The content in today's blog is taken from Jean-Pierre Tignol's Galois' Theory of Algebraic Equations.

(1) Ferrari started out with a quadratic equation of the following form:

x

^{4}+ bx

^{3}+ cx

^{2}+ dx + e = 0

(2) Then, he borrowed a method from Girolamo Cardano (see here) and set y = x + (b/4) which means x = y - (b/4).

(3) Substituting x = y - (b/4) gives us:

(y - [b/4])

^{4}+ b(y - [b/4])

^{3}+ c(y - [b/4])

^{2}+ d(y - [b/4]]) + e = 0

Working this through (see details here), gives us:

y

^{4}+ py

^{2}+ qy + r = 0

where:

p = c - 6(b/4)

^{2}

q = d - (b/2)c + (b/2)

^{3}

r = e - (b/4)c + (b/4)

^{2}c - 3(b/4)

^{4}

(4) Now if we move the q and r term over we get:

y

^{4}+ py

^{2}= -qy -r

This helps us because if we add (p/2)

^{2}to both sides, we can complete the square to get:

(y

^{2}+ p/2)

^{2}= -qy -r + (p/2)

^{2}

(5) Now, if we add a value u to our square, we get the following:

(y

^{2}+ p/2 + u)

^{2}= -qy -r + (p/2)

^{2}+ 2uy

^{2}+ pu + u

^{2}

This may not appear to help us but it gives us a possibility for completing the square on the other side.

(6) For example, if we could find a value u such that:

-qy -r + (p/2)

^{2}+ 2uy

^{2}+ pu + u

^{2}= ([√2uy - q/(2√2u)]

^{2}

Then we have found a solution (see step #11 if this point is not clear)

(7) Now:

([√2uy - q/(2√2u)]

^{2}= 2u

^{2}y

^{2}- 2*(√2uy)(q/(2√2u)) + q

^{2}/(8u) =

= 2u

^{2}y

^{2}- q/u + q

^{2}/(8u)

(8) So this equation works only if:

-qy -r + (p/2)

^{2}+ 2uy

^{2}+ pu + u

^{2}= 2uy

^{2}- qy + q

^{2}/(8u)

That is if:

-r + (p/2)

^{2}+ pu + u

^{2}= q

^{2}/(8u)

(9) But if we multiply 8u to all sides and then subtract by q

^{2}, we get:

8u

^{3}+ 8pu

^{2}+ (2p

^{2}- 8r)u - q

^{2}= 0

(10) This is the general cubic equation which we can solve using Cardano's formula so we know that u can take on this value.

(11) This then gives us:

(y

^{2}+ p/2 + u)

^{2}= ([√2uy - q/(2√2u)]

^{2}.

(12) This can we solve using the quadratic equation. The four solutions can be found from:

(y

^{2}+ p/2 + u)

^{2 }= ± [√2uy - q/(2√2u)]

I should probably note that we assumed that u ≠ 0. I will talk about the situation were u = 0 in my next blog where I provide a proof for Ferrari's solution.

References

- Jean-Pierre Tignol, Galois' Theory of Algebraic Equations, World Scientific, 2002