Rational Inequality Solver

Math Algebra • Inequalities

Written by STEM Calculators Team Published December 12, 2025 Updated February 24, 2026

Enter numerator and denominator coefficients (up to quadratic), pick the inequality, set \(c\), then press Calculate.

Inequality sign: Right-hand side constant \(c\): Options:

Show algebraic steps Show sign chart Show equivalent “no-fraction” form

Numerator \(P(x) = a_2 x^2 + a_1 x + a_0\)

\(a_2\): \(a_1\): \(a_0\):

Denominator \(Q(x) = b_2 x^2 + b_1 x + b_0\) (not identically zero)

\(b_2\): \(b_1\): \(b_0\):

Critical points: real zeros of \(R(x)\) (possible equality points) and zeros of \(Q(x)\) (excluded; vertical asymptotes/holes).

Ready

Enter a rational inequality and press Calculate.

Graph of \(y=\dfrac{P(x)}{Q(x)}\) (blue) and \(y=c\) (orange). Green segments on the \(x\)-axis show the solution; red dashed lines mark vertical asymptotes.

Number line: green segments are solution intervals; closed dots are included endpoints; open dots are excluded; red dashed ticks mark asymptotes.

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Theory: Solving Rational Inequalities

Sign-chart method (robust and visual)

Problem. Solve \(\displaystyle \frac{P(x)}{Q(x)} \,\square\, c\) with \(\square \in \{<,\le,>,\ge\}\) and \(Q\not\equiv 0\).

1) Move to zero on the right.

\[ \frac{P(x)}{Q(x)} \,\square\, c \iff \frac{P(x)-c\,Q(x)}{Q(x)} \,\square\, 0. \]

Let \(R(x)=P(x)-c\,Q(x)\). The sign of the fraction is the product of the signs of \(R\) and \(Q\).

2) Critical points and domain.

Zeros of \(R\): the fraction can be \(0\) (include them only for \(\le,\ge\) and when \(Q\neq 0\)).
Zeros of \(Q\): the fraction is undefined; exclude them (vertical asymptotes or removable holes).

List the real zeros of \(R\) and \(Q\). They split \(\mathbb{R}\) into open intervals on which the sign of \(\dfrac{R}{Q}\) is constant.

3) Build a sign chart.

Pick one test point in each open interval and compute \(\operatorname{sign}R\), \(\operatorname{sign}Q\), hence \(\operatorname{sign}\!\left(\dfrac{R}{Q}\right)\). Shade intervals where the inequality holds: positive for \(>\) or \(\ge\); negative for \(<\) or \(\le\). For non-strict cases, add isolated roots of \(R\) with \(Q\neq 0\) as closed endpoints.

Multiplicity hint: crossing a root of odd multiplicity flips the sign; even multiplicity preserves it. The sample-point method captures this automatically.

4) Read the solution.

Combine the shaded intervals and equality points; exclude all \(Q(x)=0\). Express as a union of intervals.

Example.

\(\displaystyle \frac{x^2-1}{x+3}\ge 0\). Here \(R(x)=x^2-1=(x-1)(x+1)\), \(Q(x)=x+3\). Critical points: \(x=-3\) (denominator), \(x=\pm1\) (numerator).

\[ (-\infty,-3),\quad (-3,-1),\quad (-1,1),\quad (1,\infty). \]

Testing \(-4,-2,0,2\) gives signs \((-),(+),(-),(+)\). With \(\ge\), we keep the “\(+\)” intervals and include the zeros of \(R\) where \(Q\neq 0\): solution \(({-}3,-1]\cup[1,\infty)\).

Algebraic (compound) form without fractions

Starting from \(\dfrac{R}{Q}\,\square\,0\), you may write an equivalent compound statement that avoids dividing by a sign-changing quantity:

\[ \bigl(R(x)\ \sigma_1\ 0 \ \text{and}\ Q(x) > 0\bigr)\quad\text{or}\quad \bigl(R(x)\ \sigma_2\ 0 \ \text{and}\ Q(x) < 0\bigr),\quad Q(x)\neq 0, \]

where: \((\sigma_1,\sigma_2)= (<,>)\) if \(\square\) is “\(<\)”; \((\le,\ge)\) if “\(\le\)”; \((>,<)\) if “\(>\)”; \((\ge,\le)\) if “\(\ge\)”.

This form is logically clean for proofs and matches what the calculator shows in the steps.

Notes & edge cases

\(R\equiv 0\): then \(\dfrac{P}{Q}\equiv c\). The inequality holds on the whole domain \(Q\neq 0\) for \(\le,\ge\), and never for \(<,>\).
No real critical points: the sign is constant on \(\mathbb{R}\setminus\{Q=0\}\); testing a single value suffices.
Vertical asymptote vs hole: if a real \(x_0\) is a zero of \(Q\) but not of \(R\), the graph has a vertical asymptote at \(x_0\); if both \(R(x_0)=Q(x_0)=0\), the point is removable (a hole). In both cases \(x_0\) is excluded.

Frequently Asked Questions

How do you solve a rational inequality like P(x)/Q(x) ≥ c?

Move everything to one side by rewriting it as (P(x) - cQ(x))/Q(x) ≥ 0. Then find where the numerator is zero and where the denominator is zero, and use sign analysis on the intervals between those points.

Why are values where Q(x) = 0 excluded from the solution?

When Q(x) = 0 the rational expression is undefined, so those x-values cannot be included. These points correspond to vertical asymptotes or removable holes on the graph.

When are endpoints included in a rational inequality solution?

Endpoints from zeros of the transformed numerator R(x) can be included only for ≤ or ≥, and only if the denominator is not zero at that point. Any point where Q(x) = 0 is excluded even if it also makes the numerator zero.

What is a sign chart and how does it determine the solution intervals?

A sign chart records the sign of the numerator and denominator on each interval split by critical points. The sign of R(x)/Q(x) is positive or negative on each interval, and you keep the intervals where that sign matches the chosen inequality.

What kinds of rational inequalities can this solver handle?

It is designed for inequalities where P(x) and Q(x) are polynomials up to degree 2 and the right-hand side is a constant c. The solver analyzes real critical points and returns the solution as a union of real-number intervals.

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