# Inequalities – Square inequality with absolute value – Exercise 1866

Exercise

Solve the inequality:

$$|x^2-1|\leq 1$$

$$-\sqrt{2}\leq x \leq \sqrt{2}$$

Solution

$$|x^2-1|\leq 1$$

By absolute value definition, the inequality is equivalent to this inequality:

$$-1\leq x^2-1\leq 1$$

And this inequality is equivalent to these two the intersection between the inequalities:

$$-1\leq x^2-1 \text{ and } x^2-1\leq 1$$

Therefore, instead of solving the original inequality, we solve both inequalities and intersect their solutions. Let’s start with the first inequality:

$$-1\leq x^2-1$$

Solve the inequality:

$$0\leq x^2$$

We got an inequality that exists for every x, which means its solution is all x.

Moving on to the second inequality:

$$x^2-1\leq 1$$

Solve the inequality:

$$x^2-2\leq 0$$

It is a square inequality. Its roots are

$$x_{1,2}=-\sqrt{2}, \sqrt{2}$$

Therefore, the solution of the inequality is

$$-\sqrt{2}\leq x\leq \sqrt{2}$$

We intersect both solutions (“and”). That is, all x and

$$-\sqrt{2}\leq x\leq \sqrt{2}$$

The result of the intersection is

$$-\sqrt{2}\leq x\leq \sqrt{2}$$