Quadratic Simultaneous Equations | 二次联立方程

📚 Quadratic Simultaneous Equations | 二次联立方程

In Edexcel A Level Mathematics, quadratic simultaneous equations usually combine one linear equation with one quadratic equation, such as a straight line intersecting a parabola or a circle. The key skill is to substitute one equation into the other, turning the system into a single quadratic equation that can be solved using standard methods.

在爱德思 A Level 数学中，二次联立方程通常由一条直线和一个二次方程组成，例如直线与抛物线或圆相交。核心技巧是将一个方程代入另一个方程，把方程组转化为一个一元二次方程，再用标准方法求解。

1. What Are Quadratic Simultaneous Equations? | 什么是二次联立方程？

A quadratic simultaneous equation system contains at least one equation of degree 2. The most common Edexcel exam form is one linear equation and one quadratic equation, for example y = 2x + 1 and y = x² + 3x – 5. You may also see a circle equation such as x² + y² = 25 paired with a straight line.

二次联立方程组至少包含一个二次方程。Edexcel 考试中最常见的形式是一条直线方程配一个二次方程，例如 y = 2x + 1 与 y = x² + 3x – 5。也可能出现圆方程 x² + y² = 25 与直线联立。

Common pairs: linear + parabola; linear + circle; two quadratic equations (less common).
常见组合：直线 + 抛物线；直线 + 圆；两个二次方程（较少见）。

2. General Strategy: Substitution Method | 通用策略：代入法

The standard method is substitution. Rearrange the linear equation to make one variable the subject, then substitute this expression into the quadratic equation. This gives a quadratic in a single variable, which can be solved by factorising, completing the square, or using the quadratic formula.

标准方法是代入法。先把线性方程整理成以某个变量为主语的表达式，再将该表达式代入二次方程。这样就得到一个一元二次方程，可以通过因式分解、配方法或求根公式求解。

After solving for the first variable, substitute back to find the corresponding value of the other variable. Always write final answers as coordinate pairs such as (x, y) unless the question asks for variables separately.

求出第一个变量后，代回原式求出另一个变量的对应值。除非题目另有要求，最终答案应写成坐标对，例如 (x, y)。

For any quadratic equation ax² + bx + c = 0, the quadratic formula is:

对于任意一元二次方程 ax² + bx + c = 0，求根公式为：

x = (-b ± √(b² – 4ac)) / (2a)

3. Worked Example: Line and Parabola | 例题：直线与抛物线

Solve the simultaneous equations y = 2x + 1 and y = x² + x – 3.

求解联立方程 y = 2x + 1 与 y = x² + x – 3。

Substitute y = 2x + 1 into the quadratic equation:

将 y = 2x + 1 代入二次方程：

x² + x – 3 = 2x + 1

Rearrange to form a standard quadratic:

整理为标准二次方程：

x² – x – 4 = 0

This does not factorise neatly, so apply the quadratic formula with a = 1, b = -1, c = -4:

该方程不易因式分解，因此使用求根公式，其中 a = 1，b = -1，c = -4：

x = (1 ± √17) / 2

Now substitute each x value into y = 2x + 1:

现在将每个 x 值代入 y = 2x + 1：

y = 2 + √17 or y = 2 – √17

The solutions are ((1 + √17)/2, 2 + √17) and ((1 – √17)/2, 2 – √17).

解为 ((1 + √17)/2, 2 + √17) 和 ((1 – √17)/2, 2 – √17)。

4. Worked Example: Line and Circle | 例题：直线与圆

Solve the simultaneous equations x² + y² = 25 and y = x + 1.

求解联立方程 x² + y² = 25 与 y = x + 1。

Substitute y = x + 1 into the circle equation:

将 y = x + 1 代入圆的方程：

x² + (x + 1)² = 25

Expand (x + 1)² carefully:

仔细展开 (x + 1)²：

x² + x² + 2x + 1 = 25

Simplify and rearrange:

化简并整理：

2x² + 2x – 24 = 0

Divide by 2:

除以 2：

x² + x – 12 = 0

Factorise:

因式分解：

(x – 3)(x + 4) = 0

So x = 3 or x = -4. Substitute back into y = x + 1:

因此 x = 3 或 x = -4。代回 y = x + 1：

When x = 3, y = 4; when x = -4, y = -3.

The solutions are (3, 4) and (-4, -3). Geometrically, the line cuts the circle at these two points.

解为 (3, 4) 和 (-4, -3)。从几何上看，直线与圆相交于这两点。

5. Using the Discriminant to Determine Number of Solutions | 利用判别式判断解的数量

After substitution, the system reduces to a quadratic equation ax² + bx + c = 0. The discriminant Δ = b² – 4ac tells you how many real solutions exist.

代入后，方程组转化为一元二次方程 ax² + bx + c = 0。判别式 Δ = b² – 4ac 可以判断实数解的个数。

Discriminant Δ	Real solutions	Geometric meaning
Δ > 0	2	Two intersections
Δ = 0	1 repeated	Tangent
Δ < 0	0	No intersection

In exam problems, you may be asked to find the range of a constant so that a line is tangent to a curve, or to show that a line does not meet a given circle.

在考试题中，你可能需要求某个常数的范围，使直线与曲线相切，或证明直线与给定圆不相交。

6. Geometric Interpretation: Tangents, Secants and No Intersection | 几何解释：切线、割线与无交点

When a line and a quadratic curve are solved simultaneously, the number of real roots equals the number of intersection points. Two real roots mean the line is a secant; one repeated root means the line is a tangent; no real roots means the line does not touch the curve.

当直线与二次曲线联立求解时，实数根的个数等于交点的个数。两个不同实数根表示直线是割线；一个重根表示直线是切线；没有实数根表示直线与曲线不相交。

To find a tangent condition, set the discriminant of the resulting quadratic to zero. For example, find k such that y = 2x + k is tangent to y = x².

求切线条件时，可将代入后所得二次方程的判别式设为零。例如，求 k 使 y = 2x + k 与 y = x² 相切。

x² = 2x + k → x² – 2x – k = 0

Here a = 1, b = -2, c = -k, so Δ = 4 + 4k. For a tangent, Δ = 0, giving k = -1.

这里 a = 1，b = -2，c = -k，所以 Δ = 4 + 4k。相切时 Δ = 0，得到 k = -1。

7. Hidden Quadratics After Rearranging | 整理后出现的隐藏二次方程

Some simultaneous equations do not look quadratic at first, but become quadratic after substitution. For example, y = x – 1 and xy = 6. The second equation contains a product xy, but substitution reveals a quadratic.

有些联立方程一开始看起来不像二次方程，但代入后会变成二次方程。例如 y = x – 1 与 xy = 6。第二个方程含乘积 xy，但代入后会显现出二次方程。

Substitute y = x – 1 into xy = 6:

将 y = x – 1 代入 xy = 6：

x(x – 1) = 6 → x² – x – 6 = 0

Factorise:

因式分解：

(x – 3)(x + 2) = 0

So x = 3 or x = -2. Substitute back into y = x – 1 to get y = 2 or y = -3. The solutions are (3, 2) and (-2, -3).

因此 x = 3 或 x = -2。代回 y = x – 1，得到 y = 2 或 y = -3。解为 (3, 2) 和 (-2, -3)。

8. Systems with Two Quadratic Equations | 两个二次方程的情形

Although less common at AS Level, two quadratic equations can be solved by elimination. For example, solve x² + y² = 25 and x² – y² = 7.

虽然在 AS 阶段较少见，但两个二次方程可以通过消元法求解。例如，求解 x² + y² = 25 与 x² – y² = 7。

Add the two equations to eliminate y²:

将两式相加以消去 y²：

2x² = 32 → x² = 16 → x = ±4

Subtract the second equation from the first to eliminate x²:

用第一式减第二式以消去 x²：

2y² = 18 → y² = 9 → y = ±3

Combining the signs gives four solutions: (4, 3), (4, -3), (-4, 3) and (-4, -3). Each pair satisfies both original equations.

组合符号得到四个解：(4, 3)、(4, -3)、(-4, 3) 和 (-4, -3)。每一对都满足原方程组。

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One common error is forgetting to substitute back to find the other coordinate after solving for x. Always use the linear equation to find the matching y value.

一个常见错误是在求出 x 后忘记代回求另一个坐标。应始终利用线性方程

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Quadratic Simultaneous Equations | 二次联立方程