JC2 H2 Maths Bukit Timah | Rational Inequalities

Did you know? A denominator can be zero at the same x-value where a simplified inequality seems to say “equal to zero”, yet that point is still forbidden. For example, (x−1)(x−4)/(x−2)≥0 has numerator roots 1 and 4, and a denominator zero at 2. Solving the sign pattern gives [1,2) ∪ [4,∞). The endpoints 1 and 4 may be included because the expression equals zero there. The point 2 can never be included because the original expression is undefined there.

This JC2 H2 Mathematics Tutor Bukit Timah guide answers an exact A-Level 9758 question: how do you solve rational inequalities without including poles, losing sign changes, or reversing an inequality incorrectly? The method begins with the excluded denominator values, then uses a factor sign chart. There is a full three-critical-point example, a repeated-factor exception, and 32 original practice clinics whose domain and endpoint decisions can be checked algebraically.

The 2027 SEAB H2 Mathematics syllabus 9758 includes solving inequalities of the form f(x)/g(x)>0 with linear or quadratic numerator and denominator expressions, and graphing rational functions. This lesson uses inequalities with the four comparison symbols as practice but keeps each rational function within that low-degree scope. The datasets are mathematical constructions, not real pupil marks or actual examination extracts.

Quick answer: mark zeros and poles before testing signs

Given a rational inequality N(x)/D(x) > 0, first solve D(x)=0 and exclude those x-values from the domain. Next find numerator zeros N(x)=0. Put all distinct critical points in increasing order on a number line. Determine the sign of the expression in each interval, using factored terms, then choose the intervals satisfying the required inequality. For a non-strict comparison ≥0 or ≤0, include genuine numerator zeros in the domain but never include denominator zeros.

Numerator root: makes the fraction zero if the denominator is nonzero there.
Denominator root: makes the fraction undefined and is always excluded.
Strict > or <: do not include zeros even where the denominator is valid.
Non-strict ≥ or ≤: numerator zeros can be included, but poles cannot.
Simple factor: crossing its root changes the sign of that factor.
Repeated even factor: the sign need not flip on crossing that root.
Do not blindly multiply by the denominator: its unknown sign may reverse the inequality; a sign chart is safer.

Complete worked example: three critical points

Solve (x−1)(x−4)/(x−2) > 0. The numerator is zero at x=1 and x=4. The denominator is zero at x=2, which is a pole and must be excluded. These points divide the real line into four open intervals: (−∞,1), (1,2), (2,4) and (4,∞). In the far-right interval, all three factors are positive, so the rational expression is positive. Cross x=4, which changes one numerator factor’s sign, so the expression becomes negative. Crossing x=2 reverses the denominator factor’s sign, making it positive in (1,2). Crossing x=1 makes it negative in (−∞,1).

For the strict inequality, the positive intervals are (1,2)∪(4,∞). Now change only the comparison symbol to ≥0. The numerator zeros x=1 and x=4 satisfy equality and are in the domain, so they may be included. The pole x=2 remains forbidden. The new answer is [1,2)∪[4,∞). This changed-case check isolates endpoint logic from the rest of the computation.

Why clearing a denominator can reverse the answer

A common shortcut is to start with (x−1)/(x−2)>0 and multiply both sides by x−2. But if x−2 is negative, multiplying reverses the inequality; if it is positive, it does not. Assuming one case without checking the sign loses half of the solution. The correct sign analysis shows the ratio is positive when both numerator and denominator are negative (x<1), or both are positive (x>2). Thus the solution is (−∞,1)∪(2,∞).

Multiplying by (x−2)² would be non-negative for all permitted x, but that changes the algebra and still requires x≠2. The more transparent approach at H2 level is to factor the expression, mark critical points and check signs. This prevents the subtle errors that arise when a denominator’s sign is silently assumed.

Numerator zero is not always a permitted x-intercept

Consider [(x−1)(x−2)]/(x−2)≥0. The original domain excludes x=2 even though the factor cancels for every other input. On the permitted domain, the expression equals x−1. Therefore the inequality requires x≥1 while retaining x≠2, giving [1,2) ∪ (2,∞). The simplified line has a missing point at x=2. That missing input cannot be restored by cancellation.

This also illustrates a distinction between a genuine numerator zero and a cancelled common zero. x=1 is a valid zero, so it belongs to the ≥0 solution. x=2 is a denominator zero in the original function, so it is not even a candidate for membership. The related hole versus vertical asymptote guide owns that graph question; the present article owns the inequality endpoint decision.

Repeated roots do not always flip signs

Solve (x−1)²/(x−3)>0. For x≠1 the squared numerator is positive, while at x=1 it equals zero. The denominator is positive for x>3 and negative for x<3. The required strict-positive region is therefore (3,∞). Crossing the double numerator root x=1 does not change the expression from negative to positive; its sign stays negative on either side up to the pole at 3. Treating every listed root as a sign-flip point would create false intervals.

If the comparison changes to ≤0, then every permitted x<3 satisfies the inequality: most have a negative expression and x=1 gives zero. The solution is (−∞,3). The pole at 3 is excluded. In this example the double root is included for the non-strict inequality even though the surrounding sign does not reverse.

The interval table is a quick independent check

Interval for (x−1)(x−4)/(x−2)	Numerator sign	Denominator sign	Fraction sign
x<1	(−)(−)=+	−	−
1<x<2	(+) (−)=−	−	+
2<x<4	(+) (−)=−	+	−
x>4	(+) (+)=+	+	+

Every interval contains no critical point, so the sign is constant there for this factored rational expression. You can choose any convenient test value within an interval, or reason from the signs of the factors. A graphing calculator can validate the pattern, but a graph that only marks the roots without explaining a forbidden denominator value does not fully justify the solution.

A four-factor ratio gives five sign intervals

Now solve (x+3)(x−3)/[(x+2)(x−1)] >0. The sorted critical points are −3 (numerator), −2 (denominator), 1 (denominator), and 3 (numerator). The rational expression has positive leading behaviour as |x| grows large. Because all four factors are simple and distinct, its sign flips at each critical point. The positive intervals are (−∞,−3)∪(−2,1)∪(3,∞).

If the inequality becomes ≥0, include the zeros at −3 and 3, but keep −2 and 1 excluded. This gives (−∞,−3]∪(−2,1)∪[3,∞). The middle interval remains open on both sides because its endpoints are poles. This pair of answers is a useful template for a four-critical-point sign chart.

Why equality belongs to numerator roots, not to poles

For any real input in the fraction’s domain, N(x)/D(x)=0 exactly when N(x)=0 and D(x)≠0. If D(x)=0, the fraction is undefined, regardless of whether N(x)=0 as well. The comparison symbol ≥ does not change that definition. An undefined expression is not zero; it cannot satisfy any ordinary real inequality at that input.

This logic is particularly valuable after algebraic cancellation, where an excluded input may disappear from the simplified formula. Students should preserve the original domain in the margin before cancelling factors. The final answer is then the intersection of the simplified inequality’s candidate solution with that original domain.

One-minute Bukit Timah diagnostic

Show the ratio (x−1)/(x−2) and ask for its sign for x<1, 12. The signs are positive, negative and positive. Ask whether x=1 belongs to ≥0: yes. Ask whether x=2 belongs: no, never. Next square the numerator, (x−1)²/(x−2), and ask whether the sign flips at x=1: no. These short changed-case questions isolate the exact misconception without requiring a long worksheet.

Thirty-two original rational-inequality clinics

Each clinic uses two distinct numerator roots and two distinct denominator roots, all simple, so its sign alternates at each critical point. The questions vary the four roots’ ordering and the inequality symbol. The initial interval sign is positive on the far right because numerator and denominator both have degree two with positive leading coefficients. Worked answers include interval notation with the correct open or closed endpoints. These are original educational exercises rather than official past-year questions.

1. Four critical points: -10, -9, -7, -3

Question. Solve the original rational inequality (x+10)(x+7)/[(x+9)(x+3)] >0. The numerator roots are -10 and -7; the denominator zeros are -9 and -3. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -10, -9, -7, -3. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-10) ∪ (-9,-7) ∪ (-3,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -9 and -3 remain excluded under all circumstances.

Check and transfer. The sign switches at each simple numerator root or pole, but an excluded pole remains open under every comparison. Change only the comparison from >0 to ≥0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

2. Four critical points: -6, -2, 1, 4

Question. Solve the original rational inequality (x+6)(x+2)/[(x−1)(x−4)] ≥0. The numerator roots are -6 and -2; the denominator zeros are 1 and 4. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -6, -2, 1, 4. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-6] ∪ [-2,1) ∪ (4,∞). Because the comparison includes equality, the numerator roots -6 and -2 may be included when they border the selected intervals. The denominator zeros 1 and 4 remain excluded under all circumstances.

Check and transfer. Equality is allowed only when the numerator is zero and the original denominator is nonzero. Change only the comparison from ≥0 to >0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

3. Four critical points: -11, -3, 3, 11

Question. Solve the original rational inequality (x+3)(x−3)/[(x+11)(x−11)] <0. The numerator roots are -3 and 3; the denominator zeros are -11 and 11. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -11, -3, 3, 11. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-11,-3) ∪ (3,11). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -11 and 11 remain excluded under all circumstances.

Check and transfer. Do not multiply through an unsignposted denominator without tracking where its sign reverses the inequality. Change only the comparison from <0 to ≤0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

4. Four critical points: -5, 0, 5, 8

Question. Solve the original rational inequality x(x−8)/[(x+5)(x−5)] ≤0. The numerator roots are 0 and 8; the denominator zeros are -5 and 5. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -5, 0, 5, 8. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is (-5,0] ∪ (5,8]. Because the comparison includes equality, the numerator roots 0 and 8 may be included when they border the selected intervals. The denominator zeros -5 and 5 remain excluded under all circumstances.

Check and transfer. The same four critical values can lead to different endpoint brackets when the comparison is strict or non-strict. Change only the comparison from ≤0 to <0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

5. Four critical points: -10, 2, 3, 11

Question. Solve the original rational inequality (x−3)(x−11)/[(x+10)(x−2)] >0. The numerator roots are 3 and 11; the denominator zeros are -10 and 2. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -10, 2, 3, 11. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-10) ∪ (2,3) ∪ (11,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -10 and 2 remain excluded under all circumstances.

Check and transfer. Factor order does not change the underlying function: sign intervals depend on the numerical ordering of the roots. Change only the comparison from >0 to ≥0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

6. Four critical points: -5, -1, 6, 9

Question. Solve the original rational inequality (x+5)(x−9)/[(x+1)(x−6)] ≥0. The numerator roots are -5 and 9; the denominator zeros are -1 and 6. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -5, -1, 6, 9. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-5] ∪ (-1,6) ∪ [9,∞). Because the comparison includes equality, the numerator roots -5 and 9 may be included when they border the selected intervals. The denominator zeros -1 and 6 remain excluded under all circumstances.

Check and transfer. A numerator root need not lie in a positive interval, but if it gives zero and the comparison is inclusive it can be included. Change only the comparison from ≥0 to >0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

7. Four critical points: -7, 0, 9, 10

Question. Solve the original rational inequality (x+7)(x−9)/[x(x−10)] <0. The numerator roots are -7 and 9; the denominator zeros are 0 and 10. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -7, 0, 9, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-7,0) ∪ (9,10). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros 0 and 10 remain excluded under all circumstances.

Check and transfer. An open endpoint at a pole is a domain condition, not a convention that the inequality symbol may override. Change only the comparison from <0 to ≤0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

8. Four critical points: -11, -2, 0, 5

Question. Solve the original rational inequality (x+11)(x+2)/[x(x−5)] ≤0. The numerator roots are -11 and -2; the denominator zeros are 0 and 5. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -11, -2, 0, 5. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is [-11,-2] ∪ (0,5). Because the comparison includes equality, the numerator roots -11 and -2 may be included when they border the selected intervals. The denominator zeros 0 and 5 remain excluded under all circumstances.

Check and transfer. Check a sample x in the far-right interval; its sign is positive here because all four factors are positive. Change only the comparison from ≤0 to <0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

9. Four critical points: -8, 7, 9, 10

Question. Solve the original rational inequality (x−7)(x−9)/[(x+8)(x−10)] >0. The numerator roots are 7 and 9; the denominator zeros are -8 and 10. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -8, 7, 9, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-8) ∪ (7,9) ∪ (10,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -8 and 10 remain excluded under all circumstances.

Check and transfer. When two simple factors cross, the sign switches each time; repeated factors would require a different analysis. Change only the comparison from >0 to ≥0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

10. Four critical points: -9, -8, -5, -3

Question. Solve the original rational inequality (x+8)(x+3)/[(x+9)(x+5)] ≥0. The numerator roots are -8 and -3; the denominator zeros are -9 and -5. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -9, -8, -5, -3. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-9) ∪ [-8,-5) ∪ [-3,∞). Because the comparison includes equality, the numerator roots -8 and -3 may be included when they border the selected intervals. The denominator zeros -9 and -5 remain excluded under all circumstances.

Check and transfer. A sorted number line is safer than guessing signs from the expanded numerator coefficients. Change only the comparison from ≥0 to >0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

11. Four critical points: -11, -3, -2, 8

Question. Solve the original rational inequality (x+2)(x−8)/[(x+11)(x+3)] <0. The numerator roots are -2 and 8; the denominator zeros are -11 and -3. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -11, -3, -2, 8. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-11,-3) ∪ (-2,8). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -11 and -3 remain excluded under all circumstances.

Check and transfer. The sign switches at each simple numerator root or pole, but an excluded pole remains open under every comparison. Change only the comparison from <0 to ≤0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

12. Four critical points: -4, 1, 2, 5

Question. Solve the original rational inequality (x+4)(x−5)/[(x−1)(x−2)] ≤0. The numerator roots are -4 and 5; the denominator zeros are 1 and 2. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -4, 1, 2, 5. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is [-4,1) ∪ (2,5]. Because the comparison includes equality, the numerator roots -4 and 5 may be included when they border the selected intervals. The denominator zeros 1 and 2 remain excluded under all circumstances.

Check and transfer. Equality is allowed only when the numerator is zero and the original denominator is nonzero. Change only the comparison from ≤0 to <0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

13. Four critical points: -11, -9, 4, 7

Question. Solve the original rational inequality (x+11)(x−4)/[(x+9)(x−7)] >0. The numerator roots are -11 and 4; the denominator zeros are -9 and 7. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -11, -9, 4, 7. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-11) ∪ (-9,4) ∪ (7,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -9 and 7 remain excluded under all circumstances.

Check and transfer. Do not multiply through an unsignposted denominator without tracking where its sign reverses the inequality. Change only the comparison from >0 to ≥0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

14. Four critical points: -11, -5, -4, 7

Question. Solve the original rational inequality (x+11)(x+5)/[(x+4)(x−7)] ≥0. The numerator roots are -11 and -5; the denominator zeros are -4 and 7. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -11, -5, -4, 7. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-11] ∪ [-5,-4) ∪ (7,∞). Because the comparison includes equality, the numerator roots -11 and -5 may be included when they border the selected intervals. The denominator zeros -4 and 7 remain excluded under all circumstances.

Check and transfer. The same four critical values can lead to different endpoint brackets when the comparison is strict or non-strict. Change only the comparison from ≥0 to >0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

15. Four critical points: -6, 3, 6, 10

Question. Solve the original rational inequality (x−3)(x−6)/[(x+6)(x−10)] <0. The numerator roots are 3 and 6; the denominator zeros are -6 and 10. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -6, 3, 6, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-6,3) ∪ (6,10). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -6 and 10 remain excluded under all circumstances.

Check and transfer. Factor order does not change the underlying function: sign intervals depend on the numerical ordering of the roots. Change only the comparison from <0 to ≤0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

16. Four critical points: -10, -6, -1, 10

Question. Solve the original rational inequality (x+6)(x−10)/[(x+10)(x+1)] ≤0. The numerator roots are -6 and 10; the denominator zeros are -10 and -1. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -10, -6, -1, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is (-10,-6] ∪ (-1,10]. Because the comparison includes equality, the numerator roots -6 and 10 may be included when they border the selected intervals. The denominator zeros -10 and -1 remain excluded under all circumstances.

Check and transfer. A numerator root need not lie in a positive interval, but if it gives zero and the comparison is inclusive it can be included. Change only the comparison from ≤0 to <0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

17. Four critical points: -7, -6, 4, 5

Question. Solve the original rational inequality (x−4)(x−5)/[(x+7)(x+6)] >0. The numerator roots are 4 and 5; the denominator zeros are -7 and -6. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -7, -6, 4, 5. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-7) ∪ (-6,4) ∪ (5,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -7 and -6 remain excluded under all circumstances.

Check and transfer. An open endpoint at a pole is a domain condition, not a convention that the inequality symbol may override. Change only the comparison from >0 to ≥0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

18. Four critical points: -7, -4, 1, 9

Question. Solve the original rational inequality (x+7)(x−9)/[(x+4)(x−1)] ≥0. The numerator roots are -7 and 9; the denominator zeros are -4 and 1. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -7, -4, 1, 9. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-7] ∪ (-4,1) ∪ [9,∞). Because the comparison includes equality, the numerator roots -7 and 9 may be included when they border the selected intervals. The denominator zeros -4 and 1 remain excluded under all circumstances.

Check and transfer. Check a sample x in the far-right interval; its sign is positive here because all four factors are positive. Change only the comparison from ≥0 to >0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function’s domain.

19. Four critical points: -9, -1, 4, 8

Question. Solve the original rational inequality (x+9)(x−4)/[(x+1)(x−8)] <0. The numerator roots are -9 and 4; the denominator zeros are -1 and 8. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -9, -1, 4, 8. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-9,-1) ∪ (4,8). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -1 and 8 remain excluded under all circumstances.

Check and transfer. When two simple factors cross, the sign switches each time; repeated factors would require a different analysis. Change only the comparison from <0 to ≤0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

20. Four critical points: -8, -4, -2, 2

Question. Solve the original rational inequality (x+8)(x+4)/[(x+2)(x−2)] ≤0. The numerator roots are -8 and -4; the denominator zeros are -2 and 2. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -8, -4, -2, 2. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is [-8,-4] ∪ (-2,2). Because the comparison includes equality, the numerator roots -8 and -4 may be included when they border the selected intervals. The denominator zeros -2 and 2 remain excluded under all circumstances.

Check and transfer. A sorted number line is safer than guessing signs from the expanded numerator coefficients. Change only the comparison from ≤0 to <0; the open sign intervals remain the same, but genuine numerator zeros may enter or leave the answer. The poles never enter. This is the precise distinction between changing an inequality symbol and changing the function's domain.

21. Four critical points: -1, 1, 3, 5

Question. Solve the original rational inequality (x−1)(x−3)/[(x+1)(x−5)] >0. The numerator roots are 1 and 3; the denominator zeros are -1 and 5. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -1, 1, 3, 5. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-1) ∪ (1,3) ∪ (5,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -1 and 5 remain excluded under all circumstances.

22. Four critical points: -9, 0, 6, 8

Question. Solve the original rational inequality x(x−8)/[(x+9)(x−6)] ≥0. The numerator roots are 0 and 8; the denominator zeros are -9 and 6. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -9, 0, 6, 8. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-9) ∪ [0,6) ∪ [8,∞). Because the comparison includes equality, the numerator roots 0 and 8 may be included when they border the selected intervals. The denominator zeros -9 and 6 remain excluded under all circumstances.

23. Four critical points: -10, 1, 2, 11

Question. Solve the original rational inequality (x−2)(x−11)/[(x+10)(x−1)] <0. The numerator roots are 2 and 11; the denominator zeros are -10 and 1. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -10, 1, 2, 11. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-10,1) ∪ (2,11). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -10 and 1 remain excluded under all circumstances.

24. Four critical points: -10, -9, -7, -3

Question. Solve the original rational inequality (x+10)(x+3)/[(x+9)(x+7)] ≤0. The numerator roots are -10 and -3; the denominator zeros are -9 and -7. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -10, -9, -7, -3. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is [-10,-9) ∪ (-7,-3]. Because the comparison includes equality, the numerator roots -10 and -3 may be included when they border the selected intervals. The denominator zeros -9 and -7 remain excluded under all circumstances.

25. Four critical points: -6, -2, 1, 4

Question. Solve the original rational inequality (x+6)(x−1)/[(x+2)(x−4)] >0. The numerator roots are -6 and 1; the denominator zeros are -2 and 4. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -6, -2, 1, 4. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-6) ∪ (-2,1) ∪ (4,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -2 and 4 remain excluded under all circumstances.

26. Four critical points: -11, -3, 3, 11

Question. Solve the original rational inequality (x+11)(x+3)/[(x−3)(x−11)] ≥0. The numerator roots are -11 and -3; the denominator zeros are 3 and 11. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -11, -3, 3, 11. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-11] ∪ [-3,3) ∪ (11,∞). Because the comparison includes equality, the numerator roots -11 and -3 may be included when they border the selected intervals. The denominator zeros 3 and 11 remain excluded under all circumstances.

27. Four critical points: -5, 0, 5, 8

Question. Solve the original rational inequality x(x−5)/[(x+5)(x−8)] <0. The numerator roots are 0 and 5; the denominator zeros are -5 and 8. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -5, 0, 5, 8. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-5,0) ∪ (5,8). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -5 and 8 remain excluded under all circumstances.

28. Four critical points: -10, 2, 3, 11

Question. Solve the original rational inequality (x−2)(x−11)/[(x+10)(x−3)] ≤0. The numerator roots are 2 and 11; the denominator zeros are -10 and 3. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -10, 2, 3, 11. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is (-10,2] ∪ (3,11]. Because the comparison includes equality, the numerator roots 2 and 11 may be included when they border the selected intervals. The denominator zeros -10 and 3 remain excluded under all circumstances.

29. Four critical points: -5, -1, 6, 9

Question. Solve the original rational inequality (x−6)(x−9)/[(x+5)(x+1)] >0. The numerator roots are 6 and 9; the denominator zeros are -5 and -1. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -5, -1, 6, 9. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for >0. The solution set is (−∞,-5) ∪ (-1,6) ∪ (9,∞). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -5 and -1 remain excluded under all circumstances.

30. Four critical points: -7, 0, 9, 10

Question. Solve the original rational inequality (x+7)(x−10)/[x(x−9)] ≥0. The numerator roots are -7 and 10; the denominator zeros are 0 and 9. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the positive sign.

Exact worked answer. The sorted critical values are -7, 0, 9, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the positive intervals for ≥0. The solution set is (−∞,-7] ∪ (0,9) ∪ [10,∞). Because the comparison includes equality, the numerator roots -7 and 10 may be included when they border the selected intervals. The denominator zeros 0 and 9 remain excluded under all circumstances.

31. Four critical points: -11, -2, 0, 5

Question. Solve the original rational inequality (x+11)x/[(x+2)(x−5)] <0. The numerator roots are -11 and 0; the denominator zeros are -2 and 5. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -11, -2, 0, 5. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for <0. The solution set is (-11,-2) ∪ (0,5). Because this is a strict comparison, even valid numerator roots giving zero are excluded. The denominator zeros -2 and 5 remain excluded under all circumstances.

32. Four critical points: -8, 7, 9, 10

Question. Solve the original rational inequality (x+8)(x−7)/[(x−9)(x−10)] ≤0. The numerator roots are -8 and 7; the denominator zeros are 9 and 10. First mark all four values on a number line in ascending order and explicitly exclude the denominator zeros before any cancellation or multiplication. The next task is to identify the intervals where the quotient has the negative sign.

Exact worked answer. The sorted critical values are -8, 7, 9, 10. On the far-right interval, all four factors are positive, so the quotient is positive; as each simple root or pole is crossed to the left, one factor changes sign. Select the negative intervals for ≤0. The solution set is [-8,7] ∪ (9,10). Because the comparison includes equality, the numerator roots -8 and 7 may be included when they border the selected intervals. The denominator zeros 9 and 10 remain excluded under all circumstances.

Ten quick practice answers

(x−1)/(x−2)>0: (−∞,1)∪(2,∞).
(x−1)/(x−2)≥0: (−∞,1]∪(2,∞).
(x−1)/(x−2)<0: (1,2).
(x−1)/(x−2)≤0: [1,2).
(x−1)(x−4)/(x−2)>0: (1,2)∪(4,∞).
The same expression ≥0: [1,2)∪[4,∞).
(x−1)²/(x−3)>0: (3,∞).
(x−1)²/(x−3)≤0: (−∞,3).
(x−1)(x−2)/(x−2)≥0: [1,2)∪(2,∞); x=2 stays excluded.
(x+3)(x−3)/[(x+2)(x−1)]>0: (−∞,−3)∪(−2,1)∪(3,∞).

Five-day revision and small-group audit

Day one: distinguish numerator zeros from denominator zeros and mark them on a number line. Day two: solve simple ratios of two linear factors without multiplying across an unknown sign. Day three: introduce two quadratic-factor products and practise finding five sign intervals. Day four: change strict comparisons into inclusive comparisons, adding zeros while never adding poles. Day five: study repeated factors and cancelled common factors, where ordinary sign-flip and domain assumptions need extra care.

In a three-learner tutorial, one student factors numerator and denominator, another constructs the sign chart, and the third audits endpoints against the original domain. Rotate these roles. Progress is apparent when the student can explain why an excluded pole remains open even in a ≥0 problem, rather than relying on a calculator-only inequality answer.

Frequently asked questions

Can a denominator zero be included for ≥0?

No. A rational expression is undefined where its denominator is zero. Inclusivity applies to permitted numerator zeros, not forbidden inputs.

Why is it dangerous to multiply by x−a?

Its sign changes across x=a. Multiplying an inequality by a negative number reverses its direction, so a blind multiplication can lose a sign case.

What if a factor cancels?

The simplified expression is valid only on the original domain. Retain any excluded input from the denominator even if the factor disappears algebraically.

Does a repeated root flip the sign?

A root of even multiplicity generally does not flip the factor’s sign. Use a test point or sign chart rather than assuming every critical value reverses the inequality.

Are numerator zeros always in the answer?

Only when the inequality includes equality and the denominator is nonzero there. For strict inequalities, zeros are excluded.

Can a calculator replace the factor check?

A graph can support the sign interpretation, but the domain exclusions and exact endpoints still need mathematical justification where requested.

Does 2027 H2 Mathematics include rational inequalities?

Yes. The syllabus includes solving inequalities involving linear and quadratic rational expressions and graphical methods.

What is the fastest independent check?

Choose one test value from each sign interval and substitute into the original factored expression, then audit all excluded denominator values.

Related H2 guides and official curriculum

Circular Seating: When Rotations Count as the Same
Rational Inequalities: Why Denominator Zeros Stay Open: this guide
Maclaurin by Implicit Differentiation: First Four Terms
Recurrence Sequences: Fixed Points vs Actual Terms

For related mathematical ownership, see the Hole or Vertical Asymptote? guide, Bukit Timah H2 Modulus, Inequalities and Graphs, and JC2 Bukit Timah tuition. The immutable Clementi small-group Mathematics tutoring reference concerns the tuition format at another level.

Official source: SEAB Singapore–Cambridge H2 Mathematics 9758 syllabus for 2027, Equations and Inequalities §1.3. All cases are original fictional mathematical exercises, not official past-year examination items or records of student performance.

JC2 H2 Mathematics Tutor Bukit Timah | Rational Inequalities: Why Denominator Zeros Stay Open