# Thread: Solution of Cubic

1. ## Solution of Cubic

Can someone help me find the exact solutions to the equation

$\displaystyle x^3 + 2x^2 - 9x + 3 = 0$?

I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ($\displaystyle t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.

A big part of my problem here is that using the rational root theorem leads me to try $\displaystyle x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).

Obviously I've put a good amount of thought into this problem; somebody please help me out!

2. Originally Posted by English Major
Can someone help me find the exact solutions to the equation

$\displaystyle x^3 + 2x^2 - 9x + 3 = 0$?

I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ($\displaystyle t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.

A big part of my problem here is that using the rational root theorem leads me to try $\displaystyle x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).

Obviously I've put a good amount of thought into this problem; somebody please help me out!
Is there any particular reason why you want exact solutions, as opposed to approximate ones?

3. Originally Posted by English Major
Can someone help me find the exact solutions to the equation

$\displaystyle x^3 + 2x^2 - 9x + 3 = 0$?

I've used a grapher to find approximate solutions at the three places where it crosses the x-axis, and tried my best to work through Cardano's solution ($\displaystyle t^3 -\frac{31}{3}t + \frac{259}{27} = 0$) but I just can't finish it.

A big part of my problem here is that using the rational root theorem leads me to try $\displaystyle x = \pm 1, \pm 3$, none of which work, so all three real solutions (it crosses three times, so no complex-imaginary roots) must be irrational. I recognize it obviously must be possible, but I'm not used to three irrational roots giving such a nice, simplistic equation. These three irrationals must multiply to be perfectly 3, add to be perfectly -2, and something else (what?) to be perfectly -9 (or +9? I can't remember).

Obviously I've put a good amount of thought into this problem; somebody please help me out!
For:

$\displaystyle x^3+a_1x^2+a_2 x+a_3=0$

Put:

$\displaystyle Q=\frac{3a_1-a_1^2}{9}$

and:

$\displaystyle R=\frac{9a_1a_2-27a_3-2a_1^3}{54}$

We know in the case under consideration that the roots are real, as also the discriminant

$\displaystyle D=Q^3+R^2<0$

tells us (since $\displaystyle Q=\frac{-31}{9}$ and $\displaystyle R=\frac{-259}{54}$).

In which case Tartaglia-Cardano formula reduces to:

$\displaystyle \theta=\arccos(-R/\sqrt{-Q^3})$

$\displaystyle x_1=2\sqrt{-Q}\cos(\theta/3)$

$\displaystyle x_1=2\sqrt{-Q}\cos(\theta/3+2\pi/3)$

$\displaystyle x_3=2\sqrt{-Q}\cos(\theta/3+4\pi/3)$

RonL

4. Originally Posted by mr fantastic
Is there any particular reason why you want exact solutions, as opposed to approximate ones?
Honestly, it's for my own understanding- I always believed that for polynomials with rational coefficients, any irrational solutions would come in conjugate pairs from quadratic formula use. You could use three cube root zeros to make a polynomial with a rational constant term, but I don't see how they would balance each other out for rational coefficients in other terms.

The short answer is, I'm interested to see the form the solutions take, and to use that to expand my own knowledge. Thanks for any help you can provide.

5. Originally Posted by CaptainBlack
For:

$\displaystyle x^3+a_1x^2+a_2 x+a_3=0$

(snipped)

In which case Tartaglia-Cardano formula reduces to:

$\displaystyle \theta=\arccos(-R/\sqrt{-Q^3})$

$\displaystyle x_1=2\sqrt{-Q}\cos(\theta/3)$

$\displaystyle x_1=2\sqrt{-Q}\cos(\theta/3+2\pi/3)$

$\displaystyle x_3=2\sqrt{-Q}\cos(\theta/3+4\pi/3)$

RonL
Okay, I'll admit: I was not expecting the solutions to include trig (though the $\displaystyle 2\pi/3$ and $\displaystyle 4\pi/3$ imply to me that this may have something to do with roots of unity, being equally spaced around a unit circle. True?)
Can you recommend a place to read up on this type of solution? The wikipedia article I linked to provides similar, but distinctly different solution methods. Incidentally, looking through my article, I'm noticing that the method I was trying "fails for the case of three real roots," which I wish I'd noticed earlier.
Thanks!

6. Originally Posted by English Major
Okay, I'll admit: I was not expecting the solutions to include trig (though the $\displaystyle 2\pi/3$ and $\displaystyle 4\pi/3$ imply to me that this may have something to do with roots of unity, being equally spaced around a unit circle. True?)
Can you recommend a place to read up on this type of solution? The wikipedia article I linked to provides similar, but distinctly different solution methods. Incidentally, looking through my article, I'm noticing that the method I was trying "fails for the case of three real roots," which I wish I'd noticed earlier.
Thanks!
The form of solution given in my earlier post is just another way of writing the
solution in terms of the Chebyshev cube root.

RonL