📚 Solving Systems of Quadratic Equations in Two Variables | 二元二次方程解法与题型精讲
Systems of equations involving two variables where at least one equation is quadratic appear frequently across IGCSE, A-Level, and IB curricula. This article systematically breaks down solution strategies, classification of curves, and worked examples to help you master this essential exam topic.
涉及两个变量且至少有一个方程为二次的方程组，在IGCSE、A-Level和IB课程中频繁出现。本文系统梳理解法策略、曲线类型分类及典型例题，帮助你彻底掌握这一重要考点。
1. What Is a System of Quadratic Equations? | 什么是二元二次方程组？
A system of two equations in two variables \(x\) and \(y\) is called a “quadratic system” if at least one of the equations is of degree 2. The general form of such a system can be written as:
含有两个变量 \(x\) 和 \(y\) 的方程组，若其中至少有一个方程是二次的，则称为“二元二次方程组”。其一般形式可写为：
ax² + bxy + cy² + dx + ey + f = 0
paired with either another quadratic equation or a linear equation. The most common exam scenario is one linear equation combined with one quadratic equation, solvable via substitution.
它既可以与另一个二次方程配对联立，也可以与一个线性方程联立。考试中最常见的题型是一个线性方程配一个二次方程，通过代入法求解。
2. Classification of Systems | 方程组的分类
Based on the degree of each equation, quadratic systems in two variables can be classified into three main categories for exam purposes:
根据每个方程的次数，二元二次方程组在考试中主要可分为以下三类：
- Type I: One linear equation + one quadratic equation — most common, requires substitution.
- Type I（第一类）： 一个线性方程 + 一个二次方程 — 最常见，需用代入法求解。
- Type II: Both equations quadratic, sharing symmetrical structure (e.g., sum and product forms).
- Type II（第二类）： 两个均为二次方程且结构对称（如和与积的形式）。
- Type III: One quadratic equation reducible via factoring into linear factors, yielding multiple linear paths.
- Type III（第三类）： 其中一个二次方程可因式分解为两个线性因子，从而转化为多条线性路径。
| Type | Form | Method |
| I | Linear + Quadratic | Substitution |
| II | Quadratic + Quadratic | Elimination or substitution |
| III | Factorisable quadratic | Factor and split cases |
3. The Substitution Method | 代入法详解
The substitution method is the most important tool for solving Type I systems. The procedure follows four clear steps:
代入法是求解第一类方程组最重要的工具，其步骤清晰，共分四步：
Step 1: Rearrange the linear equation to express one variable in terms of the other (e.g., y = mx + c).
步骤一：将线性方程改写，用其中一个变量表示另一个变量（如 y = mx + c）。
Step 2: Substitute this expression into the quadratic equation, eliminating one variable.
步骤二：将此表达式代入二次方程，消去一个变量。
Step 3: Solve the resulting quadratic equation in one variable using factorisation, completing the square, or the quadratic formula.
步骤三：利用因式分解、配方法或求根公式求解所得的一元二次方程。
Step 4: Substitute each solution back into the linear equation to find the corresponding values of the other variable.
步骤四：将每一个解代回线性方程，求出对应的另一个变量的值。
4. Worked Example 1: Line and Parabola | 例题一：直线与抛物线
Solve the following system of equations:
解下列方程组：
y = x² − 3x + 2
y = 2x − 1
Solution: Since the second equation already expresses y in terms of x, equate the two right-hand sides:
解：第二个方程已用 x 表示 y，直接令两式右边相等：
x² − 3x + 2 = 2x − 1
Bring all terms to one side:
将所有项移到一边：
x² − 5x + 3 = 0
This quadratic does not factor nicely, so apply the quadratic formula:
此二次方程不易因式分解，故应用求根公式：
x = [5 ± √(25 − 12)] ⁄ 2 = (5 ± √13) ⁄ 2
Substituting these x-values into y = 2x − 1 gives:
将这两个 x 值代回 y = 2x − 1，得到：
y = 4 ± √13
Thus the two intersection points are ((5+√13)/2, 4+√13) and ((5−√13)/2, 4−√13).
因此两条曲线的交点分别为 ((5+√13)/2, 4+√13) 与 ((5−√13)/2, 4−√13)。
Key takeaway: when both equations are already of the form y = …, simply equate them and solve.
关键收获：当两个方程都已写成 y = … 的形式时，只需将它们相等联立求解。
5. Worked Example 2: Circle and Line | 例题二：圆与直线
Find the points of intersection of the circle x² + y² = 25 and the line x + y = 7.
求圆 x² + y² = 25 与直线 x + y = 7 的交点。
Solution: From the line, express y = 7 − x. Substitute into the circle equation:
解：由直线方程得 y = 7 − x。代入圆的方程：
x² + (7 − x)² = 25
Expand and simplify:
展开并化简：
x² + 49 − 14x + x² = 25
2x² − 14x + 24 = 0
Divide the entire equation by 2:
等式两边同时除以 2：
x² − 7x + 12 = 0 → (x − 3)(x − 4) = 0
Hence x = 3 or x = 4. Substitute back into y = 7 − x to get y = 4 or y = 3 respectively. The solutions are (3,4) and (4,3).
因此 x = 3 或 x = 4。代回 y = 7 − x 得 y = 4 或 y = 3。解为 (3,4) 和 (4,3)。
This example demonstrates how factorisation simplifies the process — always check for factorable quadratics before resorting to the quadratic formula.
该例题展示了因式分解如何简化过程——在动用求根公式之前，务必先检查二次方程是否可因式分解。
6. Discriminant: Intersection Type | 判别式：判定交点类型
The discriminant Δ = b² − 4ac of the resulting quadratic tells us the nature of the intersection between the line and the curve:
代入后所得一元二次方程的判别式 Δ = b² − 4ac，可以判定直线与曲线的交点性质：
- Δ > 0: Two distinct real intersection points — the line cuts the curve.
- Δ > 0（判别式大于零）： 两个不同的实交点 — 直线穿过曲线。
- Δ = 0: Exactly one real intersection point — the line is tangent to the curve.
- Δ = 0（判别式等于零）： 恰好一个实交点 — 直线与曲线相切。
- Δ < 0: No real intersection — the line does not touch the curve at all.
- Δ < 0（判别式小于零）： 无实交点 — 直线与曲线不相交。
This is a powerful shortcut: in many exam questions, you are asked only to show that a line is tangent to a curve. Setting the discriminant to zero gives you an equation to solve for unknown parameters.
这是一种强有力的捷径：许多考题只要求证明直线与曲线相切。令判别式为零即可得到关于未知参数的方程。
7. Worked Example 3: Tangent Condition | 例题三：相切条件
Find the value of k for which the line y = 3x + k is tangent to the parabola y = x² + 2x + 1.
求 k 的值，使得直线 y = 3x + k 与抛物线 y = x² + 2x + 1 相切。
Solution: Equate the two expressions:
解：令两式相等：
x² + 2x + 1 = 3x + k
x² − x + (1 − k) = 0
For tangency, the discriminant must be zero:
相切要求判别式为零：
Δ = (−1)² − 4 × 1 × (1 − k) = 0
1 − 4 + 4k = 0 → 4k = 3 → k = 3⁄4
Therefore, the line y = 3x + 3/4 is tangent to the parabola.
因此，直线 y = 3x + 3/4 与该抛物线相切。
Notice how elegantly the discriminant handles a question that would otherwise require graphing or calculus.
请注意，判别式巧妙地解决了若用图像法或微积分法则会更为繁琐的问题。
8. Symmetric Systems (Type II) | 对称方程组（第二类题型）
When both equations are quadratic but symmetric in x and y, a common technique is to introduce new variables:
当两个方程均为二次但关于 x 和 y 对称时，常用技巧是引入新变量：
S = x + y and P = xy
Here is a typical example. Solve:
以下是一个典型例题。求解：
x² + y² = 10
xy = 3
Recall that x² + y² = (x + y)² − 2xy, so:
注意 x² + y² = (x + y)² − 2xy，因此：
(x + y)² − 2 × 3 = 10 → (x + y)² = 16
Hence x + y = 4 or x + y = −4. Along with xy = 3, we form two linear-quadratic pairs:
于是 x + y = 4 或 x + y = −4。结合 xy = 3 可构成两个线性-二次方程组：
Case 1: x + y = 4, xy = 3 → t² − 4t + 3 = 0 → (t−1)(t−3) = 0 → solutions (1,3) and (3,1).
情况一：x + y = 4, xy = 3 → t² − 4t + 3 = 0 → (t−1)(t−3) = 0 → 解为 (1,3) 和 (3,1)。
Case 2: x + y = −4, xy = 3 → t² + 4t + 3 = 0 → (t+1)(t+3) = 0 → solutions (−1,−3) and (−3,−1).
情况二：x + y = −4, xy = 3 → t² + 4t + 3 = 0 → (t+1)(t+3) = 0 → 解为 (−1,−3) 和 (−3,−1)。
Total: four ordered pairs satisfy the system.
该方程组共有四组有序解。
9. Factorisable Quadratic Systems (Type III) | 可因式分解方程组（第三类题型）
Consider a system where one quadratic equation factors into linear terms:
考虑一个二次方程可因式分解为线性因子的方程组：
x² − y² = 0
x² + y² = 4
Factor the first equation:
对第一个方程进行因式分解：
(x − y)(x + y) = 0
This gives two cases:
这给出两种情况：
Case 1: y = x. Substitute into x² + y² = 4:
情况一：y = x。代入 x² + y² = 4：
2x² = 4 → x² = 2 → x = ±√2, y = ±√2
Case 2: y = −x. Substitute similarly:
情况二：y = −x。同样代入：
2x² = 4 → x = ±√2, y = ∓√2
Thus the four solutions are (±√2, ±√2) and (±√2, ∓√2). The factorisation approach reduces the degree of the system dramatically.
因此四组解为 (±√2, ±√2) 与 (±√2, ∓√2)。因式分解法大幅降低了方程组的次数，使求解更直接。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
In exams, students frequently lose marks on quadratic systems due to several recurring errors. Here are the most important pitfalls and tips to avoid them:
在考试中，学生常因一些反复出现的错误而在二次方程组上失分。以下是常见的陷阱与应对技巧：
- Mistake — Substituting the wrong expression: Always double-check which equation is linear and rearrange that one for substitution.
- 错误 — 代错了表达式： 始终确认哪个方程是线性的，并从它出发改写以进行代入。
- Mistake — Losing solutions when dividing: Never divide both sides of an equation by a variable expression that could equal zero; factor instead.
- 错误 — 除以可能为零的表达式导致失根： 切勿在等式两边同除以一个可能等于零的变量表达式；应改用因式分解。
- Mistake — Not writing final answers as coordinate pairs: In question contexts involving curves, always present solutions as (x,y) ordered pairs.
- 错误 — 未将最终答案写成坐标对： 涉及曲线的题目中，请将解写成 (x,y) 有序对。
- Tip — Check the discriminant first: Before solving fully, compute Δ to know how many intersections to expect.
- 技巧 — 先算判别式： 完整求解之前，先计算 Δ 以判断交点个数。
- Tip — Verify your answers: Substitute both x and y values back into the original equations to confirm they satisfy both.
- 技巧 — 回代验证： 将 x 与 y 的解代回原方程组，确认它们同时满足两个方程。
11. Summary of Methods | 解法总结
The table below summarises the best approach for each type of quadratic system in two variables:
下表总结了不同类型二元二次方程组的最佳解法策略：
| System Type | Recommended Method | Output |
| Linear + Quadratic | Substitution from linear | 0, 1, or 2 points |
| Quadratic + Quadratic (symmetric) | Set S = x + y, P = xy | Up to 4 points |
| Factorable quadratic | Factor into linear cases | Split into simpler systems |
| Parameter/tangency problems | Discriminant Δ = 0 | Solve for unknown parameter |
Within A-Level and IGCSE exams, the most frequently examined combination is the line-with-parabola or line-with-circle system. Master the substitution method, recognise symmetric structures, and use the discriminant as a diagnostic tool. With consistent practice, this topic becomes one of the most reliable scoring areas in the algebra section.
在A-Level与IGCSE考试中，最常考的组合是直线配抛物线或直线配圆的方程组。熟练掌握代入法、识别对称结构，并善用判别式进行预判，足以将此话题变为代数板块中最稳定的得分项。加上持续练习，此类问题将不再成为失分点。
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