用Cardano法解三次方程x³-15x-4=0的解生成疑问

Great question! The short answer is: you absolutely can generate all three roots using Cardano's method—you just need to account for all branches of the complex cube root, not just the principal one. Alternatively, once you have one root (from the principal cube roots), you can use polynomial factorization to find the other two, which is often simpler in practice. Let's break this down:

1. Why the Initial Result Only Gives One Root #

When you apply Cardano's method to $x^3 -15x -4=0$, you start by setting $x=u+v$, leading to the system:

• $u^3 + v^3 = 4$
• $uv = 5$ (since for the cubic $x^3 + px + q=0$, we have $uv = -p/3$; here $p=-15$, so $-p/3=5$)

Taking the principal cube roots gives $u=\sqrt[3]{2+11i}=2+i$ and $v=\sqrt[3]{2-11i}=2-i$, so $x=(2+i)+(2-i)=4$—this is the obvious real root. But complex numbers have three distinct cube roots, not just one. Ignoring the other two branches is why you only saw one solution initially.

2. Using All Cube Root Branches to Find the Other Roots #

For any complex number $z$, its cube roots are $z^{1/3}$, $z^{1/3}\omega$, and $z^{1/3}\omega2$, where $\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ is a primitive 3rd root of unity (satisfies $\omega^3=1$ and $1+\omega+\omega^2=0$).

Step 1: Find all cube roots of $2+11i$ and $2-11i$ #

• Principal root of $2+11i$: $u_1=2+i$ (since $(2+i)^3=2+11i$)
• Other roots: $u_2=u_1\omega=(2+i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$, $u_3=u_1\omega^2=(2+i)\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$
• Principal root of $2-11i$: $v_1=2-i$ (conjugate of $u_1$)
• Other roots: $v_2=v_1\omega=(2-i)\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)$, $v_3=v_1\omega^2=(2-i)\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)$

Step 2: Pair roots correctly (must satisfy $uv=5$) #

We can't pair arbitrary roots—we need $u \cdot v=5$ to satisfy the original system. This means:

• Pair $u_2$ with $v_3$: $u_2 + v_3 = (2+i)\omega + (2-i)\omega^2$
  Using $\omega+\omega^2=-1$ and $\omega-\omega^2=\sqrt{3}i$, we simplify:
  $$
  2(\omega+\omega^2) + i(\omega-\omega^2) = 2(-1) + i(\sqrt{3}i) = -2 - \sqrt{3}
  $$
• Pair $u_3$ with $v_2$: $u_3 + v_2 = (2+i)\omega^2 + (2-i)\omega$
  Similarly:
  $$
  2(\omega^2+\omega) + i(\omega^2-\omega) = 2(-1) + i(-\sqrt{3}i) = -2 + \sqrt{3}
  $$

These are exactly the two missing roots you mentioned!

3. The Simpler Alternative: Factorization #

Once you have the real root $x=4$, you can perform polynomial division or use synthetic division to factor the cubic:
$$
x^3 -15x -4 = (x-4)(x^2 +4x +1)
$$
Then solve the quadratic equation $x^2+4x+1=0$ using the quadratic formula:
$$
x = \frac{-4 \pm \sqrt{16-4}}{2} = -2 \pm \sqrt{3}
$$
This method is often faster for finding the remaining roots, especially if you don't want to work with complex cube root branches.

So to sum up: Cardano's method can generate all three roots if you consider all cube root branches, but factorization is a practical shortcut once you have one root.

内容的提问来源于stack exchange，提问作者user465188