Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are a cornerstone of IGCSE Mathematics. They appear in almost every exam paper, whether as direct solving questions, factorising exercises, or worded problems. Understanding how to manipulate and solve quadratics with confidence is essential for success in both Paper 1 and Paper 2.

二次方程是 IGCSE 数学的基石。几乎每一份试卷都会出现它们，无论是直接求解、因式分解练习，还是文字应用题。自信地处理和解二次方程，是在 Paper 1 和 Paper 2 中取得好成绩的关键。

1. What Is a Quadratic Equation? | 什么是二次方程

A quadratic equation is any equation that can be rearranged into the standard form shown below. The variable is x, and the letters a, b and c represent constants, with the condition that a is not equal to zero.

二次方程是任何可以整理成如下标准形式的方程。变量为 x，字母 a、b、c 代表常数，条件是 a 不等于零。

ax² + bx + c = 0, a ≠ 0

The highest power of x is 2, which is why the equation is called “quadratic”. For example, x² − 5x + 6 = 0 and 2x² + 7x − 3 = 0 are both quadratic equations, while x³ − 1 = 0 is not.

x 的最高次数是 2，因此这种方程被称为”二次”。例如，x² − 5x + 6 = 0 和 2x² + 7x − 3 = 0 都是二次方程，而 x³ − 1 = 0 不是。

2. Expanding and Factorising | 展开与因式分解

Before solving quadratic equations, you must be fluent in expanding double brackets and factorising quadratic expressions. These algebraic skills are tested directly and are also the foundation for solving.

在解二次方程之前，你必须熟练地将双重括号展开并对二次式进行因式分解。这些代数技能既是直接考点，也是求解的基础。

To expand two brackets, multiply each term in the first bracket by each term in the second bracket:

展开两个括号时，用第一个括号中的每一项乘以第二个括号中的每一项：

(x + 3)(x − 2) = x² + x − 6

To factorise a monic quadratic like x² + bx + c, find two numbers that multiply to give c and add to give b. For example, for x² + 5x + 6, the numbers 2 and 3 multiply to 6 and add to 5.

对于形如 x² + bx + c 的二次式，找出两个数，使它们相乘等于 c，相加等于 b。例如，x² + 5x + 6 中，2 和 3 相乘为 6，相加为 5。

x² + 5x + 6 = (x + 2)(x + 3)

3. Solving by Factorisation | 因式分解法求解

If a quadratic expression factorises neatly, this is the fastest solving method. The key principle is that if the product of two factors equals zero, then at least one of the factors must be zero.

如果二次式能干净地因式分解，这是最快的求解方法。关键原理是：如果两个因式的乘积等于零，那么至少有一个因式等于零。

If A × B = 0, then A = 0 or B = 0

Consider the equation x² − 7x + 12 = 0. The numbers −3 and −4 multiply to 12 and add to −7, so the factorised form is (x − 3)(x − 4) = 0. Setting each factor to zero gives x = 3 or x = 4.

考虑方程 x² − 7x + 12 = 0。−3 和 −4 相乘为 12，相加为 −7，因此因式分解形式为 (x − 3)(x − 4) = 0。令每个因式为零，得到 x = 3 或 x = 4。

(x − 3)(x − 4) = 0 → x = 3 or x = 4

Always check your solution by substituting back into the original equation. For x = 3, we get 9 − 21 + 12 = 0, which is correct; for x = 4, we get 16 − 28 + 12 = 0, also correct.

始终通过代回原方程来检验解。对于 x = 3，代入得 9 − 21 + 12 = 0，正确；对于 x = 4，代入得 16 − 28 + 12 = 0，同样正确。

4. The Quadratic Formula | 求根公式法

When a quadratic cannot be factorised easily, or when you are not sure whether it factorises, the quadratic formula always works. This formula is provided in the IGCSE formula sheet, but you must know how to apply it accurately.

当二次式不易因式分解，或你不确定它是否能因式分解时，求根公式永远有效。这个公式在 IGCSE 公式表中会给出，但你必须知道如何准确运用它。

x = (−b ± √(b² − 4ac)) / (2a)

For a quadratic written in the form ax² + bx + c = 0, identify a, b and c, substitute them into the formula, then simplify carefully.

对于形式为 ax² + bx + c = 0 的二次式，先确定 a、b、c，将它们代入公式，然后仔细化简。

Consider 2x² − 5x + 1 = 0. Here a = 2, b = −5 and c = 1. Substituting into the formula:

考虑 2x² − 5x + 1 = 0。这里 a = 2，b = −5，c = 1。代入公式：

x = (5 ± √((−5)² − 4 × 2 × 1)) / (2 × 2) = (5 ± √17) / 4

This gives two solutions: x = (5 + √17) / 4 ≈ 2.28 and x = (5 − √17) / 4 ≈ 0.22. Be careful with negative signs when b is negative, and round only at the final step.

这给出两个解：x = (5 + √17) / 4 ≈ 2.28 和 x = (5 − √17) / 4 ≈ 0.22。当 b 为负数时要注意符号，并且只在最后一步进行四舍五入。

5. Completing the Square | 配方法

Completing the square rewrites a quadratic in the form p(x + q)² + r. This method is especially useful for finding the turning point of a parabola and for solving equations that do not factorise.

配方法将二次式改写为 p(x + q)² + r 的形式。这种方法在求抛物线顶点以及解不能因式分解的方程时特别有用。

For a monic quadratic, the rule is simple: x² + bx can be rewritten as (x + b/2)² − (b/2)². You take half of the coefficient of x, square it, and then subtract that square to keep the expression equal.

对于首项系数为 1 的二次式，规则很简单：x² + bx 可以改写为 (x + b/2)² − (b/2)²。取 x 系数的一半，将其平方，然后减去该平方以保持等式成立。

Solve x² + 6x + 2 = 0 by completing the square. Since b = 6, half of it is 3, so x² + 6x = (x + 3)² − 9. Substituting back:

用配方法解 x² + 6x + 2 = 0。因为 b = 6，其一半为 3，所以 x² + 6x = (x + 3)² − 9。代回原方程：

(x + 3)² − 9 + 2 = 0 → (x + 3)² = 7

Taking square roots gives x + 3 = ±√7, so x = −3 ± √7. Notice the ± sign is essential: forgetting it loses one of the two solutions.

开平方得 x + 3 = ±√7，因此 x = −3 ± √7。注意 ± 符号至关重要：忘记它会丢失一个解。

6. The Discriminant | 判别式

The discriminant is the expression inside the square root of the quadratic formula, b² − 4ac. It tells us how many real roots a quadratic equation has without fully solving it.

判别式是求根公式中根号内的表达式 b² − 4ac。它告诉我们二次方程有多少个实数根，而无需完整求解。

If b² − 4ac > 0, the equation has two distinct real roots.

如果 b² − 4ac > 0，方程有两个不同的实数根。
If b² − 4ac = 0, the equation has exactly one repeated root (a double root).

如果 b² − 4ac = 0，方程恰有一个重根（二重根）。
If b² − 4ac < 0, the equation has no real roots.

如果 b² − 4ac < 0，方程没有实数根。

For example, for x² − 4x + 4 = 0, the discriminant is 16 − 16 = 0, so there is exactly one root: (x − 2)² = 0, giving x = 2. Exam questions often ask you to “find the value of k for which the equation has two equal roots”; use the condition b² − 4ac = 0 to form an equation in k.

例如，对于 x² − 4x + 4 = 0，判别式为 16 − 16 = 0，因此恰好有一个根：(x − 2)² = 0，得到 x = 2。考试常问”求 k 的值，使该方程有两个相等的根”；此时用条件 b² − 4ac = 0 建立关于 k 的方程。

7. Graphs of Quadratic Functions | 二次函数图像

The graph of y = ax² + bx + c is a parabola. The sign of a controls the shape, and the roots of the equation ax² + bx + c = 0 correspond to where the graph crosses the x-axis.

y = ax² + bx + c 的图像是一条抛物线。a 的符号控制开口方向，方程 ax² + bx + c = 0 的根对应图像与 x 轴的交点。

If a > 0, the parabola opens upwards and has a minimum point.

如果 a > 0，抛物线开口向上，有最小值点。
If a < 0, the parabola opens downwards and has a maximum point.

如果 a < 0，抛物线开口向下，有最大值点。
The x-coordinate of the turning point (vertex) is x = −b/(2a).

顶点（驻点）的 x 坐标为 x = −b/(2a)。
Setting x = 0 gives the y-intercept: c.

令 x = 0 得到 y 截距：c。

For y = x² − 4x + 3, the discriminant is 16 − 12 = 4 > 0, so there are two x-intercepts. Factorising gives (x − 1)(x − 3) = 0, so the graph crosses the x-axis at x = 1 and x = 3. The vertex occurs at x = 4/2 = 2, and substituting gives y = 4 − 8 + 3 = −1. The graph is a U-shape with vertex at (2, −1).

对于 y = x² − 4x + 3，判别式为 16 − 12 = 4 > 0，因此有两个 x 截距。因式分解得 (x − 1)(x − 3) = 0，图像在 x = 1 和 x = 3 处穿过 x 轴。顶点在 x = 4/2 = 2 处，代入得 y = 4 − 8 + 3 = −1。图像为 U 形，顶点在 (2, −1)。

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

Many exam questions require you to translate a real-world situation into a quadratic equation. The strategy is always: define a variable, form the equation, solve it, and interpret the answers in context.

许多考试题要求你将现实情境转化为二次方程。策略始终是：设变量、建立方程、求解，并在实际背景中解释答案。

Example: A rectangle has area 24 cm². Its length is 2 cm greater than its width. Find the dimensions of the rectangle.

例：一个长方形面积为 24 平方厘米，长比宽多 2 厘米。求该长方形的尺寸。

Let the width be x cm, then the length is (x + 2) cm. The area gives x(x + 2) = 24, which expands to x² + 2x = 24. Rearranging:

设宽为 x 厘米，则长为 (x + 2) 厘米。由面积得 x(x + 2) = 24，展开为 x² + 2x = 24。整理得：

x² + 2x − 24 = 0 → (x + 6)(x − 4) = 0

So x = −6 or x = 4. Since a length cannot be negative, we reject x = −6. The width is 4 cm, and the length is 6 cm. Always reject solutions that do not make sense in the context.

因此 x = −6 或 x = 4。由于长度不能为负，舍去 x = −6。宽为 4 厘米，长为 6 厘米。始终舍去在情境中无意义的解。

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

Quadratic equations are a high-scoring topic, but small errors can cost marks. Here are the most common pitfalls and how to avoid them.

二次方程是一个高得分话题，但小错误可能丢分。以下是最常见的陷阱以及如何避免它们。

Setting the equation to zero first: you cannot factorise or apply the formula correctly unless the equation is in the form ax² + bx + c = 0. Move all terms to one side before solving.

首先将方程整理为零：除非方程是 ax² + bx + c = 0 的形式，否则无法正确因式分解或套用公式。务必先将所有项移到一边。
Checking signs in factorisation: in x² − 5x + 6 = 0, the factors are (x − 2)(x − 3), not (x + 2)(x + 3). Test your brackets by expanding them mentally.

检查因式分解中的符号：在 x² − 5x + 6 = 0 中，因式为 (x − 2)(x − 3)，而非 (x + 2)(x + 3)。用心算展开来检验你的括号。
Including ± when taking square roots: if (x + 3)² = 7, then x = −3 ± √7. Omitting the negative root loses half of the solutions.

开平方时保留 ±：如果 (x + 3)² = 7，则 x = −3 ± √7。遗漏负根会丢失一半解。
Reading the question: if the question asks for answers to 2 decimal places, apply the rounding at the very end. If it asks for exact answers, leave square roots in surd form.

看清题目要求：如果题目要求保留两位小数，在最后一步才四舍五入。如果要求精确答案，保留根号形式。
Rejecting invalid solutions in word problems: negative lengths, times or distances are not physically possible. State that you reject the extraneous root.

在应用题中舍去无效解：负的长度、时间或距离在物理上不可能。明确写出你舍去了这个多余根。

Finally, practise each method until it becomes automatic. Factorisation is fastest when it works, the formula is a reliable fallback, and completing the square is essential for graph-sketching questions.

最后，练习每种方法直到它们变得自动化。因式分解在可行时最快，求根公式是可靠的备选方案，而配方法对于图像作图类问题必不可少。

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Solving Quadratic Equations | 解二次方程