1. 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?  2.

3. 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  4. 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.  5. You can just divide your whole formula by a and it won't change the zeros.  6.   7. Originally Posted by Megabrain i declare this the most embarissing moment in megabrains history on here   8. 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(3uv-p) = 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.  9. If you have a graphing calculator (such as a TI-83+) than it's pretty easy--all you must do is input the equation of the polynomial (in this case, ) into the calculator, graph it, and than find the x-intercept. 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 x-intecept to a greater degree after every computation.  Bookmarks
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