Solving Quadratic Equations | 解一元二次方程

📚 Solving Quadratic Equations | 解一元二次方程

A quadratic equation is one of the most important topics in IGCSE Mathematics. It appears in almost every exam paper, either as a direct question or as the method needed to solve a harder problem. In this article, we will build a complete toolkit: expansion, factorisation, the quadratic formula, completing the square, the discriminant and graph sketching, with worked examples at every stage.

一元二次方程是 IGCSE 数学中最核心的考点之一，几乎每份考卷都会出现，既可能直接考查，也可能是解决更复杂问题的必备工具。本文将系统构建完整工具箱：展开、因式分解、求根公式、配方法、判别式以及函数图象，并在每一阶段配以例题详解。

1. What Is a Quadratic Equation? | 什么是一元二次方程？

A quadratic equation in one variable is an equation that can be written in the standard form ax² + bx + c = 0, where a ≠ 0, and a, b and c are constants. The highest power of the variable is 2, which is why it is called ‘quadratic’ (from the Latin ‘quadratus’, meaning square).

一元二次方程是指可以写成标准形式 ax² + bx + c = 0 的方程，其中 a ≠ 0，a、b、c 为常数。变量的最高次数为 2，因此称为”二次”方程（quadratic 源自拉丁语 quadratus，意为”平方”）。

For example, 2x² − 5x + 3 = 0 is quadratic, while x³ − 2x = 0 is not, because its highest power is 3. A quadratic equation normally has two solutions, called roots; sometimes the two roots are equal, and sometimes the roots are not real numbers. You will see all three cases in this article.

例如，2x² − 5x + 3 = 0 是二次方程，而 x³ − 2x = 0 就不是，因为它最高次数为 3。一元二次方程通常有两个解，称为根；有时两个根相等，有时根不是实数。这三种情况本文将逐一讨论。

2. Expanding and Simplifying | 展开与化简

Before solving quadratics, you must be fluent in expanding products of binomials such as (x + 3)(x − 5). Use the rule: multiply each term in the first bracket by each term in the second bracket, then collect like terms.

在求解二次方程之前，必须熟练展开二项式乘积，例如 (x + 3)(x − 5)。规则是：用第一个括号中的每一项乘以第二个括号中的每一项，再合并同类项。

(x + 3)(x − 5) = x² − 5x + 3x − 15 = x² − 2x − 15

A useful special case is the difference of two squares: (a + b)(a − b) = a² − b². For instance, (x + 4)(x − 4) = x² − 16. This pattern appears frequently in IGCSE papers, especially when simplifying algebraic fractions.

一个常用的特例是平方差公式：(a + b)(a − b) = a² − b²。例如 (x + 4)(x − 4) = x² − 16。这个模式在 IGCSE 考试中频繁出现，尤其是在化简代数分式时。

Remember that expanding is the reverse of factorising; fluency in one skill directly strengthens the other. Practise until expansion is automatic, because every solving method below depends on it.

请记住，展开是因式分解的逆运算；熟练其中一项技能会直接强化另一项。请练习到展开成为本能反应，因为下面每一种求解方法都依赖它。

3. Factorising Quadratics | 因式分解

To factorise x² + bx + c, find two integers whose product is c and whose sum is b. For example, x² + 7x + 12 = (x + 3)(x + 4), because 3 + 4 = 7 and 3 × 4 = 12. Check by expanding: (x + 3)(x + 4) = x² + 7x + 12.

要对 x² + bx + c 进行因式分解，需要找到两个整数，使它们的乘积为 c、和为 b。例如，x² + 7x + 12 = (x + 3)(x + 4)，因为 3 + 4 = 7 且 3 × 4 = 12。验证：(x + 3)(x + 4) = x² + 7x + 12。

Watch the signs carefully. To factorise x² − x − 20, you need two numbers whose product is −20 and whose sum is −1. The numbers are −5 and 4, so x² − x − 20 = (x − 5)(x + 4). A sign error here will produce a completely different expansion.

要特别注意符号。分解 x² − x − 20 时，需要找两个数，使乘积为 −20、和为 −1。这两个数是 −5 和 4，所以 x² − x − 20 = (x − 5)(x + 4)。符号出错会导致完全不同的展开结果。

When the coefficient of x² is not 1, such as 2x² + 7x + 3, you must consider the factors of a and c together. A systematic method is to multiply a × c, then find two numbers whose product is ac and whose sum is b. Here, ac = 6 and b = 7, so the two numbers are 6 and 1:

当 x² 的系数不为 1 时，例如 2x² + 7x + 3，就必须同时考虑 a 和 c 的因数。系统的方法是先计算 a × c，再找两个数，使它们的乘积为 ac、和为 b。此处 ac = 6，b = 7，所以这两个数是 6 和 1：

2x² + 7x + 3 = 2x² + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (2x + 1)(x + 3)

This ‘split the middle term’ technique is reliable and examiners reward clear working. If the quadratic has a common factor, always take it out first, for example 3x² + 6x = 3x(x + 2).

这种”拆分中间项”的技巧非常可靠，考官也会欣赏清晰的书写过程。如果二次式含有公因式，务必先提取出来，例如 3x² + 6x = 3x(x + 2)。

4. Solving by Factorisation | 用因式分解求解

If a quadratic equation can be factorised, solving is quick and accurate. The key idea is the zero product property: if A × B = 0, then A = 0 or B = 0. This property is true because zero can only be produced by multiplying two factors when at least one of them is zero.

如果一元二次方程可以因式分解，求解既快速又准确。关键思想是零乘积性质：若 A × B = 0，则 A = 0 或 B = 0。该性质成立，是因为只有当一个因子为零时，乘积才可能为零。

Solve x² − 5x + 6 = 0. First factorise: (x − 2)(x − 3) = 0. Then set each factor to zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The two roots are x = 2 and x = 3.

解方程 x² − 5x + 6 = 0。先因式分解：(x − 2)(x − 3) = 0。再令每个因式为零：x − 2 = 0 得 x = 2，x − 3 = 0 得 x = 3。所以两个根为 x = 2 和 x = 3。

Always rearrange the equation into the form ax² + bx + c = 0 before factorising. A common error is attempt to factorise an expression that is not set equal to zero, or to treat the constant term as independent. For example, x² + 3x = 10 must first become x² + 3x − 10 = 0 before factorising.

在因式分解之前，务必先把方程整理成 ax² + bx + c = 0 的形式。常见错误是直接对没有等于零的表达式进行因式分解，或者忽略常数项。例如，x² + 3x = 10 必须先变形为 x² + 3x − 10 = 0，再因式分解。

5. The Quadratic Formula | 求根公式

Not every quadratic can be factorised easily, especially when the roots are irrational. The quadratic formula works for every quadratic equation and is printed in the IGCSE formula booklet:

并非所有一元二次方程都能轻松因式分解，尤其是当根为无理数时。求根公式适用于所有一元二次方程，并且印在 IGCSE 公式表中：

x = [−b ± √(b² − 4ac)] / 2a

To use it, write the equation in the form ax² + bx + c = 0, identify a, b and c carefully, then substitute. For example, solve 2x² + 3x − 5 = 0. Here a = 2, b = 3, c = −5:

使用时，先把方程写成 ax² + bx + c = 0 的形式，仔细确定 a、b、c 的值，然后代入。例如，解 2x² + 3x − 5 = 0，其中 a = 2，b = 3，c = −5：

x = [−3 ± √(9 − 4 × 2 × (−5))] / (2 × 2) = [−3 ± √49] / 4 = (−3 ± 7) / 4

Therefore x = (−3 + 7) / 4 = 1, or x = (−3 − 7) / 4 = −2.5. The two roots are x = 1 and x = −2.5. Always write the substitution clearly; this catches sign errors and earns method marks even if the final answer is wrong.

因此 x = (−3 + 7) / 4 = 1，或 x = (−3 − 7) / 4 = −2.5。两个根为 x = 1 和 x = −2.5。务必清晰写出代入过程；这样即使最终答案有误，也能获得方法分，并且能帮助发现符号错误。

6. Completing the Square | 配方法

Completing the square rewrites ax² + bx + c in the form a(x + p)² + q. For x² + bx + c, take half of b, write (x + b/2)², then subtract (b/2)² to keep the expression unchanged.

配方法将 ax² + bx + c 改写为 a(x + p)² + q 的形式。对于 x² + bx + c，取 b 的一半，写成 (x + b/2)²，再减去 (b/2)²，以保证表达式与原式恒等。

For example, x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8. This form reveals the turning point of the parabola and is also used to solve equations:

例如，x² + 6x + 1 = (x + 3)² − 9 + 1 = (x + 3)² − 8。这种形式能直接显示抛物线的顶点，也用于求解方程：

(x + 3)² − 8 = 0 → (x + 3)² = 8 → x + 3 = ±√8 → x = −3 ± 2√2

When a ≠ 1, factor a out of the first two terms first. For example, 2x² + 8x + 5 = 2(x² + 4x) + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3. This method is particularly useful for finding maximum and minimum values, because the minimum value of (x + p)² is 0.

当 a ≠ 1 时，先从前两项中提取 a。例如，2x² + 8x + 5 = 2(x² + 4x) + 5 = 2(x + 2)² − 8 + 5 = 2(x + 2)² − 3。这种方法在求最大值和最小值时特别有用，因为 (x + p)² 的最小值为 0。

7. The Discriminant | 判别式

The expression b² − 4ac is called the discriminant, often written as Δ. It tells us the nature of the roots without actually solving the equation. This is one of the most frequently tested ideas in IGCSE algebra.

表达式 b² − 4ac 称为判别式，通常记作 Δ。它可以在不解方程的情况下判断根的性质。这是 IGCSE 代数中最常考的知识点之一。

If Δ > 0, there are two distinct real roots. 若 Δ > 0，方程有两个不相等的实数根。
If Δ = 0, there is one repeated real root. 若 Δ = 0，方程有一个重根（两个相等实根）。
If Δ < 0, there are no real roots (two complex roots). 若 Δ < 0，方程没有实数根（有两个复数根）。

For example, the equation x² + 4x + k = 0 has Δ = 16 − 4k. For two distinct real roots we need 16 − 4k > 0, which gives k < 4. For one repeated root, k = 4; for no real roots, k > 4.

例如，方程 x² + 4x + k = 0 的判别式为 Δ = 16 − 4k。若要有两个

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Solving Quadratic Equations | 解一元二次方程