Solving Cubic Equations - General Method

  • Thread starter Thread starter TheDestroyer
  • Start date Start date
AI Thread Summary
To solve the cubic equation ax^3 + bx^2 + cx + d = 0, the general method involves reducing it to a simpler form without the x^2 term. This is achieved by substituting x with y + y0 to eliminate the quadratic term. The resulting reduced cubic equation can be solved using the relationship x^3 + mx - n = 0, where m and n are derived from the original coefficients. The solutions for a and b can yield up to three distinct values for x, as each cube root provides multiple complex solutions. This method effectively allows for solving all cubic equations systematically.
TheDestroyer
Messages
401
Reaction score
1
Hi guyz,

How can I solve the equation with the form :

ax^3 + bx^2 + cx + d = 0

I want the general way to solve allllllllll cubic equations..

Thanks
 
Mathematics news on Phys.org
The post above is very good but here it is in a nutshell:

We know that (a- b)3= a3-3a2b+ 3ab2- b3 and that 3ab(a-b)= 3a2b- 3ab so:

(a-b)3+ 3ab(a-b)= a3- b3
(the middle terms cancel).

That is: if we pick any two numbers a, b and let x= a- b, m= 3ab and n= a3- b3, then x3+ mx= n.

What about the other way? If we are given m and n, can we solve for a and b (and so find x)?

Yes, we can. From m= 3ab, we have b= m/(3a). Put that into
a3- b3= n and we have a3- m3/(33a3)= n

Multiply both sides of the equation by a3 and we have the (6th degree!) equation
a6- (m/3)3= naa3.

But if we let u= a3, this reduces to a quadratic equation for u: u2- nu- (m/3)3= 0.

We can solve that by the quadratic formula:
u= a3= (n +/- √(n2+ 4(m/3)3))/2= (n/2)+/- √((n/2)2+ (m/3)3).

Since a3- b3= n,
b3= a3- n
= (-n/2)+/- √((n/2)2+ (m/3)3).

Solving for a and b,

a= (((n/2)+/- √((n/2)2+ (m/3)3))1/3
b= ((-(n/2)+/- √((n/2)2+ (m/3)3))1/3

and, finally, x= a-b.


Notice that the equation was x3+ mx- n= 0 which has no x2. This is the "reduced" cubic.

To solve a general cubic, x3+ ax2+ bx+ c= 0,
"Shift" the variable: replace x by y+ y0 in this equation and then choose y0 to make the coefficient of y2 equal to 0.
 
Thanks Guyz,

Thanks Guyz, I hope these equations are true, because i can't check them right now :)

I'm very thankful :):)
 
?

Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?
 


Originally posted by TheDestroyer
Hey Guyz, Does the equation "imhereyeah" posted solves with 3 values of x? or 1 only?

Yes. Note that the "a" and "b" in that reply are derived from cube roots. So "a" and "b" each have 3 possible (complex) values giving a total of 9 combinations in the solution (though at most 3 will be distinct).
 
Last edited:

Thread 'What Exactly is Dirac’s Delta Function? - Insight'

Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
View full post »

Thread 'Fermat's Last Theorem'

Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
View full post »
Thread 'Imaginary Pythagorus'

Thread 'Imaginary Pythagorus'

I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...
View full post »

Similar threads

I Quadratic, cubic, quartic, quintic equations
Replies
16
Views
4K
MHB Understanding Cubic Equation Formula for Polynomials of Degree Three
Replies
9
Views
3K
B Are Inverse Problems More Complex Than Direct Problems in Mathematics?
Replies
13
Views
3K
I Why was the substitution x = y - b/3a used in Cardano's Cubic Formula?
Replies
10
Views
3K
I Can an inverse function of a special cubic function be found?
Replies
15
Views
2K
Problem of solving the cubic function
Replies
4
Views
2K
MHB Finding Min Value of $\dfrac{|b|+|c|}{a}$ from Roots of Cubic Equations
Replies
4
Views
1K
MHB Polynomial Challenge: Show Real Roots >1 Exist
Replies
1
Views
1K
MHB Prove No Integers Solve $ax^3+bx^2+cx+d=1$ for x=19,2 for x=62
Replies
3
Views
2K
B Understanding the relationship between ratios and fractions
Replies
8
Views
2K

Hot Threads

Recent Insights

Back
Top