Product and rational inequalities | Algebra 2

Signs are all that matter

When is $$(x + 4)(x - 1)$$ negative? You don't need to multiply it out. A product is negative exactly when its factors have opposite signs, and positive when they have the same sign. So all you need is the sign of each factor.

A quotient $\frac{A}{B}$ follows the same sign rules as the product $A \cdot B$ , with one extra condition: the denominator can never be zero.

The sign chart method

ExampleSolve $$(x + 4)(x - 1) < 0$$

Critical values: $$x = -4$$ and $$x = 1$$ .

Factor	$$x < -4$$	$$-4 < x < 1$$	$$x > 1$$
$$x + 4$$	$$-$$	$$+$$	$$+$$
$$x - 1$$	$$-$$	$$-$$	$$+$$
product	$$+$$	$$-$$	$$+$$

We want negative:

-4 < x < 1 \qquad (-4, 1)

ExampleSolve $\frac{x - 3}{x + 2} \ge 0$

Critical values: $$3$$ (numerator) and $$-2$$ (denominator).

Factor	$$x < -2$$	$$-2 < x < 3$$	$$x > 3$$
$$x - 3$$	$$-$$	$$-$$	$$+$$
$$x + 2$$	$$-$$	$$+$$	$$+$$
quotient	$$+$$	$$-$$	$$+$$

We want positive or zero. The quotient is zero at $$x = 3$$ , so $$3$$ is included. At $$x = -2$$ it's undefined, so $$-2$$ is excluded.

x < -2 \quad \text{or} \quad x \ge 3 \qquad (-\infty, -2) \cup [3, \infty)

Don't multiply by the denominator

With rational equations you clear the denominators. With rational inequalities you can't, because the denominator contains $$x$$ : it's positive for some values and negative for others, and you don't know whether to flip the sign.

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

\frac{x + 5}{x - 1} - 2 \le 0 \quad \Rightarrow \quad \frac{x + 5 - 2(x - 1)}{x - 1} \le 0 \quad \Rightarrow \quad \frac{7 - x}{x - 1} \le 0

Critical values: $$7$$ and $$1$$ .

Factor	$$x < 1$$	$$1 < x < 7$$	$$x > 7$$
$$7 - x$$	$$+$$	$$+$$	$$-$$
$$x - 1$$	$$-$$	$$+$$	$$+$$
quotient	$$-$$	$$+$$	$$-$$

x < 1 \quad \text{or} \quad x \ge 7

Check $$x = 0$$ : $\frac{5}{-1} = -5 \le 2$ . True. The shortcut would have lost this whole interval.

Practice

Solve $$(2 - x)(x + 3) > 0$$ .

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Critical values: $$2$$ and $$-3$$ .

Factor	$$x < -3$$	$$-3 < x < 2$$	$$x > 2$$
$$2 - x$$	$$+$$	$$+$$	$$-$$
$$x + 3$$	$$-$$	$$+$$	$$+$$
product	$$-$$	$$+$$	$$-$$

Positive: $$-3 < x < 2$$ .

Practice

Solve $\frac{4}{x + 1} < 1$ .

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\frac{4}{x + 1} - 1 < 0 \quad \Rightarrow \quad \frac{4 - (x + 1)}{x + 1} < 0 \quad \Rightarrow \quad \frac{3 - x}{x + 1} < 0

Factor	$$x < -1$$	$$-1 < x < 3$$	$$x > 3$$
$$3 - x$$	$$+$$	$$+$$	$$-$$
$$x + 1$$	$$-$$	$$+$$	$$+$$
quotient	$$-$$	$$+$$	$$-$$

x < -1 \quad \text{or} \quad x > 3

Quick check

What is the solution of $\frac{x}{x + 3} < 0$ ?

Quick check

Why can't you solve $\frac{1}{x - 2} > 3$ by writing $$1 > 3(x - 2)$$ ?

Factor	$\(x < -4\)$	$\(-4 < x < 1\)$	$\(x > 1\)$
$\(x + 4\)$	$\(-\)$	$\(+\)$	$\(+\)$
$\(x - 1\)$	$\(-\)$	$\(-\)$	$\(+\)$
product	$\(+\)$	$\(-\)$	$\(+\)$

Factor	$\(x < -2\)$	$\(-2 < x < 3\)$	$\(x > 3\)$
$\(x - 3\)$	$\(-\)$	$\(-\)$	$\(+\)$
$\(x + 2\)$	$\(-\)$	$\(+\)$	$\(+\)$
quotient	$\(+\)$	$\(-\)$	$\(+\)$

Factor	$\(x < 1\)$	$\(1 < x < 7\)$	$\(x > 7\)$
$\(7 - x\)$	$\(+\)$	$\(+\)$	$\(-\)$
$\(x - 1\)$	$\(-\)$	$\(+\)$	$\(+\)$
quotient	$\(-\)$	$\(+\)$	$\(-\)$

Factor	$\(x < -3\)$	$\(-3 < x < 2\)$	$\(x > 2\)$
$\(2 - x\)$	$\(+\)$	$\(+\)$	$\(-\)$
$\(x + 3\)$	$\(-\)$	$\(+\)$	$\(+\)$
product	$\(-\)$	$\(+\)$	$\(-\)$

Factor	$\(x < -1\)$	$\(-1 < x < 3\)$	$\(x > 3\)$
$\(3 - x\)$	$\(+\)$	$\(+\)$	$\(-\)$
$\(x + 1\)$	$\(-\)$	$\(+\)$	$\(+\)$
quotient	$\(-\)$	$\(+\)$	$\(-\)$