📚 Investigation 4 - Sum and Product of Roots | 探究4：根的和与积
One of the most elegant and powerful connections in algebra is the relationship between the roots of a polynomial and its coefficients. Rather than solving an equation by factoring or using the quadratic formula, we can often extract useful information about the roots – such as their sum and product – directly from the coefficients. These relationships, known as Vieta's formulas, form the core of this investigation. From simple quadratics to higher-degree polynomials, the sums and products of roots uncover symmetries that simplify problem-solving, curve sketching and even proof construction.
代数中最优美且有力的联系之一就是多项式的根与其系数之间的关系。我们往往无需通过因式分解或求根公式去解方程，就可以直接从系数中提取根的有用信息——例如根的和与积。这些关系被称为韦达公式，是本次探究的核心。从简单的二次方程到高次多项式，根的和与积揭示了丰富的对称性，能够极大地简化问题求解、曲线作图甚至证明构造。
1. Introduction to Root-Coefficient Relationships | 根与系数关系简介
Every polynomial equation can be expressed in terms of its roots through factorisation. For a quadratic ax² + bx + c = 0 with roots α and β, we can write a(x − α)(x − β) = 0 s to answer questions about the roots without ever finding the actual values of α and β. The same principle extends to cubic, quartic and general n-th degree polynomials.
每一个多项式方程都可以通过因式分解用根表示。对于有根 α 和 β 的二次方程 ax² + bx + c = 0，我们可以写成 a(x − α)(x − β) = 0。展开乘积并比较系数便得到两个简单但深刻的关系式：α + β = −b/a 且 αβ = c/a。借助这两个关系，我们无需真正求出 α 和 β 的具体数值，就能解答关于根的各种问题。同样的原理可以推广到三次、四次乃至一般的 n 次多项式。
2. Quadratic Equations: The Standard Form | 二次方程的标准形式
Consider the general quadratic equation ax² + bx + c = 0, where a ≠ 0. Suppose the two roots are x₁ and x₂. The polynomial can be factored as a(x − x₁)(x − x₂) = 0 x₂. Since this must be identical to ax² + bx + c, the coefficients of the corresponding powers of x must match. This matching is the essence of Vieta's approach.
考虑一般二次方程 ax² + bx + c = 0，其中 a ≠ 0。设两个根为 x₁ 和 x₂。该多项式可因式分解为 a(x − x₁)(x − x₂) = 0。展开乘积得 a[x² − (x₁ + x₂)x + x₁x₂] = ax² − a(x₁ + x₂)x + a x₁x₂。该式必须与 ax² + bx + c 完全相同，因此对应幂次项的系数必须相等。这种匹配正是韦达方法的核心思想。
3. Deriving Sum and Product for Quadratics | 推导二次方程的根和与积
Equating the linear coefficients gives b = −a(x₁ + x₂), so x₁ + x₂ = −b/a. Equating the constant terms gives c = a x₁x₂, so x₁x₂ = c/a istency makes the sum and product relationships a reliable tool even when we cannot easily factor the quadratic.
令一次项系数相等得 b = −a(x₁ + x₂)，故 x₁ + x₂ = −b/a。令常数项相等得 c = a x₁x₂，故 x₁x₂ = c/a。这些公式对所有二次方程都成立，无论根是实数还是复数。当有重根时同样适用：若 x₁ = x₂ = r，则 2r = −b/a 且 r² = c/a。这种一致性使根和与积的关系成为可靠工具，即便不方便对二次式进行因式分解。
4. Cubic Equations: Sum of Roots | 三次方程的根之和
For a cubic equation ax³ + bx² + cx + d = 0 with a ≠ 0, let the three roots be x₁, x₂, x₃. The factorised form is a(x − x₁)(x − x₂)(x − x₃) = 0. Expanding the product of the three linear factors yields x³ − (x₁ + x₂ + x₃)x² + (x₁x₂ + x₂x₃ + x₃x₁)x − x₁x₂x₃. Multiplying by a and comparing with ax³ + bx² + cx + d gives the sum of roots immediately: x₁ + x₂ + x₃ = −b/a. Observe the alternating sign pattern that begins with the sum.
对于三次方程 ax³ + bx² + cx + d = 0（a ≠ 0），设三个根为 x₁, x₂, x₃。因式分解形式为 a(x − x₁)(x − x₂)(x − x₃) = 0。展开这三个一次因式的乘积得 x³ − (x₁ + x₂ + x₃)x² + (x₁x₂ + x₂x₃ + x₃x₁)x − x₁x₂x₃。乘以 a 后与 ax³ + bx² + cx + d 比较系数，直接得到根的和：x₁ + x₂ + x₃ = −b/a。注意从和开始出现的正负号交替模式。
5. Cubic Equations: Sum of Pairwise Products and Product | 三次方程的两两乘积和与乘积
Continuing the comparison, the coefficient of x gives the sum of all products of roots taken two at a time: x₁x₂ + x₂x₃ + x₃x₁ = c/a. The constant term gives the product of all three roots: x₁x₂x₃ = −d/a mple, if x³ − 6x² + 11x − 6 = 0, then sum = 6, sum of pairwise products = 11, product = 6.
继续比较系数，x 的系数给出每两个根乘积的和：x₁x₂ + x₂x₃ + x₃x₁ = c/a。常数项给出三个根的乘积：x₁x₂x₃ = −d/a。注意正负号交替：根的和带负号，两两积和带正号，三乘积带负号。在处理高次多项式时，这一模式至关重要。例如方程 x³ − 6x² + 11x − 6 = 0，则有和 = 6，两两积和 = 11，乘积 = 6。
6. General Polynomial Patterns | 一般多项式的规律
For an n-th degree polynomial aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0 with roots x₁, x₂, …, xₙ, the coefficients are linked to the elementary symmetric sums of the roots. The sum of all roots is Σ xᵢ = −aₙ₋₁/aₙ. The sum of all products of two distinct roots is Σ xᵢxⱼ = aₙ₋₂/aₙ (for i < j). The sum of products of three distinct roots is Σ xᵢxⱼxₖ = −aₙ₋₃/aₙ, and so on, ending with the product of all roots Π xᵢ = (−1)ⁿ a₀/aₙ. Every symmetric sum follows a strict alternating sign convention.
对于 n 次多项式 aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₁x + a₀ = 0，其根为 x₁, x₂, …, xₙ，系数便与根的初等对称和有关。所有根的和为 Σ xᵢ = −aₙ₋₁/aₙ。所有两两不同根的乘积之和为 Σ xᵢxⱼ = aₙ₋₂/aₙ (i < j)。所有三个不同根的乘积之和为 Σ xᵢxⱼxₖ = −aₙ₋₃/aₙ，以此类推，最终所有根的乘积 Π xᵢ = (−1)ⁿ a₀/aₙ。每一个对称和都遵守严格的正负号交替规则。
7. Vieta's Formulas for Any Degree | 任意次数的韦达公式
Formally, if P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ = aₙ(x − x₁)(x − x₂)…(x − xₙ), then for each k = 1, 2, …, n, the sum of all possible products of k distinct roots, denoted eₖ, satisfies eₖ = (−1)ᵏ aₙ₋ₖ / aₙ. Here e₁ = Σ xᵢ, e₂ = Σ xᵢxⱼ, …, eₙ = x₁x₂… of all roots multiplied by the sign factor. These formulas are collectively known as Vieta's formulas, named after François Viète, and are widely used in algebra, number theory and beyond.
严格来说，若 P(x) = aₙxⁿ + aₙ₋₁xⁿ⁻¹ + … + a₀ = aₙ(x − x₁)(x − x₂)…(x − xₙ)，则对每个 k = 1, 2, …, n，所有可能的 k 个不同根的乘积之和（记作 eₖ）满足 eₖ = (−1)ᵏ aₙ₋ₖ / aₙ。其中 e₁ = Σ xᵢ，e₂ = Σ xᵢxⱼ，…，eₙ = x₁x₂…xₙ。当 k = n 时，便得到带符号因子的全根之积。这些公式统称为韦达公式，以弗朗索瓦·韦达的名字命名，广泛应用于代数、数论等领域。
8. Symmetric Polynomials and Sign Patterns | 对称多项式与符号模式
The alternating sign pattern arises because each factor (x − xᵢ) introduces a negative sign before the root. When we expand the product, the sum of roots appears with a minus, the sum of double products with a plus, triple products with a minus, and so forth. This pattern can be remembered by noting that the coefficient of xⁿ⁻¹ gives −(sum), the coefficient of xⁿ⁻² gives +(sum of pairwise products), and so on, with the constant term giving (−1)ⁿ times the product. Recognising this pattern allows quick mental extraction of root-coefficient links without detailed expansion.
正负号交替模式源于每个因式 (x − xᵢ) 在根前面引入了一个负号。展开乘积时，根的和出现负号，两两积和为正号，三积和为负号，依次类推。记忆要点：xⁿ⁻¹ 的系数为 −(和)，xⁿ⁻² 的系数为 +(两两积和)，以此类推，常数项为 (−1)ⁿ 乘以全乘积。识别这一模式后，无需完整展开便能快速提取根与系数的联系。
9. Using Sum and Product to Find Unknown Coefficients | 利用根的和与积求未知系数
the sum is 3r = −p and the product is 2r² = 12, so r² = 6. We can then find p = −3r = −3(±√6). This approach avoids solving the quadratic directly. In cubic equations, knowing the sum of two roots or a relation between roots lets us set up a system of equations using Vieta's formulas to find the remaining root or the unknown coefficients.
典型的 IB 风格题目会给出含未知系数的二次方程，如 x² + px + 12 = 0，并告知一根是另一根的两倍。设根为 r 和 2r，则和 3r = −p，积 2r² = 12，于是 r² = 6，p = −3r = −3(±√6)。这一方法避开了直接求解二次方程。在三次方程中，若知道某两根之和或根之间的某种关系，便可利用韦达公式建立
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