Linear and quadratic inequalitiesAQA A-Level Maths: Revision notes

An inequality compares two expressions using $<$, $\le$, $>$ or $\ge$. Solve a linear inequality like an equation, with one crucial difference: multiplying or dividing both sides by a negative number reverses the inequality sign. Brackets are expanded first, and fractions are removed by multiplying every term by a positive common denominator (which does not change the sign). Example: $3(2x-1)<5x+4\Rightarrow6x-3<5x+4\Rightarrow x<7$. Example with a fraction: $\frac{2x-1}{3}-\frac{x+2}{4}>1$; multiplying by 12 gives $4(2x-1)-3(x+2)>12$, so $5x>22$ and $x>\frac{22}{5}$. Example with a negative coefficient: $5-2x\ge11\Rightarrow-2x\ge6\Rightarrow x\le-3$.

A solution is a set of values. Write it as an inequality, as set notation, or using and/or.

• 'and': both conditions hold, e.g. $x\ge-3$ and $x<7$, written as one chain $-3\le x<7$. In set notation, $\{x:-3\le x<7\}$.
• 'or': either condition holds, e.g. $x<-1$ or $x>4$. In set notation, $\{x:x<-1\}\cup\{x:x>4\}$. A chain such as $4<x<-1$ is impossible and must never be written. A solution which has two separate pieces always needs 'or'. When you need values satisfying two inequalities at once, take the intersection (the overlap) of the two solution sets.

To solve a quadratic inequality, use these steps.

1. Rearrange so that one side is $0$, e.g. $x^2+6x-40<0$.
2. Solve the equation to find the critical values: $(x+10)(x-4)=0$ gives $x=-10$ and $x=4$.
3. Sketch the graph. With a positive $x^2$ coefficient it is a $\cup$-shape.
4. Read off the part of the $x$-axis where the curve is above ($>0$) or below ($<0$) the axis. For a $\cup$-shaped graph with roots $\alpha<\beta$: $f(x)<0$ gives the inside region $\alpha<x<\beta$ ('and'); $f(x)>0$ gives the outside region $x<\alpha$ or $x>\beta$ ('or'). So $x^2+6x-40<0$ gives $-10<x<4$.

Solve $2x^2-5x-12\ge0$. Factorising, $(2x+3)(x-4)\ge0$ with critical values $x=-\frac32$ and $x=4$. The graph is a $\cup$-shape and we need it on or above the axis, so $x\le-\frac32$ or $x\ge4$. The inequality is $\ge$, so the critical values are included. If the $x^2$ coefficient is negative, multiply through by $-1$ and reverse the sign (or sketch an $\cap$-shape instead). For instance, $-x^2+3x+10>0$ becomes $x^2-3x-10<0$, so $(x-5)(x+2)<0$ and $-2<x<5$. If the quadratic does not factorise, use the quadratic formula for the critical values and leave surds exact if asked.

When two conditions apply, solve each one, then find where both are true. Example: $f(x)\ge0$ gives $x\le-\frac32$ or $x\ge4$, and $f(x)<3x+12$ gives $-2<x<6$. Both hold for $-2<x\le-\frac32$ or $4\le x<6$. Context problems add a restriction, e.g. a length must satisfy $x>0$. Always apply it at the end. Special cases: $(x-3)^2\ge0$ is true for all $x$; $(x-3)^2>0$ is true for all $x$ except $3$; $(x-3)^2\le0$ is true only when $x=3$; and $(x-3)^2<0$ has no solutions. A quadratic such as $x^2+4$ is always positive. Never divide by an expression containing $x$, such as dividing $x^2>3x$ by $x$, because $x$ might be negative and that would lose solutions. Instead write $x^2-3x>0$ and factorise to $x(x-3)>0$, giving $x<0$ or $x>3$.