10.2 Solving Inequalities

To solve inequalities, we can work line by line and reason by equivalence, much as we would when solving equations. Just be very careful when multiplying or dividing by any quantity as its algebraic sign may affect the direction of the inequality.

Example 10.10.

Find the values of $x\in\mathbb{R}$ for which $2\,x^{2}+x\geq 6$ .

Observe that

\begin{array}[]{lll}&2\,x^{2}+x\geq 6&\cr\Leftrightarrow&2\,x^{2}+x-6\geq 0&% \cr\Leftrightarrow&\left(2\,x-3\right)\,\left(x+2\right)\geq 0&\cr% \Leftrightarrow&(2\,x-3\geq 0\,{\text{ and }}\,x+2\geq 0)\,{\text{ or }}\,(2\,% x-3\leq 0\,{\text{ and }}\,x+2\leq 0)&\cr\Leftrightarrow&(x\geq\frac{3}{2}\,{% \text{ and }}\,x\geq-2)\,{\text{ or }}\,(x\leq\frac{3}{2}\,{\text{ and }}\,x% \leq-2)&\cr\Leftrightarrow&x\geq\frac{3}{2}\,{\text{ or }}\,x\leq-2.&\cr\end{array}

In the above example, notice:

•

all steps are genuinely reversible, and so retain equivalence;
•

we avoid multiplying or dividing which might require us to change $\geq$ to $\leq$ when multiplying by a negative term;
•

we use the rule

$AB\geq 0\iff((A\geq 0\mbox{ and }B\geq 0)\mbox{ or }(A\leq 0\mbox{ and }B\leq 0% )),$

which can be derived from Proposition 10.7. We then have to simplify the resulting systems of inequalities, e.g. removing redundant inequalities such as

$(x\geq\frac{3}{2}\,{\text{ and }}\,x\geq-2)\iff x\geq\frac{3}{2}.$

Example 10.11.

Find the values of $a\in\mathbb{R}$ for which $-x^{2}+a\,x+a-3<0$ for all $x\in\mathbb{R}$ .

Observe that

\begin{array}[]{lll}&-x^{2}+a\,x+a-3<0&\cr\Leftrightarrow&a-3<x^{2}-a\,x&\cr% \Leftrightarrow&a-3<{\left(x-\frac{a}{2}\right)}^{2}-\frac{a^{2}}{4}&\cr% \Leftrightarrow&\frac{a^{2}}{4}+a-3<{\left(x-\frac{a}{2}\right)}^{2}&\cr% \Leftrightarrow&a^{2}+4\,a-12<4\,{\left(x-\frac{a}{2}\right)}^{2}&\cr% \Leftrightarrow&\left(a-2\right)\,\left(a+6\right)<4\,{\left(x-\frac{a}{2}% \right)}^{2}.&\end{array}

This inequality is required to be true for all $x$ . So it must be true when the right-hand side takes its minimum value. This happens for $x=a/2$ . Hence, we require

\begin{array}[]{lll}&\left(a-2\right)\,\left(a+6\right)<0&\cr\Leftrightarrow&(% a-2<0\,{\text{ and }}\,a+6>0)\,{\text{ or }}\,(a-2>0\,{\text{ and }}\,a+6<0)&% \cr\Leftrightarrow&(a<2\,{\text{ and }}\,a>-6)\,{\text{ or }}\,(a>2\,{\text{ % and }}\,a<-6)&\cr\Leftrightarrow&(-6<a\,{\text{ and }}\,a<2)\,{\text{ or }}\,% \mathbf{False}&\cr\Leftrightarrow&-6<a<2.&\cr\end{array}

Example 10.12.

Find the minimum value of the positive integer $a>0$ , for which the equation $x^{2}+(a-2)x+a=0$ has real roots. (Taken from Finnish national exam, Q9 of 22 March 1971.)

Observe that

\begin{array}[]{lll}&x^{2}+\left(a-2\right)\,x+a=0&\cr\Leftrightarrow&{\left(x% +\frac{a-2}{2}\right)}^{2}-{\left(\frac{a-2}{2}\right)}^{2}+a=0&\cr% \Leftrightarrow&{\left(x+\frac{a-2}{2}\right)}^{2}=\frac{{\left(a-2\right)}^{2% }}{4}-a.&\end{array}

This has real roots if and only if

\begin{array}[]{lll}&\frac{{\left(a-2\right)}^{2}}{4}-a\geq 0&\cr% \Leftrightarrow&a^{2}-4\,a+4-4\,a\geq 0&\cr\Leftrightarrow&a^{2}-8\,a+4\geq 0&% \cr\Leftrightarrow&{\left(a-4\right)}^{2}-16+4\geq 0&\cr\Leftrightarrow&{\left% (a-4\right)}^{2}\geq 12&\cr\Leftrightarrow&a-4\geq\sqrt{12}\,{\text{ or }}\,a-% 4\leq-\sqrt{12}.&\end{array}

Ignoring the negative solution, we obtain $a\geq\sqrt{12}+4$ . Using the assumption that $a$ is an integer, this means that $a\geq 8$ .

Example 10.13.

Solve

\frac{(2x+1)(1-x)}{x^{2}(x+2)}>0.

We find the critical values of $x$ , where one of the sub-expressions changes sign, are $-2$ , $-1/2$ , $0$ and $1$ and form a table of signs for each sub-expression.

	$x<-2$	$-2<x<-\frac{1}{2}$	$-\frac{1}{2}<x<0$	$0<x<1$	$1<x$
$2x+1$	-	-	+	+	+
$1-x$	+	+	+	+	-
$x^{2}$	+	+	+	+	+
$x+2$	-	+	+	+	+
Expression	+	-	+	+	-

The solution is therefore $x<-2,-\frac{1}{2}<x<0$ , or $0<x<1$ .

Note, the above method is simple, reliable but requires systematic work.

You can double-check by plotting the graph, although numerical experiments are rarely considered as rigorous as a careful proof. See Figure 10.2.

Figure 10.2: The function

\frac{(2x+1)(1-x)}{x^{2}(x+2)}

for

\in\mathbb{R}

10.2.1 Inequalities involving the Modulus Function

Recall the definition of the absolute value (or modulus) function that we were introduced to in Block 3 Functions.

Definition (Absolute Value (Modulus) Function).

The modulus or absolute value function is defined by $|\cdot|:\mathbb{R}\to[0,\infty)$ where

|x|=\begin{cases}x,&\text{if }x\geq 0,\\ -x,&\text{if }x<0.\end{cases}

Exercise 10.14.

Prove that $|xy|=|x||y|$ for all $x,y\in\mathbb{R}$ .

Solution (please try for yourself before looking)

We consider the following cases.

Case 1: $x,y\geq 0$ Then $xy\geq 0$ which means $|xy|=xy=|x||y|$ .

Case 2: $x\geq 0$ and $y<0$ . Then $xy\leq 0$ , $|x|=x$ and $|y|=-y$ . Hence, $|xy|=-xy=|x||y|$ .

Case 3: $x,y<0$ . Then $xy>0$ , $|x|=-x$ and $|y|=-y$ . Thus, $|xy|=xy=|x||y|$ .

Note that the case $x<0$ and $y\geq 0$ follows from Case 2 by symmetry.

If we want to solve an inequality involving the absolute value, we can simplify it by considering cases.

Example 10.15.

Suppose we wish to solve the inequality

|x+1|+|x-2|\leqslant 4.

The values within the modulus function, $x+1$ and $x-2$ , change their algebraic sign at the values $-1$ and $2$ so these will be our critical values that define the boundaries of our cases.

Case 1: $x<-1$ . Then we have $x+1<0$ and $x-2<0$ . So our inequality becomes $(-(x+1))+(-(x-2))\leqslant 4$ which is equivalent to $-2x-3\leqslant 0$ and gives $x\geqslant-3/2$ . So from this case, we know the inequality is satisfied for all $x$ satisfying $x<-1$ and $x\geq-3/2$ i.e. for $-3/2\leq x<-1$ .

Case 2: $-1\leq x<2$ . Then we have $x+1\geq 0$ and $x-2<0$ . So our inequality becomes $(x+1)+(-(x-2))\leqslant 4$ which is equivalent to $3\leq 4$ which is always true. So the inequality is satisfied whenever $-1\leq x<2$ .

Case 3: $x\geq 2$ . Then we have $x+1,x-2\geq 0$ . So our inequality becomes $(x+1)+(x-2)\leqslant 4$ which is equivalent to $2x\leq 5$ and gives $x\leq 5/2$ . So this case tells us the inequality is satisfied when $2\leq x\leq 5/2$ .

Combining all three cases (i.e. taking the union of the three intervals), our inequality is satisfied if and only if $-3/2\leq x\leq 5/2$ .