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I was a bit surprised that there is a general formula for the roots of a quartic equation, so I decided to test Wikipedia's version of it myself. To my surprise, I have arrived at a correct answer only once in about five attempts, using only integer coefficients of factorizable polynomials.

I am testing the formula for the equation $(x-1)(x-2)(x-3)(x-4)=0$, which should obviously return 1, 2, 3, and 4. However, my calculations tell a different story:

  • Expanding gives $x^4 - 10x^3 + 35x^2 - 50x + 24$. Therefore, $a = -10, b = 35, c = -50, d = 24$.
  • $u$ is therefore equal to $\frac{3(-10)^2 - 8(35)}{12} = \frac{300 - 280}{12} = \frac{20}{12} = \frac{5}{3}$.
  • $\Delta_0$ is $35^2 - 3(-10)(-50) + 12(24) = 1225 - 1500 + 288 = 13$.
  • $\Delta_1$ is $2(35^3) - 9(-10)(35)(-50) + 27(-10)^2(24) + 27(-50)^2 - 72(35)(24)$, which works out to 70.
  • The discriminant is $-\frac{\Delta_1^2 - 4\Delta_0^3}{27} = -\frac{-3888}{27} = 144$. Because it is positive, the equation must have either four real roots or four complex roots.
  • $Q$ is equal to $\sqrt[3]{\frac{70 + \sqrt{-3888}}{2}} = \sqrt[3]{\frac{70 + 36i\sqrt{3}}{2}} = \sqrt[3]{35 + 18i\sqrt{3}}$.
  • $v$ is $\frac{\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2 + 13}{3\sqrt[3]{35 + 18i\sqrt{3}}}$. After rationalizing the denominator it becomes $\frac{1 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$.
  • $w$ might make things even more complicated, but mercifully its numerator (and therefore value) is zero.

Placing these values into one of the final expressions for the formula gives

$$\frac{1}{4}\left(10 + 2\sqrt{\frac{6 + \sqrt[3]{35 - 18i\sqrt{3}}}{3}} \pm 2\sqrt{\frac{9 - \sqrt[3]{35 - 18i\sqrt{3}}}{3}}\right),$$

a number which is neither rational nor even fully real, and the other two roots are the same.

I expected that the error would become self-evident as I typed this question (which took a while, as you can imagine) — but it has not. So what is wrong with these calculations?

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    $\begingroup$ Why shouldn't that number (with the suitable determination of the cube roots, prescribed by the rules) be rational? After all, $(1+i)+(1-i)$ is rational. $\endgroup$ – egreg May 19 '13 at 16:18
  • $\begingroup$ Did you put your final expression into a calculator to see that it is not an integer? It can be difficult to simplify expressions like these. $\endgroup$ – Samuel May 19 '13 at 16:19
  • $\begingroup$ @egreg Admittedly I used only the real-valued root for the cube root when I calculated this with Wolfram Alpha, but trying the principal root yields the same thing. $\endgroup$ – Lee Sleek May 19 '13 at 16:21
  • $\begingroup$ @Samuel I used Wolfram Alpha. $\endgroup$ – Lee Sleek May 19 '13 at 16:21
  • $\begingroup$ @LeeSleek Your expression for $v$ after rationalizing the denominator is incorrect, could you show steps there? $\endgroup$ – Cocopuffs May 19 '13 at 16:34
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My error was made in rationalizing the denominator of $v$. I multiplied both parts of the fraction by $Q^2$, but rendered the fraction as

$$\frac{35 + 18i\sqrt{3} + 13\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{3(35 + 18i\sqrt{3})}$$

(The first $35 + 18i\sqrt{3}$ should be $(35 + 18i\sqrt{3})\sqrt[3]{35 + 18i\sqrt{3}}$.) Following from this, we can multiply both parts of the fraction by $35 - 18i\sqrt{3}$:

$$\frac{2197\sqrt[3]{35 + 18i\sqrt{3}} + 13(35 - 18i\sqrt{3})\left(\sqrt[3]{35 + 18i\sqrt{3}}\right)^2}{6561}$$

And, simplifying, we find that the true value of $v$ is

$$\frac{\sqrt[3]{35 + 18i\sqrt{3}} + \sqrt[3]{35 - 18i\sqrt{3}}}{3}$$

Or, to put it more simply, $\frac{7}{3}$. (This is what Wolfram Alpha evaluates as the decimal — I have no idea how to obtain this from the cube roots; any help in that respect will be appreciated.)

The expressions for the roots are then

$$\frac{1}{2}(5 + 2 \pm 1), \frac{1}{2}(5 - 2 \pm 1)$$

Which yield the expected 1, 2, 3, 4.

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    $\begingroup$ As mentioned below, $(-1 - 2i\sqrt{3})^3 = 35 + 18i\sqrt{3}$. You can find this by noting that $|35 + 18i\sqrt{3}|^2 = 2197 = 13^3$, and so an element $x+iy\sqrt{3}$ of $\mathbb{Z}[\sqrt{-3}]$ that is a cube root of $35 + 18i\sqrt{3}$ must have absolute value $\sqrt{13}$; that is, $x^2 + 3y^2 = 13$. There aren't many choices to test. $\endgroup$ – Cocopuffs May 20 '13 at 9:15
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Everytime there is a complicated cube root or square root, there is a simplification that can occur.

For example, you can compute $(35+18i\sqrt 3)^{1/3}$ approximately in $\Bbb C$, or you can magically realize that $(-1-2i\sqrt 3)^3 = 35+18i\sqrt 3$ and so you can pick $Q = -1-2i\sqrt 3$ (don't try to find this cube root algebraically).

This should help you with the rest, and feel free to see what happens when you replace $Q$ with one of the other two cube roots.

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