Relationship between the Roots of a Cubic Equation and its Coefficients | 三次方程的根与系数关系

📚 Relationship between the Roots of a Cubic Equation and its Coefficients | 三次方程的根与系数关系

For a quadratic equation, the sum and product of the roots are well known. For a cubic equation, we can extend these ideas to obtain three separate relations connecting the roots with the coefficients. These relations are often called Vieta’s formulas, and they allow us to work with the roots without solving the cubic explicitly.

对于二次方程，根的和与积广为人知。对于三次方程，我们可以将这些思想推广，得到三个分别联系根与系数的关系式。这些关系通常称为“韦达定理”，它们使我们无需显式求解三次方程，也能研究根的性质。

1. General Form of a Cubic | 三次方程的一般形式

Let a cubic equation be written as \( ax^3 + bx^2 + cx + d = 0 \), where \( a \neq 0 \) and \( a, b, c, d \) are real (or complex) constants. We denote its three roots by \( \alpha, \beta, \gamma \). By the fundamental theorem of algebra, every cubic has exactly three roots when counted with multiplicity.

设三次方程为 \( ax^3 + bx^2 + cx + d = 0 \)，其中 \( a \neq 0 \)，且 \( a, b, c, d \) 为实数（或复数）常数。我们用 \( \alpha, \beta, \gamma \) 表示它的三个根。根据代数基本定理，计入重数后，每个三次方程恰好有三个根。

2. Vieta’s Formulas for a Cubic: Sum of Roots | 三次方程的韦达公式（一）：根之和

If \( \alpha, \beta, \gamma \) are the roots, then the first relation is:

若 \( \alpha, \beta, \gamma \) 是根，则第一个关系式为：

α + β + γ = − b / a

In words, the sum of the three roots equals the negative of the coefficient of \( x^2 \) divided by the coefficient of \( x^3 \).

用语言表述：三个根之和等于 \( x^2 \) 的系数除以 \( x^3 \) 的系数后取相反数。

3. Sum of Products of Roots Taken Two at a Time | 两两乘积之和

The second relation involves all possible pairwise products of the roots:

第二个关系式涉及所有可能的根的成对乘积：

αβ + βγ + γα = c / a

This is the sum of the three products obtained by choosing two different roots at a time, and it equals the coefficient of \( x \) divided by the coefficient of \( x^3 \).

这是每次选取两个不同根所得三个乘积之和，它等于 \( x \) 的系数除以 \( x^3 \) 的系数。

4. Product of All Three Roots | 三个根的乘积

The third relation gives the product of all three roots:

第三个关系式给出三个根的乘积：

αβγ = − d / a

Notice the minus sign. This is because a cubic has an odd degree, so the constant term appears with a negative sign after normalising the leading coefficient to 1.

注意这里有一个负号。这是因为三次方程的次数为奇数，在将首项系数化为 1 后，常数项带有一个负号。

5. Derivation from Factorization | 从因式分解推导

Suppose the cubic has roots \( \alpha, \beta, \gamma \). Then it can be factorised as:

设三次方程有根 \( \alpha, \beta, \gamma \)，则它可以分解为：

a(x − α)(x − β)(x − γ) = 0

Expanding the product gives:

展开这个乘积得到：

a[ x³ − (α+β+γ)x² + (αβ+βγ+γα)x − αβγ ] = 0

Comparing coefficients with \( ax^3 + bx^2 + cx + d \) immediately yields the three Vieta formulas above.

将它与 \( ax^3 + bx^2 + cx + d \) 比较系数，立即得到上述三个韦达公式。

6. Monic Cubic and the Reduced Forms | 首一三次方程及简化形式

If \( a = 1 \), the cubic becomes \( x^3 + bx^2 + cx + d = 0 \), and the relations simplify to:

若 \( a = 1 \)，三次方程变为 \( x^3 + bx^2 + cx + d = 0 \)，此时关系式简化为：

α + β + γ = − b
αβ + βγ + γα = c
αβγ = − d

These simplified versions are often easier to remember and are frequently used in AQA exam questions.

这些简化形式更容易记忆，并且在 AQA 考试题目中经常使用。

7. Worked Example 1: Using Relations Without Solving | 例 1：不解方程直接使用关系

The roots of \( 2x^3 + 6x^2 – 4x + 10 = 0 \) are \( \alpha, \beta, \gamma \). Find \( \alpha + \beta + \gamma \), \( \alpha\beta + \beta\gamma + \gamma\alpha \), and \( \alpha\beta\gamma \).

已知方程 \( 2x^3 + 6x^2 – 4x + 10 = 0 \) 的根为 \( \alpha, \beta, \gamma \)。求 \( \alpha + \beta + \gamma \)、\( \alpha\beta + \beta\gamma + \gamma\alpha \) 和 \( \alpha\beta\gamma \)。

Here \( a = 2, b = 6, c = -4, d = 10 \). Hence:

这里 \( a = 2, b = 6, c = -4, d = 10 \)。因此：

α + β + γ = − 6/2 = − 3

αβ + βγ + γα = − 4/2 = − 2

αβγ = − 10/2 = − 5

Notice that we did not need to solve the cubic at all. This is the power of Vieta’s formulas.

注意我们完全不需要求解这个三次方程。这正是韦达公式的强大之处。

8. Worked Example 2: Using One Known Root | 例 2：已知一个根时求解其他关系

Given that \( x = 1 \) is a root of \( 2x^3 – 3x^2 – 11x + 6 = 0 \), find the other two roots.

已知 \( x = 1 \) 是方程 \( 2x^3 – 3x^2 – 11x + 6 = 0 \) 的一个根，求另外两个根。

Let the roots be \( 1, \alpha, \beta \). Then using Vieta’s formulas:

设三个根为 \( 1, \alpha, \beta \)。利用韦达公式：

\( 1 + \alpha + \beta = -(-3)/2 = 1.5 \)
\( 1\cdot\alpha + 1\cdot\beta + \alpha\beta = -11/2 = -5.5 \)

From the first equation, \( \alpha + \beta = 0.5 \). From the second, \( \alpha + \beta + \alpha\beta = -5.5 \), so \( \alpha\beta = -6 \).

由第一个方程得 \( \alpha + \beta = 0.5 \)。由第二个方程得 \( \alpha + \beta + \alpha\beta = -5.5 \)，因此 \( \alpha\beta = -6 \)。

Thus we need two numbers whose sum is \( 0.5 \) and product is \( -6 \). Solving gives \( \alpha = 1.5 \) or \( -4 \), and \( \beta = -4 \) or \( 1.5 \). So the roots are \( 1, 1.5, -4 \).

于是我们需要两个数，其和为 \( 0.5 \)，积为 \( -6 \)。解得 \( \alpha = 1.5 \) 或 \( -4 \)，\( \beta = -4 \) 或 \( 1.5 \)。所以三个根为 \( 1, 1.5, -4 \)。

9. Transforming Roots: Reciprocal Roots | 根的变换：取倒数

Suppose \( \alpha, \beta, \gamma \) are the roots of \( ax^3 + bx^2 + cx + d = 0 \). What cubic has roots \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma} \)?

设 \( \alpha, \beta, \gamma \) 是 \( ax^3 + bx^2 + cx + d = 0 \) 的根。什么三次方程以 \( \frac{1}{\alpha}, \frac{1}{\beta}, \frac{1}{\gamma} \) 为根？

Using the product relation, \( \alpha\beta\gamma = -d/a \). The reciprocal roots have:

利用乘积关系，\( \alpha\beta\gamma = -d/a \)。倒根具有：

Sum: \( \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma} = \frac{c/a}{-d/a} = -c/d \)
Pairwise sum: \( \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} + \frac{1}{\gamma\alpha} = \frac{\alpha+\beta+\gamma}{\alpha\beta\gamma} = \frac{-b/a}{-d/a} = b/d \)
Product: \( \frac{1}{\alpha\beta\gamma} = -a/d \)

Therefore the new cubic is \( x^3 + \frac{c}{d}x^2 + \frac{b}{d}x + \frac{a}{d} = 0 \), or \( dx^3 + cx^2 + bx + a = 0 \).

因此新的三次方程为 \( x^3 + \frac{c}{d}x^2 + \frac{b}{d}x + \frac{a}{d} = 0 \)，即 \( dx^3 + cx^2 + bx + a = 0 \)。

10. Forming a Cubic with Related Roots | 构造具有相关根的三次方程

If the roots of a new cubic are \( \alpha + k, \beta + k, \gamma + k \), we can compute the new sum, pairwise sum and product directly by substitution.

如果新三次方程的根为 \( \alpha + k, \beta + k, \gamma + k \)，我们可以通过代入直接计算新的和、两两乘积之和以及积。

For example, if the original roots satisfy \( \alpha+\beta+\gamma = S_1 \), \( \alpha\beta+\beta\gamma+\gamma\alpha = S_2 \), \( \alpha\beta\gamma = P \), then the new roots have:

例如，若原根满足 \( \alpha+\beta+\gamma = S_1 \)，\( \alpha\beta+\beta\gamma+\gamma\alpha = S_2 \)，\( \alpha\beta\gamma = P \)，则新根具有：

Sum = S₁ + 3k

Pairwise sum = S₂ + 2kS₁ + 3k²

Product = P + kS₂ + k²S₁ + k³

These expressions allow you to write down the new cubic in \( x \) without finding the roots explicitly.

这些表达式使您无需显式求根即可直接写出关于 \( x \) 的新三次方程。

11. Common Pitfalls and Exam Tips | 常见易错点与考试提示

Sign errors: The sum is \( -b/a \), not \( b/a \). The product is \( -d/a \), not \( d/a \). Always check the signs.
中文对照：符号错误： 根之和是 \( -b/a \) 而不是 \( b/a \)；根之积是 \( -d/a \) 而不是 \( d/a \)。务必检查符号。
Pairwise sum: Remember there are three terms: \( \alpha\beta + \beta\gamma + \gamma\alpha \). Missing one is a common mistake.
中文对照：两两乘积之和： 记住共有三项：\( \alpha\beta + \beta\gamma + \gamma\alpha \)。漏项是常见错误。
Monic simplification: When \( a = 1 \), the formulas are cleaner. If \( a \neq 1 \), divide the whole equation by \( a \) first.
中文对照：首一化： 当 \( a = 1 \) 时公式更简洁。若 \( a \neq 1 \)，可先将整个方程除以 \( a \)。
Complex roots: For a cubic with real coefficients, complex roots occur in conjugate pairs. Vieta’s formulas still hold.
中文对照：复数根： 对于实系数三次方程，复数根成对共轭出现。韦达公式依然成立。

12. Conclusion | 总结

The relationship between the roots and coefficients of a cubic is a compact and powerful tool. With just these three relations, you can solve many problems involving symmetric functions of the roots, construct new polynomials, and analyse equations without performing long algebraic expansions.

三次方程的根与系数之间的关系简洁而强大。仅凭这三个关系式，你就可以解决许多涉及根对称函数的问题、构造新的多项式，并在不进行冗长代数展开的情况下分析方程。

Remember the core formulas:

请牢记核心公式：

α + β + γ = −b/a

αβ + βγ + γα = c/a

αβγ = −d/a

Master these, and cubic root problems become a matter of substitution rather than solving.

掌握这些公式后，三次方程根的问题就变成了代入运算，而非求解方程。

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Relationship between the Roots of a Cubic Equation and its Coefficients | 三次方程的根与系数关系