Vieta’s Formulas and Their Applications | 韦达定理及其应用

📚 Vieta’s Formulas and Their Applications | 韦达定理及其应用

Vieta’s formulas, named after the French mathematician François Viète, establish a fundamental connection between the roots of a polynomial and its coefficients. These formulas are among the most powerful tools in algebra, frequently tested in board examinations and competitive mathematics.

韦达定理以法国数学家弗朗索瓦·韦达命名，它建立了多项式根与系数之间的基本联系。这些公式是代数中最强大的工具之一，在考试和数学竞赛中频繁出现。

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1. Vieta’s Formulas for Quadratic Equations | 一元二次方程的韦达定理

For a quadratic equation in the standard form ax² + bx + c = 0 (a ≠ 0), if its two roots are α and β, then Vieta’s formulas state:

对于标准形式 ax² + bx + c = 0（a ≠ 0）的一元二次方程，若其两个根为 α 和 β，则韦达定理表述为：

α + β = −b/a, αβ = c/a

The sum of the roots equals the negative of the coefficient of x divided by the coefficient of x², and the product of the roots equals the constant term divided by the coefficient of x².

两根之和等于一次项系数除以二次项系数的相反数，两根之积等于常数项除以二次项系数。

For example, for the equation 2x² − 8x + 6 = 0, we have α + β = −(−8)/2 = 4 and αβ = 6/2 = 3. Indeed, solving gives x = 1 and x = 3, and 1 + 3 = 4, 1 × 3 = 3.

例如，对于方程 2x² − 8x + 6 = 0，有 α + β = −(−8)/2 = 4，αβ = 6/2 = 3。实际上，解方程得 x = 1 和 x = 3，且 1 + 3 = 4，1 × 3 = 3。

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2. Derivation and Verification | 推导与验证

Vieta’s formulas can be derived from the factored form. Since α and β are roots, we can write ax² + bx + c = a(x − α)(x − β). Expanding the right side:

韦达定理可以从因式分解形式推导。由于 α 和 β 是根，可写为 ax² + bx + c = a(x − α)(x − β)。展开右边：

a(x − α)(x − β) = a[x² − (α + β)x + αβ]

Comparing coefficients with ax² + bx + c, we obtain −a(α + β) = b and aαβ = c, which gives α + β = −b/a and αβ = c/a.

与 ax² + bx + c 比较系数，得 −a(α + β) = b 和 aαβ = c，故 α + β = −b/a，αβ = c/a。

This derivation works because two polynomials are identical if and only if their corresponding coefficients are equal. This principle is essential to understanding Vieta’s formulas.

该推导成立的原因在于：两个多项式相等当且仅当对应系数相等。这一原理是理解韦达定理的关键。

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3. Vieta’s Formulas for Cubic Equations | 一元三次方程的韦达定理

For a cubic equation ax³ + bx² + cx + d = 0 (a ≠ 0) with roots α, β, γ, Vieta’s formulas extend naturally:

对于一元三次方程 ax³ + bx² + cx + d = 0（a ≠ 0）的根 α、β、γ，韦达定理自然推广为：

α + β + γ = −b/a

αβ + βγ + γα = c/a

αβγ = −d/a

The sum of the roots, the sum of pairwise products, and the product of the roots are all expressed in terms of the coefficients, with alternating signs.

根之和、两两乘积之和以及根之积都可以用系数表达，符号交替变化。

For example, for x³ − 6x² + 11x − 6 = 0, we have α + β + γ = 6, αβ + βγ + γα = 11, and αβγ = 6. The roots are 1, 2, and 3, which indeed satisfy these relations.

例如，对于 x³ − 6x² + 11x − 6 = 0，有 α + β + γ = 6，αβ + βγ + γα = 11，αβγ = 6。其根为 1、2、3，确实满足这些关系。

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4. General Formula for Polynomials of Degree n | n 次多项式的一般公式

For a polynomial of degree n, P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ (aₙ ≠ 0) with roots r₁, r₂, …, rₙ, Vieta’s formulas generalize to:

对于 n 次多项式 P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀（aₙ ≠ 0），若根为 r₁, r₂, …, rₙ，则韦达定理推广为：

r₁ + r₂ + … + rₙ = −aₙ₋₁/aₙ

r₁r₂ + r₁r₃ + … + rₙ₋₁rₙ = aₙ₋₂/aₙ

r₁r₂r₃ + … = −aₙ₋₃/aₙ

The k-th elementary sum of the roots equals (−1)ᵏ times the ratio of the coefficient of xⁿ⁻ᵏ to the leading coefficient aₙ.

根的 k 个基本对称和等于 (−1)ᵏ 乘以 xⁿ⁻ᵏ 的系数与首项系数 aₙ 之比。

Note the alternating signs: the sum of products of k roots takes the sign (−1)ᵏ. This pattern is crucial for solving problems correctly.

注意符号交替：k 个根乘积之和的符号为 (−1)ᵏ。这一规律对正确解题至关重要。

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5. Key Application: Constructing Equations from Roots | 核心应用：由根构造方程

One of the most common applications is constructing a quadratic equation when the sum and product of its roots are known. If α and β are roots with sum S and product P, the equation is:

最常见的应用之一是在已知两根之和 S 与两根之积 P 时构造二次方程。若 α 和 β 为根，方程为：

x² − Sx + P = 0

Because x² − (α + β)x + αβ = (x − α)(x − β) = 0, any quadratic equation with leading coefficient 1 and roots α, β must be x² − Sx + P = 0.

因为 x² − (α + β)x + αβ = (x − α)(x − β) = 0，任何首项系数为 1 且根为 α、β 的二次方程必然是 x² − Sx + P = 0。

For example, the equation whose roots are 2 and −5 is x² − (2 + (−5))x + 2 × (−5) = x² + 3x − 10 = 0.

例如，以 2 和 −5 为根的方程为 x² − (2 + (−5))x + 2 × (−5) = x² + 3x − 10 = 0。

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6. Key Application: Evaluating Symmetric Expressions | 核心应用：计算对称表达式

Vieta’s formulas allow us to evaluate symmetric functions of roots without explicitly solving for the roots. For example, consider α² + β². Using (α + β)² = α² + 2αβ + β², we get:

韦达定理使我们无需显式解根即可计算根的对称函数。例如，考虑 α² + β²。利用 (α + β)² = α² + 2αβ + β²，可得：

α² + β² = (α + β)² − 2αβ

Similarly, the sum of reciprocals can be computed as:

类似地，倒数之和可以计算为：

1/α + 1/β = (α + β)/(αβ)

These transformations are invaluable when the roots are irrational or complex, making direct computation difficult.

当根为无理数或复数时，这些变换尤为重要，因为直接计算非常困难。

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7. Worked Example: Sum and Product with Irrational Roots | 典型例题：无理根的和与积

Consider the equation x² − 6x + 2 = 0, whose roots are α and β. Without solving, find the value of α² + β² and 1/α + 1/β.

考虑方程 x² − 6x + 2 = 0，其根为 α 和 β。不解方程，求 α² + β² 与 1/α + 1/β 的值。

From Vieta’s formulas, α + β = 6 and αβ = 2. Then:

根据韦达定理，α + β = 6，αβ = 2。于是：

α² + β² = 6² − 2 × 2 = 36 − 4 = 32

1/α + 1/β = 6/2 = 3

Indeed, the roots are 3 ± √7, and direct calculation confirms α² + β² = (3 + √7)² + (3 − √7)² = 9 + 6√

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