📚 Cubic Equations with Related Roots | 相关根的三次方程
In AQA A-Level Mathematics, cubic equation questions often go beyond finding the three roots of one equation. You may be given a cubic with roots α, β and γ, then asked to construct a new cubic whose roots are related to α, β and γ - for example 2α, α + 1, 1/α or α². This article explains the systematic methods, the Vieta formulas behind them, and the common exam techniques that turn a long-looking question into a short algebraic routine.
在 AQA A-Level 数学中，三次方程题经常不止是求一个方程的三个根。题目可能给出一个以 α、β、γ 为根的三次方程，然后要求构造一个新的三次方程，其根与 α、β、γ 相关，例如 2α、α + 1、1/α 或 α²。本文讲解系统方法、背后的韦达公式，以及把看似冗长的题目变成简短代数流程的常见考试技巧。
1. Core Vieta Formulas for Cubic Equations | 三次方程的韦达公式
roots α, β and γ, the relationships between the roots and coefficients are known as Vieta's formulas. They are the foundation of every related-roots question.
对于三次方程 a x³ + b x² + c x + d = 0，设根为 α、β、γ，根与系数之间的关系称为韦达公式。它们是所有相关根问题的基础。
| Symmetric sum | Expression | Value |
|---|---|---|
| Sum of roots | S₁ = α + β + γ | −b/a |
| Pairwise sum | S₂ = αβ + βγ + γα | c/a |
| Product of roots | S₃ = αβγ | −d/a |
You should memorise these three symmetric sums, because every related-roots question uses them. Notice the alternating signs: the sum has a minus sign, the pairwise sum has a plus sign, and the product has a minus sign again when the cubic is written in standard form.
你应该熟记这三个对称和，因为每个相关根问题都会用到它们。注意符号交替：和带负号，两两和带正号，乘积在标准形式下又带负号。
d S₃ = 6. This quick check confirms the signs and helps you avoid coefficient slips later.
例如，若 x³ − 6x² + 11x − 6 = 0 的根为 1、2、3，则 S₁ = 6，S₂ = 11，S₃ = 6。这个快速检验能确认符号，并帮助你在后续避免系数错误。
2. Why Related-Root Questions Appear | 为什么考相关根
Questions with related roots test two skills at once: whether you can connect the roots of one polynomial to its coefficients, and whether you can reverse the process to build a new polynomial from a required set of sums.
相关根问题同时考查两种能力：能否把一个多项式的根与系数联系起来，以及能否反向根据指定和构造新多项式。
In AQA papers, the original cubic is often simple to factorise or has simple Vieta sums. You then use those sums to find the corresponding sums for the new roots. The examiner is not usually trying to make you solve a difficult cubic; the difficulty lies in organisation and accurate algebra.
在 AQA 试卷中，原三次方程通常容易因式分解，或者韦达和很简单。然后你利用这些和求出新根对应的和。考官通常不是要你解一个困难的三次方程；难点在于条理和准确的代数运算。
Once you have the new symmetric sums T₁, T₂ and T₃, the required polynomial can be written directly. This reverse construction is the key idea: from roots to coefficients, and then from new coefficients back to a new equation.
一旦你求出新对称和 T₁、T₂、T₃，所需多项式可以直接写出。这种反向构造是关键思想：从根到系数，再从新系数回到新方程。
3. General Strategy: Build the New Cubic from T₁, T₂, T₃ | 通用策略：由 T₁、T₂、T₃ 构造新三次方程
ic cubic equation is obtained from the standard factorised form (x − p)(x − q)(x − r) = 0.
设新根为 p、q、r。定义 T₁ = p + q + r，T₂ = pq + qr + rp，T₃ = pqr。则新的首一三次方程可由标准因式形式 (x − p)(x − q)(x − r) = 0 得到。
x³ − T₁x² + T₂x − T₃ = 0
This formula is the target of every related-roots construction. Your job is to express T₁, T₂ and T₃ in terms of the original S₁, S₂ and S₃, without ever finding α, β or γ individually.
这个公式是每个相关根构造题的目标。你的任务是用原 S₁、S₂、S₃ 表示 T₁、T₂、T₃，而无需分别求出 α、β、γ。
A reliable exam routine is: write down the original S₁, S₂, S₃; identify the transformation from α, β, γ to the new roots; compute T₁ first, then T₂, then T₃; substitute into the formula; and finally check the degree and leading coefficient.
可靠的考试流程是：写下原 S₁、S₂、S₃；识别从 α、β、γ 到新根的变换；先算 T₁，再算 T₂，再算 T₃；代入公式；最后检查次数和首项系数。
4. Type 1: Scaled Roots kα, kβ, kγ | 类型一：缩放根 kα、kβ、kγ
If the new roots are kα, kβ and kγ, then each symmetric sum is scaled by a power of k. The sum is multiplied by k, the pairwise sum by k², and the product by k³.
如果新根为 kα、kβ、kγ，则每个对称和按 k 的幂缩放。和乘以 k，两两和乘以 k²，乘积乘以 k³。
T₁ = kS₁, T₂ = k²S₂, T₃ = k³S₃
e new sums are T₁ = 12, T₂ = 44 and T₃ = 48, so the new cubic is x³ − 12x² + 44x − 48 = 0.
例如，若 x³ − 6x² + 11x − 6 = 0 的根为 1、2、3，新根为 2α、2β、2γ，则新根为 2、4、6。新和为 T₁ = 12，T₂ = 44，T₃ = 48，所以新三次方程为 x³ − 12x² + 44x − 48 = 0。
Notice that this method avoids solving anything. You only use the original symmetric sums and the scaling rule. This is exactly what AQA expects in a related-roots question.
注意这种方法不需要解方程。你只需使用原对称和与缩放规则。这正是 AQA 在相关根题中期望的做法。
5. Type 2: Shifted Roots α + k, β + k, γ + k | 类型二：平移根 α + k、β + k、γ + k
If the new roots are α + k, β + k and γ + k, the new sums are found by expanding. The sum becomes S₁ + 3k, and the other sums pick up cross terms.
如果新根为 α + k、β + k、γ + k，新和通过展开得到。和变为 S₁ + 3k，其他和会出现交叉项。
T₁ = S₁ + 3k
T₂ = S₂ + 2kS₁ + 3k²
T₃ = S₃ + kS₂ + k²S₁ + k³
These formulas are worth remembering, but they can also be rebuilt quickly from (α + k) + (β + k) + (γ + k), and from the pairwise and product expansions. In an exam, a quick expansion is often safer than a half-remembered formula.
这些公式值得记忆，但也可以从 (α + k) + (β + k) + (γ + k) 以及两两和、乘积展开快速重建。考试中，快速展开往往比记错一半的公式更安全。
uting gives T₁ = 6 + 3 = 9, T₂ = 11 + 12 + 3 = 26, T₃ = 6 + 11 + 6 + 1 = 24, which agrees.
简单检验：取 α = 1，β = 2，γ = 3，k = 1。则 S₁ = 6，S₂ = 11，S₃ = 6。新根为 2、3、4，所以 T₁ = 9，T₂ = 26，T₃ = 24。代入得 T₁ = 6 + 3 = 9，T₂ = 11 + 12 + 3 = 26，T₃ = 6 + 11 + 6 + 1 = 24，完全一致。
6. Type 3: Reciprocal Roots 1/α, 1/β, 1/γ | 类型三：倒数根
If the new roots are 1/α, 1/β and 1/γ, you must assume αβγ ≠ 0. The new symmetric sums are obtained by dividing the original sums by the product S₃.
如果新根为 1/α、1/β、1/γ，必须假设 αβγ ≠ 0。新对称和通过把原和除以乘积 S₃ 得到。
T₁ = S₂/S₃, T₂ = S₁/S₃, T₃ = 1/S₃
inal is a x³ + b x² + c x + d = 0, the reciprocal-root equation is d x³ + c x² + b x + a = 0, up to an overall non-zero multiplier.
还有一个很漂亮的捷径：根为倒数的方程可以通过反转原三次方程的系数得到。若原方程为 a x³ + b x² + c x + d = 0，则倒数根方程为 d x³ + c x² + b x + a = 0，允许相差一个非零常数倍。
− 1/6 = 0, or 6x³ − 11x² + 6x − 1 = 0. Reversing coefficients gives −6x³ + 11x² − 6x + 1 = 0, which is the same after multiplying by −1.
例如，x³ − 6x² + 11x − 6 = 0 的根为 1、2、3。倒数根为 1、1/2、1/3。新和为 T₁ = 11/6，T₂ = 1，T₃ = 1/6，得到 x³ − (11/6)x² + x − 1/6 = 0，即 6x³ − 11x² + 6x − 1 = 0。反转系数得 −6x³ + 11x² − 6x + 1 = 0，乘以 −1 后相同。
7. Type 4: Squared Roots α², β², γ² | 类型四：平方根
If the new roots are α², β² and γ², use the standard symmetric identities. These identities convert sums of squares and products of squares into the original S₁, S₂ and S₃.
如果新根为 α²、β²、γ²，使用标准对称恒等式。这些恒等式把平方和与平方积转化为原 S₁、S₂、S₃。
T₁ = S₁² − 2S₂
T₂ = S₂² − 2S₁S₃
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