Converting sets into intervals with variables I am trying to convert the set to interval form where center is $c$ and the radius is $r$. I have not had any problem doing the first example but I cannot find a way to get the $c$ value for the second question. How would I be able to solve the $c$ for the second question as well as turning it into interval notation?


*

*$ \left|x^2 - 1\right| > 2 $

*$ \left|x^2 - 2x\right| > 2 $

Work

*

*$ \left|x^2 - 1\right| > 2 $
$c= 1$ and $r = 2$
$$(-\infty, \sqrt{c-r})\cup(\sqrt{c+r}, +\infty)$$
$$(-\infty, \sqrt{1-2})\cup(\sqrt{1+2}, +\infty)$$
$$(-\infty, \sqrt{-1})\cup(\sqrt3, +\infty)$$
So the intervals are between $(-\infty, -\sqrt3)\cup(\sqrt3, +\infty)$ for example 1.


*$ \left|x^2 - 2x\right| > 2 $
$c = ?$ and $r = 2$
 A: It is ideal to solve these problems directly.
First problem
Note that for the first problem, you have to find all $x\in\mathbb{R}$ such that
$$ |x^2 -1| > 2 \tag{1}$$
Let $x\in\mathbb{R}$ be a real number that satisfies (1). By definition of the modulus sign, this condition is equivalent to the statement that
$$
\begin{eqnarray}
&x^2 - 1 > 2 \tag{1a}\\
\text{or } &x^2 - 1 < -2. \tag{1b}
\end{eqnarray}
$$
What remains is for us to solve each of these inequalities, and find all the real values that satisfy (1a) or (1b). Indeed, all the $x$ values that satisfy (1a) are such that
$$
x^2 > 3 \quad \iff \quad x > \sqrt{3} \text{ or }x<-\sqrt{3}. 
$$
Regarding the second inequality (1b) $ x^2 < -1 $, no real number satisfies this constraint, and thus the only real values that satisfy the initial inequality $|x^2 - 1| > 2$ are such that $x\in(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty)$, i.e.
$$
|x^2 - 1| > 2 \quad \iff \quad x\in(-\infty,-\sqrt{3})\cup(\sqrt{3},\infty).
$$
Second problem:
For the second problem, one needs simply to apply the same procedure. We have to find all $x\in\mathbb{R}$ such that
$$
|x^2 - 2x| > 2. \tag{2}
$$
Assume that $x\in\mathbb{R}$ is a real number that satisfies (2). Then by definition of the modulus sign, (2) is equivalent that $x$ satisfies either of the following conditions:
$$
\begin{eqnarray}
&x^2 - 2x > 2 \tag{2a} \\
\text{or } &x^2 - 2x < -2. \tag{2b}
\end{eqnarray}
$$
Solving the first inequality (2a), we find that
$$
x^2 - 2x > 2 \quad \iff \quad (x - 1)^2  > 3 \quad \iff \quad x-1 > \sqrt{3} \text{ or } x - 1 < -\sqrt{3}
$$
That is, $x > 1 + \sqrt{3}$ or $x < 1 - \sqrt{3}$.
A similar approach can be done for the second inequality (2b). Indeed, we need to find all the $x$ reals such that
$$
x^2 - 2x < -2.
$$
Indeed, we have that
$$
x^2 - 2x < -2 \quad \iff \quad x^2 - 2x + 2 < 0 \quad \iff \quad (x-1)^2 + 1 < 0,
$$
but this is a sum of two positive real numbers, and so no real numbers that satisfy this second inequality exist.
In conclusion, we have that
$$
|x^2 - 2x| > 2 \quad \iff \quad x\in(-\infty,1-\sqrt{3})\cup(1+\sqrt{3},\infty).
$$
