Simultaneous Linear Inequalities

Lessons

Senior 3 · Mathematics

Simultaneous Linear Inequalities

4 lessons

Lesson 2 of 4

Tables of signs: product and quotient inequalities

25 min

When is a product negative?

Take the expression $(x - 1) (x - 4)$ and try some values of $x$ :

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$x - 1$	$- 1$	$0$	$1$	$2$	$3$	$4$
$x - 4$	$- 4$	$- 3$	$- 2$	$- 1$	$0$	$1$
$(x - 1) (x - 4)$	$4$	$0$	$- 2$	$- 2$	$0$	$4$

The product is negative only between 1 and 4, where the two factors have opposite signs. Its sign can change only where a factor is zero. The graph of $y = (x - 1) (x - 4)$ shows the same thing: the curve lies below the $x$ -axis exactly for $1 < x < 4$ .

The inequality (x − 1)(x − 4) < 0 asks where the curve is below the x-axis.

This lesson gives two exact methods for inequalities such as $(x - 1) (x - 4) < 0$ and $\frac{x + 1}{x - 3} \ge 0$ : splitting into cases (the analytical method) and the table of signs.

Definition

Binomial of the first degree: an expression $a x + b$ with $a \ne 0$ . Its root (or zero) is the value of $x$ that makes it zero:

ax + b = 0 \quad\Longleftrightarrow\quad x = -\frac{b}{a}.

Example: $2 x - 6$ has root $x = 3$ ; $5 - x$ has root $x = 5$ .

Important

The sign of $a x + b$ . To the right of its root, $a x + b$ has the same sign as $a$ . To the left of its root, it has the opposite sign. At the root it is 0.

Why: with $x_0 = -\frac{b}{a}$ , you can write $a x + b = a (x - x_{0})$ . To the right of the root, $x - x_{0} > 0$ , so the product has the sign of $a$ ; to the left, $x - x_{0} < 0$ , so the sign is reversed.

Definition

Table of signs: a table that lists the roots of the factors in increasing order across the top, and shows the sign ( $+$ , $-$ or $0$ ) of each factor, and of the whole expression, in each interval between them.

Example: the table of signs of the binomials $2 x - 6$ (here $a = 2 > 0$ ) and $5 - x$ (here $a = - 1 < 0$ ):

$x$	$x < 3$	$3$	$3 < x < 5$	$5$	$x > 5$
$2 x - 6$	$-$	$0$	$+$	$+$	$+$
$5 - x$	$+$	$+$	$+$	$0$	$-$

Method 1: the analytical method (cases)

Use the rule of signs. A product $A B$ is

positive when $A$ and $B$ have the same sign: ( $A > 0$ and $B > 0$ ) or ( $A < 0$ and $B < 0$ );
negative when they have opposite signs: ( $A > 0$ and $B < 0$ ) or ( $A < 0$ and $B > 0$ ).

Each case is a system of two inequalities in one unknown, which you solved in Lesson 1. The answer is the union of the two cases. A quotient $\frac{A}{B}$ follows the same sign rule, but $B \ne 0$ .

Worked example 1: the analytical method

Solve $(x + 2)(x - 5) \le 0$ .

Solution A product is $\le 0$ when the factors have opposite signs, or one of them is zero.

Case 1: $x + 2 \ge 0$ and $x - 5 \le 0$ , so $x \ge -2$ and $x \le 5$ : that is, $-2 \le x \le 5$ .

Case 2: $x + 2 \le 0$ and $x - 5 \ge 0$ , so $x \le -2$ and $x \ge 5$ : no number does both, so $\varnothing$ .

Union of the cases: $S = [- 2, 5]$ . Check $x = 0$ : $(2)(-5) = -10 \le 0$ ✓. Check $x = 6$ , outside: $(8) (1) = 8$ , not $\le 0$ ✓.

Method 2: the table of signs

Move every term to one side, so the other side is 0.
Write the expression as a product or quotient of first-degree binomials.
Find the root of each binomial and write the roots in increasing order.
Fill in the sign of each factor in each interval, using the rule above.
Multiply the signs down each column to get the sign of the whole expression. For a quotient, mark the root of the denominator "undefined".
Read off the intervals with the sign you want. Include a root only if the inequality allows 0 ( $\leq \le$ or $\geq \ge$ ) and it is not a root of the denominator.

Worked example 2: a product

Solve $(2x - 3)(4 - x) \ge 0$ .

Solution Roots: $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$ , and $4 - x = 0 \Rightarrow x = 4$ .

$x$	$x < \frac{3}{2}$	$\frac{3}{2}$	$\frac{3}{2} < x < 4$	$4$	$x > 4$
$2 x - 3$	$-$	$0$	$+$	$+$	$+$
$4 - x$	$+$	$+$	$+$	$0$	$-$
$(2 x - 3) (4 - x)$	$-$	$0$	$+$	$0$	$-$

The product is $\ge 0$ in the middle interval and at both roots: $S = \left[\frac{3}{2}, 4\right]$ .

Check $x = 2$ : $(1)(2) = 2 \ge 0$ ✓.

Worked example 3: a quotient

Solve $\frac{x + 1}{x - 3} \ge 0$ .

Solution Roots: $x = - 1$ (numerator) and $x = 3$ (denominator, where the quotient is undefined).

$x$	$x < - 1$	$- 1$	$- 1 < x < 3$	$3$	$x > 3$
$x + 1$	$-$	$0$	$+$	$+$	$+$
$x - 3$	$-$	$-$	$-$	$0$	$+$
$\frac{x + 1}{x - 3}$	$+$	$0$	$-$	undefined	$+$

The quotient is $\ge 0$ for $x \le -1$ (the 0 at $- 1$ is allowed) and for $x > 3$ (3 is never allowed):

S = {]}{-}\infty, -1] \;\cup\; ]3, +\infty[.

Warning

Never multiply both sides by the denominator. In $\frac{x + 1}{x - 3} \ge 0$ , multiplying by $x - 3$ keeps the sign only when $x - 3 > 0$ . You do not know its sign, so you cannot tell whether to reverse. Use the table of signs instead.

Never include the root of a denominator, even with $\leq \le$ or $\geq \ge$ : there the expression has no value.

Worked example 4: make one side zero first

Solve $\frac{x + 5}{x - 1} \le 2$ .

Solution Subtract 2 and write one fraction:

\frac{x + 5}{x - 1} - 2 \le 0 \;\Longrightarrow\; \frac{x + 5 - 2(x - 1)}{x - 1} \le 0 \;\Longrightarrow\; \frac{7 - x}{x - 1} \le 0.

$x$	$x < 1$	$1$	$1 < x < 7$	$7$	$x > 7$
$7 - x$	$+$	$+$	$+$	$0$	$-$
$x - 1$	$-$	$0$	$+$	$+$	$+$
$\frac{7 - x}{x - 1}$	$-$	undefined	$+$	$0$	$-$

So $S = {]}{-}\infty, 1[\; \cup\; [7, +\infty[$ .

Check: $x = 7$ gives $\frac{12}{6} = 2 \le 2$ ✓; $x = 2$ gives $\frac{7}{1} = 7$ , which is not $\le 2$ ✓, so the middle interval is rightly left out.

Tip

Test one number per interval. After you fill in the table, substitute one easy number from each interval into the original inequality. If one check disagrees with your table, a sign is wrong somewhere.

Khan Academy: Quadratic inequalities. Watch how the quadratic is factorised into two binomials and how the signs of the factors decide where the product is positive or negative. This is the analytical method of this lesson.

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Practice

Solve each inequality. Use the analytical method for question 2 and a table of signs for the rest. The last three are harder.

Draw the table of signs of (a) $3 x - 6$ (b) $5 - x$ , each on its own.
$(x - 2) (x + 3) > 0$
$(x + 4)(x - 1) \le 0$
$(3 - x) (2 x + 1) < 0$
$\frac{x - 5}{x + 2} \ge 0$
$\frac{2x - 1}{x - 3} < 0$
$\frac{x + 3}{x - 2} \le 1$
$x^{2} < 5 x$

Show answersHide answers

(a) root 2: $-$ for $x < 2$ , 0 at 2, $+$ for $x > 2$ . (b) root 5: $+$ for $x < 5$ , 0 at 5, $-$ for $x > 5$ .
Same signs: ( $x > 2$ and $x > - 3$ ) or ( $x < 2$ and $x < - 3$ ), so $x > 2$ or $x < - 3$ : $]{-}\infty, -3[\; \cup\; ]2, +\infty[$ .
$[- 4, 1]$ .
Roots $-\frac{1}{2}$ and 3; the product is negative outside them: $x < -\frac{1}{2}$ or $x > 3$ .
$x < - 2$ or $x \ge 5$ : $]{-}\infty, -2[\; \cup\; [5, +\infty[$ .
$\frac{1}{2} < x < 3$ .
$\frac{x + 3 - (x - 2)}{x - 2} = \frac{5}{x - 2} \le 0$ . The numerator is always positive, so $x - 2 < 0$ : $x < 2$ .
$x^{2} - 5 x < 0$ , so $x (x - 5) < 0$ : $0 < x < 5$ .

Sources and credits

Written by Ganzaa.
Video: Quadratic inequalities, Khan Academy, embedded from YouTube.

The tables of signs, the graph, the worked examples, the practice items and the quiz questions were written by Ganzaa.

1When is a product negative?

2Method 1: the analytical method (cases)

3Method 2: the table of signs

When is a product negative?

Method 1: the analytical method (cases)

Method 2: the table of signs