Linear and quadratic inequalitiesEdexcel International A Level Maths: Revision notes

Solve a linear inequality like an equation, with one rule: multiplying or dividing by a negative number reverses the inequality sign. Example: $3x-7>5x+1\Rightarrow-2x>8\Rightarrow x<-4$. A double inequality such as $-3<2x+1\le9$ is solved by doing the same operation to all three parts, giving $-2<x\le4$. For two conditions joined by 'and', the solution is where the sets overlap; for 'or', it is the union of the sets. Solutions can be written as inequalities or in set notation, e.g. $\{x:x<-4\}\cup\{x:x>-3\}$.

Method for $px^2+qx+r\ge0$ (or $>$, $<$, $\le$):

1. Rearrange so one side is $0$.
2. Solve the equation to find the critical values.
3. Sketch the parabola and read off where it is above or below the $x$-axis. For $x^2-2x-8=(x-4)(x+2)$: $f(x)<0$ between the roots, $-2<x<4$; $f(x)>0$ outside, $x<-2$ or $x>4$. Include the critical values only for $\le$ or $\ge$. Always write the 'outside' answer as two separate inequalities joined by 'or'; never as $4<x<-2$.

Expand brackets first, then collect everything on one side. For $x^2-4x+7<x+1$, subtract $x+1$: $x^2-5x+6<0$, so $(x-2)(x-3)<0$ and $2<x<3$. For $px^2+qx+r<ax+b$, move every term to one side so the quadratic is compared with $0$, keeping the $x^2$ coefficient positive if you can (multiplying by $-1$ reverses the sign).

Solving $f(x)<g(x)$ means finding the $x$-values for which the graph of $y=f(x)$ lies below the graph of $y=g(x)$. The $x$-coordinates of the intersection points are the critical values. For $y=x^2-4x+7$ and $y=x+1$, they meet at $x=2$ and $x=3$; the curve is below the line between them, so $x^2-4x+7<x+1$ for $2<x<3$. The solution is a set of $x$-values only, not coordinates.

Never multiply an inequality by an expression such as $x$ whose sign you do not know. Instead multiply by $x^2$, which is positive for $x\neq0$. Example: $\frac6x<x+1$, $x\neq0$. Multiply by $x^2$: $6x<x^2(x+1)$, so $x(x^2+x-6)>0$, i.e. $x(x+3)(x-2)>0$. The critical values are $-3$, $0$, $2$; the cubic is positive for $x>2$ and $-3<x<0$. So $\frac6x<x+1$ for $-3<x<0$ or $x>2$. In general, $\frac ax<b$ becomes $ax<bx^2$. Always state $x\neq0$.