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So I am now researching for the cardano method and I do not understand where did the cubic discriminant come from..... It must be from the cardano method.....

Also in this video 2 min 21 sec https://youtu.be/DCdIeS4ls-g?t=141 it says the three solution are A+B, wA+w^2B, w^2A+wB. I doesnt make sense...... I thought the three solutions must be A+B,w(A+B),w2(A+B)..... and I also dont understand where did these long equations come from....

Finally, in the wiki page.

https://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

The last part says if p is not 0, q = 0. u= (p/3)^1/2 v = (P/3)^1/2

and the three roots are t = u + v, t= wu - p/3wu , t= u/w = (-p)^1/2 , t = -(-p)^1/2...... why.....?

I am sorry if it was unclear. I will get more reputation and then i can upload photos :)

thank you so much for reading and input will be appreciated :)

Tommy Lassa
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  • I just explained this in a walkthru answer here :-) The point is that the validity of Cardano's method depends on the product $AB$ having a fixed value. The other quantity $A^3+B^3$ does not change when you multiply $A$ and $B$ by any powers of $\omega$, but you need to be careful not to change the value of $AB$. So if you multiply $A$ by $\omega$, you need to multiply $B$ by $\omega^{-1}=\omega^2$ – Jyrki Lahtonen Jul 23 '15 at 16:23

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Hint:

Let us consider the depressed form, $$y=x^3+px+q.$$

By canceling its derivative $$y'=3x^2+p,$$ we have two extrema (maximum then minimum) at $\mp\sqrt{-\dfrac p3}$, provided $p$ is negative.

The values of the function at the extrema are $$\pm\sqrt{-\frac{p^3}{27}}\mp p\sqrt{-\dfrac p3}+q=\pm2\sqrt{-\dfrac{p^3}{27}}+q.$$

When the value of the maximum is positive and that of the minimum is negative, there are three real roots, otherwise one:

$$-\sqrt{{-\frac{p^3}{27}}}+\frac q2>0\text{ and }\sqrt{{-\frac{p^3}{27}}}+\frac q2<0\iff \frac{p^3}{27}+\frac{q^2}4>0.$$ enter image description here

This is how the discriminant appears.

  • Sorry I dont get it. // for this thing "-\sqrt{{-\frac{p^3}{27}}}+\frac q2>0\land \sqrt{{-\frac{p^3}{27}}}+\frac q2<0\iff \frac{p^3}{27}+\frac{q^2}4>0.", what is the ^ doing there ? from wiki, I found that the discriminant for cubic is 18abcd - 4b^3d+ b^2c^2-4ac^3-27a^2d^2.... I really dont understand where it came from... sorry – Tommy Lassa Jul 23 '15 at 17:26
  • I said depressed form. –  Jul 23 '15 at 18:10
  • Hi, why does p have to be negative? Cant t be a complex number? – Tommy Lassa Jul 27 '15 at 15:40
  • For real extrema and real roots, no. –  Jul 27 '15 at 16:01
  • I am sorry. In that case, when we have two extrema, the cubic function will be symmetrical? I mean, if the value of function at the extrema are +/- (square root(-p/3)), the function will be symmetrical? – Tommy Lassa Jul 28 '15 at 10:22
  • A cubic is always symmetrical around its inflection point. Always. –  Jul 28 '15 at 10:34
  • Hi, so when p is positive, the cubic must have 1 real and 1 pair complex conjugate? – Tommy Lassa Jul 28 '15 at 11:14
  • how about when the cubic has 2 double and 1 distinct root? In that case, the discriminant will equal to 0? But I dont see where it comes from... sorry – Tommy Lassa Jul 28 '15 at 20:04
  • Try and understand the picture. –  Jul 28 '15 at 20:37
  • I really tried. I didnt get how you jump from saying max is positive min is negative to the discriminant equation. ( the last part. ) as when the discriminant is positive there will be 1 real solution, which the max is not positive and min is not negative. So they dont match. I dont get it. :( – Tommy Lassa Jul 28 '15 at 21:03
  • $$-a<b<a\leftrightarrow b^2>a^2$$ –  Jul 29 '15 at 07:21