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:

1. $\sqrt{x^2 - 1} > - x$
2. $x^2 - 1 > x^2$ (squared both sides)
3. $-1 > 0$

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?

• in step two, when you square $-x$, the outcome is $x^2$, not $-x^2$ Apr 25 '14 at 8:04
• point taken, but the answer still differs from Wolfram's. Apr 25 '14 at 8:05
• you can't square on both side if $x \geq 1$, because $\sqrt{x^2-1}$ is positive and $-x$ negative. Square function is decreasing on negative numbers, increasing on positive ones but you can't "automatically" compare a priori the square of a positive and a negative (think of $-2 < 1$ and $-0.5 < 1$ but $(-2)^2 = 4 > 1$ and $(-0.5)^2 = 0.25 < 1$). For the solution, please refer to JB King's anwser
– yago
Apr 25 '14 at 8:24

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:

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

2. 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 $x\ge1$.

• @JBKing Really? Is $(1/4)^2-1\ge0$? I don't think so. I have set $x^2-1\ge0$ to begin with, so that I don't have to bother with checking whether the square root exists. Apr 25 '14 at 9:12
• If you are putting things together, then state it that way instead of leaving things open to interpretation. Apr 25 '14 at 9:18

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.

• To answer your concerns from your recently-deleted comment, rep doesn't decay, but you can also lose rep if someone who has upvoted you deletes their account. If you look at your rep page, that's what happened with you. Cheers. Jun 8 '14 at 19:40

Let's break this down into 3 cases:

1. $x\geq1$. 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.

2. $-1< x<1$. 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 $i>0$?

3. $x\leq-1$. 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.

• Thank you, but my question is about solving it algebraically. Why can't i apply the rules of algebra to this problem? Apr 25 '14 at 8:25
• Thank you. I understand now. Apr 25 '14 at 13:13