How do you clear fractions in an inequality?

Multiply every term by a positive common denominator; the sign does not change.

-2x\ge6, so x\le-3 (sign reversed).

What does 'and' mean in a solution such as x\ge-3 and x<7?

Both conditions hold; write as a single chain, -3\le x<7.

What does 'or' mean in a solution?

Either condition holds, giving two separate pieces, e.g. x<-1 or x4.

What are the four steps for a quadratic inequality?

Rearrange to 0, find critical values, sketch, read off the region.

For a \cup-shaped quadratic with roots \alpha<\beta, what is the solution of f(x)<0?

\alpha<x<\beta (between the roots, 'and').

For a \cup-shaped quadratic with roots \alpha<\beta, what is the solution of f(x)0?

x<\alpha or x\beta (outside the roots, 'or').

Why must you not divide x^23x by x?

x could be negative, which reverses the sign and loses the solutions x<0; use x(x-3)0.

How do you find where two inequalities are both true?

Take the intersection (overlap) of their solution sets, e.g. using number lines.

What must you do at the end of a context problem?

Apply any restriction from the context, e.g. a length must satisfy x0.

Linear and quadratic inequalitiesAQA A-Level Maths: Flashcards

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What happens to an inequality sign when you multiply or divide by a negative number?

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What happens to an inequality sign when you multiply or divide by a negative number?: It reverses: $<$ becomes $>$ and $\le$ becomes $\ge$ .
How do you clear fractions in an inequality?: Multiply every term by a positive common denominator; the sign does not change.
Solve $5-2x\ge11$ .: $-2x\ge6$ , so $x\le-3$ (sign reversed).
What does 'and' mean in a solution such as $x\ge-3$ and $x<7$ ?: Both conditions hold; write as a single chain, $-3\le x<7$ .
What does 'or' mean in a solution?: Either condition holds, giving two separate pieces, e.g. $x<-1$ or $x>4$ .
Write $-3\le x<7$ in set notation.: $\{x:-3\le x<7\}$
What are the four steps for a quadratic inequality?: Rearrange to $0$ , find critical values, sketch, read off the region.
For a $\cup$ -shaped quadratic with roots $\alpha<\beta$ , what is the solution of $f(x)<0$ ?: $\alpha<x<\beta$ (between the roots, 'and').
For a $\cup$ -shaped quadratic with roots $\alpha<\beta$ , what is the solution of $f(x)>0$ ?: $x<\alpha$ or $x>\beta$ (outside the roots, 'or').
Solve $x^2-4x-12<0$ .: $(x-6)(x+2)<0$ , so $-2<x<6$ .
Why must you not divide $x^2>3x$ by $x$ ?: $x$ could be negative, which reverses the sign and loses the solutions $x<0$ ; use $x(x-3)>0$ .
What is the solution of $(x-3)^2>0$ ?: All real $x$ except $x=3$ .
How do you find where two inequalities are both true?: Take the intersection (overlap) of their solution sets, e.g. using number lines.
What must you do at the end of a context problem?: Apply any restriction from the context, e.g. a length must satisfy $x>0$ .

Exam questions on Linear and quadratic inequalities

The inequality $3(2x-1)<5x+4$ is to be solved, together with the inequality $2x+1\ge-5$ .
Write the solution set in part (b) using set notation, and state how many integers it contains.2 marks
A rectangular garden has width $x$ metres and length $(x+6)$ metres.
The perimeter of the garden must be at least $20$ m as well. Find the range of values of $x$ for which both the perimeter and the area conditions are met.2 marks
Let $f(x)=2x^2-5x-12$ .
Solve $f(x)\ge0$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).