = -2$.

Step three: sketch the parabola

Draw a quick parabola through the two critical values on the $x$-axis. If the coefficient of $x^2$ is positive the parabola opens upward (a valley): it is below the axis (negative) between the roots and above the axis (positive) outside the roots. If the coefficient is negative the parabola opens downward and the regions swap.

Step four: read off the solution

Match the region to the inequality sign:

• For an upward parabola, "$< 0$" (below the axis) gives the inside region $\text{root}_1 < x < \text{root}_2$.
• For an upward parabola, "$> 0$" (above the axis) gives the outside region $x < \text{root}_1$ or $x > \text{root}_2$.

Use $\le$ or $\ge$ (and include the critical values) when the inequality is "or equal to".

A number-line alternative

If you prefer not to sketch, mark the critical values on a number line and test one value in each of the three intervals by substituting it into the factorised form. The sign you get tells you whether that whole interval satisfies the inequality.

Examples in context

Example 1. A safe range of speeds. Suppose a stopping-distance model gives $d = v^2 - 12v + 27$ metres and a road requires $d \le 0$ to be safe within markings (a simplified model). Factorising, $(v - 3)(v - 9) \le 0$, so the safe range is $3 \le v \le 9$. The inequality turns a modelling condition directly into an allowed interval.

Example 2. Linking to the discriminant. A question may ask for the values of $k$ for which $x^2 + kx + 4 = 0$ has two distinct real roots. The condition $b^2 - 4ac > 0$ becomes $k^2 - 16 > 0$, itself a quadratic inequality, solved as $k < -4$ or $k > 4$. So solving inequalities is the final step of many discriminant problems.

Try this

Q1. Solve $(x - 1)(x - 5) < 0$. [2 marks]

• Cue. Upward parabola, want below the axis (inside): $1 < x < 5$.

Q2. Solve $x^2 - 9 \ge 0$. [2 marks]

• Cue. Factorise to $(x - 3)(x + 3) \ge 0$; outside region: $x \le -3$ or $x \ge 3$.

Q3. Solve $x^2 - 2x - 3 \le 0$. [3 marks]

• Cue. Factorise to $(x - 3)(x + 1) \le 0$; inside region: $-1 \le x \le 3$.