Inequalities

Why you cannot solve an inequality like an equation

One rule separates inequalities from equations:

$3 < 5$ is true. Multiply both sides by $-1$ and $-3 < -5$ is false; it has to become $-3 > -5$.

That is manageable when you can see the number. It becomes dangerous when you are multiplying by something algebraic, because you do not know its sign. This is the single biggest source of lost marks in the topic, and the whole method below is built to avoid it.

Quadratic inequalities

Never test values at random and never guess. The method is always the same three steps.

1. Rearrange so that one side is zero.
2. Factorise and find the critical values — the values that make each bracket zero.
3. Sketch the parabola and read off the region the question asks for.

For $x^2 - x - 6 > 0$: factorising gives $(x-3)(x+2) > 0$, so the critical values are $x = 3$ and $x = -2$. The coefficient of $x^2$ is positive, so the parabola opens upwards, dipping below the axis between the roots and rising above it on either side. We want where it is above, so

$$x < -2 \quad \text{or} \quad x > 3$$

Rational inequalities

For something like $\dfrac{x-1}{x+3} \geq 0$, the tempting move is to multiply both sides by $(x+3)$. Do not: you do not know whether $x+3$ is positive or negative, so you do not know whether the sign should flip.

Two safe routes:

Multiply by the square of the denominator. $(x+3)^2$ is positive for every $x$ except $x = -3$, so multiplying by it never changes the direction. The inequality becomes

$$(x-1)(x+3) \geq 0$$

which is an ordinary quadratic inequality, giving $x \leq -3$ or $x \geq 1$.

Or rearrange to a single fraction against zero and use a sign table. Either is acceptable.

When the right-hand side is not zero, move everything to one side first:

$$\frac{x+4}{x-1} \geq 2 \quad\Longrightarrow\quad \frac{x+4}{x-1} - 2 \geq 0 \quad\Longrightarrow\quad \frac{6-x}{x-1} \geq 0$$

and then proceed as above. Subtracting first is what keeps you safe; multiplying first is what gets the answer wrong.

Working backwards from a solution set

If you are told the solution of a quadratic inequality is $-2 < x < 5$, then the critical values are $-2$ and $5$, so the brackets are $(x+2)$ and $(x-5)$, and the expression is

$$(x+2)(x-5) = x^2 - 3x - 10$$

The solution lies between the roots, so the parabola must be below the axis there, which means the inequality is $x^2 - 3x - 10 < 0$.