General Quartic Solution

**topsquark** · August 7th 2006, 03:28 PM

Okay, this is related to a question of Quick's. I am proud to say that I was able to work backwards and reconstruct how and why Cardano's method for solving a general cubic works (so I no longer have to have that pesky formula memorized).

I have the solution to a general quartic polynomial, but I haven't been able to work that one backwards to figure out how it works. Would anyone be willing to show me the method of how to solve a general quartic?

(If it's too long to write out I'll understand!

)
(And if it depends directly on Galois theory, please forget that I asked because I'll never understand it!

)

-Dan

**topsquark** · August 7th 2006, 04:04 PM

I should mention that the following is the solution I have for a quartic: (as, for all I know, different forms exist...)

Solve $a_0x^4 + a_1x^3 + a_2x^2 + a_3x + a_4 = 0$ .

Transform this equation to $y^4 + ay^2 + by + c = 0$ by the usual transformation $x = y - \frac{a_1}{4a_0}$ .

Denote the three roots of the equation
$z^3 + 2az^2 + (a^2 - 4c)z - b^2 = 0$ (Note: I hope this is right as I had to correct the coefficient of the z^2 term in my book.)
as $\alpha _1$ , $\alpha _2$ , and $\alpha _3$ .

Then the 4 solutions to the y equation above will be:
$y_1 = \frac{1}{2} \left ( \sqrt{\alpha_1} + \sqrt{\alpha_2} + \sqrt{\alpha_3} \right )$

$y_2 = \frac{1}{2} \left ( \sqrt{\alpha_1} - \sqrt{\alpha_2} - \sqrt{\alpha_3} \right )$

$y_3 = \frac{1}{2} \left ( - \sqrt{\alpha_1} + \sqrt{\alpha_2} - \sqrt{\alpha_3} \right )$

$y_4 = \frac{1}{2} \left ( - \sqrt{\alpha_1} - \sqrt{\alpha_2} + \sqrt{\alpha_3} \right )$

-Dan

**ThePerfectHacker** · August 7th 2006, 06:41 PM

Those who solved these nasty equations would be familar with this: Some solutions have complicated radical form, yet they can be greatly simplified. Thus, I am curious, was there ever a theory devoloped to simplify these nested radicals over the rational field?

**Rebesques** · August 11th 2006, 11:08 AM

Don't think so. But why do that really?

Until the 50's, there was not enough computing power around to simplify stuff, and ever since then, if you want to find the root of some equation, you go with numerical analysis!

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