How to solve this algebraically This seems like a really stupid question to ask here...
I'm trying to solve $\sqrt{x^2 - 1} + x > 0$.
When I try this happens:


*

*$\sqrt{x^2 - 1} > - x$

*$x^2 - 1 > x^2$ (squared both sides) 

*$-1 > 0$


However, on Wolfram Alpha, it says that the answer is $x \ge 1$.
It seems to me that basic rules of algebra are simply breaking down here... what am I doing wrong?
 A: You have a first condition to satisfy, that is
$$
x^2-1\ge0
$$
so that the square root exists. Thus, from now on, we assume this condition on $x$ holds.
After this preliminary, let's rewrite the inequation as
$$
\sqrt{x^2-1}>-x
$$
There are two cases:


*

*if $-x<0$, the inequality is clearly satisfied, because $\sqrt{x^2-1}\ge0$ by definition;

*if $-x\ge0$, we can square both sides, because inequalities between positive numbers is preserved squaring or taking the square root.
In case 2 we get the false inequality $-1>0$, so the second case doesn't provide solutions; hence we remain with only the first case, that is, the system
\begin{cases}
x^2-1\ge0\\
-x<0
\end{cases}
that is satisfied for $x\ge1$.
A: Let's break this down into 3 cases:

*

*$x\geq1$. In this case, the square root will evaluate to at least zero and the other term is positive, thus the sum is greater than zero.


*$-1< x<1$. In this case, the square root doesn't evaluate to a real value and thus I'm not sure how this should be handled. Is $i>0$?


*$x\leq-1$. In this case, the $x$ term will be larger and thus the sum is negative as one could plug in $x=-1,-2,$ or $-3$ to see this.
The squaring function with inequalities has some rules to consider:
$1>-2$ but $1<4$ as a relatively simple example to consider. In multiplying one side by a negative value, you change the inequality direction potentially. Another example would be $5>-2$ and $25>4$ so it isn't a sure thing unless one uses absolute values.
Ask Dr. Math notes the following that may be useful:

The inequality has to be turned around when you multiply by a negative
number, since that essentially flips the whole number line around,
reversing the order of everything.  But squaring or raising to a
higher power doesn't work that way; you can't always obtain an
equivalent inequality at all.
For example, suppose you know that a < b.  Perhaps a is -2 and b is 1;
or perhaps a is -1 and b is 2.  What happens when we square each of
these numbers?
-2 < 1  becomes  4 > 1;   the square of -2 is GREATER than the
square of 1
-1 < 2  becomes  1 < 4;   the square of -1 is LESS than the square
of 2
So whether the inequality is reversed depends on the specific numbers
you have, in particular on their absolute values.  We can't say
something like
if a < b, then a^2 < b^2
All we can say is
if |a| < |b|, then a^2 < b^2
So your question is in a sense invalid: yes, we can determine whether
the square of this is less than the square of that, but the result is
not derived in any way from the given inequality.  We aren't really
turning anything around at all.

Thus, before you square, consider $\sqrt{x^2-1}$ and $x$ in absolute values which changes things considerably on your second line.
A: Indeed, look at the domain of your square root function. That might shed some light. Infact, if you look at the graph of $(\sqrt{2}x + 1)(\sqrt{2}x -1) = 2x^2 - 1$ and look at the condition that the inequality must be strictly positive E.g the inequality won't be satisfied for negative $x$ coupled with the domain condition. I think this is because the $|\sqrt{x^2-1}|\leq|x|$, You will see very clearly whats going on. Anyways, it will be your job to verify. 
