How do you find roots of cubic equations?

[uob_student]

hello

how can i finding roots to cubics??

explain by example

 

[Tide]

Perhaps you have an example in mind?

 

[mezarashi]

Oh how I just love these open-ended questions ;P

 

[Fermat]

If you want to find approximate vaues, use Newton's method of approximation. (you can google for it).
For a more analytical approach try here

 

[uob_student]

Perhaps you have an example in mind?

x^3+(5*x^2)+(3*x)+9=0

 

[uob_student]

If you want to find approximate vaues, use Newton's method of approximation. (you can google for it).
For a more analytical approach try here

no i do not want Newton's method of approximation

i want (kardan) method but i am not sure about the spelling of (kardan)

 

[Tide]

Cardan's formula gives

$$x = -\frac {\left( 179 - 9 \sqrt {345} \right)^{1/3}}{3}-\frac {\left( 179 + 9 \sqrt {345} \right)^{1/3}}{3} - \frac {5}{3}$$

for the real root. The other two are complex. And, no, I did not do it by hand! :)

 

[HallsofIvy]

If a and b are any two numbers then
(a- b)³= a³-3a²b+ 3ab²- b³
3ab(a-b)= 3a²b- 3ab²

so (a-b)³+ 3ab(a-b)= a³- b³.

In particular, if we let x= a-b, m= 3ab, and n= a³- b³, that says that x³+ mx= n. That is, we can pick any two numbers a, b and right down a cubic equation that has x= a- b as a root.
The question is, can we go the other way: given m and n, can we find a and b so we can write x= a-b as a solution.
The answer to that question is "Yes, we can"!

Since m= 3ab, b= m/3a. Putting that int n= a³- b³, we have $$n= a^3- \frac{m^3}{3^3a^3}$$.
Multiplying both sides of the equation by a³, we have
$$na^3= a^6- (\frac{m}{3})^3$$
which looks worse but is just a quadratic equation in a³:
$$(a^3)^2- n(a^3)- (\frac{m}{3})^3$$.
Use the quadratic formula to solve that
$$a^3= \frac{n +/- \sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}$$
$$a^3= \frac{n}{2} +/- \sqrt{\left(\frac{n}{2}<br /> \right)^2- \left(\frac{m}{3}\right)^3}$$

Since a³- b³= n, solving for b³ gives
$$a^3= -\frac{n}{2} +/- \sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^3}$$

Finding the cube root of each of those, then subtracting to get x= a- b gives the formula that Tide cited.

Warning- applying that formula is really, really hard!

 

[uob_student]

thanks