Domain of $h(x) = \log_2 (x^2+4)$ I am setting the $x^2+4 > 0$ and solving, but this leaves me with a question I can't seem to understand.
The resource I'm using is saying the domain is $(-\infty, \infty)$; $x$ is a set of all Real numbers.
I don't understand how, considering I would need to square a negative number during my search for what works, maybe I'm missing a key piece of information on rules of domain here?
Steps:
\begin{align*}
x^2 + 4 & \geq 0\\
x^2 & \geq -4\\
x & \geq \sqrt{-4}
\end{align*}
This is where I'm stuck.
 A: \begin{align}&x^2 + 4  > 0 \\\iff{}& x^2 > -4\\\iff{}& x\in\mathbb R,\end{align} therefore, $\log_2 (x^2+4)$ has domain $\mathbb R.$
Note that technically, both directions of argument are required:

*

*although $\boldsymbol{x > 0}\implies x>-7,$
the domain of $\log (\boldsymbol x)$ is not $[-7,\infty);$


*although $x>3\implies \boldsymbol{x > 0},$
the domain of $\log (\boldsymbol x)$ is not $[3,\infty).$

\begin{align*}
x^2 & \geq -4\\
x & \geq \sqrt{-4}\end{align*}

Incidentally, $\:x^2\ge9\:$ does not imply that $\:x\ge\sqrt9;$ take for example $x=-5.$
In fact, when $c$ is real, \begin{gather}x^2\ge c^2\;\iff\; x\le-|c| \quad\text{or}\quad |c|\le x,\tag1\\x^2\ge9\;\iff\; x\le-\sqrt9 \quad\text{or}\quad \sqrt9\le x.\end{gather} In any case, since $(-4)$ isn't the square of a real number, equivalence $(1)$ doesn't hold.
A: The argument of this logarithm is always positive, whatever the value of x. You don't need so solve the inequality $x^2 + 4 > 0$. Therefore, the domain is in fact $(-\infty, \infty)$.
