A rational inequality is an inequality of the form \(\frac{P(x)}{Q(x)} > 0\), where \(P(x)\) and \(Q(x)\) are polynomials, and the \(>\) sign can be replaced by \(<\), \(\ge\), or \(\le\). These inequalities are solved using the interval method: move everything to the left side, factor, mark the roots of the numerator and denominator on the number line, and determine the signs. Below is the definition, a complete algorithm, the rule for excluded points, root multiplicity, solved examples, and a practice tool with checks.
What is a Rational Inequality
A rational inequality is an inequality where both sides are expressed as rational expressions, meaning polynomials and fractions of polynomials. After moving everything to one side, it always reduces to the form
where \(P(x)\) and \(Q(x)\) are polynomials, and \(\vee\) represents one of the signs \(<\), \(>\), \(\le\), or \(\ge\). The right side must be zero; this condition is essential for the interval method to work.
Rational inequalities are divided into two types. If the variable appears only in the numerator, the inequality is called integer rational: there's nothing to divide by, and any value of \(x\) is valid. If the variable is in the denominator, the inequality is fractional-rational (also called a fractional inequality or an inequality with a fraction) — and then the restriction \(Q(x) \neq 0\) immediately arises.
Why You Can't Multiply Both Sides by the Denominator
The strongest temptation when solving an inequality with a fraction is to 'get rid of the denominator' by multiplying both sides by it. This is permissible with equations, but not with inequalities. The reason is simple: when multiplying by a negative number, the inequality sign flips, and the sign of the expression \(Q(x)\) is unknown in advance — it depends on \(x\).
Let's look at the inequality \(\frac{5}{x-3} < 1\). Multiply both sides by \(x - 3\), as if it were positive: we get \(5 < x - 3\), which means \(x > 8\). Let's check the answer by substituting \(x = 0\): \(\frac{5}{0-3} = -\frac{5}{3} < 1\) — the inequality is true, so zero is also a solution, but it's not included in the answer \(x > 8\). The answer is lost.
The correct approach is to move everything to the left and combine into a single fraction:
In the last step, we multiplied both sides by \(-1\) and flipped the sign — this is allowed because \(-1\) is always negative. Then, the interval method applies, and the answer is \(x \in (-\infty; 3) \cup (8; +\infty)\) — which includes the lost zero.
It's easiest to remember this: never multiply both sides by an expression with a variable; you can always move terms. Moving terms does not change the inequality sign.
Interval Method: Step-by-Step Algorithm
The interval method relies on one property: between two consecutive roots of a rational expression, its sign remains constant. Therefore, it's sufficient to find all points where the numerator or denominator equals zero, use these points to divide the number line, and determine the sign in each interval using a test point or by considering root multiplicity.
Excluded and Included Points: Domain and Inequality Sign
An error in one parenthesis can turn a correct solution into an incorrect answer, so it's worth learning the rule about points separately. Exactly two considerations apply.
First is the domain of permissible values. A root of the denominator is always excluded, regardless of the inequality sign. At this point, the fraction simply doesn't exist: division by zero is not allowed, so the number cannot be a solution, even if the sign is non-strict.
Second is the strict or non-strict sign. Roots of the numerator make the fraction equal to zero. With a non-strict sign (\(\le\) or \(\ge\)), equality to zero is acceptable, so the point is included (filled in) and part of the answer with a square bracket. With a strict sign (\(<\) or \(>\)), zero is not acceptable, and the point is excluded.
For example, for the inequality \(\frac{x-6}{x+1} \le 0\), the point \(x = 6\) is included (numerator root, non-strict sign), while the point \(x = -1\) is excluded (denominator root). The answer is \(x \in (-1; 6]\).
| What kind of point | How to mark |
|---|---|
| Denominator root, Q(x) = 0 | excluded always, for any sign — the point is not in the domain |
| Numerator root, strict sign < or > | excluded the fraction is zero, but a strictly different sign is required |
| Numerator root, non-strict sign ≤ or ≥ | included equality to zero is acceptable, the point is included in the answer |
Root Multiplicity: Where the Sign Changes and Where It Doesn't
The signs on the intervals don't necessarily alternate — this is the most common misconception in this topic. Alternation only occurs when all factors are raised to the first power. The rule is more precisely stated in terms of root multiplicity — the exponent of the corresponding factor.
When crossing a root of odd multiplicity (power 1, 3, 5), the sign of the expression changes. When crossing a root of even multiplicity (power 2, 4), the sign does not change: a factor raised to an even power is non-negative and doesn't affect the sign of the expression. Multiplicity is counted the same way for both the numerator and the denominator.
Consider \((x-9)^2(x+3) > 0\). The roots are \(x = 9\) with multiplicity 2 and \(x = -3\) with multiplicity 1. To the right of nine, the expression is positive. Crossing \(9\) — even multiplicity, the sign remains the same, so the interval \((-3; 9)\) also has a plus sign. Crossing \(-3\) — odd multiplicity, the sign changes to minus. The answer for the strict inequality is: \(x \in (-3; 9) \cup (9; +\infty)\), the point \(9\) is excluded because the expression is zero there. For the non-strict inequality \((x-9)^2(x+3) \ge 0\), the answer is different: \(x \in [-3; +\infty)\) — nine is included and 'glues' two intervals together.
| Root Multiplicity | Sign Change When Crossing |
|---|---|
| Odd: (x − a), (x − a)3 | changes |
| Even: (x − a)2, (x − a)4 | does not change |
Number Line Example: Assigning Signs
Let's put it all together for the inequality \(\frac{x+7}{(x-2)(x-5)} < 0\). Roots of the numerator: \(x = -7\). Roots of the denominator: \(x = 2\) and \(x = 5\), which also define the domain. All three factors are to the first power, so the signs will alternate. A test point \(x = 10\) in the rightmost interval gives \(\frac{17}{8 \cdot 5} > 0\) — we start with a plus and move left, changing the sign at each point.
Examples of Solving Rational Inequalities
Example 1. Simple Fractional Inequality
Solve \(\frac{x-3}{x+5} \ge 0\).
Step 1. Zero is already on the right, and it's a single fraction — no need to move or combine.
Step 2. Domain (ODZ). \(x + 5 \neq 0\), which means \(x \neq -5\).
Step 3. Roots. The numerator is zero when \(x = 3\), the denominator when \(x = -5\).
Step 4. Points on the number line. \(x = 3\) is included (numerator root, non-strict sign), \(x = -5\) is excluded (denominator root).
Step 5. Signs. Test point \(x = 4\): \(\frac{1}{9} > 0\) — plus on the right. Both factors are to the first power, so signs alternate: minus on \((-5; 3)\), plus on \((-\infty; -5)\). Check: for \(x = -6\), we get \(\frac{-9}{-1} = 9 > 0\) — consistent.
Step 6. Answer. We need the sign 'plus or zero': \(x \in (-\infty; -5) \cup [3; +\infty)\).
Example 2. Not Zero on the Right — Move It
Solve \(\frac{2x+1}{x-4} \le 1\).
Step 1. Move one to the left. Multiplying by \(x - 4\) is not allowed, as the sign of this expression is unknown:
Step 2. Combine into a single fraction.
Step 3. Domain (ODZ). \(x \neq 4\).
Step 4. Points. \(x = -5\) — numerator root, non-strict sign, included. \(x = 4\) — denominator root, excluded.
Step 5. Signs. For \(x = 5\), the fraction is \(\frac{10}{1} > 0\). So, plus on \((4; +\infty)\), minus on \((-5; 4)\), plus on \((-\infty; -5)\).
Step 6. Answer. \(x \in [-5; 4)\). Check boundary points: at \(x = -5\), the fraction is zero — suitable for \(\le\); at \(x = 4\), the fraction is undefined — not suitable.
Example 3. Even Multiplicity Root and Isolated Point
Solve \(\frac{(x-2)^2(x+6)}{x-9} \ge 0\).
Step 1. Domain (ODZ). \(x \neq 9\).
Step 2. Roots and multiplicities. \(x = 2\) — multiplicity 2 (even), \(x = -6\) — multiplicity 1, \(x = 9\) — multiplicity 1 in the denominator.
Step 3. Points. \(x = 2\) and \(x = -6\) are included (numerator roots with non-strict sign), \(x = 9\) is excluded.
Step 4. Signs. Test point \(x = 10\): numerator is positive, denominator is 1 — plus. Moving left. Crossing \(9\) (odd multiplicity), the sign changes: minus on \((2; 9)\). Crossing \(2\) (even multiplicity), the sign does not change: minus on \((-6; 2)\) as well. Crossing \(-6\) (odd multiplicity), the sign changes: plus on \((-\infty; -6)\).
Step 5. Answer. Select intervals with a 'plus' sign plus all points where the expression equals zero. The point \(x = 2\) lies within a minus interval, but the expression itself is zero there, so it's included separately:
Such an 'isolated' point in the answer is a sure sign of an even multiplicity root with a non-strict sign.
Example 4. Simplifying a Fraction: The Domain Trap
Solve \(\frac{25x^2-10x+1}{5x^2+9x-2} \le 0\).
Step 1. Factor. The numerator is a perfect square: \(25x^2-10x+1 = (5x-1)^2\). For the denominator, calculate the discriminant: \(D = 81 + 40 = 121\), roots \(x_1 = \frac{1}{5}\) and \(x_2 = -2\), so \(5x^2+9x-2 = (5x-1)(x+2)\).
Step 2. Domain (ODZ) — before any simplification. \(5x - 1 \neq 0\) and \(x + 2 \neq 0\), meaning \(x \neq \frac{1}{5}\) and \(x \neq -2\).
Step 3. Simplify. \(\frac{(5x-1)^2}{(5x-1)(x+2)} = \frac{5x-1}{x+2}\), but only if \(x \neq \frac{1}{5}\) — simplification does not restore an excluded point.
Step 4. Solve \(\frac{5x-1}{x+2} \le 0\). Roots: \(\frac{1}{5}\) and \(-2\). At \(x = 1\), the fraction is \(\frac{4}{3} > 0\). So, plus on \((\frac{1}{5}; +\infty)\), minus on \((-2; \frac{1}{5})\), plus on \((-\infty; -2)\).
Step 5. Reapply the domain (ODZ). The interval \((-2; \frac{1}{5}]\) would fit the sign requirement, but the point \(\frac{1}{5}\) is excluded: in the original inequality, the denominator \(5x^2+9x-2\) becomes zero at \(x = \frac{1}{5}\), and the fraction is undefined there.
Answer: \(x \in \left(-2; \frac{1}{5}\right)\) — both parentheses are round.
Example 5. Quadratic Trinomial in the Numerator
Solve \(\frac{x^2-7x+10}{x+4} < 0\).
Step 1. Factor the numerator. By Vieta's formulas, the sum of roots is 7 and the product is 10, so \(x^2-7x+10 = (x-2)(x-5)\). The inequality becomes \(\frac{(x-2)(x-5)}{x+4} < 0\).
Step 2. Domain (ODZ). \(x \neq -4\).
Step 3. Points. All three points are excluded: \(-4\) is a denominator root, \(2\) and \(5\) are numerator roots with a strict sign.
Step 4. Signs. Test point \(x = 6\): \(\frac{4 \cdot 1}{10} > 0\). All factors are to the first power, so signs alternate: plus on \((5; +\infty)\), minus on \((2; 5)\), plus on \((-4; 2)\), minus on \((-\infty; -4)\).
Step 5. Answer. We need the strict minus sign: \(x \in (-\infty; -4) \cup (2; 5)\). Check by substitution: at \(x = -5\), we get \(\frac{70}{-1} < 0\); at \(x = 3\), we get \(\frac{-2}{7} < 0\) — both intervals are correct.
Common Mistakes When Solving Rational Inequalities
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Multiplying both sides by the denominator: from $\frac{3}{x-8} > 1$, they get $3 > x - 8$ and write $x < 11$.
The sign of the expression \(x - 8\) is unknown, and multiplying by a negative number flips the inequality sign. Check: \(x = 0\) gives \(\frac{3}{-8} < 1\) — zero is not a solution, yet it fell into the answer \(x < 11\). Correct: \(\frac{3-(x-8)}{x-8} > 0\), which means \(\frac{11-x}{x-8} > 0\), leading to \(x \in (8; 11)\).
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Including the denominator root because the sign is non-strict: for $\frac{x-1}{x-7} \ge 0$, they write the answer with a square bracket at seven.
A denominator root is always excluded, regardless of the sign: the fraction is undefined at that point. The correct answer is: \(x \in (-\infty; 1] \cup (7; +\infty)\) — one has a square bracket, seven has a parenthesis.
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Simplifying the fraction and losing the domain (ODZ): from $\frac{(x+3)^2}{(x+3)(x-5)} \le 0$, they get $\frac{x+3}{x-5} \le 0$ and include the point $x = -3$ in the answer.
The domain (ODZ) must be written down before simplification, based on the original denominator: \(x \neq -3\) and \(x \neq 5\). The simplified fraction \(\frac{x+3}{x-5}\) is non-positive on \([-3; 5)\) — at \(x = 0\), it equals \(-0.6\), which satisfies the \(\le\) sign. However, the point \(-3\) is excluded by the domain, so the answer is \((-3; 5)\), not \([-3; 5)\).
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Assuming signs always alternate, and for $\frac{(x-1)^2}{x+4} > 0$, they put a minus sign to the right of one.
Alternation is only valid for factors raised to the first power. The root \(x = 1\) has multiplicity 2, so the sign does not change when crossing it. The expression is positive on the entire interval \((-4; +\infty)\), except at one itself, where it is zero. The answer is: \(x \in (-4; 1) \cup (1; +\infty)\).
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Changing the inequality sign when moving a term: from $\frac{x}{x-2} \le 3$, they get $\frac{x}{x-2} - 3 \ge 0$.
Moving a term from one side to the other does not change the inequality sign — only the sign of the term itself changes. The sign flips only when multiplying or dividing both sides by a negative number. Correct: \(\frac{x}{x-2} - 3 \le 0\), then \(\frac{x-3(x-2)}{x-2} \le 0\), which means \(\frac{6-2x}{x-2} \le 0\).
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Marking only the numerator roots on the number line, and checking the denominator 'later': for $\frac{x+11}{x^2-16} > 0$, they put only one point, $-11$.
All roots must be marked on the number line at once — both numerator and denominator roots, otherwise the intervals will be incorrect. Here, \(x^2-16 = (x-4)(x+4)\), so there are three points: \(-11\), \(-4\), and \(4\), not just one. All are excluded (strict sign), the answer is: \(x \in (-11; -4) \cup (4; +\infty)\).
Questions and Answers
What is a rational inequality?
It's an inequality where both sides are rational expressions (polynomials and fractions of polynomials). After moving everything to the left side, it reduces to the form \(\frac{P(x)}{Q(x)} > 0\), where \(P(x)\) and \(Q(x)\) are polynomials, and \(>\) can be replaced by \(<\), \(\le\), or \(\ge\).
How do you solve rational inequalities?
Using the interval method in six steps: move everything to the left side so that zero remains on the right; combine into a single fraction and factor the numerator and denominator; write down the domain (ODZ, \(Q(x) \neq 0\)); mark the numerator and denominator roots on the number line; assign signs to the intervals using a test point and root multiplicity; write the intervals with the required sign in the answer.
How does a fractional-rational inequality differ from an integer rational one?
In an integer rational inequality, the variable appears only in the numerator, e.g., \((x-1)(x+4) > 0\), and the domain (ODZ) includes all real numbers. In a fractional-rational inequality, the variable is in the denominator, e.g., \(\frac{x-1}{x+4} > 0\), and the restriction \(x \neq -4\) arises. The algorithm is the same — only the excluded points change.
How do you solve inequalities with fractions?
The same way as any rational inequality: first, move everything to the left side and combine into a single fraction, rather than trying to eliminate the denominator. For example, \(\frac{x}{x+2} \ge 1\) becomes \(\frac{x-(x+2)}{x+2} \ge 0\), which is \(\frac{-2}{x+2} \ge 0\). The numerator is a constant negative value, so the fraction is non-negative only when the denominator is negative: \(x + 2 < 0\). The answer is: \(x \in (-\infty; -2)\).
Why can't you multiply both sides of an inequality by the denominator?
Because the sign of the denominator depends on \(x\): multiplying by a positive expression preserves the inequality sign, while multiplying by a negative one flips it. Since you don't know in advance which case applies, some solutions are lost and others are incorrectly added. You can always move terms; moving terms does not change the inequality sign.
How do you find the domain (ODZ) in an inequality?
For a rational inequality, the domain (ODZ) consists of all values of \(x\) for which the denominator is not zero. Set the denominator equal to zero, solve the equation, and exclude the found numbers. You must write down the domain (ODZ) based on the original denominator, before simplifying the fraction — otherwise, excluded points will be lost.
Which points are excluded, and which are included?
Denominator roots are always excluded, regardless of the inequality sign. Numerator roots are included for non-strict signs (\(\le\) or \(\ge\)) and excluded for strict signs (\(<\) or \(>\)). In the answer notation, an excluded point corresponds to a parenthesis, and an included point to a bracket.
What is root multiplicity and why is it needed in the interval method?
Multiplicity is the power to which a factor is raised in its factorization. When crossing a root of odd multiplicity, the sign of the expression changes; when crossing a root of even multiplicity, the sign remains the same. This is why signs don't always simply alternate: for \((x-9)^2(x+3)\), the sign does not change at point \(9\).
What grade level are rational inequalities taught?
In 9th grade, during algebra lessons, after quadratic inequalities and factoring polynomials. The interval method is later used for exponential, logarithmic, and irrational inequalities, as well as in OGE and profile ЕГЭ tasks.
