B5 - Linear and quadratic inequalities

Rearranging linear inequalities

The equation $x^2-x-6=0$ has two solutions. Can the inequality $x^2-x-6\le0$ have infinitely many? At $x=0$, the expression equals $-6$, so this value satisfies the inequality even though it is not a root. To find every allowed value, first make sure your rearrangements preserve a comparison.

For instance, $-2x+3>7$ becomes $-2x>4$, then $x<-2$. The last sign reverses: on a number line, multiplying by a negative number reverses left and right. The true comparison $2<5$ becomes $-4>-10$ when multiplied by $-2$.

An inequality states that one expression is smaller or larger than another. The signs $<$ and $>$ are strict and exclude equality; $\le$ and $\ge$ allow equality. Thus $x<2$ excludes $2$, while $x\le2$ includes it. A solution must make the original comparison true.

When denominators are positive constants, their positive common multiple clears the fractions without reversing the inequality. Keep each complete numerator in brackets until you expand.

Worked example 2 changes only the coefficient of $x$ in the right numerator, from $1$ to $5$. Collecting now gives a negative coefficient, so the final division changes the direction.

These examples use constant denominators. Multiplying an inequality by an expression such as $x-2$ is unsafe unless you know its sign: it is negative on one side of $2$, positive on the other and zero at $2$.

The next question changes the same numerator coefficient once more. Check the direction as carefully as the arithmetic.

Quadratic signs and critical values

A quadratic's roots locate where its graph meets the horizontal axis; the graph or the signs of its factors then show which intervals satisfy an inequality.

For $(x+2)(x-3)$, try $x=-3$, $x=0$ and $x=4$. The products are $6$, $-6$ and $6$. These test values lie in the three intervals separated by $-2$ and $3$. A factor cannot change sign inside one of these intervals without passing through zero, so one test value determines each interval's sign.

Always collect the inequality onto one side before choosing intervals. The leading coefficient matters: an upward-opening quadratic is positive outside two distinct roots, while a downward-opening quadratic has the opposite sign pattern. These sign rules are not in the formula booklet: learn how to recover them from a sketch or factors.

The graph gives the same argument visually. Its shaded portion is on or below the horizontal axis, and the closed number-line endpoints include equality.

Worked example 4 changes only the sign of the entire quadratic. Its roots stay fixed, but every nonzero height changes sign.

For a strict comparison, omit the roots: for instance, $-x^2+4x-3>0$ becomes $-(x-1)(x-3)>0$, giving $1<x<3$. Equality at either root does not meet $>0$.

A quadratic inequality can also allow every real number, or none. Since a real square is non-negative, $(x-1)^2\ge0$ holds for all real $x$, whereas $(x-1)^2<0$ has no solution. A repeated root does not cause a sign change. With no real roots, the sign is constant too: $x^2+1>0$ always, while $x^2+1\le0$ never. These conclusions come from the expressions themselves; they are not supplied in the formula booklet.

Now change the constant and linear terms of a downward-opening quadratic. Establish its new critical values before selecting the outside intervals.

Writing solution sets

Set notation records which values are included, so an interval cannot be mistaken for a list of its endpoints or a list of integers.

The answer $-2\le x\le3$ includes $\dfrac12$, $\sqrt2$ and every other real number between the bounds. By contrast, $\{-2,-1,0,1,2,3\}$ lists exactly six numbers. Braces mean a set, and the commas in a listed set separate its individual elements.

An interval contains every real number between its bounds. A square bracket includes a finite endpoint and a round bracket excludes it. The following notation is not supplied in the formula booklet: learn its meaning.

Infinity is not a real endpoint that can be included. Thus $x<2$ is $(-\infty,2)$, and $x\ge2$ is $[2,\infty)$. The empty set, $\varnothing$, records no solutions; all real numbers are $\mathbb R$.

The symbol $\in$ means “is an element of”, so $0\in S$ and $4\notin S$. If $T=\{-2,-1,0,1,2,3\}$, then every element of $T$ belongs to $S$, so $T\subseteq S$. It is a proper subset, written $T\subset S$, because $S$ also has elements outside $T$, such as $\dfrac12$. The notation $\subseteq$ allows equality; $\subset$ here does not.

For a finite set, $n(A)$ counts its elements. Here $n(T)=6$, while $n(\varnothing)=0$; the real interval $S$ has infinitely many elements.

The domain matters when deciding what “all” means. The standard number sets are:

For instance, $\dfrac12\in\mathbb Q$ but $\dfrac12\notin\mathbb Z$. Restricting $-2\le x\le3$ to integers gives the six-element set $T$; keeping a real domain gives the whole interval $S$.

The universal set $U$ contains all the values under discussion. If $U=\{0,1,2,3,4\}$ and $A=\{1,3\}$, the values of $U$ outside $A$ form its complement, $A'=\{0,2,4\}$. In general, $A'=\{x\in U:x\notin A\}$; the complement depends on $U$. For real inequality solutions, $U$ is usually $\mathbb R$, unless the question restricts it. These number-set, membership, subset and complement conventions are not supplied in the formula booklet.

Read each option below as a claim about every real value, including the endpoints and non-integers.

Combining conditions with and and or

Several inequalities can describe one solution set: “and” keeps their overlap, while “or” keeps values allowed by either condition.

If $x\le3$ and $x>1$, a value must pass both tests, giving $1<x\le3$. If the conditions use “or”, the value need pass only one. A value passing both is still included: mathematical “or” is inclusive.

Solve each condition separately first. Then compare their number-line intervals; do not try to join the unsolved algebra by a new inequality sign.

Worked example 6 changes only the joining word to “or”. The two individual sets are unchanged, but now you keep their union.

A union need not be one continuous interval. For example, $x^2-x-6\le0$ or $x<-4$ gives $(-\infty,-4)\cup[-2,3]$; the gap stays. Conversely, $x<-2$ and $x>3$ has no solution, since a number cannot meet both demands.

For the next question, the second condition moves to the other side of the quadratic interval. Decide whether any gap remains in the union.

Quick Recap

• Reverse an inequality when multiplying or dividing by a negative number.
• For a quadratic, find the critical values and use signs or a graph to select whole intervals.
• Strict comparisons exclude equality; square interval brackets include finite endpoints.
• “And” takes an intersection; inclusive “or” takes a union.

The opening inequality has infinitely many solutions: every real number in $[-2,3]$ satisfies $x^2-x-6\le0$, while only the two endpoints solve the corresponding equation.