what's the solution for $x > \sqrt{4}$?

Question

I'm just starting pre calculus and I have trouble understanding the solutions to inequalities involving a radical sign.

For example, what's the solution to $x > \sqrt{4}$? Intuitively, the answer appears to be $x > +2$

but what about the negative root for $4$? If we consider this root, then $x > -2$ would also be a possible solution.

(this is my first post here: I couldn't figure out how to get the radical sign)

Answer 1

$\sqrt{4}$ will ALWAYS mean $+\sqrt{4}$. By definition, $\sqrt{x}$ is the least NONNEGATIVE number $y$ such that $y^2=x$.

Answer 2

As @avid pointed out, for real numbers, $\sqrt{4}=2$. Therefore, $x>\sqrt{4}\implies x>2$.

The source of confusion often originates from a related equation, $x^2>4$.

Here, we have $x^2>4\implies (x-2)(x+2)>0\implies$ either $x>2$ or $x<-2$, which can be concisely written $|x|>2$.

Another way to see this latter inequality is to write $x^2=|x|^2>4\implies |x|>\sqrt{4}=2$ and we reach the same result!