Proof of the Solution to the General Quartic Equation

In today's blog, I provide the proof for the solution of the general quartic equation. If you would like to see how Lodovico Ferrari found his solution, see here.

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

Lemma 1: Depressed Quartic Equation

If:

x⁴ + bx³ + cx² + dx + e = 0

Then there exists y,p,q,r such that:

y⁴ + py² + qy + r = 0

where:

y = x + (b/4)

p = c - 6(b/4)²

q = d - (b/2)c + (b/2)³

r = e - (b/4)d + (b/4)²c - 3(b/4)⁴

Proof:

(1) Let y = x + (b/4) so that x = y - (b/4)

(2) We note that (see here for details on the Binomial Theorem if needed):

(u - v)² = u² - 2uv + v²

(u - v)³ = u³ - 3u²v + 3uv² -v³

(u - v)⁴ = u⁴ - 4u³v + 6u²v² -4uv³ + v⁴

(3) This then gives us:

(y - [b/4])⁴ = y⁴ - by³ + (3/8)b²y² -b³y/16 + (b/4)⁴

b(y - [b/4])³ = by³ - (3/4)b²y² + (3/16)b³y - b⁴/64

c(y - [b/4])² = cy² - bcy/2 + c(b/4)²

d(y - [b/4]) = dy - bd/4

(4) Putting this all together gives us:

x⁴ + bx³ + cx² + dx + e =

= y⁴ + (3/8)b²y² -b³y/16 + (b/4)⁴ - (3/4)b²y² + (3/16)b³y - b⁴/64 + cy² - bcy/2 + c(b/4)² + dy - bd/4 +e =

= y⁴ + [c-(3/8)b²]y² + [ (1/8)b³ -bc/2 + d]y + [e - bd/4 + c(b/4)² - 3b⁴/256 ]

QED

Lemma 2: Solution for y⁴ + py² + r = 0

If y⁴ + py² + r = 0

Then:

y = ± √ (1/2)[-p ± √p² - 4r

Proof:

(1) y⁴ + py² + r = 0

(2) (y²)² + p(y²) + r = 0

(3) Using the quadratic equation (see here):

y² = (1/2)[-p ± √p² - 4r

(4) Solving for y², we get:

y = ± √ (1/2)[-p ± √p² - 4r

QED

Lemma 3:

If:

y⁴ + py² + qy + r = 0 and q ≠ 0

Then there exists u ≠ 0 such that:

(y² + p/2 + u)² = [√2uy - q/(2√2u)]²

Proof:

(1) y⁴ + py² + qy + r = 0

(2) y⁴ + py² + (p/2)² = -qy -r + (p/2)²

(3) (y² + p/2)² = -qy -r + (p/2)²

(4) Let u be the solution of the following equation:

8u³ + 8pu² + (2p² -8r)u - q² = 0.

We know that this solution exists from the general cubic equation (see here)

(5) We know that u ≠ 0 since if u = 0, then q = 0 but q ≠ 0.

(6) So, we can rearrange our equation in step #4 to:

8u³ + 8pu² + (2p² -8r)u = q²

And after dividing both sides by 8u, we get:

u² + pu + (p/2)² - r = q²/8u.

(7) Now, we know that:

(y² + p/2 + u)² = -qy -r + (p/2)² + 2uy² + pu + u²

(8) So, using step #6, we have:

-qy -r + (p/2)² + 2uy² + pu + u² =

= -qy + 2uy² + q²/8u =

= [√2uy - q/(2√2u)]²

QED

Thereom: General Quartic Equation

If:

ax⁴ + bx³ + cx² + dx + e = 0

p = (c/a) - 6([b/a]/4)²

q = (d/a) - ([b/a]/2)c + ([b/a]/2)³

r = (e/a) - ([b/a]/4)d + ([b/a]/4)²c - 3([b/a]/4)⁴

s = ([b/a]/4)

u = the solution to the cubic equation: 8u³ + 8pu² + (2p² -8r)u - q² = 0.

Then:

if q = 0,

x = ± √ (1/2)[-p ± √p² - 4r - s

if q ≠ 0, then:

x = ± (1/2) [√2uy + √2u - 2p - 4u + 2q/(√2u)] - s

or

x = ± (1/2) [-√2uy + √2u - 2p - 4u - 2q/(√2u)] - s

Proof:

(1) ax⁴ + bx³ + cx² + dx + e = 0

(2) If we divide all sides by a, we get:

x⁴ + (b/a)x³ + (c/a)x² + (d/a)x + (e/a) = 0

(3) Using Lemma 1 above, we get:

y⁴ + py² + qy + r = 0

where:

y = x + ([b/a]/4)

p = (c/a) - 6([b/a]/4)²

q = (d/a) - ([b/a]/2)c + ([b/a]/2)³

r = (e/a) - ([b/a]/4)d + ([b/a]/4)²c - 3([b/a]/4)⁴

(4) If q=0, then using Lemma 2 above, we have:

y = ± √ (1/2)[-p ± √p² - 4r

(5) If q ≠ 0, then using Lemma 3 above, we have:

(y² + p/2 + u)² = [√2uy - q/(2√2u)]²

where

u is the solution to:

8u³ + 8pu² + (2p² -8r)u - q² = 0.

We can find this solution using the cubic equation (see here).

(6) This then gives us:

y² + p/2 + u = ± √2uy - q/(2√2u).

(7) This then gives us four solutions:

(a) Case I: y² + p/2 + u = + √2uy - q/(2√2u).

In this case:

y² - √2uy + [p/2 + u + q/(2√2u)] = 0

Using the quadratic equation:

y = ± (1/2) [√2uy + √2u - 2p - 4u - 2q/(√2u)]

(b) Case I: y² + p/2 + u = - √2uy + q/(2√2u).

In this case:

y² + √2uy + [p/2 + u - q/(2√2u)] = 0

Using the quadratic equation:

y = ± (1/2) [-√2uy + √2u - 2p - 4u + 2q/(√2u)]

(8) Since x = y - ([b/a]/4), we have the following solutions:

(a) If q=0, then:

x = ± √ (1/2)[-p ± √p² - 4r - ([b/a]/4)

(b) If q ≠ 0, then:

x = ± (1/2) [√2uy + √2u - 2p - 4u - 2q/(√2u)] - ([b/a]/4)

or

x = ± (1/2) [-√2uy + √2u - 2p - 4u + 2q/(√2u)] - ([b/a]/4)

QED

References

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