# |x^2 - 2| < 1 Please can anyone help? How to solve this absolute value problem?

Mar 15, 2018

#### Answer:

The solution is $x \in \left(- \sqrt{3} , - 1\right) \cup \left(1 , \sqrt{3}\right)$

#### Explanation:

$| {x}^{2} - 2 | < 1$

There are $2$ solutions

${x}^{2} - 2 < 1$ and ${x}^{2} - 2 > - 1$

${x}^{2} - 3 < 0$,$\implies$ $x \in \left(- \sqrt{3} , \sqrt{3}\right)$

${x}^{2} - 1 > 0$, $\implies$, $x \in \left(- \infty , - 1\right) \cup \left(1 , + \infty\right)$

You can prove theses solutions by a sign chart.

Combining the $2$ solutions

$x \in \left(- \sqrt{3} , - 1\right) \cup \left(1 , \sqrt{3}\right)$

graph{|x^2-2|-1 [-6.24, 6.25, -3.117, 3.126]}

Mar 15, 2018

See below.

#### Explanation:

Squaring both sides

${\left({x}^{2} - 2\right)}^{2} < 1 \Rightarrow {\left({x}^{2} - 2\right)}^{2} - 1 < 0 \Rightarrow \left({x}^{2} - 2 - 1\right) \left({x}^{2} - 2 + 1\right) < 0$ or

$\left({x}^{2} - 3\right) \left({x}^{2} - 1\right) < 0$ which is equivalent to

$\left\{\begin{matrix}{x}^{2} - 3 < 0 \\ {x}^{2} - 1 > 0\end{matrix}\right. \Rightarrow 1 < {x}^{2} < 3$

the set

$\left\{\begin{matrix}{x}^{2} - 3 > 0 \\ {x}^{2} - 1 < 0\end{matrix}\right. \Rightarrow \left\{{x}^{2} < 1\right\} \cap \left\{{x}^{2} > 3\right\}$

is an extraneous solution, is not feasible and was included due to the squaring operation.

Finally

$1 < {x}^{2} < 3$