Mastering Quadratic Equations & Functions | 二次方程与二次函数精讲

📚 Mastering Quadratic Equations & Functions | 二次方程与二次函数精讲

Quadratic equations and functions form one of the most important pillars of the IGCSE Mathematics syllabus. From solving equations by factorisation to interpreting parabolic graphs, this topic appears in almost every examination paper, often across multiple sections. In this comprehensive revision guide, we will break down every key concept, method, and common pitfall to help you secure full marks.

二次方程与二次函数是 IGCSE 数学考纲中最核心的支柱之一。无论是通过因式分解解方程，还是解读抛物线图像，这一主题几乎出现在每份试卷中，且常常横跨多个板块。在本篇系统性复习指南中，我们将逐一拆解所有关键概念、解题方法与常见失分点，助你冲击满分。

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

A quadratic equation in one variable is any equation that can be written in the general form ax² + bx + c = 0, where a, b, and c are constants, and a ≠ 0. The condition a ≠ 0 is essential; if a = 0, the equation becomes linear.

一元二次方程是任何可以写成一般形式 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。条件 a ≠ 0 至关重要；若 a = 0，方程将退化为一次方程。

For example, 2x² – 5x + 3 = 0 is a quadratic equation with a = 2, b = -5, and c = 3. The highest power of the variable is always 2. The solutions to a quadratic equation are also called its roots or zeros.

例如，2x² – 5x + 3 = 0 是一个二次方程，其中 a = 2，b = -5，c = 3。变量的最高次数始终为 2。二次方程的解也称为它的根或零点。

2. Solving by Factorisation | 因式分解法

Factorisation is often the fastest method when the quadratic expression can be factored into two linear binomials. The principle relies on the zero product property: if the product of two factors equals zero, then at least one of the factors must be zero.

当二次表达式可以分解为两个一次二项式时，因式分解法通常是最快的解法。其原理基于零乘积性质：若两个因式的乘积为零，则至少有一个因式为零。

Consider the equation x² – 7x + 12 = 0. We look for two numbers that multiply to +12 and add to -7. These numbers are -3 and -4. Thus, we factor: (x – 3)(x – 4) = 0. Setting each factor to zero gives x = 3 or x = 4.

考虑方程 x² – 7x + 12 = 0。我们要寻找两个数，它们的乘积为 +12，和为 -7。这两个数是 -3 和 -4。因此分解为：(x – 3)(x – 4) = 0。令每个因式为零，得 x = 3 或 x = 4。

x² – 7x + 12 = 0 → (x – 3)(x – 4) = 0 → x = 3 或 x = 4

When the coefficient of x² is not 1, the factorisation becomes slightly more involved. Take 2x² + 7x + 3 = 0. We look for two numbers whose product equals 2 × 3 = 6 and whose sum equals 7. Those numbers are 6 and 1. We rewrite the middle term: 2x² + 6x + x + 3, then factor by grouping: 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0. Hence x = -½ or x = -3.

当 x² 的系数不为 1 时，因式分解会稍显复杂。以 2x² + 7x + 3 = 0 为例。我们要找两个数，乘积等于 2 × 3 = 6，和等于 7。这两个数是 6 和 1。我们改写中间项：2x² + 6x + x + 3，然后分组分解：2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3) = 0。因此 x = -½ 或 x = -3。

3. The Quadratic Formula | 求根公式

Not every quadratic equation can be factored easily. When factorisation is difficult or impossible, we turn to the quadratic formula. For any quadratic equation ax² + bx + c = 0, the roots are given by:

并非所有二次方程都能轻松因式分解。当因式分解困难或不可行时，我们使用求根公式。对于任意二次方程 ax² + bx + c = 0，其根由下式给出：

x = (-b ± √(b² – 4ac)) / 2a

This formula is derived by completing the square on the general form, and it works for every quadratic equation. The symbol ± indicates that there are two possible solutions: one with the plus sign and one with the minus sign.

该公式通过对一般形式进行配方法推导而来，适用于所有二次方程。符号 ± 表示有两个可能的解：一个取加号，一个取减号。

For example, solve 2x² – 3x – 5 = 0 using the formula. Here a = 2, b = -3, c = -5. Substituting into the formula gives x = (3 ± √(9 + 40)) / 4 = (3 ± √49) / 4 = (3 ± 7) / 4. Thus x = 10/4 = 2.5 or x = -4/4 = -1.

例如，使用求根公式解 2x² – 3x – 5 = 0。这里 a = 2，b = -3，c = -5。代入公式得 x = (3 ± √(9 + 40)) / 4 = (3 ± √49) / 4 = (3 ± 7) / 4。因此 x = 10/4 = 2.5 或 x = -4/4 = -1。

4. Completing the Square | 配方法

Completing the square is another powerful technique that transforms a quadratic expression into a perfect square plus a constant. The general approach rewrites x² + bx as (x + b/2)² – (b/2)². This method is especially useful for finding the turning point of a quadratic graph and for solving equations without factorisation.

配方法是另一种强大的技巧，它将二次表达式转化为一个完全平方加上一个常数。一般做法是将 x² + bx 改写为 (x + b/2)² – (b/2)²。这种方法在求二次图像的顶点以及不解因式分解而求解方程时尤为实用。

Consider x² + 8x + 15 = 0. We complete the square for x² + 8x first, writing it as (x + 4)² – 16. The equation becomes (x + 4)² – 16 + 15 = 0, which simplifies to (x + 4)² = 1. Taking square roots gives x + 4 = ±1, so x = -3 or x = -5.

考虑 x² + 8x + 15 = 0。我们先对 x² + 8x 配方，写作 (x + 4)² – 16。方程变为 (x + 4)² – 16 + 15 = 0，即 (x + 4)² = 1。两边开平方得 x + 4 = ±1，所以 x = -3 或 x = -5。

x² + 8x + 15 = 0 → (x + 4)² = 1 → x = -3 或 x = -5

When the coefficient of x² is not 1, factor it out first. For example, 2x² + 4x – 6 = 0 becomes 2(x² + 2x – 3) = 0. Completing the square inside the brackets: 2[(x + 1)² – 4] = 0, giving 2(x + 1)² = 8, so (x + 1)² = 4, hence x = 1 or x = -3.

当 x² 的系数不为 1 时，先将其提出。例如，2x² + 4x – 6 = 0 变为 2(x² + 2x – 3) = 0。在括号内配方：2[(x + 1)² – 4] = 0，即 2(x + 1)² = 8，所以 (x + 1)² = 4，因此 x = 1 或 x = -3。

5. The Discriminant: b² – 4ac | 判别式：b² – 4ac

The expression b² – 4ac, known as the discriminant, reveals the nature of the roots without actually solving the equation. Let Δ represent the discriminant:

表达式 b² – 4ac 称为判别式，它可以在不实际求解方程的情况下揭示根的性质。令 Δ 表示判别式：

If Δ > 0: two distinct real roots (the graph crosses the x-axis at two points). / 若 Δ > 0：有两个不等实根（图像与 x 轴交于两点）。
If Δ = 0: exactly one repeated real root (the graph touches the x-axis at exactly one point). / 若 Δ = 0：恰有一个重根（图像与 x 轴相切于一点）。
If Δ < 0: no real roots (the graph does not intersect the x-axis). / 若 Δ < 0：没有实根（图像不与 x 轴相交）。

For instance, the equation x² – 4x + 4 = 0 has a discriminant of (-4)² – 4(1)(4) = 16 – 16 = 0, so it has exactly one root: x = 2. In contrast, x² – 4x + 7 = 0 has Δ = 16 – 28 = -12, so it has no real solutions.

例如，方程 x² – 4x + 4 = 0 的判别式为 (-4)² – 4(1)(4) = 16 – 16 = 0，因此它只有一个根：x = 2。相比之下，x² – 4x + 7 = 0 的 Δ = 16 – 28 = -12，因此没有实数解。

6. The Graph of a Quadratic Function | 二次函数的图像

The graph of a quadratic function y = ax² + bx + c is a curve called a parabola. The sign of the leading coefficient a determines the direction of the opening. If a > 0, the parabola opens upwards (U-shaped) and has a minimum point. If a < 0, it opens downwards (∩-shaped) and has a maximum point.

二次函数 y = ax² + bx + c 的图像是一条称为抛物线的曲线。首项系数 a 的符号决定了开口方向。若 a > 0，抛物线向上开口（U 形），有最小值点。若 a < 0，抛物线向下开口（∩ 形），有最大值点。

The coordinates of the turning point can be found by completing the square. Written in the form y = a(x – h)² + k, the vertex is at (h, k). Alternatively, the x-coordinate of the vertex is given by x = -b / (2a), and the y-coordinate is found by substituting this value back into the original equation.

顶点的坐标可以通过配方法求得。将函数写成 y = a(x – h)² + k 的形式后，顶点为 (h, k)。另一种方法是，顶点的 x 坐标为 x = -b / (2a)，将 x 值代回原方程即可求得 y 坐标。

顶点 x 坐标：x = -b / (2a)

7. Sketching Parabolas | 抛物线作图

To sketch a quadratic graph accurately, you need to identify the following key features in order:

要准确绘制二次函数图像，你需要依次确定以下关键特征：

The y-intercept: the point where the graph crosses the y-axis, found by setting x = 0, giving y = c. / y 截距：图像与 y 轴的交点，令 x = 0，得 y = c。
The x-intercepts: the roots of the equation, found by setting y = 0 and solving. / x 截距：方程的根，令 y = 0 并求解。
The turning point: the vertex of the parabola. / 顶点：抛物线的最高点或最低点。
The axis of symmetry: the vertical line x = h that passes through the vertex. / 对称轴：通过顶点的垂直线 x = h。

For example, to sketch y = x² – 2x – 3: the y-intercept is (0, -3). Solving x² – 2x – 3 = 0 gives (x – 3)(x + 1) = 0, so x-intercepts are (3, 0) and (-1, 0). The vertex has x-coordinate x = -(-2)/(2·1) = 1, and substituting x = 1 gives y = 1 – 2 – 3 = -4. So the vertex is (1, -4). With a = 1 > 0, the parabola opens upwards.

例如，绘制 y = x² – 2x – 3 的图像：y 截距为 (0, -3)。解 x² – 2x – 3 = 0 得 (x – 3)(x + 1) = 0，因此 x 截距为 (3, 0) 和 (-1, 0)。顶点的 x 坐标为 x = -(-2)/(2·1) = 1，代入 x = 1 得 y = 1 – 2 – 3 = -4。所以顶点为 (1, -4)。由于 a = 1 > 0，抛物线向上开口。

8. Quadratic Inequalities | 二次不等式

Solving a quadratic inequality such as x² – 5x + 6 > 0 requires a combination of factorisation and graphical reasoning. First, factorise the quadratic: (x – 2)(x – 3) > 0. The roots are x = 2 and x = 3. Since the parabola opens upwards, the expression is positive outside the roots and negative between them.

解二次不等式（如 x² – 5x + 6 > 0）需要结合因式分解与图像推理。首先分解： (x – 2)(x – 3) > 0。根为 x = 2 和 x = 3。由于抛物线向上开口，表达式在两根之外为正，在两根之间为负。

Therefore, x² – 5x + 6 > 0 holds when x < 2 or x > 3. For the inequality x² – 5x + 6 ≤ 0, the solution is 2 ≤ x ≤ 3. Always remember to reverse the inequality sign when multiplying or dividing both sides by a negative number.

因此，x² – 5x + 6 > 0 在 x < 2 或 x > 3 时成立。对于不等式 x² – 5x + 6 ≤ 0，解为 2 ≤ x ≤ 3。切记：当两边同时乘以或除以负数时，不等号方向必须反转。

9. Sum and Product of Roots | 根的和与积

For any quadratic equation ax² + bx + c = 0 with roots α and β, there are two important relationships that appear frequently in examination problems:

对于任意以 α 和 β 为根的二次方程 ax² + bx + c = 0，存在两个在考试中频繁出现的重要关系：

α + β = -b / a

α × β = c / a

These formulae allow you to solve problems where roots are given but the equation is unknown. For example, if α + β = 5 and αβ = 6, then the quadratic equation is x² – 5x + 6 = 0. Notice the sign pattern: the sum is the negative of the x coefficient, and the product equals the constant term.

这些公式允许你在已知根但未知方程的情况下解题。例如，若 α + β = 5 且 αβ = 6，则二次方程为 x² – 5x + 6 = 0。注意符号规律：和等于 x 系数的相反数，积等于常数项。

10. Word Problems Involving Quadratics | 二次方程应用题

Examiners love to test quadratics in realistic contexts. Common types include the area of a rectangle with a fixed perimeter, projectile motion under gravity, and consecutive integer problems. The key skill is translating the worded condition into a correct algebraic equation.

考官喜欢在实际情境中考查二次方程。常见题型包括：给定周长求矩形面积、重力下的抛体运动、连续整数问题等。关键技能是将文字条件转化为正确的代数方程。

Example: A rectangle has a length of (x + 5) cm and a width of x cm. Its area is 36 cm². Form and solve a quadratic equation. The equation is x(x + 5) = 36, which expands to x² + 5x – 36 = 0. Factorising gives (x + 9)(x – 4) = 0, so x = 4 (rejecting x = -9 since length cannot be negative).

例题：一个矩形的长为 (x + 5) cm，宽为 x cm，面积为 36 cm²。建立并求解二次方程。方程为 x(x + 5) = 36，展开得 x² + 5x – 36 = 0。因式分解得 (x + 9)(x – 4) = 0，所以 x = 4（舍弃 x = -9，因为长度不能为负）。

11. Simultaneous Equations with One Quadratic | 含一个二次方程的联立方程组

IGCSE extended papers often require solving a pair of simultaneous equations where one is linear and one is quadratic. The solution method is always substitution: express one variable from the linear equation and substitute into the quadratic equation, yielding a single quadratic in one variable.

IGCSE 扩展卷中经常要求解一个线性方程与一个二次方程组成的联立方程组。解法永远是代入消元法：从线性方程中表示出一个变量，代入二次方程，得到一个一元二次方程。

Example: Solve y = 2x + 1 and y = x² – 2x + 2. Substituting gives 2x + 1 = x² – 2x + 2, which rearranges to x² – 4x + 1 = 0. Using the quadratic formula, x = (4 ± √(16 – 4)) / 2 = (4 ± √12) / 2 = 2 ± √3. Then calculate the corresponding y-values from the linear equation.

例题：解方程组 y = 2x + 1 和 y = x² – 2x + 2。代入得 2x + 1 = x² – 2x + 2，整理为 x² – 4x + 1 = 0。使用求根公式，x = (4 ± √(16 – 4)) / 2 = (4 ± √12) / 2 = 2 ± √3。然后从线性方程中求得相应的 y 值。

Graphically, the solutions represent the intersection points between a straight line and a parabola. If the resulting quadratic has no real roots, the line and the parabola do not intersect.

从图像来看，解代表直线与抛物线的交点。若所得二次方程无实根，则直线与抛物线不相交。

12. Common Mistakes and Exam Tips | 常见错误与考试技巧

Avoiding common pitfalls is half the battle in the examination room. The following list highlights frequent errors and best practices:

避免常见失分点是在考场上取胜的一半。以下列表突出常见错误与最佳做法：

Common Mistake \| 常见错误	Correction \| 正确做法
Forgetting that a must not be 0. / 忽略 a ≠ 0 的前提。	Always check the leading coefficient before applying quadratic methods. / 在应用二次方法前始终检查首项系数。
Losing negative signs when substituting into the formula. / 代入公式时丢失负号。	Write down a, b, c explicitly, then substitute carefully. / 先明确写出 a、b、c，再仔细代入。
Cancelling a variable term from both sides incorrectly. / 错误地在等式两边同时约去含变量项。	Never divide both sides by x — you may lose a root at x = 0. / 切勿两边同除以 x —— 你可能会丢失 x = 0 这个根。
Sketching a parabola without computing the vertex. / 绘图时不计算顶点。	Always include the turning point and axis of symmetry in your sketch. / 绘图时务必包含顶点和对称轴。

In the exam, if a quadratic does not factorise easily, do not waste time — switch to the quadratic formula. Always check your answers by substituting back into the original equation. For graph sketching questions, label all intercepts and the vertex clearly, and draw the curve smoothly.

在考试中，如果二次式不易因式分解，不要浪费时间——立即改用求根公式。始终通过代回原方程来检验答案。对于作图题，务必清楚标注所有截距和顶点，并平滑地绘制曲线。

By mastering these twelve areas, you now possess a complete toolkit for tackling any quadratic question on your IGCSE Mathematics paper. Keep practising with past papers, memorise the quadratic formula, and trust your preparation.

通过掌握这十二个板块，你现在已拥有解决 IGCSE 数学试卷上任何二次方程问题的完整工具包。坚持练习历年真题，牢记求根公式，相信你的备考成果。

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Mastering Quadratic Equations & Functions | 二次方程与二次函数精讲