Hi!
I am only 18 year old, so you must excuse me if this question is very stupid, but I've been wondering a long time how to solve a cubic equation. Could anyone explain this to me?

Hi!
I am only 18 year old, so you must excuse me if this question is very stupid, but I've been wondering a long time how to solve a cubic equation. Could anyone explain this to me?
oh thats far from a stupid question, but its not easy to answer since its a very complex way of solving it.
http://planetmath.org/?op=getobj&from=objects&id=1407
there is the formulas to solve it, as you see they arent small and nice
No but if you take care it's just a "plug in the values and turn the handle situation".
Do you happen to know if there is a solution also to ax^3+bx^2+cx+d=0?
Since the link you point to shows only the solution to x^3.
You can just divide your whole formula by a and it won't change the zeros.
i declare this the most embarissing moment in megabrains history on hereOriginally Posted by Megabrain
I wouldn't call the derivation of the solution of the cubic equation complex  its easy enough as long as you know the trick
So lets consider a monic cubic equation x<sup>3</sup> + a x<sup>2</sup> + b x + c = 0 (which covers all possible cubic equations as Zelos said). By using the substitution y = x  a/3 we can rewrite this equation in the reduced form y<sup>3</sup> + p y + q = 0 (so we have got rid of the first haggle, the x^2 term). Now any solution to this equation can be written in the forum y = v  u where v and u are real numbers (and there are an infinite amount to choose from). So lets substitute into our equation, and find that we get the new equation
(v<sup>3</sup>  u<sup>3</sup> + q) + v(p  3uv) + u(3uvp) = 0
Now this equation is true if u<sup>3</sup>  v<sup>3</sup> = q and 3uv=p and we can solve this simultaneous equations by substitution
We find that u<sup>6</sup>  q u<sup>3</sup>  1/27 p<sup>3</sup> = 0 which is a quadratic equation in u<sup>3</sup>! We can then use the quadratic equation to see that
u = (q/2 ± Sqrt(q<sup>2</sup>/2 + p<sup>3</sup>/27))^(1/3)
You then just trace your variables back to the original x and you get the equations Zelos posted.
If you have a graphing calculator (such as a TI83+) than it's pretty easyall you must do is input the equation of the polynomial (in this case, ) into the calculator, graph it, and than find the xintercept. If not this, than a relatively good way to do so is using Newton's method if you have studied calculus of one variable well. The process involves continually finding the tangent line to a curve at some point such that you approach the xintecept to a greater degree after every computation.
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