Rational Inequalities

➗ Rational Inequalities

Part 1 of 7 — Why Rational Inequalities Are Different

Topics in This Part

🔑 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 $0$):

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

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.

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

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 inequality, that move is dangerous.

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.

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 undefined here. These are always excluded (open dot $\circ$), no matter the inequality sign.

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.

Find the Critical Numbers 🧮

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

1) Numerator zero: $x = \,?$ 2) Denominator zero (undefined): $x = \,?$ 3) 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) and the denominator (undefined, always excluded).
• 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.

➗ 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: $\dfrac{x^2 - 4}{x^2 - x - 6}$

Factor both:

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

• 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: $-2,\; 2,\; 3$.

💡 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 $x^2-1 = (x-1)(x+1)$.

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

➗ 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:

1. Get one side $= 0$ (Part 5 covers this; here the fraction is already vs. $0$).
2. Factor the numerator and denominator completely.
3. Mark critical numbers on a number line (open $\circ$ for denominator zeros, dot type for numerator zeros set by the inequality).
4. Test one point in each interval — record the sign of the fraction there.
5. 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: $(-\infty,-2)$, $(-2,1)$, $(1,\infty)$.

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.

➗ 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)

The key takeaways:

• 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

$\infty$ 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 $[1,\infty)$.
• $x=-2$ is a denominator zero → exclude it: keep the parenthesis.

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

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 $3$:

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

➗ 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 $\dfrac{5}{x} \ge 2$, where a fraction is compared with a number instead of $0$. The first move is always the same: get a single rational expression on one side and $0$ on the other. Only then do you build the sign chart.

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 $x = 0$.

⚠️ 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:

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 → exclude (parenthesis).

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

💡 Sanity check $x=1$: $\dfrac{5}{1} = 5 \ge 2$ ✓. And $x=4$: $\dfrac{5}{4} = 1.25 \not\ge 2$ ✓ (correctly excluded).

Several Factors in One Expression

Example: $\dfrac{2 - x}{x(x-1)} < 0$

This inequality already has one rational expression on the left and $0$ on the right, so go straight to the critical numbers. Set every factor of the numerator and the denominator equal to $0$:

Three critical numbers split the number line into four intervals: $(-\infty, 0)$, $(0, 1)$, $(1, 2)$ and $(2, \infty)$. The sign of the expression can change only at these three numbers, so one test value per interval tells you the sign of the whole interval.

⚠️ Don't forget a denominator factor! Both $x$ and $x-1$ create undefined points, and each one is a boundary on the sign chart even though it can never be in the solution.

💡 AP Precalculus rational inequalities give you one rational expression, or one fraction against a constant; you will not need to combine two fractions over a common denominator.

Concept Check 🎯

Finish the Sign Chart 🔽

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

➗ 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:

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.

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 $\ge$ → include → $[4,\infty)$.
• $x=-2$: undefined → exclude.

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

💡 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 $x=5$, 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.

➗ 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

⚠️ 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 $(0,125)$ and $(125,\infty)$:

• $(0,125)$, test $x=100$: $\dfrac{500-400}{100} = \dfrac{100}{100} = +$ → not $<0$.
• $(125,\infty)$, test $x=200$: $\dfrac{500-800}{200} = \dfrac{-300}{200} = -$ → satisfies $<0$. ✓

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

1. Is one side $0$? If not, move and combine — don't cross-multiply.
2. Factor top and bottom; list all critical numbers in order.
3. Mark denominator zeros as always open.
4. Build the sign chart with test points (or count negatives).
5. 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.