2.15 Solving inequalities

An inequality compares two quantities that need not be equal. Solving it means finding every value of $x$ for which the comparison is true, so the answer is usually a whole interval or a union of intervals rather than a list of numbers. Inequalities describe conditions such as "the profit is at least zero", "the concentration stays below a safe level" or "the curve lies above the line".

Linear and quadratic inequalities can be solved with a few rules and a sketch. For polynomials of higher degree and for rational functions, the same rules need more care, because multiplying by an expression whose sign changes can reverse an inequality on part of the number line without warning. Critical values and sign diagrams give a reliable method that works in every case.

Inequalities and their solutions

An inequality uses one of four symbols. A strict inequality uses $<$ or $>$; a non-strict inequality uses $\le$ or $\ge$ and also allows equality.

The difference matters at the boundary of a solution. The solution of $x^2 < 4$ is $-2 < x < 2$, while the solution of $x^2 \le 4$

Writing a solution set

A solution can be written with inequalities or in interval notation. A round bracket means the endpoint is excluded; a square bracket means it is included; $\infty$ always takes a round bracket. The symbol $\cup$ (union) joins separate intervals.

The IB uses outward-facing square brackets, $]a, b[$, for open intervals; round brackets $(a, b)$ mean the same thing and are equally acceptable. With inequalities, a union is written with the word "or": "$x < 3$ or $x \ge 7$

Rules for manipulating inequalities

An inequality is preserved when

• the same quantity is added to or subtracted from both sides;

• both sides are multiplied or divided by the same positive number.

It is reversed when both sides are multiplied or divided by the same negative number: from $-2x > 6$, dividing by $-2$ gives $x < -3$.

When the multiplier is an expression such as $x$ or $(x - 3)$, its sign depends on $x$: positive for some values and negative for others. Multiplying by it would keep the inequality for some values of $x$ and reverse it for others, so a single answer cannot be correct.

Solving inequalities graphically

The inequality $g(x) \ge f(x)$ is true for exactly those values of $x$ where the graph of $y = g(x)$ is on or higher than the graph of $y = f(x)$

So a graphical solution has three steps:

1. Draw both graphs (or the single graph of $y = g(x) - f(x)$).

2. Find the $x$-coordinates of the intersection points, and any vertical asymptotes.

3. Read off the intervals where the graph of $g$ is higher than the graph of $f$

Moving everything to one side gives an equivalent form: $g(x) \ge f(x)$ is the same as $g(x) - f(x) \ge 0$, which is true where the graph of $y = g(x) - f(x)$

Using a GDC

A graphic display calculator (GDC) draws both graphs and finds their intersections with its intersection tool, or the zeros of $g(x) - f(x)$ with its zero (root) tool. This handles inequalities that are hard or impossible to solve algebraically.

Critical values and sign diagrams

The analytic method rests on one fact: a polynomial or rational expression can only change sign at a value where it is zero or undefined. Between two such values the graph is unbroken and never meets the $x$-axis, so the expression keeps the same sign throughout the interval.

The critical values split the number line into intervals. One test value in each interval, or the signs of the factors, gives the sign of $h(x)$ on the whole interval.

The method

1. Rearrange the inequality so that one side is $0$: $h(x) > 0$, $h(x) \ge 0$, and so on.

2. Factorise $h(x)$ fully, writing it as a single fraction if it is rational.

3. Find the critical values: the zeros of the numerator and of the denominator.

Polynomial inequalities

For a polynomial in factorised form, the sign on each interval comes from the signs of the factors. With the leading coefficient positive, the sign in the right-most interval is always $+$. Moving left, the sign changes at each simple root.

A polynomial that is not yet factorised is first factorised using the factor theorem.

Repeated roots

At a root of even multiplicity the factor $(x - a)^2$ does not change sign, so neither does the polynomial: the graph touches the axis and turns back. The sign is the same on both sides of such a critical value.

Rational inequalities

Rational inequalities are where multiplying by an unknown expression most often goes wrong. Two safe methods exist.

Method 1: bring everything to one side

Subtract so that one side is $0$, combine into a single fraction, and draw a sign diagram. The critical values are the zeros of the numerator and of the denominator. A zero of the denominator is never part of the solution, because the expression is undefined there.

Method 2: multiply by the square of the denominator

A square is never negative, so multiplying both sides by $(x - 3)^2$, which is positive for every $x \neq 3$, keeps the direction of the inequality. For $\frac{x + 1}{x - 3} \le 2$

Collect everything on one side and factorise, taking out the common factor rather than expanding:

So $(x - 3)(x - 7) \ge 0$, which is a quadratic inequality with solution $x \le 3$ or $x \ge 7$

Both methods give the same answer. Method 1 is usually shorter for a single fraction; Method 2 is convenient when the denominators are linear and the resulting polynomial factorises easily.

Fractions on both sides

When the inequality compares two fractions, Method 1 combines them over a common denominator. The denominators then appear as factors in the sign diagram.

A denominator that is always positive

If the denominator has no real zeros and a positive leading coefficient, it is positive for every $x$, and multiplying by it is safe. For example,

since $x^2 + 1 > 0$ for all $x$. So $(x - 3)(x + 2) \ge 0$

The same reasoning applies whenever the sign of the multiplier is known from the context or the domain. If the question states $x > 0$, multiplying by $x$ keeps the inequality.

Harder rational inequalities

Sometimes the numerator has a repeated root, and the same care with even multiplicity applies as for polynomials.

Combining graphical and analytic methods

A sketch of both sides shows the shape of the solution; algebra then gives the exact boundaries. When the intersections cannot be found exactly, a GDC gives them to 3 significant figures.

Inequalities in context

In a modelling problem the domain often fixes the sign of an expression, which makes multiplying safe, and the answer must be interpreted: a number of items is a whole number, a time is positive.

Exam-style questions on inequalities

The question that follows combines polynomial factorisation with a rational inequality, as is typical of Paper 1 (no GDC).

Self-check

• To do:

  I can solve an inequality graphically, using a GDC to find intersection points or zeros.

• To do:

  I can explain why multiplying by an expression of unknown sign can lose or add solutions.

• To do:

  I can find the critical values of a polynomial or rational inequality.

• To do:

  I can draw a sign diagram, including at roots of even multiplicity.

• To do:

  I can solve polynomial inequalities after factorising with the factor theorem.

• To do:

  I can solve rational inequalities by bringing everything to one side, or by multiplying by a square.

• To do:

  I can decide which critical values are included, and always exclude zeros of a denominator.

• To do:

  I can write a solution set with inequalities or interval notation and interpret it in context.