Spivak Calc Ch 1 Problem 4(iii) I would like to follow up on a similar question.
Find all numbers  $x$ for which $$5 - {x}^{2} < -2$$
Omitting the intermediate steps, we get $${x}^{2} > 7$$
Now, I would like to rigidly proceed.
My approach was ${x} > \pm  \sqrt{7}$, which correctly gives me ${x} > \sqrt{7}$, but following my logic would also give me ${x} > - \sqrt{7}$, which is clearly incorrect.
So, assuming at this point I have no knowledge of quadratic functions, or plotting the points on a number line, or upward, downward facing graphs,  can someone guide to the correct answer of $x > \sqrt{7}$ or $x < -\sqrt{7}$
Thanks in advance.
 A: What is needed is a standard way of attacking a problem similar to $x^2 > 7.$
Note 1:
If $0 < r,s, k$ then
$r < s \iff r^k < s^k.$ 
This holds for all positive real numbers $r,s, k$. 
This assertion is based on the idea that for $x > 0, f(x) = x^k$ 
is a strictly increasing function. 
The assertion is basic to Calculus.  In the pre-Calculus world, you can simply consider what the graph of $f(x) = x^k ~: ~0 < x,k ~: ~$ looks like.
Note 2:
If you have an inequality, and you multiply both sides of the inequality by a negative number, then you reverse the direction of the inequality.
For example, if $0 < a,b$, then you have that 
$a < b \iff (-a) > (-b).$
Note 3:
When comparing a variable $x$ to the expression $\sqrt{x^2}$, you have the convention that the square root function always unambiguously refers to a non-negative value.
Thus, if $x \geq 0$, then $x = \sqrt{x^2}.$ 
If $x < 0$, then $(-x) = \sqrt{x^2}.$

Consider the problem $x^2 > 7.$ 
Based on note 1, this implies that
$$\sqrt{x^2} > \sqrt{7}.\tag 1$$
You now have to break the problem into two cases, depending on whether $x \geq 0$, or $x < 0$.  Each case becomes a separate inequality to be solved.

$\underline{\text{Case 1:} ~x \geq 0}.$ 
By Note 3, inequality (1) above becomes 
$x > \sqrt{7}.$
This means that in order to find values of $x$ that satisfy the inequality under Case 1, two separate constraints must be simultaneously satisfied:

*

*$x \geq 0.$

*$x > \sqrt{7}$.

If the 2nd constraint above is satisfied, then the 1st constraint above will automatically be satisfied.  Therefore, the Case 1 solution is that 
$x > \sqrt{7}$.

$\underline{\text{Case 2:} ~x < 0}.$ 
By Note 3, inequality (1) above becomes 
$(-x) > \sqrt{7}.$ 
Using Note 2 above, the constraint may be re-expressed as 
$x  = (-1) \times (-x) < (-1) \times \sqrt{7} = -\sqrt{7}.$ 
This simplifies to $x < -\sqrt{7}.$
This means that in order to find values of $x$ that satisfy the inequality under Case 2, two separate constraints must be simultaneously satisfied:

*

*$x < 0.$

*$x < -\sqrt{7}$.

If the 2nd constraint above is satisfied, then the 1st constraint above will automatically be satisfied.  Therefore, the Case 2 solution is that 
$x < -\sqrt{7}$.

Therefore, when the values of $x$ that satisfy Case 1 are combined with the values of $x$ that satisfy Case 2, you have that
$$ x > \sqrt{7} ~~\text{or}~~ x < -\sqrt{7}.$$
A: $X^2<7 \implies (x+\sqrt{7})(x-\sqrt{7}) >0$
$AB>0 \implies (1) A>0 \& B>0$. OR (2) A<0 & B<0$
So (1): $x>-\sqrt{7} ~\& ~ x>\sqrt{7} \implies  x> \sqrt{7}$.
(2): $x<-\sqrt{7} ~\&~ x <\sqrt{7} \implies x <-\sqrt{7.}$
Finally the solution is $x \in (-\infty;-\sqrt{7}) \cup (\sqrt{7}, \infty)$ Or alternatively $x<-\sqrt{7}$ or $x>\sqrt{7}$.
A: In general, in real analysis, $$x^2>p^2\iff |x|>|p|.$$
Therefore, $$x^2>7\\\iff x^2>(\sqrt7)^2\\\iff|x|>|\sqrt7|=\sqrt7\\\iff x<-\sqrt7\quad\text{or}\quad x>\sqrt7.$$
Alternatively, $$x^2>7\\\iff x^2>(-\sqrt7)^2\\\iff|x|>|-\sqrt7|=\sqrt7\\\iff x<-\sqrt7\quad\text{or}\quad x>\sqrt7.$$
