Algebraic and modulus inequalitiesEdexcel International A Level Further Maths: Revision notes

To solve an inequality such as $x^2-5x+6<0$, find the critical values (where the two sides are equal), here $x=2$ and $x=3$, then decide which regions satisfy the inequality by sketching the graph or testing a value in each region. The quadratic $(x-2)(x-3)$ is negative between its roots, so the solution is $2<x<3$. For a positive quadratic the solution of '$>0$' is outside the roots: $x<2$ or $x>3$. Multiplying or dividing both sides by a negative number reverses the inequality sign. Inequations such as $x(x-1)\ge2$ are rearranged to have $0$ on one side first.

Never multiply both sides by an expression whose sign you do not know, such as $x-3$: it could be negative. Two safe methods: (1) multiply both sides by the square of the denominator, which is positive; (2) move everything to one side, form a single fraction, and use a sign diagram. For $\frac{x}{x-3}>2$, multiplying by $(x-3)^2$ gives $x(x-3)>2(x-3)^2$, so $(x-3)(6-x)>0$ and $3<x<6$. The critical values are the roots of the equation ($x=6$) and the values making a denominator zero ($x=3$), which are never part of the solution.

Multiply by $(x-a)^2(x-b)^2$ to get $(x-a)(x-b)^2>x(x-a)^2(x-b)$, then bring everything to one side and factorise $(x-a)(x-b)$. Example: $\frac{1}{x-1}>\frac{x}{x-4}$ gives $(x-1)(x-4)\left[(x-4)-x(x-1)\right]>0$, which is $(x-1)(x-4)\left(x^2-2x+4\right)<0$. The quadratic $x^2-2x+4=(x-1)^2+3$ is always positive, so the condition is $(x-1)(x-4)<0$, and $1<x<4$. When the numerator quadratic has real roots, they join the asymptotes as critical values, giving up to four critical values and a sign diagram: for $\frac{1}{x-5}>\frac{x}{x-8}$ the result is $(x-2)(x-4)(x-5)(x-8)<0$, so $2<x<4$ or $5<x<8$.

$|f(x)|$ is the non-negative value of $f(x)$. For $g(x)\ge0$: $|f(x)|<g(x)\iff-g(x)<f(x)<g(x)$, and $|f(x)|>g(x)\iff f(x)>g(x)$ or $f(x)<-g(x)$. Example: $|2x-3|<x+2$ means $-(x+2)<2x-3<x+2$, giving $\frac13<x<5$. A second method finds the critical values by solving $|f(x)|=g(x)$ in both cases ($f=g$ and $f=-g$), then testing regions. Example: $|x^2-1|>2(x+1)$. Solving $x^2-1=2x+2$ gives $x=3$ or $-1$; solving $1-x^2=2x+2$ gives $(x+1)^2=0$, so $x=-1$. Testing $x=0$ (false), $x=-2$ and $x=4$ (true) gives $x<-1$ or $x>3$. A graph sketch of $y=|f(x)|$ and $y=g(x)$ shows where one lies above the other. Squaring is valid only when both sides are non-negative.

Give solutions as intervals with strict or non-strict signs matching the original inequality: $>$ and $<$ exclude the end values; $\ge$ and $\le$ include them, except where the expression is undefined. For $\frac{x}{x-3}\le2$ the answer is $x<3$ or $x\ge6$, not $x\le3$. Use 'or' for separate intervals, never a single chain such as $x<3<6$. Always test at least one value from each region, because a mistake in sign is easy to make and cheap to catch.