Introduction

An inequality is used when a range of values satisfies a condition.

Example 1 In many contexts it makes more sense to use an interval of values rather than a specific value. E.g. a nutritionist may recommend that a person aim consume between 1200 and 1800 mg sodium per day; rather than exactly 1500 mg per day.

Example 2 We use an inequality to express each of the domain and range of a function:

Note that an empty point us used for ‘less than’ or ‘greater than’ but not equal to; a solid point is used to represent ‘less than or equal to‘ or ‘greater than or equal to‘.

Intervals on the number line

A single interval or inequality can be represented on a number line. Here are some examples:

Original Author Lew.W.S Original Applet link

Solve Linear Inequalities

Example 3 An inequality such as $2x+5 \le 13$ behaves in a very similar way to solving an equation. However instead of finding one single solution that satisfies the equation, we find the interval of solutions that satisfy the equation.

$\begin{align*}2x+5 &\le 13 \\[10 pt]2x &\le 8 \\[10 pt]x &\le 4\end{align}$

Any value of $x$ , as long as it is less than or equal to 4, will satisfy this inequality.

Let’s try $x=3$ : $2(3)+5=11$ It is true that $11 \le 13$ so this is correct.

Let’s try $x=4$ : $2(4)+5=13$ It is true that $13 \le 13$ so this is correct.

Let’s try $x=4.1$ This is not in our solution interval. $2(4.1)+5=13.2$ . This is not less than or equal to 13, and so 4.1 is not a solution.

To understand this result visually, we graph $y=2x+5$ (blue) and $y=13$ (green). Notice that the $y$ on the blue line is less than (below) the $y$ on the green line for all $x$ values up to $x=4$ .

Multiply or Divide by Negative

Example 4 An inequality can be solved in the same way as an equation, however we must take care when multiplying by a negative number.

First, let’s divide by a positive number. Solve

$\begin{align*}2x &> 10\\[10 pt]x&>5\end{align}$

Any value of $x$ greater than $5$ will satisfy $2x$ is greater than $10$ .

Example 5 Now let’s see what happens when both sides are negative. Solve

$-2x>-10$

Unlike the equations $2x=10$ and $-2x=-10$ ; the inequalities $2x>10$ and $-2x>-10$ do not yield the same solution.

Let’s solve:

$\begin{align*}-2x&>-10 \\[10 pt]0&>2x-10\\[10 pt]10&>2x\\[10 pt]5 &>x\\[10 pt]x&<5\end{align}$

The inequality $-2x>-10$ has solution $x<5$ .

The inequality $2x>10$ has solution $x>5$ .

When we multiply or divide by a negative number, the inequality sign is reversed.

Let’s solve again:

$\begin{align*}-2x &> -10\quad\quad\text{Multiply by -1, flip inequality}\\[10 pt]2x&<10\\[10 pt]x&<5\end{align}$

Compare these results graphically:

Your turn:

applet link

Solve Quadratic Inequalities

One way to solve a quadratic inequality is to sketch a graph of the parabola, paying particular attention to the $x$ -intercepts (critical values).

First, rearrange the inequality to the form $ax^2+bx+c\le 0$ or $ax^2+bx+c \ge 0$ . Then solve the corresponding equation $ax^2+bx+c=0$ . The solution interval(s) will one of:

a single interval between the two roots, $r_1\le x \le r_2$
two intervals, left of $r_1$ and right of $r_2$ : $x\le r_1$ or $x \ge r_2$
no solutions. The parabola does not cut the $x-axis$ and no values of $x$ satisfy the original equation;
$x \in \bbmath{R}$ . The parabola does not cut the $x-axis$ and all values of $x$ satisfy the original equation.

Sketch a graph of the related parabola to determine the correct solution set.

Example 6

Solve $x^2-3x-10\le 0$

Let’s factor to find $x$ -intercepts:

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

The $x$ intercepts satisfy $x-5=0$ or $x+2=0$ , that is, $x=5$ and $x=-2$ . Plot these two points and sketch an ‘upwards’ parabola.

Looking at the graph $y=x^2-3x-10$ we see that the $y$ coordinate is less than zero on the interval highlighted in red.

$x^2-3x-10&\le 0\Rightarrow -2\le x &\le 5$

Example 7

Solve $-3(x-4)^2+13\le 0$

We are given the vertex form of this parabola. Let’s plot the key features:

Vertex has coordinates $(4, 13)$

$y$ -intercept has coordinates $x=0$ , $y=-3(0-4)^2+13=-3(16)+13=-35$ .

To calculate $x$ -intercepts, set $y=0$ and solve. It is possible to solve this directly when the parabola is given in vertex form, as follows:

$\begin{align*}-3(x-4)^2+13&=0\\[10 pt]-3(x-4)^2&=-13\\[10 pt]3(x-4)^2&=13\\[10 pt](x-4)^2&=\dfrac{13}{3}\\[10 pt]x-4&=\pm \sqrt{\dfrac{13}{3}}\\[10 pt]x&=4\pm \sqrt{\dfrac{13}{3}}\\[10 pt]x=1.91, \,\, x&=6.08\end{align}$