A doubt in quadratic inequality 
While I was doing a problem I came upon this:
  $$(a^2-4)>0$$
  $$(a-2)(a+2)>0$$

Now I thought it will be 
$a>2 $ or $a>-2$
but it was $a<-2$ or $a>2$
Can u explain me why it is so?
Thank you
 A: Now the inequality is valid if * $(a-2)$ and $(a+2)$ are both positive or are both negative.
Now
$$ (a-2)>0 ~\hbox{and}~ (a+2)> 0 \Rightarrow  a>2$$
$$ (a-2)<0 ~\hbox{and}~ (a+2)< 0 \Rightarrow  a<-2$$
A: Hint
$$xy>0\iff (x>0\land y>0)\lor(x<0\land y<0)$$
A: If $x, y \in \mathbb{R}$ and $xy > 0$, either $x, y > 0$ or $x, y < 0$. As $(a-2)(a+2) > 0$, then either $a - 2 > 0$ and $a + 2 > 0$ which occurs when $a > 2$, or $a - 2 < 0$ and $a + 2 < 0$ which occurs when $a < -2$.
Alternatively, given $a^2 - 4 > 0$ we have $a^2 > 4$ so $\sqrt{a^2} > 2$ because $\sqrt{x}$ is an increasing function. As $|a| = \sqrt{a^2}$ we see that $|a| > 2$ so $a < -2$ or $a > 2$.
A: $$(a+2)(a-2)\gt0$$
means that either "$a+2$ and $a-2$ are positive" or "$a+2$ and $a-2$ are negative", so? 
A: Others have already answered this question, but I'd like to give a hint.
If your answer differs from the expected one, find a solution that gives different results for a quick check.
You say a > -2 or a > 2; another answer is a < -2 or a > 2. Where do these solutions differ? Most obviously, when a = 0.
Replace a with 0, and what do you get? 0*0 - 4 = -4 , therefore 0 is not a part of solution, therefore your answer is invalid. When you know that you've made a mistake, you may go on to actually searching for the error. Otherwise you may waste time trying to fix the right answer: textbooks are not immune from errors.
