5. Inequalities


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


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:

  1. a single interval between the two roots, r_1\le x \le r_2
  2. two intervals, left of r_1 and right of r_2: x\le r_1 or x \ge r_2
  3. no solutions. The parabola does not cut the x-axis and no values of x satisfy the original equation;
  4. 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:


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}

Plot these five points and draw the parabola.

We can now read that the values of -3(x-4)^2+13\le 0  when x\le1.91 or x\ge 6.08.

Example 8

Solve x^2-2x+7< 7x-13

Let’s begin by rearranging so that one side is zero:

    \begin{align*}x^2-2x+7 &< 7x-13 \\[10 pt]x^2-9x+7 &< -13\\[10 pt]x^2-9x+20&<0\end{align}

If it is a good day, the quadratic expression will factor:


Sketch the parabola, paying attention to the x-intercepts to see the solution interval  is 4 <x<5.


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