Mastering Mathematics Smartly by Wee Wen Shih

Dealing with modulus (I)

This article is contributed by Mr Steven Lee. Thank you!

Generally, modulus inequalities involving negative numbers could pose a challenge for students. I'd like to list down a few cases and share on how to tackle such questions.

Case 1: A negative number on one side of the inequality.

Example 1: Solve |x| < -1 => No solution (because |x|>=0).

Example 2: Solve |x| > -1 => x can take any real value as |x| >= 0 > -1.

Case 2: Negative numbers on both sides of the inequality.

Example: Solve -5 < |x| < -1 => No solution (because |x|>=0).

Case 3: A negative and a positive/non-negative number on both sides of the inequality.

Example 1: Solve -5 < |x| < 3. We look at 3 methods.

Method 1

Break up the inequalities into 2 parts: -5 < |x| < 3 => |x| > -5 and |x| < 3.
For |x| > -5, x can take any real value as |x| >= 0.

For |x| < 3 => -3 < x < 3.
Taking the intersection of "x can take any real value" and

"-3 < x < 3", we conclude that -3 < x < 3.

Method 2

Alternatively, we may consider 2 cases:

a) For x >= 0, -5 < |x| < 3 => -5 < x < 3. But x >= 0, so 0 <= x < 3.

b) For x < 0, -5 < |x| < 3 => -5 < -x < 3=> -3 < x < 5. But x < 0, so -3 < x < 0.

Combining the 2 parts, we conclude that -3 < x < 3.

Method 3

This method would first require you to be aware that |x| > -5 => x can take any real value as |x| >= 0.

So, -5 < |x| < 3 => x can take any real value and |x| < 3.

Hence, the intersection is -3 < x < 3.
A fast way to look at it is to always 'ignore' the part where the modulus is greater than a negative value and just work out the inequality for |x| < 3.

Example 2: Solve -5 <= |x| <= 0.

By Method 3, we have |x| >= -5 => x can take any real value as |x| >= 0 and |x| <= 0.
But |x| >= 0, so |x| <= 0 => |x| = 0 => x = 0.
Therefore, we conclude that x can take any real value as |x| >= 0 and x = 0. Hence, x = 0.

Deal with modulus (II)

How does one simplify an inequality like $\frac{1}{3} < |x| < 1$ ?

First, we break it up into $|x| > \frac{1}{3}$ and $|x| < 1$ .

Then, we draw a number line diagram (try to do it!) and take the intersection between $x < -\frac{1}{3}$ or $x > \frac{1}{3}$
and $-1 < x < 1$ .

Finally, we obtain $-1 < x < -\frac{1}{3}$ or $\frac{1}{3} < x < 1$ .

Excluding solutions

An inequality may involve rational expressions. In such a case, it is a good practice to exclude certain solutions at the start. For example, consider $\frac{1}{2x - 1} > 1 - 5x$ . We will remind ourselves, in writing, that $x \ne \frac{1}{2}$ .

Common substitutions

Deducing the solution of another inequality that looks similar to the previous one often gets a student stuck. The approach is to make use of a suitable substitution for the variable (say $x$ ). Let's look at some common substitutions below:

$-x$ ,
$x^2$ ,
$\frac{1}{x}$ ,
$\text{e}^x$ ,
$\text{ln} x$ ,
$|x|$ ,
$\sqrt{x}$ ,
in some cases, the substitution may not be straightforward, like $\text{ln}x - 1$ .

Solving inequalities with GC

Question asked by a forumer:

Response:

Checking process: Verify that the inequality is indeed satisfied if one takes any value greater than 10.5.

Completing the square

Some questions require the student to complete the square of some expression and then apply the result to solve an inequality.

Let's consider a simple example: Show, without the use of a calculator, that $x^2 - 2x + 2$ is always positive for all real values of x. Hence, solve the inequality $\frac{x^2 - 2x + 2}{x^2 - 3} \leq 0$ .

Firstly, we have $x^2 - 2x + 2 = (x - 1)^2 + 1 \ge 1$ , so the expression is always positive for all $x \in \mathbb{R}$ (shown).
Next, $\frac{x^2 - 2x + 2}{x^2 - 3} \leq 0 \Rightarrow (x^2 - 2x + 2)(x^2 - 3) \leq 0$ , as a result of multiplying both sides by $(x^2 - 3)^2$ .
By the earlier result, we can conclude that $x^2 - 3 < 0$ . We drop the equality sign because $x^2 - 3 \ne 0$ , since it occurs as a denominator in the original inequality.
Finally, the solution is $-\sqrt{3} < x < \sqrt{3}$ .

Thus, we see that the act of completing a square has the effect of simplifying an inequality. Do learn this practical strategy well.

Graphical or algebraic method?

A student asked me this question: Solve $\frac{3x+7}{2\left|x\right|+1} < 2$ .

Now what is interesting about this problem is that it does not specify which method is to be used to solve the inequality. Thus the student has to decide between the graphical and the algebraic method, or he/she will get stumped by the question, not knowing what to do, in the worst situation.

In this case, the graphical method is more suitable than the algebraic method. If we take the algebraic approach, we will have to deal with $x$ and $\left|x\right|$ in the expression, leading to complications. Let's go into details of the graphical method.

First, we note that $2\left|x\right| + 1 \ge 0$ , so we can cross-multiply to give us $3x+7 < 4\left|x\right| + 2$ which simplifies to $\frac{3x+5}{4} < \left|x\right|$ . We will then sketch the graphs of $y=\left|x\right|$ and $y=\frac{3x+5}{4}$ , from which the points of intersection may be found easily.

We shall now look at several variations of this question. Consider this variation: Solve $\frac{3\left|x\right|+7}{2\left|x\right|+1}<2$ .

First, we observe that both numerator and denominator are positive, so we can cross-multiply and solve it via the algebraic approach, i.e.
$3\left|x\right|+7 < 4\left|x\right|+2$
$\Rightarrow\left|x\right| > 5$
$\Rightarrow x < -5\:\:\text{or}\:\:x > 5$ .

Consider another variation: Solve $\frac{3x^2+7}{2\left|x\right|+1} < 2$ .

This may be solved with both methods. In the graphical approach, cross-multiplication and simplification will lead us to $\frac{3x^2+5}{4}<\left|x\right|$ . One will find that both graphs do not intersect at all.

In the algebraic method, we shall apply the substitution $y=\left|x\right|$ , leading to the following steps:
$\frac{3y^2+7}{2y+1}<2$
$\Rightarrow 3y^2-4y+5<0$ ,
from which we can conclude that $y$ (and therefore $\left|x\right|$ and thus $x$ ) has no real roots because the discriminant $(-4)^2-4(3)(5)=-44<0$ .

Finally, I leave you to ponder on this problem: Solve $\frac{3x+7}{2\left|x\right|-1}<2$ .