§ Galois theory perspective of the quadratic equation I found this quite delightful the first time I saw it, so I wanted to record
it ever since.
Let x 2 + b x + c x^2 + bx + c x 2 + b x + c equate it to the roots:
x 2 + b x + c = ( x − p ) ( x − q ) x 2 + b x + c = x 2 − x ( p + q ) + p q − ( p + q ) = b ; p q = 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}
x 2 + b x + c = ( x − p ) ( x − q ) x 2 + b x + c = x 2 − x ( p + q ) + p q − ( p + q ) = b ; p q = c We want to extract the values of b b b c c c the symmetric functions:
( p + q ) 2 = b 2 ( p − q ) 2 = ( p + q ) 2 − 4 p q = b 2 − 4 c
(p + q)^2 = b^2
(p - q)^2 = (p + q)^2 - 4pq = b^2 - 4c
( p + q ) 2 = b 2 ( p − q ) 2 = ( p + q ) 2 − 4 p q = b 2 − 4 c Hence we get that
p − q = ± b 2 − 4 c
p - q = \pm\sqrt{b^2 - 4c}
p − q = ± b 2 − 4 c From this, we can solve for p , q p, q p , q p = ( ( p + q ) + ( p − q ) ) / 2 = ( − b ± b 2 − 4 c ) / 2
p = ((p + q) + (p - q))/2 = (-b \pm \sqrt{b^2 - 4c})/2
p = ( ( p + q ) + ( p − q ) ) / 2 = ( − b ± b 2 − 4 c ) / 2 § Galois theory for cubics § Galois theory for bi-quadratics § Galois theory for quintics § References