The following is taken from the translation of Galois' Memoir by Harold M. Edwards found in his book Galois Theory. The proof itself is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.
Lemma 1: An irreducible equation g(x) cannot have a root in common with a rational equation h(x) without dividing it.
(1) Let g(x) be a polynomial with coefficients in a given field K that is irreducible over K.
(2) Let h(x) be a polynomial with coefficients in the field K.
(3) Let r be a root for both h(x),g(x) so that h(r)=0 and g(r)=0
(4) Assume that g(x) does not divide h(x).
(5) Then g(x),h(x) are relatively prime since g(x) is irreducible [see Lemma 1, here]
(6) Then (see Corollary 3.1, here), there exists polynomials A(x), B(x) such that A, B all have coefficients in K and:
1 = A(x)g(x) + B(x)h(x)
(7) Since g(r)=0 and h(r)=0, it follows that:
1 = A(r)*0 + B(r)*0
which is impossible.
(8) So we have a contradiction and we reject our assumption in step #4.