Inequalities – Square inequality with absolute value – Exercise 1866

Exercise

Solve the inequality:

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

Final Answer

Show final answer

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

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

And that’s the final answer.

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