Is $\sqrt[n]{x} = 0$ an algebraic equation?

Question

According to the definition for algebraic equation on wikipedia, https://en.wikipedia.org/wiki/Algebraic_equation, it is an equation of the form $P(x,y) = 0$. But I've read that an algebraic equation is one that involves the four elementary operations plus roots finitely many times, here for reference: https://www.quora.com/What-is-the-difference-between-algebraic-and-transcendental-function. So, can algebraic equation have roots or not?

Answer 1

An algebraic equation is an equation in the form $$ P(x,y,\cdots) = 0, $$ in which $P$ is a polynomial in the variables $x,y, \cdots$. An algebraic function is a solution of a polynomial equation, e.g. $$ P(x,y) = x^2y(x)^2 + xy(x) + x = 0, $$ which has as solutions $$ y = -\frac{-x + \sqrt{x^2-4x^3}}{2x^2} \ \ \mathrm{and} \ \ y = -\frac{-x - \sqrt{x^2-4x^3}}{2x^2}. $$ See that you have both rational and negative powers.

Any polynomial in the form $$ P(x) = a_n x^n + \cdots a_1 x + a_0 = 0 $$ has a constant solution, which is also an algebraic function.