What happens when square root is performed in inequalities? Simplify: $x^2 > 1$.
My solution: Taking square root on both sides:
$±x > ±1$
So my results are:


*

*$x > 1$

*$x > -1$

*$-x > 1$ $\implies$ $(-1 > x)$

*$-x > -1$ $\implies$ $(1 > x) $


But I strongly feel this is wrong. What is wrong here?
A step-by-step explanation will help me.
 A: You can just go from $x^2 > 1$ directly to $|x| > |1|$. 
Now obviously $|1| = 1$, so $|x| > 1$, therefore either $x > 1$ or $x < -1$.
A: When you apply square root operation to both sides of inequation like $f^2(x) < c$ or $f^2(x) > c$, then $c < 0$ gives you no solutions for $f^2(x) < c$ and any $x \in \mathscr D(f)$ for $f^2(x) > c$.
If $c \geq 0$, then $f^2(x) < c \rightarrow |f(x)| < \sqrt{c}$ and $f^2(x) > c \rightarrow |f(x)| > \sqrt{c}$ (same for $\leq$ and $\geq$).
In your case $c = 1$ and $f(x) = x$, so $f^2(x) > c \rightarrow |x| > \sqrt{c}, c = 1$, so $|x| > 1$, or $x \in (-\infty,1)\cup(1,+\infty)$.
A: As $\sqrt\cdot$ is increasing and $\sqrt{(\cdot)^2}=|\cdot|$,
$$a^2>b^2\iff|a|>|b|.$$
A: We need $x^2-1=(x-1)(x+1)>0$
As the product is positive, multiplicands should either be both  positive or both negative.
First we consider the scenario where they are both positive:
If $x-1>0\iff x>1\  \ \ \ (1)$
and $x+1>0\iff x>-1\  \ \ \ (2)$
$(1),(2)\implies x>$max$(1,-1)$
Test the negative case similarly 
A: In solving quadratic inequalities, it must be the case that the right hand side of the equation is zero. So we have:
$$x^2-1>0.$$
Then we find the critical numbers, these are values of $x$ that will make our inequality above zero.
We have $x=1$ and $x=-1$ as critical numbers.
My critical numbers then partitioned my real number line into 3 parts/subintervals. Namely:
$$(-\infty,-1),\quad(-1,1)\quad\mbox{and}\quad(1,\infty).$$
In each part it is advisable to get a test value, a number that lies on the subintervals and substitute it on the LHS above and we must note the sign, this are the sign of the subintervals relative to our inequality. 
If $x=-2$ we have positive sign so for all $x$ in the first subinterval we have $x^2-1$ is positive.
If $x=0$ we have a negative sign so for all $x$ in the second subinterval $x^2-1$ is negative.
Lastly if $x=2$ we have a positive sign, so for all $x$ in the third subinterval we have $x^2-1$ is positive.
Originally we have $x^2-1>0$ this means that $x^2-1$ is positive so our answer is $$(-\infty,-1)\cup (1,+\infty).$$
This is the general way to solve quadratic inequality. 
