Modulus Function issue. 
I'm asked to solve $$|x+1|>x^2-5$$

My attempt, 
For  my basic inequality skill, this is a very easy question. 
Since $$|x|=\left\{\begin{matrix}
x, x \geq0 & \\ 
 -x,x<0& 
\end{matrix}\right.$$
Then for $x<-1$
$x^2+x-4<0$
$\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$
Then for $x\geq-1$
$x+1 >x^2-5$
$x^2-x-6 <0$
$-2 < x<3$
So combine the ranges, 
I got $\frac{-1-\sqrt{17}}{2}<x<3$
So, a senior told me that basically my solution is correct, but instead of $x< -1$, I should write $x \leq -1$. Why? According to the definition of modulus function which shows that I'm correct. But why he said that actually the = sign can be for both. So, he asked me don't be bothered too much by this trivial issue. I really don't understand why he said so. Can anyone explain it for me? Thanks in advance.
 A: The senior person is both right and wrong. Right because $|x|=-x$ is indeed true for $x\le0$, but wrong because the case of $x=0$ is already handled (by $x\ge0$) and needn't be repeated.

The discussion can be made as follows:


*

*$x+1\ge0\to x+1>x^2-5\to-2<x<3\to-1\le x<3$,

*$x+1<0\to -(x+1)>x^2-5\to-\dfrac{\sqrt{17}+1}2<x<\dfrac{\sqrt{17}-1}2\to-\dfrac{\sqrt{17}+1}2<x<-1$
This is essentially what you did.
A: It's $x+1>x^2-5$ or $x+1<-x^2+5$  without any additional cases.
because if $x^2-5<0$ then the inequality is obviously true. 
We get $-2<x<3$ or $\frac{-1-\sqrt{17}}{2}<x<\frac{-1+\sqrt{17}}{2}$, which gives your answer:
$$\frac{-1-\sqrt{17}}{2}<x<3$$
I think your solution is right, but your way is bad.
For example, solve the following inequality:
$$|x^3+x-1|>2-x$$
By your way we need to solve two inequalities: $x^3+x-1\geq0$ and $x^3+x-1<0$.
I don't say that it's hard (sometimes it's just impossible!), I say that it's not necessary.
Indeed, $|x|>a\Leftrightarrow x>a$ or $x<-a$ for all $a\in\mathbb R$ 
because for $a<0$ the inequality $|x|>a$ is obvious.
Thus, $|x^3+x-1|>2-x$ gives $x^3+x-1>2-x$ or $x^3+x-1<-2+x$,
which is $x^3+2x-3>0$ or $x^3+1<0$, which is
$x^3-x^2+x^2-x+3x-3>0$ or $(x+1)(x^2-x+1)<0$, which is
$(x-1)(x^2+x+3)>0$ or $x<-1$, which gives the answer:
$$(-\infty,-1)\cup(1,+\infty)$$
