Solve quadratic inequalities by factorising and reasoning about the sign of the quadratic between and beyond its roots

What this dot point is asking

SEAB wants you to solve inequalities such as $x^2 - x - 6 > 0$, giving the answer as a range of $x$. The skill is to find where the quadratic equals zero, then decide on which side of those roots the quadratic has the sign you want, using the shape of the parabola.

The answer

Step one: rearrange to compare with zero

Always move everything to one side so the inequality reads (quadratic) compared with $0$. So $2x^2 + 5x \leq 3$ becomes $2x^2 + 5x - 3 \leq 0$. The sign reasoning only works against zero.

Step two: find the roots

Factorise (or use the formula) to find where the quadratic equals zero. These critical values split the number line into regions.

Step three: use the parabola's shape

A quadratic $ax^2 + \dots$ with $a > 0$ is a U-shape: negative between its roots and positive outside them. With $a < 0$ it is an upside-down U: positive between and negative outside. A quick sketch of the parabola cutting the $x$-axis at the roots makes the sign obvious.

Reading off the answer

• For $> 0$ (or $\geq 0$) with an upward parabola, take the regions outside the roots.
• For $< 0$ (or $\leq 0$) with an upward parabola, take the region between the roots.

Use strict inequalities ($<, >$) when the original is strict, and inclusive ($\leq, \geq$) when the roots themselves are allowed.

A number-line picture

A quick number line with the two roots marked makes the regions concrete: test one value in each of the three regions (left of both roots, between them, right of both) in the factorised expression and note its sign. Shade the regions whose sign matches the inequality, and read off the answer.

When the quadratic does not factorise

If the quadratic has no rational factors, find the roots with the quadratic formula and use those (possibly surd) values as the region boundaries. If the discriminant is negative there are no real roots, so the parabola keeps one sign everywhere, and the inequality is either always true or never true.

Examples in context

Example 1. Region of profit. If profit is modelled by $P = -x^2 + 8x - 7$, solving $P > 0$ gives the range of output for which the firm makes money, namely $1 < x < 7$, a between-the-roots answer for a downward parabola.

Example 2. Discriminant conditions. Finding the values of $k$ for which a quadratic has real roots leads to an inequality such as $k^2 + 2k - 15 > 0$; solving that quadratic inequality in $k$ gives the allowed range, which is why this technique reappears in discriminant problems.

Try this

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

• Cue. Between the roots: $-5 < x < 2$.

Q2. Solve $x^2 \geq 9$. [2 marks]

• Cue. $x^2 - 9 \geq 0$, so $x \leq -3$ or $x \geq 3$.

Q3. Solve $x^2 - 4x + 3 < 0$. [3 marks]

• Cue. $(x - 1)(x - 3) < 0$, so $1 < x < 3$.