How to arithmetically determine values of x that satisfies $2^x > x^2$

Question

Without using guessing and checking or graphing the equations, is there a method to determine values of x that satisfies $2^x > x^2$?

Answer 1

$$ 2^x > x^2 $$

We quickly check that $2^x = x^2$ holds for $x = 2$ and $x = 4$. But $2^x$ is an exponential curve which gets closer and closer to the $x$-axis in the negative $x$-direction, while $x^2$ is a parabola which rises up to infinity in the negative $x$-direction. Also, $2^0>0^2$. Hence we conclude that the equation $2^x -x^2 = 0$ has 3 real roots, one of which is negative, say $\epsilon$.

$2^xx^2$ in $(\epsilon,2)$, $2^xx^2$ in $(4,\infty)$. For your requirement,

$$ 2^x > x^2 \ \forall x \in (\epsilon,2)\cup (4,\infty). $$

What's left is to find $\epsilon$.

EDIT: This solution is based on the nature of the curves.

UPDATE: It turns out that $\epsilon ≈ -0.77$.