Question

Formulas to find roots of a cubic equation

Answer 1

The "Cubic Formula"

Introduction.

Knowledge of the quadratic formula is older than the Pythagorean Theorem. Solving a cubic equation, on the other hand, was the first major success story of Renaissance mathematics in Italy. The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna.

Our objective is to find a real root of the cubic equation 

ax3+bx2+cx+d=0.


The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. The solution proceeds in two steps. First, the cubic equation is "depressed"; then one solves the depressed cubic.

Depressing the cubic equation.

This trick, which transforms the general cubic equation into a new cubic equation with missing x2-term is due to Nicolò Fontana Tartaglia (1500-1557). We apply the substitution 




to the cubic equation, to obtain: 




Multiplying out and simplifying, we obtain the "depressed" cubic 



Let's try this for the example 

2x3-30x2+162x-350=0.


Our substitution will be x=y+5; expanding and simplifying, we obtain the depressed cubic equation 

y3+6y-20=0.


Solving the depressed cubic.

We are left with solving a depressed cubic equation of the form 

y3+Ay=B.


How to do this had been discovered earlier by Scipione del Ferro (1465-1526).

We will find s and t so that 

3st=A(1)s3-t3=B.(2)


It turns out that y=s-t will be a solution of the depressed cubic. Let's check that: Replacing A, B and y as indicated transforms our equation into 

(s-t)3+3st (s-t)=s3-t3.


This is true since we can simplify the left side by using the binomial formula to: 

(s3-3s2t+3st2-t3)+(3s2t-3st2)=s3-t3.


How can we find s and t satisfying (1) and (2)? Solving the first equation for s and substituting into (2) yields: 




Simplifying, this turns into the "tri-quadratic" equation 




which using the substitution u=t3 becomes the quadratic equation 




From this, we can find a value for u by the quadratic formula, then obtain t, afterwards s and we're done.

Let's do the computation for our example 

y3+6y=20.


We need s and t to satisfy 

3st=6(3)s3-t3=20.(4)


Solving for s in (3) and substituting the result into (4) yields: 




which multiplied by t3 becomes 

t6+20t3-8=0.


Using the quadratic formula, we obtain that 




We will discard the negative root, then take the cube root to obtain t: 



By Equation (4), 



Our solution y for the depressed cubic equation is the difference of s and t: 



The solution to our original cubic equation 

2x3-30x2+162x-350=0


is given by 



Concluding remarks.

I will not discuss a slight problem you can encounter, if you follow the route outlined. What problem am I talking about?

Shortly after the discovery of a method to solve the cubic equation, Lodovico Ferrari (1522-1565), a student of Cardano, found a similar method to solve the quartic equation.

This section is loosely based on a chapter in the book Journey Through Genius by William Dunham.







don't mind it is according to my maths book

Answer 2

For the general cubic equation of the form:

$f(x) = a {x}^{3} + b {x}^{2} + cx + d = 0$

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$x_{1,2,3} = \frac{ - b + 3a. \omega _{k}. \sqrt[3]{u} + 3a.( \omega _{k} ) ^{2} . \sqrt[3]{v} }{3a}$

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where

$u,v = \frac{9abc - 2 {b}^{3} - 27 {a}^{2}d \: ± \: 3ai \sqrt{3Δ} }{54 {a}^{3} }$

$i = \sqrt{ - 1}$

$\omega _{k}∈[1, \frac{ - 1 + \sqrt{3} }{2} , \frac{ - 1 - \sqrt{3} }{2} ]$

Cubic Discriminant:

$Δ = - 27 {a}^{2} {d}^{2} + 18abcd - 4a {c}^{3} - 4 {b}^{3} d + { b}^{2} {c}^{2}$