Math Revisited: Cubic Equations

Almost everyone who's taken intermediate algebra would have been exposed to quadratic equations, e.g.  of the form:

ax² + bx + c = 0

which are then solved, either by factoring and solving for x, or - very often - by using the quadratic formula:

x = [-b + {b² - 4ac}^1/2]/ 2a

But what about solving basic cubic equations? Here we solve equations of the form:


x³ – px² + qx – r = 0


Example:

x³ – x² - x + 2 = 0

What I’ll show here, or rather demonstrate, is an algorithm for simple solution provided the numerical coefficient of the cubic term is 1.

We begin by writing: p = 1, q = -1, r = -2

And let:

D =  [(- 4p³ r – 27r² + 18 pqr – 4 q³ + p² q²)] ^1/2

With:

x1 = [p³ –9/2 (pq – 3r) – 3/2 x (-3)^1/2 x D ]^1/3

x2 = (p² – 3q)/ x1


The ROOTS are then – in turn:

a1 = 1/3(p + r2 x x1 + r x x2)

WHERE:

r = - 1/2  + (-3)^1/2/2 x (-3)^1/2  D

r2 = -1/2 – (-3)^1/2/2


Last TWO roots:

a2 = 1/3( p + r2 x x1+ r x x2)

a3 = 1/3( p + r x x1 + rx r2 x x2)


For the original eqn. in question, one obtains:

D = 7.681i

x1 = 0.578 + 1.001i

x2 = 1.73 – 2.997i

and

a1 = 1.103 - 0.665i

a2 = 1.103 + 0.665i

a3 = -1.206

IF the Roots are correct THEN:

a1 + a2 + a3 = 1

and:

(a1 x a2) + (a 2 x a3) + (a3 x a1) = q

AND

a1 x a2 x a3 = r

AND:

(a1 – a2) x (a2 – a3)x (a3 – a1) = D


We will just check the first: a1 + a2 + a3 = 1


We have then:

[1.103 - 0.665i] + [1.103 + 0.665i] + (-1.206) = 2.206 - 1.206 = 1

Energetic readers can check the others!