finding roots to cubics

[uob_student]

hello

how can i finding roots to cubics??

explain by example

:smile:

[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 (http://mathforum.org/dr.math/faq/faq.cubic.equations.html)

[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 (http://mathforum.org/dr.math/faq/faq.cubic.equations.html)


no i do not want newton's method of approximation :smile:

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}
\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]

:rolleyes: thanks