I found this quite delightful the first time I saw it, so I wanted to record it ever since.

Let x2+bx+cx^2 + bx + c be a quadratic. Now to apply galois theory, we first equate it to the roots:

x2+bx+c=(x−p)(x−q)x2+bx+c=x2−x(p+q)+pq−(p+q)=b;pq=c \begin{aligned} &x^2 + bx + c = (x - p)(x-q) &x^2 + bx + c = x^2 - x(p + q) + pq &-(p + q) = b; pq = c \end{aligned}

We want to extract the values of bb and cc from this. To do so, consider the symmetric functions:

(p+q)2=b2(p−q)2=(p+q)2−4pq=b2−4c (p + q)^2 = b^2 (p - q)^2 = (p + q)^2 - 4pq = b^2 - 4c

Hence we get that

p−q=±b2−4c p - q = \pm\sqrt{b^2 - 4c}

From this, we can solve for p,qp, q, giving us:

p=((p+q)+(p−q))/2=(−b±b2−4c)/2 p = ((p + q) + (p - q))/2 = (-b \pm \sqrt{b^2 - 4c})/2

§ Galois theory for cubics

§ Galois theory for bi-quadratics

§ Galois theory for quintics

§ References