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Rational Inequalities

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Rational Inequalities - Complete Interactive Lesson

Part 1: Why Rational Inequalities Are Different

➗ Rational Inequalities

Part 1 of 7 — Why Rational Inequalities Are Different

Topics in This Part

Section
What Is a Rational Inequality?
The Cross-Multiplication Trap
Two Kinds of "Critical" Points

🔑 Key Concept: A rational inequality compares a ratio of polynomials to $0$ — for example $\dfrac{x-1}{x+2} > 0$ . The whole challenge is that a fraction can flip sign in two ways: when its top hits zero, or when its bottom hits zero. Tracking both is what this lesson is about.

What Is a Rational Inequality?

A rational inequality is an inequality involving a rational expression — a polynomial divided by a polynomial — compared with $0$ (or that can be rearranged to compare with $0$ ):

$\frac{x-1}{x+2} > 0, \qquad \frac{x^2-4}{x-3} \le 0, \qquad \frac{5}{x} \ge 2$

The Cross-Multiplication Trap

With an equation like $\dfrac{x-1}{x+2} = 0$ you might clear the fraction by multiplying both sides by $x+2$ . With an , that move is .

Concept Check 🎯

Two Kinds of "Critical" Points

A rational expression can only change sign at a place where it equals $0$ or where it is undefined. These are the critical numbers (also called boundary points):

Zeros of the numerator → the expression equals $0$ here. These may be included (closed dot $\bullet$ ) if the inequality is $\le$ or $\ge$ .
Zeros of the denominator → the expression is here. These are (open dot ), no matter the inequality sign.

Find the Critical Numbers 🧮

For $\dfrac{x+5}{x-3}$ :

1) Numerator zero: $x = \,?$ 2) Denominator zero (undefined): How many intervals do these two points create on the number line?

Sort the Critical Numbers 🔽

For $\dfrac{x-1}{x+2}$ , the critical numbers are $1$ and $-2$ . Order them on the number line and identify each.

Part 1 Recap

A rational inequality compares $\dfrac{P(x)}{Q(x)}$ with $0$ .
Never cross-multiply by a variable expression — the sign is unknown.
Critical numbers come from the numerator ( $=0$ , maybe included) the denominator (undefined, always excluded).

Part 2: Finding Every Critical Number

➗ Rational Inequalities

Part 2 of 7 — Finding Every Critical Number

🔑 The Goal: Before any sign chart, you must list all the values where the numerator is $0$ and all the values where the denominator is $0$ . Miss one and your number line is wrong.

Factor First, Then Read Off the Zeros

The fastest route to the critical numbers is to factor the top and bottom completely, then set each factor to $0$ .

Example: $\dfrac{x^2 - 4}{x^2 - x - 6}$

Part 3: The Sign-Chart Method

➗ Rational Inequalities

Part 3 of 7 — The Sign-Chart Method

🔑 The Core Technique: Plot the critical numbers, pick a test point inside each interval, plug it into the factored fraction, and record whether the result is $+$ or $-$ . The sign is constant across each whole interval, so one test point decides the entire interval.

The Five Steps

To solve a rational inequality:

Get one side $= 0$ (Part 5 covers this; here the fraction is already vs. $0$ ).

Part 4: Endpoints & Writing the Solution

➗ Rational Inequalities

Part 4 of 7 — Endpoints & Writing the Solution

🔑 The Finishing Move: Once the sign chart is built, choosing the right intervals is easy. The subtle part is the endpoints — when to use $[\;]$ (included) vs. $(\;)$ (excluded), and how to write the answer in clean interval notation.

Endpoint Rules (memorize these)

Critical number is a…	Strict $<,\,>$

Part 5: Getting One Side to Zero

➗ Rational Inequalities

Part 5 of 7 — Getting One Side to Zero

🔑 The Setup Step: Real problems rarely arrive as "fraction $> 0$ ." You'll see things like $\dfrac{5}{x} \ge 2$ or . The : move to one side and combine into a single fraction compared with . Only then do you build the sign chart.

Part 6: Multiplicity & Repeated Factors

➗ Rational Inequalities

Part 6 of 7 — Multiplicity & Repeated Factors

🔑 The Subtlety: Usually the sign flips every time you cross a critical number. But a squared (even-power) factor like $(x-2)^2$ does not flip the sign — it just touches $0$ and bounces back. Knowing when the sign holds vs. flips saves you from a wrong chart.

Even vs. Odd Multiplicity

A factor's multiplicity is its exponent. As you cross its critical number:

Multiplicity

Part 7: Applications, Mixed Practice & Mastery Check

➗ Rational Inequalities

Part 7 of 7 — Applications, Mixed Practice & Mastery Check

You can now (1) find every critical number, (2) build a sign chart, (3) handle the endpoints, (4) clear fractions to one side, and (5) account for repeated factors. Let's apply it and then prove mastery.

Quick Reference

Step	What to do
1. One side $=0$	Move all terms left; combine over a common denominator. Never cross-multiply.
2. Factor	Factor numerator and denominator fully.
3. Critical numbers	Numerator zeros + denominator zeros, in increasing order.
4. Sign chart	Test one point per interval (or count negative factors).
5. Select & endpoints	Match the inequality's sign. Numerator zeros included only if $\le/\ge$ ; denominator zeros excluded.

A solution is the set of all $x$ -values that make the statement true. The answer is almost always a union of intervals, not a single number.

Expression	Top is $0$ when…	Bottom is $0$ when…
$\dfrac{x-1}{x+2}$	$x = 1$	$x = -2$
$\dfrac{x^2-4}{x-3}$	$x = 2,\,-2$
$\dfrac{5}{x}$	never	$x = 0$

⚠️ Domain first: Wherever the denominator is $0$ , the expression is undefined. Those $x$ -values can never be part of the solution, no matter what.

Why? Because multiplying both sides of an inequality by a negative number flips the inequality sign — and we usually don't know the sign of $x+2$ (it depends on $x$ !).

$\frac{x-1}{x+2} > 0 \;\;\overset{?}{\Longrightarrow}\;\; x - 1 > 0 \quad\text{(WRONG — assumes } x+2>0)$

⚠️ Never multiply an inequality by a variable expression unless you know its sign. Instead, we'll get one side to $0$ and read the sign of the fraction directly. That's the sign-chart method, coming in Part 3.

For $\dfrac{x-1}{x+2}$ , the critical numbers are $x = 1$ (a zero) and $x = -2$ (undefined). They split the number line into three pieces:

$(-\infty,\,-2),\qquad (-2,\,1),\qquad (1,\,\infty)$

💡 With $n$ critical numbers you get $n+1$ intervals to test. The sign of the fraction is constant on each interval — it can only switch at a critical number.

They cut the number line into intervals where the sign is constant.

Next, in Part 2, we'll nail down exactly how to find every critical number, including from factored quadratics.

\frac{x ^{2} - 4}{x ^{2} - x - 6}

$\frac{x^2-4}{x^2-x-6} = \frac{(x-2)(x+2)}{(x-3)(x+2)}$

Numerator $= 0$ : $\;x = 2$ or $x = -2$ .
Denominator $= 0$ : $\;x = 3$ or $x = -2$ .

⚠️ Watch the overlap! $x=-2$ makes both top and bottom $0$ . Because it makes the denominator $0$ , the expression is undefined there — it is always excluded, even though it's also a numerator zero. The denominator wins.

Concept Check 🎯

Listing and Ordering the Critical Numbers

Once you have every zero of the top and bottom, plot them in increasing order on a number line and mark each dot type:

$\bullet$ closed = included (numerator zero with $\le$ or $\ge$ )
$\circ$ open = excluded (denominator zero always; or any zero with strict $<$ or $>$ )

Example: $\dfrac{(x-2)(x+2)}{(x-3)(x+2)} \le 0$

Critical numbers in order: $-2,\; 2,\; 3$ .

$x$ -value	Source	Dot
$-2$	denominator $0$ (undefined)	$\circ$ open
$2$	numerator , and allows equality

💡 The dot type doesn't change which intervals solve the inequality — it only decides whether the endpoints themselves get included.

Classify the Endpoints 🔽

You're solving $\dfrac{x-4}{x+1} \ge 0$ . Decide each endpoint's dot type.

Collect All Critical Numbers 🧮

For $\dfrac{x^2 - 5x + 6}{x^2 - 1}$ , factor and list the critical numbers in increasing order.

Hint: $x^2-5x+6 = (x-2)(x-3)$ and .

1) Smallest critical number $= \,?$ 2) Next $= \,?$ 3) Next $= \,?$ 4) Largest $= \,?$

Factor the numerator and denominator completely.

Mark critical numbers on a number line (open

\circ

for denominator zeros, dot type for numerator zeros set by the inequality).

Test one point in each interval — record the sign of the fraction there.

Select the intervals whose sign matches the inequality, then attach the right endpoints.

Worked Example: $\dfrac{x-1}{x+2} > 0$

Critical numbers: $x = 1$ (zero), $x = -2$ (undefined). Intervals: $(-\infty,-2)$ , $(-2,1)$ , $(1,\infty)$ .

Interval	Test $x$	$x-1$	$x+2$	$\dfrac{x-1}{x+2}$
$(-\infty,-2)$	$-3$	$-$	$-$	$+$

We want $> 0$ (positive), so the solution is $(-\infty,-2)\,\cup\,(1,\infty)$ . Both endpoints are open: $-2$ is undefined, and $1$ is excluded because $>$ is strict.

Concept Check 🎯

A Faster Way to Read Signs

You don't have to recompute the whole fraction at each test point. Just count the negative factors:

An even number of negative factors → the fraction is positive $(+)$ .
An odd number of negative factors → the fraction is negative $(-)$ .

Example: $\dfrac{(x-1)(x-4)}{x+2}$ at $x = 0$

$x-1 = -1$ → negative
$x-4 = -4$ → negative
$x+2 = 2$ → positive

Two negative factors (even) → the fraction is positive at $x=0$ . ✓

💡 This factor-counting trick is the heart of the sign chart. As you cross each single critical number left to right, exactly one factor flips sign, so the overall sign of the fraction usually flips too.

Count the Negatives 🧮

For $\dfrac{(x+3)(x-1)}{(x-5)}$ , evaluate the sign at $x = 0$ .

1) How many of the three factors are negative at $x=0$ ? $\,?$ 2) Is the fraction positive or negative there? (type "positive" or "negative") $\,?$

Build the Sign Chart 🔽

For $\dfrac{x+1}{x-2}$ , critical numbers are $-1$ and $2$ . Fill in the sign of the fraction on each interval using a test point.

A denominator zero is always excluded — even with $\le$ or $\ge$ . You can't include an undefined point.
A numerator zero is included only when the inequality allows equality ( $\le$ or $\ge$ ), because there the fraction equals $0$ .

⚠️ The single most common mistake is writing $[\,]$ around a denominator zero. Undefined points get a parenthesis $(\,)$ every time.

Interval Notation Quick Reference

Notation	Meaning
$(a, b)$	$a < x < b$ (both ends excluded)
$[a, b]$	$a \le x \le b$ (both ends included)
$[a, b)$	$a \le x < b$ (left included, right not)
$(-\infty, a)$	all $x < a$
$\cup$	"or" — union of separate pieces

$\infty$ and $-\infty$ are never included, so they always get a parenthesis.

Worked Example: $\dfrac{x-1}{x+2} \ge 0$

Same sign chart as before ( $+,\,-,\,+$ across $-2$ and $1$ ). We want $\ge 0$ , so we take the positive intervals and the points where the fraction $=0$ .

Positive intervals: $(-\infty,-2)$ and $(1,\infty)$ .
$x=1$ is a numerator zero with $\ge$ → include it: the right piece becomes .

$\boxed{(-\infty,\,-2)\,\cup\,[1,\,\infty)}$

Concept Check 🎯

Full Worked Example, Start to Finish

Solve $\dfrac{x-3}{x+1} \le 0$ .

Step 1 — Critical numbers: numerator zero $x=3$ ; denominator zero $x=-1$ .

Step 2 — Sign chart across $-1$ and $3$ :

Interval	Test $x$	Sign of $\dfrac{x-3}{x+1}$
$(-$

Step 3 — Select: we want $\le 0$ (negative or zero) → the middle interval $(-1,3)$ .

Step 4 — Endpoints: $x=3$ is a numerator zero with $\le$ → include $\Rightarrow$ bracket. $x=-1$ is undefined → exclude $\Rightarrow$ parenthesis.

$\boxed{(-1,\,3]}$

Endpoint Decisions 🧮

A sign chart gives the negative interval between a numerator zero at $x = 6$ and a denominator zero at $x = 10$ . For each inequality, decide whether $x=6$ is included.

Type 1 for included (bracket) or 0 for excluded (parenthesis).

1) With $\dfrac{P(x)}{Q(x)} \le 0$ , is $x=6$ included? $\,?$ 2) With $\dfrac{P(x)}{Q(x)} < 0$ , is $x=6$ included? $\,?$ 3) Is $x=10$ (denominator zero) ever included? $\,?$

Write the Solution 🔽

The sign chart for $\dfrac{x-3}{x+1}$ is $+,\,-,\,+$ across critical numbers $-1$ (undefined) and $3$ (zero). Now answer for the inequality $\dfrac{x-3}{x+1} \ge 0$ .

\frac{1}{x - 1} < \frac{2}{x}

first move is always the same

Move, Don't Cross-Multiply

Example: $\dfrac{5}{x} \ge 2$

Wrong instinct: multiply both sides by $x$ . (We don't know the sign of $x$ !)

Right move: subtract $2$ and combine over a common denominator:

$\frac{5}{x} - 2 \ge 0 \;\Longrightarrow\; \frac{5}{x} - \frac{2x}{x} \ge 0 \;\Longrightarrow\; \frac{5 - 2x}{x} \ge 0$

Now it's a clean rational inequality vs. $0$ . Critical numbers: numerator $5-2x=0 \Rightarrow x = \tfrac{5}{2}$ ; denominator .

⚠️ A frequent error is to "simplify" $5 - 2x$ and forget it's now the numerator of a fraction. Keep the single-fraction form until the sign chart is done.

Combine Into One Fraction 🧮

Rewrite $\dfrac{5}{x} \ge 2$ as $\dfrac{5-2x}{x} \ge 0$ , then find its critical numbers.

1) Numerator zero (enter as the fraction $5/2$ ): $x = \,?$ 2) Denominator zero: $x = \,?$

Finish the Example

We had $\dfrac{5-2x}{x} \ge 0$ with critical numbers $0$ and $\tfrac{5}{2}$ . Sign chart:

Interval	Test $x$	$5-2x$	$x$	$\dfrac{5-2x}{x}$

We want $\ge 0$ (positive or zero) → the middle interval $(0,\,\tfrac{5}{2})$ , plus the zero.

$x=\tfrac{5}{2}$ is a numerator zero with $\ge$ → include (bracket).
$x=0$ is undefined → (parenthesis).

$\boxed{\left(0,\;\tfrac{5}{2}\right]}$

💡 Sanity check $x=1$ : $\dfrac{5}{1} = 5 \ge 2$ ✓. And : ✓ (correctly excluded).

Two Fractions: Same Idea

Example: $\dfrac{1}{x-1} < \dfrac{2}{x}$

Move everything left and use the common denominator $x(x-1)$ :

$\frac{1}{x-1} - \frac{2}{x} < 0 \;\Longrightarrow\; \frac{x - 2(x-1)}{x(x-1)} < 0 \;\Longrightarrow\; \frac{x - 2x + 2}{x(x-1)} < 0 \;\Longrightarrow\; \frac{-x + 2}{x(x-1)} < 0$

So $\dfrac{2 - x}{x(x-1)} < 0$ . Critical numbers: numerator $x=2$ ; denominators and .

⚠️ Don't forget a denominator factor! Both $x$ and $x-1$ create undefined points. Three critical numbers here means four intervals to test.

Concept Check 🎯

Finish the Two-Fraction Problem 🔽

Solve $\dfrac{2-x}{x(x-1)} < 0$ . Critical numbers in order: $0,\,1,\,2$ . Fill in each interval's sign (test points $x=-1,\,0.5,\,1.5,\,3$ ).

Why? An even power is never negative, so it can't change the fraction's sign — it only contributes a $0$ at that point.

Example: $\dfrac{(x-1)^2 (x-4)}{x+2}$

Crossing left to right, the sign flips at $x=-2$ (mult. 1), holds at $x=1$ (mult. 2, even — bounce), and flips at $x=4$ (mult. 1).

Concept Check 🎯

Worked Example With a Repeated Factor

Solve $\dfrac{(x-1)^2 (x-4)}{x+2} \ge 0$ .

Critical numbers in order: $-2$ (undefined), $1$ (numerator zero, mult. 2), $4$ (numerator zero, mult. 1).

Start at the far right $(4,\infty)$ and work left, flipping only at odd-multiplicity points.

Interval	Sign	Reason
$(4,\infty)$	$+$	test $x=5$ : $\frac{(+) (+)}{(+}$

We want $\ge 0$ : positive intervals $(-\infty,-2)$ and $(4,\infty)$ , plus zeros.

$x=1$ : numerator zero, fraction $=0$ , and $\ge$ → include the single point $\{1\}$ .
$x=4$ : numerator zero with → → .

$\boxed{(-\infty,\,-2)\,\cup\,\{1\}\,\cup\,[4,\,\infty)}$

💡 The lone point $\{1\}$ surprises students. The fraction equals $0$ there, $0 \ge 0$ is true, so $1$ is a solution even though both neighboring intervals are negative.

Multiplicity Drill 🧮

For $\dfrac{(x+3)^2 (x-1)}{x-5}$ , classify the behavior at each critical number.

1) At $x=-3$ , does the sign flip? (type "yes" or "no") $\,?$ 2) At $x=1$ , does the sign flip? (type "yes" or "no") $\,?$ 3) At , does the sign flip? (type "yes" or "no")

Apply It 🔽

Using the worked sign chart $+,\,-,\,-,\,+$ (left to right across $-2$ , $1$ , $4$ ) for $\dfrac{(x-1)^2(x-4)}{x+2} \le 0$ , choose the solution pieces.

⚠️ Two recurring traps: (1) cross-multiplying by a variable, and (2) bracketing a denominator zero. Avoid both and you'll be right almost every time.

A Real Application: Average Cost

A company's average cost per unit (in dollars) for producing $x$ units is

$A(x) = \frac{2x + 500}{x}, \qquad x > 0.$

Question: For how many units is the average cost below $6 per unit?

Set up the inequality and move everything to one side:

$\frac{2x+500}{x} < 6 \;\Longrightarrow\; \frac{2x+500}{x} - 6 < 0 \;\Longrightarrow\; \frac{2x+500 - 6x}{x} < 0 \;\Longrightarrow\; \frac{500 - 4x}{x} < 0$

Critical numbers: numerator $500-4x=0 \Rightarrow x = 125$ ; denominator $x=0$ . Since the context requires $x>0$ , test on and :

$(0,125)$ , test $x=100$ : $\dfrac{500-400}{100} = \dfrac{100}{100} = +$ → not .

So the average cost dips below $6 once more than 125 units are produced: $x > 125$ .

Application Drill 🧮

Using $A(x) = \dfrac{2x+500}{x}$ from above:

1) At exactly $x = 125$ units, what is the average cost in dollars? $\,?$ 2) The cost falls below $6 when $x$ is greater than what number? $\,?$

Before You Start the Final Practice

A quick pre-flight checklist for every rational inequality:

🔑 The 5-Point Check

Is one side $0$ ? If not, move and combine — don't cross-multiply.

Factor top and bottom; list all critical numbers in order.

Mark denominator zeros as always open.

Build the sign chart with test points (or count negatives).

Select matching intervals; decide endpoints one at a time.

💡 If your final answer brackets a denominator zero, you made an error — back up and re-check step 5.

Mixed Practice 🎯

Exit Quiz ✅

Answer all three to finish the lesson.

x^{2} - 1 = (x - 1) (x + 1)

x=-2

is a denominator zero → exclude it: keep the parenthesis.

\dfrac{5}{4} = 1.25 \not\ge 2

x(x−1)x−2(x−1)<

\frac{( + ) ( + )}{( + )} = +

x=-2

: undefined → exclude.

(125,\infty)

, test

x=200

\dfrac{500-800}{200} = \dfrac{-300}{200} = -

→ satisfies

<0

. ✓

Numerator zero (value $=0$ )	exclude $\circ$	include $\bullet$
Denominator zero (undefined)	exclude $\circ$	exclude $\circ$

Odd ( $1, 3, \dots$ )	$(x-2),\ (x-2)^3$	sign flips ( $+\leftrightarrow-$ )
Even ( $2, 4, \dots$ )	$(x-2)^2,\ (x-2)^4$

$(-\infty,0)$	$-1$	$7\,(+)$	$-1\,(-)$	$-$
$(0,\,5/2)$	$1$	$3\,(+)$	$1\,(+)$
$(5/2,\infty)$	$3$	$-1\,(-)$	$3\,(+)$

Rational Inequalities

Rational Inequalities - Complete Interactive Lesson

Part 1: Why Rational Inequalities Are Different

➗ Rational Inequalities

Topics in This Part

What Is a Rational Inequality?

The Cross-Multiplication Trap

Two Kinds of "Critical" Points

Part 1 Recap

Part 2: Finding Every Critical Number

➗ Rational Inequalities

Factor First, Then Read Off the Zeros

Example: x2−4x2−x−6\dfrac{x^2 - 4}{x^2 - x - 6}

Part 3: The Sign-Chart Method

➗ Rational Inequalities

The Five Steps

Part 4: Endpoints & Writing the Solution

➗ Rational Inequalities

Endpoint Rules (memorize these)

Part 5: Getting One Side to Zero

➗ Rational Inequalities

Part 6: Multiplicity & Repeated Factors

➗ Rational Inequalities

Even vs. Odd Multiplicity

Part 7: Applications, Mixed Practice & Mastery Check

➗ Rational Inequalities

Quick Reference

Listing and Ordering the Critical Numbers

Example: (x−2)(x+2)(x−3)(x+2)≤0\dfrac{(x-2)(x+2)}{(x-3)(x+2)} \le 0(x−3)(x+2)(x−2)(x+2)​

Worked Example: x−1x+2>0\dfrac{x-1}{x+2} > 0x+2x−1​>0

A Faster Way to Read Signs

Example: (x−1)(x−4)x+2\dfrac{(x-1)(x-4)}{x+2}x+2(x−1)(x−4)​ at x=0x = 0x=0

Interval Notation Quick Reference

Worked Example: x−1x+2≥0\dfrac{x-1}{x+2} \ge 0x+2x−1​≥0

Full Worked Example, Start to Finish

Move, Don't Cross-Multiply

Example: 5x≥2\dfrac{5}{x} \ge 2x5​≥2

Finish the Example

Two Fractions: Same Idea

Example: 1x−1<2x\dfrac{1}{x-1} < \dfrac{2}{x}x−11​<x2​

Example: (x−1)2(x−4)x+2\dfrac{(x-1)^2 (x-4)}{x+2}x+2(x−1)2(x−4)​

Worked Example With a Repeated Factor

A Real Application: Average Cost

Before You Start the Final Practice

Example: $\dfrac{x^2 - 4}{x^2 - x - 6}$

Example: $\dfrac{(x-2)(x+2)}{(x-3)(x+2)} \le 0$

Worked Example: $\dfrac{x-1}{x+2} > 0$

Example: $\dfrac{(x-1)(x-4)}{x+2}$ at $x = 0$

Worked Example: $\dfrac{x-1}{x+2} \ge 0$

Example: $\dfrac{5}{x} \ge 2$

Example: $\dfrac{1}{x-1} < \dfrac{2}{x}$

Example: $\dfrac{(x-1)^2 (x-4)}{x+2}$