Algebraic inequalitiesEdexcel A-Level Further Maths: Revision notes

You may add or subtract the same quantity on both sides, and multiply or divide by a positive quantity, without changing the inequality sign. Multiplying or dividing by a negative quantity reverses the sign: $-2x<6\Rightarrow x>-3$. When the multiplier contains $x$, its sign is not known. Either split into cases ($x$ such that the factor is positive, then negative) or, better, multiply by a square, which is always positive. An inequation is just an inequality that contains an unknown, such as $x^2-5x+6>0$. Never cancel a factor containing $x$ unless you know its sign.

To solve $p(x)>0$ or $p(x)<0$, move everything to one side and factorise. The critical values are the roots of the factors. These split the number line into intervals, and in each interval $p(x)$ keeps one sign, which you find from a sketch or sign table. Example: $(x-2)(x+1)(x+7)>0$ has critical values $-7,-1,2$. For large $x$ the cubic is positive, and the sign alternates across each simple root: positive for $x>2$, negative on $(-1,2)$, positive on $(-7,-1)$, negative for $x<-7$. The solution is $-7<x<-1$ or $x>2$. A repeated root such as $(x+1)^2$ does not change sign at $-1$. Use $\ge$ or $\le$ if the root itself satisfies the inequality.

For an inequality with $x$ in a denominator, such as $\frac{1}{x-a}>\frac{x}{x-b}$, two reliable methods are:

• Multiply by the square of the denominators, here $(x-a)^2(x-b)^2$, which is positive, so the sign is unchanged. This produces a polynomial inequality.
• Move everything to one side, combine into one fraction and find the critical values from the numerator and the denominator. Worked example: $\frac{1}{x-3}>\frac{x}{x-2}$. Multiplying by $(x-3)^2(x-2)^2$ gives $(x-3)(x-2)^2>x(x-3)^2(x-2)$, so $(x-2)(x-3)\big[(x-2)-x(x-3)\big]>0$, which is $(x-2)(x-3)(x^2-4x+2)<0$. Critical values: $2-\sqrt2,\ 2,\ 3,\ 2+\sqrt2$. The quartic is negative between the first two and between the last two, so $2-\sqrt2<x<2$ or $3<x<2+\sqrt2$.

The modulus $|f(x)|$ is the non-negative value of $f(x)$. Key facts, for a constant $k>0$: $|f(x)|<k\iff-k<f(x)<k$ and $|f(x)|>k\iff f(x)<-k$ or $f(x)>k$. When the right-hand side is an expression $g(x)$, think about its sign. If $g(x)<0$ then $|f|>g$ is always true and $|f|<g$ never. Where $g(x)\geq0$ you can square both sides: $|f|>g\iff f^2>g^2$. Worked example: $|x^2-1|>2(x+1)$. For $x<-1$ the right-hand side is negative, so every such $x$ is a solution. For $x\ge-1$ both sides are non-negative, so square: $(x^2-1)^2>4(x+1)^2$, so $(x+1)^2\big[(x-1)^2-4\big]>0$, i.e. $(x+1)^3(x-3)>0$. With $x\ge-1$ this gives $x>3$ ($x=-1$ is excluded because both sides are 0). Together: $x<-1$ or $x>3$.

• Test a value from each region, and test each critical value if it could be included.
• Use strict inequalities when the original is strict, and exclude values that make a denominator zero.
• Write the solution as ranges, e.g. $x<-1$ or $x>3$, not as $3<x<-1$.
• If the question says hence, build on the previous result; the complement of a solution is often what is needed (for $|f|\geq g$ take the complement of $|f|<g$).