Linear and quadratic inequalitiesEdexcel A-Level Maths: Revision notes

Solve a linear inequality like an equation, with one extra rule: multiplying or dividing by a negative number reverses the inequality sign. Collect the $x$ terms on the side that keeps the coefficient positive if you can, which avoids the reversal altogether. Example: $7-2x<3x+1\Rightarrow 6<5x\Rightarrow x>\frac65$. Here $x$ stays on the side with a positive coefficient, so there is no reversal. Brackets are expanded first, and fractions are cleared by multiplying every term by a positive common denominator. Example: $\frac{x+1}{3}\ge\frac{x}{2}-1\Rightarrow2(x+1)\ge3x-6\Rightarrow x\le8$.

Never divide by $x$ or cancel an $x$ from both sides. Instead: (1) rearrange to $px^2+qx+r\ \square\ 0$; (2) find the critical values by solving $px^2+qx+r=0$; (3) sketch the parabola and read off the correct region. Example: $x^2-3x-10<0$. Critical values from $(x-5)(x+2)=0$ are $x=-2$ and $x=5$. The parabola opens upwards, so it is below the axis between the roots: $-2<x<5$. For $x^2-3x-10\ge0$ the answer is $x\le-2$ or $x\ge5$. A squared term is simple to handle directly: $x^2\le16$ gives $-4\le x\le4$, and $x^2>9$ gives $x<-3$ or $x>3$.

A solution between two roots is a single interval, written with 'and' (or a double inequality): $-2<x<5$ means $x>-2$ and $x<5$. A solution outside the roots is two separate regions, joined with or: $x<-2$ or $x>5$. Never write $x<-2$ and $x>5$, because no number satisfies both. Set notation: $\{x:-2<x<5\}$, or in interval form $(-2,5)$. For 'or': $\{x:x<-2\}\cup\{x:x>5\}$. Combining two conditions means taking the intersection: for $x>\frac65$ and $x\le4$ the answer is $\left\{x:\frac65<x\le4\right\}$.

If an inequality has $x$ in a denominator you cannot multiply by it directly, because it may be negative. Multiply by the square of the denominator, which is always positive, so the sign never changes. Example: $\frac{6}{x+2}<3$ with $x\ne-2$. Multiply by $(x+2)^2$: $6(x+2)<3(x+2)^2\Rightarrow3x(x+2)>0$. Critical values $x=-2$ and $x=0$, so $x<-2$ or $x>0$. Check with a test value: $x=1$ gives $2<3$, true; $x=-1$ gives $6<3$, false. Note that $x=-2$ is never included, because the denominator is zero there.

Solving $f(x)<g(x)$ means finding where the curve $y=f(x)$ is below the curve or line $y=g(x)$. Find the intersections by solving $f(x)=g(x)$, then read off the interval. Example: the curve $y=x^2-3x-10$ is below the line $y=2x+4$ when $x^2-5x-14<0$, that is $(x-7)(x+2)<0$, so $-2<x<7$. To show a region, graph the boundary and shade the side that satisfies the inequality. Use a dotted line (or curve) for $<$ or $>$, and a solid line for $\le$ or $\ge$. For $y>x+1$, draw $y=x+1$ dotted and shade above it. For $y\le ax^2+bx+c$, draw the curve solid and shade below it.