I've decided to study calculus on my own, and I've started working on "A First Course in Calculus" by Serge Lang, 5th edition. Now I'm just reading the chapter on preliminaries, and there is a section about inequalities. In the exercises on p.13, there is a problem which I can solve, but I'm not quite sure that I'm doing it right.

The problem is:

$|x^{2}-2|\leqslant 1$

I am solving it in the following way:

1) Suppose that $x^{2}-2 \geqslant 0$, then $|x^{2}-2|=x^{2}-2$

First, I find the values of x for which $x^{2}-2$ is non-negative: $x^{2}-2 \geqslant 0$, and obtain $x \geqslant \sqrt{2}$ or $x \leqslant -\sqrt{2}$

Second, I solve the inequality $x^{2}-2 \leqslant 1$ and obtain $-\sqrt{3} \leqslant x \leqslant \sqrt{3}$

The solution for non-negative values of $x^{2}-2$ is thus $-\sqrt{3}\leqslant x \leqslant -\sqrt{2}$ or $\sqrt{2} \leqslant x \leqslant \sqrt{3}$

2) Suppose that $x^{2}-2<0$, then $|x^{2}-2|=-(x^{2}-2)=2-x^{2}$

First, I find the values of x for which $x^{2}-2$ is negative: $x^{2}-2<0$, and obtain $-\sqrt{2} < x < \sqrt{2}$

Second, I solve the inequality $2-x^{2} \leqslant 1$ and obtain $x \leqslant -1$ or $x \geqslant 1$

The solution for negative values of $x^{2}-2<0$ is thus $-\sqrt{2} < x \leqslant -1$ or $1 \leqslant x < \sqrt{2}$

Combining the solutions for non-negative and negative values of $x^{2}-2<0$, I obtain the final answer:

$-\sqrt{3} \leqslant x \leqslant -1$ or $1 \leqslant x \leqslant \sqrt{3}$

This is the correct answer.

And here I go to the question itself: is it necessary to find the values of x for which $x^{2}-2<0$ and $x^{2}-2>0$, or can this step be skipped? If I just find the intersection of solutions of two inequalties, $x^{2}-2 \leqslant 1$ and $2-x^{2} \leqslant 1$, I still get the same answer. But I feel that skipping this step would be wrong, and that maybe it can lead to incorrect answers when I deal with more complex problems of this type.

I also have a second, though smaller, question. When we speak about the solution of an inequality, and we have several intervals, do we use the conjunction "or", or "and"? That is, is it

$-\sqrt{3} \leqslant x \leqslant -1$ or $1 \leqslant x \leqslant \sqrt{3}$


$-\sqrt{3} \leqslant x \leqslant -1$ and $1 \leqslant x \leqslant \sqrt{3}$?

Lang uses "or" in the "Answers" section, but it just seems to me that we speak about a set of solutions here, so maybe "and" would be more appropriate.


For the main question, the answer is yes. It is possible for the answer to change when you consider the condition (in general problems).

For the smaller question, the answer is OR. You can easily test this by trying out values.

  • $\begingroup$ So do I get it right that, methodologically, when I solve inequalities with absolute values, I always need to calculate the intervals of x for the non-negative and negative values of the expression? $\endgroup$ – Vladimir May 11 '14 at 15:49
  • $\begingroup$ If you use the definition of $|.|$ in the way you did above, yes. There are of course other ways of solving it, which may not require that step. For e.g., we could work out the solution using $\iff (x^2-2)^2 \le 1 \iff (x-1)(x+1)(x-\sqrt3)(x+\sqrt3) \le 0$ which will give you the same solution by comparing intervals where the product can change signs. Which way you use depends on the problem and what you are comfortable with. $\endgroup$ – Macavity May 11 '14 at 15:59
  • $\begingroup$ Thank you. But could you please explain why you use $(x^{2}-2)^{2}$ instead of $|x^{2}-2|$ in your answer? $\endgroup$ – Vladimir May 11 '14 at 16:08
  • $\begingroup$ Because another possible way of using absolute value is $|x| = \sqrt{x^2}$. Hence $|x| \le 1 \iff x^2 \le = 1^2=1$. In this problem, then you can get some nice factorisations going, of the form $a^2-b^2 = (a-b)(a+b)$. $\endgroup$ – Macavity May 11 '14 at 16:12

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