Additional Mathematics | How to Solve Rational Inequalities Without Crossing a Forbidden Value

A rational inequality is not solved safely by multiplying both sides by a denominator whose sign is unknown. First identify where the numerator is zero and where the denominator is zero. Mark every critical value on a number line, test the sign of the fraction in each interval, then include valid zeros and exclude all forbidden denominator values.

For example, [(x − 1)(x + 2)]/(x − 3) ≥ 0 has solution −2 ≤ x ≤ 1 or x > 3. The point x = 3 is never allowed because the denominator vanishes there, even if a sign chart seems to change from negative to positive. This guide covers strict and inclusive inequalities, repeated factors, cancelling expressions and how to tell a zero from a vertical asymptote or removable hole.

Why ordinary cross-multiplication can fail

When an inequality involves (x − 1)/(x − 3), the denominator changes sign at x = 3. Multiplying both sides by x − 3 without checking its sign may preserve or reverse the inequality depending on the unknown x.

For equations, multiplying by a non-zero expression preserves equality. For inequalities, multiplying by a positive number preserves direction, while multiplying by a negative number reverses it. An expression involving x can be either.

A sign chart avoids this trap. It keeps the rational expression intact, locates each zero and forbidden value, and identifies intervals on which its sign is stable.

The guiding idea is simple: determine the sign of the numerator and denominator separately, then combine the signs. Do not cross a point where the denominator is zero while reasoning about a continuous interval, because the original fraction is undefined there.

Find zeros and exclusions before choosing a sign

Consider R(x) = [(x − 1)(x + 2)]/(x − 3). The numerator vanishes at x = 1 and x = −2. These can be solutions to a non-strict inequality because the fraction equals zero there, provided the denominator is non-zero.

The denominator vanishes at x = 3. This input is excluded from the domain for every inequality involving the original fraction. It cannot be included simply because the expression changes sign nearby.

Thus the three critical values are −2, 1 and 3. Put them in ascending order on a number line, and use each to divide the real line into intervals.

This first classification is vital. A numerator zero and a denominator zero are not interchangeable: one may give a valid output zero, while the other gives no defined output at all.

A complete sign chart for the opening problem

Solve [(x − 1)(x + 2)]/(x − 3) ≥ 0. Use intervals x < −2, −2 < x < 1, 1 < x < 3 and x > 3.

At x = −3, both numerator factors are negative, so their product is positive, while the denominator is negative; the fraction is negative. At x = 0, the numerator factors have opposite signs and the denominator is negative, giving a positive fraction.

At x = 2, both numerator factors are positive but the denominator is negative, so the fraction is negative. At x = 4, all three factors are positive, giving a positive fraction.

Thus the positive intervals are (−2, 1) and (3, ∞). Since the question uses ≥, include the numerator zeros −2 and 1. Exclude 3. The final solution is [−2, 1] ∪ (3, ∞).

Why one test point per interval is enough

A rational expression with polynomial numerator and denominator is continuous within an interval containing no zero of the numerator and no zero of the denominator.

Its sign cannot change on that interval without passing through zero or becoming undefined. Therefore, after the critical values have been correctly identified, one convenient test point in each interval determines its sign throughout that interval.

Choose simple values such as −3, 0, 2 and 4 for the opening example. The numerical magnitude of the fraction is not important; only whether it is positive or negative matters.

This is more systematic than sampling many arbitrary x-values. The efficiency comes from proving that the intervals contain no other sign-changing boundaries, not from hoping a few selected numbers happen to be representative.

A numerator zero may be included or excluded

Suppose the inequality is strictly positive: [(x − 1)(x + 2)]/(x − 3) > 0. The positive intervals are the same as before, but values giving a fraction of exactly zero are no longer acceptable.

Therefore the solution becomes (−2, 1) ∪ (3, ∞). The endpoints −2 and 1 are open because the expression equals zero there.

For a non-strict condition ≥ 0, the endpoints −2 and 1 are included, provided neither is forbidden by the denominator. The value x = 3 is excluded in both cases.

Students should apply inequality strictness after determining the sign intervals. Copying the same square or round brackets from a previous example can change the solution set even when the algebraic sign chart was correct.

A denominator zero is never an included endpoint

At x = 3 in R(x) = [(x − 1)(x + 2)]/(x − 3), the numerator is 2 × 5 = 10, while the denominator is zero. The rational expression is undefined.

The inequality R(x) ≥ 0 has no truth value at that x because R(3) is not a real number. It cannot be declared positive, negative or zero.

This remains true regardless of how close the fraction becomes to a chosen sign on either side. A denominator zero is an exclusion in the domain, not an ordinary boundary where the function takes the value zero.

Mark poles with open circles or vertical divider marks in a sign table. Never turn them into closed endpoints for a ≥ or ≤ inequality.

Another example with one numerator and one denominator factor

Solve (x − 2)/(x + 1) ≤ 0. The numerator is zero at x = 2, and the denominator is zero at x = −1.

For x < −1, both numerator and denominator are negative, so the ratio is positive. For −1 < x < 2, the numerator is negative while the denominator is positive, so the ratio is negative. For x > 2, both are positive, so the ratio is positive.

The desired non-positive region is between −1 and 2. Include x = 2, where the fraction is exactly zero, but exclude x = −1, where it is undefined.

Thus the answer is (−1, 2]. This simple example is an excellent diagnostic for whether a student distinguishes numerator zeros from forbidden denominators.

Why multiplying by x + 1 is dangerous

Take the inequality (x − 2)/(x + 1) ≤ 0. A careless student might multiply by x + 1 and write x − 2 ≤ 0, producing x ≤ 2. That answer wrongly includes many x-values less than −1.

For x < −1, the denominator is negative. Multiplying an inequality by it must reverse the sign, not preserve it. For x > −1, the denominator is positive, and the original direction is preserved.

It is possible to solve by cases on x + 1, but the sign chart is usually clearer and less repetitive. It handles both denominator signs and the excluded point at x = −1 together.

A quick counterexample exposes the false result: at x = −2, the fraction (−4)/(−1) = 4 is positive, so it does not satisfy ≤ 0, although −2 ≤ 2.

Repeated roots do not always change the sign

Consider F(x) = (x − 1)²/(x + 2). Its numerator is a perfect square and therefore non-negative for all real x, equalling zero only at x = 1.

Because the denominator has sign matching x + 2, the fraction is negative for x < −2 and positive for x > −2 except at the numerator zero where it equals zero.

The sign does not flip as x passes through x = 1. The squared factor has even multiplicity and remains non-negative on both sides.

This helps students read sign charts with repeated factors: crossing a simple odd-multiplicity zero changes the sign of that factor, while crossing an even-multiplicity zero generally does not. A repeated root may still be included in a non-strict inequality because the whole fraction can equal zero there.

Solve a repeated-numerator-factor inequality

For F(x) = (x − 1)²/(x + 2) ≥ 0, the denominator must be positive unless the numerator is zero. It is positive for x > −2, and the numerator is non-negative throughout that region.

Thus the solution is x > −2, or (−2, ∞). The zero at x = 1 is included within this interval, while the forbidden x = −2 is not.

For the strict inequality F(x) > 0, the point x = 1 must be excluded because the fraction is exactly zero there. The answer becomes (−2, 1) ∪ (1, ∞).

The key is to distinguish sign changes from zeros. Even when the sign remains positive across a repeated root, a strict inequality still cannot include the value where the expression equals zero.

The inequality can have an isolated zero solution

For the same F(x), consider F(x) ≤ 0. The fraction is negative on x < −2 and zero at x = 1.

Therefore the solution is (−∞, −2) ∪ {1}. The value x = 1 is an isolated included point rather than part of a non-positive interval around it.

This is a useful case that students may miss if they only shade regions labelled negative in the sign chart and forget valid numerator zeros. The solution of a non-strict rational inequality can contain isolated points.

When converting the answer to interval notation, use {1} or state x = 1 separately. Do not incorrectly write a closed interval extending through positive values between −2 and 1.

Cancellation does not restore a forbidden point

Consider G(x) = [(x − 1)(x − 4)]/[(x − 1)(x + 2)]. Algebraically this simplifies to (x − 4)/(x + 2), but only for x ≠ 1 and x ≠ −2.

The cancellation removes a common factor from the formula, not the original domain restriction. At x = 1 the original expression has zero divided by zero and is undefined, even though the simplified formula would produce a finite number.

This excluded point is commonly described as a removable hole. It must still be excluded from any inequality solution based on the original expression.

A domain note before cancellation protects this information. Write x ≠ 1, −2, then simplify and solve using the reduced formula while carrying both exclusions into the final answer.

A removable hole inside the inequality solution

Solve G(x) < 0, where G(x) = [(x − 1)(x − 4)]/[(x − 1)(x + 2)]. Work on its original domain x ≠ 1, −2.

The simplified sign expression (x − 4)/(x + 2) is negative for −2 < x < 4. But the original function is also undefined at x = 1, which lies within that interval.

Therefore the correct solution is (−2, 1) ∪ (1, 4). Writing (−2, 4) would silently include an invalid input.

This is an especially valuable A-Math habit: whenever a common factor is cancelled, the cancelled zero remains a domain exclusion. An algebraically simpler expression can describe the same values only on the original allowed inputs.

Vertical asymptotes and holes have different local behaviour

In a rational function, an uncancelled denominator zero with nonzero numerator often produces vertical-asymptote behaviour: the magnitude can grow very large as x approaches the forbidden input.

By contrast, if a common factor cancels and a finite simplified expression remains, the original function may have a removable hole at that input. The value is still excluded, but nearby function values can approach a finite limit.

These features can matter when sketching a rational graph and interpreting the sign near a boundary. However, neither vertical asymptotes nor holes may be included as valid x-values in an inequality answer.

For sign charts, the essential information is that each remains an excluded input. The graph’s local shape provides additional understanding rather than changing the domain rule.

An always-positive denominator simplifies the analysis

Suppose H(x) = (x − 3)/(x² + 1). Since x² + 1 is strictly positive for every real x, the fraction has the same sign as x − 3.

Therefore H(x) ≥ 0 exactly when x ≥ 3. There are no forbidden real denominator values because x² + 1 never vanishes over the real numbers.

This shows that a rational inequality does not always require a long sign table. First ask whether the denominator’s sign is known for all allowed inputs. If it is always positive, the sign comparison reduces immediately to the numerator.

The safest method is not necessarily the longest. It is the one that justifies the denominator’s behaviour before simplifying the inequality.

A denominator square is positive except at its zero

Consider J(x) = (x − 3)/(x − 1)². The squared denominator is positive for every x ≠ 1, and the expression is undefined at x = 1.

Thus on its domain J(x) has the same sign as x − 3. For J(x) ≥ 0, the answer is x ≥ 3. For J(x) < 0, the answer is x < 3 with x ≠ 1, or (−∞, 1) ∪ (1, 3).

Students may forget the excluded x = 1 because the denominator is a square and therefore not negative. But positivity away from a zero does not mean the denominator exists at the zero.

This is a useful distinction between the sign of a factor and the domain of the rational expression containing that factor.

Solve a product over a quadratic denominator

Consider K(x) = (x + 1)/[(x − 2)(x + 3)] and require K(x) > 0. The critical values are −3 and 2 from the denominator, and −1 from the numerator.

On x < −3, choose −4: the numerator is negative and the denominator has two negative factors, so the ratio is negative. On −3 < x < −1, choose −2: the numerator is negative and the denominator has one negative and one positive factor, so the ratio is positive.

On −1 < x < 2, choose 0: numerator positive and denominator negative, so ratio negative. On x > 2, all factors are positive, so ratio positive.

The strict solution is (−3, −1) ∪ (2, ∞). The numerator zero −1 is excluded by strictness, and the denominator zeros −3 and 2 are always excluded.

Ordering critical values prevents missed intervals

A sign chart should begin with all distinct critical values in ascending order. For K(x), these are −3, −1 and 2. They split the real line into four intervals.

A common mistake is to list the numerator zero first and then the denominator roots in the order they appear in the formula. That can produce an incorrect interval table, especially when negative numbers are involved.

Sort the values numerically before testing signs. Mark whether each is a numerator zero, denominator zero, or a zero that appeared before cancellation and remains excluded.

This small organisational decision turns a complicated rational expression into a sequence of intervals whose signs are straightforward to check.

An inequality with a squared numerator is not automatically positive

Suppose the numerator is (x − 1)². That factor is non-negative, but the sign of the entire rational expression still depends on the denominator.

For (x − 1)²/(x + 2), the fraction is negative when x < −2, despite the squared numerator being positive. The denominator is negative there.

The correct general rule is: a quotient is positive when numerator and denominator have the same sign and negative when their signs differ. A perfect-square factor contributes non-negativity but does not determine the sign of the whole quotient on its own.

Students should analyse every factor that can change sign, not simply look for one square and assume the expression is always positive.

A short difference between sign-chart tests and actual roots

Critical values are places where a numerator vanishes or a denominator is undefined. They are not all roots of the inequality’s associated equality.

For R(x) = [(x − 1)(x + 2)]/(x − 3), the fraction equals zero at −2 and 1. At 3 it is undefined, so 3 is not a root of R(x) = 0.

All three belong on the sign chart because they split the domain into different regions, but they have different roles in the final answer.

A student who says “the roots are −2, 1 and 3” has already lost the distinction between output zero and forbidden input. Precise vocabulary helps prevent a later closed endpoint at a pole.

A four-step inspection before finalising the solution

First read the original domain and list denominator zeros. Second factor numerator and denominator where possible, carrying exclusions through any cancellations.

Third order all critical values and test a point in each interval, accounting for repeated factors and known-positive expressions. Fourth combine intervals according to the requested sign and decide which numerator zeros are included.

Before submitting, inspect every boundary one last time. Numerator zeros may be included for ≥ or ≤; denominator zeros never are. A cancelled denominator root remains excluded even when the reduced formula is finite there.

This is a compact way to preserve logical correctness while working quickly under examination conditions.

Practice set with answers

Question one: solve [(x − 1)(x + 2)]/(x − 3) ≥ 0. Answer: [−2, 1] ∪ (3, ∞).

Question two: solve (x − 2)/(x + 1) ≤ 0. Answer: (−1, 2].

Question three: solve (x − 1)²/(x + 2) > 0. Answer: (−2, 1) ∪ (1, ∞).

Question four: solve (x − 3)/(x² + 1) ≥ 0. Answer: x ≥ 3.

Question five: solve [(x − 1)(x − 4)]/[(x − 1)(x + 2)] < 0. Answer: (−2, 1) ∪ (1, 4).

Question six: solve (x + 1)/[(x − 2)(x + 3)] > 0. Answer: (−3, −1) ∪ (2, ∞).

For each, state the original excluded inputs before forming the sign chart. This one habit prevents many otherwise correct algebraic calculations from producing an invalid interval.

Parents, lesson diagnosis and syllabus scope

Rational inequalities combine algebraic factors, sign reasoning and domain restrictions. Their exact syllabus status and difficulty should be checked against the learner’s school curriculum; advanced rational-sign charts may be enrichment for some Additional Mathematics pathways.

For Singapore examination references, use SEAB’s 2026 O-Level 4049 list, 2027 SEC G3 K341 list or 2027 SEC G2 K232 list as applicable.

A student who gets intervals wrong may need work on factor signs, ordering negative numbers or preserving excluded values. These are different prerequisites. In eduKateSG’s maximum three-student tutorials, generally 1.5 hours weekly near Sixth Avenue MRT where placement is suitable, tutors can address the first broken decision before assigning further practice.