In short

A rational inequality like \frac{(x-1)(x+2)}{(x-3)} \geq 0 asks where a ratio of polynomials is positive, negative, or zero. You solve it by finding all critical points — the zeros of the numerator and the zeros of the denominator — marking them on the number line, and applying the sign scheme (wavy curve). The only new rule: denominator zeros are always excluded from the solution, because division by zero is undefined.

A courier company charges \frac{500}{x} rupees per parcel when x parcels are shipped together. A customer wants the per-parcel cost to stay below ₹40. That means solving \frac{500}{x} < 40, which rearranges to \frac{500 - 40x}{x} < 0. Both the numerator and the denominator can change sign, and x = 0 is off-limits. You need a method that handles all of this cleanly.

Polynomial inequalities involve a single polynomial: find the roots, draw the wavy curve, read off the sign. Rational inequalities involve a fraction \frac{p(x)}{q(x)}, which introduces a new complication — the denominator can be zero, creating points where the expression is undefined, not just zero. The sign scheme still works, but you need to treat these undefined points with care.

The core idea: a fraction's sign depends on numerator and denominator

A fraction \frac{A}{B} is positive when A and B have the same sign, and negative when they have opposite signs. So the sign of \frac{p(x)}{q(x)} changes at every point where either p(x) or q(x) changes sign — that is, at every root of p and every root of q.

These points — all the roots of the numerator and the denominator — are called the critical points of the rational expression. They play exactly the same role that roots play in polynomial inequalities: they divide the number line into intervals of constant sign.

Sign of a fraction depends on numerator and denominator signsA table showing four cases: numerator positive and denominator positive gives a positive fraction; numerator negative and denominator positive gives negative; numerator positive and denominator negative gives negative; numerator negative and denominator negative gives positive. Sign of p(x)/q(x) p > 0, q > 0 + p > 0, q < 0 − p < 0, q > 0 − p < 0, q < 0 +
A fraction is positive when numerator and denominator share the same sign, and negative when they have opposite signs.

The key insight: you do not need to track the numerator and denominator separately. The ratio \frac{p(x)}{q(x)} has the same sign as the product p(x) \cdot q(x), because multiplying numerator and denominator by q(x) gives \frac{p(x) \cdot q(x)}{[q(x)]^2}, and the denominator [q(x)]^2 is always positive (when it is nonzero). So the sign scheme for the rational expression is identical to the sign scheme for the product of all its linear factors — the same wavy curve you already know.

The method: critical points and the sign scheme

Critical points method for rational inequalities

To solve \frac{p(x)}{q(x)} > 0 (or \geq, <, \leq):

1. Move everything to one side so the other side is 0. Factor the numerator and denominator completely.

2. List all critical points: the roots of p(x) (where the expression equals zero) and the roots of q(x) (where the expression is undefined).

3. Mark them on the number line. Use open circles for denominator roots (always excluded) and use open or filled circles for numerator roots depending on strict or non-strict inequality.

4. Apply the wavy curve: determine the sign in the rightmost interval, then alternate at each critical point of odd multiplicity and keep the same sign at each critical point of even multiplicity.

5. Read off the solution intervals.

The sign-change rule at each critical point depends on multiplicity, exactly as in polynomial inequalities. A factor (x - a)^k causes a sign change when k is odd, and no change when k is even — regardless of whether a is a numerator root or a denominator root.

Step-by-step: a first example

Take \frac{x - 1}{x + 2} > 0. The numerator is zero at x = 1. The denominator is zero at x = -2. Both are simple (multiplicity 1).

Mark -2 and 1 on the number line. For large positive x, both factors are positive, so the fraction is positive in (1, \infty).

Moving left: at x = 1 (odd multiplicity), the sign flips to negative in (-2, 1). At x = -2 (odd multiplicity), the sign flips to positive in (-\infty, -2).

The inequality is strict (> 0), so exclude both critical points. The solution is (-\infty, -2) \cup (1, \infty).

Sign scheme for (x-1)/(x+2) greater than 0A number line with critical points at x = -2 (open circle, undefined) and x = 1 (open circle, excluded by strict inequality). Signs: positive for x less than -2, negative between -2 and 1, positive for x greater than 1. Solution is the union of the two positive intervals. (x − 1)/(x + 2) > 0 −∞ +∞ −2 (undef.) 1 (zero) + − + x ∈ (−∞, −2) ∪ (1, ∞)
The sign scheme for $\frac{x-1}{x+2}$. Both critical points are excluded: $x = 1$ because the inequality is strict, and $x = -2$ because the expression is undefined there.

The trap: never cross-multiply blindly

The most common error in rational inequalities is cross-multiplying as if you were solving an equation. When you see \frac{x-1}{x+2} > 0 and multiply both sides by (x+2), you assume (x+2) is positive. But (x+2) could be negative — and multiplying by a negative number flips the inequality.

Cross-multiplying splits the problem into two cases (one where the denominator is positive, one where it is negative). The sign scheme avoids this entirely: it handles both cases at once. Use the sign scheme.

Why cross-multiplying is dangerous in rational inequalitiesTwo paths shown: the wrong way (cross-multiply, get x minus 1 greater than 0, answer x greater than 1 which is incomplete) and the right way (sign scheme, which gives both intervals of the correct solution). Cross-multiply (wrong) (x−1)/(x+2) > 0 → x − 1 > 0 ??? → x > 1 Misses (−∞, −2) Assumed x + 2 > 0 Sign scheme (correct) Critical pts: −2, 1 Signs: +, −, + Solution: both + intervals (−∞, −2) ∪ (1, ∞) No sign assumption needed
Cross-multiplying by $(x+2)$ is only valid when $x + 2 > 0$. If you forget the other case ($x + 2 < 0$), you lose half the solution. The sign scheme handles both cases automatically.

Handling non-strict inequalities: when is equality allowed?

For \frac{p(x)}{q(x)} \geq 0, the expression equals zero when p(x) = 0 (numerator root), and it is undefined when q(x) = 0 (denominator root). So:

This is the one place where rational inequalities differ from polynomial inequalities. In a polynomial inequality p(x) \geq 0, you include all roots. In a rational inequality, you include only numerator roots.

Repeated factors and multiplicity

Repeated factors in the numerator or denominator follow the same multiplicity rule. Consider \frac{(x-1)^2}{(x+3)^3} \leq 0.

The critical points: x = 1 (numerator, even multiplicity 2) and x = -3 (denominator, odd multiplicity 3).

For large x, the expression is positive (both factors are positive and the overall sign is +). Moving left from (1, \infty):

The expression is \leq 0 in (-\infty, -3). At x = 1, the expression equals zero (numerator root), so include it. At x = -3, the expression is undefined, so exclude it.

Solution: (-\infty, -3) \cup \{1\}.

Sign scheme for (x-1)^2 / (x+3)^3 less than or equal to 0Number line with critical points at x = -3 (open circle, denominator root, odd multiplicity) and x = 1 (filled circle, numerator root, even multiplicity). Signs: negative for x less than -3, positive between -3 and 1, positive for x greater than 1. The sign does not change at x = 1 (even multiplicity). Solution is x in negative infinity to -3, union the isolated point 1. (x − 1)² / (x + 3)³ ≤ 0 −∞ +∞ −3 (denom, odd) 1 (numer, even) − + + no flip x ∈ (−∞, −3) ∪ {1} {1} included: numerator zero satisfies ≤ 0
At the even-multiplicity numerator root $x = 1$, the sign does not change — but the expression equals zero there, so it is included under $\leq 0$. The denominator root $x = -3$ is always excluded.

Interactive: explore a rational inequality

Drag the slider to change the value of x and watch the sign of \frac{(x-2)(x+1)}{(x-4)} update. The number line highlights positive and negative regions. Notice how the sign flips at each critical point.

Interactive sign explorer for (x-2)(x+1)/(x-4)A coordinate plane with the graph of y = (x-2)(x+1)/(x-4). A draggable point moves along the curve, and a readout shows the current x value and the sign of the expression. A vertical dashed line marks the asymptote at x = 4. x y x = 4 ↔ drag the red point
Move the red point along the curve. The expression is positive when the curve is above the $x$-axis and negative when below. At $x = 4$, the expression is undefined — the curve shoots off to $\pm\infty$.

Two worked examples

Example 1: Solve $\frac{(x - 3)(x + 1)}{(x - 5)} \leq 0$

Step 1. Identify the critical points. Numerator roots: x = 3 and x = -1 (both simple). Denominator root: x = 5 (simple).

Why: critical points are where the expression is zero (numerator roots) or undefined (denominator roots). Both types can cause sign changes.

Step 2. Determine the sign in the rightmost interval. For large x, all three factors are positive: (x-3) > 0, (x+1) > 0, (x-5) > 0. The expression is positive in (5, \infty).

Why: the leading behaviour of the rational expression is \frac{x^2}{x} = x, which is positive for large x.

Step 3. Apply the sign scheme moving left. All critical points have odd multiplicity (1), so the sign flips at every one:

  • (5, \infty): positive
  • (3, 5): negative (flip at x = 5)
  • (-1, 3): positive (flip at x = 3)
  • (-\infty, -1): negative (flip at x = -1)

Why: each simple factor changes sign once at its root, causing the overall expression to flip.

Step 4. Read the solution. The inequality is \leq 0, so take the negative intervals. Include numerator roots (where the expression is zero), but exclude the denominator root x = 5.

  • (-\infty, -1]: negative interval plus the numerator root x = -1.
  • [3, 5): negative interval plus the numerator root x = 3, but exclude x = 5.
x \in (-\infty, -1] \cup [3, 5)

Step 5. Verify. At x = -2: \frac{(-5)(-1)}{(-7)} = \frac{5}{-7} < 0. At x = 0: \frac{(-3)(1)}{(-5)} = \frac{-3}{-5} > 0. At x = 4: \frac{(1)(5)}{(-1)} = -5 < 0. At x = 6: \frac{(3)(7)}{(1)} = 21 > 0. All consistent.

Result: x \in (-\infty, -1] \cup [3, 5).

Number line solution of (x-3)(x+1)/(x-5) less than or equal to 0A number line with critical points at -1 (filled), 3 (filled), and 5 (open, denominator root). Signs from left to right: negative, positive, negative, positive. Solution intervals are highlighted: negative infinity to -1 inclusive, and 3 inclusive to 5 exclusive. (x − 3)(x + 1) / (x − 5) ≤ 0 −∞ +∞ −1 3 5 (undef.) − + − + x ∈ (−∞, −1] ∪ [3, 5) x = −1 and x = 3 included (≤); x = 5 excluded (undefined)
The solution to $\frac{(x-3)(x+1)}{(x-5)} \leq 0$. Filled circles at $x = -1$ and $x = 3$ mean those roots are included (the expression is zero there). The open circle at $x = 5$ means it is always excluded — the expression is undefined there.

The test points confirm every interval. Notice the asymmetry in the solution boundaries: -1 and 3 get square brackets (included), but 5 gets a round bracket (excluded). This is the signature of rational inequalities — numerator roots and denominator roots are treated differently.

Example 2: Solve $\frac{x + 2}{x^2 - 4x + 3} \geq 0$

Step 1. Factor the denominator: x^2 - 4x + 3 = (x-1)(x-3). The inequality becomes \frac{x+2}{(x-1)(x-3)} \geq 0.

Why: factoring reveals the critical points. You cannot apply the sign scheme to an unfactored expression.

Step 2. List the critical points. Numerator: x = -2 (simple). Denominator: x = 1 and x = 3 (both simple). Four intervals: (-\infty, -2), (-2, 1), (1, 3), (3, \infty).

Why: three critical points create four intervals. The sign is constant within each interval.

Step 3. Sign in the rightmost interval. For large x: (x+2) > 0, (x-1) > 0, (x-3) > 0. So the expression is positive in (3, \infty).

Step 4. Alternate signs (all multiplicities are odd):

  • (3, \infty): positive
  • (1, 3): negative
  • (-2, 1): positive
  • (-\infty, -2): negative

The inequality \geq 0 selects the positive intervals. Include the numerator root x = -2 (expression is zero). Exclude denominator roots x = 1 and x = 3.

x \in [-2, 1) \cup (3, \infty)

Result: x \in [-2, 1) \cup (3, \infty).

Sign scheme for (x+2)/((x-1)(x-3)) greater than or equal to 0Number line with critical points at x = -2 (filled, numerator), x = 1 (open, denominator), x = 3 (open, denominator). Signs from left to right: negative, positive, negative, positive. Solution intervals highlighted: from -2 inclusive to 1 exclusive, and from 3 exclusive to positive infinity. (x + 2) / ((x − 1)(x − 3)) ≥ 0 −∞ +∞ −2 (numer) 1 (denom) 3 (denom) − + − + x ∈ [−2, 1) ∪ (3, ∞) x = −2 included (zero); x = 1 and x = 3 excluded (undefined)
Three critical points, four intervals. The filled circle at $x = -2$ and open circles at $x = 1, 3$ reflect the different roles of numerator and denominator roots under the $\geq$ inequality.

The mixed brackets — [-2, 1) has one square and one round — tell the full story: x = -2 makes the expression zero (allowed by \geq), while x = 1 makes the denominator zero (never allowed). The graph of this function would show a vertical asymptote at x = 1 and another at x = 3, with the curve above the x-axis exactly in the highlighted intervals.

Common confusions

Going deeper

If you can factor the numerator and denominator, list critical points, apply the sign scheme with correct multiplicities, and remember to exclude denominator roots, you have the complete method for rational inequalities. What follows extends the technique.

Rational inequalities that do not start at zero

Many problems arrive in a form like \frac{x+1}{x-2} \leq 3. The first step is always to bring everything to one side:

\frac{x+1}{x-2} - 3 \leq 0 \implies \frac{x + 1 - 3(x-2)}{x - 2} \leq 0 \implies \frac{-2x + 7}{x - 2} \leq 0

Now the critical points are x = \frac{7}{2} (numerator) and x = 2 (denominator), and the sign scheme proceeds normally. Skipping this rearrangement step — and instead trying to "reason" about when \frac{x+1}{x-2} is bigger than 3 — is the source of most errors on exams.

Why the sign of \frac{p}{q} equals the sign of p \cdot q

Here is the precise justification. For any real number x where q(x) \neq 0:

\frac{p(x)}{q(x)} = \frac{p(x) \cdot q(x)}{[q(x)]^2}

The denominator [q(x)]^2 is strictly positive (it is a nonzero number squared). So the sign of the fraction equals the sign of p(x) \cdot q(x). This is why the sign scheme for the ratio \frac{p}{q} is identical to the sign scheme for the product p \cdot q — you can treat every factor the same, whether it comes from the numerator or the denominator.

Connection to polynomial inequalities

Every polynomial inequality p(x) > 0 is a special case of a rational inequality with denominator q(x) = 1. Going the other direction, every rational inequality \frac{p(x)}{q(x)} > 0 has the same sign pattern as the polynomial inequality p(x) \cdot q(x) > 0, with the extra constraint that the roots of q are excluded. The sign scheme is one unified tool that handles both.

Where this leads next