📚 Quadratic Simultaneous Equations | 二次联立方程组
Quadratic simultaneous equations involve a system where at least one of the equations is of degree two. In A-Level Edexcel Mathematics, the most common scenario pairs a linear equation with a quadratic equation, allowing you to find the intersection points of a straight line and a curve such as a parabola or circle. These problems extend GCSE algebraic methods and introduce the discriminant as a tool for analysing the number of solutions.
二次联立方程组是指至少含有一个二次方程的方程组。在 Edexcel A-Level 数学中，最常见的情形是一个一次方程与一个二次方程联立，从而求出直线与抛物线、圆等曲线的交点。这些问题在 GCSE 代数方法的基础上延伸，并引入判别式作为分析解个数的工具。
1. What Are Quadratic Simultaneous Equations? | 什么是二次联立方程组？
nvolves a linear equation and a quadratic equation, such as y = mx + c and y = ax² + bx + c. You may also encounter a circle equation x² + y² = r² paired with a linear equation, or even two quadratic equations.
二次联立方程组是由两个（或更多）方程组成的系统，其中至少一个方程含有平方项。最简单的形式包含一个一次方程和一个二次方程，例如 y = mx + c 与 y = ax² + bx + c。你也可能遇到圆的方程 x² + y² = r² 与一条直线联立，甚至两个二次方程联立的情况。
The primary goal is to find the set of (x, y) values that satisfy all equations simultaneously. This process mirrors finding the coordinates where the graphs of the equations intersect.
主要目标是求出一组 (x, y) 值，使所有方程同时成立。这一过程相当于求出方程图像交点的坐标。
2. The Substitution Method | 代入法
The most reliable technique for solving a linear-quadratic system is substitution. Rearrange the linear equation to isolate one variable – typically y or x – and substitute this expression into the quadratic equation. This converts the problem into a single quadratic equation in one unknown.
解一次与二次联立方程组最可靠的方法是代入法。将一次方程变形，使其中一个变量（通常是 y 或 x）单独出现在等号一边，然后将该表达式代入二次方程。这样就把问题转化为了一个关于单个未知数的一元二次方程。
For example, given y = 2x – 1 and x² + y² = 10, you replace y in the second equation with 2x – 1 to obtain x² + (2x – which can be solved by factorising, completing the square or using the quadratic formula.
例如，给定 y = 2x – 1 和 x² + y² = 10，将第二个方程中的 y 替换为 2x – 1，得到 x² + (2x – 1)² = 10。展开并化简后，你会得到一个关于 x 的二次方程，可以通过因式分解、配方法或求根公式求解。
3. Elimination in Special Cases | 特殊情况下的消元法
Elimination is less common but can be useful when both equations contain the same squared terms with identical coefficients. If you have x² + y² in both equations, subtracting one from the other removes the quadratic parts, leaving a linear relationship. This new linear equation can then be used with substitution.
消元法较少使用，但当两个方程含有相同系数且相同的平方项时可能很有效。如果两个方程中都出现 x² + y²，将两式相减就可消去二次项，得到一个一次关系式。然后可将该一次方程与代入法结合使用。
For instance, given x² + y² + 2x = 9 and x² + y² – ich simplifies to x + 2y = 4. You then substitute this into one of the original equations.
例如，给定 x² + y² + 2x = 9 和 x² + y² – 4y = 1，用第一个方程减去第二个得到 2x + 4y = 8，化简为 x + 2y = 4。然后将该一次关系代入原方程之一继续求解。
4. Forming a Quadratic Equation in One Variable | 转化为一元二次方程
Expand any brackets carefully, collect like terms and bring everything to one side. At this stage, always double-check your expansion – a sign error here will cascade through the whole solution.
代入后，必须将得到的方程整理为标准二次形式 ax² + bx + c = 0。仔细展开括号，合并同类项，并将所有项移到等号一边。在这一步，务必反复检查展开过程——一个符号错误会导致整个解答完全错误。
dratic with an x² coefficient that is not 1. For example, substituting y = 3x + 2 into 2x² – y + 4 = 0 gives 2x² – (3x + 2) + 4 = 0, which simplifies to 2x² – 3x + 2 = 0. You can then apply the quadratic formula directly.
得到二次项系数不为 1 的方程是完全正常的。例如，将 y = 3x + 2 代入 2x² – y + 4 = 0 得到 2x² – (3x + 2) + 4 = 0，化简为 2x² – 3x + 2 = 0。然后可以直接使用求根公式。
x = [ –b ± √b² – 4ac ] / (2a)
5. Using the Discriminant to Determine Number of Solutions | 利用判别式判断解的个数
Before solving fully, the discriminant Δ = b² – n intersecting a line with a curve, this corresponds to the number of intersection points.
在完全求解之前，由最终二次方程得到的判别式 Δ = b² – 4ac 能告诉你存在多少个实数解 (x, y)。当探究直线与曲线相交时，这直接对应于交点的个数。
| Discriminant value | Geometric meaning | Number of (x,y) pairs |
| Δ > 0 | Two distinct intersection points | 2 |
| Δ = 0 | Line is tangent to curve (one repeated point) | 1 (repeated) |
| Δ < 0 | No intersection (line misses curve) | 0 |
This is particularly useful in questions that ask 'show that the line does not meet the curve' or 'find the condition for tangency'. You simply set up the combined quadratic and examine its discriminant.
这在要求“证明直线不与曲线相交”或“求相切的条件”等题型中特别有用。你只需整理出合并后的一元二次方程，然后考察其判别式即可。
6. Solving Worked Example 1: Line and Parabola | 例题1：直线与抛物线
Solve the simultaneous equations: y = x + 2 and y = x² – 4x + 7.
解联立方程组：y = x + 2 和 y = x² – 4x + 7。
Step 1: Substitute the linear expression into the quadratic. x + 2 = x² – 4x + 7. Step 2: Rearrange to zero: 0 = x² – 5x + 5, so x² – 5x + 5 = 0. Step 3: Use the quadratic formula with a = 1, b = –5, c = 5. x = [5 ± √25 – 20] / 2 = [5 ± √5] / 2. Step 4: For each x, find y using y = x + 2. When x = (5 + √5)/2, y = (9 + √5)/2; when x = (5 – √5)/2, y = (9 – √5)/2. Express answers as exact coordinates.
步骤1：将一次式代入二次式。x + 2 = x² – 4x + 7。步骤2：移项使右边为零：0 = x² – 5x + 5，即 x² – 5x + 5 = 0。步骤3：使用求根公式，a = 1，b = –5，c = 5。x = [5 ± √25 – 20] / 2 = [5 ± √5] / 2。步骤4：对每个 x，用 y = x + 2 求 y。当 x = (5 + √5)/2 时，y = (9 + √5)/2；当 x = (5 – √5)/2 时，y = (9 – √5)/2。答案要用精确坐标形式表示。
7. Solving Worked Example 2: Line and Circle | 例题2：直线与圆
Solve the system: x² + y² = 13 and y = x + 1.
解方程组：x² + y² = 13 和 y = x + 1。
the circle equation. x² + (x + 1)² = 13. Step 2: Expand: x² + x² + 2x + 1 = 13 → 2x² + 2x – 12 = 0. Step 3: Divide by 2: x² + x – 6 = 0. Step 4: Factorise: (x + 3)(x – 2) = 0, giving x = –3 or x = 2. Step 5: Find corresponding y: when x = –3, y = –2; when x = 2, y = 3. Solutions: (–3, –2) and (2, 3).
步骤1：将 y = x + 1 代入圆的方程。x² + (x + 1)² = 13。步骤2：展开：x² + x² + 2x + 1 = 13 → 2x² + 2x – 12 = 0。步骤3：除以2：x² + x – 6 = 0。步骤4：因式分解：(x + 3)(x – 2) = 0，得 x = –3 或 x = 2。步骤5：求相应的 y：当 x = –3 时，y = –2；当 x = 2 时，y = 3。解为 (–3, –2) 和 (2, 3)。
8. Solving Two Quadratic Equations | 解两个二次方程组成的方程组
When both equations are quadratic, such as y = x² – 2x and x = y² – rst equation, y is already expressed in terms of x, so substitute into the second: x = (x² – 2x)² – 2(x² – lify nicely, perhaps after spotting a symmetry or using the fact that y = x is a solution.
当两个方程都是二次方程时，例如 y = x² – 2x 和 x = y² – 2y，通常仍使用代入法。第一个方程已经将 y 表示为了 x 的式子，代入第二个方程：x = (x² – 2x)² – 2(x² – 2x)。这会得到一个四次方程，可能可以因式分解为低次多项式。在 Edexcel A-Level 考试中，这类题目往往经过精心设计，可以通过发现对称性或利用 y = x 是解这一事实来简化求解。
y² + 2xy, which allows you to find xy and then solve a quadratic whose roots are x and y.
对于对称方程组，如 x² + y² = A 和 x + y = B，另一个有效方法是利用恒等式 (x + y)² = x² + y² + 2xy，这样就能求出 xy，然后构造一个以 x 和 y 为根的一元二次方程来求解。
9. Interpreting Solutions Graphically | 从图形上解释解
Every solution pair (x, y) corresponds to an intersection point on the Cartesian plane. A line can cut a parabola at two points, touch it at one point (tangent) or miss it entirely. For a circle, a line can be a secant, a tangent or not intersect. Visualising these helps you check whether your algebraic solutions are reasonable.
每一组解 (x, y) 都对应着直角坐标平面上的一个交点。一条直线可以与抛物线相交于两点、相切于一点（切线），或者完全不相交。对于圆来说，直线可以是割线、切线或者不相交。把这些图形想象出来有助于你检查代数解是否合理。
Sketching the curves and the line quickly on a diagram can also guide you when the algebra yields surds or complex numbers. If the discriminant is negative, your sketch should confirm that no real intersection exists.
快速画出曲线和直线的示意图也可以在代数得到无理数或复数解时为你提供指引。若判别式为负，草图将能证实不存在实数交点。
10. Common Mistakes and How to Avoid Them | 常见错误及避免方法
Forgetting to find both x and y: After solving for x, always substitute back into the linear equation (not the quadratic) to get y correctly.
忘记同时求出 x 和 y：解出 x 后，一定要代回一次方程（而非二次方程）来正确求出 y。
Expanding (mx + c)² incorrectly: Remember (ax + b)² = a²x² + 2abx + b², not a²x² + b².
错误地展开 (mx + c)²：记住 (ax + b)² = a²x² + 2abx + b²，而不是 a²x² + b²。
Sign errors when moving terms: Always use brackets when substituting negative expressions.
移项时的符号错误：代入含有负号的表达式时始终使用括号。
Presenting approximate coordinates when exact answers are required: Edexcel A-Level often wants the final solutions in surd form.
在要求精确答案时给出了近似坐标：Edexcel A-Level 考试常要求用根式（surd 形式）给出最终解。
11. Exam Tips for Edexcel A-Level | Edexcel A-Level 考试技巧
In Edexcel A-Level Pure Mathematics, quadratic simultaneous equations typically appear as 5–7 mark questions within the Algebra and Functions section. The mark scheme rewards a clear substitution step, correct expansion, simplification to a standard quadratic, and accurate solving using factorisation or the formula.
在 Edexcel A-Level 纯数中，二次联立方程组通常作为代数和函数部分的 5–7 分题出现。评分标准重视清晰的代入步骤、正确的展开、化简为标准二次式，以及准确使用因式分解或求根公式求解。
Always check the discriminant when a question asks 'determine the number of intersection points' or 'find the exact value of k for which the line is tangent' and solve for k. Show all working; even if the final answer is wrong, you can earn method marks.
当问题要求“确定交点个数”或“求使直线相切的 k 的精确值”时，一定要检查判别式。建立合并后的一元二次方程，将 Δ 用 k 表示，令 Δ = 0 求切点条件，然后解出 k。展示所有步骤；即使最终答案错误，也能获得方法分。
12. Practice Summary | 练习总结
To master quadratic simultaneous equations, practise a variety of pairings: line and parabola, line and circle, and two quadratic curves. Focus on fluency in algebraic manipulation and interpreting the discriminant. With confidence in these techniques, you will be well-prepared to handle any simultaneous equation problem that appears on your Edexcel A-Level exam.
要掌握二次联立方程组，你需要练习各种组合形式：直线与抛物线、直线与圆，以及两条二次曲线。重点在于熟练进行代数操作并会解读判别式。一旦对这些技巧充满信心，你就能从容应对 Edexcel A-Level 考试中出现的任何联立方程问题。
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