Why are both roots real if and only if $p^2 - 4q \ge 0$?

Question

Let,

$$x^2 + px + q = (x - p/2)^2 + q - p^2/4 = \frac{4q - p^2}{4}.$$ This is zero if and only if

$$(x-p/2)^2 = \frac{p^2 - 4q}{4}.$$

Why are both roots real if and only if $$p^2 - 4q \ge 0$$?

Answer 1

It should be a comment, but I make it a quick answer. Reason is: $\left(x-\frac{p}{2}\right)^2 \ge 0$.

Answer 2

You know $p/2$ is real. So $x$ is real iff $x-p/2$ is real. And you know that when $x$ is a root then $x-p/2$ is the square root of $(p^2-4q)/4$.