Inequalities – Square inequality with absolute value – Exercise 1866


Solve the inequality:

|x^2-1|\leq 1

Final Answer

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


|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}

And that’s the final answer.

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