Is the Golden ratio transcendental?

Relevant wiki: Transcendental Numbers

We have that $\phi = 1 + \dfrac{1}{1 + \dfrac{1}{1 + \ddots}}$ . We can now rewrite $\phi$ as $x = 1 + \dfrac{1}{x} \implies x^2 - x -1 = 0$ . We need not evaluate $\phi$ as it is a root ( $x = \phi$ ) of a quadratic equation with integer coefficients and hence, it is not a transcendental number.