Solving cubic equation?

Question

how to solve cubic equation of the form:
ax^3+bx^2+cx+d=0 ?

is there any fixed formula that can be used to solve ANY cubic equation?

Answer 1

Look up cubic equation. It is a standard formula that does the same thing as the quadratic does for squared equations.............

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Answer 2

It's quite complex in comparision with the quatratic once.... There is a long-length formula for this...

Now, the concept behind this formula is applying some linear transformation to the equation to end up with the form K*X^3 + L*X + N = 0 (can be done with all cubic equations). Then, for the solutions one proceeds using the roots of unity.

Specifically, we factorize x^3 - 3*a*b + a^3 + b^3, i.e.
x^3 - 3*a*b + a^3 + b^3 = (x + a + b)*(x + a*o +b*o^2)*(x + b*o +a*o^2)
where o = -1/2 + i*sqrt(3)/2 i.e a complex third root of unity

Now, the complex roots of x^3 - 3*a*b + a^3 + b^3= 0 are obviously: x = - (a + b), x = - (a*o +b*o^2) and x = - (b*o + a*o^2) SOLUTIONS (I)
Then, we compare:
x^3 - 3*a*b + a^3 + b^3= 0
with K*X^3 + L*X + N = 0 or equivalently X^3 + (L/K)*X + (N/K) = 0. From this we obtain:
L/K = - 3*a*b and N/K = a^3 + b^3. We proceed by solving this simulatneous equation to find a, b in terms of K, L, M and substitute to SOLUTIONS (I) to get some new solutions. To find the final solutions of our initial equation we apply the inverse initial linear tranformation to the solutions.... and voila the general cubic equation is solved...

As you realize solving your equation requires a lot lof of algebraic calculations . . . that's why the cubic formula is much much bigger than the quatratic...

Answer 3

Yes, there is Cardan's Cubic Formula, but it is kind of hard to actually use. If you want to see it, view the links in the source. You can imagine it is not exactly practical.

Answer 4

Yes there is, but it is very complicated.

The solution of ax3+bx2+cx+d=0 is found at :Mouse over link.
http://www.math.vanderbilt.edu/~schectex/courses/cubic

Answer 5

you can either use the horner diagram or put where x= y - b/3a
reform the equasition to this form x^3=px + q
and then use this
x= CRT(q/2 + SQRT(D)) + CRT(Q/2- SQRT(D)) where
D=(q/2)^2 - (p/3)^2
SQRT= square root
CRT= cubic root

P.S. if I were you I would put my faith with Horner

Answer 6

not that I am aware of. Above the second degree it pretty much by hook or crook. Generally you test for rational roots and use synthetic division to locate real ones.

Answer 7

I'm pretty sure the answer for x is 0; I graphed it on my graphing calc, and that's what I got.