How to find the zero roots of $e^{-x^2}$?

Question

Square root of negative number doesn't exist, so there is no x for which y is zero?

Answer 1

Recall that $\exp x > 0$ for any real $x$. Thus no such $x$ such that $\exp (-x^2) = 0$ exists.

If you want complex roots, remember that $\exp z = \cos z + i \sin z$. The complex exponential will be zero when both the sine and cosine are zero for the same $z$.

Answer 2

$f: x\mapsto e^{-x^2} $ is even and strictly decreasing at $(0,+\infty)$ and $$\lim_{x\to+\infty}f (x)=0$$ $$\implies 0 <f (x)\leq f (0)=1$$

thus no roots.

Answer 3

The exponential is nowhere zero because it obeys a power law $\exp(x+y)=\exp(x)\exp(y)$. Assume $\exp(a)=0$ then $1=\exp(0)=\exp(a-a)=\exp(a)\exp(-a)=0$, contradiction.