Week 1 · Day 3 Lesson 1.5–1.6 About 14 minutes Free

Quadratic and rational inequalities

By the end: Solve quadratic and rational inequalities with the “between or outside the roots” rule, and write the answer in interval notation with the right brackets.

The “between or outside the roots” trick appeared at Question 3 and again at Questions 10–12 in all five sittings we studied. The method below is the whole topic.

Understand it

A quadratic inequality asks where a parabola is above or below the $x$ -axis. Everything depends on two things: where it crosses the axis (the roots) and which way it opens.

Quadratic inequalities, step by step.

Move everything to one side so the other side is $0$ .
Make the coefficient of $x^2$ positive. If it is negative, multiply by $-1$ and flip the inequality sign.
Factor, or use the quadratic formula, to find the roots $x_1<x_2$ .
With a positive leading coefficient, the parabola opens upward: it is below the axis between the roots and above it outside the roots. So “ $<0$ ” means between, and “ $>0$ ” means outside.
For $\le$ or $\ge$ the roots are included (square brackets). For $<$ or $>$ they are not (round brackets).

If the quadratic has no real roots, the parabola never touches the axis. With a positive leading coefficient it is above the axis for every $x$ , so “ $>0$ ” holds for all real numbers and “ $<0$ ” has no solution. If it has a double root $a$ , it only touches the axis there: $(x-a)^2>0$ holds for every $x\ne a$ , $(x-a)^2\ge0$ for every $x$ , $(x-a)^2\le0$ only at $x=a$ , and $(x-a)^2<0$ never.

Rational inequalities. Never multiply both sides by an expression containing $x$ unless you know its sign. Instead:

Move everything to one side, so the other side is $0$ .
Combine into a single fraction $\dfrac{P(x)}{Q(x)}$ and factor the top and the bottom.
$\dfrac{P}{Q}>0$ has the same solutions as $P\cdot Q>0$ , and $\dfrac{P}{Q}<0$ the same as $P\cdot Q<0$ , so you can use the quadratic rule on the product.
A root of the denominator is never allowed, so its bracket is always round, even when the inequality is $\le$ or $\ge$ . A root of the numerator is included only for $\le$ and $\ge$ .
Take a minus sign out of a factor such as $1-x=-(x-1)$ , then multiply both sides by $-1$ and flip the sign: $\dfrac{1-x}{x+2}\ge0$ becomes $\dfrac{x-1}{x+2}\le0$ . (Multiplying the top and the bottom by $-1$ changes nothing, so it never flips the sign.)

Check yourself. Pick one number from each region of your answer and test it in the original inequality. It takes ten seconds and catches reversed brackets and flipped signs.

Essential formulas

(x-x_1)(x-x_2)<0\iff x_1<x<x_2\qquad (x-x_1)(x-x_2)>0\iff x<x_1\ \text{or}\ x>x_2

Here

x_1<x_2

are the roots and the leading coefficient is positive. For

\le

and

\ge

the roots belong to the solution.

\frac{x-a}{x-b}\ge0\iff (x-a)(x-b)\ge0\ \text{and}\ x\ne b

The numerator root

a

is allowed with

\ge

. The denominator root

b

is never allowed, so it always has a round bracket.

ax^2+bx+c\ \text{with}\ \Delta=b^2-4ac<0\ \text{and}\ a>0\ \Rightarrow\ ax^2+bx+c>0\ \text{for all}\ x

\Delta<0

means no real roots. The parabola stays above the axis, so “

>0

” is every real number and “

<0

” has no solution.

(x-a)^2>0\iff x\ne a\qquad (x-a)^2\le0\iff x=a

A double root (

\Delta=0

): the parabola only touches the axis at

a

. So

(x-a)^2\ge0

holds for every

x

, and

(x-a)^2<0

has no solution.

See it

Roots −1 and 3. Below the x-axis (y < 0) for −1 < x < 3. Above the x-axis (y > 0) for x < −1 or x > 3.

(x − 1)/(x + 2) ≤ 0 on −2 < x ≤ 1. The point −2 is never allowed; the point 1 is allowed.

Worked examples

Worked example 1

Solve $2x^2-x-3<0$ .

The leading coefficient is positive and the right side is already $0$ . Factor: $2x^2-x-3=(2x-3)(x+1)$ .
The roots are $x=-1$ and $x=\tfrac32$ .
The inequality is “ $<0$ ”, so the solution is between the roots. The inequality is strict, so both ends are round.
Check with $x=0$ : $2\cdot0-0-3=-3<0$ , which is true.

Answer: $\left(-1,\tfrac32\right)$

Worked example 2

Solve $-x^2+2x+3\ge0$ .

The leading coefficient is negative. Multiply by $-1$ and flip the sign: $x^2-2x-3\le0$ .
Factor: $(x-3)(x+1)\le0$ , with roots $-1$ and $3$ .
“ $\le0$ ” means between the roots, and $\le$ includes the roots.
Check with $x=0$ in the original: $-0+0+3=3\ge0$ , which is true.

Answer: $[-1,3]$

Worked example 3

Solve $\dfrac{x-3}{x+1}\ge2$ .

Do not cross-multiply: the sign of $x+1$ is unknown. Move everything to the left: $\dfrac{x-3}{x+1}-2\ge0$ .
Combine: $\dfrac{x-3-2(x+1)}{x+1}=\dfrac{-x-5}{x+1}\ge0$ .
The numerator is $-x-5=-(x+5)$ . Multiply both sides by $-1$ and flip the sign: $\dfrac{x+5}{x+1}\le0$ .
Roots: $x=-5$ from the numerator (allowed, because of $\le$ ) and $x=-1$ from the denominator (never allowed). “ $\le0$ ” means between: $-5\le x<-1$ .
Check $x=-3$ , inside the answer: $\dfrac{-6}{-2}=3\ge2$ is true. Check $x=0$ , outside it: $\dfrac{-3}{1}=-3\ge2$ is false. Both agree with the answer.

Answer: $[-5,-1)$

Try it yourself

Work it on paper first. Open a hint only when you are stuck, then compare with the solution.

Solve $\dfrac{1-x}{x+2}\ge0$ . Then, if the solution set of $x^2+bx+c<0$ is $(-1,3)$ , find $b$ and $c$ .

Hint 1 of 3

Write $1-x=-(x-1)$ , then multiply both sides by $-1$ and flip: $\dfrac{x-1}{x+2}\le0$ .

Hint 2 of 3

Which root is allowed and which is never allowed?

Hint 3 of 3

For the second part, the solution “between $-1$ and $3$ ” tells you the roots, and the leading coefficient is $1$ .

Show the step-by-step solution

Write $1-x=-(x-1)$ , so the inequality is $-\dfrac{x-1}{x+2}\ge0$ . Multiply both sides by $-1$ and flip the sign: $\dfrac{x-1}{x+2}\le0$ .
The numerator root $x=1$ is allowed (it gives $0$ ). The denominator root $x=-2$ is never allowed. “ $\le0$ ” means between the roots, so $-2<x\le1$ .
Check $x=0$ , inside the answer: $\dfrac{1}{2}\ge0$ is true. Check $x=2$ , outside it: $\dfrac{-1}{4}\ge0$ is false. Both agree with the answer.
For the second part, “ $<0$ between $-1$ and $3$ ” with leading coefficient $1$ means $x^2+bx+c=(x+1)(x-3)=x^2-2x-3$ . So $b=-2$ and $c=-3$ .

Answer: $(-2,1]$ ; $b=-2$ , $c=-3$ .

Common mistakes

Reversing between and outside. Say it aloud: “less than zero is between, greater than zero is outside”, for a positive leading coefficient only.
Forgetting to flip the sign when multiplying or dividing by a negative number, for example when turning $-x^2+2x+3\ge0$ into $x^2-2x-3\le0$ .
Giving a denominator root a square bracket. A value that makes the denominator $0$ is never allowed.
Writing $\mathbb{R}$ for $(x-3)^2>0$ . At $x=3$ the left side is $0$ , which is not $>0$ , so the answer is every $x\ne3$ .
Cross-multiplying a rational inequality without knowing the sign of the denominator. Move everything to one side instead.
Writing “between” answers as a union such as $(-\infty,-1)\cup(3,+\infty)$ . Between is one interval, and outside is two pieces joined with $\cup$ .

Exam tips

Rule of thumb: positive leading coefficient, “ $<0$ ” is between the roots, “ $>0$ ” is outside. A negative factor flips it. A denominator root is always open.
Make the $x^2$ coefficient positive first, then solve. This avoids most sign errors.
Test $x=0$ (if it is not a root) in your final answer and in the original inequality. If they disagree, something flipped.
When the options differ only in brackets, decide each endpoint on its own: numerator roots follow the sign ( $\le$ , $\ge$ ), denominator roots are always round.

In short

Quadratic: positive leading coefficient, then “ $<0$ ” is between the roots, “ $>0$ ” is outside.
Double root $a$ : $(x-a)^2>0$ is every $x\ne a$ ; $(x-a)^2\le0$ is only $x=a$ .
Rational: move everything to one side, combine into one fraction, factor, and treat $\dfrac PQ$ like $P\cdot Q$ .
A denominator root is never included. A negative factor flips the sign.

Knowledge check

5 questions on this lesson, marked as you go. Take them in ACE CSCA with a free account: passing them completes the lesson, and your progress is saved on every device.

All of Week 1 is free: the eight lessons, the daily questions and the Week 1 test.

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