Quadratic inequalities

By the end of this lesson

By the end you can solve quadratic inequalities such as $x^2 - x - 6 > 0$ by finding the critical values and reading the answer off a quick sketch.

What you're looking for

An equation like $x^2 - x - 6 = 0$ has a couple of answers. An inequality like $x^2 - x - 6 > 0$ asks a different question: for which values of $x$ is the left side positive? The answer is a whole range of numbers, not just one or two.

The trick is to think of $y = x^2 - x - 6$ as a graph. " $> 0$ " means "where is the graph above the $x$ -axis?" and " $< 0$ " means "where is it below?"

The graph can only switch from above to below by crossing the axis, at the roots. Those roots are called the critical values: they are the boundaries of the answer. (Below, $\alpha$ and $\beta$ , the Greek letters alpha and beta, stand for the smaller and larger critical value.)

The three-step method

Rearrange so one side is 0 (a positive $x^2$ coefficient makes life easier).
Find the critical values: solve the equation $= 0$ by factorising, completing the square or the formula.
Sketch the parabola through those roots and read off where it's above the $x$ -axis ( $> 0$ ) or below it ( $< 0$ ).

The sketch is 5 seconds of work and it's what stops you picking the wrong region.

Worked example: Outside or between?

Solve $x^2 - x - 6 > 0$ .

Have a go first?

Why outside-or-between always works

When there are two distinct critical values $\alpha < \beta$ , a ∪-shaped parabola comes down from above, dips below the axis between its roots, and goes back up. So with a positive $x^2$ coefficient:

$> 0$ (or $\geqslant 0$ ): two separate intervals, outside the roots.
$< 0$ (or $\leqslant 0$ ): one interval, a "sandwich" $\alpha < x < \beta$ (or $\alpha \leqslant x \leqslant \beta$ ).

Don't memorise that blindly, though. One sketch and you can see it. (No critical values, or only one? See the "go deeper" note below.)

Quick check

Multiple choiceQuestion 1 · Recall

Answer the question above to continue.