MathsISC Class 11 · Linear and Quadratic Inequalities

Quadratic and Rational Inequalities by the Method of Intervals

Updated 17 October 2026 · Fact-checked

To solve a quadratic or rational inequality, move everything to one side, factorise, mark the critical points on the number line and test the sign in each interval (the method of intervals). A factor appearing an even number of times does not change the sign, and zeros of denominators are always excluded.

Start with + on the right, flip sign at each critical point, keep the + intervals for ≥ 0, and always leave out the denominator zero.

Illustration drawn for QuestionBuddy.

Understand Quadratic and Rational Inequalities: Method of Intervals

Quadratic inequalities. To solve $a x^{2} + b x + c > 0$ (or <, ≥, ≤): (1) factorise to $a (x - p) (x - q)$ with $p < q$ ; (2) mark p and q on the number line, which split it into three intervals; (3) for $a > 0$ the quadratic is positive outside the roots and negative between them; for $a < 0$ the reverse. So $(x - p) (x - q) < 0$ gives $p < x < q$ , and $(x - p) (x - q) > 0$ gives $x < p$ or $x > q$ .

Discriminant cases. If $D < 0$ and $a > 0$ , the quadratic is always positive, so $> 0$ holds for all real x and $< 0$ has no solution. If $D = 0$ , $a (x - p)^{2} \geq 0$ always; the inequality $> 0$ holds for $x \neq = p$ .

Method of intervals (wavy curve). For a product or quotient of linear factors, mark all the critical points (zeros of numerator and denominator) in increasing order. Start from the right: for large positive x with the leading coefficient positive, the expression is positive; the sign alternates when passing a critical point of odd multiplicity and stays the same through a point of even multiplicity.

Rational inequalities. Never multiply by an expression whose sign is unknown. Instead move all terms to one side, combine into a single fraction $\frac{f ( x )}{g ( x )}$ , and analyse the signs. Include zeros of the numerator for ≥ or ≤, but always exclude zeros of the denominator.

Presenting the answer. Write in interval notation with correct brackets and check a test value in each interval.

Quadratic and Rational Inequalities: Method of Intervals: formulas to remember

Quadratic with roots p < q: $$(x - p)(x - q) < 0 ⟺ p < x < q$ | $(x - p)(x - q) > 0 ⟺ x < p$ or $x > q$$; For a > 0.
Perfect square: $$(x - p)^2 \geq 0$ always | $(x - p)^2 > 0 ⟺ x \neq p$$; No sign change at p.
Rational: $$\frac{f(x)}{g(x)} ≥ 0$: critical points from f and g; exclude zeros of g$; Sign chart.
Repeated factor: $E v e n p o w er : s i g n u n c han g e d ∣ O dd p o w er : s i g n c han g es$; At the critical point.

Quadratic and Rational Inequalities: Method of Intervals in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Safe speed range

Suppose a road's rules need the speed v (in km/h) to satisfy $v^{2} - 100 v + 2400 < 0$ for fuel-efficient driving in a test model.

Find the speed range.

Try it first, then show the solutionSolution

Factorise: $v^{2} - 100 v + 2400 = (v - 40) (v - 60)$ .
The product is negative between the roots: $40 < v < 60$ .
Check: v = 50 gives $2500 - 5000 + 2400 = - 100 < 0$ .

Answer: 40 < v < 60 km/h.

Memory hook: A positive quadratic is negative between its roots.

Tricks to remember Quadratic and Rational Inequalities: Method of Intervals

Short memory aids. Say each one aloud once, then use it on a question the same day.

Between roots negative
For a > 0, the quadratic is below zero between its roots.
Use it when: Quick solving.
Rational means no multiply
Move to one side and combine; do not cross-multiply.
Use it when: Rational inequalities.

Quadratic and Rational Inequalities: Method of Intervals: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Interval of a quadratic · 1 mark · MCQBoard level

The solution of $x^{2} - 5 x + 6 < 0$ is:
(a) $(2, 3)$
(b) $(- \infty, 2) \cup (3, \infty)$
(c) $[2, 3]$
(d) $(- 3, - 2)$

Try it first, then show the solutionSolution

$(x - 2) (x - 3) < 0$ between the roots.

Answer: (a) $(2, 3)$

Example 2: Quadratic inequality · 3 marks · Short answerBoard level

Solve $x^{2} - x - 12 \geq 0$ .

Try it first, then show the solutionSolution

Factorise: $(x - 4) (x + 3) \geq 0$ . (1 mark)
Critical points −3 and 4; the product is positive outside the roots and zero at them. (1 mark)
Solution: $x \leq - 3$ or $x \geq 4$ , that is $(- \infty, - 3] \cup [4, \infty)$ . (1 mark)

Answer: (−∞, −3] ∪ [4, ∞)

Example 3: Rational inequality with a product · 4 marks · Short answerHarder board level

Solve $\frac{( x - 1 ) ( x + 2 )}{x - 3} \geq 0$ .

Try it first, then show the solutionSolution

Critical points: −2, 1 (zeros of the numerator) and 3 (zero of the denominator, excluded). (1 mark)
For x > 3 all three factors are positive: positive. The sign alternates: negative on (1, 3), positive on (−2, 1), negative for x < −2. (1.5 marks)
Include −2 and 1 (the expression is 0 there); exclude 3. (0.5 mark)
Solution: $[- 2, 1] \cup (3, \infty)$ . (1 mark)

Answer: [−2, 1] ∪ (3, ∞)

Example 4: Fraction compared with a constant · 4 marks · Short answerHarder board level

Solve $\frac{x + 1}{x - 2} \leq 3$ .

Try it first, then show the solutionSolution

Move to one side: $\frac{x + 1}{x - 2} - 3 \leq 0$ . (1 mark)
Combine: $\frac{x + 1 - 3 x + 6}{x - 2} = \frac{7 - 2 x}{x - 2} \leq 0$ , equivalent to $\frac{2 x - 7}{x - 2} \geq 0$ . (1.5 marks)
Critical points 2 (excluded) and 7/2; positive for x > 7/2 and x < 2, negative between. (1 mark)
Solution: $(- \infty, 2) \cup [\frac{7}{2}, \infty)$ . (0.5 mark)

Answer: (−∞, 2) ∪ [7/2, ∞)

Example 5: Repeated factors and no real roots · 5 marks · Long answerHOTS / competency

Solve $(x - 1)^{2} (x + 2) > 0$ and $x^{2} + 2 x + 5 > 0$ . For which values of k does $x^{2} + k x + 4 = 0$ have real roots?

Try it first, then show the solutionSolution

$(x - 1)^{2} \geq 0$ and equals 0 only at x = 1, so we need $x + 2 > 0$ and $x \neq = 1$ : $(- 2, 1) \cup (1, \infty)$ . (2 marks)
$x^{2} + 2 x + 5 = (x + 1)^{2} + 4 > 0$ for every real x: solution ℝ. (1.5 marks)
Real roots need discriminant $k^{2} - 16 \geq 0$ , so $k \leq - 4$ or $k \geq 4$ . (1.5 marks)

Answer: (−2, 1) ∪ (1, ∞); ℝ; k ≤ −4 or k ≥ 4.

Where students lose marks in Quadratic and Rational Inequalities: Method of Intervals

The slips examiners see most often, and what to write instead.

Mistake: Using the wrong region for a quadratic
Writing between the roots for x² > 9.
Do this: For a > 0: "> 0" is outside the roots; "< 0" is between them.
Usually costs: 2 marks.
Mistake: Cross-multiplying in a rational inequality
Multiplying by a denominator of unknown sign.
Do this: Move terms to one side and use a sign chart.
Usually costs: 2 marks.
Mistake: Including a zero of the denominator
Writing [3 for a rational function.
Do this: Always exclude zeros of the denominator.
Usually costs: 1 to 2 marks.

Exam tips

Factorise completely before using a sign chart.
List all critical points in order.
Test one value in each interval.
Check the end points against the original inequality.

Practice questions on Quadratic and Rational Inequalities: Method of Intervals

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The solution of $x^{2} > 9$ is:
(a) $x > 3$
(b) $- 3 < x < 3$
(c) $x < - 3$ or $x > 3$
(d) $x < 3$
Hint
Factorise x² − 9.
Answer
(c) $x < - 3$ or $x > 3$
Q2 · 2 marks · Short answerBoard level
Solve $2 x^{2} - 7 x + 3 < 0$ .
Hint
Factorise.
Answer
$(2 x - 1) (x - 3) < 0$ gives $\frac{1}{2} < x < 3$ .
Q3 · 3 marks · Short answerHarder board level
Solve $\frac{x - 2}{x + 3} > 0$ .
Hint
Signs of numerator and denominator.
Answer
Critical points −3 (excluded) and 2 (excluded since strict). Positive for $x < - 3$ or $x > 2$ : $(- \infty, - 3) \cup (2, \infty)$ .
Q4 · 3 marks · Short answerHarder board level
Solve $x^{2} - 4 x + 4 \leq 0$ and explain.
Hint
Perfect square.
Answer
$(x - 2)^{2} \leq 0$ only at $x = 2$ . The solution set is {2}.
Q5 · 3 marks · Short answerHarder board level
Solve $\frac{x ^{2} - 3 x + 2}{x ^{2} - x - 6} < 0$ .
Hint
Factorise both.
Answer
$\frac{( x - 1 ) ( x - 2 )}{( x - 3 ) ( x + 2 )} < 0$ ; critical points −2, 1, 2, 3. Negative on $(- 2, 1)$ and $(2, 3)$ .
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the profit of a stall (in Rs) from selling n items is $P = - n^{2} + 30 n - 200$ .
(i) Factorise P. (1 mark)
(ii) Find the values of n for which the stall makes a profit (P > 0). (1 mark)
(iii) Find the n for maximum profit. (1 mark)
(iv) Find the maximum profit. (1 mark)
Hint
Factorise the negative quadratic.
Answer
(i) $P = - (n - 10) (n - 20)$ . (ii) $P > 0$ for $10 < n < 20$ . (iii) $n = 15$ (midpoint). (iv) $P (15) = - 225 + 450 - 200 = 25$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.(x − p)(x − q) < 0?Show answer
p < x < q.
2.Zeros of denominators?Show answer
Always excluded.
3.Even power factor?Show answer
No sign change.
4.x² + bx + c with D < 0 and a > 0?Show answer
Always positive.

Quadratic and Rational Inequalities: Method of Intervals: frequently asked questions

What is the wavy curve method?

A sign chart drawn as a wavy line through the critical points, passing above and below the number line to show where the expression is positive or negative.

How do I handle a negative leading coefficient?

Multiply by −1 (and reverse the sign) to make it positive before factorising.

Why is the solution of (x − 2)² ≤ 0 only {2}?

A square is never negative, so it can only equal zero, at x = 2.