Topic drillAbout 15 minutesTested in all 5 recent real CSCA papers (10 questions)

Quadratic and Rational Inequalities

A quadratic inequality asks which side of the x-axis a parabola is on. See why "greater than: outside, less than: inside" works, then handle a<0, double roots, no real roots and rational inequalities — with CSCA exam examples, the most common mistakes and a mastery check.

This page is written in Chinese and English (the two CSCA exam languages); you are reading the English version.

1Warm-up

Start with a real exam question, on instinct:

Real examSource: CSCA Math June 2026 Past Paper

The solution set of the quadratic inequality

3(x + 4)(x - 2) < 0

2The idea: read which side of the x-axis the parabola is on

Solving $ax^2+bx+c>0$ asks: for which $x$ is the parabola $y=ax^2+bx+c$ above the x-axis? " $<0$ " asks where it is below. The roots of the equation are exactly where the parabola meets the axis.

Move the roots, flip the opening and change the sign; the bold part of the x-axis is the solution set.

Root 1: -1Root 2: 3

$x^2-2x-3\;>\;0$

$\Leftrightarrow\; (x+1)(x-3)\;>\;0$

Solution set: $(-\infty,-1)\cup(3,+\infty)$

a>0 with two roots: “>” takes the outside, “<” the inside

With $a>0$ the parabola opens up: outside the roots it is above the axis, between them below — hence "greater than: outside; less than: inside". With $\geq$ or $\leq$ , the roots themselves count too, so the endpoints become closed.

With $a<0$ it opens down and everything reverses. Rather than learn a second rule, multiply by $-1$ first to make $a$ positive, flipping the sign, and keep using the one rule above.

With no real root ( $\Delta<0$ ) the whole parabola is on one side, so the answer is $\mathbb{R}$ or $\varnothing$ . With a double root ( $\Delta=0$ ) it just touches the axis: $\{x\mid x\neq x_0\}$ , $\mathbb{R}$ , $\{x_0\}$ or $\varnothing$ — all visible at a glance.

Rational inequalities work the same way: $\frac{f(x)}{g(x)}>0$ has the same solutions as $f(x)\cdot g(x)>0$ (a quotient has the sign of the product). With $\geq$ or $\leq$ add one rule: a root of the denominator is never included.

34 steps

1. Make $a$ positive
If the leading coefficient is negative, multiply by $-1$ and flip the sign. For a fraction, move everything to one side first: $\frac{f(x)}{g(x)}$ compared with $0$ .
2. Find the roots
Factor or use the formula; $\Delta<0$ means no real root, $\Delta=0$ a double root.
3. Read the graph
Opening up: $>0$ is outside the roots, $<0$ between them. For double or no roots, just look.
4. Fix the endpoints
With equality the roots are included (closed), otherwise open; a denominator's root is always open.

Four-step practice

Pick an inequality. At each step, choose first, then check.

Step 1Step 1, make $a$ positive: is the leading coefficient positive?

Read all 5 worked examples in full

$x^2-x-6>0$

Step 1Step 1, make $a$ positive: is the leading coefficient positive?
$a=1>0$ : the parabola opens up.
✗ Multiply both sides by $-1$ : $a=1$ is already positive; multiplying by $-1$ would only force you to flip the sign.
Step 2Step 2, the roots of the matching equation:
$(x+2)(x-3)=0$ : roots $-2,\ 3$ .
✗ $x=2,\ x=-3$ : $x^2-x-6=(x+2)(x-3)$ , so the roots are $-2$ and $3$ — mind the signs: $x+2=0$ gives $x=-2$ .
✗ No real root: $\Delta=1+24=25>0$ : two roots.
Step 3Step 3, read the graph: which part?
Greater than: outside.
✗ between the roots:An upward parabola is above the axis outside the roots — "greater than: outside".
Step 4Step 4, the endpoints:
Strict: open intervals.
✗ Include them (closed):The inequality is strict; at a root the value is $0$ , which is not $>0$ .

Result: $x\in(-\infty,-2)\cup(3,+\infty)$

$-x^2+2x+3\geq 0$

Step 1Step 1, make $a$ positive: is the leading coefficient positive?
It becomes $x^2-2x-3\leq 0$ .
✗ Multiply by $-1$ : $x^2-2x-3\geq 0$ :Multiplying by a negative number flips the sign: $\geq$ becomes $\leq$ . This is the most common mistake on these questions.
✗ Leave it: $a=-1<0$ : it opens down. Make $a$ positive first so one rule covers every case.
Step 2Step 2, the roots of the matching equation:
$(x+1)(x-3)\leq 0$ .
✗ $x=1,\ x=-3$ : $x^2-2x-3=(x+1)(x-3)$ : roots $-1$ and $3$ .
Step 3Step 3, read the graph: which part?
Less than: inside.
✗ outside the roots:It is now $\leq 0$ : less than takes the inside.
Step 4Step 4, the endpoints:
With equality: closed.
✗ Exclude them (open):With $\leq$ , a root gives $0$ , which satisfies it: include it.

Result: $[-1,3]$

$x^2-4x+4>0$

Step 1Step 1, make $a$ positive: is the leading coefficient positive?
Opens up.
✗ Multiply by $-1$ : $a=1$ is already positive.
Step 2Step 2, the roots of the matching equation:
$(x-2)^2>0$ , double root $x=2$ .
✗ Two roots $x=\pm 2$ : $x^2-4x+4=(x-2)^2$ , $\Delta=0$ : a single double root.
✗ No real root: $\Delta=16-16=0$ : one double root.
Step 3Step 3, read the graph: which part?
It only touches the axis at $x=2$ .
✗ $\mathbb{R}$ :At $x=2$ , $(x-2)^2=0$ , which is not $>0$ .
✗ $\varnothing$ :The parabola only touches the axis at $x=2$ and is above it everywhere else.
Step 4Step 4, the endpoints:
Solution: $\{x\mid x\neq 2\}$ .
✗ Write $(2,+\infty)$ :The part $x<2$ satisfies it too.

Result: $\{x\mid x\neq 2\}=(-\infty,2)\cup(2,+\infty)$

$\frac{x-1}{x+2}\leq 0$

Step 1Step 1, convert: the fraction inequality is equivalent to
A quotient has the sign of the product; then drop the denominator's root.
✗ $(x-1)(x+2)\leq 0$ :At $x=-2$ the denominator is $0$ — with $\leq$ you must remove the denominator's root separately.
✗ $x-1\leq 0$ :You cannot multiply by $x+2$ — it may be negative. A quotient and a product have the same sign: use the product.
Step 2Step 2, the roots of the matching equation:
Roots $-2,\ 1$ .
✗ $x=-1,\ x=2$ : $x-1=0$ gives $1$ , $x+2=0$ gives $-2$ .
Step 3Step 3, read the graph: which part?
Less than: inside.
✗ outside the roots: $\leq 0$ : inside.
Step 4Step 4, the endpoints:
Numerator roots follow the equality sign; denominator roots are always open.
✗ Both closed: $x=-2$ makes the denominator $0$ : never included.
✗ Both open:At $x=1$ the fraction is $0$ , which satisfies $\leq 0$ .

Result: $(-2,1]$

$\frac{x+2}{x-1}\geq 2$

Step 1Step 1, move everything to one side:
$\frac{x+2}{x-1}-2=\frac{4-x}{x-1}\geq 0$ .
✗ $x+2\geq 2(x-1)$ :Do not multiply by $x-1$ — its sign is unknown. Move the $2$ over: $\frac{x+2}{x-1}-2=\frac{4-x}{x-1}$ .
Step 2Step 2, make the coefficient of $x$ positive:
$\frac{x-4}{x-1}\leq 0$ .
✗ $\frac{x-4}{x-1}\geq 0$ :Multiplying the numerator by $-1$ flips the sign.
Step 3Step 3, read the graph: which part?
Between $1$ and $4$ .
✗ Outside $1$ and $4$ : $\leq 0$ : inside.
Step 4Step 4, the endpoints:
$(1,4]$ .
✗ $1$ closed, $4$ open:Reversed: $x=1$ zeroes the denominator (never included); $x=4$ makes the fraction $0$ (included).

Result: $(1,4]$

4Worked examples

Try it yourself, then open the steps

Real examSource: CSCA Math January 2026 Past Paper

The solution set of the inequality

x ^ { 2 } - x - 2 > 0

is ().

Step 1Make $a$ positive: $a=1>0$ already. Roots: $x^2-x-2=(x+1)(x-2)=0$ gives $-1,\ 2$ .
Step 2Graph: opens up, $>0$ takes the outside.
Step 3Strict, so endpoints are excluded: $\{x\mid x<-1\ \text{or}\ x>2\}$ — answer A.

Try it yourself, then open the steps

MockSource: Math Mock Exam 1

Find the solution set of the inequality

-x^2+3x-5>0

Step 1Make $a$ positive: $a=-1<0$ , so multiply by $-1$ and flip: $x^2-3x+5<0$ .
Step 2Roots: $\Delta=9-20=-11<0$ , none. An upward parabola with no real root lies entirely above the axis.
Step 3Nothing is below the axis, so the answer is $\varnothing$ — D.

Try it yourself, then open the steps

Real examSource: CSCA Math January 2026 Past Paper

The solution set of the rational inequality

\frac{2x+1}{2-x}\leq0

is ( ).

Step 1Make it positive: the denominator $2-x$ has a negative $x$ coefficient; multiply the whole thing by $-1$ : $\frac{2x+1}{x-2}\geq 0$ (sign flipped).
Step 2As a product: $(2x+1)(x-2)\geq 0$ with $x\neq 2$ ; roots $-\frac12,\ 2$ ; $\geq 0$ takes the outside.
Step 3Endpoints: $-\frac12$ is a numerator root (included with $\geq$ ); $2$ is a denominator root (never). Answer $(-\infty,-\frac12]\cup(2,+\infty)$ — B.

5Common mistakes

Answer data: CSCA.app students' answers over the 90 days before 2026-10-21.

1. Not flipping the sign after multiplying by a negative

What the data shows:On the mock question $\frac{3x-1}{2-x}\geq 1$ , fewer than four in ten students get it right, and C is one of the most common wrong answers: they rewrite $2-x$ as $x-2$ without flipping the sign. On $-x^2+3x-5>0$ , about one in ten choose $\mathbb{R}$ for the same reason.

How to avoid it:Whenever you multiply or divide by a negative — including rewriting a factor like $2-x$ or $3-2x$ as $x-2$ or $2x-3$ — flip the sign. Only then apply the outside/inside rule.

Check yourself

MockSource: Math Mock Exam 1

Find the solution set of the inequality

\frac{3x-1}{2-x}\geq1

2. Including the root of the denominator

What the data shows:On the June 2026 real paper, $\frac{2}{x-2}\leq 0$ : nearly one student in five chooses $(-\infty,2]$ , including $x=2$ , where the denominator is $0$ .

How to avoid it:With $\geq$ or $\leq$ , a numerator root follows the equality sign, but a denominator root is never included — the expression is undefined there.

Check yourself

Real examSource: CSCA Math June 2026 Past Paper

The solution set of the rational inequality

\frac{2}{x-2} \leq 0

3. Mixing up outside and inside

What the data shows:On the warm-up real-exam question, more than one in five students take the outside for "less than"; on a mock question asking for the domain of $\log_{\frac12}(x^2-4x+3)$ , about a quarter take the inside for "greater than".

How to avoid it:Do not rely on the words alone — picture an upward parabola: below the axis ( $<0$ ) is always the part between the roots; above ( $>0$ ) is outside.

Check yourself

MockSource: Math Mock Exam 1

Find the domain of the function

y=\log_{\frac12}(x^2-4x+3)

6Practice

Question 1 of 6

MockSource: Math Mock Exam 7

The solution set of the inequality

x^{2} - 3x + 2 < 0

0 of 6 right in this set

7Mastery check

5 questions; 4 right means you have it. No solutions until the end.

Learn first

Quadratic Functions and Equations

Concept and formulas

Solving Inequalities

Learn next

Sets and Set Operations Function Domains

Chapter practice

Want more? Go to practice

1Warm-up

2The idea: read which side of the x-axis the parabola is on

34 steps

Four-step practice

x2−x−6>0x^2-x-6>0x2−x−6>0

−x2+2x+3≥0-x^2+2x+3\geq 0−x2+2x+3≥0

x2−4x+4>0x^2-4x+4>0x2−4x+4>0

x−1x+2≤0\frac{x-1}{x+2}\leq 0x+2x−1​≤0

x+2x−1≥2\frac{x+2}{x-1}\geq 2x−1x+2​≥2

4Worked examples

5Common mistakes

1. Not flipping the sign after multiplying by a negative

2. Including the root of the denominator

3. Mixing up outside and inside

6Practice

7Mastery check

Learn first

Concept and formulas

Learn next

Chapter practice

$x^2-x-6>0$

$-x^2+2x+3\geq 0$

$x^2-4x+4>0$

$\frac{x-1}{x+2}\leq 0$

$\frac{x+2}{x-1}\geq 2$