A rational inequality involves a rational expression (fraction with polynomials) and an inequality sign.

Examples:

• $\frac{x + 1}{x - 2} > 0$
• $\frac{x^2 - 4}{x + 3} \leq 0$
• $\frac{2x}{x^2 - 9} \geq 1$

Key Differences from Polynomial Inequalities

Critical values come from TWO sources:

1. Zeros of the numerator (where the expression equals 0)
2. Zeros of the denominator (vertical asymptotes - where expression is undefined)

IMPORTANT: Values that make the denominator zero are NEVER included in the solution, even with $\leq$ or $\geq$.

Solution Strategy

Step-by-Step Process

Step 1: Move everything to one side

Get the inequality in the form $\frac{P(x)}{Q(x)} \diamond 0$ where $\diamond$ is $<$, $>$, $\leq$, or $\geq$.

Warning: Never multiply both sides by the denominator (you don't know if it's positive or negative).

Step 2: Find critical values

• Numerator zeros: Set $P(x) = 0$ and solve
• Denominator zeros: Set $Q(x) = 0$ and solve

Step 3: Create intervals

The critical values divide the number line into regions.

Step 4: Make a sign chart

Test a point from each interval to determine the sign of the rational expression.

Step 5: Identify solution intervals

• For $> 0$ or $\geq 0$: positive intervals
• For $< 0$ or $\leq 0$: negative intervals

Step 6: Check endpoints

• Include numerator zeros if using $\leq$ or $\geq$
• Never include denominator zeros (always undefined)

Sign Chart for Rational Expressions

For $\frac{(x - a)(x - b)}{(x - c)}$ where $a < b < c$:

Note: At $x = c$, the expression is undefined (vertical asymptote).

Common Mistake to Avoid

WRONG: Multiplying by the denominator without considering its sign

Example of wrong approach: $\frac{x + 1}{x - 2} > 0$ Wrong: Multiply by $(x - 2)$ to get $x + 1 > 0$

Why it's wrong: If $x - 2 < 0$, multiplying reverses the inequality!

RIGHT: Use sign analysis on the rational expression as-is.

Converting to Standard Form

If the inequality is not in the form $\frac{P(x)}{Q(x)} \diamond 0$, rearrange:

Example: $\frac{x}{x - 1} \geq 2$

Move everything to one side: $\frac{x}{x - 1} - 2 \geq 0$

Get common denominator: $\frac{x - 2(x - 1)}{x - 1} \geq 0$ $\frac{x - 2x + 2}{x - 1} \geq 0$ $\frac{-x + 2}{x - 1} \geq 0$

Now analyze this rational expression.

Special Considerations

Numerator Zeros vs. Denominator Zeros

• Numerator zero: Expression equals 0 (may be included in solution)
• Denominator zero: Expression is undefined (never in solution)

Use different notation:

• Open circle ○ for denominator zeros (excluded)
• Closed circle ● for numerator zeros with $\leq$ or $\geq$ (included)

Multiple Factors

Apply the same multiplicity rules as with polynomials:

• Odd multiplicity: sign changes
• Even multiplicity: sign stays the same

Simplifying First

Be careful: Canceling common factors can eliminate critical values!

Example: $\frac{(x - 2)(x + 1)}{x - 2} \geq 0$

If you cancel $(x - 2)$, you get $x + 1 \geq 0$, which gives $x \geq -1$.

But: $x = 2$ must be excluded (makes original denominator 0).

Correct answer: $[-1, 2) \cup (2, \infty)$

Graphical Interpretation

The graph of $y = \frac{P(x)}{Q(x)}$ has:

• Zeros at numerator zeros (crosses or touches x-axis)
• Vertical asymptotes at denominator zeros

Solution to $\frac{P(x)}{Q(x)} > 0$: where graph is above x-axis Solution to $\frac{P(x)}{Q(x)} < 0$: where graph is below x-axis

Denominator zero: $x - 2 = 0 \Rightarrow x = 2$

Critical values: $x = -1, 2$

Step 2: Create intervals

The critical values divide the number line: $(-\infty, -1)$, $(-1, 2)$, $(2, \infty)$

Step 3: Make a sign chart

Step 4: Identify where expression $> 0$

We need positive intervals:

• $(-\infty, -1)$: positive ✓
• $(2, \infty)$: positive ✓

Step 5: Check endpoints

• At $x = -1$: Expression equals 0, but we have $>$ (strict), so exclude
• At $x = 2$: Expression is undefined, so exclude

Answer: $(-\infty, -1) \cup (2, \infty)$

Verification:

• At $x = -2$: $\frac{-2+1}{-2-2} = \frac{-1}{-4} = \frac{1}{4} > 0$ ✓
• At $x = 0$: $\frac{0+1}{0-2} = \frac{1}{-2} = -\frac{1}{2} < 0$ ✓
• At $x = 3$: $\frac{3+1}{3-2} = \frac{4}{1} = 4 > 0$ ✓