IB Mathematics: Vieta's Formulas for Quadratic Equations | IB数学：二次方程根与系数关系（韦达定理）

📚 IB Mathematics: Vieta’s Formulas for Quadratic Equations | IB数学：二次方程根与系数关系（韦达定理）

Quadratic equations are a cornerstone of the IB Mathematics curriculum, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI) courses. Among the most elegant and frequently tested results is Vieta’s Theorem, which establishes a direct connection between the roots of a quadratic equation and its coefficients. This article provides a comprehensive, exam-focused exploration of this topic.

二次方程是IB数学课程的核心内容之一，在分析与方法（AA）和应用与解释（AI）两门课程中都会出现。其中最为精妙且高频考查的结论之一，便是韦达定理——它建立了二次方程根与系数之间的直接联系。本文将围绕这一主题，提供全面且紧扣考点的系统讲解。

1. The Standard Form of a Quadratic Equation | 二次方程的标准形式

A quadratic equation in one variable is typically written in the standard form ax² + bx + c = 0, where a, b, c are real coefficients and a ≠ 0. The two solutions (roots) of this equation are denoted by α (alpha) and β (beta).

一元二次方程通常写成标准形式 ax² + bx + c = 0，其中 a、b、c 为实数系数，且 a ≠ 0。该方程的两个解（根）记为 α（阿尔法）和 β（贝塔）。

The roots can be found explicitly using the quadratic formula:

根可以用求根公式显式求出：

x = (−b ± √(b² − 4ac)) / 2a

The expression under the square root, Δ = b² − 4ac, is called the discriminant and determines the nature of the roots.

根号内的表达式 Δ = b² − 4ac 称为判别式，它决定了根的性质。

If Δ > 0: two distinct real roots | 若 Δ > 0：两个不相等的实数根
If Δ = 0: one repeated real root | 若 Δ = 0：两个相等的实数根（重根）
If Δ < 0: two complex conjugate roots | 若 Δ < 0：一对共轭复数根

2. Deriving the Sum of Roots | 推导两根之和

Using the quadratic formula, the two roots are:

利用求根公式，两个根分别为：

α = (−b + √Δ) / 2a, β = (−b − √Δ) / 2a

Adding these two expressions, the square-root terms cancel:

将两个表达式相加，根号项相互抵消：

α + β = (−b + √Δ − b − √Δ) / 2a = −2b / 2a = −b / a

This gives us the first of Vieta’s formulas: the sum of the roots is −b/a. Notice that this result holds regardless of whether the roots are real or complex, as long as they are counted with multiplicity.

由此得到韦达定理的第一条公式：两根之和为 −b/a。注意，无论根是实数还是复数（按重数计），这一结论始终成立。

3. Deriving the Product of Roots | 推导两根之积

Multiplying the two roots together yields:

将两个根相乘，得到：

αβ = [(−b + √Δ)(−b − √Δ)] / (4a²)

Using the difference of squares identity (p + q)(p − q) = p² − q²:

利用平方差公式 (p + q)(p − q) = p² − q²：

αβ = [b² − (Δ)] / (4a²) = [b² − (b² − 4ac)] / (4a²) = 4ac / 4a² = c / a

Thus, the product of the roots is c/a. Together with the sum formula, these two results constitute Vieta’s Theorem for quadratic equations:

因此，两根之积为 c/a。与两根之和的公式合并，就构成了二次方程的韦达定理：

α + β = −b/a, αβ = c/a

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

IB students studying Higher Level Mathematics will also encounter Vieta’s formulas extended to cubic equations. For a cubic ax³ + bx² + cx + d = 0 with roots α, β and γ:

IB高级水平（HL）的学生还会遇到韦达定理在三次方程中的推广。对于三次方程 ax³ + bx² + cx + d = 0，设其三个根为 α、β 和 γ：

α + β + γ = −b/a

αβ + βγ + γα = c/a

αβγ = −d/a

Notice the alternating signs: for an odd-degree term, the sum carries a negative sign; for an even-degree term, a positive sign. This pattern continues for higher-degree polynomials.

注意符号的交替规律：奇数次项的根之和取负号，偶数次项的根两两乘积之和取正号。这一规律可以推广到更高次的多项式。

5. Constructing a Quadratic Equation from Given Roots | 由已知根构造二次方程

A direct and powerful application of Vieta’s theorem is constructing a quadratic equation when the roots are known. If the roots are α and β, then the monic quadratic (leading coefficient 1) is:

韦达定理一个直接而强大的应用是：当已知根时，可以构造出二次方程。若根为 α 和 β，则首项系数为1的二次方程为：

x² − (α + β)x + αβ = 0

More generally, any scalar multiple k(x² − Sx + P) = 0, where S = α + β and P = αβ, represents the same equation. This is particularly useful in IB problems that ask for a new equation whose roots are functions of the original roots.

更一般地，k(x² − Sx + P) = 0（其中 S = α + β，P = αβ）的任意非零倍数都表示同一方程。这在IB题目中特别有用，尤其是当要求构造一个根为原根函数的全新方程时。

For example, given α and β as roots of ax² + bx + c = 0, an equation with roots 1/α and 1/β has:

例如，已知 α 和 β 是 ax² + bx + c = 0 的根，则以 1/α 和 1/β 为根的方程为：

S_new = 1/α + 1/β = (α + β)/(αβ) = (−b/a)/(c/a) = −b/c

P_new = (1/α)(1/β) = 1/(αβ) = a/c

Thus the new equation is cx² + bx + a = 0 — a beautiful reciprocal relationship.

因此新方程为 cx² + bx + a = 0 —— 这是非常优美的倒数对称关系。

6. Evaluating Symmetric Expressions of Roots | 求根的对称表达式的值

Many IB examination questions require evaluating expressions involving α and β without explicitly solving for them. The key technique is expressing the target expression in terms of the sum (α + β) and product (αβ).

许多IB考题要求在不解出 α 和 β 的前提下，计算含有 α 和 β 的表达式。核心技巧是将目标表达式转化为只含两根之和 (α + β) 与两根之积 (αβ) 的形式。

The most frequently tested identities include:

最常考的恒等式包括：

Expression \| 表达式	In terms of S = α + β and P = αβ \| 用 S = α + β 和 P = αβ 表示
α² + β²	S² − 2P
α³ + β³	S³ − 3PS
α² − β²	(α − β)(α + β)
(α − β)²	S² − 4P = Δ/a²
1/α + 1/β	S/P

It is essential to master these identities, as they appear consistently in both Paper 1 (no calculator) and Paper 2 (calculator allowed) across IB exams.

务必熟练掌握这些恒等式，因为它们在IB考试中经常出现，无论是Paper 1（不能使用计算器）还是Paper 2（允许使用计算器）。

7. The Discriminant and the Difference of Roots | 判别式与两根之差

The discriminant Δ = b² − 4ac not only determines the nature of the roots but also provides a direct formula for the difference between the roots. From the quadratic formula:

判别式 Δ = b² − 4ac 不仅决定了根的性质，还直接给出了两根之差的公式。由求根公式：

α − β = (√Δ) / a

Squaring both sides gives (α − β)² = Δ/a², which is consistent with the identity (α − β)² = (α + β)² − 4αβ. This connection is frequently tested in IB problems that require establishing relationships between coefficients and root differences.

两边平方得到 (α − β)² = Δ/a²，这与恒等式 (α − β)² = (α + β)² − 4αβ 一致。这一联系在IB题目中经常出现，特别是在需要建立系数与根差之间关系的题中。

For example, if a problem states that the roots differ by 2, we can immediately write:

例如，若题目说两根之差为2，我们可直接写出：

(α − β)² = 4 = (b² − 4ac)/a²

This gives a direct equation in the coefficients, often allowing us to solve for an unknown parameter.

这样就得到了一个关于系数的直接方程，通常能用来求解未知参数。

8. IB Exam Question Patterns | IB考试题型分析

Vieta’s formulas appear in several recurring IB question formats. Understanding these patterns is key to efficient exam preparation.

韦达定理在IB考试中有几种反复出现的题型。理解这些模式是高效备考的关键。

Finding unknown coefficients: Given the roots (or a relationship between them), determine the values of coefficients. For instance, if one root of x² − 5x + k = 0 is 2, find k.
求未知系数：已知根（或根之间的关系），确定系数的值。例如，若 x² − 5x + k = 0 的一个根是2，求 k。

Forming new equations: Given a quadratic equation, form another quadratic whose roots are transformations of the original roots (e.g., doubled, reciprocals, squared).
构造新方程：已知一个二次方程，构造另一个二次方程，使其根是原根的变换（例如：加倍、取倒数、平方）。

Parameter conditions: Determine the range of a parameter for which the roots satisfy certain conditions, such as being real, equal, or of opposite signs.
参数条件：确定参数的取值范围，使得根满足某些条件，如为实数、相等或异号。

In the IB, these questions often appear as part (a), (b), (c) structured problems worth 5–9 marks total, testing both algebraic manipulation and conceptual understanding.

在IB考试中，这类题目通常以 (a)、(b)、(c) 分步设问的形式出现，总分值5–9分，考查代数操作能力和概念理解。

9. Common Pitfalls and How to Avoid Them | 常见错误及规避方法

Even strong students lose marks on Vieta’s theorem problems due to careless errors. Here are the most common pitfalls to watch for:

即使是优秀学生也可能因粗心错误在韦达定理题上丢分。以下是最常见的陷阱，务必留意：

Sign errors: The sum of roots is −b/a, not b/a. Always check the sign carefully, especially when b is negative.
符号错误：两根之和是 −b/a，而不是 b/a。务必仔细检查符号，尤其是当 b 为负数时。

Forgetting a ≠ 0: Vieta’s formulas are only valid for genuine quadratic equations. If a = 0, the equation degenerates to a linear equation.
忘记 a ≠ 0：韦达定理只对真正的二次方程成立。若 a = 0，方程退化为一次方程。

Confusing coefficients: In ax² + bx + c = 0, the coefficient of x is b and the constant term is c. Mixing these up leads to entirely incorrect answers.
混淆系数：在 ax² + bx + c = 0 中，x 的系数是 b，常数项是 c。混淆它们会导致答案完全错误。

Not simplifying expressions: When evaluating symmetric expressions, always reduce to the simplest form before substituting numerical values.
不化简表达式：计算对称表达式时，务必先化简到最简形式，再代入数值。

10. Worked Example: A Complete Solution | 典型例题完整解析

Let us work through a classic IB-style problem step by step.

让我们逐步解析一道经典的IB风格题目。

Problem: The quadratic equation 2x² + 4x − 3 = 0 has roots α and β. Find: (a) α + β; (b) αβ; (c) α² + β²; (d) the quadratic equation with roots 2α and 2β.

题目：二次方程 2x² + 4x − 3 = 0 的根为 α 和 β。求：(a) α + β；(b) αβ；(c) α² + β²；(d) 以 2α 和 2β 为根的二次方程。

Solution: Here a = 2, b = 4, c = −3.

解答：这里 a = 2，b = 4，c = −3。

(a) By Vieta’s theorem:

(a) 根据韦达定理：

α + β = −b/a = −4/2 = −2

(b) The product is:

(b) 两根之积为：

αβ = c/a = −3/2

α² + β² = (−2)² − 2(−3/2) = 4 + 3 = 7

(d) For new roots 2α and 2β:

(d) 对于新根 2α 和 2β：

Sum = 2α + 2β = 2(α + β) = −4

Product = (2α)(2β) = 4αβ = −6

Therefore the new equation is x² − (−4)x + (−6) = x² + 4x − 6 = 0.

因此新方程为 x² − (−4)x + (−6) = x² + 4x − 6 = 0。

11. Advanced Application: Roots with Given Conditions | 进阶应用：满足给定条件的根

In higher-level IB problems, Vieta’s formulas are combined with constraints on the roots. A common format is: “Find the value(s) of k such that the equation x² + kx + 9 = 0 has roots that are equal.”

在高级IB题目中，韦达定理常与根的约束条件结合使用。常见题型如：”求 k 的值，使得方程 x² + kx + 9 = 0 的两个根相等。”

For equal roots, α = β, so the discriminant must be zero:

若两根相等，则 α = β，所以判别式必须为零：

Δ = k² − 4(1)(9) = k² − 36 = 0

k = ±6

Alternatively, using Vieta’s formulas: if α = β = r, then 2r = −k and r² = 9, giving r = ±3 and k = −2r = ∓6. Both approaches yield the same result, and using either method in the IB exam is perfectly acceptable.

或者用韦达定理：若 α = β = r，则 2r = −k 且 r² = 9，得 r = ±3，k = −2r = ∓6。两种方法结果一致，在IB考试中使用任意一种方法均可。

For two roots with opposite signs (αβ < 0), the condition immediately becomes c/a < 0, providing a quick inequality to solve for a parameter.

若两根异号（αβ < 0），条件立即化为 c/a < 0，从而可直接解关于参数的不等式。

12. Practice Problems with Answers | 练习题与答案

Test your understanding with these IB-style problems. Attempt each fully before checking the answer.

用以下IB风格的题目检验你的理解。请先独立完成，再核对答案。

Problem 1: The roots of 3x² − 6x + 2 = 0 are α and β. Find α² + β².

练习1：方程 3x² − 6x + 2 = 0 的根为 α 和 β。求 α² + β²。

Answer: α + β = 2, αβ = 2/3, so α² + β² = 4 − 4/3 = 8/3.

答案：α + β = 2，αβ = 2/3，故 α² + β² = 4 − 4/3 = 8/3。

Problem 2: Find a quadratic equation whose roots are 3 and −5.

练习2：求一个根为 3 和 −5 的二次方程。

Answer: S = −2, P = −15, so x² + 2x − 15 = 0.

答案：S = −2，P = −15，故 x² + 2x − 15 = 0。

Problem 3: The equation x² + px + q = 0 has roots α and β. Express α²β + αβ² in terms of p and q.

练习3：方程 x² + px + q = 0 的根为 α 和 β。用 p 和 q 表示 α²β + αβ²。

Answer: α²β + αβ² = αβ(α + β) = q(−p) = −pq.

答案：α²β + αβ² = αβ(α + β) = q(−p) = −pq。

Problem 4: If the roots of 2x² − 3x + k = 0 differ by 1, find k.

练习4：若 2x² − 3x + k = 0 的两根之差为1，求 k。

Answer: (α − β)² = 1 = (b² − 4ac)/a² = (9 − 8k)/4. Thus 9 − 8k = 4, so k = 5/8.

答案：(α − β)² = 1 = (b² − 4ac)/a² = (9 − 8k)/4。故 9 − 8k = 4，所以 k = 5/8。

Problem 5: Construct a quadratic equation with roots α + 1 and β + 1, given that α and β are roots of x² − 4x + 1 = 0.

练习5：已知 α 和 β 是 x² − 4x + 1 = 0 的根，构造以 α + 1 和 β + 1 为根的二次方程。

Answer: New sum = (α + 1) + (β + 1) = 4 + 2 = 6; new product = (α + 1)(β + 1) = αβ + (α + β) + 1 = 1 + 4 + 1 = 6. Equation: x² − 6x + 6 = 0.

答案：新和 = (α + 1) + (β + 1) = 4 + 2 = 6；新积 = (α + 1)(β + 1) = αβ + (α + β) + 1 = 1 + 4 + 1 = 6。方程为：x² − 6x + 6 = 0。

Vieta’s formulas are more than a shortcut—they embody a deep structural relationship between algebra and geometry. Mastering them not only secures marks in IB examinations but also builds a foundation for calculus, complex numbers, and polynomial theory at the university level. Regular practice with the identities and question patterns above will ensure you approach any Vieta-related problem with confidence.

韦达定理不仅仅是一种捷径——它体现了代数学与几何学之间深刻的结构性联系。熟练掌握它不仅能在IB考试中稳定拿分，更为大学阶段的微积分、复数和多项式理论打下坚实的基础。通过反复练习上述恒等式和题型，你将在面对任何与韦达定理相关的题目时充满自信。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

IB Mathematics: Vieta’s Formulas for Quadratic Equations | IB数学：二次方程根与系数关系（韦达定理）